complex differences and difference equations
TRANSCRIPT
Guest Editors Zong-Xuan Chen Kwang Ho Shon and Zhi-Bo Huang
Complex Differences and Difference Equations
Abstract and Applied Analysis
Complex Differences and Difference Equations
Abstract and Applied Analysis
Complex Differences and Difference Equations
Guest Editors Zong-Xuan Chen Kwang Ho Shonand Zhi-Bo Huang
Copyright copy 2014 Hindawi Publishing Corporation All rights reserved
This is a special issue published in ldquoAbstract and Applied Analysisrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited
Editorial Board
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Contents
Complex Differences and Difference Equations Zong-Xuan Chen Kwang Ho Shon and Zhi-Bo HuangVolume 2014 Article ID 124843 1 page
Blow-Up Analysis for a Quasilinear Parabolic Equation with Inner Absorption and NonlinearNeumann Boundary Condition Zhong Bo Fang and Yan ChaiVolume 2014 Article ID 289245 8 pages
Some Properties on Complex Functional Difference Equations Zhi-Bo Huang and Ran-Ran ZhangVolume 2014 Article ID 283895 10 pages
The Regularity of Functions on Dual Split Quaternions in Clifford AnalysisJi Eun Kim and Kwang Ho ShonVolume 2014 Article ID 369430 8 pages
Unicity of Meromorphic Functions Sharing Sets withTheir Linear Difference PolynomialsSheng Li and BaoQin ChenVolume 2014 Article ID 894968 7 pages
A ComparisonTheorem for Oscillation of the Even-Order Nonlinear Neutral Difference EquationQuanxin ZhangVolume 2014 Article ID 492492 5 pages
Difference Equations and Sharing Values Concerning Entire Functions andTheir DifferenceZhiqiang Mao and Huifang LiuVolume 2014 Article ID 584969 6 pages
Admissible Solutions of the Schwarzian Type Difference Equation Baoqin Chen and Sheng LiVolume 2014 Article ID 306360 5 pages
Statistical Inference for Stochastic Differential Equations with Small NoisesLiang Shen and Qingsong XuVolume 2014 Article ID 473681 6 pages
On the Deficiencies of Some Differential-Difference Polynomials Xiu-Min Zheng and Hong Yan XuVolume 2014 Article ID 378151 12 pages
On Growth of Meromorphic Solutions of Complex Functional Difference Equations Jing LiJianjun Zhang and Liangwen LiaoVolume 2014 Article ID 828746 6 pages
Unicity of Entire Functions concerning Shifts and Difference Operators Dan Liu Degui Yangand Mingliang FangVolume 2014 Article ID 380910 5 pages
On Positive Solutions and Mann Iterative Schemes of aThird Order Difference Equation Zeqing LiuHeng Wu Shin Min Kang and Young Chel KwunVolume 2014 Article ID 470181 16 pages
Algebroid Solutions of Second Order Complex Differential Equations Lingyun Gao and Yue WangVolume 2014 Article ID 123049 4 pages
EditorialComplex Differences and Difference Equations
Zong-Xuan Chen1 Kwang Ho Shon2 and Zhi-Bo Huang1
1School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Zong-Xuan Chen chzxvipsinacom
Received 12 August 2014 Accepted 12 August 2014 Published 22 December 2014
Copyright copy 2014 Zong-Xuan Chen et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
In more recent years activity in the area of the complexdifferences and the complex difference equations has fleetlyincreased
This journal has set up a column of this special issue Wewere pleased to invite the interested authors to contributetheir original research papers as well as good expositorypapers to this special issue that will make better improvementon the theory of complex differences and difference equa-tions
In this special issue many good results are obtainedDifference equations are widely applied to mathematical
physics economics and chemistry In this special issue Z-B Huang and R-R Zhang J Li et al and L Gao and YWang investigate the growth a Borel exceptional value ofmeromorphic solutions to different types of higher ordernonliear difference equations respectively B Chen and S Liinvestigate the Schwarzian type difference equation D Liuet al Z Mao and H Liu and S Li and B Chen investigateunicity of meromorphic functions concerning different typesof difference operators Recently many difference analoguesof the classic Nevanlinna theory are obtained
In this special issue X-M Zheng and H Y Xu obtaina differential difference analogue of Valiron-Mohonko theo-rem Related topics with complex difference J E Kim andK H Shon investigate the regularity of functions on dualsplit quaternions in Clifford analysis and the tensor productrepresentation of polynomials of weak type in a DF-spaceQ Zhang and Z Liu et al investigate different types ofreal difference equations respectively L Shen and Q Xuinvestigate stochastic differential equations
This special issue stimulates the continuing efforts to thecomplex differences and the complex difference equations
Zong-XuanChenKwangHo ShonZhi-BoHuang
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 124843 1 pagehttpdxdoiorg1011552014124843
Research ArticleBlow-Up Analysis for a Quasilinear Parabolic Equation withInner Absorption and Nonlinear Neumann Boundary Condition
Zhong Bo Fang and Yan Chai
School of Mathematical Sciences Ocean University of China Qingdao 266100 China
Correspondence should be addressed to Zhong Bo Fang fangzb7777hotmailcom
Received 24 February 2014 Revised 11 April 2014 Accepted 11 April 2014 Published 30 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Z B Fang and Y Chai This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We investigate an initial-boundary value problem for a quasilinear parabolic equation with inner absorption and nonlinearNeumann boundary condition We establish respectively the conditions on nonlinearity to guarantee that 119906(119909 119905) exists globally orblows up at some finite time 119905lowast Moreover an upper bound for 119905lowast is derived Under somewhat more restrictive conditions a lowerbound for 119905lowast is also obtained
1 Introduction
We are concerned with the global existence and blow-upphenomenon for a quasilinear parabolic equation with non-linear inner absorption term
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
minus 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (1)
subjected to the nonlinear Neumann boundary and initialconditions
(|nabla119906|119901
+ 1)
120597119906
120597]= 119892 (119906) (119909 119905) isin 120597Ω times (0 119905
lowast
) (2)
119906 (119909 0) = 1199060(0) ge 0 119909 isin Ω (3)
whereΩ is a bounded star-shaped region of 119877119873 (119873 ge 2) withsmooth boundary 120597Ω ] is the unit outward normal vectoron 120597Ω 119901 ge 0 119905lowast is the blow-up time if blow-up occurs orelse 119905lowast
= +infin the symbol 119894 denotes partial differentiationwith respect to 119909
119894 119894 = 1 2 119873 the repeated index indicates
summation over the index and nabla is gradient operatorMany physical phenomena and biological species theo-
ries such as the concentration of diffusion of some non-Newton fluid through porous medium the density of somebiological species and heat conduction phenomena havebeen formulated as parabolic equation (1) (see [1ndash3]) Thenonlinear Neumann boundary condition (2) can be physi-cally interpreted as the nonlinear radial law (see [4 5])
In the past decades there have been many works dealingwith existence and nonexistence of global solutions blow-upof solutions bounds of blow-up time blow-up rates blow-up sets and asymptotic behavior of solutions to nonlinearparabolic equations see the books [6ndash8] and the surveypapers [9ndash11] Specially we would like to know whether thesolution blows up and at which time when blow-up occursA variety of methods have been used to study the problemabove (see [12]) and in many cases these methods used toshow that solutions blow up often provide an upper boundfor the blow-up time However lower bounds for blow-uptime may be harder to be determined For the study ofthe initial boundary value problem of a parabolic equationwith homogeneous Dirichlet boundary condition see [1314] Payne et al [13] considered the following quasilinearparabolic equation
119906119905= div (120588|nabla119906|2nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (4)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω To get the lower bound for the blow-up time the authorsassumed that 120588 is a positive 1198621 function which satisfies
120588 (119904) + 1199041205881015840
(119904) gt 0 119904 gt 0 (5)
The lower bound for the blow-up time of solution to (4) withRobin boundary condition was obtained in [15] where 120588 is
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 289245 8 pageshttpdxdoiorg1011552014289245
2 Abstract and Applied Analysis
also assumed to satisfy the condition (5) However under thisboundary condition the best constant of Sobolev inequalityused in [13] is no longer applicable They imposed suitableconditions on 119891 and 120588 and determined a lower bound forthe blow-up time if blow-up occurs and determined whenblow-up cannot occur Marras and Vernier Piro [14] studiedthe nonlinear parabolic problem with time dependent coeffi-cients
1198961(119905) div (119892 (|nabla119906|2nabla119906)) + 119896
2(119905) 119891 (119906) = 119896
3(119905) 119906119905
(119909 119905) isin Ω times (0 119905lowast
)
(6)
where Ω is a bounded domain in 119877119873 with smooth boundary
120597Ω Under some conditions on the data and geometry of thespatial domain they obtained upper and lower bounds of theblow-up time Moreover the sufficient conditions for globalexistence of the solution were derived
For the study of the initial boundary value problem ofa parabolic equation with Robin boundary condition werefer to [15ndash19] Li et al [16] investigated the problem of thenonlinear parabolic equation
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
+ 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (7)
where Ω is a bounded domain in 1198773 with smooth boundary
120597ΩThey derived the lower bound for the blow-up timewhenthe blow-up occurs Clearly |nabla119906|119901 + 1 does not satisfy thecondition (5) Enache [17] discussed the quasilinear parabolicproblem
119906119905= (119892 (119906) 119906
119894)119894
+ 119891 (119906) (8)
where Ω is a bounded domain in 119877119873
(119873 ge 2) with smoothboundary 120597Ω By virtue of a first-order differential inequalitytechnique they showed the sufficient conditions to guaranteethat the solution 119906(119909 119905) exists globally or blows up Inaddition a lower bound for the blow-up time when blow-up occurs was also obtained Ding [18] studied the nonlinearparabolic problem
(119887 (119906))119905= nabla sdot (119892 (119906) nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (9)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω They derived conditions on the data which guaranteethe blow-up or the global existence of the solution A lowerbound on blow-up time when blow-up occurs was alsoobtained For the problem of the nonlinear nonlocal porousmedium equation we read the paper of Liu [19]
Recently for the problems with nonlinear Neumannboundary conditions Payne et al [20] studied the semilinearheat equation with inner absorption term
119906119905= Δ119906 minus 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (10)
They established conditions on nonlinearity to guarantee thatthe solution119906(119909 119905) exists for all time 119905 gt 0 or blows up at somefinite time 119905lowast Moreover an upper bound for 119905lowast was derivedUnder somewhat more restrictive conditions a lower bound
for 119905lowast was derivedThereafter they considered the quasilinearparabolic equation
119906119905= nabla sdot (|nabla119906|
119901
nabla119906) (119909 119905) isin Ω times (0 119905lowast
) (11)
and they showed that blow-up occurs at some finite timeunder certain conditions on the nonlinearities and the dataupper and lower bounds for the blow-up time were derivedwhen blow-up occurs see [21] Liu et al The authors [22 23]studied the reaction diffusion problem with nonlocal sourceand inner absorption terms or with local source and gradientabsorption terms Very recently Fang et al [24] consideredlower bounds estimate for the blow-up time to nonlocalproblemwith homogeneousDirichlet orNeumann boundarycondition
Motivated by the above work we intend to study theglobal existence and the blow-up phenomena of problem (1)ndash(3) and the results of the semilinear equations are extended tothe quasilinear equations Unfortunately the techniques usedfor semilinear equation to analysis of blow-up phenomenaare no longer applicable to our problem As a consequenceby using the suitable techniques of differential inequalitieswe establish respectively the conditions on the nonlinearities119891 and 119892 to guarantee that 119906(119909 119905) exists globally or blows upat some finite time If blow-up occurs we derive upper andlower bounds of the blow-up time
The rest of our paper is organized as follows In Section 2we establish conditions on the nonlinearities to guaranteethat 119906(119909 119905) exists globally In Section 3 we show the condi-tions on data forcing the solution 119906(119909 119905) to blow up at somefinite time 119905
lowast and obtain an upper bound for 119905lowast A lower
bound of blow-up time under some assumptions is derivedin Section 4
2 The Global Existence
In this sectionwe establish the conditions on the nonlinearity119891 and nonlinearity 119892 to guarantee that 119906(119909 119905) exists globallyWe state our result as follows
Theorem 1 Assume that the nonnegative functions 119891 and 119892
satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(12)
where 1198961gt 0 119896
2ge 0 119904 gt 1 2119904 lt 119902 + 1 and 119904 minus 1 lt 119901 lt 119902 minus 1
Then the (nonnegative) solution 119906(119909 119905) of problem (1)-(3) doesnot blow up that is 119906(119909 119905) exists for all time 119905 gt 0
Proof Set
Ψ (119905) = int
Ω
1199062
119889119909 (13)
Abstract and Applied Analysis 3
Similar to Theorem 21 in [20] we get
Ψ1015840
(119905) le 21205752
int
Ω
1199062119904
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
+
21198962119873
1205880
int
Ω
119906119904+1
119889119909
minus2int
Ω
|nabla119906|119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 1198681+ 1198682
(14)
where 120575 = 1198962(119904 + 1)1198892120588
0 1205880
= min119909isin120597Ω
(119909 sdot ]) 119889 =
max119909isin120597Ω
|119909| and
1198681le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
times 1198601|Ω|(119902minus119904)(119902+1)
minus 1198602(int
Ω
119906119902+1
119889119909)
(119902minus119904)(119902+1)
(15)
where 1198601= 21205752
120572120576(120572minus1)120572 119860
2= 1198961minus 21205752
(1 minus 120572)120576 120572 = (119902 + 1 minus
2119904)(119902 minus 119904) lt 1 120576 gt 0Next we estimate 119868
2= (211989621198731205880) intΩ
119906119904+1
119889119909minus2 intΩ
|nabla119906|119901+2
119889119909 minus 1198961intΩ
119906119902+1
119889119909 Since
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
= (
119901
2
+ 1)
2
119906119901
|nabla119906|2
(16)
it follows from Holder inequality that
int
Ω
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
119889119909 le (
119901
2
+ 1)
2
(int
Ω
|nabla119906|119901+2
119889119909)
2(119901+2)
times (int
Ω
119906119901+2
119889119909)
119901(119901+2)
(17)
Furthermore we have
int
Ω
119906119901+2
119889119909 le [
(119901 + 2)2
41205821
]
(1199012)+1
int
Ω
|nabla119906|119901+2
119889119909 (18)
which follows from (17) and membrane inequality
1205821int
Ω
1205962
119889119909 le int
Ω
|nabla120596|2
119889119909 (19)
where 1205821is the first eigenvalue in the fixed membrane
problem
Δ120596 + 120582120596 = 0 120596 gt 0 in Ω 120596 = 0 on 120597Ω (20)
Combining 1198682and (18) we have
1198682le
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 2[
41205821
(119901 + 2)2]
(1199012)+1
times int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
=
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 3[
41205821
(119901 + 2)2
]
(1199012)+1
int
Ω
119906119901+2
119889119909
+
[
41205821
(119901 + 2)2]
(1199012)+1
int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 11986821+ 11986822
(21)
Making use of Holder inequality we obtain
int
Ω
119906119904+1
119889119909 le (int
Ω
119906119901+2
119889119909)
(119904+1)(119901+2)
|Ω|(119901minus119904+1)(119901+2)
(22)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119904+1
119889119909)
2(119904+1)
|Ω|(119904minus1)(119904+1)
(23)
Combining (21) (22) with (23) we get
11986821
le (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
(24)
with
1198611=
21198962119873
1205880
1198612= 3[
41205821
(119901 + 2)2]
(1199012)+1
|Ω|minus(119901minus119904+1)2
(25)
Applying Holder inequality we obtain
int
Ω
119906119901+2
119889119909 le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
|Ω|(119902minus119901minus1)(119902+1)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119902+1
119889119909)
2(119902+1)
|Ω|(119902minus1)(119902+1)
(26)
It follows from (26) that
11986822
le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(27)
where
1198621= [
41205821
(119901 + 2)2
]
(1199012)+1
|Ω|(119902minus119901minus1)(119902+1)
1198622= 1198961|Ω|(1minus119902)(119902minus119901minus1)2(119902+1)
(28)
4 Abstract and Applied Analysis
Combining (14) (15) (21) and (24) with (27) we obtain
Ψ1015840
(119905) le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
1198601minus 1198602Ψ(119905)(119902minus119904)2
+ (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
+ (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(29)
with
1198601= 1198601|Ω|(119902minus119904)(119902+1)
1198602= 1198602|Ω|(1minus119902)(119902minus119904)2(119902+1)
(30)
We conclude from (29) that Ψ(119905) is decreasing in each timeinterval on which we obtain
Ψ (119905) ge max(1198601
1198602
)
2(119902minus119904)
(
1198611
1198612
)
2(119901minus119904+1)
(
1198621
1198622
)
2(119902minus119901minus1)
(31)
so that Ψ(119905) remains bounded for all time under theconditions in Theorem 1 This completes the proof ofTheorem 1
3 Blow-Up and Upper Bound of 119905lowast
In this section Ω needs not to be star-shaped We establishthe conditions to assure that the solution of (1)ndash(3) blowsup at finite time 119905lowast and derive an upper bound for 119905lowast Moreprecisely we establish the following result
Theorem 2 Let 119906(119909 119905) be the classical solution of problem (1)-(3) Assume that the nonnegative and integrable functions 119891and 119892 satisfy
120585119891 (120585) le 2 (1 + 120572) 119865 (120585) 120585 ge 0
120585119892 (120585) ge 2 (1 + 120573)119866 (120585) 120585 ge 0
(32)
with
119865 (120585) = int
120585
0
119891 (120578) 119889120578 119866 (120585) = int
120585
0
119892 (120578) 119889120578 (33)
where 120572 ge 0
120573 ge max (119901
2
120572) (34)
Moreover assume that Φ(0) ge 0 with
Φ (119905) = 2int
120597Ω
119866 (119906) 119889119878 minus int
Ω
|nabla119906|2
(1 +
2
119901 + 2
|nabla119906|119901
)119889119909
minus 2int
Ω
119865 (119906) 119889119909
(35)
Then the solution 119906(119909 119905) of problem (1)-(3) blows up at somefinite time 119905lowast lt 119879 with
119879 =
Ψ (0)
2120573 (1 + 120573)Φ (0)
120573 gt 0 (36)
where Ψ(119905) is defined in (13) If 120573 = 0 we have 119879 = infin
Proof We compute
Ψ1015840
(119905) = 2int
Ω
119906119906119905119889119909 = 2int
Ω
119906 [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 2int
120597Ω
119906 (|nabla119906|119901
+ 1)
120597119906
120597]119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
= 2int
120597Ω
119906119892 (119906) 119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
(37)
Making use of the hypotheses stated inTheorem 2 we have
Ψ1015840
(119905) ge 2 (1 + 120573)Φ (119905) (38)
Differentiating (35) we derive
Φ1015840
(119905) = 2int
120597Ω
119892 (119906) 119906119905119889119878 minus int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
minus 2int
Ω
119891 (119906) 119906119905119889119909
(39)
Integrating the identity nabla sdot (119906119905(|nabla119906|119901
+1)nabla119906) = 119906119905nabla sdot ((|nabla119906|
119901
+
1)nabla119906) + (12)(|nabla119906|119901
+ 1)(|nabla119906|2
)119905overΩ we get
int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
= 2int
Ω
nabla sdot (119906119905(|nabla119906|119901
+ 1) nabla119906) 119889119909
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1) nabla119906 sdot ]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
(40)
Substituting (40) into (39) we have
Φ1015840
(119905) = 2int
Ω
1199062
119905119889119909 gt 0 (41)
whichwithΦ(0) gt 0 implyΦ(119905) gt 0 for all 119905 isin (0 119905lowast
) Makinguse of the Schwarz inequality we obtain
2 (1 + 120573)Ψ1015840
Φ le (Ψ1015840
(119905))
2
= 4(int
Ω
119906119906119905119889119909)
2
le 2Ψ (119905)Φ1015840
(119905)
(42)
Abstract and Applied Analysis 5
Multiplying the above inequality by Ψminus2minus120573 we deduce
(ΦΨminus(1+120573)
)
1015840
ge 0 (43)
Arguing as in Theorem 31 in [20] we find
119905lowast
le 119879 =
1
2120573 (1 + 120573)
(Ψ (0))minus120573
=
Ψ (0)
2120573 (1 + 120573)Φ (0)
(44)
valid for 120573 gt 0 If 120573 = 0 we have
Ψ (119905) ge Ψ (0) 1198902119872119905 (45)
valid for 119905 gt 0 implying that 119905lowast = infin This completes theproof of Theorem 2
4 Lower Bounds for 119905lowast
In this section under the assumption that Ω is a star shapeddomain in 119877
3 convex in two orthogonal directions we seek alower bound for the blow-up time 119905lowast Now we state the resultas follows
Theorem 3 Let 119906(119909 119905) be the nonnegative solution of problem(1)-(3) and 119906(119909 119905) blows up at 119905lowast moreover the nonnegativefunctions 119891 and 119892 satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(46)
with 1198961gt 0 119896
2gt 0 119902 gt 1 119904 gt 1 119902 lt 119904 Define
120593 (119905) = int
Ω
119906119899(119904minus1)
119889119909 (47)
where 119899 is a parameter restricted by the condition
119899 gt max 4 2
119904 minus 1
(48)
Then 120593(119905) satisfies inequality
1205931015840
(119905) le Γ (120593) (49)
for some computable function Γ(120593) It follows that 119905lowast is boundedfrom below We have
119905lowast
ge int
infin
120593(0)
119889120578
Γ (120578)
119889120578 (50)
Proof Differentiating (47) and making use of the boundarycondition (2) together with the conditions (46) we have
1205931015840
(119905) = 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119906119905119889119909
= 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
times [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 119899 (119904 minus 1) int
120597Ω
119906119899(119904minus1)minus1
(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119891 (119906) 119889119909
le 1198962119899 (119904 minus 1) int
120597Ω
119906(119899+1)(119904minus1)
119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(51)
Applying inequality (27) in [20] to the first term on the righthand side of (51) we have
int
120597Ω
119906(119899+1)(119904minus1)
119889119878 le
3
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
(119899 + 1) (119904 minus 1) 119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
(52)
6 Abstract and Applied Analysis
Substituting (52) into (51) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(53)
Making use of arithmetic-geometric mean inequality wederive
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909 le
120583
2
int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
+
1
2120583
int
Ω
119906(119899+2)(119904minus1)
119889119909
(54)
for all 120583 gt 0 Choose 120583 gt 0 such that
1198962119899 (119899 + 1) (119904 minus 1)
2
119889120583
21205880
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] = 0 (55)
We rewrite (53) as
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(56)
Using Holder inequality we get
int
Ω
119906119899(119904minus1)
119889119909 le (int
Ω
119906119899(119904minus1)+119902minus1
119889119909)
119899(119904minus1)(119899(119904minus1)+119902minus1)
times |Ω|(119902minus1)(119899(119904minus1)+119902minus1)
(57)
Combining (56) with (57) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
=
31198962119899 (119904 minus 1)
1205880
1198691(119905)
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
1198692(119905)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(58)
where
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
1198692(119905) = int
Ω
119906(119899+2)(119904minus1)
119889119909
120596 (119905) = int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
(59)
Using Sobolev type inequality (A5) derived by Payne et al[21] we obtain
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
le
3
1205880
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 +
(119899 + 1) (119904 minus 1)
3
times(1 +
119889
1205880
)int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
32
(60)
We now make use of Holder inequality to bound the secondintegral on the right hand side of (60) as follows
int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
le (int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909)
(119901+1)(119901+2)
1205961(119901+2)
(61)
with
1205751=
(119899 minus 2) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(62)
Abstract and Applied Analysis 7
Wenote that 1205751lt 1 for 119899 gt (3119901minus2(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Using again Holderrsquosinequality we obtain
int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909
le 1205932(119899+1)(1minus120575
1)3119899
|Ω|1minus(2(119899+1)(1minus120575
1)3119899)
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 le 1205932(119899+1)3119899
|Ω|1minus(2(119899+1)3119899)
(63)
where |Ω| = intΩ
119889119909 is the volume of Ω Substituting (61) and(63) in (60) we obtain the following inequality
1198691(119905) le 119888
11205932(119899+1)3119899
+ 1198882120593(2(119899+1)(1minus120575
1)3119899)((119901+1)(119901+2))
times 1205961(119901+2)
32
le 1198881120593(119899+1)119899
+ 1198882120593((119899+1)(1minus120575
1)119899)((119901+1)(119901+2))
12059632(119901+2)
(64)
where 1198881 1198882are computable positive constants Note that the
last inequality in (64) follows from Holder inequality underthe particular form (119886 + 119887)
32
le radic2(11988632
+ 11988732
) Similarly wecan bound 119869
2and get
1198692(119905) le 119888
3120593(119899+2)119899
+ 1198884120593((119899+2)(1minus120575
2)119899)((119901+1)(119901+2))
12059632(119901+2)
(65)
where 1198883 1198884are computable positive constants
1205752=
(119899 minus 4) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(66)
Wenote that 1205752lt 1 for 119899 gt (3119901minus4(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Inserting (64) and (65)in (58) we arrive at
1205931015840
(119905) le1198891120593(119899+1)119899
+1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
+ 1198893120593(119899+2)119899
+1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(67)
where 120582 = (119901 + 1)(119901 + 2) 1198893and
119889119895(119895 = 1 2 4) are
computable positive constants Next we want to eliminatethe quantity 120596(119905) in inequality (67) By using the followinginequality
120593120572
120596120573
= (120574120596)120573
120593120572(1minus120573)
120574120573(1minus120573)
1minus120573
le 120574120573120596 + (1 minus 120573) 120574120573(120573minus1)
+ (1 minus 120573) 120574120573(120573minus1)
120593120572(1minus120573)
(68)
valid for 0 lt 120573 lt 1 where 120574 is an arbitrary positive constantthen we have
1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
le 1205741120596 (119905) + 119889
2120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
le 1205742120596 (119905) + 119889
4120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(69)
with arbitrary positive constants 1205741 1205742and computable
positive constants 1198892 1198894 Substitute (69) in (67) and choose
the arbitrary (positive) constants 1205741 1205742such that 120574
1+1205742minus119899(119904minus
1)[119899(119904 minus 1) minus 1] = 0 We obtain
1205931015840
(119905) le1198891120593(119899+1)119899
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(70)
We eliminate the last term in (70) by using the followinginequality
120593(119899+1)119899
= 119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(2119899minus1)(119904minus1)((2119899minus1)(119904minus1)+119904minus119902)
times 119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(119904minus119902)((2119899minus1)(119904minus1)+119904minus119902)
le
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
+
119904 minus 119902
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(71)
valid for 119902 lt 119904 and arbitrary119898 gt 0 and choose119898 such that
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
1198891119898 minus 119896
1119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
= 0
(72)
Then (70) can be rewritten as
1205931015840
(119905) le 11988911205933
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(73)
Integrating (73) over [0 119905] we conclude
119905lowast
ge int
infin
120593(0)
119889120578
times (11988911205783
+ 1198892120578(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120578(119899+2)119899
+ 1198894120578(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
)
minus1
(74)
This completes the proof of Theorem 3
8 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Authorsrsquo Contribution
All authors contributed equally to the paper and read andapproved the final paper
Acknowledgments
This work is supported by the Natural Science Foundationof Shandong Province of China (ZR2012AM018) and theFundamental Research Funds for the Central Universities(no 201362032) The authors would like to deeply thank allthe reviewers for their insightful and constructive comments
References
[1] J Bebernes and D Eberly Mathematical Problems from Com-bustion Theory vol 83 of Applied Mathematical SciencesSpringer New York NY USA 1989
[2] C V Pao Nonlinear Parabolic and Elliptic Equations PlenumPress New York NY USA 1992
[3] J L Vazquez The Porous Medium Equations MathematicalTheory Oxford University Press Oxford UK 2007
[4] J Filo ldquoDiffusivity versus absorption through the boundaryrdquoJournal of Differential Equations vol 99 no 2 pp 281ndash305 1992
[5] H A Levine and L E Payne ldquoNonexistence theorems forthe heat equation with nonlinear boundary conditions and forthe porous medium equation backward in timerdquo Journal ofDifferential Equations vol 16 pp 319ndash334 1974
[6] B Straughan Explosive Instabilities in Mechanics SpringerBerlin Germany 1998
[7] A A Samarskii V A Galaktionov S P Kurdyumov and A PMikhailov Blow-Up in Quasilinear Parabolic Equations vol 19of de Gruyter Expositions in Mathematics Walter de GruyterBerlin Germany 1995
[8] PQuittner andP Souplet Superlinear Parabolic Problems Blow-Up Global Existence and Steady States Birkhauser AdvancedTexts Birkhauser Basel Switzerland 2007
[9] C Bandle and H Brunner ldquoBlowup in diffusion equations asurveyrdquo Journal of Computational andAppliedMathematics vol97 no 1-2 pp 3ndash22 1998
[10] V A Galaktionov and J L Vazquez ldquoThe problem of blow-up in nonlinear parabolic equationsrdquo Discrete and ContinuousDynamical Systems vol 8 no 2 pp 399ndash433 2002
[11] H A Levine ldquoThe role of critical exponents in blowup theo-remsrdquo SIAM Review vol 32 no 2 pp 262ndash288 1990
[12] H A Levine ldquoNonexistence of global weak solutions to someproperly and improperly posed problems of mathematicalphysics the method of unbounded Fourier coefficientsrdquoMath-ematische Annalen vol 214 pp 205ndash220 1975
[13] L E Payne G A Philippin and P W Schaefer ldquoBlow-upphenomena for some nonlinear parabolic problemsrdquoNonlinearAnalysis Theory Methods amp Applications vol 69 no 10 pp3495ndash3502 2008
[14] MMarras and S Vernier Piro ldquoOn global existence and boundsfor blow-up time in nonlinear parabolic problems with time
dependent coefficientsrdquo Discrete and Continuous DynamicalSystems vol 2013 pp 535ndash544 2013
[15] Y Li Y Liu and C Lin ldquoBlow-up phenomena for some non-linear parabolic problems under mixed boundary conditionsrdquoNonlinear Analysis Real World Applications vol 11 no 5 pp3815ndash3823 2010
[16] Y Li Y Liu and S Xiao ldquoBlow-up phenomena for some non-linear parabolic problems under Robin boundary conditionsrdquoMathematical and Computer Modelling vol 54 no 11-12 pp3065ndash3069 2011
[17] C Enache ldquoBlow-up phenomena for a class of quasilinearparabolic problems under Robin boundary conditionrdquo AppliedMathematics Letters vol 24 no 3 pp 288ndash292 2011
[18] J Ding ldquoGlobal and blow-up solutions for nonlinear parabolicequations with Robin boundary conditionsrdquo Computers ampMathematics with Applications vol 65 no 11 pp 1808ndash18222013
[19] Y Liu ldquoBlow-up phenomena for the nonlinear nonlocal porousmedium equation under Robin boundary conditionrdquo Comput-ers amp Mathematics with Applications vol 66 no 10 pp 2092ndash2095 2013
[20] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary conditon Irdquo Zeitschrift fur Angewandte Mathematikund Physik vol 61 no 6 pp 999ndash1007 2010
[21] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary condition IIrdquoNonlinear Analysis Theory Methods ampApplications vol 73 no 4 pp 971ndash978 2010
[22] Y Liu ldquoLower bounds for the blow-up time in a non-local reac-tion diffusion problem under nonlinear boundary conditionsrdquoMathematical and ComputerModelling vol 57 no 3-4 pp 926ndash931 2013
[23] Y Liu S Luo and Y Ye ldquoBlow-up phenomena for a parabolicproblem with a gradient nonlinearity under nonlinear bound-ary conditionsrdquo Computers amp Mathematics with Applicationsvol 65 no 8 pp 1194ndash1199 2013
[24] Z B Fang R Yang and Y Chai ldquoLower bounds estimate for theblow-up time of a slow diffusion equation with nonlocal sourceand inner absorptionrdquo Mathematical Problems in Engineeringvol 2014 Article ID 764248 6 pages 2014
Research ArticleSome Properties on Complex Functional Difference Equations
Zhi-Bo Huang12 and Ran-Ran Zhang3
1 School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Physics and Mathematics University of Eastern Finland PO Box 111 80101 Joensuu Finland3Department of Mathematics Guangdong University of Education Guangzhou 510303 China
Correspondence should be addressed to Zhi-Bo Huang huangzhiboscnueducn
Received 15 January 2014 Accepted 13 March 2014 Published 24 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z-B Huang and R-R ZhangThis is an open access article distributed under theCreativeCommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited
We obtain some results on the transcendental meromorphic solutions of complex functional difference equations of the formsum120582isin119868120572120582(119911)(prod
119899
119895=0119891(119911 + 119888
119895)120582119895) = 119877(119911 119891 ∘119901) = ((119886
0(119911) +119886
1(119911)(119891 ∘119901)+ sdot sdot sdot + 119886
119904(119911) (119891 ∘119901)
119904
)(1198870(119911) + 119887
1(119911) (119891 ∘119901)+ sdot sdot sdot + 119887
119905(119911) (119891 ∘119901)
119905
))where 119868 is a finite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119899) are distinct complex constants 119901(119911) is
a polynomial and 120572120582(119911) (120582 isin 119868) 119886
119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative to 119891(119911)
We further investigate the above functional difference equation which has special type if its solution has Borel exceptional zero andpole
1 Introduction and Main Results
In this paper a meromorphic function means meromorphicin the whole complex plane C For a meromorphic function119910(119911) let 120590(119910) be the order of growth and 120583(119910) the lowerorder of119910(119911) Further let 120582(119910) (resp 120582(1119910)) be the exponentof convergence of the zeros (resp poles) of 119910(119911) We alsoassume that the reader is familiar with the fundamentalresults and the standard notations of Nevanlinna theory ofmeromorphic functions (see eg [1]) Given a meromorphicfunction 119910(119911) we call a meromorphic function 119886(119911) a smallfunction relative to 119910(119911) if 119879(119903 119886(119911)) = 119878(119903 119910) = 119900(119879(119903 119910))as 119903 rarr infin possibly outside of an exceptional set of finitelogarithmic measure Moreover if 119877(119911 119910) is rational in 119910(119911)with small functions relative to 119910(119911) as its coefficients we usethe notation 119889 = deg
119910119877(119911 119910) for the degree of 119877(119911 119910) with
respect to119910(119911) Inwhat follows we always assume that119877(119911 119910)is irreducible in 119910(119911)
Meromorphic solutions of complex difference equationshave recently gained increasing interest due to the problemof integrability of difference equations This is related to theactivity concerning Painleve differential equations and theirdiscrete counterparts in the last decades Ablowitz et al [2]considered discrete equations to be delay equations in thecomplex plane This allowed them to analyze these equations
with the methods from complex analysis In regard to relatedpapers concerning a more general class of complex differenceequations we may refer to [3ndash5] These papers mainly dealtwith equations of the form
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119877 (119911 119891) (1)
where 119869 is a collection of all nonempty subsets of1 2 119899 119888
119895(119895 isin 119869) are distinct complex constants 119891(119911) is
a transcendental meromorphic function 120572119869(119911) (119869 isin 119869) are
small functions relative to119891(119911) and119877(119911 119891) is a rational func-tion in 119891(119911) with small meromorphic coefficients Moreoverif the right-hand side of (1) is essentially like the compositefunction 119890 ∘119891 of 119891(119911) and a rational function 119890(119911) Laine et alreversed the order of composition that is they considered thecomposite function 119891 ∘ 119890 of 119891(119911) and a rational function 119890(119911)which resulted in a complex functional difference equationThe following theorem [5 Theorem 28] gives an example
Theorem A (see [5 Theorem 28]) Suppose that 119891(119911) is atranscendental meromorphic solution of equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (2)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 283895 10 pageshttpdxdoiorg1011552014283895
2 Abstract and Applied Analysis
where 119901(119911) is a polynomial of degree 119896 ge 2 Moreover oneassumes that the coefficients 120572
119869(119911) are small functions relative
to 119891(119911) and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (3)
where 120572 = (log 119899)(log 119896)At this point we briefly introduce some notations used in
this paper A difference monomial of a meromorphic function119891(119911) is defined as
119891(119911)1205820119891(119911 + 119888
1)1205821
sdot sdot sdot 119891(119911 + 119888119899)120582119904
=
119904
prod
119895=0
119891(119911 + 119888119895)
120582119895
(4)
where 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119904) are distinct
constants and 120582119895(119895 = 0 1 119904) are natural numbers A
difference polynomial 119867(119911 119891(119911)) of a meromorphic function119891(119911) a finite sum of difference monomials is defined as
119867(119911 119891 (119911)) = sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
) (5)
where 119868 is a finite set of multi-indexes 120582 = (1205820 1205821 120582
119899)
120572120582(119911) (120582 isin 119868) are small functions relative to 119891(119911) The degree
and the weight of the difference polynomial (5) respectively aredefined as
deg119891(119867) = max
120582isin119868
119899
sum
119895=0
120582119895
120581119891(119867) = max
120582isin119868
119899
sum
119895=1
120582119895
(6)
Consequently 120581119891(119867) le deg
119891(119867) For instance the degree and
the weight of the difference polynomial 1198912(119911)119891(119911minus1)119891(119911+1)+119891(119911)119891(119911 + 1)119891(119911 + 2) + 119891
2
(119911 minus 1)119891(119911 + 2) respectively arefour and three Moreover a difference polynomial (5) is said tobe homogeneous with respect to 119891(119911) if the degree sum119899
119895=0120582119895of
each monomial in the sum of (5) is nonzero and the same forall 120582 isin 119868
In the following we proceed to prove generalizations ofTheorem A and investigate some new results for the first timeWe permit more general expressions on both sides of (1)
Theorem 1 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(7)
where119867(119911 119891(119911)) is defined as (5) 119901(119911) = 119889119896119911119896
+ sdot sdot sdot+1198891119911+119889
0
is a polynomial with constant coefficients 119889119896( = 0) 119889
1 1198890
and of the degree 119896 ge 2 and 119886119894(119911) (119894 = 0 1 119904) and
119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative
to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Set 119889 = max119904 119905 If 119896119889 le
(119899 + 1)deg119891(119867) then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (8)
where 120572 = (log(119899 + 1) + log deg119891(119867) minus log119889)(log 119896)
Similar to the proof of Theorem 1 we easily obtain thefollowing result which is a generation of Theorem A
Theorem 2 Let 119888119894isin C (119894 = 1 2 119899) be distinct
constants and 119891(119911) be a transcendental meromorphic solutionof equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895))
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(9)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119896119889 =
119896max119904 119905 le 119899 then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (10)
where 120572 = (log 119899 minus log 119889)(log 119896)
We then proceed to consider the distribution of zeros andpoles of meromorphic solutions of (7) The following resultindicates that solutions having Borel exceptional zeros andpoles appear only in special situations
Theorem 3 Let 1198880= 0 let 119888
119894isin C 0 (119894 = 1 2 119899) be
distinct constants and let 119891(119911) be a finite order transcendentalmeromorphic solution of equation
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(11)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 1
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 If
max120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (12)
then (11) is either of the form119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(13)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (14)
where 120572 isin C 0 is some constant
Abstract and Applied Analysis 3
Example 4 119891(119911) = cos 119911 solves difference equation
4119891(119911)2
119891(119911 + 120587)2
= 119891(2119911)2
+ 2119891 (2119911) + 1 (15)
Here 119901(119911) = 2119911 Clearly 120582(1119891) = 0 lt 1 = 120582(119891) = 120590(119891) Thisexample shows that condition (12) is necessary and cannot bereplaced by
min120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (16)
Moreover we obtain a result parallel toTheorem 54 in [6]for the difference case
Theorem 5 Suppose that the equation119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
=
119888 (119911)
119891(119911)119898 119898 isin N (17)
has a meromorphic solution of finite order where 1198880= 0 119888
119895isin
C 0 (119895 = 1 2 119899) are distinct constants and 119888(119911) is anontrivialmeromorphic function If119891(119911)has only finitelymanypoles then119891(119911) = 119863(119911)119890119864(119911) where119863(119911) is a rational functionand119864(119911) is a polynomial if and only if 119888(119911) = 119866(119911)119890119872(119911) where119866(119911) is a rational function and119872(119911) is a polynomial
Example 6 Difference equation
119891(119911)2
119891 (119911 + 1) 119891 (119911 minus 1) = (
1
1199114(1199112minus 1)
1198906119911
) sdot
1
119891(119911)2
(18)
of the type (17) is solved by 119891(119911) = 119890119911119911 Here 119891(119911) = 119890119911119911and 119888(119911) = 11989061199111199114(1199112 minus 1) satisfy Theorem 5
As an application of Theorem 3 we obtain the following
Theorem 7 Let 119888 isin C 0 and let 119891(119911) be a finite ordertranscendental meromorphic solution of equation
119891 (119911 + 119888) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(19)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 Then 119891(119911) has at most one Borel exceptional value
If the degree 119896 of polynomial 119901(119911) is 1 in Theorem 7 theresult does not hold For example we have the following
Example 8 119891(119911) = tan 119911 solves difference equation
119891 (119911 + 1) =
119891 (119911) + tan 11 minus (tan 1) 119891 (119911)
(20)
of the type (19) Obviously 119891(119911) has two Borel exceptionalvalues plusmn119894
If we remove the assumption max120582(119891) 120582(1119891) lt 120590(119891)used in Theorem 3 we obtain a result similar to Theorem 12in [4]
Theorem 9 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891)
=
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905
(21)
where 119867(119911 119891(119911)) is defined as (5) and 119886119894(119911) (119894 = 0 1 119904)
and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions
relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119889 = max119904 119905 gt
(119899 + 1) deg119891(119867) then 120590(119891) = infin
In fact the following examples show that the assertion ofTheorem 9 does not remain valid identically if 119889 le (119899 +1)deg
119891(119867)
Example 10 119891(119911) = exp119890119911119911 solves the difference equation
(119911 minus 120587119894) (119911 + log 2 minus 120587119894) 119891 (119911) 119891 (119911 minus 120587119894) 119891 (119911 + log 2 minus 120587119894)
+ (119911 + log 8) 119891 (119911 + log 8) =1 + 119911
11
119891(119911)10
1199113119891(119911)
2
(22)
Clearly 119889 = 10 lt (3+1) sdot 3 = (119899+1) deg119891(119867) and 120590(119891) = infin
Example 11 119891(119911) = tan 119911 satisfies the difference equation
119891(119911 +
120587
4
) + 119891(119911 minus
120587
4
) =
4119891 (119911)
1 minus 119891(119911)2
(23)
Obviously 119889 = 2 lt (2+1)times1 = (119899+1) deg119891(119867) and120590(119891) = 1
Example 12 (see [7 pages 103ndash106] and [8 page 8]) Thefollowing difference equation
119891 (119911 + 1) = 120572119891 (119911) (1 minus 119891 (119911)) 120572 = 0 (24)
derives from a well-known discrete logistic model in biologyIt has been proved that all other meromorphic solutions areof infinite order apart from the constant solutions 119891(119911) equiv 0and 119891(119911) = (120572 minus 1)120572 For instance (24) has one-parameterfamilies of entire solutions of infinite order
119891 (119911) =
1
2
(1 minus exp (119860119890119911 log 2)) 119860 isin C 0 120572 = 2
119891 (119911) = sin2 (119861119890119911 log 2) 119861 isin C 0 120572 = 4(25)
Here 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867)
Example 13 119891(119911) = 119911 solves the difference equation
119891 (119911 + 1) =
1 minus 119891(119911)2
minus1199112minus 119911 + 1 + 119891(119911)
2 (26)
We get 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867) and 120590(119891) = 0
If the difference polynomial in the left-hand side of (21)is homogeneous we further obtain the following theorem
4 Abstract and Applied Analysis
Theorem 14 Let 119891(119911) be a transcendental meromorphic solu-tion of (21) where 119867(z 119891(119911)) is defined as (5) and 119886
119894(119911) (119894 =
0 1 119904) and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Suppose
that 119867(119911 119891) is homogeneous and has at least one differencemonomial of type
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
(120582119895isin N
+ 119895 = 0 1 119899) (27)
If 119889 = max119904 119905 gt 3deg119891(119867) then 120590(119891) = infin
2 Proof of Theorem 1
We need some preliminaries to proveTheorem 1
Lemma 15 (see [9 Lemma 4]) Let 119891(119911) be a transcendentalmeromorphic function and let 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot + 1198891119911 +
1198890(119889119896= 0)be a polynomial of degree 119896 Given 0 lt 120575 lt |119889
119896|
denote ] = |119889119896| + 120575 and 120583 = |119889
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C cup infin one has for all 119903 ge 1199030gt 0
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 ∘ 119901) le 119896119899 (]119903119896 119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 ∘ 119901) le 119873(]119903119896 119886 119891)
+ 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 ∘ 119901) le (1 + 120576) 119879 (]119903119896 119891) (28)
Lemma16 (see [10Theorem B16]) Given distinctmeromor-phic functions 119891
1 119891
119899 let 119869 denote the collection of all
nonempty subsets of 1 2 119899 and suppose that 120572119869isin C for
each 119869 isin 119869 Then
119879(119903sum
119869
120572119869(prod
119895isin119869
119891119895)) le
119899
sum
119896=1
119879 (119903 119891119896) + 119874 (1) (29)
By denoting 119891119894+1= 119891(119911 + 119888
119894)120582119894(119894 = 0 1 119899) below it is
an easy exercise to prove the following result from Lemma 16
Lemma 17 Let 119891(119911) be a meromorphic function let 119868 be afinite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) and let 120572
120582(119911)
be small functions relative to 119891(119911) for all 120582 isin 119868 Then thecharacteristic function of the difference polynomial (5) satisfies
119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
))
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(30)
where 119862 = max|1198881| |119888
2| |119888
119899|
Lemma18 (see [11 Lemma5]) Let119892(119903) andℎ(119903) bemonotonenondecreasing functions on [0infin) such that 119892(119903) le ℎ(119903) for all119903 notin 119864 cup [0 1] where 119864 sub (1infin) is a set of finite logarithmicmeasure Let 120572 gt 1 be a given constant Then there exists an1199030= 119903
0(120572) gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 ge 119903
0
Lemma 19 (see [12 Lemma 3]) Let 120595(119903) be a function of119903 (119903 ge 119903
0) positive and bounded in every finite interval
(i) Suppose that 120595(120583119903119898) le 119860120595(119903) + 119861 (119903 ge 1199030) where
120583 (120583 gt 0)119898 (119898 gt 1)119860 (119860 ge 1) and 119861 are constantsThen 120595(119903) = 119874((log 119903)120572) with 120572 = (log119860)(log119898)unless 119860 = 1 and 119861 gt 0 and if 119860 = 1 and 119861 gt 0 thenfor any 120576 gt 0 120595(119903) = 119874((log 119903)120576)
(ii) Suppose that (with the notation of (i)) 120595(120583119903119898) ge119860120595(119903) (119903 ge 119903
0) Then for all sufficiently large values of
119903 120595(119903) ge 119870(log 119903)120572 with 120572 = (log119860)(log119898) for somepositive constant 119870
Proof of Theorem 1 For any 120576 (0 lt 120576 lt 1) we may applyValiron-Mohonrsquoko lemma Lemmas 15 and 17 and (5) and (7)to conclude that119889 (1 minus 120576) 119879 (120583119903
119896
119891)
le 119889119879 (119903 119891 ∘ 119901) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905)
+ 119878 (119903 119891)
= 119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
)) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(31)holds for all sufficiently large 119903 possibly outside of anexceptional set of finite logarithmic measure where 119862 =max|119888
1| |119888
2| |119888
119899| and 120583 is defined as Lemma 15 Now we
may apply Lemma 18 to deal with the exceptional set andconclude that for every 120578 gt 1 there exists an 119903
0gt 0 such
that119889 (1 minus 120576) 119879 (120583119903
119896
119891) le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (120578119903 119891)
(32)holds for all 119903 ge 119903
0 Denote 120596 = 120578119903 Then (32) can be written
in the form
119879(
120583
120578119896
120596119896
119891) le
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
119879 (120596 119891) (33)
Since 119889119896 le (119899 + 1)deg119891(119867) we get ((119899 + 1)deg
119891(119867)(1 +
120576))(119889(1 minus 120576)) gt 1 for all 0 lt 120576 lt 1 Thus we now applyLemma 19(i) to conclude that
119879 (119903 119891) = 119874 ((log 119903)120572+120576)
120572 =
log ((119899 + 1) deg119891(119867) (1 + 120576) 119889 (1 minus 120576))
log 119896
=
log (119899 + 1) + log deg119891(119867) minus log119889
log 119896+ 119900 (1)
(34)
The proof of Theorem 1 is completed
Abstract and Applied Analysis 5
3 Proof of Theorems 3 and 5
We again need some preliminaries
Lemma 20 (see [13 Theorem 15]) Suppose that 119891119895(119911) (119895 =
1 2 119899) (119899 ge 2) are meromorphic functions and 119892119895(119911) (119895 =
1 2 119899) are entire functions satisfying the following condi-tions
(1) sum119899119895=1119891119895(119911)119890
119892119895(119911)
= 0
(2) 119892119895(119911) minus 119892
119896(119911) are not constants for 1 le 119895 lt 119896 le 119899
(3) For 1 le 119895 le 119899 1 le ℎ lt 119896 le 119899
119879 (119903 119891119895) = 119900 119879 (119903 119890
119892ℎminus119892119896) (119903 997888rarr +infin 119903 notin 119864) (35)
where 119864 sub (1 +infin) is of finite linear measure or finitelogarithmic measure
Then 119891119895(119911) equiv 0 (119895 = 1 2 119899)
Lemma21 (see [14Theorem4]) Let119865(119911)119875119899(119911) 119875
0(119911) be
polynomials such that 1198651198751198991198750equiv 0 and then every finite order
transcendental meromorphic solution 119891(119911) of equation
119875119899(119911) 119891 (119911 + 119899) + sdot sdot sdot + 119875
1(119911) 119891 (119911 + 1) + 119875
0(119911) 119891 (119911) = 119865 (119911)
(36)
satisfies 120582(119891) = 120590(119891) ge 1
Remark 22 Replacing 119895 by 119888119895(119895 = 1 2 119899) where 119888
119895(119895 =
1 2 119899) are distinct nonzero complex constants Lemma 21remains valid
Proof of Theorem 3 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed withnonzero poles of 119891(119911) Since max120582(119891) 120582(1119891) lt 120590(119891) thenℎ(119911) = 119911
120591
119902(119911) is an entire function such that
120590 (ℎ) = 120582(
1
119891
) lt 120590 (119891) lt +infin (37)
and 119892(119911) = ℎ(119911)119891(119911) is a transcendental entire function with
119879 (119903 119892) = 119879 (119903 119891) + 119878 (119903 119891) 120590 (119892) = 120590 (119891)
120582 (119892) = 120582 (119891)
(38)
If 119902(119911) is a polynomial we obtain quickly that 120590(ℎ ∘ 119901) =0 lt 120590(119892 ∘ 119901) Otherwise we conclude from the last assertionof Lemma 15 (37) and (38) that
120590 (ℎ ∘ 119901) = 119896120590 (ℎ) = 119896120582(
1
119891
) lt 119896120590 (119892) = 120590 (119892 ∘ 119901) (39)
Therefore
119879 (119903 ℎ ∘ 119901) = 119878 (119903 119892 ∘ 119901) (40)
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (11) we concludethat
(ℎ ∘ 119901)119904minus119905
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
(
119899
prod
119895=0
119892(119911 + 119888119895)
120582119895
)
= (1198860(119911) (ℎ ∘ 119901)
119904
+ 1198861(119911) (ℎ ∘ 119901)
119904minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119886119904(119911) (119892 ∘ 119901)
119904
)
times (1198870(119911) (ℎ ∘ 119901)
119905
+ 1198871(119911) (ℎ ∘ 119901)
119905minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119887119905(119911) (119892 ∘ 119901)
119905
)
minus1
(41)
Obviously it follows from (37)ndash(40) and Lemma 15 that
119879(119903
1
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 (ℎ ∘ 119901)119904minus119905
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 119886119906(119911) (ℎ ∘ 119901)
119904minus119906
) = 119878 (119903 119892 ∘ 119901) 119906 = 0 1 119904
119879 (119903 119887V (119911) (ℎ ∘ 119901)119905minusV) = 119878 (119903 119892 ∘ 119901) V = 0 1 119905
(42)
Denoting119860(119911) = (ℎ ∘ 119901)119904minus119905prod119899119895=0ℎ(119911 + 119888
119895)120582119895 we get from (42)
that119879 (119903 119860) = 119878 (119903 119892 ∘ 119901) (43)
Since zeros and poles are Borel exceptional values of 119891(119911) by(12) we may apply a result due to Whittaker see [15 Satz134] to deduce that 119891(119911) is of regular growth Thus we useLemma 15 and (12) again to get
119879(119903
1198911015840
119891
) = 119873(119903 119891) + 119873(119903
1
119891
) + 119878 (119903 119891) = 119878 (119903 119892 ∘ 119901)
(44)
Similarly if we set 119861(119911) = 119860(119911)(prod119899
119895=0119892(119911 + 119888
119895)120582119895) we
also deduce from the lemma of the logarithmic derivativeLemma 15 (12) (38) and (43) that
119879(119903
1198611015840
119861
) = 119879(119903
1198601015840
119860
+
119899
sum
119895=0
120582119895
1198921015840
(119911 + 119888119895)
119892 (119911 + 119888119895)
) = 119878 (119903 119892 ∘ 119901)
(45)Denoting 119865(119911) = 119892 ∘ 119901
119875 (119911 119865) =
1198860(119911)
119886119904(119911)
(ℎ ∘ 119901)119904
+
1198861(119911)
119886119904(119911)
(ℎ ∘ 119901)119904minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119904
119876 (119911 119865) =
1198870(119911)
119887119905(119911)
(ℎ ∘ 119901)119905
+
1198871(119911)
119887119905(119911)
(ℎ ∘ 119901)119905minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119905
(46)
6 Abstract and Applied Analysis
Therefore we deduce from Lemma 15 and (42) that thecoefficients of 119875(119911 119865) and119876(119911 119865) are small functions relativeto 119865(119911) Thus (41) can be written in the form
119887119905(119911)
119886119904(119911)
119861 (119911) =
119875 (119911 119865)
119876 (119911 119865)
= 119906 (119911 119865) (47)
Denoting
120595 (119911) =
1198651015840
(119911)
119865 (119911)
119880 (119911) =
1199061015840
(119911 119865)
119906 (119911 119865)
(48)
we get 119879(119903 119880) = 119878(119903 119892 ∘ 119901) from (45) and (47) Wealso conclude from the lemma of logarithmic derivativeLemma 15 and (12) that
119879 (119903 120595) = 119879(119903
1198651015840
119865
) = 119898(119903
1198651015840
119865
) + 119873(119903
1198651015840
119865
)
le 119873 (119903 119865) + 119873(119903
1
119865
) + 119878 (119903 119865)
= 119873 (119903 119892 ∘ 119901) + 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(]1199031198961
119892
) + 119878 (119903 119892 ∘ 119901) = 119878 (119903 119892 ∘ 119901)
(49)
where ] is defined as Lemma 15Since
1198751015840
119876 minus 1198751198761015840
1198762
= 1199061015840
= 119880119906 =
119880119875
119876
(50)
we conclude that
1198751015840
119876 minus 1198751198761015840
= 119880119875119876 (51)
Now writing 1198651015840 = 120595119865 in (51) regarding then (51) as analgebraic equation in119865with coefficients of growth 119878(119903 119865) andcomparing the leading coefficients we deduce that
(119904 minus 119905) 120595 = 119880 (52)
By integrating both sides of the last equality above weconclude that
119906 (119911 119865) = 120572119865(119911)119904minus119905
(53)
for some 120572 isin C 0 Therefore by combining therepresentations of 119865 119861 119860 119892 with (53) we conclude that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
(54)
If 119904119905 = 0 we deduce from (11) and (54) that
120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(55)
From this we get that 119877(119911 119891 ∘ 119901) is not irreducible in 119891 ∘ 119901a contradiction Thus 119905 = 0 or 119904 = 0 Therefore we deducefrom (54) that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(56)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (57)
The proof of Theorem 3 is completed
Proof of Theorem 5 Assume first that 119891(119911) = 119863(119911)119890119864(119911)
where 119863(119911) is a rational function and 119864(119911) is a polynomialOne can see from (17) that
119888 (119911) = 119891(119911)119898
119899
prod
119894=0
119891(119911 + 119888119894)120582119894
= [119863(119911)119898
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
] 119890119898119864(119911)+sum
119899
119894=0120582119894119864(119911+119888
119894)
= 119866 (119911) 119890119872(119911)
(58)
where 119866(119911) = 119863(119911)119898prod119899119894=0119863(119911 + 119888
119894)120582119894 is rational and119872(119911) =
119898119864(119911) + sum119899
119894=0120582119894119864(119911 + 119888
119894) is a polynomial
Suppose next that 119888(119911) = 119866(119911)119890119872(119911) where 119866(119911) is arational function and 119872(119911) is a polynomial Since 119891(119911) hasonly finitely many poles we conclude from (17) that
119873(119903
1
119891
) le 119873(119903
1
119891119898
) = 119873(119903
prod119899
119894=0119891(119911 + 119888
119894)120582119894
119888 (119911)
)
= 119874 (log 119903)
(59)
Thus 119891(119911) has only finitely many zeros and poles and119891(119911) = 119863(119911)119890
119864(119911) where119863(119911) is rational and 119864(119911) is an entirefunction In the followingwe only prove119864(119911) is a polynomialNow substituting119891(119911) = 119863(119911)119890119864(119911) and 119888(119911) = 119866(119911)119890119872(119911) into(17) we get
119899
prod
119894=0
119863(119911 + 119888119894)120582119894 exp (120582
119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(60)
(
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
) exp(119899
sum
119894=0
120582119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(61)
Thus we deduce from Lemma 20 that two exponents in (61)cancel each other to a constant 120591 isin C such that
119899
sum
119894=0
120582119894119864 (119911 + 119888
119894) = 119872 (119911) minus 119898119864 (119911) + 120591 (62)
Abstract and Applied Analysis 7
that is119899
sum
119894=1
120582119894119864 (119911 + 119888
119894) + (120582
0+ 119898)119864 (119911) = 119872 (119911) + 120591 (63)
Suppose that 119864(119911) is not a polynomial If 119864(119911) is a transcen-dental entire function of finite order we get from Lemma 21Remark 22 and (63) that 120590(119864) ge 1 Otherwise 119864(119911) is atranscendental entire function of infinite order These bothshow that 120590(119891) = infin contradicting the assumption that119891(119911) is finite order Thus 119864(119911) is a polynomial The proof ofTheorem 5 is completed
4 Proof of Theorem 7
Lemmas 23 and 25 reveal some properties of the maximalmodule of the polynomial in composite function 119891 ∘ 119901 with ameromorphic function 119891(119911) and a polynomial 119901(119911) whichare useful for proving the existence of Borel exceptionalvalue of finite order meromorphic solutions of functionaldifference equation of type (19)
Lemma 23 Let 119892(119911) be a nonconstant entire function of order120590(119892) = 120590 lt infin Suppose that 120572
119895(119911) (119895 = 1 2 119898) are small
meromorphic functions relative to 119892(119911) Then there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 120572119895)
119872 (119903 119892)
997888rarr 0 119895 = 1 2 119898 (64)
hold simultaneously for all 119903 isin 119864 as 119903 rarr infin where the lowerlogarithmic density of set 119864 is defined by
logdens (119864) = lim inf119903rarrinfin
int[1119903]cap119864
(119889119905119905)
log 119903 (65)
Remark 24 The proof of Lemma 23 is similar to the proof ofLemma 24 and Remark 25 in [16] Here we omit it
Lemma 25 Let 119891(119911) be a finite order transcendental mero-morphic function satisfying (12) and 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot+1198891119911+
1198890is a polynomial with constant coefficients 119889
119896( = 0) 119889
1 1198890
and of the degree 119896 ge 1 Suppose that
119867(119911) = 119886119899(119911) (119891 ∘ 119901)
119899
+ 119886119899minus1(119911) (119891 ∘ 119901)
119899minus1
+ sdot sdot sdot + 1198861(119911) (119891 ∘ 119901) + 119886
0(119911)
(66)
is a polynomial in 119891 ∘119901 where 119899 (ge 1) is a positive integer and119886119899(119911) ( equiv 0) 119886
119899minus1(119911) 119886
1(119911) 119886
0(119911) are small meromorphic
functions relative to 119891(119911) Then there exists a set 1198641of lower
logarithmic density 1 such that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (67)
for all 119903 isin 1198641as 119903 rarr infin where 0 lt 120583 lt |119889
119896| Hence119867(119911) equiv
0
Proof of Lemma 25 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed
with the nonzero poles of 119891(119911) Since 119891(119911) satisfies (12) thenℎ(119911) = 119911
120591
119902(119911) is an entire function Thus 119892(119911) = ℎ(119911)119891(119911) isentire and (37) (38) and (40) also hold
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (66) we con-clude that
119867(119911) = 119886119899(119911) sdot
(119892 ∘ 119901)119899
(ℎ ∘ 119901)119899+ 119886
119899minus1(119911) sdot
(119892 ∘ 119901)119899minus1
(ℎ ∘ 119901)119899minus1
+ sdot sdot sdot
+ 1198861(119911) sdot
(119892 ∘ 119901)
(ℎ ∘ 119901)
+ 1198860(119911)
=
119886119899(119911)
(ℎ ∘ 119901)119899(119892 ∘ 119901)
119899
times [1 +
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
(119892 ∘ 119901)minus1
+ sdot sdot sdot
+
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
(119892 ∘ 119901)1minus119899
+
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
(119892 ∘ 119901)minus119899
]
(68)
We note from Lemma 15 and (40) that
119879(119903
119886119899(119911)
(ℎ ∘ 119901)119899) = 119878 (119903 119892 ∘ 119901)
119879(119903
119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911)
) = 119878 (119903 119892 ∘ 119901)
for 119895 = 0 1 119899 minus 1
(69)
Therefore we deduce from Lemma 23 that there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 119886119899(119911) (ℎ ∘ 119901)
119899
)
119872 (119903 119892 ∘ 119901)
997888rarr 0
119872(119903 119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911))
119872 (119903 119892 ∘ 119901)
997888rarr 0
(119895 = 0 1 119899 minus 1)
(70)
Moreover according to the choosing of 119864 in the proofof Lemma 23 we know that 119886
119895(119911)(ℎ ∘ 119901)
119899minus119895
119886119899(119911) for
8 Abstract and Applied Analysis
119895 = 0 1 119899 minus 1 have no zeros and poles for all |119911| = 119903 isin 119864Thus we conclude from (68) and (70) that for any 120576 gt 0
119872(119903119867) ge 119872(119903 119892 ∘ 119901)119899minus120576
times [1 minus
100381610038161003816100381610038161003816100381610038161003816
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus1
minus sdot sdot sdot
minus
1003816100381610038161003816100381610038161003816100381610038161003816
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
1003816100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)1minus119899
minus
100381610038161003816100381610038161003816100381610038161003816
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus119899
]
ge (1 minus 120576)119872(119903 119892 ∘ 119901)119899minus120576
(71)
and so
log+119872(119903119867) ge (119899 minus 32
120576) log+119872(119903 119892 ∘ 119901) (72)
for all |119911| = 119903 isin 119864 and |119892 ∘ 119901(119911)| = 119872(119903 119892 ∘ 119901)Therefore we deduce from Lemma 15 and (38) that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (73)
for all |119911| = 119903 isin 1198641= 119864 cap (119903
0 +infin) where 119903
0gt 0 It is
obvious that 1198641has lower logarithmic density 1 The proof of
Lemma 25 is completed
Proof of Theorem 7 Suppose that 119891(119911) has two finite Borelexceptional values 119886 and 119887 ( = 0 119886) For the case where oneof 119886 and 119887 is infinite we can use a similar method to prove itSet
119892 (119911) =
119891 (119911) minus 119886
119891 (119911) minus 119887
(74)
Then 120590(119892) = 120590(119891) and
120582 (119892) = 120582 (119891 minus 119886) lt 120590 (119892) 120582 (
1
119892
) = 120582 (119891 minus 119887) lt 120590 (119892)
(75)
It follows from (74) that
119891 (119911) =
119886 minus 119887119892 (119911)
1 minus 119892 (119911)
(76)
Now substituting (76) into (19) we conclude that
119892 (119911 + 119888) = (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119886
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
times (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119887
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
minus1
= ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
))
times ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
))
minus1
(77)
Since 1198860(119911) + 119886
1(119911)(119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911)(119891 ∘ 119901)
119904 and 1198870(119911) +
1198871(119911)(119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911)(119891 ∘ 119901)
119905 are irreducible in 119891 ∘ 119901 weconclude that at least one of the following three inequalitiesholds that is
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(78)
Thus we deduce fromTheorem 3 that
119892 (119911 + 119888) = 119888 (119911) (119892 ∘ 119901)119897
(79)
where 119888(119911) is meromorphic function satisfying 119879(119903 119888) =119878(119903 119892) and 119897 isin Z Clearly 119897 = 0 and 119892(119911) is of regular growthfrom (75) see [15 Staz 134] Therefore 120590(119888) lt 120590(119892)
If 119897 ge 1 we conclude from (77) and (79) that
(minus1)119904+119905
119888 (119911)(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)(119892 ∘ 119901)119904+119905+119897
+ sdot sdot sdot + (minus
119904
sum
119894=0
119886119894(119911) 119886
119894
+ 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
) = 0
(80)
Abstract and Applied Analysis 9
Thus we deduce from Lemma 25 that (80) is a contradictionIf 119897 le minus1 we use the same method as above to getanother contradiction Therefore 119891(119911) has at most one Borelexceptional value The proof of Theorem 7 is completed
5 Proof of Theorems 9 and 14
We first recall two lemmas
Lemma 26 (see [17 Lemma 21]) Let 119891(119911) be a nonconstantmeromorphic function 119904 gt 0 120572 lt 1 and 119865 sub R+ the set of all119903 such that
119879 (119903 119891) le 120572119879 (119903 + 119904 119891) (81)
If the logarithmicmeasure of119865 is infinite that isint119865
(119889119905119905) = infinthen 119891(119911) is of infinite order of growth
Lemma 27 (see [18 Corollary 26] and [19 Corollary 22])Let 119891(119911) be a meromorphic function of finite order and let 119888 isinC Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (82)
for all 119903 outside of a possible exceptional set of finite logarithmicmeasure
Proof of Theorem 9 For any 120576 (0 lt 120576 lt (119889 minus (119899 +
1)deg119891(119867))(119889 + (119899 + 1) deg
119891(119867))) we may apply Valiron-
Mohonrsquoko lemma Lemma 17 (5) and (21) to conclude that
119889 (1 minus 120576) 119879 (119903 119891)
le 119889119879 (119903 119891) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905)
= 119879 (119903119867 (119911 119891 (119911))) le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891)
+ 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(83)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote
120572 =
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
(84)
Then 120572 lt 1 since 0 lt 120576 lt (119889 minus (119899 + 1)deg119891(119867))(119889 + (119899 +
1)deg119891(119867)) and 119889 gt (119899 + 1)deg
119891(119867) Thus
119879 (119903 119891) le 120572119879 (119903 + 119862 119891) (85)
holds for all 119903 in a set with infinite logarithmic measureTherefore we deduce from Lemma 26 and (85) that 120590(119891) =infin The proof of Theorem 9 is completed
Proof of Theorem 14 Assume contrary to the assertion that119891(119911) is meromorphic of finite order Taking into account theassumption that119867(119911 119891(119911)) is homogeneous we deduce fromLemma 27 that
119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) = 119878 (119903 119891) (86)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote 119862 = max1le119894le119899|119888119894| Since 119867(119911) is homogeneous
and has at least one difference monomial of typeprod119899119895=0119891(119911 +
119888119895)120582119895 we immediately conclude that by looking at pole
multiplicities summing over |119911| le 119903 and integratinglogarithmically
119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
)
le 120581119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
(87)
Therefore
119879 (119903119867 (119911 119891 (119911)))
= 119898 (119903119867 (119911 119891 (119911))) + 119873 (119903119867 (119911 119891 (119911)))
le 119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119898(119903 119891(119911)deg119891(119867)
)
+ 119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119873(119903 119891(119911)deg119891(119867)
)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
))
+ 119879 (119903 119891(119911)deg119891(119867)
) + 119878 (119903 119891)
le 3deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(88)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure The remainder can be proven by asimilar method in Theorem 9 The proof of Theorem 14 iscompleted
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful to the referees for their helpfulsuggestions to improve this paper The first author alsothanks Professor Ilpo laine and Professor Risto Korhonen for
10 Abstract and Applied Analysis
their valuable suggestion to the present paper Research issupported by National Natural Science Foundation of China(nos 11171119 and 11171121) and Guangdong National NaturalScience Foundation (no S2012040006865)
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[3] BGrammaticos T Tamizhmani A Ramani andKMTamizh-mani ldquoGrowth and integrability in discrete systemsrdquo Journal ofPhysics A vol 34 no 18 pp 3811ndash3821 2001
[4] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of Malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
[7] WGKelley andAC PetersonDifference Equations AcademicPress Boston Mass USA 1991
[8] I Laine and C C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[9] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[10] V I Gromak I Laine and S Shimomura Painleve DifferentialEquations in the Complex Plane vol 28 Walter de GruyterBerlin Germany 2002
[11] G G Gundersen ldquoFinite order solutions of second orderlinear differential equationsrdquo Transactions of the AmericanMathematical Society vol 305 no 1 pp 415ndash429 1988
[12] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[13] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions vol 557 Kluwer Academic Publishers Group Dor-drecht The Netherlands 2003
[14] Z X Chen and K H Shon ldquoOn growth of meromorphicsolutions for linear difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 619296 6 pages 2013
[15] G Jank and L Volkmann Einfuhrung in die Theorie derganzen und meromorphen Funktionen mit Anwendungen aufDifferentialgleichungen Birkhauser Basel Switzerland 1985
[16] J Wang ldquoGrowth and poles of meromorphic solutions of somedifference equationsrdquo Journal of Mathematical Analysis andApplications vol 379 no 1 pp 367ndash377 2011
[17] R G Halburd and R J Korhonen ldquoFinite-order meromorphicsolutions and the discrete Painleve equationsrdquoProceedings of theLondon Mathematical Society vol 94 no 2 pp 443ndash474 2007
[18] Y M Chiang and S J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[19] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
Research ArticleThe Regularity of Functions on Dual Split Quaternionsin Clifford Analysis
Ji Eun Kim and Kwang Ho Shon
Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Kwang Ho Shon khshonpusanackr
Received 28 January 2014 Accepted 2 April 2014 Published 17 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 J E Kim and K H Shon This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
This paper shows some properties of dual split quaternion numbers and expressions of power series in dual split quaternions andprovides differential operators in dual split quaternions and a dual split regular function on Ω sub C2 times C2 that has a dual splitCauchy-Riemann system in dual split quaternions
1 Introduction
Hamilton introduced quaternions extending complex num-bers to higher spatial dimensions in differential geometry (see[1]) A set of quaternions can be represented as
H = 119911 = 1199090+ 1199091119894 + 1199092119895 + 1199093119896 119909119898
isin R 119898 = 0 1 2 3
(1)
where 1198942
= 1198952
= 1198962
= minus1 119894119895119896 = minus1 and R denotes the set ofreal numbers Cockle [2] introduced a set of split quaternionsas
S = 119911 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 119909119898
isin R 119898 = 0 1 2 3
(2)
where 1198902
1= minus1 119890
2
2= 1198902
3= 1 and 119890
111989021198903
= 1 Aset of split quaternions is noncommutative and containszero divisors nilpotent elements and nontrivial idempotents(see [3 4]) Previous studies have examined the geometricand physical applications of split quaternions which arerequired in solving split quaternionic equations (see [5 6])Inoguchi [7] reformulated the Gauss-Codazzi equations informs consistent with the theory of integrable systems in theMinkowski 3-space for split quaternion numbers
A dual quaternion can be represented in a form reflect-ing an ordinary quaternion and a dual symbol Because
dual-quaternion algebra is constructed from real eight-dimensional vector spaces and an ordered pair of quater-nions dual quaternions are used in computer vision appli-cations Kenwright [8] provided the characteristics of dualquaternions and Pennestrı and Stefanelli [9] examined someproperties by using dual quaternions Son [10 11] offeredan extension problem for solutions of partial differentialequations and generalized solutions for the Riesz system Byusing properties of Hamilton operators Kula and Yayli [4]defined dual split quaternions and gave some properties ofthe screw motion in the Minkowski 3-space showing thatHhas a rotation with unit split quaternions in H and a scalarproduct that allows it to be identifiedwith the semi-Euclideanspace for split quaternion numbers
It was shown (see [12 13]) that any complex-valued har-monic function 119891
1in a pseudoconvex domain 119863 of C2 times C2
C being the set of complex numbers has a conjugate function1198912in 119863 such that the quaternion-valued function 119891
1+ 1198912119895 is
hyperholomorphic in119863 and gave a regeneration theorem in aquaternion analysis in view of complex and Clifford analysisIn addition we [14 15] provided a new expression of thequaternionic basis and a regular function on reduced quater-nions by associating hypercomplex numbers 119890
1and 119890
2 We
[16] investigated the existence of hyperconjugate harmonicfunctions of an octonion number system and we [17 18]obtained some regular functions with values in dual quater-nions and researched an extension problem for properties
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 369430 8 pageshttpdxdoiorg1011552014369430
2 Abstract and Applied Analysis
of regular functionswith values in dual quaternions and someapplications for such problems
This paper provides a regular function and some prop-erties of differential operators in dual split quaternions Inadditionwe research some equivalent conditions forCauchy-Riemann systems and expressions of power series in dual splitquaternions from the definition of dual split regular on anopen set Ω sub C2 times C2
2 Preliminaries
A dual number 119860 has the form 119886 + 120576119887 where 119886 and 119887 are realnumbers and 120576 is a dual symbol subject to the rules
120576 = 0 0120576 = 1205760 = 0 1120576 = 1205761 = 120576 1205762
= 0 (3)
and a split quaternion 119902 isin S is an expression of the form
119902 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 (4)
where 119909119898
isin R (119898 = 0 1 2 3) and 119890119903(119903 = 1 2 3) are split
quaternionic units satisfying noncommutativemultiplicationrules (for split quaternions see [1])
1198902
1= minus1 119890
2
2= 1198902
3= 1
11989011198902= minus11989021198901= 1198903 119890
21198903= minus11989031198902= minus1198901
11989031198901= minus11989011198903= 1198902
(5)
Similarly a dual split quaternion 119911 can be written as
D (S) = 119911 | 119911 = 1199010+ 1205761199011 119901119903isin S 119903 = 0 1 (6)
which has elements of the following form
119911 = (1199090+ 11990911198901) + (119909
2+ 11990931198901) 1198902
+ 120576 (1199100+ 11991011198901) + (119910
2+ 11991031198901) 1198902
= (1199110+ 11991111198902) + 120576 (119911
2+ 11991131198902)
= 1199010+ 1205761199011
(7)
where 1199010= 1199110+ 11991111198902and 119901
1= 1199112+ 11991131198902are split quaternion
components 1199110= 1199090+ 11990911198901 1199111= 1199092+ 11990931198901 1199112= 1199100+ 11991011198901
and 1199113= 1199102+ 11991031198901are usual complex numbers and 119909
119898 119910119898
isin
R (119898 = 0 1 2 3) The multiplication of split quaternionicunits with a dual symbol is commutative 120576119890
119903= 119890119903120576 (119903 =
1 2 3) However by properties of split quaternionic unit
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 0 1)
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 2 3)
119890119903119901119896
= 119901119896119890119903 119890119903119901119896= 119901(119896119903)
119890119903
(119903 = 1 2 3 119896 = 0 1)
(8)
where
119901(01)
= 1199110minus 11991111198902= 1199090+ 11990911198901minus 11990921198902minus 11990931198903
119901(02)
= 1199110+ 11991111198902= 1199090minus 11990911198901+ 11990921198902minus 11990931198903
119901(03)
= 1199110minus 11991111198902= 1199090minus 11990911198901minus 11990921198902+ 11990931198903
119901(11)
= 1199112minus 11991131198902= 1199100+ 11991011198901minus 11991021198902minus 11991031198903
119901(12)
= 1199112+ 11991131198902= 1199100minus 11991011198901+ 11991021198902minus 11991031198903
119901(13)
= 1199112minus 11991131198902= 1199100minus 11991011198901minus 11991021198902+ 11991031198903
(9)
with 1199110
= 1199090minus 11990911198901 1199111= 1199092minus 11990931198901 1199112
= 1199100minus 11991011198901 and
1199113= 1199102minus 11991031198901 For instance
11989021199010= 1198902(1199090+ 11990911198901+ 11990921198902+ 11990931198903)
= (1199090minus 11990911198901+ 11990921198902minus 11990931198903) 1198902= 119901(02)
1198902
11989011199011= 1198901(1199100+ 11991011198901+ 11991021198902+ 11991031198903)
= (1199100+ 11991011198901minus 11991021198902minus 11991031198903) 1198901= 119901(11)
1198901
(10)
Because of the properties of the eight-unit equality the addi-tion and subtraction of dual split quaternions are governedby the rules of ordinary algebra Here the symbol 119901
(119896119903)is used
by just enumerating 119903 and 119896 not 119903 times 119896 For example119901(22)
= 1199014and 119901
22= 1199014
For any two elements 119911 = 1199010+ 1205761199011and 119908 = 119902
0+ 1205761199021
of D(S) where 1199020= sum3
119903=0119904119903119890119903and 119902
1= sum3
119903=0119905119903119890119903are split
quaternion components and 119904119903 119905119903isin R (119903 = 0 1 2 3) their
noncommutative product is given by
119911119908 = (1199010+ 1205761199011) (1199020+ 1205761199021) = 11990101199020+ 120576 (119901
01199021+ 11990111199020)
(11)
The conjugation 119911lowast of 119911 and the corresponding modulus 119911119911lowast
inD(S) are defined by
119911lowast
= 119901lowast
0+ 120576119901lowast
1
119911119911lowast
= 119911lowast
119911 = 1199010119901lowast
0+ 120576 (119901
0119901lowast
1+ 1199011119901lowast
0)
= (11991101199110minus 11991111199111) + 2120576 (119911
01199112minus 11991111199113)
=
1
sum
119903=0
(1199092
119903minus 1199092
119903+2) + 120576 (119909
119903119910119903minus 119909119903+2
119910119903+2
)
(12)
where 119901lowast0= 1199110minus 11991111198902and 119901
lowast
1= 1199112minus 11991131198902
Lemma 1 For all 119911 isin D(S) and 119899 isin N = 1 2 3 wehave
119911119899
= 119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0 (13)
Abstract and Applied Analysis 3
Proof If 119899 = 1 then (13) is trivial Now suppose that thisholds for some 119899 isin N Then as desired
119911119899+1
= 119911119911119899
= 119911(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119901119899+1
0+ 120576
119899
sum
119896=1
119901119899minus119896+1
01199011119901119896minus1
0+ 1205761199011119901119899
0
= 119901119899+1
0+ 120576
119899+1
sum
119896=1
119901119899+1minus119896
01199011119901119896minus1
0
(14)
By the principle of mathematical induction (13) holds for all119899 isin N
Let Ω be an open subset of C2 times C2 Then the function119891 Ω rarr D(S) can be expressed as
119891 (119911) = 119891 (1199010 1199011) = 1198910(1199010 1199011) + 1205761198911(1199010 1199011) (15)
where the component functions 119891119903 Ω rarr S (119903 = 0 1) are
split quaternionic-valued functions The component func-tions 119891
119903(119903 = 0 1) are
1198910(1199010 1199011) = 1198910(1199110 1199111 1199112 1199113)
= 1198920(1199110 1199111 1199112 1199113) + 1198921(1199110 1199111 1199112 1199113) 1198902
1198911(1199010 1199011) = 1198911(1199110 1199111 1199112 1199113)
= 1198922(1199110 1199111 1199112 1199113) + 1198923(1199110 1199111 1199112 1199113) 1198902
(16)
where 119892119896
= 1199062119896
+ 1199062119896+1
1198901(119896 = 0 1) and 119892
119896= V2119896minus4
+
V2119896minus3
1198901(119896 = 2 3) are complex-valued functions and 119906
119903and
V119903(119903 = 0 1 2 3) are real-valued functionsNow we let differential operators 119863
1and 119863
2be defined
onD(S) as
1198631= 119863(11)
+ 120576119863(12)
1198632= 119863(21)
+ 120576119863(22)
(17)
Then the conjugate operators119863lowast1and119863
lowast
2are
119863lowast
1= 119863lowast
(11)+ 120576119863lowast
(12) 119863
lowast
2= 119863lowast
(21)+ 120576119863lowast
(22) (18)
where
119863(11)
=
120597
1205971199110
+
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
119863(12)
=
120597
1205971199112
+
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
120597
1205971199102
1198902+
120597
1205971199103
1198903)
119863(21)
=
120597
1205971199110
+
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
1
2
120597
1205971199092
1198902minus
1
2
120597
1205971199093
1198903)
119863(22)
=
120597
1205971199112
+
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
1
2
120597
1205971199102
1198902minus
1
2
120597
1205971199103
1198903)
(19)
119863lowast
(11)=
120597
1205971199110
minus
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
120597
1205971199092
1198902minus
120597
1205971199093
1198903)
119863lowast
(12)=
120597
1205971199112
minus
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
120597
1205971199102
1198902minus
120597
1205971199103
1198903)
119863lowast
(21)=
120597
1205971199110
minus
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
1
2
120597
1205971199092
1198902+
1
2
120597
1205971199093
1198903)
119863lowast
(22)=
120597
1205971199112
minus
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
(20)
act on D(S) These operators are called correspondingCauchy-Riemann operators in D(S) where 120597120597119911
119903and
120597120597119911119903(119903 = 0 1 2 3) are usual differential operators used in
the complex analysis
Remark 2 From the definition of differential operators onD(S)
119863119903119891 = (119863
(1199031)+ 120576119863(1199032)
) (1198910+ 1205761198911)
= 119863(1199031)
1198910+ 120576 (119863
(1199031)1198911+ 119863(1199032)
1198910)
119863lowast
119903119891 = (119863
lowast
(1199031)+ 120576119863lowast
(1199032)) (1198910+ 1205761198911)
= 119863lowast
(1199031)1198910+ 120576 (119863
lowast
(1199031)1198911+ 119863lowast
(1199032)1198910)
(21)
where 119903 = 1 2
Definition 3 LetΩ be an open set inC2 timesC2 A function 119891 =
1198910+ 1205761198911is called an 119871
119903(resp 119877
119903)-regular function (119903 = 1 2)
onΩ if the following two conditions are satisfied
(i) 119891119896(119896 = 0 1) are continuously differential functions
onΩ and
(ii) 119863lowast119903119891(119911) = 0 (resp 119891(119911)119863lowast
119903= 0) onΩ (119903 = 1 2)
In particular the equation 119863lowast
1119891(119911) = 0 of Definition 3 is
equivalent to
119863lowast
(11)1198910= 0 119863
lowast
(12)1198910+ 119863lowast
(11)1198911= 0 (22)
4 Abstract and Applied Analysis
In addition
1205971198920
1205971199110
minus
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1205971198923
1205971199111
minus
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1205971198922
1205971199111
minus
1205971198920
1205971199113
= 0
(23)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1205971199062
1205971199092
minus
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1205971199062
1205971199093
+
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1205971199060
1205971199092
minus
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
minus
1205971199060
1205971199093
+
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
120597V2
1205971199092
minus
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1205971199062
1205971199103
+
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1205971199060
1205971199102
minus
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
120597V0
1205971199092
minus
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
minus
1205971199060
1205971199103
+
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
minus
120597V0
1205971199093
+
120597V1
1205971199092
= 0
(24)
The above systems (23) and (24) are corresponding Cauchy-Riemann systems inD(S) Similarly the equation119863
lowast
2119891(119911) =
0 of Definition 3 is equivalent to
119863lowast
(21)1198910= 0 119863
lowast
(22)1198910+ 119863lowast
(21)1198911= 0 (25)
Then
1205971198920
1205971199110
minus
1
2
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1
2
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1
2
1205971198923
1205971199111
minus
1
2
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1
2
1205971198922
1205971199111
minus
1
2
1205971198920
1205971199113
= 0
(26)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
+
1
2
1205971199062
1205971199093
+
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
+
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
+
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1
2
1205971199062
1205971199103
+
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
+
1
2
120597V2
1205971199093
+
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
+
1
2
1205971199060
1205971199103
+
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
+
1
2
120597V0
1205971199093
+
1
2
120597V1
1205971199092
= 0
(27)
The above systems (26) and (27) are corresponding Cauchy-Riemann systems inD(S)
On the other hand the equation 119891(119911)119863lowast
1= 0 of
Definition 3 is equivalent to
1198910119863lowast
(11)= 0 119891
0119863lowast
(12)= minus1198911119863lowast
(11) (28)
Abstract and Applied Analysis 5
Then
1198920
120597
1205971199110
= 1198921
120597
1205971199111
1198921
120597
1205971199110
= 1198920
120597
1205971199111
1198920
120597
1205971199112
minus 1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+ 1198923
120597
1205971199111
1198921
120597
1205971199112
minus 1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+ 1198922
120597
1205971199111
(29)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(30)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
=
1205971199062
1205971199092
+
1205971199063
1205971199093
1205971199061
1205971199090
+
1205971199060
1205971199091
= minus
1205971199062
1205971199093
+
1205971199063
1205971199092
1205971199062
1205971199090
+
1205971199063
1205971199091
=
1205971199060
1205971199092
minus
1205971199061
1205971199093
1205971199063
1205971199090
minus
1205971199062
1205971199091
=
1205971199060
1205971199093
+
1205971199061
1205971199092
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
= minus
120597V0
1205971199090
+
120597V1
1205971199091
+
120597V2
1205971199092
+
120597V3
1205971199093
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1205971199062
1205971199103
minus
1205971199063
1205971199102
= minus
120597V1
1205971199090
minus
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1205971199060
1205971199102
+
1205971199061
1205971199103
= minus
120597V2
1205971199090
minus
120597V3
1205971199091
+
120597V0
1205971199092
minus
120597V1
1205971199093
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1205971199060
1205971199103
minus
1205971199061
1205971199102
= minus
120597V3
1205971199090
+
120597V2
1205971199091
+
120597V0
1205971199093
+
120597V1
1205971199092
(31)
Similarly the equation 119891(119911)119863lowast
2= 0 of Definition 3 is
equivalent to
1198910119863lowast
(21)= 0 119891
0119863lowast
(22)= minus1198911119863lowast
(21) (32)
Then
1198920
120597
1205971199110
=
1
2
1198921
120597
1205971199111
1198921
120597
1205971199110
=
1
2
1198920
120597
1205971199111
1198920
120597
1205971199112
minus
1
2
1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+
1
2
1198923
120597
1205971199111
1198921
120597
1205971199112
minus
1
2
1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+
1
2
1198922
120597
1205971199111
(33)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(34)
Concretely the system is obtained as follows
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1
2
1205971199062
1205971199093
minus
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
+
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
minus
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
minus
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
minus
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1
2
1205971199062
1205971199103
minus
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
1
2
120597V2
1205971199093
minus
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
+
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1
2
1205971199060
1205971199103
minus
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
minus
120597V2
1205971199091
minus
1
2
120597V0
1205971199093
minus
1
2
120597V1
1205971199092
= 0
(35)
From the systems (24) (27) (31) and (35) the equations119863lowast
119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 = 1 2) are different
Therefore the equations 119863lowast119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 =
1 2) should be distinguished as 119871119903-regular functions (119903 =
1 2) and 119877119903-regular functions (119903 = 1 2) on Ω respectively
Now the properties of the 119871119903-regular function (119903 = 1 2) with
values inD(S) are considered
3 Properties of 119871119903-Regular Functions (119903 = 1 2)
with Values in D(S)
We consider properties of a 119871119903-regular functions (119903 = 1 2)
with values inD(S)
Theorem 4 Let Ω be an open set in C2 times C2 and let 119891 =
1198910+ 1205761198911= (1198920+11989211198902)+ 120576(119892
2+11989231198902) be an 119871
1-regular function
defined on Ω Then
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (36)
6 Abstract and Applied Analysis
Proof By the system (23) we have
1198631119891 = 119863
(11)1198910+ 120576 (119863
(12)1198910+ 119863(11)
1198911)
= (
1205971198920
1205971199110
+
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+ 2
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+ 2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+ 2
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+ 2
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+ 2
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+ 2
1205971198922
1205971199111
) 1198902
= 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(37)
Therefore we obtain
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (38)
Theorem5 LetΩ be an open set inC2timesC2 and119891 = 1198910+1205761198911=
(1198920+11989211198902) + 120576(119892
2+11989231198902) be an 119871
2-regular function defined on
Ω Then
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (39)
Proof By the system (26) we have
1198632119891 = 119863
(21)1198910+ 120576 (119863
(22)1198910+ 119863(21)
1198911)
= (
1205971198920
1205971199110
+
1
2
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1
2
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(40)
Therefore we obtain the following equation
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (41)
Abstract and Applied Analysis 7
Proposition 6 From properties of differential operators thefollowing equations are obtained
119863(1119903)
119901119903minus1
= 2 119863(2119903)
119901119903minus1
= 1
119863lowast
(1119903)119901119903minus1
= minus1 119863lowast
(2119903)119901119903minus1
= 0
119863lowast
(1119903)119901lowast
119903minus1= 2 119863
lowast
(2119903)119901lowast
119903minus1= 1
119863(1199031)
1199011= 119863lowast
(1199031)1199011= 119863(1199031)
119901lowast
1= 119863lowast
(1199031)119901lowast
1
= 119863(1199032)
1199010= 119863lowast
(1199032)1199010= 119863(1199032)
119901lowast
0
= 119863lowast
(1199032)119901lowast
0= 0 (119903 = 1 2)
(42)
Proof By properties of the power of dual split quaternionsand derivatives on D(S) the following derivatives areobtained
119863(11)
1199010=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090+ 11990911198901+ 11990921198902+ 11990931198903) = 2
119863lowast
(22)1199011=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
times (1199100+ 11991011198901+ 11991021198902+ 11991031198903) = 0
119863(11)
119901lowast
0=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090minus 11990911198901minus 11990921198902minus 11990931198903) = minus1
(43)
The other equations are calculated using a similar methodand the above equations are obtained
Theorem 7 LetΩ be an open set in C2 timesC2 and let 119891(119911) be afunction on Ω with values inD(S) Then the power 119911119899 of 119911 inD(S) is not an 119871
1-regular function but an 119871
2-regular function
on Ω where 119899 isin N
Proof From the definition of the 119871119903-regular function (119903 =
1 2) on Ω and Proposition 6 we may consider whether thepower 119911119899 of 119911 in D(S) satisfies the equation 119863
lowast
119903119911119899
= 0 (119903 =
1 2) Since119863lowast(11)
1199010= 2
119863lowast
1119911119899
= (119863lowast
(11)+ 120576119863lowast
(12))(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119863lowast
(11)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(11)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(12)119901119899
0) = 0
(44)
Hence the power 119911119899 of 119911 is not 1198711-regular onΩ On the other
hand from the equations in Proposition 6 we have119863lowast(21)
1199010=
0119863lowast(21)
1199011= 0 and119863
lowast
(22)1199010= 0 Then
119863lowast
2119911119899
= 119863lowast
(21)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(21)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(22)119901119899
0) = 0
(45)
Therefore by the definition of the 119871119903-regular function (119903 =
1 2) onΩ a power 119911119899 of 119911 is 1198712-regular onΩ
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The second author was supported by the Basic ScienceResearch Program through the National Research Founda-tion of Korea (NRF) funded by the Ministry of Science ICTand Future Planning (2013R1A1A2008978)
References
[1] I L Kantor andA S SolodovnikovHypercomplex Numbers AnElementary Introduction to Algebras Springer New York NYUSA 1989
[2] J Cockle ldquoOn systems of algebra involving more than oneimaginaryrdquo PhilosophicalMagazine III vol 35 no 238 pp 434ndash435 1849
[3] M Ozdemir and A A Ergin ldquoRotations with unit timelikequaternions in Minkowski 3-spacerdquo Journal of Geometry andPhysics vol 56 no 2 pp 322ndash336 2006
[4] L Kula and Y Yayli ldquoSplit quaternions and rotations in semiEuclidean spaceE4
2rdquo Journal of the KoreanMathematical Society
vol 44 no 6 pp 1313ndash1327 2007[5] D C Brody and E-M Graefe ldquoOn complexified mechanics
and coquaternionsrdquo Journal of Physics A Mathematical andTheoretical vol 44 no 7 article 072001 2011
[6] I Frenkel and M Libine ldquoSplit quaternionic analysis andseparation of the series for SL(2R) and SL(2C)SL(2R)rdquoAdvances in Mathematics vol 228 no 2 pp 678ndash763 2011
[7] J-i Inoguchi ldquoTimelike surfaces of constant mean curvature inMinkowski 3-spacerdquo Tokyo Journal of Mathematics vol 21 no1 pp 141ndash152 1998
[8] B Kenwright ldquoA beginners guide to dual-quaternions whatthey are how they work and how to use them for 3D characterhierarchiesrdquo in Proceedings of the 20th International Conferenceson Computer Graphics Visualization and Computer Vision pp1ndash10 2012
[9] E Pennestrı and R Stefanelli ldquoLinear algebra and numericalalgorithms using dual numbersrdquo Multibody System Dynamicsvol 18 no 3 pp 323ndash344 2007
[10] L H Son ldquoAn extension problem for solutions of partialdifferential equations in R119899rdquo Complex Variables Theory andApplication vol 15 no 2 pp 87ndash92 1990
[11] L H Son ldquoExtension problem for functions with values in aClifford algebrardquo Archiv der Mathematik vol 55 no 2 pp 146ndash150 1990
[12] J Kajiwara X D Li and K H Shon ldquoRegeneration in complexquaternion and Clifford analysisrdquo in International Colloquiumon Finite or Infinite DimensionalComplex Analysis and ItsApplications vol 2 of Advances in Complex Analysis and ItsApplications no 9 pp 287ndash298 Kluwer Academic PublishersHanoi Vietnam 2004
[13] J Kajiwara X D Li and K H Shon ldquoFunction spaces incomplex and Clifford analysisrdquo in International Colloquium
8 Abstract and Applied Analysis
on Finite Or Infinite Dimensional Complex Analysis and ItsApplications vol 14 of Inhomogeneous Cauchy Riemann Systemof Quaternion andCliffordAnalysis in Ellipsoid pp 127ndash155HueUniversity Hue Vietnam 2006
[14] J E Kim S J Lim and K H Shon ldquoRegular functions withvalues in ternary number system on the complex Cliffordanalysisrdquo Abstract and Applied Analysis vol 2013 Article ID136120 7 pages 2013
[15] J E Kim S J Lim and K H Shon ldquoRegularity of functions onthe reduced quaternion field in Clifford analysisrdquo Abstract andApplied Analysis vol 2014 Article ID 654798 8 pages 2014
[16] S J Lim and K H Shon ldquoHyperholomorphic fucntions andhyperconjugate harmonic functions of octonion variablesrdquoJournal of Inequalities andApplications vol 2013 article 77 2013
[17] S J Lim and K H Shon ldquoDual quaternion functions and itsapplicationsrdquo Journal of Applied Mathematics vol 2013 ArticleID 583813 6 pages 2013
[18] J E Kim S J Lim and K H Shon ldquoTaylor series of functionswith values in dual quaternionrdquo Journal of the Korean Society ofMathematical Education BmdashThePure and AppliedMathematicsvol 20 no 4 pp 251ndash258 2013
Research ArticleUnicity of Meromorphic Functions Sharing Sets withTheir Linear Difference Polynomials
Sheng Li and BaoQin Chen
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to BaoQin Chen chenbaoqin chbq126com
Received 23 January 2014 Accepted 20 March 2014 Published 13 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 S Li and B Chen This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We mainly investigate the unicity of meromorphic functions sharing two or three sets with their linear difference polynomials andprove some results
1 Introduction and Main Results
In this paper we assume the reader is familiar with the fun-damental results and the basic notations of the Nevanlinnatheory of meromorphic functions (see eg [1ndash3]) Let 119891(119911)be meromorphic in the whole plane We use the notation120588(119891) to denote the order of growth of the meromorphicfunction 119891(119911) In addition we denote by 119878(119903 119891) any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin outside of apossible exceptional set of finite logarithmic measure We saythat a meromorphic function 119886(119911) is a small function of 119891(119911)provided that 119879(119903 119886) = 119878(119903 119891) Let 119878(119891) be the set of all smallfunctions of 119891(119911)
For a set 119878 sub 119878(119891) we define the following
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 counting multiplicities
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 ignoring multiplicities
(1)
Let 119891 and 119892 be meromorphic functions If 119864119891(119878) = 119864
119892(119878)
and 119864119891(119878) = 119864
119892(119878) respectively then we say that 119891 and 119892
share a set 119878 CM and IM respectivelyFurthermore let 119888 be a nonzero complex constant We
define the shift of 119891(119911) by 119891(119911 + 119888) and define the differenceoperators of 119891(119911) by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911)
Δ119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911)) 119899 isin N 119899 ge 2
(2)
Theunicity theory ofmeromorphic functions sharing setsis an important topic of the uniqueness theory First of all werecall the following theorem given by Li and Yang in [4]
Theorem A (see [4]) Let 119898 ge 2 and let 119899 gt 2119898 + 6 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0Then for any two nonconstant meromorphic functions 119891 and119892 the conditions 119864
119891(119878) = 119864
119892(119878) and 119864
119891(infin) = 119864
119892(infin)
imply 119891 = 119892
Yi and Lin considered the case 119898 = 1 with the conditionthat two meromorphic functions share three sets and got theresult as follows
Theorem B (see [5]) Let 1198781= 120596 120596
119899
+ 119886120596119899minus1
+ 119887 = 01198782= 0 and 119878
3= infin where 119886 119887 are nonzero constants such
that 120596119899 + 119886120596119899minus1 + 119887 = 0 has no repeated root and 119899(ge 4) is aninteger If for two nonconstant meromorphic functions 119891 and119892 119864119891(119878119895) = 119864
119892(119878119895) for 119895 = 1 2 3 and Θ(infin119891) gt 0 then
119891 equiv 119892
Recently a number of papers have focused on differenceanalogues of the Nevanlinna theory (see eg [6ndash9]) Inparticular there has been an increasing interest in studyingthe uniqueness problems related to meromorphic functionsand their shifts or their difference operators (see eg [10ndash16])
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 894968 7 pageshttpdxdoiorg1011552014894968
2 Abstract and Applied Analysis
In 2010 Zhang considered a meromorphic function 119891(119911)sharing sets with its shift 119891(119911 + 119888) and proved the followingresult
TheoremC (see [16]) Let119898 ge 2 and let 119899 ge 2119898+4with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596119899 + 119886120596119899minus119898 + 119887 = 0 Suppose that119891(119911) is a nonconstant meromorphic function of finite orderThen 119864
119891(119911)(119878) = 119864
119891(119911+119888)(119878) and 119864
119891(119911)(infin) = 119864
119891(119911+119888)(infin)
imply 119891(119911) equiv 119891(119911 + 119888)
For an analogue result in difference operator B Chen andZ Chen proved the following theorem in [10]
Theorem D (see [10]) Let 119898 ge 2 and let 119899 ge 2119898 + 4 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has nomultiple roots Let 119878 = 120596 | 120596119899+119886120596119899minus119898+119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying 119864
119891(119911)(119878) = 119864
Δ119888119891(119878) and 119864
119891(119911)(infin) = 119864
Δ119888119891(infin) If
119873(119903
1
Δ119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (3)
then Δ119888119891 equiv 119891(119911)
It is natural to ask what happens if the shift 119891(119911 + 119888) ordifference operatorΔ
119888119891(119911) is replaced by a general expression
of 119891(119911) such as a linear difference polynomial of 119891(119911)Here a linear difference polynomial of 119891(119911) is an expres-
sion of the form
119871 (119911 119891) = 119887119896(119911) 119891 (119911 + 119888
119896) + sdot sdot sdot + 119887
0(119911) 119891 (119911 + 119888
0) (4)
where 119887119896(119911) equiv 0 119887
0(119911) 119887
119896(119911) are small functions of
119891(119911) 1198880 119888
119896are complex constants and 119896 is a nonnegative
integerIn this paper our aim is to investigate the uniqueness
problems of linear difference polynomials of 119891(119911) In partic-ular we primarily consider the linear difference polynomial119871(119911 119891) which satisfies one of the following conditions
(i) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1
(ii) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0
119873(119903
1
119871 (119911 119891)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(5)
Corresponding to the above question we obtain thefollowing results
Theorem 1 Let 119898 ge 2 and let 119899 ge 2119898 + 4 with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite orderand 119871(119911 119891) is of the form (4) satisfying the condition in (5)If 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) and 119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) then119871(119911 119891) equiv 119891(119911)
Corollary 2 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying the following
119873(119903
1
Δ119896
119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (6)
If 119864119891(119911)
(119878) = 119864Δ119896
119888119891(119911)
(119878) and 119864119891(119911)
(infin) = 119864Δ119896
119888119891(119911)
(infin) thenΔ119896
119888119891 equiv 119891(119911)
With an additional restriction on the order of growth of119891(119911) we prove the following fact
Theorem 3 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersuch that 120588(119891) notin N If 119864
119891(119911)(119878) = 119864
119871(119911119891)(119878) and 119864
119891(119911)(infin) =
119864119871(119911119891)
(infin) then 119871(119911 119891) equiv 119891(119911)
Remark 4 Note that in Theorem 3 we do not assume thatthe linear polynomial 119871(119911 119891) satisfies the condition in (5)In fact since 120588(119891) notin N by (19) we can easily get 120588(119890ℎ(119911)) =deg(ℎ(119911)) lt 120588(119891) which implies 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Thenusing a similar method as in the proof of Theorem 1 we cancomplete the proof of Theorem 3
Now we may ask what happens if the condition119898 ge 2 inTheorem 1 is replaced by a weaker condition containing thecase 119898 = 1 or even 119898 = 0 By considering three sets we getthe following theorem
Theorem 5 Let 119899119898 be nonnegative integers such that 119899 gt 119898Let 119886 and 119887 be nonzero constants such that 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878
1= 120596 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 =1198782= infin and 119878
3= 0 Suppose that 119891(119911) is a nonconstant
meromorphic function of finite order 119871(119911 119891) is of the form (4)satisfying the condition in (5) and 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for
119895 = 1 2 3 Then one has the following(i) If119898 = 0 then 119871(119911 119891) equiv 119905119891(119911) where 119905119899 = 1(ii) If 119899 and119898 are coprime then 119871(119911 119891) equiv 119891(119911)
Remark 6 Taking 119898 = 1 in Theorem 5 we can obtain ananalogue result of Theorem B related to linear differencepolynomials
Furthermore the following result is a corollary ofTheorem 5 related to difference operators
Corollary 7 Let 119899 119898 and 119878119895 119895 = 1 2 3 be given as in
Theorem 5 Suppose that 119891(119911) is a nonconstant meromorphicfunction of finite order satisfying
119873(119903
1
119891 (119911)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (7)
and 119864119891(119911)
(119878119895) = 119864
Δ119896
119888119891(119878119895) for 119895 = 1 2 3 Then one has the
following
(i) If119898 = 0 then Δ119896119888119891 equiv 119905119891(119911) where 119905119899 = 1
(ii) If 119899 and119898 are coprime then Δ119896119888119891 equiv 119891(119911)
Finally we give some examples for our results
Abstract and Applied Analysis 3
Examples In the following let 119892(119911) be an entire functionwith period 1 such that 120588(119892) isin (1infin) N (see [17])
(1) For the case (i) of condition (5) let 1198911(119911) = 119890
21205871198941199111198912(119911) = 119892(119911)119890
2120587119894119911 1198913(119911) = 119890
2120587119894119911
119892(119911) and let119871(119911 119891
119895) = 2119891
119895(119911) minus 119891
119895(119911 + 1) Then for 119895 = 1 2 3
119871(119911 119891119895) = 119891
119895(119911) and the sum of the coefficients of
119871(119911 119891119895) is equal to 1These examples satisfyTheorems
1 and 5 but do not satisfy Theorem D
(2) For the case (ii) of condition (5) let 119891(119911) = 119890119911 log 2119892(119911)and let 119871(119911 119891) = Δ119891(119911) = 119891(119911 + 1) minus 119891(119911) Then119871(119911 119891) = Δ119891(119911) = 119891(119911) the sum of the coefficientsof 119871(119911 119891) equals 0 and
119873(119903
1
Δ119891
) = 119873(119903
1
119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(8)
This example satisfiesTheorems 1 and 5 and Corollar-ies 2 and 7
(3) For Theorem 3 let 119891(119911) = 119890119911 log 3
119892(119911) and let119871(119911 119891) = 119891(119911 + 1) minus 2119891(119911) Then 119871(119911 119891) = 119891(119911)
and the sum of the coefficients of 119871(119911 119891) equals minus1This example satisfies Theorem 3 but does not satisfyTheorem D andTheorems 1 and 5
2 Proof of Theorem 1
We need the following lemmas for the proof of Theorem 1The difference analogue of the logarithmic derivative
lemmawas given byHalburd-Korhonen [7] andChiang-Feng[6] independently We recall the following lemmas
Lemma 8 (see [7]) Let 119891(119911) be a nonconstant meromorphicfunction of finite order 119888 isin C and 120575 lt 1 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 + |119888| 119891)
119903120575
) (9)
for all 119903 outside of a possible exceptional set with finite logarith-mic measure
Lemma 9 (see [8]) Let 119888 isin C let 119899 isin N and let 119891(119911) bea meromorphic function of finite order Then for any smallperiodic function 119886(119911) isin 119878(119891) with period 119888 consider thefollowing
119898(119903
Δ119899
119888119891
119891 (119911) minus 119886 (119911)
) = 119878 (119903 119891) (10)
where the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Let 119891(119911) be a meromorphic function of finite orderNotice that if 119871(119911 119891) ( equiv 0) is of the form (4) such that
1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 then for any given complex con-
stant 119886 119871(119911 119886) = 0 This indicates that 119871(119911 119891) = 119871(119911 119891 minus 119886)
and hence
119898(119903
119871 (119911 119891)
119891 minus 119886
) = 119898(119903
119871 (119911 119891 minus 119886)
119891 minus 119886
)
le
119896
sum
119895=0
119898(119903
119887119895(119911) (119891 (119911 + 119888
119895) minus 119886)
119891 minus 119886
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(11)
With this one can easily prove Lemma 10 below by a similarreasoning as in the proof of the difference analogue of thesecond main theorem of the Nevanlinna theory in [8] byHalburd and Korhonen We omit those details
Lemma 10 Let 119888 isin C let 119891(119911) be a meromorphic function offinite order and let 119871(119911 119891) equiv 0 be of the form (4) such that1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 Let 119902 ge 2 and let 119886
1 119886
119902be distinct
complex constants Then
119898(119903 119891) +
119902
sum
119894=1
119898(119903
1
119891 minus 119886119894
)
le 2119879 (119903 119891) minus 119873lowast
(119903 119891) + 119878 (119903 119891)
(12)
where
119873lowast
(119903 119891) = 2119873 (119903 119891) minus 119873 (119903 119871 (119911 119891)) + 119873(119903
1
119871 (119911 119891)
)
(13)
and the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Remark 11 If the linear difference polynomial 119871(119911 119891) isreplaced by
119871lowast
(119911 119891) = 119887119896(119911) 119891 (119911 + 119896119888)
+ sdot sdot sdot + 1198871(119911) 119891 (119911 + 119888) + 119887
0(119911) 119891 (119911)
(14)
Lemma 10 also holds even if the distinct complex constants1198861 119886
119902are replaced by 119886
1(119911) 119886
119902(119911) which are distinct
meromorphic periodic functions with period 119888 such that 119886119894isin
119878(119891) for all 119894 = 1 119902
The following is the standardValiron-Mohonrsquoko theorem(see Theorem 225 in the book of Laine [2])
Lemma 12 (see [2]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(15)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(16)
4 Abstract and Applied Analysis
The characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (17)
where 119889 = max119901 119902
Proof of Theorem 1 Since 119891(119911) and 119871(119911 119891) shareinfin CM wesee that 119871(119911 119891) equiv 0 and119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) Then byLemma 8 we have
119879 (119903 119871 (119911 119891)) = 119898 (119903 119871 (119911 119891)) + 119873 (119903 119871 (119911 119891))
le 119898(119903
119871 (119911 119891)
119891 (119911)
)
+ 119898 (119903 119891 (119911)) + 119873 (119903 119891 (119911))
le
119896
sum
119894=0
119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
+
119896
sum
119894=0
119898(119903 119887119894(119911)) + 119879 (119903 119891 (119911))
le 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(18)
Since 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) where 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots we know that (119871(119911 119891))119899+119886(119871(119911 119891))119899minus119898+119887 and119891(119911)119899+119886119891(119911)119899minus119898
+ 119887 share 0 CM Then from this and the condition119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) there exists a polynomial ℎ(119911)such that
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
= 119890ℎ(119911)
(19)
Suppose that 119890ℎ(119911) equiv 1 Note that 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots Let 120596
1 120596
119899denote all different roots of the equation
120596119899
+ 119886120596119899minus119898
+ 119887 = 0Next we prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) We know that
119871 (119911 119891) minus 120596119894= 119887119896(119911) (119891 (119911 + 119888
119896) minus 119891 (119911))
+ sdot sdot sdot + 1198870(119911) (119891 (119911 + 119888
0) minus 119891 (119911))
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
(20)
(i) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1 we see that
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 + (119891 (119911) minus 120596119894)
(21)
Then we deduce from this (19) and Lemma 9 that
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
= 119898(119903
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
)
= 119898(119903
(119871 (119911 119891) minus 1205961) sdot sdot sdot (119871 (119911 119891) minus 120596
119899)
(119891 (119911) minus 1205961) sdot sdot sdot (119891 (119911) minus 120596
119899)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(22)
(ii) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 we have
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 minus 120596119894 (23)
From this (19) and Lemma 9 we get
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
=
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
(24)
Applying Lemma 10 to 119891(119911) we get
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
)
le 2119879 (119903 119891 (119911)) minus 119898 (119903 119891 (119911)) minus 2119873 (119903 119891 (119911))
+ 119873 (119903 119871 (119911 119891)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
= 119879 (119903 119891 (119911)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
(25)
Abstract and Applied Analysis 5
Then the assumptions in (5) (24) and (25) yield thefollowing
119879 (119903 119890ℎ(119911)
) le 119879 (119903 119891 (119911))
minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891) = 119878 (119903 119891)
(26)
To sum up we now prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Rewriting(19) we get
(119871 (119911 119891))119899minus119898
[(119871 (119911 119891))119898
+ 119886]
= [119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887 minus 119887119890minusℎ(119911)
] 119890ℎ(119911)
(27)
Denote 119865(119911) = 119891(119911)119899
+ 119886119891(119911)119899minus119898 It follows from
Lemma 12 and119898 gt 0 that
119879 (119903 119865 (119911)) = 119899119879 (119903 119891 (119911)) + 119878 (119903 119891) (28)
Hence 119878(119903 119865) = 119878(119903 119891)By (18) and (27) and applying the second main theorem
for three small target functions we deduce the following
119879 (119903 119865 (119911))
le 119873 (119903 119865 (119911)) + 119873(119903
1
119865 (119911)
)
+ 119873(119903
1
119865 (119911) + 119887 minus 119887119890minus119901(119911)
) + 119878 (119903 119865)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891(119911)119899minus119898
[119891(119911)119898
+ 119886]
)
+ 119873(119903
1
(119871 (119911 119891))119899minus119898
) + 119873(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891 (119911)
) + 119873(119903
1
119891(119911)119898
+ 119886
)
+ 119873(119903
1
119871 (119911 119891)
) + 119879(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119879 (119903 119891 (119911)) + 119879(119903
1
119891 (119911)
) + 119879(119903
1
119891(119911)119898
+ 119886
)
+ 119879(119903
1
119871 (119911 119891)
) + 119898119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (119898 + 2) 119879 (119903 119891 (119911))
+ (119898 + 1) 119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (2119898 + 3) 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(29)
By combining (28) and (29) we have
(119899 minus 2119898 minus 3) 119879 (119903 119891 (119911)) le 119878 (119903 119891) (30)
which contradicts with 119899 ge 2119898 + 4Now we turn to consider the case 119890ℎ(119911) equiv 1 Equation (19)
yields the following
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
equiv 119891(119911)119899
+ 119886119891(119911)119899minus119898
(31)
Set 120593(119911) = 119871(119911 119891)119891(119911) and we have
119891(119911)119898
(120593(119911)119899
minus 1) = minus119886 (120593(119911)119899minus119898
minus 1) (32)
If 120593(119911) is not a constant (32) can be rewritten as
119891(119911)119898
(120593 (119911) minus 1) (120593 (119911) minus 120583) sdot sdot sdot (120593 (119911) minus 120583119899minus1
)
= minus119886 (120593 (119911) minus 1) (120593 (119911) minus ]) sdot sdot sdot (120593 (119911) minus ]119899minus119898minus1) (33)
where 120583 = cos(2120587119899) + 119894 sin(2120587119899) and ] = cos(2120587(119899minus119898)) +119894 sin(2120587(119899 minus 119898))
By the assumption that 119899 and 119899 minus 119898 have no commonfactors we see that 120583 120583119899minus1 ] ]119899minus119898minus1 are differentAssume that 119911
0is a 120583119895-point of 120593(119911) of multiplicity 119906
119895gt 0
where 1 le 119895 le 119899 minus 1 Notice that
minus119886 (120593 (1199110) minus 1) (120593 (119911
0) minus ]) sdot sdot sdot (120593 (119911
0) minus ]119899minus119898minus1) (34)
is a constantThen (33) implies that 1199110is a pole of119891(119911)119898Thus
119906119895ge 119898 This yields the following for 1 le 119895 le 119899 minus 1
119898119873(119903
1
120593 (119911) minus 120583119895
) le 119873(119903
1
120593 (119911) minus 120583119895
)
le 119879 (119903 120593 (119911)) + 119878 (119903 ℎ)
(35)
Then by (35) we get
2 ge
119899minus1
sum
119895=1
Θ(120583119895
120593 (119911)) =
119899minus1
sum
119895=1
1 minus lim119903rarrinfin
119873(119903 1 (120593 (119911) minus 120583119895
))
119879 (119903 120593 (119911))
ge
119899minus1
sum
119895=1
(1 minus
1
119898
) = (119899 minus 1) (1 minus
1
119898
)
(36)
which is impossible with119898 ge 2 and 119899 ge 2119898 + 4Hence 120593(119911) is a constant Since 119891(119911) is a nonconstant
meromorphic function we deduce from (32) that 120593(119911) equiv 1This yields 119871(119911 119891) equiv 119891(119911) which completes the proof ofTheorem 1
3 Proof of Theorem 5
Since 119891(119911) is a nonconstant meromorphic function of finiteorder 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for 119895 = 1 2 3 119878
1= 120596 120596
119899
+
119886120596119899minus119898
+ 119887 = 0 1198782= infin and 119878
3= 0 we have 119871(119911 119891) equiv
0 119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) and 119873(119903 1119871(119911 119891)) = 119873(119903
1119891(119911)) and we also get (18) and (19)
6 Abstract and Applied Analysis
Since 119891(119911) and 119871(119911 119891) share 0 infin CM there exists apolynomial ℎlowast(119911) such that
119871 (119911 119891)
119891 (119911)
= 119890ℎlowast(119911)
(37)
By Lemma 8 we see that
119879(119903 119890ℎlowast(119911)
) = 119898(119903 119890ℎlowast(119911)
) = 119898(119903
119871 (119911 119891)
119891 (119911)
)
le
119896
sum
119895=0
119898(119903
119891 (119911 + 119888119895)
119891 (119911)
)
+
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(38)
As in the proof of Theorem 1 we see that 119879(119903 119890ℎ(119911)) =
119878(119903 119891) still holds in both cases (i) and (ii)Rewriting (19) we have
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
minus 119890ℎ(119911)
119891(119911)119899
minus 119886119890ℎ(119911)
119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(39)
Combining this and (37) we get
(119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
+ 119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(40)
Suppose that 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 If119898 = 0 (40) becomes
(119886 + 1) (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
= 119887 (119890ℎ(119911)
minus 1) (41)
By the condition that 119887 = 0 1198781= 120596 (119886 + 1)120596
119899
+ 119887 = 0 =
implies 119886 = minus 1It follows from (38) (41) and 119879(119903 119890ℎ(119911)) = 119878(119903 119891) that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 119887 (119890ℎ(119911)
minus 1)) = 119878 (119903 119891)
(42)
which is a contradiction since 119899 ge 1If119898 ge 1 it follows from (38) (41) and119879(119903 119890ℎ(119911)) = 119878(119903 119891)
that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 minus119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
) 119891(119911)119899minus119898
+ 119887 (119890ℎ(119911)
minus 1))
le (119899 minus 119898)119879 (119903 119891) + 119878 (119903 119891)
(43)
That is impossible
Therefore 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 Notice that 119886 119887 = 0 Using asimilar method we can prove that 119890(119899minus119898)ℎ
lowast(119911)
minus119890ℎ(119911)
equiv 0Then(40) implies that 119890ℎ(119911) equiv 1
If 119898 = 0 we have 119890119899ℎlowast(119911)
equiv 1 Obviously 119890ℎlowast(119911) is a
constant Set 119905 = 119890ℎlowast(119911) Thus by (37) we get 119871(119911 119891) equiv 119905119891(119911)
where 119905119899 = 1If 119899 and 119898 are coprime 119890119899ℎ
lowast(119911)
equiv 1 and 119890119898ℎlowast(119911)
equiv 1 implythat 119890ℎ
lowast(119911)
equiv 1 Thus by (37) we get 119871(119911 119891) equiv 119891(119911) ThusTheorem 5 is proved
Conflict of Interests
The authors declare that they have no conflict of interests
Authorsrsquo Contribution
Both authors drafted the paper and read and approved thefinal paper
Acknowledgments
The authors are grateful to the editor and referees fortheir valuable suggestions This work was supported by theNNSFC (no 11171119 11301091) the Guangdong Natural Sci-ence Foundation (no S2013040014347) and the Foundationfor Distinguished Young Talents in Higher Education ofGuangdong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 2003
[4] P Li and C-C Yang ldquoSome further results on the unique rangesets of meromorphic functionsrdquo Kodai Mathematical Journalvol 18 no 3 pp 437ndash450 1995
[5] H-X Yi and W-C Lin ldquoUniqueness theorems concerning aquestion of Grossrdquo Proceedings of the Japan Academy Series AMathematical Sciences vol 80 no 7 pp 136ndash140 2004
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[8] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[9] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
Abstract and Applied Analysis 7
[10] B Chen and Z Chen ldquoMeromorphic function sharing twosets with its difference operatorrdquo Bulletin of the MalaysianMathematical Sciences Society Second Series vol 35 no 3 pp765ndash774 2012
[11] B Chen Z Chen and S Li ldquoUniqueness theorems on entirefunctions and their difference operators or shiftsrdquo Abstract andApplied Analysis vol 2012 Article ID 906893 8 pages 2012
[12] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[13] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[14] S Li and B Chen ldquoMeromorphic functions sharing smallfunctions with their linear difference polynomialsrdquoAdvances inDifference Equations vol 2013 article 58 2013
[15] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[16] J Zhang ldquoValue distribution and shared sets of differences ofmeromorphic functionsrdquo Journal of Mathematical Analysis andApplications vol 367 no 2 pp 401ndash408 2010
[17] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleA Comparison Theorem for Oscillation of the Even-OrderNonlinear Neutral Difference Equation
Quanxin Zhang
Department of Mathematics Binzhou University Binzhou Shandong 256603 China
Correspondence should be addressed to Quanxin Zhang 3314744163com
Received 9 December 2013 Accepted 28 March 2014 Published 10 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Quanxin Zhang This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
A comparison theorem on oscillation behavior is firstly established for a class of even-order nonlinear neutral delay differenceequations By using the obtained comparison theorem two oscillation criteria are derived for the class of even-order nonlinearneutral delay difference equations Two examples are given to show the effectiveness of the obtained results
1 Introduction
Recently there have been a lot of research papers in con-nection with the oscillation of solutions of difference equa-tions with or without neutral terms The literature on theoscillation of neutral delay difference equations is growingvery fast and it can be widely applied to the reality Infact neutral delay difference equations arise in modellingof the networks containing lossless transmission lines (asin high speed computers where the lossless transmissionlines are used to interconnect switching circuits) For recentcontributions regarding the theoretical part and providingsystematic treatment of oscillation of solutions of neutraltype difference equations the readers can refer to the recentmonographs by Agarwal [1] Gyori and Ladas [2]
The oscillation behavior of the even-order nonlinearneutral differential equation
[119909 (119905) + 119901 (119905) 119909 (120591 (119905))](119899)
+ 119902 (119905) 119891 [119909 (120590 (119905))] = 0 (1)
has been established by Zhang et al [3] In this paperthe discrete analogue of the above equation is consideredWe consider the even-order nonlinear neutral differenceequation
Δ119898
(119909119899+ 119901119899119909119899minus120591
) + 119902119899119891 (119909119899minus119896
) = 0 (2)
where 119898 ge 2 is an even and 120591 119896 isin N let N denote the setof all natural numbers 119899 isin 119873(119899
0) = 119899
0 1198990+ 1 1198990+ 2
1198990is a nonnegative integer Δ denotes the forward difference
operator defined by Δ119909119899= 119909119899+1
minus 119909119899 Δ119898119909
119899= Δ119898minus1
119909119899+1
minus
Δ119898minus1
119909119899
Throughout this paper the following conditions areassumed to hold
(H1) 119901119899 is a sequence of nonnegative real number 0 le
119901119899
lt 1 and 119902119899 is a sequence of nonnegative real
number with 119902119899 being not eventually identically
equal to zero(H2) 119891 R rarr R (R = (minusinfin +infin)) is a continuous oddfunction and 119909119891(119909) gt 0 for all 119909 = 0
Before deriving themain results the following definitionsare given
Definition 1 By a solution of (2) one means a real sequence119909119899 defined for 119899 ge 119899
0minus120579 (120579 = max120591 119896)which satisfies (2)
for 119899 isin 119873(1198990)
In this paper we restrict our attention to nontrivialsolutions of (2)
Definition 2 A nontrivial solution 119909119899 of (2) is said to
be oscillatory if the terms 119909119899of the sequence are neither
eventually positive nor eventually negative Otherwise it iscalled nonoscillatory
Definition 3 An equation is said to be oscillatory if all itssolutions are oscillatory
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 492492 5 pageshttpdxdoiorg1011552014492492
2 Abstract and Applied Analysis
In 2004 Stavroulakis [4] studied the oscillatory behaviorof all solutions of first-order delay difference equation
119909119899+1
minus 119909119899+ 119901119899119909119899minus119896
= 0 (3)
and established one new oscillation criterion Thandapaniet al [5] studied the oscillatory behavior of all solutions ofsecond-order neutral delay difference equation
Δ2
(119910119899minus 119901119910119899minus119896
) minus 119902119899119891 (119910119899minus119905
) = 0 (4)
and established a number of new oscillation criteria In 2000Zhou et al [6] studied the oscillatory behavior of all solutionsof even-order neutral delay difference equation
Δ119898
(119909119899minus 119901119899119892 (119909119899minus119896
)) minus 119902119899ℎ (119909119899minus119897
) = 0 (5)
and established three new oscillation criteria under certainconditionsThe studies on oscillatory behavior of all solutionsof even-order delay difference equations we recommendreferring to [7ndash10] On the basis of the abovework we studiedthe oscillatory behavior of all solutions of (2) Firstly a com-parison theorem on oscillation behavior is established for (2)The comparison theorem changes the discriminant criteria ofthe oscillation of (2) into the oscillationrsquos discriminant criteriain the first-order nonneutral delay difference equationsThenby using the above comparison theorem we obtain someoscillation criteria for (2) and improve the well-known resultsof Ladas et al [11] Erbe and Zhang [12] and Stavroulakis [4]In particular the results are new when119898 = 2 119901
119899equiv 0
The paper is organized as follows In Section 2 a compar-ison theorem on oscillation behavior is firstly established fora class of even-order nonlinear neutral delay difference equa-tions Then the comparison theorem changes the discrimi-nant of the oscillation in the even-order nonlinear neutraldelay difference equation into the oscillationrsquos discriminantin the first-order nonneutral delay difference equations InSection 3 some oscillation criteria are obtained for the classof even-order nonlinear neutral delay difference equationby using the above comparison theorem In Section 4 twoexamples are given
2 Comparison Theorem
To obtain the comparison theorem in this section we needthe following lemmas which can be founded in [1] see alsoChen [7] andThandapani and Arul [8]
Lemma 4 Let 119906119899 be a sequence of real numbers for 119899 ge 119899
0
Let 119906119899 and Δ
119898
119906119899 be of constant sign where Δ
119898
119906119899is not
identically zero for 119899 ge 1198991 If
119906119899Δ119898
119906119899le 0 (6)
then
(i) there is a natural number 1198992
ge 1198991such that the
sequences Δ119895119906119899 119895 = 1 2 119898 minus 1 are of constant
sign for 119899 ge 1198992
(ii) there exists a number 119897 isin 0 1 119898 minus 1 with(minus1)119898minus119897minus1
= 1 such that
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 0 1 2 119897 119899 ge 119899
2
(minus1)119895minus119897
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 119897 + 1 119898 minus 1 119899 ge 119899
2
(7)
Lemma 5 Observe that under the hypotheses of Lemma 4 if119906119899 is increasing for 119899 ge 119899
0 then there exists a natural number
1198991ge 1198990such that for all 119899 ge 2
119898minus1
1198991
119906119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119906119899 (8)
where 120582119898= 12(119898minus1)
2
Theorem 6 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 If there exists a constant
120582119898= 12(119898minus1)
2
such that the first-order difference equation
Δ119911119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119911119899minus119896
= 0 (9)
is oscillatory then (2) is oscillatory
Proof Suppose that (2) has a nonoscillatory solution 119909119899
Without the loss of generality we assume that 119909119899 is an
eventually positive solution of (2) then there is a naturalnumber 119899
1ge 1198990such that 119909
119899gt 0 119909
119899minus119896gt 0 119909
119899minus120591gt 0 and
119909119899minus119896minus120591
gt 0 for all 119899 ge 1198991 Let
119910119899= 119909119899+ 119901119899119909119899minus120591
(10)
Then from (H1) and (H
2) there exists a natural number 119899
2ge
1198991such that
119910119899gt 0 Δ
119898
119910119899le 0 forall119899 ge 119899
2 (11)
By Lemma 4 there exist an integer 1198993ge 1198992and an integer
119897 (0 le 119897 le 119898) where (119898 + 119897) is an odd integer For all 119899 ge 1198993
we can get
Δ119895
119910119899gt 0 for 119895 = 1 2 119897
(minus1)(119895minus119897)
Δ119895
119910119899gt 0 for 119895 = 119897 + 1 119898 minus 1
(12)
Thus from (12) Δ119910119899
gt 0 and Δ119898minus1
119910119899
gt 0 for 119899 ge 1198993 By
Lemma 5 there exists an integer 1198994ge 1198993 For all 119899 ge 2
119898minus1
1198994
we derive
119910119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119910119899 120582
119898=
1
2(119898minus1)
2 (13)
From (10)
119909119899minus119896
= 119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
(14)
Consequently we have
Δ119898
119910119899+ 119902119899119891 (119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) = 0
for all sufficient large 119899
(15)
Abstract and Applied Analysis 3
Noting that |119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) le 0
for all sufficient large 119899
(16)
By 119910119899ge 119909119899 Δ119910119899gt 0 and 119899 minus 119896 minus 120591 le 119899 minus 119896 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) ge Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
(17)
Therefore we have
Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
le 0
for all sufficient large 119899
(18)
Now by using (13) we have that for 120582119898= 12(119898minus1)
2
119910119899minus119896
ge
120582119898
(119898 minus 1)
(119899 minus 119896)119898minus1
Δ119898minus1
119910119899minus119896
for all sufficient large 119899
(19)
Thus we get
Δ119898
119910119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) Δ119898minus1
119910119899minus119896
le 0
for all sufficient large 119899
(20)
where120582119898= 12(119898minus1)
2
Let119906119899= Δ119898minus1
119910119899 then for large enough
119899 we get
Δ119906119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119906119899minus119896
le 0 (21)
where 120582119898
= 12(119898minus1)
2
Therefore inequality (21) has aneventually positive solution By Lemma 5 in [9] (9) hasan eventually positive solution which contradicts that (9) isoscillatory This completes the proof
3 Applications of the Comparison Theorem
The following lemma is well known (see eg [2 11 12] andthe references therein)
Lemma 7 Let 119902119899 be a sequence of eventually nonnegative
real number and 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt (
119896
1 + 119896
)
1+119896
(22)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894gt 1 (23)
then the first-order difference equation
Δ119909119899+ 119902119899119909119899minus119896
= 0 (24)
is oscillatory
Thus from Theorem 6 and Lemma 7 we can obtain thefollowing results
Theorem 8 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1)(
119896
1 + 119896
)
1+119896
(25)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 2(119898minus1)
2
(119898 minus 1) (26)
then (2) is oscillatory
Proof From (25) and (26) we can obtain
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt (
119896
1 + 119896
)
1+119896
(27)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 (28)
where 120582119898
= 12(119898minus1)
2
By Lemma 7 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According toTheorem 8 we obtain Corollary 9
Corollary 9 Assume that conditions (H1) and (H
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 when 119901
119899equiv 0
119898 = 2 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(
119896
1 + 119896
)
1+119896
(29)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (30)
then the second-order difference equation
Δ2
119909119899+ 119902119899119891 (119909119899minus119896
) = 0 (31)
is oscillatory
The following lemma is given in [4 Theorem 26]
Lemma 10 Let 119902119899 be a sequence of nonnegative real numbers
and 119896 a positive integer Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(32)
4 Abstract and Applied Analysis
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus
1205722
4
(33)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus 120572
119896
(34)
then (24) is oscillatory
Thus fromTheorem 6 and Lemma 10 we can obtain thefollowing results
Theorem 11 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(35)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus
1205722
4
)
(36)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus 120572119896
)
(37)
then (2) is oscillatory
Proof From (36) and (37) we can obtain
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus
1205722
4
(38)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus 120572119896
(39)
where 120582119898
= 12(119898minus1)
2
By Lemma 10 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According to Theorem 11 we can obtain the followingcorollary
Corollary 12 Assume that conditions (1198671) and (119867
2) hold let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
For 119901119899equiv 0119898 = 2 assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(40)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(1 minus
1205722
4
) (41)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (1 minus 120572
119896
) (42)
then (31) is oscillatory
4 Examples
Example 1 Considering the equation
Δ119898
(119909119899+
1
119899
119909119899minus119897
) +
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(43)
where 119899 gt 3119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
1
119899
lt 1 119902119899=
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(44)
where 119902119899 is a positive sequence Then
119899
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
1
119894 minus 2
)
=
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
(45)
Thus
lim sup119899rarrinfin
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
=
6
119890
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1)
(46)
Therefore by Theorem 8 (43) is oscillatory
Example 2 Considering the equation
Δ119898
[119909119899+
119899 minus 1
119899
119909119899minus119897
]
+
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(47)
Abstract and Applied Analysis 5
where 119899 gt 2119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
119899 minus 1
119899
lt 1
119902119899=
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(48)
where 119902119899 is a positive sequence Denote 120572 = 414 then
119899minus1
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
119894 minus 2 minus 1
119894 minus 2
)
=
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
(49)
Thus
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
=
15
16
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1) (1 minus (
4
14
)
2
)
(50)
Therefore by Theorem 11 (47) is oscillatory
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The author sincerely thanks the reviewers for their valuablesuggestions and useful comments that have led to the presentimproved version of the original paper This work wassupported by a Grant from the Natural Science Foundationof Shandong Province of China (no ZR2013AM003) andthe Development Program in Science and Technology ofShandong Province of China (no 2010GWZ20401)
References
[1] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[2] I Gyori and G Ladas Oscillation Theory of Delay DifferentialEquations Oxford Clarendon Press Oxford UK 1991
[3] Q Zhang J Yan and L Gao ldquoOscillation behavior of even-order nonlinear neutral differential equations with variablecoefficientsrdquoComputers andMathematics withApplications vol59 no 1 pp 426ndash430 2010
[4] I P Stavroulakis ldquoOscillation criteria for first order delaydifference equationsrdquo Mediterranean Journal of Mathematicsvol 1 no 2 pp 231ndash240 2004
[5] E Thandapani R Arul and P S Raja ldquoBounded oscillationof second order unstable neutral type difference equationsrdquoJournal of Applied Mathematics and Computing vol 16 no 1-2pp 79ndash90 2004
[6] Z Zhou J Yu and G Lei ldquoOscillations for even-order neu-tral difference equationsrdquo Korean Journal of Computational ampApplied Mathematics vol 7 no 3 pp 601ndash610 2000
[7] S Chen ldquoOscillation criteria for certain even order quasilineardifference equationsrdquo Journal of AppliedMathematics and Com-puting vol 31 no 1-2 pp 495ndash506 2009
[8] EThandapani and R Arul ldquoOscillatory and asymptotic behav-ior of solutions of higher order damped nonlinear differenceequationsrdquoCzechoslovakMathematical Journal vol 49 no 1 pp149ndash161 1999
[9] G Ladas and C Qian ldquoComparison results and linearizedoscillations for higher-order difference equationsrdquo InternationalJournal of Mathematics andMathematical Sciences vol 15 no 1pp 129ndash142 1992
[10] L Yang ldquoOscillation behavior of even-order neutral differenceequations with variable coefficientsrdquo Pure and Applied Mathe-matics vol 24 no 4 pp 796ndash801 2008 (Chinese)
[11] G Ladas Ch G Philos and Y G Sficas ldquoSharp conditions forthe oscillation of delay difference equationsrdquo Journal of AppliedMathematics and Simulation vol 2 no 2 pp 101ndash111 1989
[12] L H Erbe and B G Zhang ldquoOscillation of discrete analogues ofdelay equationsrdquo Differential and Integral Equations vol 2 no3 pp 300ndash309 1989
Research ArticleDifference Equations and Sharing ValuesConcerning Entire Functions and Their Difference
Zhiqiang Mao1 and Huifang Liu2
1 School of Mathematics and Computer Jiangxi Science and Technology Normal University Nanchang 330038 China2 College of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China
Correspondence should be addressed to Huifang Liu liuhuifang73sinacom
Received 19 January 2014 Accepted 22 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z Mao and H Liu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The value distribution of solutions of certain difference equations is investigated As its applications we investigate the differenceanalogue of the Bruck conjecture We obtain some results on entire functions sharing a finite value with their difference operatorsExamples are provided to show that our results are the best possible
1 Introduction and Main Results
In this paper the term meromorphic function will meanbeing meromorphic in the whole complex plane C Itis assumed that the reader is familiar with the standardnotations and the fundamental results of the Nevanlinnatheory see for example [1ndash3] In addition we use notations120590(119891) 120582(119891) to denote the order and the exponent of conver-gence of the sequence of zeros of a meromorphic function 119891respectivelyThe notation 119878(119903 119891) is defined to be any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside aset 119864 of 119903 of finite logarithmic measure
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 isin C We say that 119891 and 119892 share 119886 CM providedthat 119891 minus 119886 and 119892 minus 119886 have the same zeros with the samemultiplicities Similarly we say that 119891 and 119892 share 119886 IMprovided that 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
The famous results in the uniqueness theory of meromor-phic functions are the 5 IM and 4 CM shared values theoremsdue to Nevanlinna [4] It shows that if two nonconstantmeromorphic functions 119891 and 119892 share five different valuesIM or four different values CM then 119891 equiv 119892 or 119891 is a linearfractional transformation of119892 Condition 4CMshared valueshave been improved to 2 CM + 2 IM by Gundersen [5]while the case 1 CM + 3 IM still remains an open problemSpecifically Bruck posed the following conjecture
Conjecture 1 (see [6]) Let 119891 be a nonconstant entire functionsatisfying the hyperorder 120590
2(119891) lt infin where 120590
2(119891) is not a
positive integer If 119891 and 1198911015840 share a finite value 119886 CM then119891 minus 119886 equiv 119888(119891
1015840
minus 119886) for some nonzero constant 119888
In [6] Bruck proved that the conjecture is true providedthat 119886 = 0 or 119873(119903 11198911015840) = 119878(119903 119891) He also gave counter-examples to show that the restriction on the growth of 119891 isnecessary
In recent years as the research on the difference ana-logues of Nevanlinna theory is becoming active lots ofauthors [7ndash11] started to consider the uniqueness of mero-morphic functions sharing values with their shifts or theirdifference operators
Heittokangas et al proved the following result which is ashifted analogue of Bruckrsquos conjecture
Theorem A (see [8]) Let 119891 be a meromorphic function of120590(119891) lt 2 and 120578 a nonzero complex number If119891(119911) and119891(119911+120578)share a finite value 119886 andinfin CM then
119891 (119911 + 120578) minus 119886
119891 (119911) minus 119886
= 120591 (1)
for some constant 120591
In [8] Heittokangas et al gave the example 119891(119911) = 1198901199112
+1
which shows that 120590(119891) lt 2 cannot be relaxed to 120590(119891) le 2
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 584969 6 pageshttpdxdoiorg1011552014584969
2 Abstract and Applied Analysis
For a nonzero complex number 120578 we define differenceoperators as
Δ120578119891 (119911) = 119891 (119911 + 120578) minus 119891 (119911)
Δ119899
120578119891 (119911) = Δ
119899minus1
120578(Δ120578119891 (119911)) 119899 isin N 119899 ge 2
(2)
Regarding the difference analogue of Bruckrsquos conjecture wemention the following results
Theorem B (see [7]) Let 119891 be a finite order transcendentalentire function which has a finite Borel exceptional value 119886 andlet 120578 be a constant such that 119891(119911 + 120578) equiv 119891(119911) If 119891(119911) andΔ120578119891(119911) share 119886 CM then
119886 = 0
119891 (119911 + 120578) minus 119891 (119911)
119891 (119911)
= 119888 (3)
for some nonzero constant 119888
Theorem C (see [11]) Let 119891 be a nonperiodic transcendentalentire function of finite order If 119891(119911) and Δ119899
120578119891(119911) share a
nonzero finite value 119886 CM then 1 le 120590(119891) le 120582(119891 minus 119886) + 1that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (4)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891 minus 119886) and119876(119911) is a polynomial with deg119876 le 120590(119860) + 1
Let 119891 be a nonperiodic transcendental entire function offinite order Theorem B shows that if a nonzero finite value119886 is shared by 119891(119911) and Δ
120578119891(119911) then 120590(119891) = 120582(119891 minus 119886)
It is obvious that the result in Theorem B is sharper thanTheoremC for 119899 = 1 In this paper we continue to investigatethe difference analogue of Bruckrsquos conjecture and obtain thefollowing result
Theorem 2 Let 119891 be a finite order entire function 119899 ge 2 aninteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share a finite value 119886 ( = 0)CM then120582(119891minus119886) = 120590(119891) ge
1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (5)
where 119860(119911) is an entire function with 1 le 120590(119860) = 120582(119891 minus 119886) =120590(119891) and 119876(119911) is a polynomial with deg119876 le 120590(119860)
Remark 3 It is obvious that Theorem 2 is sharper thanTheorem C and a supplement of Theorem B for 119899 ge 2
The discussions in Theorems C and 2 are concerning thecase that shared value 119886 = 0 When 119886 = 0 we obtain thefollowing result
Theorem 4 Let 119891 be a finite order entire function 119899 a positiveinteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share 0 CM then 1 le 120590(119891) le 120582(119891) + 1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
(6)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891) and 119876(119911)is a polynomial with deg119876 le 120590(119860) + 1
It is well known that if a finite order entire function 119891(119911)shares 119886 CM with Δ119899
120578119891(119911) then 119891(119911) satisfies the difference
equation
Δ119899
120578119891 (119911) minus 119886 = 119890
119876(119911)
(119891 (119911) minus 119886) (7)
where119876(119911) is a polynomialHence in order to prove the aboveresults we consider the value distribution of entire solutionsof the difference equation
119886119899(119911) 119891 (119911 + 119899120578) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 120578)
+ (1198860(119911) minus 119890
119876(119911)
) 119891 (119911) = 119861 (119911)
(8)
and obtain the following result
Theorem 5 Let 1198860 119886
119899minus1 119886119899( equiv 0) 119861( equiv 0) be polynomials
and let 119876 be a polynomial with degree 119898(ge 1) Then everyentire solution 119891 of finite order of (8) satisfies 120590(119891) ge 119898 and
(i) if 120590(119891) gt 1 then 120582(119891) = 120590(119891)(ii) if 120590(119891) = 1 then 120582(119891) = 120590(119891) or 119891 has only finitely
many zeros
2 Lemmas
Lemma 6 (see [12]) Let 119879 (0 +infin) rarr (0 +infin) be anondecreasing continuous function 119904 gt 0 120572 lt 1 and let119865 sub R+ be the set of all 119903 such that 119879(119903) le 120572119879(119903 + 119904) If thelogarithmic measure of 119865 is infinite then
lim119903rarrinfin
log119879 (119903)log 119903
= infin (9)
Lemma 7 (see [13]) Let 119891 be a nonconstant meromorphicfunction of finite order 120578 isin C 120575 lt 1 Then
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119900(
119879 (119903 +10038161003816100381610038161205781003816100381610038161003816 119891)
119903120575
) (10)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
(119889119903119903) lt infin
Remark 8 By Lemmas 6 and 7 we know that for a noncon-stant meromorphic function 119891 of finite order
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119878 (119903 119891) (11)
Lemma 9 (see [3]) Let 119891119895(119895 = 1 119899 + 1) and 119892
119895(119895 =
1 119899) be entire functions such that
(i) sum119899119895=1119891119895(119911)119890119892119895(119911)
equiv 119891119899+1(119911)
(ii) the order of 119891119895is less than the order of 119890119892119896 for 1 le 119895 le
119899+1 1 le 119896 le 119899 and furthermore the order of119891119895is less
than the order of 119890119892ℎminus119892119896 for 119899 ge 2 and 1 le 119895 le 119899+1 1 leℎ lt 119896 le 119899
Then 119891119895(119911) equiv 0 (119895 = 1 119899 + 1)
Abstract and Applied Analysis 3
Lemma 10 (see [14]) Let 119891 be a meromorphic function withfinite order 120590(119891) = 120590 lt 1 120578 isin C 0 Then for any given 120576 gt 0and integers 0 le 119895 lt 119896 there exists a set 119864 sub (1infin) of finitelogarithmic measure so that for all |119911| = 119903 notin 119864 ⋃ [0 1] wehave
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119896
120578119891 (119911)
Δ119895
120578119891 (119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|(119896minus119895)(120590minus1)+120576
(12)
Lemma 11 (see [15]) Let 1198860(119911) 119886
119896(119911) be entire functions
with finite order If there exists an integer 119897 (0 le 119897 le 119896) suchthat
120590 (119886119897) gt max0le119895le119896
119895 = 119897
120590 (119886119895) (13)
holds then every meromorphic solution 119891( equiv 0) of thedifference equation
119886119896(119911) 119891 (119911 + 119896) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 1) + 119886
0(119911) 119891 (119911) = 0
(14)
satisfies 120590(119891) ge 120590(119886119897) + 1
3 Proofs of Results
Proof of Theorem 5 Let 119891 be an entire solution of finite orderof (8) By Remark 8 and (8) we get
119879 (119903 119890119876
) = 119879 (119903 119890119876
minus 1198860) + 119878 (119903 119890
119876
)
le
119899
sum
119895=1
119898(119903
119891 (119911 + 119895120578)
119891 (119911)
)
+
119899
sum
119895=0
119898(119903 119886119895) + 119898(119903
119861 (119911)
119891 (119911)
) + 119878 (119903 119890119876
)
le 119879 (119903 119891) + 119878 (119903 119891) + 119878 (119903 119890119876
)
(15)
By (15) we get 120590(119891) ge 119898
Case 1 (120590(119891) gt 1) Suppose that 120582(119891) lt 120590(119891) by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that
120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891) lt 120590 (119891) = deg ℎ
2 (16)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (1198860(119911) minus 119890
119876(119911)
) ℎ1(119911) = 119861 (119911) 119890
minusℎ2(119911)
(17)
If deg ℎ2gt 119898 then by (16) we know that the order of the right
side of (17) is deg ℎ2 and the order of the left side of (17) is less
than deg ℎ2 This is a contradiction Hence deg ℎ
2= 119898 gt 1
Set119876 (119911) = 119887
119898119911119898
+ sdot sdot sdot + 1198870
ℎ2(119911) = 119888
119898119911119898
+ sdot sdot sdot + 1198880
(18)
where 119887119898( = 0) 119887
0 119888119898( = 0) 119888
0are complex numbers
By (17) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 1198860(119911) ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
+ 119861 (119911) 119890minusℎ2(119911)
(19)
Next we discuss the following two subcases
Subcase 1 (119887119898+ 119888119898=0)Then by Lemma 9 (16) and (19) we
get 119861(119911) equiv 0 ℎ1(119911) equiv 0 This is impossible
Subcase 2 (119887119898+ 119888119898= 0) Suppose that
ℎ1(119911) 119890119876(119911)minus119887
119898119911119898
+ 119861 (119911) 119890minusℎ2(119911)minus119887119898119911119898
equiv 0 (20)
Then ℎ1(119911) = minus119861(119911)119890
minusℎ2(119911)minus119876(119911) By 120590(ℎ
1) = 120582(ℎ
1) we
obtain that 119890minusℎ2(119911)minus119876(119911) is a nonzero constant Hence ℎ1(119911) is a
nonzero polynomial By (19) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
= minus1198860(119911) ℎ1(119911)
(21)
Since degℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578) = 119898 minus 1 gt 0 for 119894 = 119895 then
by Lemma 9 and (21) we get
119886119895(119911) ℎ1(119911 + 119895120578) equiv 0 (119895 = 0 1 119899) (22)
This is impossible Hence we have ℎ1(119911)119890119876(119911)minus119887
119898119911119898
+
119861(119911)119890minusℎ2(119911)minus119887119898119911119898
equiv 0 Then from the order considerationwe know that the order of the right side of (19) is 119898 andthe order of the left side of (19) is less than 119898 This is acontradiction Hence 120582(119891) = 120590(119891)
Case 2 (120590(119891) = 1)Then by 120590(119891) ge 119898 we get119898 = 1 Supposethat 119891(119911) has infinitely many zeros and 120582(119891) lt 120590(119891) by theWeierstrass factorization we get
119891 (119911) = ℎ3(119911) 119890120573119911
(23)
where 120573( = 0) is a complex number and ℎ3(119911)( equiv 0) is an
entire function such that
120590 (ℎ3) = 120582 (ℎ
3) = 120582 (119891) lt 1 (24)
Let 119876(119911) = 1198871119911 + 1198870 where 119887
1( = 0) 119887
0are complex numbers
Substituting 119891(119911) = ℎ3(119911)119890120573119911 into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ3(119911 + 119895120578) 119890
120573119895120578
+ 1198860(119911) ℎ3(119911)
= ℎ3(119911) 1198901198871119911+1198870+ 119861 (119911) 119890
minus120573119911
(25)
4 Abstract and Applied Analysis
Note that ℎ3(119911)1198901198870+ 119861(119911) equiv 0 otherwise 119891 has only finitely
many zeros If 1198871+ 120573 = 0 then the order of the right side of
(25) is 1 but the order of the left side of (25) is less than 1Thisis absurd If 119887
1+120573 = 0 then by Lemma 9 (24) and (25) we get
ℎ3(119911) equiv 0 119861(119911) equiv 0 This is impossible Hence 120582(119891) = 120590(119891)
Theorem 5 is thus completely proved
Proof of Theorem 2 Since 119891(119911) and Δ119899120578119891(119911) share 119886 CM and
119891 is of finite order then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119890119876(119911)
(26)
where 119876(119911) is a polynomial with deg119876 le 120590(119891) Now we willtake two steps to complete the proof
Step 1 We prove that 120582(119891 minus 119886) = 120590(119891)Let 119865(119911) = 119891(119911) minus 119886 then
120582 (119865) = 120582 (119891 minus 119886) 120590 (119865) = 120590 (119891) ge deg119876 (27)
and Δ119899120578119891(119911) = Δ
119899
120578119865(119911) = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119865(119911 + 119895120578) By thisand (26) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119890119876(119911)
) 119865 (119911) = 119886 (28)
Next we discuss the following three cases
Case 1 (deg119876 ge 1 and 120590(119865) gt deg119876) Then 120590(119865) gt 1 ByTheorem 5(i) (27) and (28) we get 120582(119891 minus 119886) = 120590(119891)
Case 2 (deg119876 ge 1 and 120590(119865) = deg119876) If 120590(119865) = deg119876 gt 1then byTheorem 5(i) (27) and (28) we get120582(119891minus119886) = 120590(119891) If120590(119865) = deg119876 = 1 then byTheorem 5(ii) and (27) we obtainthat 120582(119891 minus 119886) = 120590(119891) or 119865 has only finitely many zeros
If 119865 has only finitely many zeros set
119865 (119911) = ℎ1(119911) 119890119887119911
(29)
where ℎ1(119911)( equiv 0) is a polynomial and 119887( = 0) is a complex
number then substituting (29) into (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
119887119895120578
+ (minus1)119899
ℎ1(119911)
= ℎ1(119911) 119890119876(119911)
+ 119886119890minus119887119911
(30)
By (30) and Δ119899120578119865(119911) equiv 0 we know that the order of the left
side of (30) is 0 and the order of the right side of (30) is 1unless ℎ
1(119911) = minus119886 and119876(119911) = minus119887119911 In this case take it into the
left side of (30) we have (minus119886)(119890119887120578 minus 1)119899 = 0 Since all 119886 119887 and120578 are not zero it is impossible Hence we get 120582(119891minus119886) = 120590(119891)
Case 3 119876 is a complex constant Then by (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119888) 119865 (119911) = 119886 (31)
where 119888 (= 119890119876 =0) is a complex number Suppose that 120582(119865) lt120590(119865) Let 119865(119911) = ℎ
2(119911)119890ℎ3(119911) where ℎ
2(119911)( equiv 0) is an entire
function and ℎ3(119911) is a polynomial such that
120590 (ℎ2) = 120582 (ℎ
2) = 120582 (119865) lt 120590 (119865) = deg ℎ
3 (32)
Substituting 119865(119911) = ℎ2(119911)119890ℎ3(119911) into (31) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ2(119911 + 119895120578) 119890
ℎ3(119911+119895120578)minusℎ
3(119911)
+ ((minus1)119899
minus 119888) ℎ2(119911) = 119886119890
minusℎ3(119911)
(33)
Since deg(ℎ3(119911 + 119895120578) minus ℎ
3(119911)) = deg ℎ
3(119911) minus 1 (119895 = 1 119899)
by (32) we obtain that the order of the left side of (33) is lessthan deg ℎ
3and the order of the right side of (33) is deg ℎ
3
This is absurd Hence we get 120582(119891 minus 119886) = 120590(119891)
Step 2We prove that 120590(119891) ge 1Suppose that 120590(119891) lt 1 Since 119891(119911) and Δ119899
120578119891(119911) share 119886
CM then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119888 (34)
where 119888 is a nonzero constant Let 119865(119911) = 119891(119911) minus 119886 then by(34) we get
Δ119899
120578119865 (119911) = 119888119865 (119911) + 119886 (35)
Differentiating (35) we get
(Δ119899
120578119865 (119911))
1015840
= 1198881198651015840
(119911) (36)
Note that (Δ119899120578119865(119911))1015840
= Δ119899
120578(1198651015840
(119911)) and 120590(1198651015840) = 120590(119865) = 120590(119891) lt1 So by Lemma 10 and (36) we get
|119888| =
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119899
120578(1198651015840
(119911))
1198651015840(119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|119899(120590(119865)minus1)+120576
997888rarr 0 (37)
This is absurd So 120590(119891) ge 1 Theorem 2 is thus completelyproved
Proof of Theorem 4 Since 119891(119911) and Δ119899120578119891(119911) share 0 CM and
119891 is of finite order then
Δ119899
120578119891 (119911)
119891 (119911)
= 119890119876(119911)
(38)
where119876(119911) is a polynomial ByΔ119899120578119891 = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119891(119911+
119895120578) and (38) we get
119891 (119911 + 119899120578)
+
119899minus1
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119891 (119911 + 119895120578)
+ ((minus1)119899
minus 119890119876(119911)
) 119891 (119911) = 0
(39)
Abstract and Applied Analysis 5
We discuss the following two cases
Case 1 119876 is a polynomial with deg119876 = 119898 ge 1 Then byLemma 11 and (39) we get 120590(119891) ge 119898 + 1 Now we prove120590(119891) le 120582(119891) + 1 Suppose that 120590(119891) gt 120582(119891) + 1 then by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891)
120590 (119891) = deg ℎ2gt 120590 (ℎ
1) + 1
(40)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (39) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (minus1)119899
ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
(41)
If 120590(119891) gt 119898 + 1 then by (40) (41) and deg(ℎ2(119911 + 119895120578) minus
ℎ2(119911)) = deg ℎ
2(119911)minus1 (119895 = 1 119899) we obtain that the order
of the left side of (41) is deg ℎ2minus 1 and the order of the right
side of (41) is less than deg ℎ2minus 1 This is absurd
If 120590(119891) = 119898 + 1 then by (41) we get119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
minus ℎ1(119911) 119890119876(119911)
= (minus1)119899+1
ℎ1(119911)
(42)
Set
ℎ2(119911) = 119889
119898+1119911119898+1
+ sdot sdot sdot + 1198890 (43)
where 119889119898+1( = 0) 119889
0are complex numbers Then
ℎ2(119911 + 119895120578) minus ℎ
2(119911)
= 119889119898+1
(119898 + 1) 119895120578119911119898
+ sdot sdot sdot + 1198891119895120578
(119895 = 1 119899)
(44)
Now we discuss the following two subcases
Subcase 1 deg(119876(119911)minus(ℎ2(119911+119895120578)minusℎ
2(119911))) = 119898 holds for every
119895 isin 1 119899Then by (40) (42) deg(ℎ2(119911+119895120578)minusℎ
2(119911+119894120578)) =
119898 (119895 = 119894) and Lemma 9 we get ℎ1(119911) equiv 0 This is absurd
Subcase 2 There exist some 1198950isin 1 119899 such that
deg(119876(119911)minus(ℎ2(119911+1198950120578)minusℎ2(119911))) le 119898minus1Then by (44) we have
deg(119876(119911) minus (ℎ2(119911 + 119895120578) minus ℎ
2(119911))) = 119898 for 119895 = 119895
0 Merging the
term minusℎ1(119911)119890119876(119911) into ( 119899119895
0) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578)119890ℎ2(119911+1198950120578)minusℎ2(119911)
by (42) we get119899
sum
119895 = 1
119895 =1198950
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 ge 2)
(45)
or
119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 = 1) (46)
where 119860(119911) = (119899
1198950) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578) minus ℎ
1(119911)
119890119876(119911)minus(ℎ
2(119911+1198950120578)minusℎ2(119911)) satisfying 120590(119860) lt 119898 If 119899 ge 2 then
by (40) (45) deg(ℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578)) = 119898 (119895 = 119894) and
Lemma 9 we get ℎ1(119911) equiv 0 This is absurd If 119899 = 1 then by
(46) and ℎ1(119911) equiv 0 we get 119860(119911) equiv 0 By this we know that
the order of the left side of (46) is 119898 and the order of theright side of (46) is less than119898 This is absurd Hence we get120590(119891) le 120582(119891) + 1
Case 2119876 is a complex constantThen by Lemma 10 and (38)we get 120590(119891) ge 1 Now we prove 120590(119891) le 120582(119891) + 1 Supposethat 120590(119891) gt 120582(119891) + 1 If 120590(119891) gt 1 then by the similarargument to that of case 1 we get ℎ
1(119911) equiv 0 This is absurd If
120590(119891) = 1 then by (40) we get 0 le 120582(119891) lt 120590(119891) minus 1 = 0 Since120582(119891) = 0 then 120590(119891) = 120582(119891)+1Theorem 4 is thus completelyproved
4 Some Examples
The following examples show the existence of such entirefunctions which satisfy Theorems 2ndash5 Moreover Example 2shows that the result in Theorem 4 is the best possible
Example 1 Let 120578 = 1 119899 = 2 and 119891(119911) = (119889 + 1)119911 + ((1198892 minus1)1198892
)119886 where 119886( = 0) 119889( = 0 plusmn1) are constants Then 119891(119911)and Δ119899
120578119891(119911) share 119886 CM and 120590(119891) = 120582(119891 minus 119886) = 1
Example 2 Let 120578 = 1 119899 = 2 and 119891(119911) = 119890119911 Then 119891(119911) andΔ119899
120578119891(119911) share 0 CM and 120590(119891) = 1 = 120582(119891) + 1
Example 3 Let 120578 = 1 119899 = 2 and119891(119911) = 119867(119911)119890119911 where119867(119911)is an entire function with period 1 such that 120590(119867) gt 1 and120590(119867) notin N Then 119891(119911) and Δ119899
120578119891(119911) share 0 CM and 120582(119891) =
120582(119867) = 120590(119867) = 120590(119891) gt 1 (Ozawa [16] proved that for any120590 isin [1infin) there exists a period entire function of order 120590)
Example 4 The entire function 119891(119911) = 119911119890minus119911 satisfies the
difference equation
119891 (119911 + 2120578) minus 4119891 (119911 + 120578) + (4 minus 119890119911
) 119891 (119911) = minus119911 (47)
where 120578 = minus log 2 Here120590(119891) = 1 and119891 has only finitelymanyzeros
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11201195 11171119) and the NaturalScience Foundation of Jiangxi China (nos 20122BAB20101220132BAB201008)
6 Abstract and Applied Analysis
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers New York NY USA2003
[4] R Nevanlinna Le Theoreme de Picard-Borel et la Theorie desFonctions Meromorphes Gauthiers-Villars Paris France 1929
[5] G G Gundersen ldquoMeromorphic functions that share fourvaluesrdquo Transactions of the American Mathematical Society vol277 no 2 pp 545ndash567 1983
[6] R Bruck ldquoOn entire functions which share one value CM withtheir first derivativerdquo Results inMathematics vol 30 no 1-2 pp21ndash24 1996
[7] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[8] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[9] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables andElliptic Equations vol 56 pp 81ndash92 2011
[10] K Liu and L-Z Yang ldquoValue distribution of the differenceoperatorrdquo Archiv der Mathematik vol 92 no 3 pp 270ndash2782009
[11] S Li and Z Gao ldquoA note on the Bruck conjecturerdquo Archiv derMathematik vol 95 no 3 pp 257ndash268 2010
[12] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[13] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[14] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[15] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[16] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleAdmissible Solutions of the Schwarzian Type DifferenceEquation
Baoqin Chen and Sheng Li
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to Sheng Li lish lssinacom
Received 14 January 2014 Accepted 20 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 B Chen and S Li This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper is to investigate the Schwarzian type difference equation [(Δ3
119891Δ119891) minus (32) (Δ2
119891Δ119891)
2
]
119896
= 119877 (119911 119891) =
(119875(119911 119891)119876(119911 119891)) where 119877(119911 119891) is a rational function in 119891 with polynomial coefficients 119875(119911 119891) respectively 119876(119911 119891) are twoirreducible polynomials in 119891 of degree 119901 respectively 119902 Relationship between 119901 and 119902 is studied for some special case Denote119889 = max 119901 119902 Let 119891(119911) be an admissible solution of (lowast) such that 120588
2(119891) lt 1 then for 119904 (ge2) distinct complex constants 120572
1 120572
119904
119902 + 2119896sum119904
119895=1120575(120572119895 119891) le 8119896 In particular if119873(119903 119891) = 119878(119903 119891) then 119889 + 2119896sum119904
119895=1120575(120572119895 119891) le 4119896
1 Introduction and Results
Throughout this paper a meromorphic function alwaysmeans being meromorphic in the whole complex plane and119888 always means a nonzero constant For a meromorphicfunction 119891(119911) we define its shift by 119891(119911 + 119888) and define itsdifference operators by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911) Δ
119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911))
119899 isin N 119899 ge 2
(1)
In particular Δ119899119888119891(119911) = Δ
119899
119891(119911) for the case 119888 = 1 We usestandard notations of theNevanlinna theory ofmeromorphicfunctions such as 119879(119903 119891) 119898(119903 119891) and 119873(119903 119891) and as statedin [1ndash3] For a constant 119886 we define theNevanlinna deficiencyby
120575 (119886 119891) = lim inf119903rarrinfin
119898(119903 1 (119891 minus 119886))
119879 (119903 119891)
= 1 minus lim sup119903rarrinfin
119873(119903 1 (119891 minus 119886))
119879 (119903 119891)
(2)
Recently numbers of papers (see eg [4ndash12]) are devotedto considering the complex difference equations and differ-ence analogues of Nevanlinna theory Due to some idea of[13] we consider the admissible solution of the Schwarziantype difference equation
119878119896(119891) = [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(3)
where 119877(119911 119891) is a rational function in 119891 with polynomialcoefficients 119875(119911 119891) respectively119876(119911 119891) are two irreduciblepolynomials in 119891 of degree 119901 respectively 119902 Here and in thefollowing ldquoadmissiblerdquo always means ldquotranscendentalrdquo Andwe denote 119889 = max119901 119902 from now on For the existence ofsolutions of (3) we give some examples below
Examples (1) 119891(119911) = sin 120587119911 + 119911 is an admissible solution ofthe Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus8 [1198912
+ (1 minus 2119911) 119891 + 119911 (119911 minus 1)]
41198912minus 4 (2119911 + 1) 119891 + (2119911 + 1)
2
(4)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 306360 5 pageshttpdxdoiorg1011552014306360
2 Abstract and Applied Analysis
(2) 119891(119911) = (119890119911 ln 2 sin 2120587119911) + 119911 is an admissible solutionof the Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus1198912
+ 2 (119911 + 1) 119891 minus 1199112
minus 2119911
21198912minus 4 (119911 minus 1) 119891 + 2(119911 minus 1)
2 (5)
(3) Let 119891(119911) = 1199112
+ 119911 then 119891(119911) solves the Schwarziantype difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
= minus
3
2 [1198912minus 2 (119911
2minus 1) 119891 + (119911
2minus 1)2
]
(6)
This example shows that (3) may admit polynomial solutions
Considering the relationship between 119901 and 119902 in thoseexamples above we prove the following result
Theorem 1 For the Schwarzian type difference equation (3)with polynomial coefficients note the following
(i) If it admits an admissible solution 119891(119911) such that1205882(119891) lt 1 then
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (7)
In particular if119898(119903 119891) = 119878(119903 119891) then 119901 le 119902(ii) If its coefficients are all constants and it admits a
polynomial solution 119891(119911) with degree 119904 then 119904 ge 2 and119902119904 = 119901119904 + 2119896
Remark 2 From examples (1) and (2) we conjecture that119901 = 119902 inTheorem 1(i) However we cannot prove it currentlyFrom example (3) given before we see that the restriction onthe coefficients in Theorem 1(ii) cannot be omitted
For the Schwarzian differential equation
119878119896(119891) = [
119891101584010158401015840
1198911015840
minus
3
2
(
11989110158401015840
1198911015840
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(8)
where 119877(119911 119891) 119875(119911 119891) and 119876(119911 119891) are as stated beforeIshizaki [13] proved the following result (see also Theorem932 in [2])
Theorem A (see [2 13]) Let 119891(119911) be an admissible solutionof (8) with polynomial coefficients and let 120572
1 120572
119904be 119904 (ge2)
distinct complex constants Then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (9)
For the Schwarzian type difference equation (3) we provethe following result
Theorem 3 Let 119891(119911) be an admissible solution of (3) withpolynomial coefficients such that 120588
2(119891) lt 1 and let 120572
1 120572
119904
be 119904 (ge2) distinct complex constants Then
119902 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 8119896 (10)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (11)
Remark 4 From Theorem 1 under the condition 119873(119903 119891) =119878(119903 119891) in Theorem 3 we have 119889 = 119902 in (11) The behavior ofthe zeros and the poles of 119891(119911) in 119878
119896(119891) is essentially different
from that in the 119878119896(119891) We wonder whether the restriction
119873(119903 119891) = 119878(119903 119891) can be omitted or not
2 Lemmas
The following lemmaplays a very important role in the theoryof complex differential equations and difference equationsIt can be found in Mohonrsquoko [14] and Valiron [15] (see alsoTheorem 225 in the book of Laine and Yang [2])
Lemma 5 (see [14 15]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(12)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(13)
and the characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (14)
where 119889 = max119901 119902
The following two results can be found in [10] In factLemma 6 is a special case of Lemma 83 in [10]
Lemma 6 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891) (15)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 7 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120589minus120576
) = 119878 (119903 119891) (16)
possibly outside of a set of 119903 with finite logarithmic measure
From Lemma 7 we can easily get the following conclu-sion
Abstract and Applied Analysis 3
Lemma 8 Let 119891(119911) be a meromorphic function of hyper order1205882(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
Δ119899
119888119891 (119911)
119891 (119911)
) = 119878 (119903 119891)
119898(119903
Δ119896
119888119891 (119911)
Δ119895
119888119891 (119911)
) = 119878 (119903 119891) 119896 gt 119895
(17)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 9 Let 119891 be an admissible solution of (3) withcoefficients Then using the notation 119876(119911) = 119876(119911 119891(119911))
119902119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (18)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (19)
Proof We use the idea by Ishizaki [13] (see also [2]) to proveLemma 9 It follows from Lemma 8 that
119898(119903 119877) = 119898(119903 [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
)
le 119896119898(119903
Δ3
119891
Δ119891
) + 2119896119898(119903
Δ2
119891
Δ119891
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(20)
From this and Lemma 5 we get
119889119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119877) = 119873 (119903 119877) + 119878 (119903 119891) (21)
and hence
119889119879 (119903 119891) = 119873 (119903 119877) + 119878 (119903 119891) (22)
If 119889 = 119901 gt 119902 since all coefficients of 119875(119911 119891) and 119876(119911 119891)are polynomials there are at the most finitely many poles of119877(119911 119891) neither the poles of 119891(119911) nor the zeros of 119876(119911 119891)Therefore we see that
119873(119903 119877) le (119901 minus 119902)119873 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
le (119901 minus 119902) 119879 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
(23)
We obtain (18) from this and (22) immediatelyIf 119889 = 119902 ge 119901 there are at most finitely many poles of
119877(119911 119891) not the zeros of 119876(119911 119891) then
119873(119903 119877) le 119873(119903
1
119876
) + 119878 (119903 119891) (24)
Now (18) follows from (22) and (24)Notice that if 119873(119903 119891) = 119878(119903 119891) then (24) always holds
This finishes the proof of Lemma 9
3 Proof of Theorem 1
Case 1 Equation (3) admits an admissible solution 119891(119911) suchthat 1205882(119891) lt 1 Since all coefficients of119875(119911 119891) and119876(119911 119891) are
polynomials there are at the most finitely many poles of 119891(119911)that are not the poles of119875(119911 119891) and119876(119911 119891) This implies that
119873(119903 119875) = 119901119873 (119903 119891) + 119878 (119903 119891)
119873 (119903 119876) = 119902119873 (119903 119891) + 119878 (119903 119891)
(25)
From Lemma 5 we get
119879 (119903 119875) = 119901119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 119876) = 119902119879 (119903 119891) + 119878 (119903 119891)
(26)
We can deduce from (3) (25) (26) and Lemma 8 that
119901119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119875)
= 119898 (119903 119875) + 119873 (119903 119875)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891)119876)
+ 119878 (119903 119891)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891))
+ 119898 (119903 119876) + 119878 (119903 119891)
= 119901119873 (119903 119891) + 119879 (119903 119876) minus 119873 (119903 119876)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119879 (119903 119891) minus 119902119873 (119903 119891)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119898 (119903 119891) + 119878 (119903 119891)
(27)
It follows from this that
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (28)
What is more is that if119898(119903 119891) = 119878(119903 119891) then we obtain from(28) that 119901 le 119902
Case 2 The coefficients of (3) are all constants and it admitsa polynomial solution 119891(119911) with degree 119904 Set
119891 (119911) = 119886119904119911119904
+ 119886119904minus1119911119904minus1
+ sdot sdot sdot + 1198861119911 + 1198860 (29)
then
119891 (119911 + 1) = 119886119904119911119904
+ 119887119904minus1119911119904minus1
+ sdot sdot sdot + 1198871119911 + 1198870 (30)
where
119887119904minus119895
= 119886119904119862119895
119904+ 119886119904minus1119862119895minus1
119904minus1+ sdot sdot sdot + 119886
119904minus119895+11198621
119904minus119895+1+ 119886119904minus119895 (31)
From (29) and (30) we obtain that
Δ119891 = 119904119886119904119911119904minus1
+ (119887119904minus2
minus 119886119904minus2) 119911119904minus2
+ sdot sdot sdot + (1198871minus 1198861) 119911 + (119887
0minus 1198860)
(32)
4 Abstract and Applied Analysis
If 119904 = 1 then Δ2119891 = Δ3119891 equiv 0 which yields that 119875(119911 119891) equiv0 That is a contradiction to our assumption Thus 119904 ge 2
If 119904 = 2 thenΔ119891 = 21198862119911+1198862+1198861Δ2119891 = 2119886
2 andΔ3119891 equiv 0
Now from (3) we get
(minus3)119896
119876 (119911 119891) (Δ2
119891)
2119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(33)
Considering degrees of both sides of the equation above wecan see that 119902 = 119901 + 119896
If 119904 ge 3 we can deduce similarly that
Δ2
119891 = 119904 (119904 minus 1) 119886119904119911119904minus2
+ 1198751(119911)
Δ3
119891 = 119904 (119904 minus 1) (119904 minus 2) 119886119904119911119904minus3
+ 1198752(119911)
(34)
where 1198751(119911) 1198752(119911) are polynomials such that deg119875
1le 119904 minus
3 deg1198752le 119904 minus 4
Rewrite (3) as follows
119876 (119911 119891) [2Δ3
119891 sdot Δ119891 minus 3(Δ2
119891)
2
]
119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(35)
From (34) we find that the leading coefficient of 2Δ3119891 sdotΔ119891 minus 3(Δ
2
119891)
2 is
minus1198862
1199041199042
(119904 minus 1) (119904 + 1) = 0 (36)
Considering degrees of both sides of (35) we prove that119902119904 = 119901119904 + 2119896
4 Proof of Theorem 3
Firstly we consider the general case Asmentioned inRemark1 in [13] due to Jank and Volkmann [16] if (3) admits anadmissible solution then there are at most 119878(119903 119891) commonzeros of 119875(119911 119891) and 119876(119911 119891) Since all coefficients of 119876(119911 119891)are polynomials there are at the most finitely many poles of119891 that are the zeros of 119876(119911 119891) Therefore from (3) we have
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119879 (119903 119891 (119911 + 1) minus 119891 (119911)) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119878 (119903 119891)
(37)
Combining this and Lemma 9 applying the second maintheorem we get
119902
2119896
119879 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
119902
2119896
119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
1
2119896
119873(119903
1
119876
) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 4119879 (119903 119891) + 119878 (119903 119891)
(38)
Thus we prove that (10) holdsSecondly we consider the case that 119873(119903 119891) = 119878(119903 119891)
From (3) and Lemma 8 we similarly get that
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119898 (119903 Δ119891) + 119873 (119903 Δ119891) + 119878 (119903 119891)
le 119898(119903
Δ119891
119891
) + 119898 (119903 119891) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
(39)
From this and applying Lemma 9with (19) as arguing beforewe can prove that (11) holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the referees for their valuablesuggestions This work is supported by the NNSFC (nos11226091 and 11301091) the Guangdong Natural ScienceFoundation (no S2013040014347) and the Foundation forDistinguished Young Talents in Higher Education of Guang-dong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic PublishersGroupDordrechtTheNetherlands 2003
[4] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[5] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
Abstract and Applied Analysis 5
[8] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[9] R G Halburd and R J Korhonen ldquoExistence of finite-ordermeromorphic solutions as a detector of integrability in differ-ence equationsrdquo Physica D Nonlinear Phenomena vol 218 no2 pp 191ndash203 2006
[10] R G Halburd R J Korhonen and K Tohge ldquoHolomorphiccurves with shift-invariant hyperplane preimagesrdquo submitted toTransactions of the American Mathematical Society httparxivorgabs09033236
[11] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[12] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] K Ishizaki ldquoAdmissible solutions of the Schwarzian differentialequationrdquoAustralianMathematical SocietyA PureMathematicsand Statistics vol 50 no 2 pp 258ndash278 1991
[14] A Z Mohonrsquoko ldquoThe nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija no 14 pp 83ndash87 1971 (Russian)
[15] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Entomologique de France vol 59 pp 17ndash39 1931
[16] G Jank and L VolkmannMeromorphe Funktionen und Differ-entialgeichungen Birkhauser Verlag Basel Switzerland 1985
Research ArticleStatistical Inference for Stochastic DifferentialEquations with Small Noises
Liang Shen12 and Qingsong Xu1
1 School of Mathematics and Statistics Central South University Changsha Hunan 410075 China2 School of Science Linyi University Linyi Shandong 276005 China
Correspondence should be addressed to Qingsong Xu csuqingsongxu126com
Received 13 November 2013 Accepted 11 February 2014 Published 13 March 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 L Shen and Q Xu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper proposes the least squares method to estimate the drift parameter for the stochastic differential equations driven bysmall noises which is more general than pure jump 120572-stable noises The asymptotic property of this least squares estimator isstudied under some regularity conditions The asymptotic distribution of the estimator is shown to be the convolution of a stabledistribution and a normal distribution which is completely different from the classical cases
1 Introduction
Stochastic differential equations (SDEs) are being extensivelyused as a model to describe some phenomena which aresubject to random influences it has found many applicationsin biology [1] medicine [2] econometrics [3 4] finance[5] geophysics [6] and oceanography [7] Then statisticalinference for these differential equations was of great interestand became a challenging theoretical problem For a morerecent comprehensive discussion we refer to [8 9]
The asymptotic theory of parametric estimation for dif-fusion processes with small white noise based on continuoustime observations is well developed and it has been studiedby many authors (see eg [10ndash14]) There have been manyapplications of small noise in mathematical finance see forexample [15ndash18]
In parametric inference due to the impossibility ofobserving diffusions continuously throughout a time intervalit is more practical and interesting to consider asymptoticestimation for diffusion processes with small noise based ondiscrete observations There are many approaches to driftestimation for discretely observed diffusions (see eg [19ndash23]) Long [24] has started the study on parameter estimationfor a class of stochastic differential equations driven by smallstable noise 119885
119905 119905 ge 0 However there has been no study on
parametric inference for stochastic processes with small Levynoises yet
In this paper we are interested in the study of parameterestimation for the following stochastic differential equationsdriven by more general Levy noise 119871
119905 119905 ge 0 based
on discrete observations We will employ the least squaresmethod to obtain an asymptotically consistent estimator
Let (ΩF F119905ge0
P) be a basic complete filtered prob-ability space satisfying the usual conditions that is thefiltration is continuous on the right and F
0contains all
P-null sets In this paper we consider a class of stochasticdifferential equations as follows
119889119883119905
= 120579119891 (119883119905) 119889119905 + 120576119892 (119883
119905minus
) 119889119871119905 119905 isin [0 1]
119871119905
= 119886119861119905+ 119887119885119905
119883 (0) = 1199090
(1)
where 119891 R rarr R and 119892 R rarr R are known functionsand 119886 119887 are known constants Let 119861
119905 119905 ge 0 be a standard
Brownian motion and let 119885119905 119905 ge 0 be a standard 120572-stable
Levy motion independent of 119861119905 119905 ge 0 with 119885
1sim 119878120572(1 120573 0)
for 120573 isin [0 1] 1 lt 120572 lt 2Let 119883 = 119883
119905 119905 ge 0 be a real-valued stationary process
satisfying the stochastic differential equation (1) and weassume that this process is observed at regularly spaced timepoints 119905
119894= 119894119899 119894 = 1 2 119899 Assume 119883
0
119905is the solution of
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 473681 6 pageshttpdxdoiorg1011552014473681
2 Abstract and Applied Analysis
the underlying ordinary differential equation (ODE) with thetrue value of the drift parameter 120579
0
1198891198830
119905= 1205790119891 (1198830
119905) 119889119905 119883
0
0= 1199090 (2)
Then we get
119883119905119894
minus 119883119905119894minus1
= int
119905119894
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (3)
2 Preliminaries
In this paper we denote 119862 as a generic constant whose valuemay vary from place to place
The following regularity conditions are assumed to hold
(A1) The functions 119891(119909) and 119892(119909) satisfy the Lipschitzconditions that is there exists a constant 119871 gt 0 suchthat
1003816100381610038161003816119891 (119909) minus 119891 (119910)
1003816100381610038161003816+
1003816100381610038161003816119892 (119909) minus 119892 (119910)
1003816100381610038161003816
le 1198711003816100381610038161003816119909 minus 119910
1003816100381610038161003816 119909 119910 isin R
(4)
(A2) There exist constants 119872 gt 0 and 119903 ge 0 satisfying
the growth condition
119892minus2
(119909) le 119872 (1 + |119909|119903
) 119909 isin R (5)
(A3) There exists a positive constant 119873 gt 0 such that
0 lt |119892(119909)| le 119873 lt infin
(A4) For 119862
119903= 2119903minus1
or 1 119903 gt 0
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
le 119862119903(
100381610038161003816100381610038161198830
119905119894minus1
10038161003816100381610038161003816
119903
+
10038161003816100381610038161003816119883119905119894minus1
minus 1198830
119905119894minus1
10038161003816100381610038161003816
119903
) (6)
The LSE of 120579119899120576
is defined as
120579119899120576
= argmin120579
120588119899120576
(120579) (7)
where the contrast function
120588119899120576
(120579) =
119899
sum
119894=1
10038161003816100381610038161003816100381610038161003816100381610038161003816
119883119905119894
minus 119883119905119894minus1
minus 120579119891 (119883119905119894minus1
) Δ119905119894minus1
120576119892 (119883119905119894minus1
)
10038161003816100381610038161003816100381610038161003816100381610038161003816
2
(8)
Then the 120579119899120576
can be represented explicitly as follows
120579119899120576
=
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) (119883119905119894
minus 119883119905119894minus1
)
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
(9)
Based on (3) and (9) there is a special decomposition for 120579119899120576
120579119899120576
=
1205790
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119891 (119883119904) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119887120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119885119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119886120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
= 1205790
+
Φ2
(119899 120576)
Φ1
(119899 120576)
+
Φ3
(119899 120576)
Φ1
(119899 120576)
+
Φ4
(119899 120576)
Φ1
(119899 120576)
(10)
Now we give an explicit expression for 120576minus1
(120579119899120576
minus 1205790) By using
(10) we have
120576minus1
(120579119899120576
minus 1205790) =
120576minus1
Φ2
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ3
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ4
(119899 120576)
Φ1
(119899 120576)
=
Ψ2
(119899 120576)
Φ1
(119899 120576)
+
Ψ3
(119899 120576)
Φ1
(119899 120576)
+
Ψ4
(119899 120576)
Φ1
(119899 120576)
(11)
One of the important tools we will employ is the under-lying lemma (see (35) in the Lemma 32 of [24])
Lemma 1 Under conditions (A1)-(A2) one has
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816
le 120576119890119871|1205790|119905
10038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119883119904minus
) 119889119885119904
10038161003816100381610038161003816100381610038161003816
(12)
sup0le119905le1
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816997888rarr1198750 as 120576 997888rarr 0 (13)
Abstract and Applied Analysis 3
3 Asymptotic Property of the LeastSquares Estimator
Theorem 2 Under the conditions (A1)ndash(A4) as 119899 rarr
infin 120576 rarr 0 119899120576 rarr infin and 119899120576120572(120572minus1)
rarr infin one has
120576minus1
(120579119899120576
minus 1205790)
997904rArr 119886
(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904
119873
+ 119887 (((int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus (int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802)
times (int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
minus1
)
(14)
where 1198801and 119880
2are independent random variables with 120572-
stable distribution 119878120572(1 120573 0) and 119873 is an independent random
variable with standard normal distribution
The theoremwill be proved by establishing several propo-sitionsWewill consider the asymptotic behaviors ofΦ
1(119899 120576)
Ψ119894(119899 120576) 119894 = 2 3 4 respectively
Proposition 3 Under conditions (A1)ndash(A4) and 119899 rarr infin
120576 rarr 0 one has
Φ1
(119899 120576) 997888rarr119875
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904 (15)
Proof Under conditions (A1)ndash(A3) Proposition 3 can be
proved by using condition (A4) (see the proof of Proposition
33 in [24])
Proposition 4 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576 rarr infin one has
Ψ2
(119899 120576) 997888rarr1198750 (16)
Proof For 119905119894minus1
le 119905 le 119905119894 119894 = 1 2 119899
119883119905
= 119883119905119894minus1
+ int
119905
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (17)
It follows that10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le int
119905
119905119894minus1
10038161003816100381610038161205790
1003816100381610038161003816(
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816) 119889119904
+ 120576
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904
100381610038161003816100381610038161003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816119872 int
119905
119905119894minus1
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+ 119899minus1 1003816
1003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+ 119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
(18)
Using Gronwall inequality we get10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872(119905minus119905
119894minus1)
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(19)
which yields
sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872119899
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(20)
thus under conditions (A1) and (A
3)
1003816100381610038161003816Φ2
(119899 120576)1003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816
119899
sum
119894=1
119872 (1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
(119891 (119883119904) minus 119891 (119883
119905119894minus1
)) 119889119904
100381610038161003816100381610038161003816100381610038161003816
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
119899
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
4 Abstract and Applied Analysis
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
1198992
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
2
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890(|1205790|119872)119899
119899
119887120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
119899
119886120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
= Φ21
(119899 120576) + Φ22
(119899 120576) + Φ23
(119899 120576)
(21)
Then
1003816100381610038161003816Ψ2
(119899 120576)1003816100381610038161003816
le 120576minus1
Φ21
(119899 120576) + 120576minus1
Φ22
(119899 120576)
+ 120576minus1
Φ23
(119899 120576)
= Ψ21
(119899 120576) + Ψ22
(119899 120576) + Ψ23
(119899 120576)
(22)
Using (13) in Lemma 1 conditions (A1) and (A
4) we get
Ψ21
(119899 120576) rarr1198750 as 119899 rarr infin 120576 rarr 0 and 119899120576 rarr infin (see (326)
in [24]) By using the same techniques under condition (A2)
we can prove thatΨ2119895
(119899 120576) rarr1198750 119895 = 2 3 as 119899 rarr infin 120576 rarr 0
respectively
Proposition 5 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576120572(120572minus1)
rarr infin one has
Ψ3
(119899 120576)
997904rArr 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802
(23)
Proof Under conditions (A1)ndash(A3) Proposition 5 can be
proved by using condition (A4) (see the proof of Proposition
44 in [24])
Proposition 6 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 one has
Ψ4
(119899 120576) 997904rArr 119886(int
1
0
10038161003816100381610038161003816119892minus2
(1198830
119904)
100381610038161003816100381610038161198912
(1198830
119904) 119889119904)
12
119873 (24)
Proof Note that
Ψ4
(119899 120576) = 119886
119899
sum
119894=1
119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) int
119905119894
119905119894minus1
119892 (1198830
119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
)) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
))
times (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
=
5
sum
119895=1
Ψ4119895
(119899 120576)
(25)
For Ψ41
(119899 120576) let 119884119894
= int
119905119894
119905119894minus1
119892(119883119904minus
)119889119861119904 119895 = 1 119899 Then it is
easy to see that 119884119894
sim 119873(0 int
119905119894
119905119894minus1
1198922
(119883119904minus
)119889119904) and 1198841 119884
119899are
independent normal random variablesIt follows that
Ψ41
(119899 120576)
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) 119884119894
sim 119873 (0 1198862
119899
sum
119894=1
119892minus4
(1198830
119905119894minus1
) 1198912
(1198830
119905119894minus1
) int
119905119894
119905119894minus1
1198922
(119883119904minus
) 119889119904)
997904rArr 119886(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
119873
(26)
as 119899 rarr infin 120576 rarr 0
Abstract and Applied Analysis 5
For Ψ42
(119899 120576) using Markov inequality and Itorsquos isometryproperty for any given 120578 gt 0
Ψ42
(119899 120576)
le
1
120578
E[119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
]
le
119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
))
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119883119904minus
minus 1198830
119904minus
)
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [ sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119904minus
minus 1198830
119904minus
10038161003816100381610038161003816119899minus12
]
(27)
By using (13) Ψ42
(119899 120576) rarr 0 as 119899 rarr infin 120576 rarr 0Applying similar techniques to Ψ
4119895(119899 120576) 119895 = 3 4 5 we
get Ψ4119895
(119899 120576) rarr 0 119895 = 3 4 5 as 119899 rarr infin 120576 rarr 0
Now we can proveTheorem 2
Proof By using Propositions 3 4 5 6 and Slutskyrsquos theoremwe can get the conclusion
4 Example
We consider the following nonlinear SDE driven by generalLevy noises
119889119883119905
= 120579119883119905119889119905 +
120576
1 + 1198832
119905minus
119889119871119905 119905 isin [0 1] 119883
0= 1199090
(28)
where 119891(119909) = 119909 119892(119909) = 1(1 + 1199092
) 1199090and 120576 are known
constants and 120579 = 0 is an unknown parameterFor simplicity let 119909
0gt 0 120576 = 0 we get the ODE
1198891198830
119905= 12057901198830
119905119889119905 119905 isin [0 1] 119883
0
0= 1199090
(29)
and the solution
1198830
119905= 11990901198901205790119905
(30)
Then the asymptotic distribution is
119886(int
1
0
(1 + 1199092
011989021205790119904
)
2
1199092
011989021205790119904
119889119904)
minus12
119873
+ 119887
(int
1
0
(1 + 1199092
011989021205790119904
)
120572
119909120572
01198901205721205790119904
119889119904)
1120572
int
1
0
(1 + 1199092
011989021205790119904)2
1199092
011989021205790119904119889119904
119878120572
(1 120573 0)
(31)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R I Jennrich and P B Bright ldquoFitting systems of linear dif-ferential equations using computer generated exact derivativesrdquoTechnometrics vol 18 no 4 pp 385ndash392 1976
[2] R H Jones ldquoFitting multivariate models to unequally spaceddatardquo in Time Series Analysis of Irregularly Observed Data pp158ndash188 1984
[3] A R Bergstrom Statistical Inference in Continuous TimeEconomic Models vol 99 North-Holland Amsterdam TheNetherlands 1976
[4] A R Bergstrom ldquoThe history of continuous-time econometricmodelsrdquo Econometric Theory vol 4 no 3 pp 365ndash383 1988
[5] F Black and M Scholes ldquoThe pricing of options and corporateliabilitiesrdquoThe Journal of Political Economy vol 81 pp 637ndash6541973
[6] M Arato Linear Stochastic Systems with Constant CoefficientsSpringer Berlin Germany 1982
[7] R J Adler and P Meuller Stochastic Modelling on PhysicalOceanography vol 39 Springer New York NY USA 1996
[8] B P Rao B L P Rao I Statisticien B L P Rao B L P Rao andI Statistician Statistical Inference for di Usion Type ProcessesArnold London UK 1999
[9] Y A Kutoyants Statistical Inference for Ergodic Diffusion Pro-cesses Springer London UK 2004
[10] Yu A Kutoyants Parameter Estimation for Stochastic Processesvol 6 Heldermann Berlin Germany 1984
[11] Yu Kutoyants Identification of Dynamical Systems with SmallNoise Kluwer Academic Dordrecht The Netherlands 1994
[12] N Yoshida ldquoAsymptotic expansions of maximum likelihoodestimators for small diffusions via the theory of Malliavin-Watanaberdquo Probability Theory and Related Fields vol 92 no 3pp 275ndash311 1992
[13] N Yoshida ldquoConditional expansions and their applicationsrdquoStochastic Processes and their Applications vol 107 no 1 pp 53ndash81 2003
[14] M Uchida and N Yoshida ldquoInformation criteria for smalldiffusions via the theory of Malliavin-Watanaberdquo StatisticalInference for Stochastic Processes vol 7 no 1 pp 35ndash67 2004
[15] N Yoshida ldquoAsymptotic expansion of Bayes estimators for smalldiffusionsrdquo Probability Theory and Related Fields vol 95 no 4pp 429ndash450 1993
[16] A Takahashi ldquoAn asymptotic expansion approach to pricingfinancial contingent claimsrdquoAsia-Pacific FinancialMarkets vol6 no 2 pp 115ndash151 1999
6 Abstract and Applied Analysis
[17] N Kunitomo and A Takahashi ldquoThe asymptotic expansionapproach to the valuation of interest rate contingent claimsrdquoMathematical Finance vol 11 no 1 pp 117ndash151 2001
[18] A Takahashi and N Yoshida ldquoAn asymptotic expansionscheme for optimal investment problemsrdquo Statistical Inferencefor Stochastic Processes vol 7 no 2 pp 153ndash188 2004
[19] V Genon-Catalot ldquoMaximum contrast estimation for diffusionprocesses fromdiscrete observationsrdquo Statistics vol 21 no 1 pp99ndash116 1990
[20] A Gloter and M Soslashrensen ldquoEstimation for stochastic differ-ential equations with a small diffusion coefficientrdquo StochasticProcesses and their Applications vol 119 no 3 pp 679ndash6992009
[21] C F Laredo ldquoA sufficient condition for asymptotic sufficiencyof incomplete observations of a diffusion processrdquo The Annalsof Statistics vol 18 no 3 pp 1158ndash1171 1990
[22] M Uchida ldquoApproximate martingale estimating functions forstochastic differential equations with small noisesrdquo StochasticProcesses and their Applications vol 118 no 9 pp 1706ndash17212008
[23] M Soslashrensen and M Uchida ldquoSmall-diffusion asymptotics fordiscretely sampled stochastic differential equationsrdquo Bernoullivol 9 no 6 pp 1051ndash1069 2003
[24] H Long ldquoParameter estimation for a class of stochastic dif-ferential equations driven by small stable noises from discreteobservationsrdquo Acta Mathematica Scientia B vol 30 no 3 pp645ndash663 2010
Research ArticleOn the Deficiencies of Some Differential-Difference Polynomials
Xiu-Min Zheng1 and Hong Yan Xu2
1 Institute of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China2Department of Informatics and Engineering Jingdezhen Ceramic Institute Jingdezhen 333403 China
Correspondence should be addressed to Xiu-Min Zheng zhengxiumin2008sinacom
Received 1 November 2013 Accepted 4 January 2014 Published 27 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 X-M Zheng and H Y Xu This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
The characteristic functions of differential-difference polynomials are investigated and the result can be viewed as a differential-difference analogue of the classic Valiron-Mokhonrsquoko Theorem in some sense and applied to investigate the deficiencies of somehomogeneous or nonhomogeneous differential-difference polynomials Some special differential-difference polynomials are alsoinvestigated and these results on the value distribution can be viewed as differential-difference analogues of some classic results ofHayman and Yang Examples are given to illustrate our results at the end of this paper
1 Introduction
Throughout this paper we use standard notations in theNevanlinna theory (see eg [1ndash3]) Let 119891(119911) be a meromor-phic function Here and in the following the word ldquomero-morphicrdquo means being meromorphic in the whole complexplane We use normal notations 119898(119903 119891) 119879(119903 119891) 119873(119903 119891)119873(119903 1119891) 120590(119891) 120582(119891) and 120582(1119891) And we also use 120590
2(119891)
to denote the hyperorder of 119891(119911) and 120575(120572 119891) to denote theNevanlinna deficiency of 120572 with respect to 119891(119911) Moreoverwe denote by 119878(119903 119891) any real quantity satisfying 119878(119903 119891) =119900(119879(119903 119891)) as 119903 rarr infin outside of a possible exceptional set offinite logarithmic measure
Recently with some establishments of difference ana-logues of the classic Nevanlinna theory (two typical andmost important ones can be seen in [4ndash6]) there has beena renewed interest in the properties of complex differenceexpressions and meromorphic solutions of complex differ-ence equations (see eg [4ndash17]) By combining complex dif-ferentiates and complex differences we proceed in this way inthis paper
It is well known that the following Valiron-MokhonrsquokoTheorem due to Valiron [18] and A Z Mokhonrsquoko and V DMokhonrsquoko [19] is of essential importance in the theory ofcomplex differential equations and functional equations
Theorem A (see [2 3]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891 (119911)) =
sum119898
119894=0119886119894(119911) 119891(119911)
119894
sum119899
119895=0119887119895(119911) 119891(119911)
119895
(1)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) the characteristic
function of 119877(119911 119891(119911)) satisfies
119879 (119903 119877 (119911 119891 (119911))) = 119889119879 (119903 119891) + 119874 (Ψ (119903)) (2)
where 119889 = max119898 119899 and Ψ(119903) = max119894119895119879(119903 119886
119894) 119879(119903 119887
119895)
Noting that the difference analogue of Theorem A maynot hold we have obtained a result of this type in [16] byadding some additional assumptions as follows
Theorem B (see [16]) Suppose that 119875(119911 119891) is a differencepolynomial of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
1198940119891(119911 + 119888
1)1198941sdot sdot sdot 119891(119911 + 119888
119899)119894119899
(3)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 378151 12 pageshttpdxdoiorg1011552014378151
2 Abstract and Applied Analysis
containing just one monomial of degree 119889(119875) and 119891(119911) is atranscendental meromorphic function of finite order If 119891(119911)also satisfies119873(119903 119891) + 119873(119903 1119891) = 119878(119903 119891) then we have
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (4)
In this paper we consider removing the assumptionldquo119875(119911 119891) contains just one monomial of degree 119889(119875)rdquo in The-orem B and obtain a weaker result which is also generalizedinto differential-difference case The concrete result can beseen in Section 2
Next we recall a classic result concerning Picardrsquos valuesofmeromorphic functions and its derivatives due toHayman[20]
Theorem C (see [20]) Let 119891(119911) be a transcendental entirefunction Then
(a) for 119899 ge 3 and 119886 = 0 Ψ(119911) = 1198911015840(119911) minus 119886(119891(119911))119899 assumesall finite values infinitely often
(b) for 119899 ge 2Φ(119911) = 1198911015840(119911)(119891(119911))119899 assumes all finite valuesexcept possibly zero infinitely often
Corresponding difference analogues ofTheoremC can beseen in [12 17]
Theorem D (see [12 17]) Let 119891(119911) be a transcendental entirefunction of finite order and let 119888 be a nonzero complex constantThen
(a) for 119899 ge 3 and 119886 = 0Ψ1(119911) = 119891(119911+ 119888) minus 119886119891(119911)
119899 assumesall finite complex values infinitely often
(b) for 119899 ge 2 Φ1(119911) = 119891(119911 + 119888)119891(119911)
119899 assumes all finitecomplex values except possibly zero infinitely often
AfterTheoremCmany results have been obtained on thevalue distribution of differential polynomials A typical one isas follows
TheoremE (see [21 22]) Let119891 be a transcendental meromor-phic function with119873(119903 119891) +119873(119903 1119891) = 119878(119903 119891) and let Ψ bea differential polynomial in 119891 of the form
Ψ (119911) = sum119886 (119911) 119891(119911)11989701198911015840
(119911)1198971sdot sdot sdot 119891(119896)
(119911)119897119896 (5)
with no constant term Furthermore assume the degree 119899 ofΨ is greater than one and 119897
0lt 119899 0 le 119897
119894le 119899 for all 119894 = 0 Then
120575(119886 Ψ) lt 1 for all 119886 = 0infin Moreover if all the terms ofΨ havedifferent degrees at least two that is Ψ is nonhomogeneousthen 120575(119886 Ψ) le 1 minus (12119899) for all 119886 =infin
We also consider deficiencies of difference polynomialsof meromorphic functions of finite order in [16] which canbe viewed as difference analogues of Theorem E as well asgeneralizations of Theorem D
In this paper we proceed to investigate deficiencies ofdifferential-difference polynomials of meromorphic func-tions The concrete results can be seen in Section 3
Examples are given in Section 4 to illustrate our results
2 A Differential-Difference Analogue ofValiron-Mokhonrsquoko Theorem
In what follows we will consider differential-difference poly-nomials A differential-difference polynomial is a polynomialin 119891(119911) its shifts its derivatives and derivatives of its shifts(see [14]) that is an expression of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
120582001198911015840
(119911)12058201sdot sdot sdot 119891(119898)
(119911)1205820119898
times 119891(119911 + 1198881)120582101198911015840
(119911 + 1198881)12058211sdot sdot sdot 119891(119898)
(119911 + 1198881)1205821119898
sdot sdot sdot 119891(119911 + 119888119899)12058211989901198911015840
(119911 + 119888119899)1205821198991sdot sdot sdot 119891(119898)
(119911 + 119888119899)120582119899119898
= sum
120582isin119868
119886120582(119911)
119899
prod
119894=0
119898
prod
119895=0
119891(119895)
(119911 + 119888119894)120582119894119895
(6)
where 119868 is a finite set of multi-indices 120582 =
(12058200 120582
0119898 12058210 120582
1119898 120582
1198990 120582
119899119898) and 119888
0(= 0)
and 1198881 119888
119899are distinct complex constants And we assume
that the meromorphic coefficients 119886120582(119911) 120582 isin 119868 of 119875(119911 119891) are
of growth 119878(119903 119891) We denote the degree of the monomialprod119899
119894=0prod119898
119895=0119891(119895)
(119911 + 119888119894)120582119894119895 of 119875(119911 119891) by 119889(120582) = sum119899
119894=0sum119898
119895=0120582119894119895
Then we denote the degree and the lower degree of 119875(119911 119891)by
119889 (119875) = max120582isin119868
119889 (120582) 119889lowast
(119875) = min120582isin119868
119889 (120582) (7)
respectively In particular we call 119875(119911 119891) a homogeneousdifferential-difference polynomial if 119889(119875) = 119889
lowast
(119875) Other-wise 119875(119911 119891) is nonhomogeneous
In the following we assume 119889(119875) ge 1 and 119875(119911 119891) equiv
119875(119911 0)We prove a weaker differential-difference version of the
classic Valiron-MokhonrsquokoTheorem as follows
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic function and 119875(119911 119891) is a differential-difference polyno-mial of the form (6) If 119891(119911) also satisfies 120590
2(119891) lt 1 and
119873(119903 119891) + 119873(119903
1
119891
) = 119878 (119903 119891) (8)
then one has
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891))
le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
(9)
Remark 2 If119875(119911 119891) is a homogeneous differential-differencepolynomial in addition then
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (10)
Remark 3 Especially assumption (8) can be replaced by theassumption ldquomax120582(119891) 120582(1119891) lt 120590(119891)rdquo In fact if 119891(119911)satisfies max120582(119891) 120582(1119891) lt 120590(119891) then 119891(119911) is of regulargrowth and (8) holds consequently
Abstract and Applied Analysis 3
To proveTheorem 1 we need the following lemmas
Lemma 4 (see [6]) Let 119891(119911) be a nonconstant meromorphicfunction 120576 gt 0 and 119888 isin C If 120577 = 120590
2(119891) lt 1 then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120577minus120576
) (11)
for all 119903 outside of a set of finite logarithmic measure
Lemma 5 (see [6]) Let 119879 [0 +infin) rarr [0 +infin) be anondecreasing continuous function and let 119904 isin (0 +infin)If the hyperorder of 119879 is strictly less than one that islim119903rarrinfin
(log2119879(119903) log 119903) = 120577 lt 1 and 120575 isin (0 1 minus 120577) then
119879 (119903 + 119904) = 119879 (119903) + 119900 (
119879 (119903)
119903120575
) (12)
where 119903 runs to infinity outside of a set of finite logarithmicmeasure
It is shown in [23 p66] and [7 Lemma 1] that the ine-quality
(1 + 119900 (1)) 119879 (119903 minus |119888| 119891) le 119879 (119903 119891 (119911 + 119888))
le (1 + 119900 (1)) 119879 (119903 + |119888| 119891)
(13)
holds for 119888 = 0 and 119903 rarr infin And from the proof theabove relation is also true for counting function By combingLemma 5 and these inequalities we immediately deduce thefollowing lemma
Lemma 6 Let 119891(119911) be a nonconstant meromorphic functionof 1205902(119891) lt 1 and let 119888 be a nonzero complex constant Then
one has
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891)
119873 (119903 119891 (119911 + 119888)) = 119873 (119903 119891) + 119878 (119903 119891)
119873(119903
1
119891 (119911 + 119888)
) = 119873(119903
1
119891
) + 119878 (119903 119891)
(14)
Lemma7 Let119891(119911) be a transcendentalmeromorphic functionof 1205902(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then we one has
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891) + 119878 (119903 119891) (15)
Furthermore if 119891(119911) also satisfies
119873(119903 119891) = 119878 (119903 119891) (16)
then one has
119879 (119903 119875 (119911 119891)) le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (17)
Proof For 119894 = 0 1 119899 119895 = 0 1 119898 we define 119892119894119895(119911) =
119891(119895)
(119911 + 119888119894)119891(119911) We also define
119892lowast
119894119895(119911) =
119892119894119895(119911) if 1003816100381610038161003816
1003816119892119894119895(119911)
10038161003816100381610038161003816gt 1
1 if 10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816le 1
119891lowast
(119911) =
119891 (119911) if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816le 1
(18)
Thus
1003816100381610038161003816119875 (119911 119891)
1003816100381610038161003816le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119891 (119911)
1003816100381610038161003816
119889(120582)
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
120582119894119895
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(120582)
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)(
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
1003816100381610038161003816119891lowast
(119911)1003816100381610038161003816)
119889(119875)
(19)
By the definitions of 119891lowast(119911) and 119892lowast119894119895(119911) 119894 = 0 1 119899 119895 =
0 1 119898 we have
119898(119903 119891lowast
) = 119898 (119903 119891)
119898 (119903 119892lowast
119894119895) = 119898 (119903 119892
119894119895) 119894 = 0 119899 119895 = 0 119898
(20)
It follows by (19) and (20) that
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
= 119889 (119875)119898 (119903 119891)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892119894119895) + 119878 (119903 119891)
(21)
Lemmas 4 and 6 and the logarithmic derivative lemma implythat for 119894 = 0 1 119899 and 119895 = 0 1 119898
119898(119903 119892119894119895) = 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911)
)
le 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911 + 119888119894)
) + 119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
= 119878 (119903 119891 (119911 + 119888119894)) + 119878 (119903 119891) = 119878 (119903 119891)
(22)
Then (15) follows by (21) and (22)
4 Abstract and Applied Analysis
It is easy to find that
119873(119903 119875 (119911 119891)) = 119874(119873(119903 119891) +
119899
sum
119894=1
119873(119903 119891 (119911 + 119888119894)))
+ 119878 (119903 119891)
(23)
Then (16) (23) and Lemma 6 yield that
119873(119903 119875 (119911 119891)) = 119878 (119903 119891) (24)
Thus (17) follows by (15) and (24)
Lemma 8 Let 119891(119911) be a transcendental meromorphic func-tion of 120590
2(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then one has
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(25)
Proof Similar to (19) we have
100381610038161003816100381610038161003816100381610038161003816
119875 (119911 119891)
119891(119911)119889(119875)
100381610038161003816100381610038161003816100381610038161003816
le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895 1003816100381610038161003816119892 (119911)
1003816100381610038161003816
119889(119875)minus119889(120582)
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889(120582)
)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889lowast(119875)
(26)
where 119892(119911) = 1119891(119911) and
119892lowast
(119911) =
119892 (119911) if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816le 1
(27)
By the definition of 119892lowast(119911) we have 119898(119903 119892lowast) = 119898(119903 119892) =
119898(119903 1119891) Thus (20) (22) and (26) yield that
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898 (119903 119892lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(28)
that is (25)
Now we can finish the proof of Theorem 1 in the end
Proof of Theorem 1 We deduce from (8) (24) and Lemma 8that
119889 (119875) 119879 (119903 119891)
= 119879 (119903 119891119889(119875)
) le 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119873 (119903 119875 (119911 119891))
+ 119889 (119875)119873(119903
1
119891
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119879 (119903 119875 (119911 119891)) + 119878 (119903 119891)
(29)
that is
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891)) (30)
Then (9) follows by (17) and (30)
3 Deficiencies of SomeDifferential-Difference Polynomials
In the following we assume that 120572(119911)( equiv 0) is ameromorphicfunction of growth 119878(119903 119891)
In this section we will apply Theorem 1 to consider thedeficiencies of general homogeneous or nonhomogeneousdifferential-difference polynomials
Theorem 9 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)
(a) If119875(119911 119891) is a homogeneous differential-difference poly-nomial then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
= 1 120575 (120572 119875 (119911 119891)) = 0
(31)
(b) If 119875(119911 119891) is a nonhomogeneous differential-differencepolynomial with 2119889lowast(119875) gt 119889(119875) then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
ge
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
120575 (120572 119875 (119911 119891)) le 1 minus
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
lt 1
(32)
Thus 119875(119911 119891)minus120572(119911) has infinitely many zeros whether 119875(119911 119891)is homogeneous or nonhomogeneous
Abstract and Applied Analysis 5
Furthermore one considers some differential-differencepolynomials of special forms which are generalizations ofboth differential cases and difference cases that is TheoremsCndashE
Theorem 10 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 V minus 1 of growth 119878(119903 119891) If 119906V gt 119889(119875) 119906 = 1 then
1198761(119911 119891) = 119865 (119891) 119875 (119911 119891) (33)
satisfies
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(34)
Thus 1198761(119911 119891) minus 120572(119911) has infinitely many zeros
When 119865(119891) is of a special form 119891V we can deduce the
following result fromTheorem 9
Theorem 11 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6) If V isin N 1and V + 2119889lowast(119875) gt 119889(119875) then
1198762(119911 119891) = 119891
V119875 (119911 119891) (35)
satisfies 120575(120573 1198762(119911 119891)) lt 1 where 120573 isin C0Thus119876
2(119911 119891)minus
120573 has infinitely many zeros
Remark 12 On the one hand we can also applyTheorem 9 to1198761(119911 119891)with the assumption ldquo2(119889lowast(119875)+119889lowast(119865)) gt 119889(119875)+119906Vrdquo
and obtain the same result as Theorem 10 But our presentassumption ldquo119906V gt 119889(119875)rdquo has no concern with 119889lowast(119875) and119889lowast
(119865) so we think Theorem 10 is better to some extent Onthe other hand we can also apply Theorem 10 to 119876
2(119911 119891)
with the assumption ldquoV gt 119889(119875)rdquo which is stronger thanldquoV + 2119889lowast(119875) gt 119889(119875)rdquo in Theorem 11 showing Theorem 11 isbetter to some extent
Theorem 13 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 Vminus1 of growth 119878(119903 119891) If (119906minus1)119906V(2119906minus1) gt 119889(119875) 119906 = 1then
1198763(119911 119891) = 119865 (119891) + 119875 (119911 119891) (36)
satisfies
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(37)
Thus 1198763(119911 119891) minus 120572(119911) has infinitely many zeros
When 119906 = 1 one can consider some special cases asfollows
Theorem 14 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)(a) If V gt 119889(119875) + 2 ge 3 then
1198764(119911 119891) = 119891
V+ 119875 (119911 119891) (38)
satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(39)
(b) If (Vminus1)V(2Vminus1) gt 119889(119875) V ge 3 then1198764(119911 119891) satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
1
Vminus
2V minus 1V2
119889 (119875)
120575 (120572 1198764(119911 119891)) le
1
V+
2V minus 1V2
119889 (119875) lt 1
(40)
Especially it holds for V = 119889(119875) + 2 = 3(c) If V ge 119889(119875) + 2 ge 3 and 119891 also satisfies 119873(119903 1119891) =119878(119903 119891) then 119876
4(119911 119891) satisfies 120575(120572 119876
4(119911 119891)) lt 1
Especially it holds for V = 119889(119875) + 2 gt 3Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many zeros
If we assume that 119873(119903 1119891) = 119878(119903 119891) in addition thefollowing result follows immediately by Theorem 9
Theorem 15 Suppose that 119891(119911) is a transcendental mero-morphic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891)
is a differential-difference polynomial of the form (6) If2 min119889lowast(119875) V gt max119889(119875) V then 119876
4(119911 119891) satisfies
120575(120572 1198764(119911 119891)) lt 1 Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many
zeros
Remark 16 Noting that when V gt 3 (V minus 1)V(2V minus 1) leV minus 2 hold we see that the assumption ldquoV gt 119889(119875) + 2rdquo inTheorem 14(a) is weaker than the assumption ldquo(V minus 1)V(2V minus1) gt 119889(119875)rdquo in Theorem 14(b) And these assumptions inTheorem 14 have no concernwith119889lowast(119875)) thus they are differ-ent from the assumption ldquo2 min119889lowast(119875) V gt max119889(119875) Vrdquoin Theorem 15
6 Abstract and Applied Analysis
Remark 17 From the proofs behind it is easy to find that
120582 (119875 (119911 119891) minus 120572) = 120590 (119875 (119911 119891)) = 120590 (119891)
120582 (119876119894(119911 119891) minus 120572) = 120590 (119876
119894(119911 119891)) = 120590 (119891) 119894 = 1 3 4
(41)
hold respectively inTheorems 9 10 13 14(a) and (b) and 15Now we give the proofs of Theorems 9ndash15
Proof of Theorem 9 It follows byTheorem 1 that
119878 (119903 119891) = 119878 (119903 119875 (119911 119891)) (42)
We deduce from (8) (24) (25) and (42) that
119873(119903
1
119875 (119911 119891)
)
le 119873(119903
1
119891119889(119875)
) + 119873(119903
119891119889(119901)
119875 (119911 119891)
)
le 119873(119903
1
119891
) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119875 (119911 119891))
(43)
Thus an application of the second main theorem and (24)(42) and (43) imply that
119879 (119903 119875 (119911 119891)) le 119873 (119903 119875 (119911 119891)) + 119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(44)
(a) If 119889(119875) = 119889lowast(119875) then it follows by (44) that
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(45)
by which (31) holds
(b) If 2119889lowast(119875) gt 119889(119875) then we deduce from (30) and (44)that
119879 (119903 119875 (119911 119891)) le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le
119889 (119875) minus 119889lowast
(119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891))
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(46)
that is
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119878 (119903 119875 (119911 119891))
(47)
Since 2119889lowast(119875)minus119889(119875) gt 0 (32) follows immediately by (47)
Proof of Theorem 10 We deduce from (16) (17) and (24) that
119879 (119903 1198761(119911 119891)) le (119906V + 119889 (119875)) 119879 (119903 119891) + 119878 (119903 119891) (48)
119873(119903 1198761(119911 119891)) = 119874 (119873 (119903 119891)) + 119873 (119903 119875 (119911 119891)) + 119878 (119903 119891)
= 119878 (119903 119891)
(49)
hold Next we consider 119873(119903 11198761(119911 119891)) Let 119911
0be a zero of
1198761(119911 119891) and distinguish three cases
(i) 1199110is not a zero of 119865(119891) then 119911
0must be a zero of
119875(119911 119891) and
119906 le 120596(
1
1198761(119911 119891)
1199110) + (119906 minus 1) 120596(
1
119875 (119911 119891)
1199110) (50)
where 120596(119891 1199110) denotes the order of multiplicity of 119911
0or zero
according as 1199110is a pole of 119891(119911) or not
(ii) 1199110is a zero of 119865(119891) but not a pole of 119875(119911 119891) Then
119906 le 120596(
1
1198761(119911 119891)
1199110) (51)
(iii) 1199110is a zero of 119865(119891) and a pole of 119875(119911 119891) Then
119906 le 120596(
1
119865 (119891)
1199110) le 120596(
1
1198761(119911 119891)
1199110) + 120596 (119875 (119911 119891) 119911
0)
(52)
Abstract and Applied Analysis 7
(24) and (50)ndash(52) yield that
119906119873(119903
1
1198761(119911 119891)
) le 119873(119903
1
1198761(119911 119891)
)
+ (119906 minus 1)119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(53)
Then (48) (49) (53) and an application of the second maintheorem to 119876
1(119911 119891) imply that
119879 (119903 1198761(119911 119891))
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
1198761(119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 1198761(119911 119891))
le
1
119906
119873(119903
1
1198761(119911 119891)
) +
119906 minus 1
119906
119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(54)
consequently
119879 (119903 1198761(119911 119891)) le 119873(119903
1
119875 (119911 119891)
)
+
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(55)
Moreover by 119891119889(119875)119865(119891) = 119891119889(119875)
1198761(119911 119891)119875(119911 119891) (16)
(24) (25) andTheorem A we have
(119889 (119875) + 119906V)119898 (119903 119891)
= 119898(119903
119891119889(119875)
1198761(119911 119891)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) minus 119873(119903
119891119889(119875)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898(119903
1
119891
)
+ 119889 (119875) (119873(119903
1
119891
) minus 119873 (119903 119891))
+ 119873 (119903 119875 (119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
= 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898 (119903 119891)
minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(56)
consequently
119906V119898(119903 119891) le 119898 (119903 1198761(119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(57)
On the other hand the evident relation 119906V120596(119891 1199110) le
120596(119865(119891) 1199110) + 119906VsumVminus1
119895=0120596(119886119895 1199110) where the definition of
120596(119891 1199110) is given after (50) results in
119906V119873(119903 119891) le 119873 (119903 119865 (119891)) + 119878 (119903 119891)
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(58)
We deduce from (57) and (58) that
119906V119879 (119903 119891) le 119879 (119903 1198761(119911 119891)) + 119878 (119903 119891) (59)
Then (17) (55) and (59) yield that
119906V119879 (119903 119891)
le 119873(119903
1
119875 (119911 119891)
) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
le 119889 (119875) 119879 (119903 119891) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(60)
that is
(119906 minus 1) (119906V minus 119889 (119875))119906
119879 (119903 119891) le 119873(119903
1
1198761(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(61)
From (48) and (61) we deduce that
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(62)
Proof of Theorem 11 Assume to the contrary that120575(120573 119876
2(119911 119891)) = 1 Denoting
1198762(119911 119891) minus 120573 = 119891
V(119911) 119875 (119911 119891) minus 120573 = 119866 (119911) (63)
we deduce from (16) and (17) that
119873(119903
1
119866
) = 119873(119903
1
1198762(119911 119891) minus 120573
)
= 119878 (119903 1198762(119911 119891)) = 119878 (119903 119891)
(64)
8 Abstract and Applied Analysis
On the other hand (16) and (24) yield that
119873(119903 119866) = 119873 (119903 1198762(119911 119891) minus 120573) = 119878 (119903 119891) (65)
Differentiating both sides of (63) we obtain
119891Vminus1(119911) 119877 (119911 119891) = 119866
1015840
(119911) (66)
where 119877(119911 119891) = V1198911015840(119911)119875(119911 119891) + 119891(119911)1198751015840(119911 119891) Clearly wededuce from (16) and (24) that
119873(119903 119877 (119911 119891)) = 119878 (119903 119891) (67)
Moreover (64) and (65) yield that
119873(119903
1
1198661015840
) le 119873(119903
119866
1198661015840
) + 119873(119903
1
119866
)
le 119879(119903
1198661015840
119866
) + 119873(119903
1
119866
) + 119874 (1)
le 119898(119903
1198661015840
119866
) + 119873 (119903 119866) + 2119873(119903
1
119866
) + 119874 (1)
= 119878 (119903 119866) + 119878 (119903 119891) = 119878 (119903 119891)
(68)
It follows by (66)ndash(68) that
119873(119903
1
119891
) =
1
V minus 1119873(119903
119877 (119911 119891)
1198661015840
) = 119878 (119903 119891) (69)
Then (16) (69) and the fact 2(119889lowast(119875) + V) gt 119889(119875) + V implythat the assumptions of Theorem 9(b) are satisfied ThusTheorem 9(b) yields that 120575(120573 119876
2(119911 119891)) lt 1 a contradiction
Therefore we have 120575(120573 1198762(119911 119891)) lt 1
Proof of Theorem 13 We deduce from (16) (17) and (24) that
119879 (119903 1198763(119911 119891)) le max 119906V 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
= 119906V119879 (119903 119891) + 119878 (119903 119891) (70)
Denote
119867(119911) =
minus119875 (119911 119891) + 120572 (119911)
119865 (119891)
(71)
Now we estimate the poles the zeros and 1-points of119867(119911) accuratelyOn the one handwe see by (71) that the polesof 119867(119911) occur at zeros of 119865(119891) and poles of minus119875(119911 119891) + 120572(119911)which are not simultaneously 1-points of 119867(119911) and thosepoles of119867(119911)which are zeros of 119865(119891) but not simultaneouslyzeros of minus119875(119911 119891) + 120572(119911) also have multiplicities at least 119906 Onthe other handwe also see by (71) that the zeros of119867(119911) occurat zeros of minus119875(119911 119891) + 120572(119911) and poles of 119865(119891) which are notsimultaneously 1-points of 119867(119911) Moreover 1-points of 119867(119911)occur at zeros of 119876
3(119911 119891) minus 120572(119911) and occur at the common
poles zeros of 119865(119891) and minus119875(119911 119891) + 120572(119911) with the samemultiplicities Thus it follows by (16) and (24) that
119873(119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
)
le
1
119906
119873 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(72)
Then (17) (72) and the second main theorem result in
119879 (119903119867) le 119873 (119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
) + 119878 (119903119867)
le
1
119906
119879 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
le
1
119906
119879 (119903119867) + 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(73)
that is
(1 minus
1
119906
)119879 (119903119867) le 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(74)
Moreover Theorem A and (17) imply that
119906V119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119865 (119891)) = 119879(119903minus119875 (119911 119891) + 120572
119867
)
le 119889 (119875) 119879 (119903 119891) + 119879 (119903119867) + 119878 (119903 119891)
(75)
that is
(119906V minus 119889 (119875)) 119879 (119903 119891) le 119879 (119903119867) + 119878 (119903 119891) (76)
Then (74) and (76) yield that
((119906 minus 1) V minus2119906 minus 1
119906
119889 (119875))119879 (119903 119891) le 119873(119903
1
1198763(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(77)
Abstract and Applied Analysis 9
From (70) and (77) we deduce that
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(78)
To prove Theorem 14(c) we also need the followinglemma of one of Tumura-Clunie type theorems
Lemma 18 (see [24]) Let 119891(119911) be a meromorphic functionand suppose that Ψ = 119886
119899119891119899
+ sdot sdot sdot + 1198860has small meromorphic
coefficients 119886119895(119911) 119886119899(119911) equiv 0 in the sense of 119879(119903 119886
119895) = 119878(119903 119891)
Moreover assume that 119873(119903 1Ψ) + 119873(119903 119891) = 119878(119903 119891) ThenΨ = 119886
119899(119891 + (119886
119899minus1119899119886119899))119899
Proof of Theorem 14 (a) We deduce from (16) (17) and (24)that
119879 (119903 1198764(119911 119891)) le V119879 (119903 119891) + 119878 (119903 119891) (79)
Denote
119870 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) + 119875 (119911 119891) minus 120572 (119911) (80)
Differentiating both sides of (80) we obtain
V119891Vminus1(119911) 1198911015840
(119911) + 1198751015840
(119911 119891) minus 1205721015840
(119911)
= 1198701015840
(119911) = (119891V(119911) + 119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
(81)
that is
119891Vminus1(119911) ((V
1198911015840
(119911)
119891 (119911)
minus
1198701015840
(119911)
119870 (119911)
)119891 (119911))
= (119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
minus (1198751015840
(119911 119891) minus 1205721015840
(119911))
= (119875 (119911 119891) minus 120572 (119911)) (
1198701015840
(119911)
119870 (119911)
minus
1198751015840
(119911 119891) minus 1205721015840
(119911)
119875 (119911 119891) minus 120572 (119911)
)
(82)
It follows by (15)ndash(17) (24) (79) and (82) that
119898(119903 119891Vminus1)
le 119898 (119903 119875 (119911 119891) minus 120572) + 119898(119903
1198701015840
119870
)
+ 119898(119903
1198751015840
(119911 119891) minus 1205721015840
119875 (119911 119891) minus 120572
) + 119898(119903
1
(V (1198911015840119891) minus (1198701015840119870)) 119891)
le 119889 (119875)119898 (119903 119891) + 119898(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891)
+ 119873(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891) + 119878 (119903 119870) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1198701015840
119870
) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1
119870
) + 119878 (119903 119891)
(83)
that is
(V minus 119889 (119875) minus 2) 119879 (119903 119891) le 119873(1199031
1198764(119911 119891) minus 120572
) + 119878 (119903 119891)
(84)
From (79) and (84) we deduce that
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(85)
(b) It suffices to note that we may see 119891V as (1198911)V thenTheorem 14(b) follows immediately by Theorem 13
(c) By using a similar reasoning as [13 Theorem 1] wecan rearrange the expression for the differential-differencepolynomial 119875(119911 119891) by collecting together all terms havingthe same total degree and then writing 119875(119911 119891) in the form119875(119911 119891) = sum
119889(119875)
119896=0119887119896(119911)119891119896
(119911) Now each of the coefficients 119887119896(119911)
is a finite sum of products of functions of the form (119891(119895)
(119911 +
119888119894)119891(119911))
120582119894119895= (119891(119895)
(119911+119888119895)119891(119911+119888
119894))120582119894119895(119891(119911+119888
119894)119891(119911))
120582119894119895 with
each such product being multiplied by one of the originalcoefficients 119886
120582(119911) We deduce from the logarithmic derivative
lemma and Lemmas 4 and 6 that 119898(119903 119887119896) = 119878(119903 119891) Clearly
119873(119903 119887119896) = 119878(119903 119891) holds by (8) and Lemma 6Thus 119879(119903 119887
119896) =
119878(119903 119891) Denote
119871 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) +
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) minus 120572 (119911)
(86)
Assume to the contrary that 120575(120572 1198764(119911 119891)) = 1 Thus
Theorem A yields that
119873(119903
1
119871
) = 119873(119903
1
1198764(119911 119891) minus 120572
)
= 119878 (119903 1198764(119911 119891)) = 119878 (119903 119891)
(87)
Then (8) (86) (87) Lemma 18 and the assumption that V ge119889(119875) + 2 imply that 119871(119911) equiv 119891(119911)V that is
119875 (119911 119891) =
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) equiv 120572 (119911) (88)
Noting the fact that 119879(119903 119887119896) = 119878(119903 119891) and 119879(119903 120572) = 119878(119903 119891)
we deduce from Theorem A that (88) is a contradictionTherefore we have 120575(120572 119876
4(119911 119891)) lt 1
10 Abstract and Applied Analysis
4 Examples
Example 1 We consider nonhomogeneous differential-difference polynomials
1198751(119911 119891) = 119891 (119911) 119891
2
(119911 + log 4) minus 411989110158401015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 2) + 119891101584010158402 (119911 + log 3)
1198752(119911 119891) = 3119891
3
(119911) 11989110158402
(119911 + log 4)
minus 21198911015840
(119911) 119891 (119911 + log 3) 119891101584010158403 (119911 + log 2)
+ 1198914
(119911) minus 119891101584010158403
(119911)
1198753(119911 119891) = 119891 (119911) 119891
1015840
(119911 + log 2) 11989110158401015840 (119911 + log 3) minus 6119891101584010158402 (119911)(89)
and a homogeneous differential-difference polynomial
1198754(119911 119891) = 119891
101584010158403
(119911 + log 2) minus 1198911015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 3) minus 119891 (119911) 1198911015840 (119911) 11989110158401015840 (119911) (90)
where 119889(1198751) = 3 gt 2 = 119889
lowast
(1198751) 119889(119875
2) = 5 gt 3 = 119889
lowast
(1198752)
119889(1198753) = 3 gt 2 = 119889
lowast
(1198753) and 119889(119875
4) = 3 = 119889
lowast
(1198754) Clearly the
function 119891(119911) = 119890119911 satisfies (8) and 1205902(119891) = 0 lt 1 Then we
have
119889lowast
(1198751) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198751(119911 119890119911
)) =
2119903
120587
+ 119874 (1)
lt 119889 (1198751) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198752) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198752(119911 119890119911
)) =
4119903
120587
+ 119874 (1)
lt 119889 (1198752) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198753) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198753(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198753) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198754) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198754(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198754) 119879 (119903 119890
119911
) + 119874 (1)
(91)
This example shows that (9) is best possible
Example 2 Consider 119891(119911) = 119890119911 again Then the
homogeneous case 1198754(119911 119891) in Example 1 also illustrates
Theorem 9(a) And the nonhomogeneous differential-difference polynomials 119875
119894(119911 119891) 119894 = 1 2 3 in Example 1
also illustrate Theorem 9(b) where 120575(120572 1198751(119911 119891)) = 0
120575(120572 1198752(119911 119891)) le 14 lt 23 = 1 minus ((2119889
lowast
(1198752) minus 119889(119875
2))119889lowast
(1198752))
and 120575(120572 1198753(119911 119891)) le 13 lt 12 = 1 minus ((2119889
lowast
(1198753) minus
119889(1198753))119889lowast
(1198753)) Next we consider the nonhomogeneous
differential-difference polynomial
1198755(119911 119891) = 119891
1015840
(119911) 119891 (119911 + log 2) minus 1198912 (119911)
+ 1198911015840
(119911 + log 3) minus 3119891 (119911) + 1(92)
where 119889(1198755) = 2 119889
lowast
(1198755) = 0 Clearly 120575(1 119875
5(119911 119891)) =
120575(1 1198902119911
+ 1) = 1 Note that 2119889lowast(1198755) gt 119889(119875
5) fails then this
example shows that the assumption ldquo2119889lowast(119875) gt 119889(119875)rdquo cannotbe omitted inTheorem 9(b)
Example 3 We consider the differential-difference polyno-mials
1198761(119911 119891) = (119891
2
)
2
1198756(119911 119891)
= 1198914
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
1198762(119911 119891) = 119891
2
1198756(119911 119891)
= 1198912
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
(93)
and the function 119891(119911) = sin 119911 On the one hand 119873(119903 119891) =119878(119903 119891) 120590
2(119891) = 0 lt 1 and 119906V
1198761
gt 119889(1198756) and V
1198762
+ 2119889lowast
(1198756) gt
119889(1198756) hold where V
1198761
= V1198762
= 119906 = 2 and 119889(1198756) = 3 gt 2 =
119889lowast
(1198756) On the other hand 120575(120572 119876
1(119911 119891)) le 1 minus (1114) lt
1 minus (114) = 1 minus (119906 minus 1)(119906V1198761
minus 119889(119875))119906(119906V1198761
+ 119889(119875)) lt 1 and120575(120572 119876
2(119911 119891)) lt 1 hold This example shows that Theorems
10 and 11 may holdExample 4 We consider the differential-difference polyno-mials
119876(1)
4(119911 119891) = (119891
2
)
4
+ 1198757(119911 119891) = 119891
8
+ 1198757(119911 119891)
= 1198918
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(2)
4(119911 119891) = 119891
2
+ 1198757(119911 119891)
= 1198912
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(3)
4(119911 119891) = 2119891
3
+ 1198757(119911 119891)
= 21198913
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(4)
4(119911 119891) = 119891
4
+ 1198757(119911 119891)
= 1198914
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
(94)
and the function 119891(119911) = sin 119911 again On the one hand119876(1)
4(119911 119891) satisfies (119906 minus 1)119906V
119876(11)
4
(2119906 minus 1) gt 119889(1198757) and V
119876(12)
4
minus
2 gt (V119876(12)
4
minus 1)V119876(12)
4
(2V119876(12)
4
minus 1) gt 119889(1198757) respectively where
119906 = 4 V119876(11)
4
= 2 V119876(12)
4
= 8 119889(1198757) = 119889
lowast
(1198757) = 3 and
for 119894 = 2 3 4 119876(119894)4(119911 119891) satisfies 2 min119889lowast(119875
7) V119876(119894)
4
gt
max119889(1198757) V119876(119894)
4
where V119876(119894)
4
= 119894 On the other hand120575(120572 119876
(119894)
4(119911 119891)) lt 1 119894 = 1 2 3 4 hold This example shows
Abstract and Applied Analysis 11
that Theorems 13ndash15 may hold Moreover this example alsoshows the assumption ldquo119873(119903 1119891) = 119878(119903 119891)rdquo is not necessaryto Theorems 14(c) and 15 but it is regrettable for us notremoving it in our proofs
Example 5 We consider the differential-difference polyno-mials
1198771(119911 119891) = 119891
2
1198758(119911 119891)
= 1198912
(119911) (1198912
(119911 + 120587) +
1
sin2211991111989110158402
(119911 +
120587
2
))
1198772(119911 119891) = 119891
7
+ 1198759(119911 119891)
= 1198917
(119911) + sin 21199111198911015840 (119911 + 1205872
)1198912
(119911 +
120587
2
)
+ 11989110158402
(119911 +
120587
2
)119891(119911 +
3120587
2
)
+ 119911119891 (119911) 119891 (119911 +
120587
2
)
(95)
and the function 119891(119911) = 119890sin2119911 On the one hand 119877
1(119911 119891)
satisfies V1198771
+2119889lowast
(1198758) gt 119889(119875
8) and 119877
2(119911 119891) satisfies V
1198772
minus2 gt
(V1198772
minus 1)V1198772
(2V1198772
minus 1) gt 119889(1198759) respectively where V
1198771
=
2 and 119889(1198758) = 119889
lowast
(1198758) = 2 and V
1198772
= 7 and 119889(1198759) = 3 On
the other hand 120575(1198902 1198771(119911 119891)) = 120575(119890119911 119877
2(119911 119891)) = 1 hold
showing thatTheorems 11 and 14 fail Noting that the function119891(119911) = 119890
sin2119911 satisfies 1205902(119891) = 1 we know that the assumption
ldquo1205902(119891) lt 1rdquo is essential for Theorems 11 and 14 In fact it is
also essential for our other results in the whole paper but it isunnecessary to give examples one by one
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project was supported by the National Natural Sci-ence Foundation of China (11301233 and 11171119) and theNatural Science Foundation of Jiangxi Province in China(20132BAB211001 and 20132BAB211002) and Sponsored Pro-gram for Cultivating Youths of Outstanding Ability in JiangxiNormal University of China
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Group DordrechtTheNetherlands 2003
[4] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof f(z+120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[5] R G Halburd and R J Korhonen ldquoDifference analogue ofthe Lemma on the Logarithmic Derivative with applicationsto difference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[6] R G Halburd R J Korhonen and K Toghe ldquoHolomor-phic curves with shift-invariant hyper-planepreimagesrdquo Tran-sactions of the American Mathematical Society In press httparxivorgabs09033236
[7] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[8] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[9] Z X Chen ldquoComplex oscillation of meromorphic solutions forthe Pielou logistic equationrdquo Journal of Difference Equations andApplications vol 19 no 11 pp 1795ndash1806 2013
[10] Z X Chen and K H Shon ldquoFixed points of meromorphicsolutions for some difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 496096 7 pages 2013
[11] K Ishizaki and N Yanagihara ldquoWiman-Valiron method fordifference equationsrdquo Nagoya Mathematical Journal vol 175pp 75ndash102 2004
[12] I Laine and C-C Yang ldquoClunie theorems for difference andq-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] I Laine and C C Yang ldquoValue distribution of difference poly-nomialsrdquo Proceedings of the Japan Academy A vol 83 no 8 pp148ndash151 2007
[14] C-C Yang and I Laine ldquoOn analogies between nonlineardifference and differential equationsrdquo Proceedings of the JapanAcademy A vol 86 no 1 pp 10ndash14 2010
[15] R R Zhang and Z B Huang ldquoResults on difference analoguesof Valiron-Mokhonrsquoko theoremrdquoAbstract and Applied Analysisvol 2013 Article ID 273040 6 pages 2013
[16] XMZheng andZXChen ldquoOndeficiencies of somedifferencepolynomialsrdquo Acta Mathematica Sinica vol 54 no 6 pp 983ndash992 2011 (Chinese)
[17] XM Zheng and Z X Chen ldquoOn the value distribution of somedifference polynomialsrdquo Journal of Mathematical Analysis andApplications vol 397 no 2 pp 814ndash821 2013
[18] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Mathematique de France vol 59 pp 17ndash39 1931
[19] A Z Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoSibirskii Matematicheskii Zhurnal vol 15 pp 1305ndash1322 1974(Russian)
[20] W K Hayman ldquoPicard values of meromorphic functions andtheir derivativesrdquo Annals of Mathematics vol 70 no 2 pp 9ndash42 1959
[21] C-C Yang ldquoOn deficiencies of differential polynomialsrdquoMath-ematische Zeitschrift vol 116 no 3 pp 197ndash204 1970
[22] C-C Yang ldquoOn deficiencies of differential polynomials IIrdquoMathematische Zeitschrift vol 125 no 2 pp 107ndash112 1972
[23] A A Golrsquodberg and I V Ostrovskii The Distribution ofValues ofMeromorphic Functions NaukaMoscow Russia 1970(Russian)
12 Abstract and Applied Analysis
[24] E Mues and N Steinmetz ldquoThe theorem of Tumura-Clunie formeromorphic functionsrdquo Journal of the London MathematicalSociety vol 23 no 2 pp 113ndash122 1981
Research ArticleOn Growth of Meromorphic Solutions of Complex FunctionalDifference Equations
Jing Li12 Jianjun Zhang3 and Liangwen Liao1
1 Department of Mathematics Nanjing University Nanjing 210093 China2Nankai University Binhai College Tianjin 300270 China3Mathematics and Information Technology School Jiangsu Second Normal University Nanjing 210013 China
Correspondence should be addressed to Jianjun Zhang zhangjianjun1982163com
Received 29 November 2013 Accepted 13 January 2014 Published 25 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Jing Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The main purpose of this paper is to investigate the growth order of the meromorphic solutions of complex functional differenceequation of the form (sum
120582isin119868120572120582(119911)(prod
119899
]=1119891(119911 + 119888])119897120582]))(sum
120583isin119869120573120583(119911)(prod
119899
]=1119891(119911 + 119888])119898120583]
)) = 119876(119911 119891(119901(119911))) where 119868 = 120582 =
(1198971205821 1198971205822 119897
120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 = (119898
1205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two finite
index sets 119888] (] = 1 2 119899) are distinct complex numbers 120572120582(119911) (120582 isin 119868) and 120573
120583(119911) (120583 isin 119869) are small functions relative to 119891(119911)
and 119876(119911 119906) is a rational function in 119906 with coefficients which are small functions of 119891(119911) 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911]
of degree 119896 ge 1 We also give some examples to show that our results are sharp
1 Introduction and Main Results
Let 119891(119911) be a function meromorphic in the complex planeC We assume that the reader is familiar with the standardnotations and results in Nevanlinnarsquos value distribution the-ory ofmeromorphic functions such as the characteristic func-tion 119879(119903 119891) proximity function 119898(119903 119891) counting function119873(119903 119891) and the first and secondmain theorems (see eg [1ndash4]) We also use 119873(119903 119891) to denote the counting function ofthe poles of 119891(119911) whose every pole is counted only once Thenotations 120588(119891) and 120583(119891) denote the order and the lower orderof119891(119911) respectively 119878(119903 119891) denotes any quantity that satisfiesthe condition 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possiblyoutside an exceptional set of 119903 of finite linear measure Ameromorphic function 119886(119911) is called a small function of 119891(119911)or a small function relative to 119891(119911) if and only if 119879(119903 119886(119911)) =119878(119903 119891)
Recently some papers (see eg [5ndash7]) focusing on com-plex difference and functional difference equations emergedIn 2005 Laine et al [5] firstly considered the growth ofmeromorphic solutions of the complex functional differenceequations by utilizing Nevanlinna theory They obtained thefollowing result
Theorem A Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (1)
where 119869 is a collection of all subsets of 1 2 119899 119888119895rsquos are
distinct complex constants and 119901(119911) is a polynomial of degree119896 ge 2 Moreover we assume that the coefficients120572
119869(119911) are small
functions relative to 119891 and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (2)
where 120572 = log 119899 log 119896
In 2007 Rieppo [6] gave an estimation of growth ofmeromorphic solutions of complex functional equations asfollows
Theorem B Suppose that 119891 is a transcendental meromorphicfunction Let 119876(119911 119891) 119877(119911 119891) be rational functions in 119891
with small meromorphic coefficients relative to 119891 such that0 lt 119902 = deg
119891119876 le 119889 = deg
119891119877 and 119901(119911) = 119901
119896119911119896
+ 119901119896minus1
119911119896minus1
+
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 828746 6 pageshttpdxdoiorg1011552014828746
2 Abstract and Applied Analysis
sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 If 119891 is a solution of the
functional equation
119877 (119911 119891 (119911)) = 119876 (119911 119891 (119901 (119911))) (3)
then 119902119896 le 119889 and for any 120576 0 lt 120576 lt 1 there exist positive realconstants 119870
1and 119870
2such that
1198701(log 119903)120572minus120576 le 119879 (119903 119891) le 119870
2(log 119903)120572+120576 120572 =
log 119889 minus log 119902log 119896
(4)
when 119903 is large enough
Rieppo [6] also considered the growth order ofmeromor-phic solutions of functional equation (3) when 119896 = 1 and gotthe following
Theorem C Suppose that 119891 is a transcendental meromorphicsolution of (3) where 119901(119911) = 119886119911+119887 119886 119887 isin C 119886 = 0 and |119886| = 1Then
120583 (119891) = 120588 (119891) =
log119889 minus log 119902log |119886|
(5)
Two years later Zheng et al [7] extended Theorem A tomore general type and obtained a similar result of TheoremC In fact they got the following two results
TheoremD Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119876 (119911 119891 (119901 (119911))) (6)
where 119869 is a collection of all nonempty subsets of 1 2 119899119888119895(119895 = 1 119899) are distinct complex constants 119901(119911) = 119901
119896119911119896
+
119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a
rational function in 119906 of deg119906119876 = 119902(gt 0) Also suppose that
all the coefficients of (6) are small functions relative to 119891 Then119902119896 le 119899 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (7)
where 120572 = (log 119899 minus log 119902) log 119896
Theorem E Suppose that 119891 is a transcendental meromorphicsolution of (6) where 119869 is a collection of all nonempty subsetsof 1 2 119899 119888
119895(119895 = 1 119899) are distinct complex constants
119901(119911) = 119886119911 + 119887 119886 119887 isin C and 119876(119911 119906) is a rational function in 119906of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of (6)
are small functions relative to 119891(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log 119899minus log |119886|
(8)
(ii) If |119886| gt 1 then we have 119902 le 119899 and
120588 (119891) le
log 119899 minus log 119902log |119886|
(9)
(iii) If |119886| = 1 119902 gt 119899 then we have 120588(119891) = 120583(119891) = infin
In this paper we will consider a more general classof complex functional difference equations We prove thefollowing results which generalize the above related results
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic solution of the functional difference equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119901 (119911))) (10)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 =
120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin N⋃0 ] = 1 2 119899
are two finite index sets 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin
C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a rational function in 119906 ofdeg119906119876 = 119902(gt 0) Also suppose that all the coefficients of (10)
are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (11)
Then 119902119896 le 120590 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (12)
where 120572 = (log120590 minus log 119902) log 119896
Theorem 2 Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119886119911 + 119887))
(13)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 =
(1198981205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two
finite index sets 119886 119887 isin C and 119876(119911 119906) is a rational function in119906 of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of
(10) are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (14)
(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(15)
(ii) If |119886| gt 1 then we have 119902 le 120590 and
120588 (119891) le
log120590 minus log 119902log |119886|
(16)
(iii) If |119886| = 1 and 119902 gt 120590 then we have 120583(119891) = 120588(119891) = infin
Abstract and Applied Analysis 3
Next we will give some examples to show that our resultsare best in some extent
Example 3 Let 1198881= arctan 2 119888
2= minus1205874 Then it is easy to
check that 119891(119911) = tan 119911 solves the following equation
119891(119911 + 1198881)2
119891 (119911 + 1198882)
119891 (119911 + 1198881) + 119891(119911 + 119888
2)2
= (minus4119891(
119911
2
)
8
+ 8119891(
119911
2
)
7
+ 28119891(
119911
2
)
6
minus 56119891(
119911
2
)
5
minus 32119891(
119911
2
)
4
+ 56119891(
119911
2
)
3
+ 28119891(
119911
2
)
2
minus 8119891(
119911
2
) minus 4)
times (3119891(
119911
2
)
8
+ 10119891(
119911
2
)
7
+ 16119891(
119911
2
)
6
+ 122119891(
119911
2
)
5
minus 6119891(
119911
2
)
4
minus122119891(
119911
2
)
3
+16119891(
119911
2
)
2
minus10119891(
119911
2
) + 3)
minus1
(17)
Obviously we have
120583 (119891) = 120588 (119891) = 1 =
log 119902 minus log120590minus log |119886|
(18)
where 119902 = 8 120590 = 4 and 119886 = 12
Example 3 shows that the estimate in Theorem 2(i) issharp
Example 4 It is easy to check that 119891(119911) = tan 119911 satisfies theequation
119891(119911 + (1205873))2
119891 (119911 + (1205876)) minus 119891 (119911 + (1205876))
119891 (119911 + (1205873)) 119891(119911 + (1205876))2
minus 119891 (119911 + (1205873))
=
radic3119891(2119911)2
+ 4119891 (2119911) + radic3
minusradic3119891(2119911)2
+ 4119891 (2119911) minus radic3
(19)
Clearly we have
120583 (119891) = 120588 (119891) = 1 =
log120590 minus log 119902log |119886|
(20)
where 120590 = 4 119902 = 2 and 119886 = 2
Example 4 shows that the estimate in Theorem 2(ii) issharp
Example 5 119891(119911) = tan 119911 satisfies the equation of the form
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
minus(119891(1199112)2
minus 2119891 (1199112) minus 1)
3
8119891 (1199112) (119891(1199112)2
minus 1) (119891(1199112)2
+ 2119891 (1199112) minus 1)
(21)
where 120590 = 4 119902 = 6 and 119886 = 12 120588(119891) = 120583(119891) = 1 gt
log(32) log 2 = (log 119902 minus log120590) minus log |119886|
Example 5 shows that the strict inequality in Theorem 2may occur Therefore we do not have the same estimation asinTheoremC for the growth order ofmeromorphic solutionsof (13)
The following Example shows that the restriction 119902 gt 120590
in case (iii) in Theorem 2 is necessary
Example 6 Meromorphic function 119891(119911) = tan 119911 solves thefollowing equation
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
(119891 (119911) + 1)3
4119891 (119911) (1 minus 119891 (119911))
(22)
where 119886 = 1 and 4 = 120590 gt 119902 = 3 but 120588(119891) = 120583(119891) = 1
Next we give an example to show that case (iii) inTheorem 2 may hold
Example 7 Function 119891(119911) = 119911119890119890119911
satisfies the followingequation
(119911 + log 6) (119911 + log 2)5 [119891(119911 + log 4)4 + 119891 (119911 + log 4)](119911 + log 4) 119891 (119911 + log 6)
=
(119911 + log 4)3119891(119911 + log 2)6 + (119911 + log 2)6
119891 (119911 + log 2)
(23)
where 119886 = 1 and 119902 = 6 gt 5 = 120590 Obviously 120588(119891) = 120583(119891) = infin
2 Main Lemmas
In order to prove our results we need the following lemmas
Lemma 1 (see [4 8]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(24)
such that the meromorphic coefficients 119886119894(119911) 119887119895(119911) satisfy
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 1 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 1 119902
(25)
then one has
119879 (119903 119877 (119911 119891)) = max 119901 119902 sdot 119879 (119903 119891) + 119878 (119903 119891) (26)
From the proof ofTheorem 1 in [9] we have the followingestimate for the Nevanlinna characteristic
Lemma 2 Let 1198911 1198912 119891
119899be distinct meromorphic func-
tions and
119865 (119911) =
119875 (119911)
119876 (119911)
=
sum120582isin119868
120572120582(119911) 119891
1198971205821
1119891
1198971205822
2 119891
119897120582119899
119899
sum120583isin119869
120573120583(119911) 119891
1198981205831
1119891
1198981205832
2 119891
119898120583119899
119899
(27)
4 Abstract and Applied Analysis
Then
119879 (119903 119865 (119911)) le
119899
sum
]=1120590]119879 (119903 119891]) + 119878 (119903 119891) (28)
where 119868 = 120582 = (1198971205821 1198971205822 119897120582119899) | 119897
120582] isin N⋃0 ] =
1 2 119899 and 119869 = 120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin
N⋃0 ] = 1 2 119899 are two finite index sets 120590] =
max120582120583119897120582] 119898120583] (] = 1 2 119899) 120572
120582(119911) = 119900(119879(119903 119891])(120582 isin 119868))
and120573120583(119911) = 119900(119879(119903 119891])(120583 isin 119869)) hold for all ] isin 1 2 119899 and
satisfy 119879(119903 120572120582) = 119878(119903 119891) (120582 isin 119868) and 119879(119903 120573
120583) = 119878(119903 119891) (120583 isin
119869)
Lemma 3 (see [7]) Let 119888 be a complex constant Given 120576 gt 0
and a meromorphic function 119891 one has
119879 (119903 119891 (119911 plusmn 119888)) le (1 + 120576) 119879 (119903 + |119888| 119891) (29)
for all 119903 gt 1199030 where 119903
0is some positive constant
Lemma 4 (see [4]) Let 119892 (0 +infin) rarr R ℎ (0 +infin) rarr R
bemonotone increasing functions such that 119892(119903) le ℎ(119903) outsideof an exceptional set 119864 of finite linear measure Then for any120572 gt 1 there exists 119903
0gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 gt 119903
0
Lemma 5 (see [10]) Let 119891 be a transcendental meromorphicfunction and 119901(119911) = 119886
119896119911119896
+ 119886119896minus1
119911119896minus1
+ sdot sdot sdot + 1198861119911 + 1198860 119886119896
= 0be a nonconstant polynomial of degree 119896 Given 0 lt 120575 lt |119886
119896|
denote 120582 = |119886119896| + 120575 and 120583 = |119886
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C⋃infin one has
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 (119901 (119911))) le 119896119899 (120582119903119896
119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 (119901 (119911)))
le 119873 (120582119903119896
119886 119891) + 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 (119901 (119911))) le (1 + 120576) 119879 (120582119903119896
119891)
(30)
for all 119903 large enough
Lemma 6 (see [11]) Let 120601 [1199030 +infin) rarr (0 +infin) be
positive and bounded in every finite interval and suppose that120601(120583119903119898
) le 119860120601(119903) + 119861 holds for all 119903 large enough where 120583 gt 0119898 gt 1 119860 gt 1 and 119861 are real constants Then
120601 (119903) = 119874 ((log 119903)120572) (31)
where 120572 = log119860 log119898
Lemma 7 (see [6]) Let 120601 (1199030infin) rarr (1infin) where 119903
0ge 1
be a monotone increasing function If for some real constant120572 gt 1 there exists a real number 119870 gt 1 such that 120601(120572119903) ge
119870120601(119903) then
lim119903rarrinfin
log120601 (119903)log 119903
ge
log119870log120572
(32)
Lemma 8 (see [12]) Let 120601 (1infin) rarr (0infin) be a monotoneincreasing function and let 119891 be a nonconstant meromorphic
function If for some real constant 120572 isin (0 1) there exist realconstants 119870
1gt 0 and 119870
2ge 1 such that
119879 (119903 119891) le 1198701120601 (120572119903) + 119870
2119879 (120572119903 119891) + 119878 (120572119903 119891) (33)
then
120588 (119891) le
log1198702
minus log120572+ lim119903rarrinfin
log120601 (119903)log 119903
(34)
3 Proof of Theorems
Proof of Theorem 1 We assume 119891(119911) is a transcendentalmeromorphic solution of (10) Denoting 119862 =
max|1198881| |1198882| |119888
119899| According to Lemmas 1 2 and 3
and the last assertion of Lemma 5 we get that for any 1205761gt 0
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) + 119878 (119903 119891)
le 119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 1205761) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(35)
where 119903 is large enough and120583 = |119901119896|minus120575 for some 0 lt 120575 lt |119901
119896|
Since 119879(119903 + 119862 119891) le 119879(120573119903 119891) holds for 119903 large enough for120573 gt 1 we may assume 119903 to be large enough to satisfy
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120573119903 119891) (36)
outside a possible exceptional set of finite linear measure ByLemma 4 we know that whenever 120574 gt 1
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120574120573119903 119891) (37)
holds for all 119903 large enough Denote 119905 = 120574120573119903 thus theinequality (37) may be written in the form
119879(
120583
(120574120573)119896
119905119896
119891) le
120590 (1 + 1205761)
119902 (1 minus 1205761)
119879 (119905 119891) (38)
By Lemma 6 we have
119879 (119903 119891) = 119874 ((log 119903)1205721) (39)
Abstract and Applied Analysis 5
where
1205721=
log (120590 (1 + 1205761) 119902 (1 minus 120576
1))
log 119896
=
log120590 minus log 119902log 119896
+
log ((1 + 1205761) (1 minus 120576
1))
log 119896
(40)
Denoting now 120572 = (log120590 minus log 119902) log 119896 and 120576 = log((1 +
1205761)(1 minus 120576
1)) log 119896 thus we obtain the required form
Finally we show that 119902119896 le 120590 If 119902119896 gt 120590 then we have120572 lt 1 For sufficiently small 120576 gt 0 we have 120572 + 120576 lt 1 whichcontradicts with the transcendency of 119891 Thus Theorem 1 isproved
Proof of Theorem 2 Suppose 119891(119911) is a transcendental mero-morphic solution of (13) Denoting 119862 = max|119888
1| |1198882|
|119888119899|
(i) 0 lt |119886| lt 1 We may assume that 119902 gt 120590 since the case119902 le 120590 is trivial by the fact that 120583(119891) ge 0 By Lemmas1ndash3 we have for any 120576 gt 0 and 120573 gt 1
119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
le 120590 (1 + 120576) 119879 (120573119903 119891) + 119878 (119903 119891)
(41)
where 119903 is large enoughBy the last assertion of Lemma 5 and (41) we obtain that
for 120583 = |119886| minus 120575 (0 lt 120575 lt |119886| 0 lt 120583 lt 1) the followinginequality
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120573119903 119891) (42)
holds where 119903 is large enough outside of a possible set of finitelinear measure By Lemma 4 we get that for any 120574 gt 1 andsufficiently large 119903
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120574120573119903 119891) (43)
Therefore
119902 (1 minus 120576)
120590 (1 + 120576)
119879 (119903 119891) le 119879(
120574120573
120583
119903 119891) (44)
Since 120573 gt 1 120574 gt 1 0 lt 120583 lt 1 and 119902 gt 120590 we have 120573120574120583 gt 1
and 119902(1 minus 120576)120590(1 + 120576) gt 1 when 120576 is small enough UsingLemma 7 we see that
120583 (119891) ge
log 119902 (1 minus 120576) minus log120590 (1 + 120576)
log 120574120573 minus log 120583 (45)
Letting 120576 rarr 0 120575 rarr 0 120573 rarr 1 and 120574 rarr 1 we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(46)
(ii) |119886| gt 1 By the similar reasoning as is (i) we easilyobtain that
119902 (1 minus 120576) 119879 (120583119903 119891) le 119902119879 (119903 119891 (119901 (119911)))
le 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(47)
for all 119903 large enough We may select sufficiently smallnumbers 120575 gt 0 and 120576 gt 0 such that 120583 = |119886| minus 120575 gt 1 and(1120583) + 120576 lt 1 Thus we have
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891) (48)
namely
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) (49)
where 119903 is large enough possibly outside of a set of finite linearmeasure By Lemma 4 we have for any 1 lt 120574 lt 120583
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (120574119903 119891 (119911)) (50)
that is
119879 (119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879(
120574
120583
119903 119891 (119911)) (51)
holds for all sufficiently large 119903 By Lemma 8 we obtain
120588 (119891) le
log120590 minus log 119902 + log (1 + 120576) minus log (1 minus 120576)
minus log (120574120583) (52)
Letting 120576 rarr 0 120575 rarr 0 and 120574 rarr 1 we have
120588 (119891) le
log120590 minus log 119902log |119886|
(53)
(iii) |119886| = 1 and 119902 gt 120590 The proof of this case is completelysimilar as in the case in (i) In fact we set 120583 = |119886|minus120575 =
1 minus 120575 (0 lt 120575 lt 1 0 lt 120583 lt 1) Similarly we can get
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(54)
Since |119886| = 1 we have 120583(119891) = 120588(119891) = infin
6 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the anonymous ref-erees for their valuable comments and suggestions Theresearch was supported by Colonel-level topics (JSNU-ZY-01) (Jsie2012zd01) and NSF of China (11271179)
References
[1] W Cherry and Z Ye Nevanlinnarsquos Theory of Value DistributionSpringer Monographs in Mathematics Springer Berlin Ger-many 2001
[2] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[3] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Beijing China 1988
[4] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] J Rieppo ldquoOn a class of complex functional equationsrdquoAnnalesAcademiaelig Scientiarum Fennicaelig vol 32 no 1 pp 151ndash170 2007
[7] X-M Zheng Z-X Chen and J Tu ldquoGrowth of meromorphicsolutions of some difference equationsrdquoApplicable Analysis andDiscrete Mathematics vol 4 no 2 pp 309ndash321 2010
[8] A Z Mokhonrsquoko ldquoThe Nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija vol 14 pp 83ndash87 1971 (Russian)
[9] A A Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoAkademija Nauk SSSR vol 15 pp 1305ndash1322 1974
[10] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[11] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[12] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
Research ArticleUnicity of Entire Functions concerning Shifts andDifference Operators
Dan Liu Degui Yang and Mingliang Fang
Institute of Applied Mathematics South China Agricultural University Guangzhou 510642 China
Correspondence should be addressed to Mingliang Fang mlfangscaueducn
Received 29 October 2013 Revised 17 December 2013 Accepted 19 December 2013 Published 3 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Dan Liu et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We prove a unicity theorem of entire functions that share two distinct small functions with their shiftsThe corollary of the theoremconfirms the conjecture posed by Li and Gao (2011)
1 Introduction
Let119891 be a nonconstantmeromorphic function in the complexplane C We will use the standard notations in Nevanlinnatheory of meromorphic functions such as 119879(119903 119891) 119873(119903 119891)and 119898(119903 119891) (see [1 2]) The notation 119878(119903 119891) is defined to beany quantity satisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin pos-sibly outside a set of finite linear measures A meromorphicfunction 119886 is called a small function related to119891 provided that119879(119903 119886) = 119878(119903 119891)
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 be a small function related to both 119891 and 119892 Wesay that 119891 and 119892 share 119886 CM if 119891 minus 119886 and 119892 minus 119886 have thesame zeros with the same multiplicities 119891 and 119892 are said toshare 119886 IM if 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
Let119873(119903 119886) be the counting functions of all common zeroswith the same multiplicities of 119891 minus 119886 and 119892 minus 119886 If
119873(119903
1
119891 minus 119886
) + 119873(119903
1
119892 minus 119886
) minus 2119873 (119903 119886)
= 119878 (119903 119891) + 119878 (119903 119892)
(1)
then we say that 119891 and 119892 share 119886 CM almostFor a nonzero complex constant 119888 isin C we define
difference operators asΔ119888119891(119911) = 119891(119911+119888)minus119891(119911) andΔ119899
119888119891(119911) =
Δ119888(Δ119899minus1
119888119891(119911)) 119899 isin N 119899 gt 2
In 1977 Rubel and Yang [3] proved the following result
Theorem A Let 119891 be a nonconstant entire function If 119891(119911)and 1198911015840(119911) share two distinct finite values CM then 119891(119911) equiv
1198911015840
(119911)
In fact the conclusion still holds if the two CM values arereplaced by two IM values (see Gundersen [4 5] Mues andSteinmetz [6])
Recently a number of articles focused on value dis-tribution in shifts or difference operators of meromorphicfunctions (see [7ndash11]) In particular some papers studied theunicity of meromorphic functions sharing values with theirshifts or difference operators (see [12ndash14]) In 2009 Heit-tokangas et al [12] proved the following result concerningshifts
Theorem B Let 119891 be a nonconstant entire function of finiteorder 119888 isin C If119891(119911) and119891(119911+119888) share two distinct finite valuesCM then 119891(119911) equiv 119891(119911 + 119888)
In 2011 Li and Gao [14] proved the following resultconcerning difference operators
Theorem C Let 119891 be a nonconstant entire function of finiteorder 119888 isin C and let 119899 be a positive integer Suppose that 119891(119911)and Δ119899
119888119891(119911) share two distinct finite values 119886 119887 CM and one of
the following cases is satisfied(i) 119886119887 = 0(ii) 119886119887 = 0 and 120588(119891) notin 119873Then 119891(119911) equiv Δ119899
119888119891(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 380910 5 pageshttpdxdoiorg1011552014380910
2 Abstract and Applied Analysis
In [14] Li andGao conjectured that the restriction 120588(119891) notinN for the case 119886119887 = 0 can be removed In this paper weconfirm their conjecture In fact we prove the followingmoregeneral results
Theorem 1 Let 119891 be a nonconstant entire function of finiteorder let 119899 be a positive integer let 119886(119911) 119887(119911) be two distinctsmall functions related to 119891(119911) let 119898
1 1198982 119898
119899be nonzero
complex numbers and 1198881 1198882 119888
119899distinct finite values and let
119865 (119911) = 1198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119891 (119911 + 119888
119899)
(2)
If 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119865(119911)
Corollary 2 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number let 119899 be apositive integer and let 119886 119887 be two distinct finite values If 119891(119911)and Δ119899
119888119891(119911) share 119886 119887 CM then 119891(119911) equiv Δ119899
119888119891(119911)
Remark 3 Corollary 2 confirms the conjecture of Li and Gaoin [14]
Corollary 4 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number and let 119886(119911)119887(119911) be two distinct small functions related to 119891 If 119891(119911) and119891(119911 + 119888) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119891(119911 + 119888)
2 Some Lemmas
For the proof of Theorem 1 we require the following results
Lemma 5 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903
1
119891
) + 119873 (119903 119891) = 119878 (119903 119891)
119873(119903
1
119892
) + 119873 (119903 119892) = 119878 (119903 119892)
(3)
If 119891(119911) and 119892(119911) share 1 CM almost then either 119891(119911) equiv 119892(119911)
or 119891(119911)119892(119911) equiv 1
Lemma 6 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903 119891) = 119878 (119903 119891) 119873 (119903 119892) = 119878 (119903 119892) (4)
If 119891(119911) and 119892(119911) share 0 and 1 CM almost and
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 119891) + 119879 (119903 119892)
lt
2
3
(5)
where 119868 sub [0infin) is a set of infinitely linear measure then
119891 (119911) =
119886119892 (119911) + 119887
119888119892 (119911) + 119889
(6)
where 119886 119887 119888 and 119889 are constants satisfying 119886119889 minus 119887119888 = 0
Lemma 7 (see [10]) Let 119891 be a nonconstant meromorphicfunction of finite order 119888 isin C Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (7)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
119889119903119903 lt infin
In the following 119878(119903 119891) denotes any function satisfying119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside a set withfinite logarithmic measure
3 Proof of Theorem 1
We prove Theorem 1 by contradiction Suppose that 119891(119911) equiv
119865(119911) Then it follows from 119891(119911) and 119865(119911) being two distinctentire functions that 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) and infinCM By the Nevanlinna second fundamental theorem forthree small functions we have
119879 (119903 119891) le 119873 (119903 119891) + 119873(119903
1
119891 minus 119886
)
+ 119873(119903
1
119891 minus 119887
) + 119878 (119903 119891)
le 119873(119903
1
119865 minus 119886
) + 119873(119903
1
119865 minus 119887
) + 119878 (119903 119891)
le 2119879 (119903 119865) + 119878 (119903 119891)
(8)
Similarly we have 119879(119903 119865) le 2119879(119903 119891) + 119878(119903 119865) Therefore119878(119903 119891) = 119878(119903 119865)
Set
1198911(119911) =
119891 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
1198651(119911) =
119865 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
(9)
Thus 1198911(119911) 1198651(119911) share 0 1 andinfin CM almost
Obviously we have
119879 (119903 1198911) = 119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 1198651) = 119879 (119903 119865) + 119878 (119903 119891)
119878 (119903 119865) = 119878 (119903 1198651) = 119878 (119903 119891
1) = 119878 (119903 119891)
(10)
By Nevanlinnarsquos second fundamental theorem we have
119879 (119903 1198911) le 119873(119903
1
1198911
) + 119873(119903
1
1198911minus 1
) + 119873 (119903 1198911) + 119878 (119903 119891
1)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
Abstract and Applied Analysis 3
le 119873(119903
1
1198651minus 1198911
) + 119878 (119903 119891)
le 119879 (119903 1198651minus 1198911) + 119878 (119903 119891)
le 119879 (119903 119865 minus 119891) + 119878 (119903 119891)
le 119898 (119903 119865 minus 119891) + 119878 (119903 119891)
(11)Since 119865 minus 119891 = 119898
1119891(119911 + 119888
1) + 1198982119891(119911 + 119888
2) + sdot sdot sdot + 119898
119899119891(119911 +
119888119899) minus 119891(119911) = 119891(119911)[119898
1(119891(119911 + 119888
1)119891(119911)) +119898
2(119891(119911 + 119888
2)119891(119911)) +
sdot sdot sdot + 119898119899(119891(119911 + 119888
119899)119891(119911)) minus 1] thus
119898(119903 119865 minus 119891)
le 119898 (119903 119891)
+ 119898(1199031198981
119891 (119911 + 1198881)
119891 (119911)
+ sdot sdot sdot + 119898119899
119891 (119911 + 119888119899)
119891 (119911)
minus 1)
le 119898 (119903 119891) + 119878 (119903 119891)
(12)By (11) we have
119879 (119903 1198911) le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(13)
It follows that119873(119903 0) + 119873 (119903 1) = 119879 (119903 119891
1) + 119878 (119903 119891) (14)
On the other hand byNevanlinna first fundamental theoremwe have
2119879 (119903 1198911) = 119879(119903
1
1198911
) + 119879(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
(15)
So we get
119879 (119903 1198911) le 119898(119903
1
1198911
) + 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119898(119903
1
119891 minus 119886
) + 119898(119903
1
119891 minus 119887
) + 119878 (119903 119891)
(16)
Set1198861(119911) = 119898
1119886 (119911 + 119888
1) + 1198982119886 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119886 (119911 + 119888
119899)
1198871(119911) = 119898
1119887 (119911 + 119888
1) + 1198982119887 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119887 (119911 + 119888
119899)
(17)
If 1198861(119911) equiv 119887
1(119911) we can deduce by (16) that
119879 (119903 1198911) le 119898(119903
1
119891 minus 119886
+
1
119891 minus 119887
) + 119878 (119903 119891)
le 119898(119903
119865 minus 1198861
119891 minus 119886
+
119865 minus 1198871
119891 minus 119887
)
+ 119898(119903
1
119865 minus 1198861
) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(18)
If 1198861(119911) equiv 119887
1(119911) set
119871 (119865) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
119865 11988611198871
1198651015840
1198861015840
11198871015840
1
11986510158401015840
11988610158401015840
111988710158401015840
1
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(19)
Then we have
119898(119903
119865 minus 1198861
119891 minus 119886
) = 119898(119903
119865 minus 1198871
119891 minus 119887
) = 119878 (119903 119891)
119898(119903
119871 (119865)
119865 minus 1198861
) = 119898(119903
119871 (119865)
119865 minus 1198871
) = 119878 (119903 119891)
(20)
It followed from (16) that
119879 (119903 1198911) le 119898(119903
119865 minus 1198861
119891 minus 119886
) + 119898(119903
1
119865 minus 1198861
)
+ 119898(119903
119865 minus 1198871
119891 minus 119887
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
+
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119871 (119865)
) + 119878 (119903 119891)
le 119879 (119903 119871 (119865)) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
4 Abstract and Applied Analysis
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
le 119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 1198911) + 119878 (119903 119891)
(21)
By (18) and (21) we can deduce that
119879 (119903 1198911) = 119879 (119903 119865) + 119878 (119903 119891) = 119879 (119903 119865
1) + 119878 (119903 119891) (22)
It follows from (14) and (22) that
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 1198911) + 119879 (119903 119865
1)
=
1
2
lt
2
3
(23)
By Lemma 6 we have
1198911(119911) =
1198601198651(119911) + 119861
1198621198651(119911) + 119863
(24)
where 119860 119861 119862 and 119863 are complex numbers satisfying 119860119863 minus
119861119862 = 0Now we consider three cases
Case 1 Consider 119873(119903 0) = 119878(119903 1198911) Thus
119873(119903
1
1198911
) + 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (25)
Similarly we have
119873(119903
1
1198651
) + 119873 (119903 1198651) = 119878 (119903 119865
1) = 119878 (119903 119891) (26)
By Lemma 5 we get that either 1198911equiv 1198651or 11989111198651equiv 1
If 1198911equiv 1198651 we can easily deduce that 119891 equiv 119865 which is a
contradiction with our assumptionIf 11989111198651equiv 1 that is
(119891 (119911) minus 119886) (119865 (119911) minus 119886) equiv (119887 minus 119886)2
(27)
then we have
(119891 minus 119886)2
=
(119887 minus 119886)2
(119865 minus 119886) (119891 minus 119886)
(28)
From (28) we have
2119879 (119903 119891) le 119879 (119903 (119891 minus 119886)2
) + 119878 (119903 119891)
= 119879(119903
1
(119887 minus 119886)2
((119865 minus 119886) (119891 minus 119886))
) + 119878 (119903 119891)
le 119879(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
= 119873(119903
119865 minus 119886
119891 minus 119886
) + 119898(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
(119865 minus 1198861) + (119886
1minus 119886)
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
1198861minus 119886
119891 minus 119886
) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
(29)
It follows that 119879(119903 119891) le 119878(119903 119891) a contradiction
Case 2 Consider 119873(119903 1) = 119878(119903 1198911) Using the same argu-
ment as used in Case 1 we deduce that 119879(119903 119891) le 119878(119903 119891) acontradiction
Case 3 Consider 119873(119903 0) = 119878(119903 1198911) 119873(119903 1) = 119878(119903 119891
1) Since
1198911and 119865
1share 0 1 CM almost we deduce from (24) that
1198911(119911) =
(119862 + 119863) 1198651(119911)
1198621198651(119911) + 119863
(30)
If 119862 = 0 then 1198911equiv 1198651 that is 119891 equiv 119865 a contradiction
Hence 119862 = 0 Thus we have
119873(119903
1
1198651+ (119863119862)
) = 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (31)
Obviously 119863119862 = 0 119863119862 = minus 1 Thus by Nevanlinnasecond fundamental theorem and (14) we get
2119879 (119903 1198911) = 2119879 (119903 119865
1) + 119878 (119903 119891
1)
le 119873(119903
1
1198651
) + 119873(119903
1
1198651minus 1
)
+ 119873(119903
1
1198651+ (119863119862)
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119878 (119903 119891)
(32)
It follows that 119879(119903 1198911) le 119878(119903 119891
1) a contradiction Thus
we prove that 119891(119911) equiv 119865(119911) This completes the proof ofTheorem 1
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 5
Acknowledgments
Theauthors thank the referees for careful reading of the paperpointing out a gap in the previous version of this paperand giving many valuable suggestions Research is supportedby the NNSF of China (Grant no 11371149) and NSF ofGuangdong Province China (Grant no S2012010010121)
References
[1] W K Hayman Meromorphic Function Clarendon PressOxford UK 1964
[2] L Yang Value Distribution Theory Springer Berlin Germany1993
[3] L A Rubel and C C Yang Value Shared by an Entire Functionand Its Derivative Lecture Notes in Math Springer BerlinGermany 1977
[4] G G Gundersen ldquoMeromorphic functions that share finitevalues with their derivativerdquo Journal of Mathematical Analysisand Applications vol 75 no 2 pp 441ndash446 1980
[5] G G Gundersen ldquoErrata meromorphic functions that sharefinite values with their derivativerdquo Journal of MathematicalAnalysis and Applications vol 86 no 1 p 307 1982
[6] E Mues and N Steinmetz ldquoMeromorphe Funktionen die mitihrer Ableitung Werte teilenrdquo Manuscripta Mathematica vol29 no 2ndash4 pp 195ndash206 1979
[7] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[8] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoTheRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[9] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[10] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[11] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[12] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[13] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables and Elliptic Equations vol 56 no 1ndash4 pp 81ndash92 2011
[14] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[15] M L Fang ldquoUnicity theorems for meromorphic function andits differential polynomialrdquo Advances in Mathematics vol 24no 3 pp 244ndash249 1995
Research ArticleOn Positive Solutions and Mann Iterative Schemes of a ThirdOrder Difference Equation
Zeqing Liu1 Heng Wu1 Shin Min Kang2 and Young Chel Kwun3
1 Department of Mathematics Liaoning Normal University Dalian Liaoning 116029 China2Department of Mathematics and RINS Gyeongsang National University Jinju 660-701 Republic of Korea3 Department of Mathematics Dong-A University Pusan 614-714 Republic of Korea
Correspondence should be addressed to Young Chel Kwun yckwundauackr
Received 14 October 2013 Accepted 16 December 2013 Published 28 January 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Zeqing Liu et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The existence of uncountably many positive solutions and convergence of the Mann iterative schemes for a third order nonlinearneutral delay difference equation are proved Six examples are given to illustrate the results presented in this paper
1 Introduction and Preliminaries
Recently many researchers studied the oscillation nonoscil-lation and existence of solutions for linear and nonlinearsecond and third order difference equations and systemssee for example [1ndash23] and the references cited therein Bymeans of the Reccati transformation techniques Saker [18]discussed the third order difference equation
Δ3
119909119899+ 119901119899119909119899+1
= 0 forall119899 ge 1198990 (1)
and presented some sufficient conditions which ensure thatall solutions are to be oscillatory or tend to zero Utilizing theSchauder fixed point theorem Yan and Liu [22] proved theexistence of a bounded nonoscillatory solution for the thirdorder difference equation
Δ3
119909119899+ 119891 (119899 119909
119899 119909119899minus120591) = 0 forall119899 ge 119899
0 (2)
Agarwal [2] established the oscillatory and asymptotic prop-erties for the third order nonlinear difference equation
Δ3
119909119899+ 119902119899119891 (119909119899+1) = 0 forall119899 ge 1 (3)
Andruch-Sobiło andMigda [4] studied the third order lineardifference equation of neutral type
Δ3
(119909119899minus 119901119899119909120590119899) plusmn 119902119899119909120591119899= 0 forall119899 ge 119899
0 (4)
and obtained sufficient conditions which ensure that allsolutions of the equation are oscillatory Grace andHamedani[6] discussed the difference equation
Δ3
(119909119899minus 119909119899minus120591) plusmn 119902119899
1003816100381610038161003816119909119899minus120590
1003816100381610038161003816
3 sgn119909119899minus120590
= 0 forall119899 ge 0 (5)
and gave some new criteria for the oscillation of all solutionsand all bounded solutions
Our goal is to discuss solvability and convergence ofthe Mann iterative schemes for the following third ordernonlinear neutral delay difference equation
Δ3
(119909119899+ 119887119899119909119899minus120591) + Δℎ (119899 119909
ℎ1119899
119909ℎ2119899
119909ℎ119896119899
)
+119891 (119899 1199091198911119899
1199091198912119899
119909119891119896119899
) = 119888119899 forall119899 ge 119899
0
(6)
where 120591 119896 1198990isin N 119887
119899119899isinN1198990
119888119899119899isinN1198990
sub R ℎ 119891 isin 119862(N1198990
times
R119896R) ℎ119897119899119899isinN1198990
119891119897119899119899isinN1198990
sube N and
lim119899rarrinfin
ℎ119897119899= lim119899rarrinfin
119891119897119899= +infin 119897 isin 1 2 119896 (7)
By employing the Banach fixed point theorem and somenew techniques we establish the existence of uncountablymany positive solutions of (6) conceive a few Mann iter-ative schemes for approximating these positive solutionsand prove their convergence and the error estimates Sixnontrivial examples are included
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 470181 16 pageshttpdxdoiorg1011552014470181
2 Abstract and Applied Analysis
Throughout this paper we assume that Δ is the forwarddifference operator defined by Δ119909
119899= 119909119899+1
minus 119909119899 R =
(minusinfin +infin) R+ = [0 +infin) N0and N denote the sets of
nonnegative integers and positive integers respectively
N119905= 119899 119899 isin N with 119899 ge 119905 forall119905 isin N
120573 = min 1198990minus 120591 inf ℎ
119897119899 119891119897119899 1 le 119897 le 119896 119899 isin N
1198990
isin N
119867119899= max ℎ2
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
119865119899= max 1198912
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
(8)
and 119897infin
120573represents the Banach space of all real sequences
on N120573with norm
119909 = sup119899isinN120573
1003816100381610038161003816100381610038161003816
119909119899
1198992
1003816100381610038161003816100381610038161003816
lt +infin for each 119909 = 119909119899119899isinN120573
isin 119897infin
120573
119860 (119873119872) = 119909 = 119909119899119899isinN120573
isin 119897infin
120573 119873 le
119909119899
1198992
le 119872 119899 isin N120573
for any 119872 gt 119873 gt 0
(9)
It is easy to see that 119860(119873119872) is a closed and convex subsetof 119897infin120573 By a solution of (6) wemean a sequence 119909
119899119899isinN120573
witha positive integer 119879 ge 119899
0+120591+120573 such that (6) holds for all 119899 ge
119879
Lemma 1 Let 119901119905119905isinN be a nonnegative sequence and 120591 isin N
(i) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199052
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119904=119899+119894120591suminfin
119905=119904119901119905= 0
(ii) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199053
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119906=119899+119894120591suminfin
119904=119906suminfin
119905=119904119901119905= 0
Proof Note that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
(
infin
sum
119905=119899+119894120591
119901119905+
infin
sum
119905=119899+1+119894120591
119901119905+
infin
sum
119905=119899+2+119894120591
119901119905+ sdot sdot sdot )
=
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
(1 + 119905 minus 119899 minus 119894120591) 119901119905le
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
119905119901119905
=
1
1198992
(
infin
sum
119905=119899+120591
119905119901119905+
infin
sum
119905=119899+2120591
119905119901119905+
infin
sum
119905=119899+3120591
119905119901119905+ sdot sdot sdot )
le
1
1198992
infin
sum
119905=119899+120591
(1 +
119905 minus 119899 minus 120591
120591
) 119905119901119905=
1
1198992
infin
sum
119905=119899+120591
119905 minus 119899
120591
119905119901119905
le
1
1198992120591
infin
sum
119905=119899+120591
1199052
119901119905997888rarr 0 as 119899 997888rarr infin
(10)
that is
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905= 0 (11)
As in the proof of (10) we infer that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
(1 + 119905 minus 119906) 119901119905
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
119905119901119905le
1
1198992120591
infin
sum
119905=119899+120591
1199053
119901119905997888rarr 0 as 119899 997888rarr infin
(12)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905= 0 (13)
This completes the proof
2 Uncountably Many Positive Solutions andMann Iterative Schemes
In this section using the Banach fixed point theoremand Mann iterative schemes we establish the existence ofuncountably many positive solutions of (6) prove conver-gence of the Mann iterative schemes relative to these positivesolutions and compute the error estimates between theManniterative schemes and the positive solutions
Theorem 2 Assume that there exist twoconstants 119872 and 119873 with 119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896) minus 119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119875119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896) minus ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119877119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
forall (119899 119906119897 119906119897) isin N1198990
times (R+
0)
2
1 le 119897 le 119896
(14)
1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119876119899
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119882119899
forall (119899 119906119897) isin N1198990
times (R+
0) 1 le 119897 le 119896
(15)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (16)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (17)
119887119899= minus1 eventually (18)
Abstract and Applied Analysis 3
Then one has the following(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
+
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(19)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin 119860(119873119872) of (6)with lim
119899rarrinfin119911119899= +infin and has the following error estimate
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] such thatinfin
sum
119898=0
120572119898= +infin (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Firstly we show that (a) holds Put 119871 isin (119873119872) Itfollows from (16)sim(18) that there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (22)
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119873
(23)
119887119899= minus1 forall119899 ge 119879 (24)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879 119878119871119909119879 120573 le 119899 lt 119879
(25)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In light of (14) (15) (22)(23) and (25) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872)
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
4 Abstract and Applied Analysis
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
lt min 119872 minus 119871 119871 minus 119873
(26)which yield that
119878119871(119860 (119873119872)) sube 119860 (119873119872)
1003817100381710038171003817119878119871119909 minus 1198781198711199101003817100381710038171003817le 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817 forall119909 119910 isin 119860 (119873119872)
(27)
which implies that 119878119871is a contraction in 119860(119873119872) The
Banach fixed point theorem and (27) ensure that 119878119871has a
unique fixed point 119911 = 119911119899119899isinN120573
isin 119860(119873119872) that is
119911119899= 1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879
119911119899minus120591
= (119899 minus 120591)2
119871
+
infin
sum
119894=1
infin
sum
119906=119899+(119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
)
minus 119888119905] forall119899 ge 119879 + 120591
(28)which mean that119911119899minus 119911119899minus120591
= (2119899120591 minus 1205912
) 119871
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
(29)which yields thatΔ (119911119899minus 119911119899minus120591)
= 2120591119871 +
infin
sum
119904=119899
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
Δ2
(119911119899minus 119911119899minus120591)
= minusℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
+
infin
sum
119905=119899
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905] forall119899 ge 119879 + 120591
(30)
which gives that
Δ3
(119911119899minus 119911119899minus120591)
= minusΔℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
minus 119891 (119905 1199111198911119899
1199111198912119899
119911119891119896119899
) + 119888119899 forall119899 ge 119879 + 120591
(31)
which together with (24) implies that 119911 = 119911119899119899isinN120573
is apositive solution of (6) in 119860(119873119872) Note that
119873 le
119911119899
1198992
le 119872 forall119899 isin N120573 (32)
which guarantees that lim119899rarrinfin
119911119899= +infin It follows from (19)
(22) (24) (25) and (27) that for any 119898 isin N0and 119899 ge 119879
1003816100381610038161003816100381610038161003816
119909119898+1119899
1198992
minus
119911119899
1198992
1003816100381610038161003816100381610038161003816
=
1
1198992
1003816100381610038161003816100381610038161003816100381610038161003816
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)] minus 119911
119899
1003816100381610038161003816100381610038161003816100381610038161003816
le (1 minus 120572119898)
1003816100381610038161003816119909119898119899minus 119911119899
1003816100381610038161003816
1198992
+ 120572119898
1003816100381610038161003816119878119871119909119898119899minus 119878119871119911119899
1003816100381610038161003816
1198992
le (1 minus 120572119898)1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817+ 120579120572119898
1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817
le [1 minus (1 minus 120579) 120572119898]1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 119899 ge 119879
(33)
which implies that
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (34)
That is (20) holds Thus Lemma 1 (20) and (21) guaranteethat lim
119898rarrinfin119909119898= 119911
Next we show that (b) holds Let 1198711 1198712
isin
(119873119872) and 1198711=1198712 As in the proof of (a) we deduce
similarly that for each 119888 isin 1 2 there exist constants 120579119888isin
(0 1) and 119879119888ge 1198990+ 120591 + 120573 and a mapping 119878
119871119888
satisfying
Abstract and Applied Analysis 5
(22)sim(27) where 120579 119871 and 119879 are replaced by 120579119888 119871119888 and 119879
119888
respectively and the mapping 119878119871119888
has a fixed point 119911119888 =
119911119888
119899119899isinN120573
isin 119860(119873119872) which is a positive solution of (6) in119860(119873119872) with lim
119899rarrinfin119911119888
119899= +infin It follows that
119911119888
119899= 1198992
119871119888
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911119888
ℎ1119904
119911119888
ℎ2119904
119911119888
ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 119911119888
1198911119905
119911119888
1198912119905
119911119888
119891119896119905
) minus 119888119905]
forall119899 ge 119879119888
(35)
which together with (14) and (20) means that for 119899 ge
max1198791 1198792
100381610038161003816100381610038161003816100381610038161003816
1199111
119899
1198992
minus
1199112
119899
1198992
100381610038161003816100381610038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119911
1
ℎ1119904
1199111
ℎ2119904
1199111
ℎ119896119904
)
minus ℎ (119904 1199112
ℎ1119904
1199112
ℎ2119904
1199112
ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119911
1
1198911119905
1199111
1198912119905
1199111
119891119896119905
) minus 119891 (119905 1199112
1198911119905
1199112
1198912119905
1199112
119891119896119905
)
10038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
10038161199111
ℎ119897119904
minus 1199112
ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
10038161199111
119891119897119905
minus 1199112
119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
max 11987921 1198792
2
infin
sum
119894=1
infin
sum
119906=max1198791 1198792+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816minusmax 120579
1 1205792
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
(36)
which yields that
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817ge
10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
1 +max 1205791 1205792
gt 0 (37)
that is 1199111 =1199112
This completes the proof
Theorem 3 Assume that there exist two constants119872 and 119873with 119872 gt 119873 gt 0 and four nonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) and
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (38)
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (39)
119887119899= 1 eventually (40)
Then one has the following
(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
minus
infin
sum
119894=1
119879+2119894120591minus1
sum
119906=119879+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(41)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
6 Abstract and Applied Analysis
Proof Let 119871 isin (119873119872) It follows from (38)sim(40) that thereexist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (42)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)) lt min 119872 minus 119871 119871 minus 119873
(43)
119887119899= 1 forall119899 ge 119879 (44)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus119888119905] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(45)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Using (14) (15) (42) (43)and (45) we get that for each 119909 = 119909
119899119899isinN120573
119910 = 119910119899119899isinN120573
isin
119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) = 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)])
lt min 119872 minus 119871 119871 minus 119873
(46)
which imply (27) The rest of the proof is similar to the proofof Theorem 2 and is omitted This completes the proof
Theorem 4 Assume that there exist three constants 119887 119872and 119873 with (1 minus 119887)119872 gt 119873 gt 0 and four nonnega-tive sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and0 le 119887119899le 119887 lt 1 eventually (47)
Then one has the following(a) For any 119871 isin (119887119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any
1199090
= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated bythe scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871 minus 119887119899119909119898119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus 119887119879119909119898119879minus120591
minus
infin
sum
119906=119879
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(48)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
Abstract and Applied Analysis 7
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119887119872 + 119873119872) It follows from (38) (39) and(47) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = 119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119887119872 minus119873
0 le 119887119899le 119887 lt 1 forall119899 ge 119879
(49)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus 119887119899119909119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(50)for each 119909 = 119909
119899119899isinN120573
isin 119860(119873119872) In view of (14) (15) and(49) and (50) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 11987910038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le 119887119899
100381610038161003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
(119899 minus 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 minus 120591)2
1198992
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le 1198871003817100381710038171003817119909 minus 119910
1003817100381710038171003817
+
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le [119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817= 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus 119887119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus 119887119872
minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus 119887119872 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minus 119887119872 minusmin 119872 minus 119871 119871 minus 119887119872 minus119873 ge 119873
(51)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 5 Assume that there exist constants 119887 119872 and 119873with (1 + 119887)119872 gt 119873 gt 0 and four nonnegative sequences119875119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14)(15) (38) (39) and
minus1 lt 119887 le 119887119899le 0 eventually (52)
Then one has the following
(a) For any 119871 isin (119873 (1 + 119887)119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for
any 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by(48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
8 Abstract and Applied Analysis
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119873 (1 + 119887)119872) It follows from (38) (39) and(52) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (53)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 119887)119872 minus 119871 119871 minus 119873
(54)
minus1 lt 119887 le 119887119899le 0 forall119899 ge 119879 (55)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by (50) By virtue of
(15) (50) (53) and (55) we infer that for all 119909 = 119909119899119899isinN120573
119910 = 119910
119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le [minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 minus 119887119872
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 minus 119887119872 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 minus 119887119872 +min (1 + 119887)119872 minus 119871 119871 minus 119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minusmin (1 + 119887)119872 minus 119871 119871 minus 119873 ge 119873
(56)
That is (27) holds The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 6 Assume that there exist constants 119887119872 and 119873 with (1 minus 1119887)119872 gt 119873 gt 0 and fournonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882
119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899ge 119887 gt 1 eventually (57)
Then one has the following(a) For any 119871 isin ((1119887)119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any 119909
0=
1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterative sequence119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(58)
Abstract and Applied Analysis 9
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin ((1119887)119872 + 119873119872) It follows from (38) (39)and (57) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 =
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (59)
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus
1
119887
119872 minus119873
(60)
119887119899ge 119887 gt 1 forall119899 ge 119879 (61)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[(119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus 119888119905)] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(62)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In view of (14) (15)and (59)∽(62) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
+
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
+
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus
1
119887
119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
119887
119872
minus
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
119887
119872 minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
gt 119871 minus
1
119887
119872 minusmin 119872 minus 119871 119871 minus
1
119887
119872 minus119873 ge 119873
(63)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 7 Assume that there exist constants 119887 119872and 119873 with (1 + 1119887)119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899le 119887 lt minus1 eventually (64)
10 Abstract and Applied Analysis
Then one has the following
(a) For any 119871 isin (minus(1 + 1119887)119872 minus119873) there exist 120579 isin (0 1)and 119879 ge 119899
0+ 120591 + 120573 such that for any 119909
0= 1199090119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 120572119898 minus 1198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+120572119898 minus 119879
2
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(65)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (minus(1 + 1119887)119872 minus119873) It follows from (38)(39) and (64) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 = minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (66)
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 1
119887
)119872 minus 119871 119871 minus
1
119887
119872 minus119873
(67)
119887119899ge 119887 gt 1 forall119899 ge 119879 (68)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
minus1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(69)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Making use of (15) (66)(68) and (69) we conclude that10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le minus
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
minus
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le minus
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
minus
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le minus119871 minus
119872
119887
minus
1
1198871198992
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
Abstract and Applied Analysis 11
le minus119871 minus
119872
119887
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt minus119871 minus
119872
119887
+min (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 le 119872
119878119871119909119899
1198992
ge minus119871
+
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge minus119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt minus119871 minusmin (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 ge 119873
(70)
which yield (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
3 Examples
In this section we suggest six examples to explain the resultspresented in Section 2
Example 1 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus 119909119899minus120591) + Δ(
sin2119909119899minus3
1198997
) +
1
(1198999+ 21198995+ 1) (1 + 119909
2
1198992)
=
1198992
minus 2119899
1198998+ 1198993+ 1
forall119899 ge 4
(71)
where 120591 isin N is fixed Let 1198990= 4 119896 = 1 and 120573 = min4minus120591 1
and let119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= minus1 119888
119899=
1198992
minus 2119899
1198998+ 1198993+ 1
119891 (119899 119906) =
1
(1198999+ 21198995+ 1) (1 + 119906
2)
ℎ (119899 119906) =
sin21199061198997
1198911119899= 1198992
119865119899= 1198994
ℎ1119899= 119899 minus 3 119867
119899= (119899 minus 3)
2
119875119899=
2119872
1198999
119876119899=
1
1198999
119877119899=
2
1198997
119882119899=
1
1198997
forall (119899 119906) isin N1198990
timesR
(72)
It is easy to see that (14) (15) and (18) are satisfied Note that
1
1198992
infin
sum
119905=119899+120591
1199052max 119877
119905119867119905119882119905
=
1
1198992
infin
sum
119905=119899+120591
1199052max2(119905 minus 3)
2
1199057
1
1199057
=
infin
sum
119905=119899+120591
2(119905 minus 3)2
+ 1
1199055
le
2
1198992
infin
sum
119905=119899+120591
1
1199053
997888rarr 0 as 119899 997888rarr infin
1
1198992
infin
sum
119905=119899+120591
1199053max 119875
119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119905=119899+120591
1199053max2119872
1199055
1
1199059
100381610038161003816100381610038161199052
minus 2119905
10038161003816100381610038161003816
1199058+ 1199053+ 1
le
max 1 2119872
1198992
infin
sum
119905=119899+120591
1
1199052
997888rarr 0 as 119899 997888rarr infin
(73)
which together with Lemma 1 yield that (16) and (17) holdIt follows from Theorem 2 that (71) possesses uncountablymany positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 119899
0+120591+120573 such
that for each 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (19)converges to a positive solution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of(71) with lim
119899rarrinfin119911119899= +infin and has the error estimate (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] satisfying(21)
Example 2 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ 119909119899minus120591) + Δ(
sin211990931198993+1
1198993(1198996+ 2) (1 + 119909
4
21198992minus3
)
)
+
(minus1)119899
1198993
(1199091198992minus119899minus1
+ 119909(119899+1)(119899+2)
)
(11989913+ 1198995+ 1) (1 + 119909
2
1198992minus119899minus1
+ 1199092
(119899+1)(119899+2))
=
1198992
minus ln 1198991198996+ 1198995+ 1
forall119899 ge 5
(74)
where 120591 isin N is fixed Let 1198990= 5 119896 = 2 and 120573 = 5 minus 120591 and let
119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= 1 119888
119899=
1198992
minus ln 1198991198996+ 1198995+ 1
119891 (119899 119906 V) =(minus1)119899
1198993
(119906 + V)(11989913+ 1198995+ 1) (1 + 119906
2+ V2)
12 Abstract and Applied Analysis
ℎ (119899 119906 V) =sin2V
1198993(1198996+ 2) (1 + 119906
4)
1198911119899= 1198992
minus 119899 minus 1
1198912119899= (119899 + 1) (119899 + 2)
119865119899= (119899 + 1)
2
(119899 + 2)2
ℎ1119899= 21198992
minus 3
ℎ2119899= 31198993
+ 1
119867119899= (3119899
3
+ 1)
2
119875119899= 119876119899=
4
11989910
119877119899= 119882119899=
10
1198999
forall (119899 119906 V) isin N1198990
timesR2
(75)
It is clear that (14) (15) and (40) are fulfilled Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
10(31199043
+ 1)
2
1199049
10
1199049
le
160
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
160
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(76)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (77)
Observe that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max4(119905 + 1)2
(119905 + 2)2
11990510
4
11990510
1199052
minus ln 1199051199056+ 1199055+ 1
le
196
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
=
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
119905 minus 119906 + 1
1199054
le
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
196
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(78)
which yields that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (79)
Thus Theorem 3 guarantees that (74) possesses uncount-ably positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 120591 +
1198990+ 120573 such that the Mann iterative sequence 119909
119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (74) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 3 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 + 3 ln 1198992 + 4 ln 119899
119909119899minus120591)
+ Δ(
(minus1)119899 sin (119890minus119899
2|11990951198992minus3|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|11990921198993+1|
)
+
(minus1)119899
1198996(1 + 119909
2
119899minus3)
minus
1
(1198997+ 21198994minus 1) (1 + 119909
2
119899+4)
=
3(minus1)119899
1198992
911989910ln3119899
forall119899 ge 7
(80)
where 120591 isin N is fixed Let 1198990= 7 119896 = 2 119887 = 34 and
120573 = min7 minus 120591 4 and let119872 and119873 be two positive constantswith 119872 gt 4119873 and
119887119899=
1 + 3 ln 1198992 + 4 ln 119899
119888119899=
3(minus1)119899
1198992
911989910ln3119899
119891 (119899 119906 V) =(minus1)119899
1198996(1 + 119906
2)
minus
1
(1198997+ 21198994minus 1) (1 + V2)
ℎ (119899 119906 V) =(minus1)119899 sin (119890minus119899
2|119906|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|V|
1198911119899= 119899 minus 3 119891
2119899= 119899 + 4
119865119899= (119899 + 4)
2
ℎ1119899= 51198992
minus 3
ℎ2119899= 21198993
+ 1 119867119899= (2119899
3
+ 1)
2
119875119899= 119876119899=
3
1198996
119877119899= 119882119899=
2
11989910
forall (119899 119906 V) isin N1198990
timesR2
(81)
It is not difficult to verify that (14) (15) and (47) are fulfilledNote that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
2(21199043
+ 1)
2
11990410
2
11990410
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
18
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(82)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (83)
Abstract and Applied Analysis 13
Observe that1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max3(119905 + 4)2
1199056
3
1199056
100381610038161003816100381610038161003816100381610038161003816
3(minus1)119905
1199052
911990510ln3119905
100381610038161003816100381610038161003816100381610038161003816
le
12
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
le
12
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
12
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(84)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (85)
That is (38) and (39) hold Consequently Theorem 4 impliesthat (80) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((34)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (80) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 4 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 minus 51198993
2 + 61198993
119909119899minus120591) + Δ(
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119909
2
3119899minus7)
)
+
sin (119899211990931198992minus2)
(radic119899 + 14)22
=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
forall119899 ge 9
(86)
where 120591 isin N is fixed Let 1198990= 9 119896 = 1 119887 = minus56 and
120573 = 9 minus 120591 and let 119872 and 119873 be two positive constants with119872 gt 6119873 and
119887119899=
1 minus 51198993
2 + 61198993
119888119899=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
119891 (119899 119906) =
sin (1198992119906)
(radic119899 + 14)22
ℎ (119899 119906) =
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119906
2)
1198911119899= 31198992
minus 2 119865119899= (3119899
2
minus 2)
2
ℎ1119899= 3119899 minus 7
119867119899= (3119899 minus 7)
2
119875119899= 119876119899=
3
1198999
119877119899= 119882119899=
1
1198995
forall (119899 119906) isin N1198990
timesR
(87)
Obviously (14) (15) and (52) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(3119904 minus 7)2
1199045
1
1199045
le
9
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
9
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(88)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (89)
Notice that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max
3(31199052
minus 2)
2
1199059
3
1199059
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905
1199053
+ 51199052
+ 4119905 minus 2
1199059+ 1199058+ 21199055+ 1199053+ 7
100381610038161003816100381610038161003816100381610038161003816
le
27
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
27
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
27
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(90)
which gives that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (91)
That is (38) and (39) hold Thus Theorem 5 shows that(86) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (119873 (16)119872)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that the Mann
iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generatedby (48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
119860(119873119872) of (86) with lim119899rarrinfin
119911119899= +infin and has the error
estimate (20) where 120572119898119898isinN0
is an arbitrary sequencein [0 1] satisfying (21)
14 Abstract and Applied Analysis
Example 5 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ (
120587
2
+ 119899 sin 1119899
) 119909119899minus120591)
+ Δ(
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (11989921199092119899+1
))
)
+
119899 sin (119899119909119899minus2)
2 + (119899 + 5)16
=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
forall119899 ge 3
(92)
where 120591 isin N is fixed Let 1198990= 3 119896 = 1 119887 = 1205872 and
120573 = min3 minus 120591 1 and let119872 and119873 be two positive constantswith (1 minus 2120587)119872 gt 119873 and
119887119899=
120587
2
+ 119899 sin 1119899
119888119899=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
119891 (119899 119906) =
119899 sin (119899119906)2 + (119899 + 5)
16
ℎ (119899 119906) =
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (1198992119906))
1198911119899= 119899 minus 2 119865
119899= (119899 minus 2)
2
ℎ1119899= 2119899 + 1 119867
119899= (2119899 + 1)
2
119875119899= 119876119899=
1
11989914
119877119899= 119882119899=
2
1198999
forall (119899 119906) isin N1198990
timesR
(93)
Clearly (14) (15) and (61) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max2(2119904 + 1)2
1199049
2
1199049
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199047
le
18
1198992
infin
sum
119904=119899
1
1199046
997888rarr 0 as 119899 997888rarr infin
(94)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 2)2
11990514
1
11990514
10038161003816100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1cos3 (1199052 + 1)11990516+ ln 119905
10038161003816100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
11990514
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
11990513
le
1
1198992
infin
sum
119905=119899
1
11990512
997888rarr 0 as 119899 997888rarr infin
(95)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (96)
That is (38) and (39) hold Consequently Theorem 6 impliesthat (92) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((2120587)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (58) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (92) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 6 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119909119899minus120591) + Δ(
cos ((minus1)119899119890119899)
(119899 + 7)6
radic1 +1003816100381610038161003816119909119899minus2
1003816100381610038161003816
)
+
sin (1198992119909119899minus1)
1198999+ 31198995+ 21198992+ 1
=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
forall119899 ge 6
(97)
where 120591 isin N is fixed Let 1198990= 6 119896 = 1 119887 = minus2 and 120573 =
min6 minus 120591 3 and let 119872 and 119873 be two positive constantswith (12)119872 gt 119873 and
119887119899= minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119888119899=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
119891 (119899 119906) =
sin (1198992119906)1198999+ 31198995+ 21198992+ 1
Abstract and Applied Analysis 15
ℎ (119899 119906) =
cos ((minus1)119899119890119899)(119899 + 7)
6
radic1 + |119906|
1198911119899= 119899 minus 1 119865
119899= (119899 minus 1)
2
ℎ1119899= 119899 minus 2 119867
119899= (119899 minus 2)
2
119875119899= 119876119899=
1
1198997
119877119899= 119882119899=
1
1198996
forall (119899 119906) isin N1198990
timesR
(98)
Obviously (14) (15) and (64) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(119904 minus 2)2
1199046
1
1199046
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
1
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(99)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (100)
It is clear that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 1)2
1199057
1
1199057
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1
1199054
+ 41199052
+ 119905 minus 1
11990511+ 61199053+ 7119905 + 2
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
1
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(101)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (102)
That is (38) and (39) hold Consequently Theorem 7 impliesthat (97) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (minus1198722 minus119873)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that
the Mann iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (65) converges to a positive solution 119911 =
119911119899119899isinN120573
isin 119860(119873119872) of (97) with lim119899rarrinfin
119911119899= +infin and
has the error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in[0 1] satisfying (21)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This research was supported by the Science Research Foun-dation of Educational Department of Liaoning Province(L2012380)
References
[1] M H Abu-Risha ldquoOscillation of second-order linear differenceequationsrdquo Applied Mathematics Letters vol 13 no 1 pp 129ndash135 2000
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[3] R P Agarwal and J Henderson ldquoPositive solutions and nonlin-ear eigenvalue problems for third-order difference equationsrdquoComputers amp Mathematics with Applications vol 36 no 10-12pp 347ndash355 1998
[4] A Andruch-Sobiło and M Migda ldquoOn the oscillation ofsolutions of third order linear difference equations of neutraltyperdquoMathematica Bohemica vol 130 no 1 pp 19ndash33 2005
[5] Z Dosla and A Kobza ldquoGlobal asymptotic properties of third-order difference equationsrdquo Computers amp Mathematics withApplications vol 48 no 1-2 pp 191ndash200 2004
[6] S R Grace and G G Hamedani ldquoOn the oscillation of certainneutral difference equationsrdquo Mathematica Bohemica vol 125no 3 pp 307ndash321 2000
[7] J Cheng ldquoExistence of a nonoscillatory solution of a second-order linear neutral difference equationrdquo Applied MathematicsLetters vol 20 no 8 pp 892ndash899 2007
[8] LKongQKong andB Zhang ldquoPositive solutions of boundaryvalue problems for third-order functional difference equationsrdquoComputersampMathematics withApplications vol 44 no 3-4 pp481ndash489 2002
[9] I Y Karaca ldquoDiscrete third-order three-point boundary valueproblemrdquo Journal of Computational and Applied Mathematicsvol 205 no 1 pp 458ndash468 2007
[10] W-T Li and J P Sun ldquoExistence of positive solutions of BVPsfor third-order discrete nonlinear difference systemsrdquo AppliedMathematics and Computation vol 157 no 1 pp 53ndash64 2004
[11] W-T Li and J-P Sun ldquoMultiple positive solutions of BVPs forthird-order discrete difference systemsrdquo Applied Mathematicsand Computation vol 149 no 2 pp 389ndash398 2004
[12] Z Liu M Jia S M Kang and Y C Kwun ldquoBounded positivesolutions for a third order discrete equationrdquo Abstract andApplied Analysis vol 2012 Article ID 237036 12 pages 2012
[13] Z Liu S M Kang and J S Ume ldquoExistence of uncountablymany bounded nonoscillatory solutions and their iterativeapproximations for second order nonlinear neutral delay dif-ference equationsrdquo Applied Mathematics and Computation vol213 no 2 pp 554ndash576 2009
[14] Z Liu Y Xu and S M Kang ldquoBounded oscillation criteriafor certain third order nonlinear difference equations withseveral delays and advancesrdquo Computers amp Mathematics withApplications vol 61 no 4 pp 1145ndash1161 2011
[15] M Migda and J Migda ldquoAsymptotic properties of solutions ofsecond-order neutral difference equationsrdquoNonlinear Analysis
16 Abstract and Applied Analysis
Theory Methods and Applications vol 63 no 5ndash7 pp e789ndashe799 2005
[16] N Parhi ldquoNon-oscillation of solutions of difference equationsof third orderrdquoComputersampMathematics withApplications vol62 no 10 pp 3812ndash3820 2011
[17] N Parhi and A Panda ldquoNonoscillation and oscillation ofsolutions of a class of third order difference equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 1 pp213ndash223 2007
[18] S H Saker ldquoNew oscillation criteria for second-order nonlinearneutral delay difference equationsrdquo Applied Mathematics andComputation vol 142 no 1 pp 99ndash111 2003
[19] S H Saker ldquoOscillation of third-order difference equationsrdquoPortugaliae Mathematica vol 61 no 3 pp 249ndash257 2004
[20] S Stevic ldquoOn a third-order system of difference equationsrdquoApplied Mathematics and Computation vol 218 no 14 pp7649ndash7654 2012
[21] X H Tang ldquoBounded oscillation of second-order delay dif-ference equations of unstable typerdquo Computers amp Mathematicswith Applications vol 44 no 8-9 pp 1147ndash1156 2002
[22] J Yan and B Liu ldquoAsymptotic behavior of a nonlinear delaydifference equationrdquo Applied Mathematics Letters vol 8 no 6pp 1ndash5 1995
[23] Z G Zhang and Q L Li ldquoOscillation theorems for second-order advanced functional difference equationsrdquo Computers ampMathematics with Applications vol 36 no 6 pp 11ndash18 1998
Research ArticleAlgebroid Solutions of Second Order ComplexDifferential Equations
Lingyun Gao1 and Yue Wang2
1 Department of Mathematics Jinan University Guangzhou Guangdong 510632 China2 School of Information Renmin University of China Beijing 100872 China
Correspondence should be addressed to Lingyun Gao tgaolyjnueducn
Received 28 November 2013 Accepted 17 December 2013 Published 2 January 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 L Gao and Y WangThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Using value distribution theory and maximummodulus principle the problem of the algebroid solutions of second order algebraicdifferential equation is investigated Examples show that our results are sharp
1 Introduction and Main Results
We use the standard notations and results of the Nevanlinnatheory of meromorphic or algebroid functions see forexample [1 2]
In this paper we suppose that second order algebraicdifferential equation (3) admit at least one nonconstant ]-valued algebroid solution 119908(119911) in the complex plane Wedenote by 119864 a subset of [0infin) for which 119898(119864) lt infin and by119870 a positive constant where119898(119864) denotes the linearmeasureof 119864 119864 or 119870 does not always mean the same one when theyappear in the following
Let 119886119895119896(119895 = 0 1 119899 119896 = 0 1 119902
119895) be entire func-
tions without common zeroes such that 11988601199020
= 0 We put
119876119895(119911 119908) =
119902119895
sum
119896=0
119886119895119896119908119896
119902119895= deg119876119895
119908
119901 = max 119902119895+ 119895 119895 = 0 1 119899 minus 1
(1)
Some authors had investigated the problem of the exis-tence of algebroid solutions of complex differential equationsand they obtained many results ([2ndash10] etc)
In 1989 Toda [4] considered the existence of algebroidsolutions of algebraic differential equation of the form
119899
sum
119895=0
119876119895(119911 119908) (119908
1015840
)
119895
= 0 (2)
He obtained the following
Theorem A (see [4]) Let 119908(119911) be a nonconstant ]-valuedalgebroid solution of the above differential equation and all 119886
119895119896
are polynomials If 119901 lt 119899 + 119902119899 then 119908(119911) is algebraic
The purpose of this paper is to investigate algebroid solu-tions of the following second order differential equation inthe complex plane with the aid of the Nevanlinna theory andmaximum modulus principle of meromorphic or algebroidfunctions
119899
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
= 0 (3)
where 119876119895(119911 119908) = sum
119902119895
119896=0119886119895119896119908119896 119895 = 0 1 2 119899
We will prove the following two results
Theorem 1 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and all 119886
119895119896are polynomials
If 119901 le 119902119899 then 119908(119911) is algebraic 119901 = max119902
119895 119895 = 0 1 119899 minus
1
Theorem 2 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and the orders of all 119886
119895119896are
finite If 1199020gt max
1le119895le119899minus1119902119895+119895 then the following statements
are equivalent(a) 120575(infin119908) gt 0(b) 1199020= 119902119899+ 119899
(c) infin is a Picard exceptional value of 119908(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 123049 4 pageshttpdxdoiorg1011552014123049
2 Abstract and Applied Analysis
2 Some Lemmas
Lemma 3 (see [2]) Suppose that 119908(119911) 119886119894(119911) (119894 = 1 2 119901)
are meromorphic functions and 119886119901(119911) = 0 Then one has
119898(119903
119901
sum
119894=1
119886119894(119911) 119908119894
) le 119901119898 (119903 119908) +
119901
sum
119894=1
119898(119903 119886119894(119911)) + 119874 (1)
(4)
Examining proof of Lemma 45 presented in [2 pp 192-193] we can verify Lemma 4
Lemma 4 Let 119908(119911) be a transcendental algebroid functionsuch that 119908(119911) has only finite number of poles and let 119908(119911)1199081015840
(119911) and 11990810158401015840(119911) have no poles in |119911| gt 1199030 Then for some
constants 119862119894gt 0 119894 = 1 2 3 and 119903 ge 119903
1ge 1199030it holds
119872(119903 119908) le 1198621+ 1198622119903 + 11986231199032
119872(119903 11990810158401015840
) (5)
where119872(119903 119908) = max|119911|=119903
|119908(119911)|
Lemma 5 (see [11]) The absolute values of roots of equation
119911119899
+ 1198861119911119899minus1
+ sdot sdot sdot + 119886119899= 0 (6)
are bounded by
max 119899 10038161003816100381610038161198861
1003816100381610038161003816 (119899
10038161003816100381610038161198862
1003816100381610038161003816)12
(1198991003816100381610038161003816119886119899
1003816100381610038161003816)1119899
(7)
Lemma 6 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of the differential equation (3) and let 119886
119895119896be a
polynomial If 119901 lt 119899 + 119902119899 then
min 119899 119902119899minus 119901 log+119872(119903 119908)
+max 0 119902119899minus 119901 minus 119899 log+119872(119903 119908)
le +119874 (log 119903) + 119874 (1) (119903 notin 119864)
(8)
where119872(119903 119908) = max|119911|=119903
|119908(119911)| 119870 is a positive constant
Proof We first prove that the poles of 119908 are contained in thezeroes of 119886
119899119902119899
(119911)Suppose that 119911
0is a pole of 119908 of order 120591 and 119911
0is not the
zeroes of 119886119899119902119899
(119911) Then
119908 (119911) = (119911 minus 1199110)minus120591120582
1199081(119911) 119908
1(1199110) = 0infin
11990810158401015840
(119911) = (119911 minus 1199110)minus(120591+2120582)120582
1199082(119911) 119908
2(1199110) = 0infin
(9)
We rewrite differential equation (3) as follows
119886119899119902119899
(11990810158401015840
)
119899
=
119899minus1
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
(10)
It follows from (10) that
119902119899120591 + 119899 (120591 + 2120582) le 119901120591 + (119899 minus 1) (120591 + 2120582) (11)
Noting that 119901 le 119902119899 we have
120591 + 2120582 lt 0 (12)This is a contradiction
This shows that the poles of119908 are contained in the zeroesof 119886119899119902119899
(119911)We rewrite differential equation (3) as follows
119899
sum
119895=0
119876119895(119911 119908)119876
119899(119911 119908)
119899minus119895minus1
(11987611989911990810158401015840
)
119895
= 0 (13)
119872(119903119876111990810158401015840
) ge 119872(119903 119908)119902119899+119872(119903 119908
10158401015840
)
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
ge 119872(119903 119908)119902119899+
119872(119903 119908) minus 1198621minus 1198622119903
11986231199032
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
(14)
For 119895 = 0 1 119899 minus 1 we have10038161003816100381610038161003816119876119895(119911119903 119908)119876
119899(119911119903 119908)119899minus119895minus110038161003816
100381610038161003816le 119870119872(119903 119908)
119902119895+119902119899(119899minus119895minus1)
(15)
Applying Lemma 5 to (13) at 119911 = 119911119903
119872(119903119876111990810158401015840
) le 119870119872(119903 119908)maxℎ
119895 119895=01119899minus1
(16)
where ℎ119895= (119902119895+ 119902119899(119899 minus 119895 minus 1))(119899 minus 119895)
From (14) and (15) we have
119872(119903 119908)119902119899le 119870 119872(119903 119908)
119902119899minus1
+ 1199032
119872(119903 119908)maxℎ
119895 119895=01119899minus1
le 119870 119872(119903 119908)119902119899minus1
+ 1199032
119872(119903 119908)119902119899+((119901minus119902
119899)119899)
(17)Note that
ℎ119895=
119902119895+ 119902119899(119899 minus 119895 minus 1)
119899 minus 119895
lt
119901 + 119902119899(119899 minus 119895 minus 1) minus 119895
119899 minus 119895
= 119902119899+
119901 minus 119902119899
119899
+
119895 (119901 minus 119902119899minus 119899)
119899 (119899 minus 119895)
le 119902119899+
119901 minus 119902119899
119899
119895 = 0 1 119899 minus 1
(18)
Dividing the inequality (17) by119872(119903 119908)
max119902119899minus1119902119899+((119902119899minus119901)119899) we obtain for 119903 notin 119864
119872(119903 119908)min1(119902
119899minus119901)119899
le 11987041 +
1199032
119872(119903 119908)max0(119902
119899minus119901minus119899)119899
(19)
which reduces to our inequality by calculating log+ of theboth sidesmin 119899 119902
119899minus 119901 log+119872(119903 119908)
le minusmax 0 119902119899minus 119901 minus 119899 log+119872(119903 119908) + 119874 (log 119903) + 119874 (1)
(20)Lemma 6 is complete
Abstract and Applied Analysis 3
3 Proof of Theorem 1
First we consider119873(119903 119908)Let 1199110be a pole of 119908 of 120591 Let 119905 be the order of zero of
119886119899119902119899
(119911) at 1199110
(i) When the order of the pole of119876119899(119908)(119908
10158401015840
)119899 is not equal
to that of other terms of the left-hand side of (10) at 1199110 we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 (21)
that is
120591 le
(119905 minus 2119899) 120582
119902119899+ 119899 minus 119901
(22)
(ii)When the order of pole of119876119899(119908)(119908
10158401015840
)119899 is equal to that
of some term 119876119896(119908)(119908
10158401015840
)119899 of the left-hand side of (10) at 119911
0
we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 + 119899 (120591 + 2120582) (23)
that is
120591 le
119905120582
119902119899minus 119901
(24)
Combining cases (i) and (ii) we obtain
119873(119903 119908) le 1198708119873(119903
1
119886119899119902119899
) (25)
where1198708is a positive constant
Secondly by Lemma 6 we obtain
min 119899 119902119899minus 119901119898 (119903 119908)
le 119870[
119901
sum
119894=0
119898(119903 119886119894) +
119902119899minus1
sum
119896=0
119898(119903 119887119896)] + 119874 (log 119903) (119903 notin 119864)
(26)
Combining the inequalities (25) and (26) we have
119879 (119903 119908) = 119874 (log 119903) (119903 notin 119864) (27)
which shows that 119908 is an algebraic solution of (3)This completes the proof of Theorem 1
4 Proof of Theorem 2
(i) (a)rArr (b) Suppose that 120575(infin119908) gt 0 If 1199020gt 119902119899+ 119899 then
we have by (3)
1199081199020= minus
1
11988601199020
119899
sum
119895=1
119876119895(119908)119908
119895
(
11990810158401015840
119908
)
119895
minus
1199020minus1
sum
119896=0
1198860119896119908119896
(28)
Applying Lemma 3 to (28)
1199020119898(119903 119908) le (119902
0minus 1)119898 (119903 119908) +sum
119895119896
119898(119903 119886119895119896)
+ 119870119898(119903
11990810158401015840
119908
) + 119898(119903
1
11988601199020
) + 119874 (1)
(29)
Since 119908(119911) is admissible solution we have
119898(119903 119908) = 119878 (119903 119908) (30)
so that
120575 (119908infin) = 0 (31)
This is a contradiction Thus 1199020le 119902119899+ 119899
If 1199020lt 119902119899+119899 byTheorem 1119908(119911) is nonadmissibleThus
1199020= 119902119899+ 119899 (32)
(ii) (b) rArr (c) Let 1199020= 119902119899+ 119899 Then similar to the proof of
Lemma 6 we obtain that the poles of 119908(119911) are contained inthe set of 119886
119899119902119899
andinfin is a Picard exceptional value of 119908(119911)
(iii) (c) rArr (a) Let infin be a Picard exceptional value of 119908(119911)Then 120575(119908infin) = 1
5 Some Examples
Example 1 The differential equation
]21199082]minus111990810158401015840 minus [((2] + 1)1199084] + 21199082] minus ] + 1)] = 0 (33)
has a transcendental algebroid solution 119908(119911) = (tan 119911)1] Inthis case
119901 = 1199020= 4] gt 1 + 119902
1= 2] minus 1 + 1 = 2] (34)
Remark 7 Example 1 shows that the condition in Theorem 1is sharp
Example 2 Transcendental algebroid function 119908(119911) =
(sin 119911)12 is a 2-valued solution of the following differentialequation
161199086
(11990810158401015840
)
2
+ 21199085
11990810158401015840
minus (1199088
minus
1
2
1199086
+ 21199084
minus
1
2
1199082
+ 1) = 0
(35)
In this case
1199020= 8 119899 = 2 119902
1= 5 119902
2= 6 (36)
By Theorem 2 for transcendental algebroid function 119908(119911) =(sin 119911)12infin is a Picard exceptional value
Remark 8 Example 2 shows that the result in Theorem 2holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project is project Supported by National Natural ScienceFoundation (10471065) of China and NSF of GuangdongProvince (04010474)
4 Abstract and Applied Analysis
References
[1] H Yi and C C YangTheory of the Uniqueness of MeromorphicFunctions Science Press Beijing China 1995 (Chinese)
[2] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Science Press Beijing China 1988
[3] Y Z He and X Z Xiao ldquoAdmissible solutions and ordinarydifferential equationsrdquo Contemporary Mathematics vol 25 pp51ndash61 1983
[4] N Toda ldquoOn algebroid solutions of some algebraic differentialequations in the complex planerdquo Japan Academy A vol 65 no4 pp 94ndash97 1989
[5] T Chen ldquoOne class of ordinary differential equations whichpossess algebroid solutions in the complex domainrdquo ChineseQuarterly Journal of Mathematics vol 6 no 4 pp 45ndash51 1991
[6] K Katajamaki ldquoValue distribution of certain differential poly-nomials of algebroid functionsrdquo Archiv der Mathematik vol 67no 5 pp 422ndash429 1996
[7] L Gao ldquoSome results on admissible algebroid solutions ofcomplex differential equationsrdquo Indian Journal of Pure andApplied Mathematics vol 32 no 7 pp 1041ndash1050 2001
[8] L-Y Gao ldquoOn some generalized higher-order algebraic dif-ferential equations with admissible algebroid solutionsrdquo IndianJournal of Mathematics vol 43 no 2 pp 163ndash175 2001
[9] L Gao ldquoOn the growth of solutions of higher-order algebraicdifferential equationsrdquoActaMathematica Scientia B vol 22 no4 pp 459ndash465 2002
[10] L Y Gao ldquoThe growth of single-valued meromorphic solutionsand finite branch solutionsrdquo Journal of Systems Science andMathematical Sciences vol 24 no 3 pp 303ndash310 2004
[11] T Takagi Lecture on Algebra Kyoritsu Tokyo Japan 1957(Japanese)
Complex Differences and Difference Equations
Abstract and Applied Analysis
Complex Differences and Difference Equations
Guest Editors Zong-Xuan Chen Kwang Ho Shonand Zhi-Bo Huang
Copyright copy 2014 Hindawi Publishing Corporation All rights reserved
This is a special issue published in ldquoAbstract and Applied Analysisrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited
Editorial Board
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Contents
Complex Differences and Difference Equations Zong-Xuan Chen Kwang Ho Shon and Zhi-Bo HuangVolume 2014 Article ID 124843 1 page
Blow-Up Analysis for a Quasilinear Parabolic Equation with Inner Absorption and NonlinearNeumann Boundary Condition Zhong Bo Fang and Yan ChaiVolume 2014 Article ID 289245 8 pages
Some Properties on Complex Functional Difference Equations Zhi-Bo Huang and Ran-Ran ZhangVolume 2014 Article ID 283895 10 pages
The Regularity of Functions on Dual Split Quaternions in Clifford AnalysisJi Eun Kim and Kwang Ho ShonVolume 2014 Article ID 369430 8 pages
Unicity of Meromorphic Functions Sharing Sets withTheir Linear Difference PolynomialsSheng Li and BaoQin ChenVolume 2014 Article ID 894968 7 pages
A ComparisonTheorem for Oscillation of the Even-Order Nonlinear Neutral Difference EquationQuanxin ZhangVolume 2014 Article ID 492492 5 pages
Difference Equations and Sharing Values Concerning Entire Functions andTheir DifferenceZhiqiang Mao and Huifang LiuVolume 2014 Article ID 584969 6 pages
Admissible Solutions of the Schwarzian Type Difference Equation Baoqin Chen and Sheng LiVolume 2014 Article ID 306360 5 pages
Statistical Inference for Stochastic Differential Equations with Small NoisesLiang Shen and Qingsong XuVolume 2014 Article ID 473681 6 pages
On the Deficiencies of Some Differential-Difference Polynomials Xiu-Min Zheng and Hong Yan XuVolume 2014 Article ID 378151 12 pages
On Growth of Meromorphic Solutions of Complex Functional Difference Equations Jing LiJianjun Zhang and Liangwen LiaoVolume 2014 Article ID 828746 6 pages
Unicity of Entire Functions concerning Shifts and Difference Operators Dan Liu Degui Yangand Mingliang FangVolume 2014 Article ID 380910 5 pages
On Positive Solutions and Mann Iterative Schemes of aThird Order Difference Equation Zeqing LiuHeng Wu Shin Min Kang and Young Chel KwunVolume 2014 Article ID 470181 16 pages
Algebroid Solutions of Second Order Complex Differential Equations Lingyun Gao and Yue WangVolume 2014 Article ID 123049 4 pages
EditorialComplex Differences and Difference Equations
Zong-Xuan Chen1 Kwang Ho Shon2 and Zhi-Bo Huang1
1School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Zong-Xuan Chen chzxvipsinacom
Received 12 August 2014 Accepted 12 August 2014 Published 22 December 2014
Copyright copy 2014 Zong-Xuan Chen et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
In more recent years activity in the area of the complexdifferences and the complex difference equations has fleetlyincreased
This journal has set up a column of this special issue Wewere pleased to invite the interested authors to contributetheir original research papers as well as good expositorypapers to this special issue that will make better improvementon the theory of complex differences and difference equa-tions
In this special issue many good results are obtainedDifference equations are widely applied to mathematical
physics economics and chemistry In this special issue Z-B Huang and R-R Zhang J Li et al and L Gao and YWang investigate the growth a Borel exceptional value ofmeromorphic solutions to different types of higher ordernonliear difference equations respectively B Chen and S Liinvestigate the Schwarzian type difference equation D Liuet al Z Mao and H Liu and S Li and B Chen investigateunicity of meromorphic functions concerning different typesof difference operators Recently many difference analoguesof the classic Nevanlinna theory are obtained
In this special issue X-M Zheng and H Y Xu obtaina differential difference analogue of Valiron-Mohonko theo-rem Related topics with complex difference J E Kim andK H Shon investigate the regularity of functions on dualsplit quaternions in Clifford analysis and the tensor productrepresentation of polynomials of weak type in a DF-spaceQ Zhang and Z Liu et al investigate different types ofreal difference equations respectively L Shen and Q Xuinvestigate stochastic differential equations
This special issue stimulates the continuing efforts to thecomplex differences and the complex difference equations
Zong-XuanChenKwangHo ShonZhi-BoHuang
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 124843 1 pagehttpdxdoiorg1011552014124843
Research ArticleBlow-Up Analysis for a Quasilinear Parabolic Equation withInner Absorption and Nonlinear Neumann Boundary Condition
Zhong Bo Fang and Yan Chai
School of Mathematical Sciences Ocean University of China Qingdao 266100 China
Correspondence should be addressed to Zhong Bo Fang fangzb7777hotmailcom
Received 24 February 2014 Revised 11 April 2014 Accepted 11 April 2014 Published 30 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Z B Fang and Y Chai This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We investigate an initial-boundary value problem for a quasilinear parabolic equation with inner absorption and nonlinearNeumann boundary condition We establish respectively the conditions on nonlinearity to guarantee that 119906(119909 119905) exists globally orblows up at some finite time 119905lowast Moreover an upper bound for 119905lowast is derived Under somewhat more restrictive conditions a lowerbound for 119905lowast is also obtained
1 Introduction
We are concerned with the global existence and blow-upphenomenon for a quasilinear parabolic equation with non-linear inner absorption term
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
minus 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (1)
subjected to the nonlinear Neumann boundary and initialconditions
(|nabla119906|119901
+ 1)
120597119906
120597]= 119892 (119906) (119909 119905) isin 120597Ω times (0 119905
lowast
) (2)
119906 (119909 0) = 1199060(0) ge 0 119909 isin Ω (3)
whereΩ is a bounded star-shaped region of 119877119873 (119873 ge 2) withsmooth boundary 120597Ω ] is the unit outward normal vectoron 120597Ω 119901 ge 0 119905lowast is the blow-up time if blow-up occurs orelse 119905lowast
= +infin the symbol 119894 denotes partial differentiationwith respect to 119909
119894 119894 = 1 2 119873 the repeated index indicates
summation over the index and nabla is gradient operatorMany physical phenomena and biological species theo-
ries such as the concentration of diffusion of some non-Newton fluid through porous medium the density of somebiological species and heat conduction phenomena havebeen formulated as parabolic equation (1) (see [1ndash3]) Thenonlinear Neumann boundary condition (2) can be physi-cally interpreted as the nonlinear radial law (see [4 5])
In the past decades there have been many works dealingwith existence and nonexistence of global solutions blow-upof solutions bounds of blow-up time blow-up rates blow-up sets and asymptotic behavior of solutions to nonlinearparabolic equations see the books [6ndash8] and the surveypapers [9ndash11] Specially we would like to know whether thesolution blows up and at which time when blow-up occursA variety of methods have been used to study the problemabove (see [12]) and in many cases these methods used toshow that solutions blow up often provide an upper boundfor the blow-up time However lower bounds for blow-uptime may be harder to be determined For the study ofthe initial boundary value problem of a parabolic equationwith homogeneous Dirichlet boundary condition see [1314] Payne et al [13] considered the following quasilinearparabolic equation
119906119905= div (120588|nabla119906|2nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (4)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω To get the lower bound for the blow-up time the authorsassumed that 120588 is a positive 1198621 function which satisfies
120588 (119904) + 1199041205881015840
(119904) gt 0 119904 gt 0 (5)
The lower bound for the blow-up time of solution to (4) withRobin boundary condition was obtained in [15] where 120588 is
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 289245 8 pageshttpdxdoiorg1011552014289245
2 Abstract and Applied Analysis
also assumed to satisfy the condition (5) However under thisboundary condition the best constant of Sobolev inequalityused in [13] is no longer applicable They imposed suitableconditions on 119891 and 120588 and determined a lower bound forthe blow-up time if blow-up occurs and determined whenblow-up cannot occur Marras and Vernier Piro [14] studiedthe nonlinear parabolic problem with time dependent coeffi-cients
1198961(119905) div (119892 (|nabla119906|2nabla119906)) + 119896
2(119905) 119891 (119906) = 119896
3(119905) 119906119905
(119909 119905) isin Ω times (0 119905lowast
)
(6)
where Ω is a bounded domain in 119877119873 with smooth boundary
120597Ω Under some conditions on the data and geometry of thespatial domain they obtained upper and lower bounds of theblow-up time Moreover the sufficient conditions for globalexistence of the solution were derived
For the study of the initial boundary value problem ofa parabolic equation with Robin boundary condition werefer to [15ndash19] Li et al [16] investigated the problem of thenonlinear parabolic equation
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
+ 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (7)
where Ω is a bounded domain in 1198773 with smooth boundary
120597ΩThey derived the lower bound for the blow-up timewhenthe blow-up occurs Clearly |nabla119906|119901 + 1 does not satisfy thecondition (5) Enache [17] discussed the quasilinear parabolicproblem
119906119905= (119892 (119906) 119906
119894)119894
+ 119891 (119906) (8)
where Ω is a bounded domain in 119877119873
(119873 ge 2) with smoothboundary 120597Ω By virtue of a first-order differential inequalitytechnique they showed the sufficient conditions to guaranteethat the solution 119906(119909 119905) exists globally or blows up Inaddition a lower bound for the blow-up time when blow-up occurs was also obtained Ding [18] studied the nonlinearparabolic problem
(119887 (119906))119905= nabla sdot (119892 (119906) nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (9)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω They derived conditions on the data which guaranteethe blow-up or the global existence of the solution A lowerbound on blow-up time when blow-up occurs was alsoobtained For the problem of the nonlinear nonlocal porousmedium equation we read the paper of Liu [19]
Recently for the problems with nonlinear Neumannboundary conditions Payne et al [20] studied the semilinearheat equation with inner absorption term
119906119905= Δ119906 minus 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (10)
They established conditions on nonlinearity to guarantee thatthe solution119906(119909 119905) exists for all time 119905 gt 0 or blows up at somefinite time 119905lowast Moreover an upper bound for 119905lowast was derivedUnder somewhat more restrictive conditions a lower bound
for 119905lowast was derivedThereafter they considered the quasilinearparabolic equation
119906119905= nabla sdot (|nabla119906|
119901
nabla119906) (119909 119905) isin Ω times (0 119905lowast
) (11)
and they showed that blow-up occurs at some finite timeunder certain conditions on the nonlinearities and the dataupper and lower bounds for the blow-up time were derivedwhen blow-up occurs see [21] Liu et al The authors [22 23]studied the reaction diffusion problem with nonlocal sourceand inner absorption terms or with local source and gradientabsorption terms Very recently Fang et al [24] consideredlower bounds estimate for the blow-up time to nonlocalproblemwith homogeneousDirichlet orNeumann boundarycondition
Motivated by the above work we intend to study theglobal existence and the blow-up phenomena of problem (1)ndash(3) and the results of the semilinear equations are extended tothe quasilinear equations Unfortunately the techniques usedfor semilinear equation to analysis of blow-up phenomenaare no longer applicable to our problem As a consequenceby using the suitable techniques of differential inequalitieswe establish respectively the conditions on the nonlinearities119891 and 119892 to guarantee that 119906(119909 119905) exists globally or blows upat some finite time If blow-up occurs we derive upper andlower bounds of the blow-up time
The rest of our paper is organized as follows In Section 2we establish conditions on the nonlinearities to guaranteethat 119906(119909 119905) exists globally In Section 3 we show the condi-tions on data forcing the solution 119906(119909 119905) to blow up at somefinite time 119905
lowast and obtain an upper bound for 119905lowast A lower
bound of blow-up time under some assumptions is derivedin Section 4
2 The Global Existence
In this sectionwe establish the conditions on the nonlinearity119891 and nonlinearity 119892 to guarantee that 119906(119909 119905) exists globallyWe state our result as follows
Theorem 1 Assume that the nonnegative functions 119891 and 119892
satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(12)
where 1198961gt 0 119896
2ge 0 119904 gt 1 2119904 lt 119902 + 1 and 119904 minus 1 lt 119901 lt 119902 minus 1
Then the (nonnegative) solution 119906(119909 119905) of problem (1)-(3) doesnot blow up that is 119906(119909 119905) exists for all time 119905 gt 0
Proof Set
Ψ (119905) = int
Ω
1199062
119889119909 (13)
Abstract and Applied Analysis 3
Similar to Theorem 21 in [20] we get
Ψ1015840
(119905) le 21205752
int
Ω
1199062119904
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
+
21198962119873
1205880
int
Ω
119906119904+1
119889119909
minus2int
Ω
|nabla119906|119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 1198681+ 1198682
(14)
where 120575 = 1198962(119904 + 1)1198892120588
0 1205880
= min119909isin120597Ω
(119909 sdot ]) 119889 =
max119909isin120597Ω
|119909| and
1198681le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
times 1198601|Ω|(119902minus119904)(119902+1)
minus 1198602(int
Ω
119906119902+1
119889119909)
(119902minus119904)(119902+1)
(15)
where 1198601= 21205752
120572120576(120572minus1)120572 119860
2= 1198961minus 21205752
(1 minus 120572)120576 120572 = (119902 + 1 minus
2119904)(119902 minus 119904) lt 1 120576 gt 0Next we estimate 119868
2= (211989621198731205880) intΩ
119906119904+1
119889119909minus2 intΩ
|nabla119906|119901+2
119889119909 minus 1198961intΩ
119906119902+1
119889119909 Since
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
= (
119901
2
+ 1)
2
119906119901
|nabla119906|2
(16)
it follows from Holder inequality that
int
Ω
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
119889119909 le (
119901
2
+ 1)
2
(int
Ω
|nabla119906|119901+2
119889119909)
2(119901+2)
times (int
Ω
119906119901+2
119889119909)
119901(119901+2)
(17)
Furthermore we have
int
Ω
119906119901+2
119889119909 le [
(119901 + 2)2
41205821
]
(1199012)+1
int
Ω
|nabla119906|119901+2
119889119909 (18)
which follows from (17) and membrane inequality
1205821int
Ω
1205962
119889119909 le int
Ω
|nabla120596|2
119889119909 (19)
where 1205821is the first eigenvalue in the fixed membrane
problem
Δ120596 + 120582120596 = 0 120596 gt 0 in Ω 120596 = 0 on 120597Ω (20)
Combining 1198682and (18) we have
1198682le
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 2[
41205821
(119901 + 2)2]
(1199012)+1
times int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
=
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 3[
41205821
(119901 + 2)2
]
(1199012)+1
int
Ω
119906119901+2
119889119909
+
[
41205821
(119901 + 2)2]
(1199012)+1
int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 11986821+ 11986822
(21)
Making use of Holder inequality we obtain
int
Ω
119906119904+1
119889119909 le (int
Ω
119906119901+2
119889119909)
(119904+1)(119901+2)
|Ω|(119901minus119904+1)(119901+2)
(22)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119904+1
119889119909)
2(119904+1)
|Ω|(119904minus1)(119904+1)
(23)
Combining (21) (22) with (23) we get
11986821
le (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
(24)
with
1198611=
21198962119873
1205880
1198612= 3[
41205821
(119901 + 2)2]
(1199012)+1
|Ω|minus(119901minus119904+1)2
(25)
Applying Holder inequality we obtain
int
Ω
119906119901+2
119889119909 le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
|Ω|(119902minus119901minus1)(119902+1)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119902+1
119889119909)
2(119902+1)
|Ω|(119902minus1)(119902+1)
(26)
It follows from (26) that
11986822
le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(27)
where
1198621= [
41205821
(119901 + 2)2
]
(1199012)+1
|Ω|(119902minus119901minus1)(119902+1)
1198622= 1198961|Ω|(1minus119902)(119902minus119901minus1)2(119902+1)
(28)
4 Abstract and Applied Analysis
Combining (14) (15) (21) and (24) with (27) we obtain
Ψ1015840
(119905) le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
1198601minus 1198602Ψ(119905)(119902minus119904)2
+ (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
+ (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(29)
with
1198601= 1198601|Ω|(119902minus119904)(119902+1)
1198602= 1198602|Ω|(1minus119902)(119902minus119904)2(119902+1)
(30)
We conclude from (29) that Ψ(119905) is decreasing in each timeinterval on which we obtain
Ψ (119905) ge max(1198601
1198602
)
2(119902minus119904)
(
1198611
1198612
)
2(119901minus119904+1)
(
1198621
1198622
)
2(119902minus119901minus1)
(31)
so that Ψ(119905) remains bounded for all time under theconditions in Theorem 1 This completes the proof ofTheorem 1
3 Blow-Up and Upper Bound of 119905lowast
In this section Ω needs not to be star-shaped We establishthe conditions to assure that the solution of (1)ndash(3) blowsup at finite time 119905lowast and derive an upper bound for 119905lowast Moreprecisely we establish the following result
Theorem 2 Let 119906(119909 119905) be the classical solution of problem (1)-(3) Assume that the nonnegative and integrable functions 119891and 119892 satisfy
120585119891 (120585) le 2 (1 + 120572) 119865 (120585) 120585 ge 0
120585119892 (120585) ge 2 (1 + 120573)119866 (120585) 120585 ge 0
(32)
with
119865 (120585) = int
120585
0
119891 (120578) 119889120578 119866 (120585) = int
120585
0
119892 (120578) 119889120578 (33)
where 120572 ge 0
120573 ge max (119901
2
120572) (34)
Moreover assume that Φ(0) ge 0 with
Φ (119905) = 2int
120597Ω
119866 (119906) 119889119878 minus int
Ω
|nabla119906|2
(1 +
2
119901 + 2
|nabla119906|119901
)119889119909
minus 2int
Ω
119865 (119906) 119889119909
(35)
Then the solution 119906(119909 119905) of problem (1)-(3) blows up at somefinite time 119905lowast lt 119879 with
119879 =
Ψ (0)
2120573 (1 + 120573)Φ (0)
120573 gt 0 (36)
where Ψ(119905) is defined in (13) If 120573 = 0 we have 119879 = infin
Proof We compute
Ψ1015840
(119905) = 2int
Ω
119906119906119905119889119909 = 2int
Ω
119906 [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 2int
120597Ω
119906 (|nabla119906|119901
+ 1)
120597119906
120597]119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
= 2int
120597Ω
119906119892 (119906) 119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
(37)
Making use of the hypotheses stated inTheorem 2 we have
Ψ1015840
(119905) ge 2 (1 + 120573)Φ (119905) (38)
Differentiating (35) we derive
Φ1015840
(119905) = 2int
120597Ω
119892 (119906) 119906119905119889119878 minus int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
minus 2int
Ω
119891 (119906) 119906119905119889119909
(39)
Integrating the identity nabla sdot (119906119905(|nabla119906|119901
+1)nabla119906) = 119906119905nabla sdot ((|nabla119906|
119901
+
1)nabla119906) + (12)(|nabla119906|119901
+ 1)(|nabla119906|2
)119905overΩ we get
int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
= 2int
Ω
nabla sdot (119906119905(|nabla119906|119901
+ 1) nabla119906) 119889119909
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1) nabla119906 sdot ]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
(40)
Substituting (40) into (39) we have
Φ1015840
(119905) = 2int
Ω
1199062
119905119889119909 gt 0 (41)
whichwithΦ(0) gt 0 implyΦ(119905) gt 0 for all 119905 isin (0 119905lowast
) Makinguse of the Schwarz inequality we obtain
2 (1 + 120573)Ψ1015840
Φ le (Ψ1015840
(119905))
2
= 4(int
Ω
119906119906119905119889119909)
2
le 2Ψ (119905)Φ1015840
(119905)
(42)
Abstract and Applied Analysis 5
Multiplying the above inequality by Ψminus2minus120573 we deduce
(ΦΨminus(1+120573)
)
1015840
ge 0 (43)
Arguing as in Theorem 31 in [20] we find
119905lowast
le 119879 =
1
2120573 (1 + 120573)
(Ψ (0))minus120573
=
Ψ (0)
2120573 (1 + 120573)Φ (0)
(44)
valid for 120573 gt 0 If 120573 = 0 we have
Ψ (119905) ge Ψ (0) 1198902119872119905 (45)
valid for 119905 gt 0 implying that 119905lowast = infin This completes theproof of Theorem 2
4 Lower Bounds for 119905lowast
In this section under the assumption that Ω is a star shapeddomain in 119877
3 convex in two orthogonal directions we seek alower bound for the blow-up time 119905lowast Now we state the resultas follows
Theorem 3 Let 119906(119909 119905) be the nonnegative solution of problem(1)-(3) and 119906(119909 119905) blows up at 119905lowast moreover the nonnegativefunctions 119891 and 119892 satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(46)
with 1198961gt 0 119896
2gt 0 119902 gt 1 119904 gt 1 119902 lt 119904 Define
120593 (119905) = int
Ω
119906119899(119904minus1)
119889119909 (47)
where 119899 is a parameter restricted by the condition
119899 gt max 4 2
119904 minus 1
(48)
Then 120593(119905) satisfies inequality
1205931015840
(119905) le Γ (120593) (49)
for some computable function Γ(120593) It follows that 119905lowast is boundedfrom below We have
119905lowast
ge int
infin
120593(0)
119889120578
Γ (120578)
119889120578 (50)
Proof Differentiating (47) and making use of the boundarycondition (2) together with the conditions (46) we have
1205931015840
(119905) = 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119906119905119889119909
= 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
times [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 119899 (119904 minus 1) int
120597Ω
119906119899(119904minus1)minus1
(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119891 (119906) 119889119909
le 1198962119899 (119904 minus 1) int
120597Ω
119906(119899+1)(119904minus1)
119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(51)
Applying inequality (27) in [20] to the first term on the righthand side of (51) we have
int
120597Ω
119906(119899+1)(119904minus1)
119889119878 le
3
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
(119899 + 1) (119904 minus 1) 119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
(52)
6 Abstract and Applied Analysis
Substituting (52) into (51) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(53)
Making use of arithmetic-geometric mean inequality wederive
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909 le
120583
2
int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
+
1
2120583
int
Ω
119906(119899+2)(119904minus1)
119889119909
(54)
for all 120583 gt 0 Choose 120583 gt 0 such that
1198962119899 (119899 + 1) (119904 minus 1)
2
119889120583
21205880
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] = 0 (55)
We rewrite (53) as
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(56)
Using Holder inequality we get
int
Ω
119906119899(119904minus1)
119889119909 le (int
Ω
119906119899(119904minus1)+119902minus1
119889119909)
119899(119904minus1)(119899(119904minus1)+119902minus1)
times |Ω|(119902minus1)(119899(119904minus1)+119902minus1)
(57)
Combining (56) with (57) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
=
31198962119899 (119904 minus 1)
1205880
1198691(119905)
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
1198692(119905)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(58)
where
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
1198692(119905) = int
Ω
119906(119899+2)(119904minus1)
119889119909
120596 (119905) = int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
(59)
Using Sobolev type inequality (A5) derived by Payne et al[21] we obtain
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
le
3
1205880
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 +
(119899 + 1) (119904 minus 1)
3
times(1 +
119889
1205880
)int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
32
(60)
We now make use of Holder inequality to bound the secondintegral on the right hand side of (60) as follows
int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
le (int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909)
(119901+1)(119901+2)
1205961(119901+2)
(61)
with
1205751=
(119899 minus 2) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(62)
Abstract and Applied Analysis 7
Wenote that 1205751lt 1 for 119899 gt (3119901minus2(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Using again Holderrsquosinequality we obtain
int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909
le 1205932(119899+1)(1minus120575
1)3119899
|Ω|1minus(2(119899+1)(1minus120575
1)3119899)
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 le 1205932(119899+1)3119899
|Ω|1minus(2(119899+1)3119899)
(63)
where |Ω| = intΩ
119889119909 is the volume of Ω Substituting (61) and(63) in (60) we obtain the following inequality
1198691(119905) le 119888
11205932(119899+1)3119899
+ 1198882120593(2(119899+1)(1minus120575
1)3119899)((119901+1)(119901+2))
times 1205961(119901+2)
32
le 1198881120593(119899+1)119899
+ 1198882120593((119899+1)(1minus120575
1)119899)((119901+1)(119901+2))
12059632(119901+2)
(64)
where 1198881 1198882are computable positive constants Note that the
last inequality in (64) follows from Holder inequality underthe particular form (119886 + 119887)
32
le radic2(11988632
+ 11988732
) Similarly wecan bound 119869
2and get
1198692(119905) le 119888
3120593(119899+2)119899
+ 1198884120593((119899+2)(1minus120575
2)119899)((119901+1)(119901+2))
12059632(119901+2)
(65)
where 1198883 1198884are computable positive constants
1205752=
(119899 minus 4) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(66)
Wenote that 1205752lt 1 for 119899 gt (3119901minus4(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Inserting (64) and (65)in (58) we arrive at
1205931015840
(119905) le1198891120593(119899+1)119899
+1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
+ 1198893120593(119899+2)119899
+1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(67)
where 120582 = (119901 + 1)(119901 + 2) 1198893and
119889119895(119895 = 1 2 4) are
computable positive constants Next we want to eliminatethe quantity 120596(119905) in inequality (67) By using the followinginequality
120593120572
120596120573
= (120574120596)120573
120593120572(1minus120573)
120574120573(1minus120573)
1minus120573
le 120574120573120596 + (1 minus 120573) 120574120573(120573minus1)
+ (1 minus 120573) 120574120573(120573minus1)
120593120572(1minus120573)
(68)
valid for 0 lt 120573 lt 1 where 120574 is an arbitrary positive constantthen we have
1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
le 1205741120596 (119905) + 119889
2120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
le 1205742120596 (119905) + 119889
4120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(69)
with arbitrary positive constants 1205741 1205742and computable
positive constants 1198892 1198894 Substitute (69) in (67) and choose
the arbitrary (positive) constants 1205741 1205742such that 120574
1+1205742minus119899(119904minus
1)[119899(119904 minus 1) minus 1] = 0 We obtain
1205931015840
(119905) le1198891120593(119899+1)119899
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(70)
We eliminate the last term in (70) by using the followinginequality
120593(119899+1)119899
= 119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(2119899minus1)(119904minus1)((2119899minus1)(119904minus1)+119904minus119902)
times 119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(119904minus119902)((2119899minus1)(119904minus1)+119904minus119902)
le
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
+
119904 minus 119902
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(71)
valid for 119902 lt 119904 and arbitrary119898 gt 0 and choose119898 such that
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
1198891119898 minus 119896
1119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
= 0
(72)
Then (70) can be rewritten as
1205931015840
(119905) le 11988911205933
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(73)
Integrating (73) over [0 119905] we conclude
119905lowast
ge int
infin
120593(0)
119889120578
times (11988911205783
+ 1198892120578(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120578(119899+2)119899
+ 1198894120578(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
)
minus1
(74)
This completes the proof of Theorem 3
8 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Authorsrsquo Contribution
All authors contributed equally to the paper and read andapproved the final paper
Acknowledgments
This work is supported by the Natural Science Foundationof Shandong Province of China (ZR2012AM018) and theFundamental Research Funds for the Central Universities(no 201362032) The authors would like to deeply thank allthe reviewers for their insightful and constructive comments
References
[1] J Bebernes and D Eberly Mathematical Problems from Com-bustion Theory vol 83 of Applied Mathematical SciencesSpringer New York NY USA 1989
[2] C V Pao Nonlinear Parabolic and Elliptic Equations PlenumPress New York NY USA 1992
[3] J L Vazquez The Porous Medium Equations MathematicalTheory Oxford University Press Oxford UK 2007
[4] J Filo ldquoDiffusivity versus absorption through the boundaryrdquoJournal of Differential Equations vol 99 no 2 pp 281ndash305 1992
[5] H A Levine and L E Payne ldquoNonexistence theorems forthe heat equation with nonlinear boundary conditions and forthe porous medium equation backward in timerdquo Journal ofDifferential Equations vol 16 pp 319ndash334 1974
[6] B Straughan Explosive Instabilities in Mechanics SpringerBerlin Germany 1998
[7] A A Samarskii V A Galaktionov S P Kurdyumov and A PMikhailov Blow-Up in Quasilinear Parabolic Equations vol 19of de Gruyter Expositions in Mathematics Walter de GruyterBerlin Germany 1995
[8] PQuittner andP Souplet Superlinear Parabolic Problems Blow-Up Global Existence and Steady States Birkhauser AdvancedTexts Birkhauser Basel Switzerland 2007
[9] C Bandle and H Brunner ldquoBlowup in diffusion equations asurveyrdquo Journal of Computational andAppliedMathematics vol97 no 1-2 pp 3ndash22 1998
[10] V A Galaktionov and J L Vazquez ldquoThe problem of blow-up in nonlinear parabolic equationsrdquo Discrete and ContinuousDynamical Systems vol 8 no 2 pp 399ndash433 2002
[11] H A Levine ldquoThe role of critical exponents in blowup theo-remsrdquo SIAM Review vol 32 no 2 pp 262ndash288 1990
[12] H A Levine ldquoNonexistence of global weak solutions to someproperly and improperly posed problems of mathematicalphysics the method of unbounded Fourier coefficientsrdquoMath-ematische Annalen vol 214 pp 205ndash220 1975
[13] L E Payne G A Philippin and P W Schaefer ldquoBlow-upphenomena for some nonlinear parabolic problemsrdquoNonlinearAnalysis Theory Methods amp Applications vol 69 no 10 pp3495ndash3502 2008
[14] MMarras and S Vernier Piro ldquoOn global existence and boundsfor blow-up time in nonlinear parabolic problems with time
dependent coefficientsrdquo Discrete and Continuous DynamicalSystems vol 2013 pp 535ndash544 2013
[15] Y Li Y Liu and C Lin ldquoBlow-up phenomena for some non-linear parabolic problems under mixed boundary conditionsrdquoNonlinear Analysis Real World Applications vol 11 no 5 pp3815ndash3823 2010
[16] Y Li Y Liu and S Xiao ldquoBlow-up phenomena for some non-linear parabolic problems under Robin boundary conditionsrdquoMathematical and Computer Modelling vol 54 no 11-12 pp3065ndash3069 2011
[17] C Enache ldquoBlow-up phenomena for a class of quasilinearparabolic problems under Robin boundary conditionrdquo AppliedMathematics Letters vol 24 no 3 pp 288ndash292 2011
[18] J Ding ldquoGlobal and blow-up solutions for nonlinear parabolicequations with Robin boundary conditionsrdquo Computers ampMathematics with Applications vol 65 no 11 pp 1808ndash18222013
[19] Y Liu ldquoBlow-up phenomena for the nonlinear nonlocal porousmedium equation under Robin boundary conditionrdquo Comput-ers amp Mathematics with Applications vol 66 no 10 pp 2092ndash2095 2013
[20] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary conditon Irdquo Zeitschrift fur Angewandte Mathematikund Physik vol 61 no 6 pp 999ndash1007 2010
[21] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary condition IIrdquoNonlinear Analysis Theory Methods ampApplications vol 73 no 4 pp 971ndash978 2010
[22] Y Liu ldquoLower bounds for the blow-up time in a non-local reac-tion diffusion problem under nonlinear boundary conditionsrdquoMathematical and ComputerModelling vol 57 no 3-4 pp 926ndash931 2013
[23] Y Liu S Luo and Y Ye ldquoBlow-up phenomena for a parabolicproblem with a gradient nonlinearity under nonlinear bound-ary conditionsrdquo Computers amp Mathematics with Applicationsvol 65 no 8 pp 1194ndash1199 2013
[24] Z B Fang R Yang and Y Chai ldquoLower bounds estimate for theblow-up time of a slow diffusion equation with nonlocal sourceand inner absorptionrdquo Mathematical Problems in Engineeringvol 2014 Article ID 764248 6 pages 2014
Research ArticleSome Properties on Complex Functional Difference Equations
Zhi-Bo Huang12 and Ran-Ran Zhang3
1 School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Physics and Mathematics University of Eastern Finland PO Box 111 80101 Joensuu Finland3Department of Mathematics Guangdong University of Education Guangzhou 510303 China
Correspondence should be addressed to Zhi-Bo Huang huangzhiboscnueducn
Received 15 January 2014 Accepted 13 March 2014 Published 24 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z-B Huang and R-R ZhangThis is an open access article distributed under theCreativeCommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited
We obtain some results on the transcendental meromorphic solutions of complex functional difference equations of the formsum120582isin119868120572120582(119911)(prod
119899
119895=0119891(119911 + 119888
119895)120582119895) = 119877(119911 119891 ∘119901) = ((119886
0(119911) +119886
1(119911)(119891 ∘119901)+ sdot sdot sdot + 119886
119904(119911) (119891 ∘119901)
119904
)(1198870(119911) + 119887
1(119911) (119891 ∘119901)+ sdot sdot sdot + 119887
119905(119911) (119891 ∘119901)
119905
))where 119868 is a finite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119899) are distinct complex constants 119901(119911) is
a polynomial and 120572120582(119911) (120582 isin 119868) 119886
119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative to 119891(119911)
We further investigate the above functional difference equation which has special type if its solution has Borel exceptional zero andpole
1 Introduction and Main Results
In this paper a meromorphic function means meromorphicin the whole complex plane C For a meromorphic function119910(119911) let 120590(119910) be the order of growth and 120583(119910) the lowerorder of119910(119911) Further let 120582(119910) (resp 120582(1119910)) be the exponentof convergence of the zeros (resp poles) of 119910(119911) We alsoassume that the reader is familiar with the fundamentalresults and the standard notations of Nevanlinna theory ofmeromorphic functions (see eg [1]) Given a meromorphicfunction 119910(119911) we call a meromorphic function 119886(119911) a smallfunction relative to 119910(119911) if 119879(119903 119886(119911)) = 119878(119903 119910) = 119900(119879(119903 119910))as 119903 rarr infin possibly outside of an exceptional set of finitelogarithmic measure Moreover if 119877(119911 119910) is rational in 119910(119911)with small functions relative to 119910(119911) as its coefficients we usethe notation 119889 = deg
119910119877(119911 119910) for the degree of 119877(119911 119910) with
respect to119910(119911) Inwhat follows we always assume that119877(119911 119910)is irreducible in 119910(119911)
Meromorphic solutions of complex difference equationshave recently gained increasing interest due to the problemof integrability of difference equations This is related to theactivity concerning Painleve differential equations and theirdiscrete counterparts in the last decades Ablowitz et al [2]considered discrete equations to be delay equations in thecomplex plane This allowed them to analyze these equations
with the methods from complex analysis In regard to relatedpapers concerning a more general class of complex differenceequations we may refer to [3ndash5] These papers mainly dealtwith equations of the form
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119877 (119911 119891) (1)
where 119869 is a collection of all nonempty subsets of1 2 119899 119888
119895(119895 isin 119869) are distinct complex constants 119891(119911) is
a transcendental meromorphic function 120572119869(119911) (119869 isin 119869) are
small functions relative to119891(119911) and119877(119911 119891) is a rational func-tion in 119891(119911) with small meromorphic coefficients Moreoverif the right-hand side of (1) is essentially like the compositefunction 119890 ∘119891 of 119891(119911) and a rational function 119890(119911) Laine et alreversed the order of composition that is they considered thecomposite function 119891 ∘ 119890 of 119891(119911) and a rational function 119890(119911)which resulted in a complex functional difference equationThe following theorem [5 Theorem 28] gives an example
Theorem A (see [5 Theorem 28]) Suppose that 119891(119911) is atranscendental meromorphic solution of equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (2)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 283895 10 pageshttpdxdoiorg1011552014283895
2 Abstract and Applied Analysis
where 119901(119911) is a polynomial of degree 119896 ge 2 Moreover oneassumes that the coefficients 120572
119869(119911) are small functions relative
to 119891(119911) and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (3)
where 120572 = (log 119899)(log 119896)At this point we briefly introduce some notations used in
this paper A difference monomial of a meromorphic function119891(119911) is defined as
119891(119911)1205820119891(119911 + 119888
1)1205821
sdot sdot sdot 119891(119911 + 119888119899)120582119904
=
119904
prod
119895=0
119891(119911 + 119888119895)
120582119895
(4)
where 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119904) are distinct
constants and 120582119895(119895 = 0 1 119904) are natural numbers A
difference polynomial 119867(119911 119891(119911)) of a meromorphic function119891(119911) a finite sum of difference monomials is defined as
119867(119911 119891 (119911)) = sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
) (5)
where 119868 is a finite set of multi-indexes 120582 = (1205820 1205821 120582
119899)
120572120582(119911) (120582 isin 119868) are small functions relative to 119891(119911) The degree
and the weight of the difference polynomial (5) respectively aredefined as
deg119891(119867) = max
120582isin119868
119899
sum
119895=0
120582119895
120581119891(119867) = max
120582isin119868
119899
sum
119895=1
120582119895
(6)
Consequently 120581119891(119867) le deg
119891(119867) For instance the degree and
the weight of the difference polynomial 1198912(119911)119891(119911minus1)119891(119911+1)+119891(119911)119891(119911 + 1)119891(119911 + 2) + 119891
2
(119911 minus 1)119891(119911 + 2) respectively arefour and three Moreover a difference polynomial (5) is said tobe homogeneous with respect to 119891(119911) if the degree sum119899
119895=0120582119895of
each monomial in the sum of (5) is nonzero and the same forall 120582 isin 119868
In the following we proceed to prove generalizations ofTheorem A and investigate some new results for the first timeWe permit more general expressions on both sides of (1)
Theorem 1 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(7)
where119867(119911 119891(119911)) is defined as (5) 119901(119911) = 119889119896119911119896
+ sdot sdot sdot+1198891119911+119889
0
is a polynomial with constant coefficients 119889119896( = 0) 119889
1 1198890
and of the degree 119896 ge 2 and 119886119894(119911) (119894 = 0 1 119904) and
119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative
to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Set 119889 = max119904 119905 If 119896119889 le
(119899 + 1)deg119891(119867) then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (8)
where 120572 = (log(119899 + 1) + log deg119891(119867) minus log119889)(log 119896)
Similar to the proof of Theorem 1 we easily obtain thefollowing result which is a generation of Theorem A
Theorem 2 Let 119888119894isin C (119894 = 1 2 119899) be distinct
constants and 119891(119911) be a transcendental meromorphic solutionof equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895))
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(9)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119896119889 =
119896max119904 119905 le 119899 then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (10)
where 120572 = (log 119899 minus log 119889)(log 119896)
We then proceed to consider the distribution of zeros andpoles of meromorphic solutions of (7) The following resultindicates that solutions having Borel exceptional zeros andpoles appear only in special situations
Theorem 3 Let 1198880= 0 let 119888
119894isin C 0 (119894 = 1 2 119899) be
distinct constants and let 119891(119911) be a finite order transcendentalmeromorphic solution of equation
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(11)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 1
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 If
max120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (12)
then (11) is either of the form119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(13)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (14)
where 120572 isin C 0 is some constant
Abstract and Applied Analysis 3
Example 4 119891(119911) = cos 119911 solves difference equation
4119891(119911)2
119891(119911 + 120587)2
= 119891(2119911)2
+ 2119891 (2119911) + 1 (15)
Here 119901(119911) = 2119911 Clearly 120582(1119891) = 0 lt 1 = 120582(119891) = 120590(119891) Thisexample shows that condition (12) is necessary and cannot bereplaced by
min120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (16)
Moreover we obtain a result parallel toTheorem 54 in [6]for the difference case
Theorem 5 Suppose that the equation119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
=
119888 (119911)
119891(119911)119898 119898 isin N (17)
has a meromorphic solution of finite order where 1198880= 0 119888
119895isin
C 0 (119895 = 1 2 119899) are distinct constants and 119888(119911) is anontrivialmeromorphic function If119891(119911)has only finitelymanypoles then119891(119911) = 119863(119911)119890119864(119911) where119863(119911) is a rational functionand119864(119911) is a polynomial if and only if 119888(119911) = 119866(119911)119890119872(119911) where119866(119911) is a rational function and119872(119911) is a polynomial
Example 6 Difference equation
119891(119911)2
119891 (119911 + 1) 119891 (119911 minus 1) = (
1
1199114(1199112minus 1)
1198906119911
) sdot
1
119891(119911)2
(18)
of the type (17) is solved by 119891(119911) = 119890119911119911 Here 119891(119911) = 119890119911119911and 119888(119911) = 11989061199111199114(1199112 minus 1) satisfy Theorem 5
As an application of Theorem 3 we obtain the following
Theorem 7 Let 119888 isin C 0 and let 119891(119911) be a finite ordertranscendental meromorphic solution of equation
119891 (119911 + 119888) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(19)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 Then 119891(119911) has at most one Borel exceptional value
If the degree 119896 of polynomial 119901(119911) is 1 in Theorem 7 theresult does not hold For example we have the following
Example 8 119891(119911) = tan 119911 solves difference equation
119891 (119911 + 1) =
119891 (119911) + tan 11 minus (tan 1) 119891 (119911)
(20)
of the type (19) Obviously 119891(119911) has two Borel exceptionalvalues plusmn119894
If we remove the assumption max120582(119891) 120582(1119891) lt 120590(119891)used in Theorem 3 we obtain a result similar to Theorem 12in [4]
Theorem 9 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891)
=
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905
(21)
where 119867(119911 119891(119911)) is defined as (5) and 119886119894(119911) (119894 = 0 1 119904)
and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions
relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119889 = max119904 119905 gt
(119899 + 1) deg119891(119867) then 120590(119891) = infin
In fact the following examples show that the assertion ofTheorem 9 does not remain valid identically if 119889 le (119899 +1)deg
119891(119867)
Example 10 119891(119911) = exp119890119911119911 solves the difference equation
(119911 minus 120587119894) (119911 + log 2 minus 120587119894) 119891 (119911) 119891 (119911 minus 120587119894) 119891 (119911 + log 2 minus 120587119894)
+ (119911 + log 8) 119891 (119911 + log 8) =1 + 119911
11
119891(119911)10
1199113119891(119911)
2
(22)
Clearly 119889 = 10 lt (3+1) sdot 3 = (119899+1) deg119891(119867) and 120590(119891) = infin
Example 11 119891(119911) = tan 119911 satisfies the difference equation
119891(119911 +
120587
4
) + 119891(119911 minus
120587
4
) =
4119891 (119911)
1 minus 119891(119911)2
(23)
Obviously 119889 = 2 lt (2+1)times1 = (119899+1) deg119891(119867) and120590(119891) = 1
Example 12 (see [7 pages 103ndash106] and [8 page 8]) Thefollowing difference equation
119891 (119911 + 1) = 120572119891 (119911) (1 minus 119891 (119911)) 120572 = 0 (24)
derives from a well-known discrete logistic model in biologyIt has been proved that all other meromorphic solutions areof infinite order apart from the constant solutions 119891(119911) equiv 0and 119891(119911) = (120572 minus 1)120572 For instance (24) has one-parameterfamilies of entire solutions of infinite order
119891 (119911) =
1
2
(1 minus exp (119860119890119911 log 2)) 119860 isin C 0 120572 = 2
119891 (119911) = sin2 (119861119890119911 log 2) 119861 isin C 0 120572 = 4(25)
Here 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867)
Example 13 119891(119911) = 119911 solves the difference equation
119891 (119911 + 1) =
1 minus 119891(119911)2
minus1199112minus 119911 + 1 + 119891(119911)
2 (26)
We get 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867) and 120590(119891) = 0
If the difference polynomial in the left-hand side of (21)is homogeneous we further obtain the following theorem
4 Abstract and Applied Analysis
Theorem 14 Let 119891(119911) be a transcendental meromorphic solu-tion of (21) where 119867(z 119891(119911)) is defined as (5) and 119886
119894(119911) (119894 =
0 1 119904) and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Suppose
that 119867(119911 119891) is homogeneous and has at least one differencemonomial of type
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
(120582119895isin N
+ 119895 = 0 1 119899) (27)
If 119889 = max119904 119905 gt 3deg119891(119867) then 120590(119891) = infin
2 Proof of Theorem 1
We need some preliminaries to proveTheorem 1
Lemma 15 (see [9 Lemma 4]) Let 119891(119911) be a transcendentalmeromorphic function and let 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot + 1198891119911 +
1198890(119889119896= 0)be a polynomial of degree 119896 Given 0 lt 120575 lt |119889
119896|
denote ] = |119889119896| + 120575 and 120583 = |119889
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C cup infin one has for all 119903 ge 1199030gt 0
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 ∘ 119901) le 119896119899 (]119903119896 119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 ∘ 119901) le 119873(]119903119896 119886 119891)
+ 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 ∘ 119901) le (1 + 120576) 119879 (]119903119896 119891) (28)
Lemma16 (see [10Theorem B16]) Given distinctmeromor-phic functions 119891
1 119891
119899 let 119869 denote the collection of all
nonempty subsets of 1 2 119899 and suppose that 120572119869isin C for
each 119869 isin 119869 Then
119879(119903sum
119869
120572119869(prod
119895isin119869
119891119895)) le
119899
sum
119896=1
119879 (119903 119891119896) + 119874 (1) (29)
By denoting 119891119894+1= 119891(119911 + 119888
119894)120582119894(119894 = 0 1 119899) below it is
an easy exercise to prove the following result from Lemma 16
Lemma 17 Let 119891(119911) be a meromorphic function let 119868 be afinite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) and let 120572
120582(119911)
be small functions relative to 119891(119911) for all 120582 isin 119868 Then thecharacteristic function of the difference polynomial (5) satisfies
119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
))
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(30)
where 119862 = max|1198881| |119888
2| |119888
119899|
Lemma18 (see [11 Lemma5]) Let119892(119903) andℎ(119903) bemonotonenondecreasing functions on [0infin) such that 119892(119903) le ℎ(119903) for all119903 notin 119864 cup [0 1] where 119864 sub (1infin) is a set of finite logarithmicmeasure Let 120572 gt 1 be a given constant Then there exists an1199030= 119903
0(120572) gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 ge 119903
0
Lemma 19 (see [12 Lemma 3]) Let 120595(119903) be a function of119903 (119903 ge 119903
0) positive and bounded in every finite interval
(i) Suppose that 120595(120583119903119898) le 119860120595(119903) + 119861 (119903 ge 1199030) where
120583 (120583 gt 0)119898 (119898 gt 1)119860 (119860 ge 1) and 119861 are constantsThen 120595(119903) = 119874((log 119903)120572) with 120572 = (log119860)(log119898)unless 119860 = 1 and 119861 gt 0 and if 119860 = 1 and 119861 gt 0 thenfor any 120576 gt 0 120595(119903) = 119874((log 119903)120576)
(ii) Suppose that (with the notation of (i)) 120595(120583119903119898) ge119860120595(119903) (119903 ge 119903
0) Then for all sufficiently large values of
119903 120595(119903) ge 119870(log 119903)120572 with 120572 = (log119860)(log119898) for somepositive constant 119870
Proof of Theorem 1 For any 120576 (0 lt 120576 lt 1) we may applyValiron-Mohonrsquoko lemma Lemmas 15 and 17 and (5) and (7)to conclude that119889 (1 minus 120576) 119879 (120583119903
119896
119891)
le 119889119879 (119903 119891 ∘ 119901) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905)
+ 119878 (119903 119891)
= 119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
)) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(31)holds for all sufficiently large 119903 possibly outside of anexceptional set of finite logarithmic measure where 119862 =max|119888
1| |119888
2| |119888
119899| and 120583 is defined as Lemma 15 Now we
may apply Lemma 18 to deal with the exceptional set andconclude that for every 120578 gt 1 there exists an 119903
0gt 0 such
that119889 (1 minus 120576) 119879 (120583119903
119896
119891) le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (120578119903 119891)
(32)holds for all 119903 ge 119903
0 Denote 120596 = 120578119903 Then (32) can be written
in the form
119879(
120583
120578119896
120596119896
119891) le
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
119879 (120596 119891) (33)
Since 119889119896 le (119899 + 1)deg119891(119867) we get ((119899 + 1)deg
119891(119867)(1 +
120576))(119889(1 minus 120576)) gt 1 for all 0 lt 120576 lt 1 Thus we now applyLemma 19(i) to conclude that
119879 (119903 119891) = 119874 ((log 119903)120572+120576)
120572 =
log ((119899 + 1) deg119891(119867) (1 + 120576) 119889 (1 minus 120576))
log 119896
=
log (119899 + 1) + log deg119891(119867) minus log119889
log 119896+ 119900 (1)
(34)
The proof of Theorem 1 is completed
Abstract and Applied Analysis 5
3 Proof of Theorems 3 and 5
We again need some preliminaries
Lemma 20 (see [13 Theorem 15]) Suppose that 119891119895(119911) (119895 =
1 2 119899) (119899 ge 2) are meromorphic functions and 119892119895(119911) (119895 =
1 2 119899) are entire functions satisfying the following condi-tions
(1) sum119899119895=1119891119895(119911)119890
119892119895(119911)
= 0
(2) 119892119895(119911) minus 119892
119896(119911) are not constants for 1 le 119895 lt 119896 le 119899
(3) For 1 le 119895 le 119899 1 le ℎ lt 119896 le 119899
119879 (119903 119891119895) = 119900 119879 (119903 119890
119892ℎminus119892119896) (119903 997888rarr +infin 119903 notin 119864) (35)
where 119864 sub (1 +infin) is of finite linear measure or finitelogarithmic measure
Then 119891119895(119911) equiv 0 (119895 = 1 2 119899)
Lemma21 (see [14Theorem4]) Let119865(119911)119875119899(119911) 119875
0(119911) be
polynomials such that 1198651198751198991198750equiv 0 and then every finite order
transcendental meromorphic solution 119891(119911) of equation
119875119899(119911) 119891 (119911 + 119899) + sdot sdot sdot + 119875
1(119911) 119891 (119911 + 1) + 119875
0(119911) 119891 (119911) = 119865 (119911)
(36)
satisfies 120582(119891) = 120590(119891) ge 1
Remark 22 Replacing 119895 by 119888119895(119895 = 1 2 119899) where 119888
119895(119895 =
1 2 119899) are distinct nonzero complex constants Lemma 21remains valid
Proof of Theorem 3 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed withnonzero poles of 119891(119911) Since max120582(119891) 120582(1119891) lt 120590(119891) thenℎ(119911) = 119911
120591
119902(119911) is an entire function such that
120590 (ℎ) = 120582(
1
119891
) lt 120590 (119891) lt +infin (37)
and 119892(119911) = ℎ(119911)119891(119911) is a transcendental entire function with
119879 (119903 119892) = 119879 (119903 119891) + 119878 (119903 119891) 120590 (119892) = 120590 (119891)
120582 (119892) = 120582 (119891)
(38)
If 119902(119911) is a polynomial we obtain quickly that 120590(ℎ ∘ 119901) =0 lt 120590(119892 ∘ 119901) Otherwise we conclude from the last assertionof Lemma 15 (37) and (38) that
120590 (ℎ ∘ 119901) = 119896120590 (ℎ) = 119896120582(
1
119891
) lt 119896120590 (119892) = 120590 (119892 ∘ 119901) (39)
Therefore
119879 (119903 ℎ ∘ 119901) = 119878 (119903 119892 ∘ 119901) (40)
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (11) we concludethat
(ℎ ∘ 119901)119904minus119905
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
(
119899
prod
119895=0
119892(119911 + 119888119895)
120582119895
)
= (1198860(119911) (ℎ ∘ 119901)
119904
+ 1198861(119911) (ℎ ∘ 119901)
119904minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119886119904(119911) (119892 ∘ 119901)
119904
)
times (1198870(119911) (ℎ ∘ 119901)
119905
+ 1198871(119911) (ℎ ∘ 119901)
119905minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119887119905(119911) (119892 ∘ 119901)
119905
)
minus1
(41)
Obviously it follows from (37)ndash(40) and Lemma 15 that
119879(119903
1
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 (ℎ ∘ 119901)119904minus119905
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 119886119906(119911) (ℎ ∘ 119901)
119904minus119906
) = 119878 (119903 119892 ∘ 119901) 119906 = 0 1 119904
119879 (119903 119887V (119911) (ℎ ∘ 119901)119905minusV) = 119878 (119903 119892 ∘ 119901) V = 0 1 119905
(42)
Denoting119860(119911) = (ℎ ∘ 119901)119904minus119905prod119899119895=0ℎ(119911 + 119888
119895)120582119895 we get from (42)
that119879 (119903 119860) = 119878 (119903 119892 ∘ 119901) (43)
Since zeros and poles are Borel exceptional values of 119891(119911) by(12) we may apply a result due to Whittaker see [15 Satz134] to deduce that 119891(119911) is of regular growth Thus we useLemma 15 and (12) again to get
119879(119903
1198911015840
119891
) = 119873(119903 119891) + 119873(119903
1
119891
) + 119878 (119903 119891) = 119878 (119903 119892 ∘ 119901)
(44)
Similarly if we set 119861(119911) = 119860(119911)(prod119899
119895=0119892(119911 + 119888
119895)120582119895) we
also deduce from the lemma of the logarithmic derivativeLemma 15 (12) (38) and (43) that
119879(119903
1198611015840
119861
) = 119879(119903
1198601015840
119860
+
119899
sum
119895=0
120582119895
1198921015840
(119911 + 119888119895)
119892 (119911 + 119888119895)
) = 119878 (119903 119892 ∘ 119901)
(45)Denoting 119865(119911) = 119892 ∘ 119901
119875 (119911 119865) =
1198860(119911)
119886119904(119911)
(ℎ ∘ 119901)119904
+
1198861(119911)
119886119904(119911)
(ℎ ∘ 119901)119904minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119904
119876 (119911 119865) =
1198870(119911)
119887119905(119911)
(ℎ ∘ 119901)119905
+
1198871(119911)
119887119905(119911)
(ℎ ∘ 119901)119905minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119905
(46)
6 Abstract and Applied Analysis
Therefore we deduce from Lemma 15 and (42) that thecoefficients of 119875(119911 119865) and119876(119911 119865) are small functions relativeto 119865(119911) Thus (41) can be written in the form
119887119905(119911)
119886119904(119911)
119861 (119911) =
119875 (119911 119865)
119876 (119911 119865)
= 119906 (119911 119865) (47)
Denoting
120595 (119911) =
1198651015840
(119911)
119865 (119911)
119880 (119911) =
1199061015840
(119911 119865)
119906 (119911 119865)
(48)
we get 119879(119903 119880) = 119878(119903 119892 ∘ 119901) from (45) and (47) Wealso conclude from the lemma of logarithmic derivativeLemma 15 and (12) that
119879 (119903 120595) = 119879(119903
1198651015840
119865
) = 119898(119903
1198651015840
119865
) + 119873(119903
1198651015840
119865
)
le 119873 (119903 119865) + 119873(119903
1
119865
) + 119878 (119903 119865)
= 119873 (119903 119892 ∘ 119901) + 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(]1199031198961
119892
) + 119878 (119903 119892 ∘ 119901) = 119878 (119903 119892 ∘ 119901)
(49)
where ] is defined as Lemma 15Since
1198751015840
119876 minus 1198751198761015840
1198762
= 1199061015840
= 119880119906 =
119880119875
119876
(50)
we conclude that
1198751015840
119876 minus 1198751198761015840
= 119880119875119876 (51)
Now writing 1198651015840 = 120595119865 in (51) regarding then (51) as analgebraic equation in119865with coefficients of growth 119878(119903 119865) andcomparing the leading coefficients we deduce that
(119904 minus 119905) 120595 = 119880 (52)
By integrating both sides of the last equality above weconclude that
119906 (119911 119865) = 120572119865(119911)119904minus119905
(53)
for some 120572 isin C 0 Therefore by combining therepresentations of 119865 119861 119860 119892 with (53) we conclude that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
(54)
If 119904119905 = 0 we deduce from (11) and (54) that
120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(55)
From this we get that 119877(119911 119891 ∘ 119901) is not irreducible in 119891 ∘ 119901a contradiction Thus 119905 = 0 or 119904 = 0 Therefore we deducefrom (54) that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(56)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (57)
The proof of Theorem 3 is completed
Proof of Theorem 5 Assume first that 119891(119911) = 119863(119911)119890119864(119911)
where 119863(119911) is a rational function and 119864(119911) is a polynomialOne can see from (17) that
119888 (119911) = 119891(119911)119898
119899
prod
119894=0
119891(119911 + 119888119894)120582119894
= [119863(119911)119898
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
] 119890119898119864(119911)+sum
119899
119894=0120582119894119864(119911+119888
119894)
= 119866 (119911) 119890119872(119911)
(58)
where 119866(119911) = 119863(119911)119898prod119899119894=0119863(119911 + 119888
119894)120582119894 is rational and119872(119911) =
119898119864(119911) + sum119899
119894=0120582119894119864(119911 + 119888
119894) is a polynomial
Suppose next that 119888(119911) = 119866(119911)119890119872(119911) where 119866(119911) is arational function and 119872(119911) is a polynomial Since 119891(119911) hasonly finitely many poles we conclude from (17) that
119873(119903
1
119891
) le 119873(119903
1
119891119898
) = 119873(119903
prod119899
119894=0119891(119911 + 119888
119894)120582119894
119888 (119911)
)
= 119874 (log 119903)
(59)
Thus 119891(119911) has only finitely many zeros and poles and119891(119911) = 119863(119911)119890
119864(119911) where119863(119911) is rational and 119864(119911) is an entirefunction In the followingwe only prove119864(119911) is a polynomialNow substituting119891(119911) = 119863(119911)119890119864(119911) and 119888(119911) = 119866(119911)119890119872(119911) into(17) we get
119899
prod
119894=0
119863(119911 + 119888119894)120582119894 exp (120582
119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(60)
(
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
) exp(119899
sum
119894=0
120582119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(61)
Thus we deduce from Lemma 20 that two exponents in (61)cancel each other to a constant 120591 isin C such that
119899
sum
119894=0
120582119894119864 (119911 + 119888
119894) = 119872 (119911) minus 119898119864 (119911) + 120591 (62)
Abstract and Applied Analysis 7
that is119899
sum
119894=1
120582119894119864 (119911 + 119888
119894) + (120582
0+ 119898)119864 (119911) = 119872 (119911) + 120591 (63)
Suppose that 119864(119911) is not a polynomial If 119864(119911) is a transcen-dental entire function of finite order we get from Lemma 21Remark 22 and (63) that 120590(119864) ge 1 Otherwise 119864(119911) is atranscendental entire function of infinite order These bothshow that 120590(119891) = infin contradicting the assumption that119891(119911) is finite order Thus 119864(119911) is a polynomial The proof ofTheorem 5 is completed
4 Proof of Theorem 7
Lemmas 23 and 25 reveal some properties of the maximalmodule of the polynomial in composite function 119891 ∘ 119901 with ameromorphic function 119891(119911) and a polynomial 119901(119911) whichare useful for proving the existence of Borel exceptionalvalue of finite order meromorphic solutions of functionaldifference equation of type (19)
Lemma 23 Let 119892(119911) be a nonconstant entire function of order120590(119892) = 120590 lt infin Suppose that 120572
119895(119911) (119895 = 1 2 119898) are small
meromorphic functions relative to 119892(119911) Then there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 120572119895)
119872 (119903 119892)
997888rarr 0 119895 = 1 2 119898 (64)
hold simultaneously for all 119903 isin 119864 as 119903 rarr infin where the lowerlogarithmic density of set 119864 is defined by
logdens (119864) = lim inf119903rarrinfin
int[1119903]cap119864
(119889119905119905)
log 119903 (65)
Remark 24 The proof of Lemma 23 is similar to the proof ofLemma 24 and Remark 25 in [16] Here we omit it
Lemma 25 Let 119891(119911) be a finite order transcendental mero-morphic function satisfying (12) and 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot+1198891119911+
1198890is a polynomial with constant coefficients 119889
119896( = 0) 119889
1 1198890
and of the degree 119896 ge 1 Suppose that
119867(119911) = 119886119899(119911) (119891 ∘ 119901)
119899
+ 119886119899minus1(119911) (119891 ∘ 119901)
119899minus1
+ sdot sdot sdot + 1198861(119911) (119891 ∘ 119901) + 119886
0(119911)
(66)
is a polynomial in 119891 ∘119901 where 119899 (ge 1) is a positive integer and119886119899(119911) ( equiv 0) 119886
119899minus1(119911) 119886
1(119911) 119886
0(119911) are small meromorphic
functions relative to 119891(119911) Then there exists a set 1198641of lower
logarithmic density 1 such that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (67)
for all 119903 isin 1198641as 119903 rarr infin where 0 lt 120583 lt |119889
119896| Hence119867(119911) equiv
0
Proof of Lemma 25 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed
with the nonzero poles of 119891(119911) Since 119891(119911) satisfies (12) thenℎ(119911) = 119911
120591
119902(119911) is an entire function Thus 119892(119911) = ℎ(119911)119891(119911) isentire and (37) (38) and (40) also hold
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (66) we con-clude that
119867(119911) = 119886119899(119911) sdot
(119892 ∘ 119901)119899
(ℎ ∘ 119901)119899+ 119886
119899minus1(119911) sdot
(119892 ∘ 119901)119899minus1
(ℎ ∘ 119901)119899minus1
+ sdot sdot sdot
+ 1198861(119911) sdot
(119892 ∘ 119901)
(ℎ ∘ 119901)
+ 1198860(119911)
=
119886119899(119911)
(ℎ ∘ 119901)119899(119892 ∘ 119901)
119899
times [1 +
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
(119892 ∘ 119901)minus1
+ sdot sdot sdot
+
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
(119892 ∘ 119901)1minus119899
+
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
(119892 ∘ 119901)minus119899
]
(68)
We note from Lemma 15 and (40) that
119879(119903
119886119899(119911)
(ℎ ∘ 119901)119899) = 119878 (119903 119892 ∘ 119901)
119879(119903
119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911)
) = 119878 (119903 119892 ∘ 119901)
for 119895 = 0 1 119899 minus 1
(69)
Therefore we deduce from Lemma 23 that there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 119886119899(119911) (ℎ ∘ 119901)
119899
)
119872 (119903 119892 ∘ 119901)
997888rarr 0
119872(119903 119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911))
119872 (119903 119892 ∘ 119901)
997888rarr 0
(119895 = 0 1 119899 minus 1)
(70)
Moreover according to the choosing of 119864 in the proofof Lemma 23 we know that 119886
119895(119911)(ℎ ∘ 119901)
119899minus119895
119886119899(119911) for
8 Abstract and Applied Analysis
119895 = 0 1 119899 minus 1 have no zeros and poles for all |119911| = 119903 isin 119864Thus we conclude from (68) and (70) that for any 120576 gt 0
119872(119903119867) ge 119872(119903 119892 ∘ 119901)119899minus120576
times [1 minus
100381610038161003816100381610038161003816100381610038161003816
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus1
minus sdot sdot sdot
minus
1003816100381610038161003816100381610038161003816100381610038161003816
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
1003816100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)1minus119899
minus
100381610038161003816100381610038161003816100381610038161003816
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus119899
]
ge (1 minus 120576)119872(119903 119892 ∘ 119901)119899minus120576
(71)
and so
log+119872(119903119867) ge (119899 minus 32
120576) log+119872(119903 119892 ∘ 119901) (72)
for all |119911| = 119903 isin 119864 and |119892 ∘ 119901(119911)| = 119872(119903 119892 ∘ 119901)Therefore we deduce from Lemma 15 and (38) that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (73)
for all |119911| = 119903 isin 1198641= 119864 cap (119903
0 +infin) where 119903
0gt 0 It is
obvious that 1198641has lower logarithmic density 1 The proof of
Lemma 25 is completed
Proof of Theorem 7 Suppose that 119891(119911) has two finite Borelexceptional values 119886 and 119887 ( = 0 119886) For the case where oneof 119886 and 119887 is infinite we can use a similar method to prove itSet
119892 (119911) =
119891 (119911) minus 119886
119891 (119911) minus 119887
(74)
Then 120590(119892) = 120590(119891) and
120582 (119892) = 120582 (119891 minus 119886) lt 120590 (119892) 120582 (
1
119892
) = 120582 (119891 minus 119887) lt 120590 (119892)
(75)
It follows from (74) that
119891 (119911) =
119886 minus 119887119892 (119911)
1 minus 119892 (119911)
(76)
Now substituting (76) into (19) we conclude that
119892 (119911 + 119888) = (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119886
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
times (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119887
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
minus1
= ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
))
times ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
))
minus1
(77)
Since 1198860(119911) + 119886
1(119911)(119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911)(119891 ∘ 119901)
119904 and 1198870(119911) +
1198871(119911)(119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911)(119891 ∘ 119901)
119905 are irreducible in 119891 ∘ 119901 weconclude that at least one of the following three inequalitiesholds that is
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(78)
Thus we deduce fromTheorem 3 that
119892 (119911 + 119888) = 119888 (119911) (119892 ∘ 119901)119897
(79)
where 119888(119911) is meromorphic function satisfying 119879(119903 119888) =119878(119903 119892) and 119897 isin Z Clearly 119897 = 0 and 119892(119911) is of regular growthfrom (75) see [15 Staz 134] Therefore 120590(119888) lt 120590(119892)
If 119897 ge 1 we conclude from (77) and (79) that
(minus1)119904+119905
119888 (119911)(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)(119892 ∘ 119901)119904+119905+119897
+ sdot sdot sdot + (minus
119904
sum
119894=0
119886119894(119911) 119886
119894
+ 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
) = 0
(80)
Abstract and Applied Analysis 9
Thus we deduce from Lemma 25 that (80) is a contradictionIf 119897 le minus1 we use the same method as above to getanother contradiction Therefore 119891(119911) has at most one Borelexceptional value The proof of Theorem 7 is completed
5 Proof of Theorems 9 and 14
We first recall two lemmas
Lemma 26 (see [17 Lemma 21]) Let 119891(119911) be a nonconstantmeromorphic function 119904 gt 0 120572 lt 1 and 119865 sub R+ the set of all119903 such that
119879 (119903 119891) le 120572119879 (119903 + 119904 119891) (81)
If the logarithmicmeasure of119865 is infinite that isint119865
(119889119905119905) = infinthen 119891(119911) is of infinite order of growth
Lemma 27 (see [18 Corollary 26] and [19 Corollary 22])Let 119891(119911) be a meromorphic function of finite order and let 119888 isinC Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (82)
for all 119903 outside of a possible exceptional set of finite logarithmicmeasure
Proof of Theorem 9 For any 120576 (0 lt 120576 lt (119889 minus (119899 +
1)deg119891(119867))(119889 + (119899 + 1) deg
119891(119867))) we may apply Valiron-
Mohonrsquoko lemma Lemma 17 (5) and (21) to conclude that
119889 (1 minus 120576) 119879 (119903 119891)
le 119889119879 (119903 119891) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905)
= 119879 (119903119867 (119911 119891 (119911))) le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891)
+ 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(83)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote
120572 =
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
(84)
Then 120572 lt 1 since 0 lt 120576 lt (119889 minus (119899 + 1)deg119891(119867))(119889 + (119899 +
1)deg119891(119867)) and 119889 gt (119899 + 1)deg
119891(119867) Thus
119879 (119903 119891) le 120572119879 (119903 + 119862 119891) (85)
holds for all 119903 in a set with infinite logarithmic measureTherefore we deduce from Lemma 26 and (85) that 120590(119891) =infin The proof of Theorem 9 is completed
Proof of Theorem 14 Assume contrary to the assertion that119891(119911) is meromorphic of finite order Taking into account theassumption that119867(119911 119891(119911)) is homogeneous we deduce fromLemma 27 that
119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) = 119878 (119903 119891) (86)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote 119862 = max1le119894le119899|119888119894| Since 119867(119911) is homogeneous
and has at least one difference monomial of typeprod119899119895=0119891(119911 +
119888119895)120582119895 we immediately conclude that by looking at pole
multiplicities summing over |119911| le 119903 and integratinglogarithmically
119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
)
le 120581119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
(87)
Therefore
119879 (119903119867 (119911 119891 (119911)))
= 119898 (119903119867 (119911 119891 (119911))) + 119873 (119903119867 (119911 119891 (119911)))
le 119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119898(119903 119891(119911)deg119891(119867)
)
+ 119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119873(119903 119891(119911)deg119891(119867)
)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
))
+ 119879 (119903 119891(119911)deg119891(119867)
) + 119878 (119903 119891)
le 3deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(88)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure The remainder can be proven by asimilar method in Theorem 9 The proof of Theorem 14 iscompleted
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful to the referees for their helpfulsuggestions to improve this paper The first author alsothanks Professor Ilpo laine and Professor Risto Korhonen for
10 Abstract and Applied Analysis
their valuable suggestion to the present paper Research issupported by National Natural Science Foundation of China(nos 11171119 and 11171121) and Guangdong National NaturalScience Foundation (no S2012040006865)
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[3] BGrammaticos T Tamizhmani A Ramani andKMTamizh-mani ldquoGrowth and integrability in discrete systemsrdquo Journal ofPhysics A vol 34 no 18 pp 3811ndash3821 2001
[4] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of Malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
[7] WGKelley andAC PetersonDifference Equations AcademicPress Boston Mass USA 1991
[8] I Laine and C C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[9] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[10] V I Gromak I Laine and S Shimomura Painleve DifferentialEquations in the Complex Plane vol 28 Walter de GruyterBerlin Germany 2002
[11] G G Gundersen ldquoFinite order solutions of second orderlinear differential equationsrdquo Transactions of the AmericanMathematical Society vol 305 no 1 pp 415ndash429 1988
[12] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[13] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions vol 557 Kluwer Academic Publishers Group Dor-drecht The Netherlands 2003
[14] Z X Chen and K H Shon ldquoOn growth of meromorphicsolutions for linear difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 619296 6 pages 2013
[15] G Jank and L Volkmann Einfuhrung in die Theorie derganzen und meromorphen Funktionen mit Anwendungen aufDifferentialgleichungen Birkhauser Basel Switzerland 1985
[16] J Wang ldquoGrowth and poles of meromorphic solutions of somedifference equationsrdquo Journal of Mathematical Analysis andApplications vol 379 no 1 pp 367ndash377 2011
[17] R G Halburd and R J Korhonen ldquoFinite-order meromorphicsolutions and the discrete Painleve equationsrdquoProceedings of theLondon Mathematical Society vol 94 no 2 pp 443ndash474 2007
[18] Y M Chiang and S J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[19] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
Research ArticleThe Regularity of Functions on Dual Split Quaternionsin Clifford Analysis
Ji Eun Kim and Kwang Ho Shon
Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Kwang Ho Shon khshonpusanackr
Received 28 January 2014 Accepted 2 April 2014 Published 17 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 J E Kim and K H Shon This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
This paper shows some properties of dual split quaternion numbers and expressions of power series in dual split quaternions andprovides differential operators in dual split quaternions and a dual split regular function on Ω sub C2 times C2 that has a dual splitCauchy-Riemann system in dual split quaternions
1 Introduction
Hamilton introduced quaternions extending complex num-bers to higher spatial dimensions in differential geometry (see[1]) A set of quaternions can be represented as
H = 119911 = 1199090+ 1199091119894 + 1199092119895 + 1199093119896 119909119898
isin R 119898 = 0 1 2 3
(1)
where 1198942
= 1198952
= 1198962
= minus1 119894119895119896 = minus1 and R denotes the set ofreal numbers Cockle [2] introduced a set of split quaternionsas
S = 119911 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 119909119898
isin R 119898 = 0 1 2 3
(2)
where 1198902
1= minus1 119890
2
2= 1198902
3= 1 and 119890
111989021198903
= 1 Aset of split quaternions is noncommutative and containszero divisors nilpotent elements and nontrivial idempotents(see [3 4]) Previous studies have examined the geometricand physical applications of split quaternions which arerequired in solving split quaternionic equations (see [5 6])Inoguchi [7] reformulated the Gauss-Codazzi equations informs consistent with the theory of integrable systems in theMinkowski 3-space for split quaternion numbers
A dual quaternion can be represented in a form reflect-ing an ordinary quaternion and a dual symbol Because
dual-quaternion algebra is constructed from real eight-dimensional vector spaces and an ordered pair of quater-nions dual quaternions are used in computer vision appli-cations Kenwright [8] provided the characteristics of dualquaternions and Pennestrı and Stefanelli [9] examined someproperties by using dual quaternions Son [10 11] offeredan extension problem for solutions of partial differentialequations and generalized solutions for the Riesz system Byusing properties of Hamilton operators Kula and Yayli [4]defined dual split quaternions and gave some properties ofthe screw motion in the Minkowski 3-space showing thatHhas a rotation with unit split quaternions in H and a scalarproduct that allows it to be identifiedwith the semi-Euclideanspace for split quaternion numbers
It was shown (see [12 13]) that any complex-valued har-monic function 119891
1in a pseudoconvex domain 119863 of C2 times C2
C being the set of complex numbers has a conjugate function1198912in 119863 such that the quaternion-valued function 119891
1+ 1198912119895 is
hyperholomorphic in119863 and gave a regeneration theorem in aquaternion analysis in view of complex and Clifford analysisIn addition we [14 15] provided a new expression of thequaternionic basis and a regular function on reduced quater-nions by associating hypercomplex numbers 119890
1and 119890
2 We
[16] investigated the existence of hyperconjugate harmonicfunctions of an octonion number system and we [17 18]obtained some regular functions with values in dual quater-nions and researched an extension problem for properties
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 369430 8 pageshttpdxdoiorg1011552014369430
2 Abstract and Applied Analysis
of regular functionswith values in dual quaternions and someapplications for such problems
This paper provides a regular function and some prop-erties of differential operators in dual split quaternions Inadditionwe research some equivalent conditions forCauchy-Riemann systems and expressions of power series in dual splitquaternions from the definition of dual split regular on anopen set Ω sub C2 times C2
2 Preliminaries
A dual number 119860 has the form 119886 + 120576119887 where 119886 and 119887 are realnumbers and 120576 is a dual symbol subject to the rules
120576 = 0 0120576 = 1205760 = 0 1120576 = 1205761 = 120576 1205762
= 0 (3)
and a split quaternion 119902 isin S is an expression of the form
119902 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 (4)
where 119909119898
isin R (119898 = 0 1 2 3) and 119890119903(119903 = 1 2 3) are split
quaternionic units satisfying noncommutativemultiplicationrules (for split quaternions see [1])
1198902
1= minus1 119890
2
2= 1198902
3= 1
11989011198902= minus11989021198901= 1198903 119890
21198903= minus11989031198902= minus1198901
11989031198901= minus11989011198903= 1198902
(5)
Similarly a dual split quaternion 119911 can be written as
D (S) = 119911 | 119911 = 1199010+ 1205761199011 119901119903isin S 119903 = 0 1 (6)
which has elements of the following form
119911 = (1199090+ 11990911198901) + (119909
2+ 11990931198901) 1198902
+ 120576 (1199100+ 11991011198901) + (119910
2+ 11991031198901) 1198902
= (1199110+ 11991111198902) + 120576 (119911
2+ 11991131198902)
= 1199010+ 1205761199011
(7)
where 1199010= 1199110+ 11991111198902and 119901
1= 1199112+ 11991131198902are split quaternion
components 1199110= 1199090+ 11990911198901 1199111= 1199092+ 11990931198901 1199112= 1199100+ 11991011198901
and 1199113= 1199102+ 11991031198901are usual complex numbers and 119909
119898 119910119898
isin
R (119898 = 0 1 2 3) The multiplication of split quaternionicunits with a dual symbol is commutative 120576119890
119903= 119890119903120576 (119903 =
1 2 3) However by properties of split quaternionic unit
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 0 1)
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 2 3)
119890119903119901119896
= 119901119896119890119903 119890119903119901119896= 119901(119896119903)
119890119903
(119903 = 1 2 3 119896 = 0 1)
(8)
where
119901(01)
= 1199110minus 11991111198902= 1199090+ 11990911198901minus 11990921198902minus 11990931198903
119901(02)
= 1199110+ 11991111198902= 1199090minus 11990911198901+ 11990921198902minus 11990931198903
119901(03)
= 1199110minus 11991111198902= 1199090minus 11990911198901minus 11990921198902+ 11990931198903
119901(11)
= 1199112minus 11991131198902= 1199100+ 11991011198901minus 11991021198902minus 11991031198903
119901(12)
= 1199112+ 11991131198902= 1199100minus 11991011198901+ 11991021198902minus 11991031198903
119901(13)
= 1199112minus 11991131198902= 1199100minus 11991011198901minus 11991021198902+ 11991031198903
(9)
with 1199110
= 1199090minus 11990911198901 1199111= 1199092minus 11990931198901 1199112
= 1199100minus 11991011198901 and
1199113= 1199102minus 11991031198901 For instance
11989021199010= 1198902(1199090+ 11990911198901+ 11990921198902+ 11990931198903)
= (1199090minus 11990911198901+ 11990921198902minus 11990931198903) 1198902= 119901(02)
1198902
11989011199011= 1198901(1199100+ 11991011198901+ 11991021198902+ 11991031198903)
= (1199100+ 11991011198901minus 11991021198902minus 11991031198903) 1198901= 119901(11)
1198901
(10)
Because of the properties of the eight-unit equality the addi-tion and subtraction of dual split quaternions are governedby the rules of ordinary algebra Here the symbol 119901
(119896119903)is used
by just enumerating 119903 and 119896 not 119903 times 119896 For example119901(22)
= 1199014and 119901
22= 1199014
For any two elements 119911 = 1199010+ 1205761199011and 119908 = 119902
0+ 1205761199021
of D(S) where 1199020= sum3
119903=0119904119903119890119903and 119902
1= sum3
119903=0119905119903119890119903are split
quaternion components and 119904119903 119905119903isin R (119903 = 0 1 2 3) their
noncommutative product is given by
119911119908 = (1199010+ 1205761199011) (1199020+ 1205761199021) = 11990101199020+ 120576 (119901
01199021+ 11990111199020)
(11)
The conjugation 119911lowast of 119911 and the corresponding modulus 119911119911lowast
inD(S) are defined by
119911lowast
= 119901lowast
0+ 120576119901lowast
1
119911119911lowast
= 119911lowast
119911 = 1199010119901lowast
0+ 120576 (119901
0119901lowast
1+ 1199011119901lowast
0)
= (11991101199110minus 11991111199111) + 2120576 (119911
01199112minus 11991111199113)
=
1
sum
119903=0
(1199092
119903minus 1199092
119903+2) + 120576 (119909
119903119910119903minus 119909119903+2
119910119903+2
)
(12)
where 119901lowast0= 1199110minus 11991111198902and 119901
lowast
1= 1199112minus 11991131198902
Lemma 1 For all 119911 isin D(S) and 119899 isin N = 1 2 3 wehave
119911119899
= 119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0 (13)
Abstract and Applied Analysis 3
Proof If 119899 = 1 then (13) is trivial Now suppose that thisholds for some 119899 isin N Then as desired
119911119899+1
= 119911119911119899
= 119911(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119901119899+1
0+ 120576
119899
sum
119896=1
119901119899minus119896+1
01199011119901119896minus1
0+ 1205761199011119901119899
0
= 119901119899+1
0+ 120576
119899+1
sum
119896=1
119901119899+1minus119896
01199011119901119896minus1
0
(14)
By the principle of mathematical induction (13) holds for all119899 isin N
Let Ω be an open subset of C2 times C2 Then the function119891 Ω rarr D(S) can be expressed as
119891 (119911) = 119891 (1199010 1199011) = 1198910(1199010 1199011) + 1205761198911(1199010 1199011) (15)
where the component functions 119891119903 Ω rarr S (119903 = 0 1) are
split quaternionic-valued functions The component func-tions 119891
119903(119903 = 0 1) are
1198910(1199010 1199011) = 1198910(1199110 1199111 1199112 1199113)
= 1198920(1199110 1199111 1199112 1199113) + 1198921(1199110 1199111 1199112 1199113) 1198902
1198911(1199010 1199011) = 1198911(1199110 1199111 1199112 1199113)
= 1198922(1199110 1199111 1199112 1199113) + 1198923(1199110 1199111 1199112 1199113) 1198902
(16)
where 119892119896
= 1199062119896
+ 1199062119896+1
1198901(119896 = 0 1) and 119892
119896= V2119896minus4
+
V2119896minus3
1198901(119896 = 2 3) are complex-valued functions and 119906
119903and
V119903(119903 = 0 1 2 3) are real-valued functionsNow we let differential operators 119863
1and 119863
2be defined
onD(S) as
1198631= 119863(11)
+ 120576119863(12)
1198632= 119863(21)
+ 120576119863(22)
(17)
Then the conjugate operators119863lowast1and119863
lowast
2are
119863lowast
1= 119863lowast
(11)+ 120576119863lowast
(12) 119863
lowast
2= 119863lowast
(21)+ 120576119863lowast
(22) (18)
where
119863(11)
=
120597
1205971199110
+
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
119863(12)
=
120597
1205971199112
+
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
120597
1205971199102
1198902+
120597
1205971199103
1198903)
119863(21)
=
120597
1205971199110
+
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
1
2
120597
1205971199092
1198902minus
1
2
120597
1205971199093
1198903)
119863(22)
=
120597
1205971199112
+
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
1
2
120597
1205971199102
1198902minus
1
2
120597
1205971199103
1198903)
(19)
119863lowast
(11)=
120597
1205971199110
minus
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
120597
1205971199092
1198902minus
120597
1205971199093
1198903)
119863lowast
(12)=
120597
1205971199112
minus
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
120597
1205971199102
1198902minus
120597
1205971199103
1198903)
119863lowast
(21)=
120597
1205971199110
minus
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
1
2
120597
1205971199092
1198902+
1
2
120597
1205971199093
1198903)
119863lowast
(22)=
120597
1205971199112
minus
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
(20)
act on D(S) These operators are called correspondingCauchy-Riemann operators in D(S) where 120597120597119911
119903and
120597120597119911119903(119903 = 0 1 2 3) are usual differential operators used in
the complex analysis
Remark 2 From the definition of differential operators onD(S)
119863119903119891 = (119863
(1199031)+ 120576119863(1199032)
) (1198910+ 1205761198911)
= 119863(1199031)
1198910+ 120576 (119863
(1199031)1198911+ 119863(1199032)
1198910)
119863lowast
119903119891 = (119863
lowast
(1199031)+ 120576119863lowast
(1199032)) (1198910+ 1205761198911)
= 119863lowast
(1199031)1198910+ 120576 (119863
lowast
(1199031)1198911+ 119863lowast
(1199032)1198910)
(21)
where 119903 = 1 2
Definition 3 LetΩ be an open set inC2 timesC2 A function 119891 =
1198910+ 1205761198911is called an 119871
119903(resp 119877
119903)-regular function (119903 = 1 2)
onΩ if the following two conditions are satisfied
(i) 119891119896(119896 = 0 1) are continuously differential functions
onΩ and
(ii) 119863lowast119903119891(119911) = 0 (resp 119891(119911)119863lowast
119903= 0) onΩ (119903 = 1 2)
In particular the equation 119863lowast
1119891(119911) = 0 of Definition 3 is
equivalent to
119863lowast
(11)1198910= 0 119863
lowast
(12)1198910+ 119863lowast
(11)1198911= 0 (22)
4 Abstract and Applied Analysis
In addition
1205971198920
1205971199110
minus
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1205971198923
1205971199111
minus
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1205971198922
1205971199111
minus
1205971198920
1205971199113
= 0
(23)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1205971199062
1205971199092
minus
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1205971199062
1205971199093
+
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1205971199060
1205971199092
minus
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
minus
1205971199060
1205971199093
+
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
120597V2
1205971199092
minus
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1205971199062
1205971199103
+
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1205971199060
1205971199102
minus
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
120597V0
1205971199092
minus
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
minus
1205971199060
1205971199103
+
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
minus
120597V0
1205971199093
+
120597V1
1205971199092
= 0
(24)
The above systems (23) and (24) are corresponding Cauchy-Riemann systems inD(S) Similarly the equation119863
lowast
2119891(119911) =
0 of Definition 3 is equivalent to
119863lowast
(21)1198910= 0 119863
lowast
(22)1198910+ 119863lowast
(21)1198911= 0 (25)
Then
1205971198920
1205971199110
minus
1
2
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1
2
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1
2
1205971198923
1205971199111
minus
1
2
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1
2
1205971198922
1205971199111
minus
1
2
1205971198920
1205971199113
= 0
(26)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
+
1
2
1205971199062
1205971199093
+
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
+
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
+
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1
2
1205971199062
1205971199103
+
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
+
1
2
120597V2
1205971199093
+
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
+
1
2
1205971199060
1205971199103
+
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
+
1
2
120597V0
1205971199093
+
1
2
120597V1
1205971199092
= 0
(27)
The above systems (26) and (27) are corresponding Cauchy-Riemann systems inD(S)
On the other hand the equation 119891(119911)119863lowast
1= 0 of
Definition 3 is equivalent to
1198910119863lowast
(11)= 0 119891
0119863lowast
(12)= minus1198911119863lowast
(11) (28)
Abstract and Applied Analysis 5
Then
1198920
120597
1205971199110
= 1198921
120597
1205971199111
1198921
120597
1205971199110
= 1198920
120597
1205971199111
1198920
120597
1205971199112
minus 1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+ 1198923
120597
1205971199111
1198921
120597
1205971199112
minus 1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+ 1198922
120597
1205971199111
(29)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(30)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
=
1205971199062
1205971199092
+
1205971199063
1205971199093
1205971199061
1205971199090
+
1205971199060
1205971199091
= minus
1205971199062
1205971199093
+
1205971199063
1205971199092
1205971199062
1205971199090
+
1205971199063
1205971199091
=
1205971199060
1205971199092
minus
1205971199061
1205971199093
1205971199063
1205971199090
minus
1205971199062
1205971199091
=
1205971199060
1205971199093
+
1205971199061
1205971199092
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
= minus
120597V0
1205971199090
+
120597V1
1205971199091
+
120597V2
1205971199092
+
120597V3
1205971199093
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1205971199062
1205971199103
minus
1205971199063
1205971199102
= minus
120597V1
1205971199090
minus
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1205971199060
1205971199102
+
1205971199061
1205971199103
= minus
120597V2
1205971199090
minus
120597V3
1205971199091
+
120597V0
1205971199092
minus
120597V1
1205971199093
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1205971199060
1205971199103
minus
1205971199061
1205971199102
= minus
120597V3
1205971199090
+
120597V2
1205971199091
+
120597V0
1205971199093
+
120597V1
1205971199092
(31)
Similarly the equation 119891(119911)119863lowast
2= 0 of Definition 3 is
equivalent to
1198910119863lowast
(21)= 0 119891
0119863lowast
(22)= minus1198911119863lowast
(21) (32)
Then
1198920
120597
1205971199110
=
1
2
1198921
120597
1205971199111
1198921
120597
1205971199110
=
1
2
1198920
120597
1205971199111
1198920
120597
1205971199112
minus
1
2
1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+
1
2
1198923
120597
1205971199111
1198921
120597
1205971199112
minus
1
2
1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+
1
2
1198922
120597
1205971199111
(33)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(34)
Concretely the system is obtained as follows
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1
2
1205971199062
1205971199093
minus
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
+
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
minus
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
minus
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
minus
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1
2
1205971199062
1205971199103
minus
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
1
2
120597V2
1205971199093
minus
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
+
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1
2
1205971199060
1205971199103
minus
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
minus
120597V2
1205971199091
minus
1
2
120597V0
1205971199093
minus
1
2
120597V1
1205971199092
= 0
(35)
From the systems (24) (27) (31) and (35) the equations119863lowast
119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 = 1 2) are different
Therefore the equations 119863lowast119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 =
1 2) should be distinguished as 119871119903-regular functions (119903 =
1 2) and 119877119903-regular functions (119903 = 1 2) on Ω respectively
Now the properties of the 119871119903-regular function (119903 = 1 2) with
values inD(S) are considered
3 Properties of 119871119903-Regular Functions (119903 = 1 2)
with Values in D(S)
We consider properties of a 119871119903-regular functions (119903 = 1 2)
with values inD(S)
Theorem 4 Let Ω be an open set in C2 times C2 and let 119891 =
1198910+ 1205761198911= (1198920+11989211198902)+ 120576(119892
2+11989231198902) be an 119871
1-regular function
defined on Ω Then
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (36)
6 Abstract and Applied Analysis
Proof By the system (23) we have
1198631119891 = 119863
(11)1198910+ 120576 (119863
(12)1198910+ 119863(11)
1198911)
= (
1205971198920
1205971199110
+
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+ 2
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+ 2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+ 2
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+ 2
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+ 2
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+ 2
1205971198922
1205971199111
) 1198902
= 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(37)
Therefore we obtain
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (38)
Theorem5 LetΩ be an open set inC2timesC2 and119891 = 1198910+1205761198911=
(1198920+11989211198902) + 120576(119892
2+11989231198902) be an 119871
2-regular function defined on
Ω Then
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (39)
Proof By the system (26) we have
1198632119891 = 119863
(21)1198910+ 120576 (119863
(22)1198910+ 119863(21)
1198911)
= (
1205971198920
1205971199110
+
1
2
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1
2
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(40)
Therefore we obtain the following equation
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (41)
Abstract and Applied Analysis 7
Proposition 6 From properties of differential operators thefollowing equations are obtained
119863(1119903)
119901119903minus1
= 2 119863(2119903)
119901119903minus1
= 1
119863lowast
(1119903)119901119903minus1
= minus1 119863lowast
(2119903)119901119903minus1
= 0
119863lowast
(1119903)119901lowast
119903minus1= 2 119863
lowast
(2119903)119901lowast
119903minus1= 1
119863(1199031)
1199011= 119863lowast
(1199031)1199011= 119863(1199031)
119901lowast
1= 119863lowast
(1199031)119901lowast
1
= 119863(1199032)
1199010= 119863lowast
(1199032)1199010= 119863(1199032)
119901lowast
0
= 119863lowast
(1199032)119901lowast
0= 0 (119903 = 1 2)
(42)
Proof By properties of the power of dual split quaternionsand derivatives on D(S) the following derivatives areobtained
119863(11)
1199010=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090+ 11990911198901+ 11990921198902+ 11990931198903) = 2
119863lowast
(22)1199011=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
times (1199100+ 11991011198901+ 11991021198902+ 11991031198903) = 0
119863(11)
119901lowast
0=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090minus 11990911198901minus 11990921198902minus 11990931198903) = minus1
(43)
The other equations are calculated using a similar methodand the above equations are obtained
Theorem 7 LetΩ be an open set in C2 timesC2 and let 119891(119911) be afunction on Ω with values inD(S) Then the power 119911119899 of 119911 inD(S) is not an 119871
1-regular function but an 119871
2-regular function
on Ω where 119899 isin N
Proof From the definition of the 119871119903-regular function (119903 =
1 2) on Ω and Proposition 6 we may consider whether thepower 119911119899 of 119911 in D(S) satisfies the equation 119863
lowast
119903119911119899
= 0 (119903 =
1 2) Since119863lowast(11)
1199010= 2
119863lowast
1119911119899
= (119863lowast
(11)+ 120576119863lowast
(12))(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119863lowast
(11)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(11)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(12)119901119899
0) = 0
(44)
Hence the power 119911119899 of 119911 is not 1198711-regular onΩ On the other
hand from the equations in Proposition 6 we have119863lowast(21)
1199010=
0119863lowast(21)
1199011= 0 and119863
lowast
(22)1199010= 0 Then
119863lowast
2119911119899
= 119863lowast
(21)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(21)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(22)119901119899
0) = 0
(45)
Therefore by the definition of the 119871119903-regular function (119903 =
1 2) onΩ a power 119911119899 of 119911 is 1198712-regular onΩ
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The second author was supported by the Basic ScienceResearch Program through the National Research Founda-tion of Korea (NRF) funded by the Ministry of Science ICTand Future Planning (2013R1A1A2008978)
References
[1] I L Kantor andA S SolodovnikovHypercomplex Numbers AnElementary Introduction to Algebras Springer New York NYUSA 1989
[2] J Cockle ldquoOn systems of algebra involving more than oneimaginaryrdquo PhilosophicalMagazine III vol 35 no 238 pp 434ndash435 1849
[3] M Ozdemir and A A Ergin ldquoRotations with unit timelikequaternions in Minkowski 3-spacerdquo Journal of Geometry andPhysics vol 56 no 2 pp 322ndash336 2006
[4] L Kula and Y Yayli ldquoSplit quaternions and rotations in semiEuclidean spaceE4
2rdquo Journal of the KoreanMathematical Society
vol 44 no 6 pp 1313ndash1327 2007[5] D C Brody and E-M Graefe ldquoOn complexified mechanics
and coquaternionsrdquo Journal of Physics A Mathematical andTheoretical vol 44 no 7 article 072001 2011
[6] I Frenkel and M Libine ldquoSplit quaternionic analysis andseparation of the series for SL(2R) and SL(2C)SL(2R)rdquoAdvances in Mathematics vol 228 no 2 pp 678ndash763 2011
[7] J-i Inoguchi ldquoTimelike surfaces of constant mean curvature inMinkowski 3-spacerdquo Tokyo Journal of Mathematics vol 21 no1 pp 141ndash152 1998
[8] B Kenwright ldquoA beginners guide to dual-quaternions whatthey are how they work and how to use them for 3D characterhierarchiesrdquo in Proceedings of the 20th International Conferenceson Computer Graphics Visualization and Computer Vision pp1ndash10 2012
[9] E Pennestrı and R Stefanelli ldquoLinear algebra and numericalalgorithms using dual numbersrdquo Multibody System Dynamicsvol 18 no 3 pp 323ndash344 2007
[10] L H Son ldquoAn extension problem for solutions of partialdifferential equations in R119899rdquo Complex Variables Theory andApplication vol 15 no 2 pp 87ndash92 1990
[11] L H Son ldquoExtension problem for functions with values in aClifford algebrardquo Archiv der Mathematik vol 55 no 2 pp 146ndash150 1990
[12] J Kajiwara X D Li and K H Shon ldquoRegeneration in complexquaternion and Clifford analysisrdquo in International Colloquiumon Finite or Infinite DimensionalComplex Analysis and ItsApplications vol 2 of Advances in Complex Analysis and ItsApplications no 9 pp 287ndash298 Kluwer Academic PublishersHanoi Vietnam 2004
[13] J Kajiwara X D Li and K H Shon ldquoFunction spaces incomplex and Clifford analysisrdquo in International Colloquium
8 Abstract and Applied Analysis
on Finite Or Infinite Dimensional Complex Analysis and ItsApplications vol 14 of Inhomogeneous Cauchy Riemann Systemof Quaternion andCliffordAnalysis in Ellipsoid pp 127ndash155HueUniversity Hue Vietnam 2006
[14] J E Kim S J Lim and K H Shon ldquoRegular functions withvalues in ternary number system on the complex Cliffordanalysisrdquo Abstract and Applied Analysis vol 2013 Article ID136120 7 pages 2013
[15] J E Kim S J Lim and K H Shon ldquoRegularity of functions onthe reduced quaternion field in Clifford analysisrdquo Abstract andApplied Analysis vol 2014 Article ID 654798 8 pages 2014
[16] S J Lim and K H Shon ldquoHyperholomorphic fucntions andhyperconjugate harmonic functions of octonion variablesrdquoJournal of Inequalities andApplications vol 2013 article 77 2013
[17] S J Lim and K H Shon ldquoDual quaternion functions and itsapplicationsrdquo Journal of Applied Mathematics vol 2013 ArticleID 583813 6 pages 2013
[18] J E Kim S J Lim and K H Shon ldquoTaylor series of functionswith values in dual quaternionrdquo Journal of the Korean Society ofMathematical Education BmdashThePure and AppliedMathematicsvol 20 no 4 pp 251ndash258 2013
Research ArticleUnicity of Meromorphic Functions Sharing Sets withTheir Linear Difference Polynomials
Sheng Li and BaoQin Chen
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to BaoQin Chen chenbaoqin chbq126com
Received 23 January 2014 Accepted 20 March 2014 Published 13 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 S Li and B Chen This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We mainly investigate the unicity of meromorphic functions sharing two or three sets with their linear difference polynomials andprove some results
1 Introduction and Main Results
In this paper we assume the reader is familiar with the fun-damental results and the basic notations of the Nevanlinnatheory of meromorphic functions (see eg [1ndash3]) Let 119891(119911)be meromorphic in the whole plane We use the notation120588(119891) to denote the order of growth of the meromorphicfunction 119891(119911) In addition we denote by 119878(119903 119891) any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin outside of apossible exceptional set of finite logarithmic measure We saythat a meromorphic function 119886(119911) is a small function of 119891(119911)provided that 119879(119903 119886) = 119878(119903 119891) Let 119878(119891) be the set of all smallfunctions of 119891(119911)
For a set 119878 sub 119878(119891) we define the following
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 counting multiplicities
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 ignoring multiplicities
(1)
Let 119891 and 119892 be meromorphic functions If 119864119891(119878) = 119864
119892(119878)
and 119864119891(119878) = 119864
119892(119878) respectively then we say that 119891 and 119892
share a set 119878 CM and IM respectivelyFurthermore let 119888 be a nonzero complex constant We
define the shift of 119891(119911) by 119891(119911 + 119888) and define the differenceoperators of 119891(119911) by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911)
Δ119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911)) 119899 isin N 119899 ge 2
(2)
Theunicity theory ofmeromorphic functions sharing setsis an important topic of the uniqueness theory First of all werecall the following theorem given by Li and Yang in [4]
Theorem A (see [4]) Let 119898 ge 2 and let 119899 gt 2119898 + 6 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0Then for any two nonconstant meromorphic functions 119891 and119892 the conditions 119864
119891(119878) = 119864
119892(119878) and 119864
119891(infin) = 119864
119892(infin)
imply 119891 = 119892
Yi and Lin considered the case 119898 = 1 with the conditionthat two meromorphic functions share three sets and got theresult as follows
Theorem B (see [5]) Let 1198781= 120596 120596
119899
+ 119886120596119899minus1
+ 119887 = 01198782= 0 and 119878
3= infin where 119886 119887 are nonzero constants such
that 120596119899 + 119886120596119899minus1 + 119887 = 0 has no repeated root and 119899(ge 4) is aninteger If for two nonconstant meromorphic functions 119891 and119892 119864119891(119878119895) = 119864
119892(119878119895) for 119895 = 1 2 3 and Θ(infin119891) gt 0 then
119891 equiv 119892
Recently a number of papers have focused on differenceanalogues of the Nevanlinna theory (see eg [6ndash9]) Inparticular there has been an increasing interest in studyingthe uniqueness problems related to meromorphic functionsand their shifts or their difference operators (see eg [10ndash16])
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 894968 7 pageshttpdxdoiorg1011552014894968
2 Abstract and Applied Analysis
In 2010 Zhang considered a meromorphic function 119891(119911)sharing sets with its shift 119891(119911 + 119888) and proved the followingresult
TheoremC (see [16]) Let119898 ge 2 and let 119899 ge 2119898+4with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596119899 + 119886120596119899minus119898 + 119887 = 0 Suppose that119891(119911) is a nonconstant meromorphic function of finite orderThen 119864
119891(119911)(119878) = 119864
119891(119911+119888)(119878) and 119864
119891(119911)(infin) = 119864
119891(119911+119888)(infin)
imply 119891(119911) equiv 119891(119911 + 119888)
For an analogue result in difference operator B Chen andZ Chen proved the following theorem in [10]
Theorem D (see [10]) Let 119898 ge 2 and let 119899 ge 2119898 + 4 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has nomultiple roots Let 119878 = 120596 | 120596119899+119886120596119899minus119898+119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying 119864
119891(119911)(119878) = 119864
Δ119888119891(119878) and 119864
119891(119911)(infin) = 119864
Δ119888119891(infin) If
119873(119903
1
Δ119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (3)
then Δ119888119891 equiv 119891(119911)
It is natural to ask what happens if the shift 119891(119911 + 119888) ordifference operatorΔ
119888119891(119911) is replaced by a general expression
of 119891(119911) such as a linear difference polynomial of 119891(119911)Here a linear difference polynomial of 119891(119911) is an expres-
sion of the form
119871 (119911 119891) = 119887119896(119911) 119891 (119911 + 119888
119896) + sdot sdot sdot + 119887
0(119911) 119891 (119911 + 119888
0) (4)
where 119887119896(119911) equiv 0 119887
0(119911) 119887
119896(119911) are small functions of
119891(119911) 1198880 119888
119896are complex constants and 119896 is a nonnegative
integerIn this paper our aim is to investigate the uniqueness
problems of linear difference polynomials of 119891(119911) In partic-ular we primarily consider the linear difference polynomial119871(119911 119891) which satisfies one of the following conditions
(i) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1
(ii) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0
119873(119903
1
119871 (119911 119891)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(5)
Corresponding to the above question we obtain thefollowing results
Theorem 1 Let 119898 ge 2 and let 119899 ge 2119898 + 4 with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite orderand 119871(119911 119891) is of the form (4) satisfying the condition in (5)If 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) and 119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) then119871(119911 119891) equiv 119891(119911)
Corollary 2 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying the following
119873(119903
1
Δ119896
119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (6)
If 119864119891(119911)
(119878) = 119864Δ119896
119888119891(119911)
(119878) and 119864119891(119911)
(infin) = 119864Δ119896
119888119891(119911)
(infin) thenΔ119896
119888119891 equiv 119891(119911)
With an additional restriction on the order of growth of119891(119911) we prove the following fact
Theorem 3 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersuch that 120588(119891) notin N If 119864
119891(119911)(119878) = 119864
119871(119911119891)(119878) and 119864
119891(119911)(infin) =
119864119871(119911119891)
(infin) then 119871(119911 119891) equiv 119891(119911)
Remark 4 Note that in Theorem 3 we do not assume thatthe linear polynomial 119871(119911 119891) satisfies the condition in (5)In fact since 120588(119891) notin N by (19) we can easily get 120588(119890ℎ(119911)) =deg(ℎ(119911)) lt 120588(119891) which implies 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Thenusing a similar method as in the proof of Theorem 1 we cancomplete the proof of Theorem 3
Now we may ask what happens if the condition119898 ge 2 inTheorem 1 is replaced by a weaker condition containing thecase 119898 = 1 or even 119898 = 0 By considering three sets we getthe following theorem
Theorem 5 Let 119899119898 be nonnegative integers such that 119899 gt 119898Let 119886 and 119887 be nonzero constants such that 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878
1= 120596 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 =1198782= infin and 119878
3= 0 Suppose that 119891(119911) is a nonconstant
meromorphic function of finite order 119871(119911 119891) is of the form (4)satisfying the condition in (5) and 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for
119895 = 1 2 3 Then one has the following(i) If119898 = 0 then 119871(119911 119891) equiv 119905119891(119911) where 119905119899 = 1(ii) If 119899 and119898 are coprime then 119871(119911 119891) equiv 119891(119911)
Remark 6 Taking 119898 = 1 in Theorem 5 we can obtain ananalogue result of Theorem B related to linear differencepolynomials
Furthermore the following result is a corollary ofTheorem 5 related to difference operators
Corollary 7 Let 119899 119898 and 119878119895 119895 = 1 2 3 be given as in
Theorem 5 Suppose that 119891(119911) is a nonconstant meromorphicfunction of finite order satisfying
119873(119903
1
119891 (119911)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (7)
and 119864119891(119911)
(119878119895) = 119864
Δ119896
119888119891(119878119895) for 119895 = 1 2 3 Then one has the
following
(i) If119898 = 0 then Δ119896119888119891 equiv 119905119891(119911) where 119905119899 = 1
(ii) If 119899 and119898 are coprime then Δ119896119888119891 equiv 119891(119911)
Finally we give some examples for our results
Abstract and Applied Analysis 3
Examples In the following let 119892(119911) be an entire functionwith period 1 such that 120588(119892) isin (1infin) N (see [17])
(1) For the case (i) of condition (5) let 1198911(119911) = 119890
21205871198941199111198912(119911) = 119892(119911)119890
2120587119894119911 1198913(119911) = 119890
2120587119894119911
119892(119911) and let119871(119911 119891
119895) = 2119891
119895(119911) minus 119891
119895(119911 + 1) Then for 119895 = 1 2 3
119871(119911 119891119895) = 119891
119895(119911) and the sum of the coefficients of
119871(119911 119891119895) is equal to 1These examples satisfyTheorems
1 and 5 but do not satisfy Theorem D
(2) For the case (ii) of condition (5) let 119891(119911) = 119890119911 log 2119892(119911)and let 119871(119911 119891) = Δ119891(119911) = 119891(119911 + 1) minus 119891(119911) Then119871(119911 119891) = Δ119891(119911) = 119891(119911) the sum of the coefficientsof 119871(119911 119891) equals 0 and
119873(119903
1
Δ119891
) = 119873(119903
1
119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(8)
This example satisfiesTheorems 1 and 5 and Corollar-ies 2 and 7
(3) For Theorem 3 let 119891(119911) = 119890119911 log 3
119892(119911) and let119871(119911 119891) = 119891(119911 + 1) minus 2119891(119911) Then 119871(119911 119891) = 119891(119911)
and the sum of the coefficients of 119871(119911 119891) equals minus1This example satisfies Theorem 3 but does not satisfyTheorem D andTheorems 1 and 5
2 Proof of Theorem 1
We need the following lemmas for the proof of Theorem 1The difference analogue of the logarithmic derivative
lemmawas given byHalburd-Korhonen [7] andChiang-Feng[6] independently We recall the following lemmas
Lemma 8 (see [7]) Let 119891(119911) be a nonconstant meromorphicfunction of finite order 119888 isin C and 120575 lt 1 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 + |119888| 119891)
119903120575
) (9)
for all 119903 outside of a possible exceptional set with finite logarith-mic measure
Lemma 9 (see [8]) Let 119888 isin C let 119899 isin N and let 119891(119911) bea meromorphic function of finite order Then for any smallperiodic function 119886(119911) isin 119878(119891) with period 119888 consider thefollowing
119898(119903
Δ119899
119888119891
119891 (119911) minus 119886 (119911)
) = 119878 (119903 119891) (10)
where the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Let 119891(119911) be a meromorphic function of finite orderNotice that if 119871(119911 119891) ( equiv 0) is of the form (4) such that
1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 then for any given complex con-
stant 119886 119871(119911 119886) = 0 This indicates that 119871(119911 119891) = 119871(119911 119891 minus 119886)
and hence
119898(119903
119871 (119911 119891)
119891 minus 119886
) = 119898(119903
119871 (119911 119891 minus 119886)
119891 minus 119886
)
le
119896
sum
119895=0
119898(119903
119887119895(119911) (119891 (119911 + 119888
119895) minus 119886)
119891 minus 119886
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(11)
With this one can easily prove Lemma 10 below by a similarreasoning as in the proof of the difference analogue of thesecond main theorem of the Nevanlinna theory in [8] byHalburd and Korhonen We omit those details
Lemma 10 Let 119888 isin C let 119891(119911) be a meromorphic function offinite order and let 119871(119911 119891) equiv 0 be of the form (4) such that1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 Let 119902 ge 2 and let 119886
1 119886
119902be distinct
complex constants Then
119898(119903 119891) +
119902
sum
119894=1
119898(119903
1
119891 minus 119886119894
)
le 2119879 (119903 119891) minus 119873lowast
(119903 119891) + 119878 (119903 119891)
(12)
where
119873lowast
(119903 119891) = 2119873 (119903 119891) minus 119873 (119903 119871 (119911 119891)) + 119873(119903
1
119871 (119911 119891)
)
(13)
and the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Remark 11 If the linear difference polynomial 119871(119911 119891) isreplaced by
119871lowast
(119911 119891) = 119887119896(119911) 119891 (119911 + 119896119888)
+ sdot sdot sdot + 1198871(119911) 119891 (119911 + 119888) + 119887
0(119911) 119891 (119911)
(14)
Lemma 10 also holds even if the distinct complex constants1198861 119886
119902are replaced by 119886
1(119911) 119886
119902(119911) which are distinct
meromorphic periodic functions with period 119888 such that 119886119894isin
119878(119891) for all 119894 = 1 119902
The following is the standardValiron-Mohonrsquoko theorem(see Theorem 225 in the book of Laine [2])
Lemma 12 (see [2]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(15)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(16)
4 Abstract and Applied Analysis
The characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (17)
where 119889 = max119901 119902
Proof of Theorem 1 Since 119891(119911) and 119871(119911 119891) shareinfin CM wesee that 119871(119911 119891) equiv 0 and119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) Then byLemma 8 we have
119879 (119903 119871 (119911 119891)) = 119898 (119903 119871 (119911 119891)) + 119873 (119903 119871 (119911 119891))
le 119898(119903
119871 (119911 119891)
119891 (119911)
)
+ 119898 (119903 119891 (119911)) + 119873 (119903 119891 (119911))
le
119896
sum
119894=0
119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
+
119896
sum
119894=0
119898(119903 119887119894(119911)) + 119879 (119903 119891 (119911))
le 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(18)
Since 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) where 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots we know that (119871(119911 119891))119899+119886(119871(119911 119891))119899minus119898+119887 and119891(119911)119899+119886119891(119911)119899minus119898
+ 119887 share 0 CM Then from this and the condition119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) there exists a polynomial ℎ(119911)such that
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
= 119890ℎ(119911)
(19)
Suppose that 119890ℎ(119911) equiv 1 Note that 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots Let 120596
1 120596
119899denote all different roots of the equation
120596119899
+ 119886120596119899minus119898
+ 119887 = 0Next we prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) We know that
119871 (119911 119891) minus 120596119894= 119887119896(119911) (119891 (119911 + 119888
119896) minus 119891 (119911))
+ sdot sdot sdot + 1198870(119911) (119891 (119911 + 119888
0) minus 119891 (119911))
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
(20)
(i) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1 we see that
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 + (119891 (119911) minus 120596119894)
(21)
Then we deduce from this (19) and Lemma 9 that
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
= 119898(119903
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
)
= 119898(119903
(119871 (119911 119891) minus 1205961) sdot sdot sdot (119871 (119911 119891) minus 120596
119899)
(119891 (119911) minus 1205961) sdot sdot sdot (119891 (119911) minus 120596
119899)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(22)
(ii) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 we have
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 minus 120596119894 (23)
From this (19) and Lemma 9 we get
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
=
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
(24)
Applying Lemma 10 to 119891(119911) we get
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
)
le 2119879 (119903 119891 (119911)) minus 119898 (119903 119891 (119911)) minus 2119873 (119903 119891 (119911))
+ 119873 (119903 119871 (119911 119891)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
= 119879 (119903 119891 (119911)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
(25)
Abstract and Applied Analysis 5
Then the assumptions in (5) (24) and (25) yield thefollowing
119879 (119903 119890ℎ(119911)
) le 119879 (119903 119891 (119911))
minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891) = 119878 (119903 119891)
(26)
To sum up we now prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Rewriting(19) we get
(119871 (119911 119891))119899minus119898
[(119871 (119911 119891))119898
+ 119886]
= [119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887 minus 119887119890minusℎ(119911)
] 119890ℎ(119911)
(27)
Denote 119865(119911) = 119891(119911)119899
+ 119886119891(119911)119899minus119898 It follows from
Lemma 12 and119898 gt 0 that
119879 (119903 119865 (119911)) = 119899119879 (119903 119891 (119911)) + 119878 (119903 119891) (28)
Hence 119878(119903 119865) = 119878(119903 119891)By (18) and (27) and applying the second main theorem
for three small target functions we deduce the following
119879 (119903 119865 (119911))
le 119873 (119903 119865 (119911)) + 119873(119903
1
119865 (119911)
)
+ 119873(119903
1
119865 (119911) + 119887 minus 119887119890minus119901(119911)
) + 119878 (119903 119865)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891(119911)119899minus119898
[119891(119911)119898
+ 119886]
)
+ 119873(119903
1
(119871 (119911 119891))119899minus119898
) + 119873(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891 (119911)
) + 119873(119903
1
119891(119911)119898
+ 119886
)
+ 119873(119903
1
119871 (119911 119891)
) + 119879(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119879 (119903 119891 (119911)) + 119879(119903
1
119891 (119911)
) + 119879(119903
1
119891(119911)119898
+ 119886
)
+ 119879(119903
1
119871 (119911 119891)
) + 119898119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (119898 + 2) 119879 (119903 119891 (119911))
+ (119898 + 1) 119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (2119898 + 3) 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(29)
By combining (28) and (29) we have
(119899 minus 2119898 minus 3) 119879 (119903 119891 (119911)) le 119878 (119903 119891) (30)
which contradicts with 119899 ge 2119898 + 4Now we turn to consider the case 119890ℎ(119911) equiv 1 Equation (19)
yields the following
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
equiv 119891(119911)119899
+ 119886119891(119911)119899minus119898
(31)
Set 120593(119911) = 119871(119911 119891)119891(119911) and we have
119891(119911)119898
(120593(119911)119899
minus 1) = minus119886 (120593(119911)119899minus119898
minus 1) (32)
If 120593(119911) is not a constant (32) can be rewritten as
119891(119911)119898
(120593 (119911) minus 1) (120593 (119911) minus 120583) sdot sdot sdot (120593 (119911) minus 120583119899minus1
)
= minus119886 (120593 (119911) minus 1) (120593 (119911) minus ]) sdot sdot sdot (120593 (119911) minus ]119899minus119898minus1) (33)
where 120583 = cos(2120587119899) + 119894 sin(2120587119899) and ] = cos(2120587(119899minus119898)) +119894 sin(2120587(119899 minus 119898))
By the assumption that 119899 and 119899 minus 119898 have no commonfactors we see that 120583 120583119899minus1 ] ]119899minus119898minus1 are differentAssume that 119911
0is a 120583119895-point of 120593(119911) of multiplicity 119906
119895gt 0
where 1 le 119895 le 119899 minus 1 Notice that
minus119886 (120593 (1199110) minus 1) (120593 (119911
0) minus ]) sdot sdot sdot (120593 (119911
0) minus ]119899minus119898minus1) (34)
is a constantThen (33) implies that 1199110is a pole of119891(119911)119898Thus
119906119895ge 119898 This yields the following for 1 le 119895 le 119899 minus 1
119898119873(119903
1
120593 (119911) minus 120583119895
) le 119873(119903
1
120593 (119911) minus 120583119895
)
le 119879 (119903 120593 (119911)) + 119878 (119903 ℎ)
(35)
Then by (35) we get
2 ge
119899minus1
sum
119895=1
Θ(120583119895
120593 (119911)) =
119899minus1
sum
119895=1
1 minus lim119903rarrinfin
119873(119903 1 (120593 (119911) minus 120583119895
))
119879 (119903 120593 (119911))
ge
119899minus1
sum
119895=1
(1 minus
1
119898
) = (119899 minus 1) (1 minus
1
119898
)
(36)
which is impossible with119898 ge 2 and 119899 ge 2119898 + 4Hence 120593(119911) is a constant Since 119891(119911) is a nonconstant
meromorphic function we deduce from (32) that 120593(119911) equiv 1This yields 119871(119911 119891) equiv 119891(119911) which completes the proof ofTheorem 1
3 Proof of Theorem 5
Since 119891(119911) is a nonconstant meromorphic function of finiteorder 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for 119895 = 1 2 3 119878
1= 120596 120596
119899
+
119886120596119899minus119898
+ 119887 = 0 1198782= infin and 119878
3= 0 we have 119871(119911 119891) equiv
0 119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) and 119873(119903 1119871(119911 119891)) = 119873(119903
1119891(119911)) and we also get (18) and (19)
6 Abstract and Applied Analysis
Since 119891(119911) and 119871(119911 119891) share 0 infin CM there exists apolynomial ℎlowast(119911) such that
119871 (119911 119891)
119891 (119911)
= 119890ℎlowast(119911)
(37)
By Lemma 8 we see that
119879(119903 119890ℎlowast(119911)
) = 119898(119903 119890ℎlowast(119911)
) = 119898(119903
119871 (119911 119891)
119891 (119911)
)
le
119896
sum
119895=0
119898(119903
119891 (119911 + 119888119895)
119891 (119911)
)
+
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(38)
As in the proof of Theorem 1 we see that 119879(119903 119890ℎ(119911)) =
119878(119903 119891) still holds in both cases (i) and (ii)Rewriting (19) we have
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
minus 119890ℎ(119911)
119891(119911)119899
minus 119886119890ℎ(119911)
119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(39)
Combining this and (37) we get
(119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
+ 119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(40)
Suppose that 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 If119898 = 0 (40) becomes
(119886 + 1) (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
= 119887 (119890ℎ(119911)
minus 1) (41)
By the condition that 119887 = 0 1198781= 120596 (119886 + 1)120596
119899
+ 119887 = 0 =
implies 119886 = minus 1It follows from (38) (41) and 119879(119903 119890ℎ(119911)) = 119878(119903 119891) that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 119887 (119890ℎ(119911)
minus 1)) = 119878 (119903 119891)
(42)
which is a contradiction since 119899 ge 1If119898 ge 1 it follows from (38) (41) and119879(119903 119890ℎ(119911)) = 119878(119903 119891)
that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 minus119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
) 119891(119911)119899minus119898
+ 119887 (119890ℎ(119911)
minus 1))
le (119899 minus 119898)119879 (119903 119891) + 119878 (119903 119891)
(43)
That is impossible
Therefore 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 Notice that 119886 119887 = 0 Using asimilar method we can prove that 119890(119899minus119898)ℎ
lowast(119911)
minus119890ℎ(119911)
equiv 0Then(40) implies that 119890ℎ(119911) equiv 1
If 119898 = 0 we have 119890119899ℎlowast(119911)
equiv 1 Obviously 119890ℎlowast(119911) is a
constant Set 119905 = 119890ℎlowast(119911) Thus by (37) we get 119871(119911 119891) equiv 119905119891(119911)
where 119905119899 = 1If 119899 and 119898 are coprime 119890119899ℎ
lowast(119911)
equiv 1 and 119890119898ℎlowast(119911)
equiv 1 implythat 119890ℎ
lowast(119911)
equiv 1 Thus by (37) we get 119871(119911 119891) equiv 119891(119911) ThusTheorem 5 is proved
Conflict of Interests
The authors declare that they have no conflict of interests
Authorsrsquo Contribution
Both authors drafted the paper and read and approved thefinal paper
Acknowledgments
The authors are grateful to the editor and referees fortheir valuable suggestions This work was supported by theNNSFC (no 11171119 11301091) the Guangdong Natural Sci-ence Foundation (no S2013040014347) and the Foundationfor Distinguished Young Talents in Higher Education ofGuangdong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 2003
[4] P Li and C-C Yang ldquoSome further results on the unique rangesets of meromorphic functionsrdquo Kodai Mathematical Journalvol 18 no 3 pp 437ndash450 1995
[5] H-X Yi and W-C Lin ldquoUniqueness theorems concerning aquestion of Grossrdquo Proceedings of the Japan Academy Series AMathematical Sciences vol 80 no 7 pp 136ndash140 2004
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[8] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[9] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
Abstract and Applied Analysis 7
[10] B Chen and Z Chen ldquoMeromorphic function sharing twosets with its difference operatorrdquo Bulletin of the MalaysianMathematical Sciences Society Second Series vol 35 no 3 pp765ndash774 2012
[11] B Chen Z Chen and S Li ldquoUniqueness theorems on entirefunctions and their difference operators or shiftsrdquo Abstract andApplied Analysis vol 2012 Article ID 906893 8 pages 2012
[12] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[13] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[14] S Li and B Chen ldquoMeromorphic functions sharing smallfunctions with their linear difference polynomialsrdquoAdvances inDifference Equations vol 2013 article 58 2013
[15] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[16] J Zhang ldquoValue distribution and shared sets of differences ofmeromorphic functionsrdquo Journal of Mathematical Analysis andApplications vol 367 no 2 pp 401ndash408 2010
[17] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleA Comparison Theorem for Oscillation of the Even-OrderNonlinear Neutral Difference Equation
Quanxin Zhang
Department of Mathematics Binzhou University Binzhou Shandong 256603 China
Correspondence should be addressed to Quanxin Zhang 3314744163com
Received 9 December 2013 Accepted 28 March 2014 Published 10 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Quanxin Zhang This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
A comparison theorem on oscillation behavior is firstly established for a class of even-order nonlinear neutral delay differenceequations By using the obtained comparison theorem two oscillation criteria are derived for the class of even-order nonlinearneutral delay difference equations Two examples are given to show the effectiveness of the obtained results
1 Introduction
Recently there have been a lot of research papers in con-nection with the oscillation of solutions of difference equa-tions with or without neutral terms The literature on theoscillation of neutral delay difference equations is growingvery fast and it can be widely applied to the reality Infact neutral delay difference equations arise in modellingof the networks containing lossless transmission lines (asin high speed computers where the lossless transmissionlines are used to interconnect switching circuits) For recentcontributions regarding the theoretical part and providingsystematic treatment of oscillation of solutions of neutraltype difference equations the readers can refer to the recentmonographs by Agarwal [1] Gyori and Ladas [2]
The oscillation behavior of the even-order nonlinearneutral differential equation
[119909 (119905) + 119901 (119905) 119909 (120591 (119905))](119899)
+ 119902 (119905) 119891 [119909 (120590 (119905))] = 0 (1)
has been established by Zhang et al [3] In this paperthe discrete analogue of the above equation is consideredWe consider the even-order nonlinear neutral differenceequation
Δ119898
(119909119899+ 119901119899119909119899minus120591
) + 119902119899119891 (119909119899minus119896
) = 0 (2)
where 119898 ge 2 is an even and 120591 119896 isin N let N denote the setof all natural numbers 119899 isin 119873(119899
0) = 119899
0 1198990+ 1 1198990+ 2
1198990is a nonnegative integer Δ denotes the forward difference
operator defined by Δ119909119899= 119909119899+1
minus 119909119899 Δ119898119909
119899= Δ119898minus1
119909119899+1
minus
Δ119898minus1
119909119899
Throughout this paper the following conditions areassumed to hold
(H1) 119901119899 is a sequence of nonnegative real number 0 le
119901119899
lt 1 and 119902119899 is a sequence of nonnegative real
number with 119902119899 being not eventually identically
equal to zero(H2) 119891 R rarr R (R = (minusinfin +infin)) is a continuous oddfunction and 119909119891(119909) gt 0 for all 119909 = 0
Before deriving themain results the following definitionsare given
Definition 1 By a solution of (2) one means a real sequence119909119899 defined for 119899 ge 119899
0minus120579 (120579 = max120591 119896)which satisfies (2)
for 119899 isin 119873(1198990)
In this paper we restrict our attention to nontrivialsolutions of (2)
Definition 2 A nontrivial solution 119909119899 of (2) is said to
be oscillatory if the terms 119909119899of the sequence are neither
eventually positive nor eventually negative Otherwise it iscalled nonoscillatory
Definition 3 An equation is said to be oscillatory if all itssolutions are oscillatory
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 492492 5 pageshttpdxdoiorg1011552014492492
2 Abstract and Applied Analysis
In 2004 Stavroulakis [4] studied the oscillatory behaviorof all solutions of first-order delay difference equation
119909119899+1
minus 119909119899+ 119901119899119909119899minus119896
= 0 (3)
and established one new oscillation criterion Thandapaniet al [5] studied the oscillatory behavior of all solutions ofsecond-order neutral delay difference equation
Δ2
(119910119899minus 119901119910119899minus119896
) minus 119902119899119891 (119910119899minus119905
) = 0 (4)
and established a number of new oscillation criteria In 2000Zhou et al [6] studied the oscillatory behavior of all solutionsof even-order neutral delay difference equation
Δ119898
(119909119899minus 119901119899119892 (119909119899minus119896
)) minus 119902119899ℎ (119909119899minus119897
) = 0 (5)
and established three new oscillation criteria under certainconditionsThe studies on oscillatory behavior of all solutionsof even-order delay difference equations we recommendreferring to [7ndash10] On the basis of the abovework we studiedthe oscillatory behavior of all solutions of (2) Firstly a com-parison theorem on oscillation behavior is established for (2)The comparison theorem changes the discriminant criteria ofthe oscillation of (2) into the oscillationrsquos discriminant criteriain the first-order nonneutral delay difference equationsThenby using the above comparison theorem we obtain someoscillation criteria for (2) and improve the well-known resultsof Ladas et al [11] Erbe and Zhang [12] and Stavroulakis [4]In particular the results are new when119898 = 2 119901
119899equiv 0
The paper is organized as follows In Section 2 a compar-ison theorem on oscillation behavior is firstly established fora class of even-order nonlinear neutral delay difference equa-tions Then the comparison theorem changes the discrimi-nant of the oscillation in the even-order nonlinear neutraldelay difference equation into the oscillationrsquos discriminantin the first-order nonneutral delay difference equations InSection 3 some oscillation criteria are obtained for the classof even-order nonlinear neutral delay difference equationby using the above comparison theorem In Section 4 twoexamples are given
2 Comparison Theorem
To obtain the comparison theorem in this section we needthe following lemmas which can be founded in [1] see alsoChen [7] andThandapani and Arul [8]
Lemma 4 Let 119906119899 be a sequence of real numbers for 119899 ge 119899
0
Let 119906119899 and Δ
119898
119906119899 be of constant sign where Δ
119898
119906119899is not
identically zero for 119899 ge 1198991 If
119906119899Δ119898
119906119899le 0 (6)
then
(i) there is a natural number 1198992
ge 1198991such that the
sequences Δ119895119906119899 119895 = 1 2 119898 minus 1 are of constant
sign for 119899 ge 1198992
(ii) there exists a number 119897 isin 0 1 119898 minus 1 with(minus1)119898minus119897minus1
= 1 such that
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 0 1 2 119897 119899 ge 119899
2
(minus1)119895minus119897
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 119897 + 1 119898 minus 1 119899 ge 119899
2
(7)
Lemma 5 Observe that under the hypotheses of Lemma 4 if119906119899 is increasing for 119899 ge 119899
0 then there exists a natural number
1198991ge 1198990such that for all 119899 ge 2
119898minus1
1198991
119906119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119906119899 (8)
where 120582119898= 12(119898minus1)
2
Theorem 6 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 If there exists a constant
120582119898= 12(119898minus1)
2
such that the first-order difference equation
Δ119911119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119911119899minus119896
= 0 (9)
is oscillatory then (2) is oscillatory
Proof Suppose that (2) has a nonoscillatory solution 119909119899
Without the loss of generality we assume that 119909119899 is an
eventually positive solution of (2) then there is a naturalnumber 119899
1ge 1198990such that 119909
119899gt 0 119909
119899minus119896gt 0 119909
119899minus120591gt 0 and
119909119899minus119896minus120591
gt 0 for all 119899 ge 1198991 Let
119910119899= 119909119899+ 119901119899119909119899minus120591
(10)
Then from (H1) and (H
2) there exists a natural number 119899
2ge
1198991such that
119910119899gt 0 Δ
119898
119910119899le 0 forall119899 ge 119899
2 (11)
By Lemma 4 there exist an integer 1198993ge 1198992and an integer
119897 (0 le 119897 le 119898) where (119898 + 119897) is an odd integer For all 119899 ge 1198993
we can get
Δ119895
119910119899gt 0 for 119895 = 1 2 119897
(minus1)(119895minus119897)
Δ119895
119910119899gt 0 for 119895 = 119897 + 1 119898 minus 1
(12)
Thus from (12) Δ119910119899
gt 0 and Δ119898minus1
119910119899
gt 0 for 119899 ge 1198993 By
Lemma 5 there exists an integer 1198994ge 1198993 For all 119899 ge 2
119898minus1
1198994
we derive
119910119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119910119899 120582
119898=
1
2(119898minus1)
2 (13)
From (10)
119909119899minus119896
= 119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
(14)
Consequently we have
Δ119898
119910119899+ 119902119899119891 (119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) = 0
for all sufficient large 119899
(15)
Abstract and Applied Analysis 3
Noting that |119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) le 0
for all sufficient large 119899
(16)
By 119910119899ge 119909119899 Δ119910119899gt 0 and 119899 minus 119896 minus 120591 le 119899 minus 119896 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) ge Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
(17)
Therefore we have
Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
le 0
for all sufficient large 119899
(18)
Now by using (13) we have that for 120582119898= 12(119898minus1)
2
119910119899minus119896
ge
120582119898
(119898 minus 1)
(119899 minus 119896)119898minus1
Δ119898minus1
119910119899minus119896
for all sufficient large 119899
(19)
Thus we get
Δ119898
119910119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) Δ119898minus1
119910119899minus119896
le 0
for all sufficient large 119899
(20)
where120582119898= 12(119898minus1)
2
Let119906119899= Δ119898minus1
119910119899 then for large enough
119899 we get
Δ119906119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119906119899minus119896
le 0 (21)
where 120582119898
= 12(119898minus1)
2
Therefore inequality (21) has aneventually positive solution By Lemma 5 in [9] (9) hasan eventually positive solution which contradicts that (9) isoscillatory This completes the proof
3 Applications of the Comparison Theorem
The following lemma is well known (see eg [2 11 12] andthe references therein)
Lemma 7 Let 119902119899 be a sequence of eventually nonnegative
real number and 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt (
119896
1 + 119896
)
1+119896
(22)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894gt 1 (23)
then the first-order difference equation
Δ119909119899+ 119902119899119909119899minus119896
= 0 (24)
is oscillatory
Thus from Theorem 6 and Lemma 7 we can obtain thefollowing results
Theorem 8 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1)(
119896
1 + 119896
)
1+119896
(25)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 2(119898minus1)
2
(119898 minus 1) (26)
then (2) is oscillatory
Proof From (25) and (26) we can obtain
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt (
119896
1 + 119896
)
1+119896
(27)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 (28)
where 120582119898
= 12(119898minus1)
2
By Lemma 7 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According toTheorem 8 we obtain Corollary 9
Corollary 9 Assume that conditions (H1) and (H
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 when 119901
119899equiv 0
119898 = 2 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(
119896
1 + 119896
)
1+119896
(29)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (30)
then the second-order difference equation
Δ2
119909119899+ 119902119899119891 (119909119899minus119896
) = 0 (31)
is oscillatory
The following lemma is given in [4 Theorem 26]
Lemma 10 Let 119902119899 be a sequence of nonnegative real numbers
and 119896 a positive integer Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(32)
4 Abstract and Applied Analysis
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus
1205722
4
(33)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus 120572
119896
(34)
then (24) is oscillatory
Thus fromTheorem 6 and Lemma 10 we can obtain thefollowing results
Theorem 11 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(35)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus
1205722
4
)
(36)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus 120572119896
)
(37)
then (2) is oscillatory
Proof From (36) and (37) we can obtain
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus
1205722
4
(38)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus 120572119896
(39)
where 120582119898
= 12(119898minus1)
2
By Lemma 10 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According to Theorem 11 we can obtain the followingcorollary
Corollary 12 Assume that conditions (1198671) and (119867
2) hold let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
For 119901119899equiv 0119898 = 2 assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(40)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(1 minus
1205722
4
) (41)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (1 minus 120572
119896
) (42)
then (31) is oscillatory
4 Examples
Example 1 Considering the equation
Δ119898
(119909119899+
1
119899
119909119899minus119897
) +
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(43)
where 119899 gt 3119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
1
119899
lt 1 119902119899=
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(44)
where 119902119899 is a positive sequence Then
119899
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
1
119894 minus 2
)
=
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
(45)
Thus
lim sup119899rarrinfin
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
=
6
119890
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1)
(46)
Therefore by Theorem 8 (43) is oscillatory
Example 2 Considering the equation
Δ119898
[119909119899+
119899 minus 1
119899
119909119899minus119897
]
+
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(47)
Abstract and Applied Analysis 5
where 119899 gt 2119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
119899 minus 1
119899
lt 1
119902119899=
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(48)
where 119902119899 is a positive sequence Denote 120572 = 414 then
119899minus1
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
119894 minus 2 minus 1
119894 minus 2
)
=
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
(49)
Thus
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
=
15
16
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1) (1 minus (
4
14
)
2
)
(50)
Therefore by Theorem 11 (47) is oscillatory
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The author sincerely thanks the reviewers for their valuablesuggestions and useful comments that have led to the presentimproved version of the original paper This work wassupported by a Grant from the Natural Science Foundationof Shandong Province of China (no ZR2013AM003) andthe Development Program in Science and Technology ofShandong Province of China (no 2010GWZ20401)
References
[1] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[2] I Gyori and G Ladas Oscillation Theory of Delay DifferentialEquations Oxford Clarendon Press Oxford UK 1991
[3] Q Zhang J Yan and L Gao ldquoOscillation behavior of even-order nonlinear neutral differential equations with variablecoefficientsrdquoComputers andMathematics withApplications vol59 no 1 pp 426ndash430 2010
[4] I P Stavroulakis ldquoOscillation criteria for first order delaydifference equationsrdquo Mediterranean Journal of Mathematicsvol 1 no 2 pp 231ndash240 2004
[5] E Thandapani R Arul and P S Raja ldquoBounded oscillationof second order unstable neutral type difference equationsrdquoJournal of Applied Mathematics and Computing vol 16 no 1-2pp 79ndash90 2004
[6] Z Zhou J Yu and G Lei ldquoOscillations for even-order neu-tral difference equationsrdquo Korean Journal of Computational ampApplied Mathematics vol 7 no 3 pp 601ndash610 2000
[7] S Chen ldquoOscillation criteria for certain even order quasilineardifference equationsrdquo Journal of AppliedMathematics and Com-puting vol 31 no 1-2 pp 495ndash506 2009
[8] EThandapani and R Arul ldquoOscillatory and asymptotic behav-ior of solutions of higher order damped nonlinear differenceequationsrdquoCzechoslovakMathematical Journal vol 49 no 1 pp149ndash161 1999
[9] G Ladas and C Qian ldquoComparison results and linearizedoscillations for higher-order difference equationsrdquo InternationalJournal of Mathematics andMathematical Sciences vol 15 no 1pp 129ndash142 1992
[10] L Yang ldquoOscillation behavior of even-order neutral differenceequations with variable coefficientsrdquo Pure and Applied Mathe-matics vol 24 no 4 pp 796ndash801 2008 (Chinese)
[11] G Ladas Ch G Philos and Y G Sficas ldquoSharp conditions forthe oscillation of delay difference equationsrdquo Journal of AppliedMathematics and Simulation vol 2 no 2 pp 101ndash111 1989
[12] L H Erbe and B G Zhang ldquoOscillation of discrete analogues ofdelay equationsrdquo Differential and Integral Equations vol 2 no3 pp 300ndash309 1989
Research ArticleDifference Equations and Sharing ValuesConcerning Entire Functions and Their Difference
Zhiqiang Mao1 and Huifang Liu2
1 School of Mathematics and Computer Jiangxi Science and Technology Normal University Nanchang 330038 China2 College of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China
Correspondence should be addressed to Huifang Liu liuhuifang73sinacom
Received 19 January 2014 Accepted 22 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z Mao and H Liu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The value distribution of solutions of certain difference equations is investigated As its applications we investigate the differenceanalogue of the Bruck conjecture We obtain some results on entire functions sharing a finite value with their difference operatorsExamples are provided to show that our results are the best possible
1 Introduction and Main Results
In this paper the term meromorphic function will meanbeing meromorphic in the whole complex plane C Itis assumed that the reader is familiar with the standardnotations and the fundamental results of the Nevanlinnatheory see for example [1ndash3] In addition we use notations120590(119891) 120582(119891) to denote the order and the exponent of conver-gence of the sequence of zeros of a meromorphic function 119891respectivelyThe notation 119878(119903 119891) is defined to be any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside aset 119864 of 119903 of finite logarithmic measure
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 isin C We say that 119891 and 119892 share 119886 CM providedthat 119891 minus 119886 and 119892 minus 119886 have the same zeros with the samemultiplicities Similarly we say that 119891 and 119892 share 119886 IMprovided that 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
The famous results in the uniqueness theory of meromor-phic functions are the 5 IM and 4 CM shared values theoremsdue to Nevanlinna [4] It shows that if two nonconstantmeromorphic functions 119891 and 119892 share five different valuesIM or four different values CM then 119891 equiv 119892 or 119891 is a linearfractional transformation of119892 Condition 4CMshared valueshave been improved to 2 CM + 2 IM by Gundersen [5]while the case 1 CM + 3 IM still remains an open problemSpecifically Bruck posed the following conjecture
Conjecture 1 (see [6]) Let 119891 be a nonconstant entire functionsatisfying the hyperorder 120590
2(119891) lt infin where 120590
2(119891) is not a
positive integer If 119891 and 1198911015840 share a finite value 119886 CM then119891 minus 119886 equiv 119888(119891
1015840
minus 119886) for some nonzero constant 119888
In [6] Bruck proved that the conjecture is true providedthat 119886 = 0 or 119873(119903 11198911015840) = 119878(119903 119891) He also gave counter-examples to show that the restriction on the growth of 119891 isnecessary
In recent years as the research on the difference ana-logues of Nevanlinna theory is becoming active lots ofauthors [7ndash11] started to consider the uniqueness of mero-morphic functions sharing values with their shifts or theirdifference operators
Heittokangas et al proved the following result which is ashifted analogue of Bruckrsquos conjecture
Theorem A (see [8]) Let 119891 be a meromorphic function of120590(119891) lt 2 and 120578 a nonzero complex number If119891(119911) and119891(119911+120578)share a finite value 119886 andinfin CM then
119891 (119911 + 120578) minus 119886
119891 (119911) minus 119886
= 120591 (1)
for some constant 120591
In [8] Heittokangas et al gave the example 119891(119911) = 1198901199112
+1
which shows that 120590(119891) lt 2 cannot be relaxed to 120590(119891) le 2
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 584969 6 pageshttpdxdoiorg1011552014584969
2 Abstract and Applied Analysis
For a nonzero complex number 120578 we define differenceoperators as
Δ120578119891 (119911) = 119891 (119911 + 120578) minus 119891 (119911)
Δ119899
120578119891 (119911) = Δ
119899minus1
120578(Δ120578119891 (119911)) 119899 isin N 119899 ge 2
(2)
Regarding the difference analogue of Bruckrsquos conjecture wemention the following results
Theorem B (see [7]) Let 119891 be a finite order transcendentalentire function which has a finite Borel exceptional value 119886 andlet 120578 be a constant such that 119891(119911 + 120578) equiv 119891(119911) If 119891(119911) andΔ120578119891(119911) share 119886 CM then
119886 = 0
119891 (119911 + 120578) minus 119891 (119911)
119891 (119911)
= 119888 (3)
for some nonzero constant 119888
Theorem C (see [11]) Let 119891 be a nonperiodic transcendentalentire function of finite order If 119891(119911) and Δ119899
120578119891(119911) share a
nonzero finite value 119886 CM then 1 le 120590(119891) le 120582(119891 minus 119886) + 1that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (4)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891 minus 119886) and119876(119911) is a polynomial with deg119876 le 120590(119860) + 1
Let 119891 be a nonperiodic transcendental entire function offinite order Theorem B shows that if a nonzero finite value119886 is shared by 119891(119911) and Δ
120578119891(119911) then 120590(119891) = 120582(119891 minus 119886)
It is obvious that the result in Theorem B is sharper thanTheoremC for 119899 = 1 In this paper we continue to investigatethe difference analogue of Bruckrsquos conjecture and obtain thefollowing result
Theorem 2 Let 119891 be a finite order entire function 119899 ge 2 aninteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share a finite value 119886 ( = 0)CM then120582(119891minus119886) = 120590(119891) ge
1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (5)
where 119860(119911) is an entire function with 1 le 120590(119860) = 120582(119891 minus 119886) =120590(119891) and 119876(119911) is a polynomial with deg119876 le 120590(119860)
Remark 3 It is obvious that Theorem 2 is sharper thanTheorem C and a supplement of Theorem B for 119899 ge 2
The discussions in Theorems C and 2 are concerning thecase that shared value 119886 = 0 When 119886 = 0 we obtain thefollowing result
Theorem 4 Let 119891 be a finite order entire function 119899 a positiveinteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share 0 CM then 1 le 120590(119891) le 120582(119891) + 1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
(6)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891) and 119876(119911)is a polynomial with deg119876 le 120590(119860) + 1
It is well known that if a finite order entire function 119891(119911)shares 119886 CM with Δ119899
120578119891(119911) then 119891(119911) satisfies the difference
equation
Δ119899
120578119891 (119911) minus 119886 = 119890
119876(119911)
(119891 (119911) minus 119886) (7)
where119876(119911) is a polynomialHence in order to prove the aboveresults we consider the value distribution of entire solutionsof the difference equation
119886119899(119911) 119891 (119911 + 119899120578) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 120578)
+ (1198860(119911) minus 119890
119876(119911)
) 119891 (119911) = 119861 (119911)
(8)
and obtain the following result
Theorem 5 Let 1198860 119886
119899minus1 119886119899( equiv 0) 119861( equiv 0) be polynomials
and let 119876 be a polynomial with degree 119898(ge 1) Then everyentire solution 119891 of finite order of (8) satisfies 120590(119891) ge 119898 and
(i) if 120590(119891) gt 1 then 120582(119891) = 120590(119891)(ii) if 120590(119891) = 1 then 120582(119891) = 120590(119891) or 119891 has only finitely
many zeros
2 Lemmas
Lemma 6 (see [12]) Let 119879 (0 +infin) rarr (0 +infin) be anondecreasing continuous function 119904 gt 0 120572 lt 1 and let119865 sub R+ be the set of all 119903 such that 119879(119903) le 120572119879(119903 + 119904) If thelogarithmic measure of 119865 is infinite then
lim119903rarrinfin
log119879 (119903)log 119903
= infin (9)
Lemma 7 (see [13]) Let 119891 be a nonconstant meromorphicfunction of finite order 120578 isin C 120575 lt 1 Then
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119900(
119879 (119903 +10038161003816100381610038161205781003816100381610038161003816 119891)
119903120575
) (10)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
(119889119903119903) lt infin
Remark 8 By Lemmas 6 and 7 we know that for a noncon-stant meromorphic function 119891 of finite order
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119878 (119903 119891) (11)
Lemma 9 (see [3]) Let 119891119895(119895 = 1 119899 + 1) and 119892
119895(119895 =
1 119899) be entire functions such that
(i) sum119899119895=1119891119895(119911)119890119892119895(119911)
equiv 119891119899+1(119911)
(ii) the order of 119891119895is less than the order of 119890119892119896 for 1 le 119895 le
119899+1 1 le 119896 le 119899 and furthermore the order of119891119895is less
than the order of 119890119892ℎminus119892119896 for 119899 ge 2 and 1 le 119895 le 119899+1 1 leℎ lt 119896 le 119899
Then 119891119895(119911) equiv 0 (119895 = 1 119899 + 1)
Abstract and Applied Analysis 3
Lemma 10 (see [14]) Let 119891 be a meromorphic function withfinite order 120590(119891) = 120590 lt 1 120578 isin C 0 Then for any given 120576 gt 0and integers 0 le 119895 lt 119896 there exists a set 119864 sub (1infin) of finitelogarithmic measure so that for all |119911| = 119903 notin 119864 ⋃ [0 1] wehave
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119896
120578119891 (119911)
Δ119895
120578119891 (119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|(119896minus119895)(120590minus1)+120576
(12)
Lemma 11 (see [15]) Let 1198860(119911) 119886
119896(119911) be entire functions
with finite order If there exists an integer 119897 (0 le 119897 le 119896) suchthat
120590 (119886119897) gt max0le119895le119896
119895 = 119897
120590 (119886119895) (13)
holds then every meromorphic solution 119891( equiv 0) of thedifference equation
119886119896(119911) 119891 (119911 + 119896) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 1) + 119886
0(119911) 119891 (119911) = 0
(14)
satisfies 120590(119891) ge 120590(119886119897) + 1
3 Proofs of Results
Proof of Theorem 5 Let 119891 be an entire solution of finite orderof (8) By Remark 8 and (8) we get
119879 (119903 119890119876
) = 119879 (119903 119890119876
minus 1198860) + 119878 (119903 119890
119876
)
le
119899
sum
119895=1
119898(119903
119891 (119911 + 119895120578)
119891 (119911)
)
+
119899
sum
119895=0
119898(119903 119886119895) + 119898(119903
119861 (119911)
119891 (119911)
) + 119878 (119903 119890119876
)
le 119879 (119903 119891) + 119878 (119903 119891) + 119878 (119903 119890119876
)
(15)
By (15) we get 120590(119891) ge 119898
Case 1 (120590(119891) gt 1) Suppose that 120582(119891) lt 120590(119891) by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that
120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891) lt 120590 (119891) = deg ℎ
2 (16)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (1198860(119911) minus 119890
119876(119911)
) ℎ1(119911) = 119861 (119911) 119890
minusℎ2(119911)
(17)
If deg ℎ2gt 119898 then by (16) we know that the order of the right
side of (17) is deg ℎ2 and the order of the left side of (17) is less
than deg ℎ2 This is a contradiction Hence deg ℎ
2= 119898 gt 1
Set119876 (119911) = 119887
119898119911119898
+ sdot sdot sdot + 1198870
ℎ2(119911) = 119888
119898119911119898
+ sdot sdot sdot + 1198880
(18)
where 119887119898( = 0) 119887
0 119888119898( = 0) 119888
0are complex numbers
By (17) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 1198860(119911) ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
+ 119861 (119911) 119890minusℎ2(119911)
(19)
Next we discuss the following two subcases
Subcase 1 (119887119898+ 119888119898=0)Then by Lemma 9 (16) and (19) we
get 119861(119911) equiv 0 ℎ1(119911) equiv 0 This is impossible
Subcase 2 (119887119898+ 119888119898= 0) Suppose that
ℎ1(119911) 119890119876(119911)minus119887
119898119911119898
+ 119861 (119911) 119890minusℎ2(119911)minus119887119898119911119898
equiv 0 (20)
Then ℎ1(119911) = minus119861(119911)119890
minusℎ2(119911)minus119876(119911) By 120590(ℎ
1) = 120582(ℎ
1) we
obtain that 119890minusℎ2(119911)minus119876(119911) is a nonzero constant Hence ℎ1(119911) is a
nonzero polynomial By (19) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
= minus1198860(119911) ℎ1(119911)
(21)
Since degℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578) = 119898 minus 1 gt 0 for 119894 = 119895 then
by Lemma 9 and (21) we get
119886119895(119911) ℎ1(119911 + 119895120578) equiv 0 (119895 = 0 1 119899) (22)
This is impossible Hence we have ℎ1(119911)119890119876(119911)minus119887
119898119911119898
+
119861(119911)119890minusℎ2(119911)minus119887119898119911119898
equiv 0 Then from the order considerationwe know that the order of the right side of (19) is 119898 andthe order of the left side of (19) is less than 119898 This is acontradiction Hence 120582(119891) = 120590(119891)
Case 2 (120590(119891) = 1)Then by 120590(119891) ge 119898 we get119898 = 1 Supposethat 119891(119911) has infinitely many zeros and 120582(119891) lt 120590(119891) by theWeierstrass factorization we get
119891 (119911) = ℎ3(119911) 119890120573119911
(23)
where 120573( = 0) is a complex number and ℎ3(119911)( equiv 0) is an
entire function such that
120590 (ℎ3) = 120582 (ℎ
3) = 120582 (119891) lt 1 (24)
Let 119876(119911) = 1198871119911 + 1198870 where 119887
1( = 0) 119887
0are complex numbers
Substituting 119891(119911) = ℎ3(119911)119890120573119911 into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ3(119911 + 119895120578) 119890
120573119895120578
+ 1198860(119911) ℎ3(119911)
= ℎ3(119911) 1198901198871119911+1198870+ 119861 (119911) 119890
minus120573119911
(25)
4 Abstract and Applied Analysis
Note that ℎ3(119911)1198901198870+ 119861(119911) equiv 0 otherwise 119891 has only finitely
many zeros If 1198871+ 120573 = 0 then the order of the right side of
(25) is 1 but the order of the left side of (25) is less than 1Thisis absurd If 119887
1+120573 = 0 then by Lemma 9 (24) and (25) we get
ℎ3(119911) equiv 0 119861(119911) equiv 0 This is impossible Hence 120582(119891) = 120590(119891)
Theorem 5 is thus completely proved
Proof of Theorem 2 Since 119891(119911) and Δ119899120578119891(119911) share 119886 CM and
119891 is of finite order then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119890119876(119911)
(26)
where 119876(119911) is a polynomial with deg119876 le 120590(119891) Now we willtake two steps to complete the proof
Step 1 We prove that 120582(119891 minus 119886) = 120590(119891)Let 119865(119911) = 119891(119911) minus 119886 then
120582 (119865) = 120582 (119891 minus 119886) 120590 (119865) = 120590 (119891) ge deg119876 (27)
and Δ119899120578119891(119911) = Δ
119899
120578119865(119911) = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119865(119911 + 119895120578) By thisand (26) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119890119876(119911)
) 119865 (119911) = 119886 (28)
Next we discuss the following three cases
Case 1 (deg119876 ge 1 and 120590(119865) gt deg119876) Then 120590(119865) gt 1 ByTheorem 5(i) (27) and (28) we get 120582(119891 minus 119886) = 120590(119891)
Case 2 (deg119876 ge 1 and 120590(119865) = deg119876) If 120590(119865) = deg119876 gt 1then byTheorem 5(i) (27) and (28) we get120582(119891minus119886) = 120590(119891) If120590(119865) = deg119876 = 1 then byTheorem 5(ii) and (27) we obtainthat 120582(119891 minus 119886) = 120590(119891) or 119865 has only finitely many zeros
If 119865 has only finitely many zeros set
119865 (119911) = ℎ1(119911) 119890119887119911
(29)
where ℎ1(119911)( equiv 0) is a polynomial and 119887( = 0) is a complex
number then substituting (29) into (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
119887119895120578
+ (minus1)119899
ℎ1(119911)
= ℎ1(119911) 119890119876(119911)
+ 119886119890minus119887119911
(30)
By (30) and Δ119899120578119865(119911) equiv 0 we know that the order of the left
side of (30) is 0 and the order of the right side of (30) is 1unless ℎ
1(119911) = minus119886 and119876(119911) = minus119887119911 In this case take it into the
left side of (30) we have (minus119886)(119890119887120578 minus 1)119899 = 0 Since all 119886 119887 and120578 are not zero it is impossible Hence we get 120582(119891minus119886) = 120590(119891)
Case 3 119876 is a complex constant Then by (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119888) 119865 (119911) = 119886 (31)
where 119888 (= 119890119876 =0) is a complex number Suppose that 120582(119865) lt120590(119865) Let 119865(119911) = ℎ
2(119911)119890ℎ3(119911) where ℎ
2(119911)( equiv 0) is an entire
function and ℎ3(119911) is a polynomial such that
120590 (ℎ2) = 120582 (ℎ
2) = 120582 (119865) lt 120590 (119865) = deg ℎ
3 (32)
Substituting 119865(119911) = ℎ2(119911)119890ℎ3(119911) into (31) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ2(119911 + 119895120578) 119890
ℎ3(119911+119895120578)minusℎ
3(119911)
+ ((minus1)119899
minus 119888) ℎ2(119911) = 119886119890
minusℎ3(119911)
(33)
Since deg(ℎ3(119911 + 119895120578) minus ℎ
3(119911)) = deg ℎ
3(119911) minus 1 (119895 = 1 119899)
by (32) we obtain that the order of the left side of (33) is lessthan deg ℎ
3and the order of the right side of (33) is deg ℎ
3
This is absurd Hence we get 120582(119891 minus 119886) = 120590(119891)
Step 2We prove that 120590(119891) ge 1Suppose that 120590(119891) lt 1 Since 119891(119911) and Δ119899
120578119891(119911) share 119886
CM then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119888 (34)
where 119888 is a nonzero constant Let 119865(119911) = 119891(119911) minus 119886 then by(34) we get
Δ119899
120578119865 (119911) = 119888119865 (119911) + 119886 (35)
Differentiating (35) we get
(Δ119899
120578119865 (119911))
1015840
= 1198881198651015840
(119911) (36)
Note that (Δ119899120578119865(119911))1015840
= Δ119899
120578(1198651015840
(119911)) and 120590(1198651015840) = 120590(119865) = 120590(119891) lt1 So by Lemma 10 and (36) we get
|119888| =
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119899
120578(1198651015840
(119911))
1198651015840(119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|119899(120590(119865)minus1)+120576
997888rarr 0 (37)
This is absurd So 120590(119891) ge 1 Theorem 2 is thus completelyproved
Proof of Theorem 4 Since 119891(119911) and Δ119899120578119891(119911) share 0 CM and
119891 is of finite order then
Δ119899
120578119891 (119911)
119891 (119911)
= 119890119876(119911)
(38)
where119876(119911) is a polynomial ByΔ119899120578119891 = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119891(119911+
119895120578) and (38) we get
119891 (119911 + 119899120578)
+
119899minus1
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119891 (119911 + 119895120578)
+ ((minus1)119899
minus 119890119876(119911)
) 119891 (119911) = 0
(39)
Abstract and Applied Analysis 5
We discuss the following two cases
Case 1 119876 is a polynomial with deg119876 = 119898 ge 1 Then byLemma 11 and (39) we get 120590(119891) ge 119898 + 1 Now we prove120590(119891) le 120582(119891) + 1 Suppose that 120590(119891) gt 120582(119891) + 1 then by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891)
120590 (119891) = deg ℎ2gt 120590 (ℎ
1) + 1
(40)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (39) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (minus1)119899
ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
(41)
If 120590(119891) gt 119898 + 1 then by (40) (41) and deg(ℎ2(119911 + 119895120578) minus
ℎ2(119911)) = deg ℎ
2(119911)minus1 (119895 = 1 119899) we obtain that the order
of the left side of (41) is deg ℎ2minus 1 and the order of the right
side of (41) is less than deg ℎ2minus 1 This is absurd
If 120590(119891) = 119898 + 1 then by (41) we get119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
minus ℎ1(119911) 119890119876(119911)
= (minus1)119899+1
ℎ1(119911)
(42)
Set
ℎ2(119911) = 119889
119898+1119911119898+1
+ sdot sdot sdot + 1198890 (43)
where 119889119898+1( = 0) 119889
0are complex numbers Then
ℎ2(119911 + 119895120578) minus ℎ
2(119911)
= 119889119898+1
(119898 + 1) 119895120578119911119898
+ sdot sdot sdot + 1198891119895120578
(119895 = 1 119899)
(44)
Now we discuss the following two subcases
Subcase 1 deg(119876(119911)minus(ℎ2(119911+119895120578)minusℎ
2(119911))) = 119898 holds for every
119895 isin 1 119899Then by (40) (42) deg(ℎ2(119911+119895120578)minusℎ
2(119911+119894120578)) =
119898 (119895 = 119894) and Lemma 9 we get ℎ1(119911) equiv 0 This is absurd
Subcase 2 There exist some 1198950isin 1 119899 such that
deg(119876(119911)minus(ℎ2(119911+1198950120578)minusℎ2(119911))) le 119898minus1Then by (44) we have
deg(119876(119911) minus (ℎ2(119911 + 119895120578) minus ℎ
2(119911))) = 119898 for 119895 = 119895
0 Merging the
term minusℎ1(119911)119890119876(119911) into ( 119899119895
0) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578)119890ℎ2(119911+1198950120578)minusℎ2(119911)
by (42) we get119899
sum
119895 = 1
119895 =1198950
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 ge 2)
(45)
or
119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 = 1) (46)
where 119860(119911) = (119899
1198950) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578) minus ℎ
1(119911)
119890119876(119911)minus(ℎ
2(119911+1198950120578)minusℎ2(119911)) satisfying 120590(119860) lt 119898 If 119899 ge 2 then
by (40) (45) deg(ℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578)) = 119898 (119895 = 119894) and
Lemma 9 we get ℎ1(119911) equiv 0 This is absurd If 119899 = 1 then by
(46) and ℎ1(119911) equiv 0 we get 119860(119911) equiv 0 By this we know that
the order of the left side of (46) is 119898 and the order of theright side of (46) is less than119898 This is absurd Hence we get120590(119891) le 120582(119891) + 1
Case 2119876 is a complex constantThen by Lemma 10 and (38)we get 120590(119891) ge 1 Now we prove 120590(119891) le 120582(119891) + 1 Supposethat 120590(119891) gt 120582(119891) + 1 If 120590(119891) gt 1 then by the similarargument to that of case 1 we get ℎ
1(119911) equiv 0 This is absurd If
120590(119891) = 1 then by (40) we get 0 le 120582(119891) lt 120590(119891) minus 1 = 0 Since120582(119891) = 0 then 120590(119891) = 120582(119891)+1Theorem 4 is thus completelyproved
4 Some Examples
The following examples show the existence of such entirefunctions which satisfy Theorems 2ndash5 Moreover Example 2shows that the result in Theorem 4 is the best possible
Example 1 Let 120578 = 1 119899 = 2 and 119891(119911) = (119889 + 1)119911 + ((1198892 minus1)1198892
)119886 where 119886( = 0) 119889( = 0 plusmn1) are constants Then 119891(119911)and Δ119899
120578119891(119911) share 119886 CM and 120590(119891) = 120582(119891 minus 119886) = 1
Example 2 Let 120578 = 1 119899 = 2 and 119891(119911) = 119890119911 Then 119891(119911) andΔ119899
120578119891(119911) share 0 CM and 120590(119891) = 1 = 120582(119891) + 1
Example 3 Let 120578 = 1 119899 = 2 and119891(119911) = 119867(119911)119890119911 where119867(119911)is an entire function with period 1 such that 120590(119867) gt 1 and120590(119867) notin N Then 119891(119911) and Δ119899
120578119891(119911) share 0 CM and 120582(119891) =
120582(119867) = 120590(119867) = 120590(119891) gt 1 (Ozawa [16] proved that for any120590 isin [1infin) there exists a period entire function of order 120590)
Example 4 The entire function 119891(119911) = 119911119890minus119911 satisfies the
difference equation
119891 (119911 + 2120578) minus 4119891 (119911 + 120578) + (4 minus 119890119911
) 119891 (119911) = minus119911 (47)
where 120578 = minus log 2 Here120590(119891) = 1 and119891 has only finitelymanyzeros
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11201195 11171119) and the NaturalScience Foundation of Jiangxi China (nos 20122BAB20101220132BAB201008)
6 Abstract and Applied Analysis
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers New York NY USA2003
[4] R Nevanlinna Le Theoreme de Picard-Borel et la Theorie desFonctions Meromorphes Gauthiers-Villars Paris France 1929
[5] G G Gundersen ldquoMeromorphic functions that share fourvaluesrdquo Transactions of the American Mathematical Society vol277 no 2 pp 545ndash567 1983
[6] R Bruck ldquoOn entire functions which share one value CM withtheir first derivativerdquo Results inMathematics vol 30 no 1-2 pp21ndash24 1996
[7] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[8] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[9] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables andElliptic Equations vol 56 pp 81ndash92 2011
[10] K Liu and L-Z Yang ldquoValue distribution of the differenceoperatorrdquo Archiv der Mathematik vol 92 no 3 pp 270ndash2782009
[11] S Li and Z Gao ldquoA note on the Bruck conjecturerdquo Archiv derMathematik vol 95 no 3 pp 257ndash268 2010
[12] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[13] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[14] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[15] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[16] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleAdmissible Solutions of the Schwarzian Type DifferenceEquation
Baoqin Chen and Sheng Li
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to Sheng Li lish lssinacom
Received 14 January 2014 Accepted 20 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 B Chen and S Li This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper is to investigate the Schwarzian type difference equation [(Δ3
119891Δ119891) minus (32) (Δ2
119891Δ119891)
2
]
119896
= 119877 (119911 119891) =
(119875(119911 119891)119876(119911 119891)) where 119877(119911 119891) is a rational function in 119891 with polynomial coefficients 119875(119911 119891) respectively 119876(119911 119891) are twoirreducible polynomials in 119891 of degree 119901 respectively 119902 Relationship between 119901 and 119902 is studied for some special case Denote119889 = max 119901 119902 Let 119891(119911) be an admissible solution of (lowast) such that 120588
2(119891) lt 1 then for 119904 (ge2) distinct complex constants 120572
1 120572
119904
119902 + 2119896sum119904
119895=1120575(120572119895 119891) le 8119896 In particular if119873(119903 119891) = 119878(119903 119891) then 119889 + 2119896sum119904
119895=1120575(120572119895 119891) le 4119896
1 Introduction and Results
Throughout this paper a meromorphic function alwaysmeans being meromorphic in the whole complex plane and119888 always means a nonzero constant For a meromorphicfunction 119891(119911) we define its shift by 119891(119911 + 119888) and define itsdifference operators by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911) Δ
119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911))
119899 isin N 119899 ge 2
(1)
In particular Δ119899119888119891(119911) = Δ
119899
119891(119911) for the case 119888 = 1 We usestandard notations of theNevanlinna theory ofmeromorphicfunctions such as 119879(119903 119891) 119898(119903 119891) and 119873(119903 119891) and as statedin [1ndash3] For a constant 119886 we define theNevanlinna deficiencyby
120575 (119886 119891) = lim inf119903rarrinfin
119898(119903 1 (119891 minus 119886))
119879 (119903 119891)
= 1 minus lim sup119903rarrinfin
119873(119903 1 (119891 minus 119886))
119879 (119903 119891)
(2)
Recently numbers of papers (see eg [4ndash12]) are devotedto considering the complex difference equations and differ-ence analogues of Nevanlinna theory Due to some idea of[13] we consider the admissible solution of the Schwarziantype difference equation
119878119896(119891) = [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(3)
where 119877(119911 119891) is a rational function in 119891 with polynomialcoefficients 119875(119911 119891) respectively119876(119911 119891) are two irreduciblepolynomials in 119891 of degree 119901 respectively 119902 Here and in thefollowing ldquoadmissiblerdquo always means ldquotranscendentalrdquo Andwe denote 119889 = max119901 119902 from now on For the existence ofsolutions of (3) we give some examples below
Examples (1) 119891(119911) = sin 120587119911 + 119911 is an admissible solution ofthe Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus8 [1198912
+ (1 minus 2119911) 119891 + 119911 (119911 minus 1)]
41198912minus 4 (2119911 + 1) 119891 + (2119911 + 1)
2
(4)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 306360 5 pageshttpdxdoiorg1011552014306360
2 Abstract and Applied Analysis
(2) 119891(119911) = (119890119911 ln 2 sin 2120587119911) + 119911 is an admissible solutionof the Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus1198912
+ 2 (119911 + 1) 119891 minus 1199112
minus 2119911
21198912minus 4 (119911 minus 1) 119891 + 2(119911 minus 1)
2 (5)
(3) Let 119891(119911) = 1199112
+ 119911 then 119891(119911) solves the Schwarziantype difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
= minus
3
2 [1198912minus 2 (119911
2minus 1) 119891 + (119911
2minus 1)2
]
(6)
This example shows that (3) may admit polynomial solutions
Considering the relationship between 119901 and 119902 in thoseexamples above we prove the following result
Theorem 1 For the Schwarzian type difference equation (3)with polynomial coefficients note the following
(i) If it admits an admissible solution 119891(119911) such that1205882(119891) lt 1 then
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (7)
In particular if119898(119903 119891) = 119878(119903 119891) then 119901 le 119902(ii) If its coefficients are all constants and it admits a
polynomial solution 119891(119911) with degree 119904 then 119904 ge 2 and119902119904 = 119901119904 + 2119896
Remark 2 From examples (1) and (2) we conjecture that119901 = 119902 inTheorem 1(i) However we cannot prove it currentlyFrom example (3) given before we see that the restriction onthe coefficients in Theorem 1(ii) cannot be omitted
For the Schwarzian differential equation
119878119896(119891) = [
119891101584010158401015840
1198911015840
minus
3
2
(
11989110158401015840
1198911015840
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(8)
where 119877(119911 119891) 119875(119911 119891) and 119876(119911 119891) are as stated beforeIshizaki [13] proved the following result (see also Theorem932 in [2])
Theorem A (see [2 13]) Let 119891(119911) be an admissible solutionof (8) with polynomial coefficients and let 120572
1 120572
119904be 119904 (ge2)
distinct complex constants Then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (9)
For the Schwarzian type difference equation (3) we provethe following result
Theorem 3 Let 119891(119911) be an admissible solution of (3) withpolynomial coefficients such that 120588
2(119891) lt 1 and let 120572
1 120572
119904
be 119904 (ge2) distinct complex constants Then
119902 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 8119896 (10)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (11)
Remark 4 From Theorem 1 under the condition 119873(119903 119891) =119878(119903 119891) in Theorem 3 we have 119889 = 119902 in (11) The behavior ofthe zeros and the poles of 119891(119911) in 119878
119896(119891) is essentially different
from that in the 119878119896(119891) We wonder whether the restriction
119873(119903 119891) = 119878(119903 119891) can be omitted or not
2 Lemmas
The following lemmaplays a very important role in the theoryof complex differential equations and difference equationsIt can be found in Mohonrsquoko [14] and Valiron [15] (see alsoTheorem 225 in the book of Laine and Yang [2])
Lemma 5 (see [14 15]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(12)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(13)
and the characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (14)
where 119889 = max119901 119902
The following two results can be found in [10] In factLemma 6 is a special case of Lemma 83 in [10]
Lemma 6 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891) (15)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 7 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120589minus120576
) = 119878 (119903 119891) (16)
possibly outside of a set of 119903 with finite logarithmic measure
From Lemma 7 we can easily get the following conclu-sion
Abstract and Applied Analysis 3
Lemma 8 Let 119891(119911) be a meromorphic function of hyper order1205882(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
Δ119899
119888119891 (119911)
119891 (119911)
) = 119878 (119903 119891)
119898(119903
Δ119896
119888119891 (119911)
Δ119895
119888119891 (119911)
) = 119878 (119903 119891) 119896 gt 119895
(17)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 9 Let 119891 be an admissible solution of (3) withcoefficients Then using the notation 119876(119911) = 119876(119911 119891(119911))
119902119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (18)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (19)
Proof We use the idea by Ishizaki [13] (see also [2]) to proveLemma 9 It follows from Lemma 8 that
119898(119903 119877) = 119898(119903 [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
)
le 119896119898(119903
Δ3
119891
Δ119891
) + 2119896119898(119903
Δ2
119891
Δ119891
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(20)
From this and Lemma 5 we get
119889119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119877) = 119873 (119903 119877) + 119878 (119903 119891) (21)
and hence
119889119879 (119903 119891) = 119873 (119903 119877) + 119878 (119903 119891) (22)
If 119889 = 119901 gt 119902 since all coefficients of 119875(119911 119891) and 119876(119911 119891)are polynomials there are at the most finitely many poles of119877(119911 119891) neither the poles of 119891(119911) nor the zeros of 119876(119911 119891)Therefore we see that
119873(119903 119877) le (119901 minus 119902)119873 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
le (119901 minus 119902) 119879 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
(23)
We obtain (18) from this and (22) immediatelyIf 119889 = 119902 ge 119901 there are at most finitely many poles of
119877(119911 119891) not the zeros of 119876(119911 119891) then
119873(119903 119877) le 119873(119903
1
119876
) + 119878 (119903 119891) (24)
Now (18) follows from (22) and (24)Notice that if 119873(119903 119891) = 119878(119903 119891) then (24) always holds
This finishes the proof of Lemma 9
3 Proof of Theorem 1
Case 1 Equation (3) admits an admissible solution 119891(119911) suchthat 1205882(119891) lt 1 Since all coefficients of119875(119911 119891) and119876(119911 119891) are
polynomials there are at the most finitely many poles of 119891(119911)that are not the poles of119875(119911 119891) and119876(119911 119891) This implies that
119873(119903 119875) = 119901119873 (119903 119891) + 119878 (119903 119891)
119873 (119903 119876) = 119902119873 (119903 119891) + 119878 (119903 119891)
(25)
From Lemma 5 we get
119879 (119903 119875) = 119901119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 119876) = 119902119879 (119903 119891) + 119878 (119903 119891)
(26)
We can deduce from (3) (25) (26) and Lemma 8 that
119901119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119875)
= 119898 (119903 119875) + 119873 (119903 119875)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891)119876)
+ 119878 (119903 119891)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891))
+ 119898 (119903 119876) + 119878 (119903 119891)
= 119901119873 (119903 119891) + 119879 (119903 119876) minus 119873 (119903 119876)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119879 (119903 119891) minus 119902119873 (119903 119891)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119898 (119903 119891) + 119878 (119903 119891)
(27)
It follows from this that
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (28)
What is more is that if119898(119903 119891) = 119878(119903 119891) then we obtain from(28) that 119901 le 119902
Case 2 The coefficients of (3) are all constants and it admitsa polynomial solution 119891(119911) with degree 119904 Set
119891 (119911) = 119886119904119911119904
+ 119886119904minus1119911119904minus1
+ sdot sdot sdot + 1198861119911 + 1198860 (29)
then
119891 (119911 + 1) = 119886119904119911119904
+ 119887119904minus1119911119904minus1
+ sdot sdot sdot + 1198871119911 + 1198870 (30)
where
119887119904minus119895
= 119886119904119862119895
119904+ 119886119904minus1119862119895minus1
119904minus1+ sdot sdot sdot + 119886
119904minus119895+11198621
119904minus119895+1+ 119886119904minus119895 (31)
From (29) and (30) we obtain that
Δ119891 = 119904119886119904119911119904minus1
+ (119887119904minus2
minus 119886119904minus2) 119911119904minus2
+ sdot sdot sdot + (1198871minus 1198861) 119911 + (119887
0minus 1198860)
(32)
4 Abstract and Applied Analysis
If 119904 = 1 then Δ2119891 = Δ3119891 equiv 0 which yields that 119875(119911 119891) equiv0 That is a contradiction to our assumption Thus 119904 ge 2
If 119904 = 2 thenΔ119891 = 21198862119911+1198862+1198861Δ2119891 = 2119886
2 andΔ3119891 equiv 0
Now from (3) we get
(minus3)119896
119876 (119911 119891) (Δ2
119891)
2119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(33)
Considering degrees of both sides of the equation above wecan see that 119902 = 119901 + 119896
If 119904 ge 3 we can deduce similarly that
Δ2
119891 = 119904 (119904 minus 1) 119886119904119911119904minus2
+ 1198751(119911)
Δ3
119891 = 119904 (119904 minus 1) (119904 minus 2) 119886119904119911119904minus3
+ 1198752(119911)
(34)
where 1198751(119911) 1198752(119911) are polynomials such that deg119875
1le 119904 minus
3 deg1198752le 119904 minus 4
Rewrite (3) as follows
119876 (119911 119891) [2Δ3
119891 sdot Δ119891 minus 3(Δ2
119891)
2
]
119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(35)
From (34) we find that the leading coefficient of 2Δ3119891 sdotΔ119891 minus 3(Δ
2
119891)
2 is
minus1198862
1199041199042
(119904 minus 1) (119904 + 1) = 0 (36)
Considering degrees of both sides of (35) we prove that119902119904 = 119901119904 + 2119896
4 Proof of Theorem 3
Firstly we consider the general case Asmentioned inRemark1 in [13] due to Jank and Volkmann [16] if (3) admits anadmissible solution then there are at most 119878(119903 119891) commonzeros of 119875(119911 119891) and 119876(119911 119891) Since all coefficients of 119876(119911 119891)are polynomials there are at the most finitely many poles of119891 that are the zeros of 119876(119911 119891) Therefore from (3) we have
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119879 (119903 119891 (119911 + 1) minus 119891 (119911)) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119878 (119903 119891)
(37)
Combining this and Lemma 9 applying the second maintheorem we get
119902
2119896
119879 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
119902
2119896
119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
1
2119896
119873(119903
1
119876
) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 4119879 (119903 119891) + 119878 (119903 119891)
(38)
Thus we prove that (10) holdsSecondly we consider the case that 119873(119903 119891) = 119878(119903 119891)
From (3) and Lemma 8 we similarly get that
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119898 (119903 Δ119891) + 119873 (119903 Δ119891) + 119878 (119903 119891)
le 119898(119903
Δ119891
119891
) + 119898 (119903 119891) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
(39)
From this and applying Lemma 9with (19) as arguing beforewe can prove that (11) holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the referees for their valuablesuggestions This work is supported by the NNSFC (nos11226091 and 11301091) the Guangdong Natural ScienceFoundation (no S2013040014347) and the Foundation forDistinguished Young Talents in Higher Education of Guang-dong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic PublishersGroupDordrechtTheNetherlands 2003
[4] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[5] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
Abstract and Applied Analysis 5
[8] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[9] R G Halburd and R J Korhonen ldquoExistence of finite-ordermeromorphic solutions as a detector of integrability in differ-ence equationsrdquo Physica D Nonlinear Phenomena vol 218 no2 pp 191ndash203 2006
[10] R G Halburd R J Korhonen and K Tohge ldquoHolomorphiccurves with shift-invariant hyperplane preimagesrdquo submitted toTransactions of the American Mathematical Society httparxivorgabs09033236
[11] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[12] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] K Ishizaki ldquoAdmissible solutions of the Schwarzian differentialequationrdquoAustralianMathematical SocietyA PureMathematicsand Statistics vol 50 no 2 pp 258ndash278 1991
[14] A Z Mohonrsquoko ldquoThe nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija no 14 pp 83ndash87 1971 (Russian)
[15] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Entomologique de France vol 59 pp 17ndash39 1931
[16] G Jank and L VolkmannMeromorphe Funktionen und Differ-entialgeichungen Birkhauser Verlag Basel Switzerland 1985
Research ArticleStatistical Inference for Stochastic DifferentialEquations with Small Noises
Liang Shen12 and Qingsong Xu1
1 School of Mathematics and Statistics Central South University Changsha Hunan 410075 China2 School of Science Linyi University Linyi Shandong 276005 China
Correspondence should be addressed to Qingsong Xu csuqingsongxu126com
Received 13 November 2013 Accepted 11 February 2014 Published 13 March 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 L Shen and Q Xu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper proposes the least squares method to estimate the drift parameter for the stochastic differential equations driven bysmall noises which is more general than pure jump 120572-stable noises The asymptotic property of this least squares estimator isstudied under some regularity conditions The asymptotic distribution of the estimator is shown to be the convolution of a stabledistribution and a normal distribution which is completely different from the classical cases
1 Introduction
Stochastic differential equations (SDEs) are being extensivelyused as a model to describe some phenomena which aresubject to random influences it has found many applicationsin biology [1] medicine [2] econometrics [3 4] finance[5] geophysics [6] and oceanography [7] Then statisticalinference for these differential equations was of great interestand became a challenging theoretical problem For a morerecent comprehensive discussion we refer to [8 9]
The asymptotic theory of parametric estimation for dif-fusion processes with small white noise based on continuoustime observations is well developed and it has been studiedby many authors (see eg [10ndash14]) There have been manyapplications of small noise in mathematical finance see forexample [15ndash18]
In parametric inference due to the impossibility ofobserving diffusions continuously throughout a time intervalit is more practical and interesting to consider asymptoticestimation for diffusion processes with small noise based ondiscrete observations There are many approaches to driftestimation for discretely observed diffusions (see eg [19ndash23]) Long [24] has started the study on parameter estimationfor a class of stochastic differential equations driven by smallstable noise 119885
119905 119905 ge 0 However there has been no study on
parametric inference for stochastic processes with small Levynoises yet
In this paper we are interested in the study of parameterestimation for the following stochastic differential equationsdriven by more general Levy noise 119871
119905 119905 ge 0 based
on discrete observations We will employ the least squaresmethod to obtain an asymptotically consistent estimator
Let (ΩF F119905ge0
P) be a basic complete filtered prob-ability space satisfying the usual conditions that is thefiltration is continuous on the right and F
0contains all
P-null sets In this paper we consider a class of stochasticdifferential equations as follows
119889119883119905
= 120579119891 (119883119905) 119889119905 + 120576119892 (119883
119905minus
) 119889119871119905 119905 isin [0 1]
119871119905
= 119886119861119905+ 119887119885119905
119883 (0) = 1199090
(1)
where 119891 R rarr R and 119892 R rarr R are known functionsand 119886 119887 are known constants Let 119861
119905 119905 ge 0 be a standard
Brownian motion and let 119885119905 119905 ge 0 be a standard 120572-stable
Levy motion independent of 119861119905 119905 ge 0 with 119885
1sim 119878120572(1 120573 0)
for 120573 isin [0 1] 1 lt 120572 lt 2Let 119883 = 119883
119905 119905 ge 0 be a real-valued stationary process
satisfying the stochastic differential equation (1) and weassume that this process is observed at regularly spaced timepoints 119905
119894= 119894119899 119894 = 1 2 119899 Assume 119883
0
119905is the solution of
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 473681 6 pageshttpdxdoiorg1011552014473681
2 Abstract and Applied Analysis
the underlying ordinary differential equation (ODE) with thetrue value of the drift parameter 120579
0
1198891198830
119905= 1205790119891 (1198830
119905) 119889119905 119883
0
0= 1199090 (2)
Then we get
119883119905119894
minus 119883119905119894minus1
= int
119905119894
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (3)
2 Preliminaries
In this paper we denote 119862 as a generic constant whose valuemay vary from place to place
The following regularity conditions are assumed to hold
(A1) The functions 119891(119909) and 119892(119909) satisfy the Lipschitzconditions that is there exists a constant 119871 gt 0 suchthat
1003816100381610038161003816119891 (119909) minus 119891 (119910)
1003816100381610038161003816+
1003816100381610038161003816119892 (119909) minus 119892 (119910)
1003816100381610038161003816
le 1198711003816100381610038161003816119909 minus 119910
1003816100381610038161003816 119909 119910 isin R
(4)
(A2) There exist constants 119872 gt 0 and 119903 ge 0 satisfying
the growth condition
119892minus2
(119909) le 119872 (1 + |119909|119903
) 119909 isin R (5)
(A3) There exists a positive constant 119873 gt 0 such that
0 lt |119892(119909)| le 119873 lt infin
(A4) For 119862
119903= 2119903minus1
or 1 119903 gt 0
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
le 119862119903(
100381610038161003816100381610038161198830
119905119894minus1
10038161003816100381610038161003816
119903
+
10038161003816100381610038161003816119883119905119894minus1
minus 1198830
119905119894minus1
10038161003816100381610038161003816
119903
) (6)
The LSE of 120579119899120576
is defined as
120579119899120576
= argmin120579
120588119899120576
(120579) (7)
where the contrast function
120588119899120576
(120579) =
119899
sum
119894=1
10038161003816100381610038161003816100381610038161003816100381610038161003816
119883119905119894
minus 119883119905119894minus1
minus 120579119891 (119883119905119894minus1
) Δ119905119894minus1
120576119892 (119883119905119894minus1
)
10038161003816100381610038161003816100381610038161003816100381610038161003816
2
(8)
Then the 120579119899120576
can be represented explicitly as follows
120579119899120576
=
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) (119883119905119894
minus 119883119905119894minus1
)
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
(9)
Based on (3) and (9) there is a special decomposition for 120579119899120576
120579119899120576
=
1205790
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119891 (119883119904) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119887120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119885119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119886120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
= 1205790
+
Φ2
(119899 120576)
Φ1
(119899 120576)
+
Φ3
(119899 120576)
Φ1
(119899 120576)
+
Φ4
(119899 120576)
Φ1
(119899 120576)
(10)
Now we give an explicit expression for 120576minus1
(120579119899120576
minus 1205790) By using
(10) we have
120576minus1
(120579119899120576
minus 1205790) =
120576minus1
Φ2
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ3
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ4
(119899 120576)
Φ1
(119899 120576)
=
Ψ2
(119899 120576)
Φ1
(119899 120576)
+
Ψ3
(119899 120576)
Φ1
(119899 120576)
+
Ψ4
(119899 120576)
Φ1
(119899 120576)
(11)
One of the important tools we will employ is the under-lying lemma (see (35) in the Lemma 32 of [24])
Lemma 1 Under conditions (A1)-(A2) one has
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816
le 120576119890119871|1205790|119905
10038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119883119904minus
) 119889119885119904
10038161003816100381610038161003816100381610038161003816
(12)
sup0le119905le1
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816997888rarr1198750 as 120576 997888rarr 0 (13)
Abstract and Applied Analysis 3
3 Asymptotic Property of the LeastSquares Estimator
Theorem 2 Under the conditions (A1)ndash(A4) as 119899 rarr
infin 120576 rarr 0 119899120576 rarr infin and 119899120576120572(120572minus1)
rarr infin one has
120576minus1
(120579119899120576
minus 1205790)
997904rArr 119886
(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904
119873
+ 119887 (((int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus (int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802)
times (int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
minus1
)
(14)
where 1198801and 119880
2are independent random variables with 120572-
stable distribution 119878120572(1 120573 0) and 119873 is an independent random
variable with standard normal distribution
The theoremwill be proved by establishing several propo-sitionsWewill consider the asymptotic behaviors ofΦ
1(119899 120576)
Ψ119894(119899 120576) 119894 = 2 3 4 respectively
Proposition 3 Under conditions (A1)ndash(A4) and 119899 rarr infin
120576 rarr 0 one has
Φ1
(119899 120576) 997888rarr119875
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904 (15)
Proof Under conditions (A1)ndash(A3) Proposition 3 can be
proved by using condition (A4) (see the proof of Proposition
33 in [24])
Proposition 4 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576 rarr infin one has
Ψ2
(119899 120576) 997888rarr1198750 (16)
Proof For 119905119894minus1
le 119905 le 119905119894 119894 = 1 2 119899
119883119905
= 119883119905119894minus1
+ int
119905
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (17)
It follows that10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le int
119905
119905119894minus1
10038161003816100381610038161205790
1003816100381610038161003816(
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816) 119889119904
+ 120576
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904
100381610038161003816100381610038161003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816119872 int
119905
119905119894minus1
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+ 119899minus1 1003816
1003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+ 119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
(18)
Using Gronwall inequality we get10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872(119905minus119905
119894minus1)
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(19)
which yields
sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872119899
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(20)
thus under conditions (A1) and (A
3)
1003816100381610038161003816Φ2
(119899 120576)1003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816
119899
sum
119894=1
119872 (1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
(119891 (119883119904) minus 119891 (119883
119905119894minus1
)) 119889119904
100381610038161003816100381610038161003816100381610038161003816
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
119899
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
4 Abstract and Applied Analysis
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
1198992
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
2
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890(|1205790|119872)119899
119899
119887120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
119899
119886120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
= Φ21
(119899 120576) + Φ22
(119899 120576) + Φ23
(119899 120576)
(21)
Then
1003816100381610038161003816Ψ2
(119899 120576)1003816100381610038161003816
le 120576minus1
Φ21
(119899 120576) + 120576minus1
Φ22
(119899 120576)
+ 120576minus1
Φ23
(119899 120576)
= Ψ21
(119899 120576) + Ψ22
(119899 120576) + Ψ23
(119899 120576)
(22)
Using (13) in Lemma 1 conditions (A1) and (A
4) we get
Ψ21
(119899 120576) rarr1198750 as 119899 rarr infin 120576 rarr 0 and 119899120576 rarr infin (see (326)
in [24]) By using the same techniques under condition (A2)
we can prove thatΨ2119895
(119899 120576) rarr1198750 119895 = 2 3 as 119899 rarr infin 120576 rarr 0
respectively
Proposition 5 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576120572(120572minus1)
rarr infin one has
Ψ3
(119899 120576)
997904rArr 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802
(23)
Proof Under conditions (A1)ndash(A3) Proposition 5 can be
proved by using condition (A4) (see the proof of Proposition
44 in [24])
Proposition 6 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 one has
Ψ4
(119899 120576) 997904rArr 119886(int
1
0
10038161003816100381610038161003816119892minus2
(1198830
119904)
100381610038161003816100381610038161198912
(1198830
119904) 119889119904)
12
119873 (24)
Proof Note that
Ψ4
(119899 120576) = 119886
119899
sum
119894=1
119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) int
119905119894
119905119894minus1
119892 (1198830
119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
)) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
))
times (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
=
5
sum
119895=1
Ψ4119895
(119899 120576)
(25)
For Ψ41
(119899 120576) let 119884119894
= int
119905119894
119905119894minus1
119892(119883119904minus
)119889119861119904 119895 = 1 119899 Then it is
easy to see that 119884119894
sim 119873(0 int
119905119894
119905119894minus1
1198922
(119883119904minus
)119889119904) and 1198841 119884
119899are
independent normal random variablesIt follows that
Ψ41
(119899 120576)
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) 119884119894
sim 119873 (0 1198862
119899
sum
119894=1
119892minus4
(1198830
119905119894minus1
) 1198912
(1198830
119905119894minus1
) int
119905119894
119905119894minus1
1198922
(119883119904minus
) 119889119904)
997904rArr 119886(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
119873
(26)
as 119899 rarr infin 120576 rarr 0
Abstract and Applied Analysis 5
For Ψ42
(119899 120576) using Markov inequality and Itorsquos isometryproperty for any given 120578 gt 0
Ψ42
(119899 120576)
le
1
120578
E[119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
]
le
119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
))
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119883119904minus
minus 1198830
119904minus
)
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [ sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119904minus
minus 1198830
119904minus
10038161003816100381610038161003816119899minus12
]
(27)
By using (13) Ψ42
(119899 120576) rarr 0 as 119899 rarr infin 120576 rarr 0Applying similar techniques to Ψ
4119895(119899 120576) 119895 = 3 4 5 we
get Ψ4119895
(119899 120576) rarr 0 119895 = 3 4 5 as 119899 rarr infin 120576 rarr 0
Now we can proveTheorem 2
Proof By using Propositions 3 4 5 6 and Slutskyrsquos theoremwe can get the conclusion
4 Example
We consider the following nonlinear SDE driven by generalLevy noises
119889119883119905
= 120579119883119905119889119905 +
120576
1 + 1198832
119905minus
119889119871119905 119905 isin [0 1] 119883
0= 1199090
(28)
where 119891(119909) = 119909 119892(119909) = 1(1 + 1199092
) 1199090and 120576 are known
constants and 120579 = 0 is an unknown parameterFor simplicity let 119909
0gt 0 120576 = 0 we get the ODE
1198891198830
119905= 12057901198830
119905119889119905 119905 isin [0 1] 119883
0
0= 1199090
(29)
and the solution
1198830
119905= 11990901198901205790119905
(30)
Then the asymptotic distribution is
119886(int
1
0
(1 + 1199092
011989021205790119904
)
2
1199092
011989021205790119904
119889119904)
minus12
119873
+ 119887
(int
1
0
(1 + 1199092
011989021205790119904
)
120572
119909120572
01198901205721205790119904
119889119904)
1120572
int
1
0
(1 + 1199092
011989021205790119904)2
1199092
011989021205790119904119889119904
119878120572
(1 120573 0)
(31)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R I Jennrich and P B Bright ldquoFitting systems of linear dif-ferential equations using computer generated exact derivativesrdquoTechnometrics vol 18 no 4 pp 385ndash392 1976
[2] R H Jones ldquoFitting multivariate models to unequally spaceddatardquo in Time Series Analysis of Irregularly Observed Data pp158ndash188 1984
[3] A R Bergstrom Statistical Inference in Continuous TimeEconomic Models vol 99 North-Holland Amsterdam TheNetherlands 1976
[4] A R Bergstrom ldquoThe history of continuous-time econometricmodelsrdquo Econometric Theory vol 4 no 3 pp 365ndash383 1988
[5] F Black and M Scholes ldquoThe pricing of options and corporateliabilitiesrdquoThe Journal of Political Economy vol 81 pp 637ndash6541973
[6] M Arato Linear Stochastic Systems with Constant CoefficientsSpringer Berlin Germany 1982
[7] R J Adler and P Meuller Stochastic Modelling on PhysicalOceanography vol 39 Springer New York NY USA 1996
[8] B P Rao B L P Rao I Statisticien B L P Rao B L P Rao andI Statistician Statistical Inference for di Usion Type ProcessesArnold London UK 1999
[9] Y A Kutoyants Statistical Inference for Ergodic Diffusion Pro-cesses Springer London UK 2004
[10] Yu A Kutoyants Parameter Estimation for Stochastic Processesvol 6 Heldermann Berlin Germany 1984
[11] Yu Kutoyants Identification of Dynamical Systems with SmallNoise Kluwer Academic Dordrecht The Netherlands 1994
[12] N Yoshida ldquoAsymptotic expansions of maximum likelihoodestimators for small diffusions via the theory of Malliavin-Watanaberdquo Probability Theory and Related Fields vol 92 no 3pp 275ndash311 1992
[13] N Yoshida ldquoConditional expansions and their applicationsrdquoStochastic Processes and their Applications vol 107 no 1 pp 53ndash81 2003
[14] M Uchida and N Yoshida ldquoInformation criteria for smalldiffusions via the theory of Malliavin-Watanaberdquo StatisticalInference for Stochastic Processes vol 7 no 1 pp 35ndash67 2004
[15] N Yoshida ldquoAsymptotic expansion of Bayes estimators for smalldiffusionsrdquo Probability Theory and Related Fields vol 95 no 4pp 429ndash450 1993
[16] A Takahashi ldquoAn asymptotic expansion approach to pricingfinancial contingent claimsrdquoAsia-Pacific FinancialMarkets vol6 no 2 pp 115ndash151 1999
6 Abstract and Applied Analysis
[17] N Kunitomo and A Takahashi ldquoThe asymptotic expansionapproach to the valuation of interest rate contingent claimsrdquoMathematical Finance vol 11 no 1 pp 117ndash151 2001
[18] A Takahashi and N Yoshida ldquoAn asymptotic expansionscheme for optimal investment problemsrdquo Statistical Inferencefor Stochastic Processes vol 7 no 2 pp 153ndash188 2004
[19] V Genon-Catalot ldquoMaximum contrast estimation for diffusionprocesses fromdiscrete observationsrdquo Statistics vol 21 no 1 pp99ndash116 1990
[20] A Gloter and M Soslashrensen ldquoEstimation for stochastic differ-ential equations with a small diffusion coefficientrdquo StochasticProcesses and their Applications vol 119 no 3 pp 679ndash6992009
[21] C F Laredo ldquoA sufficient condition for asymptotic sufficiencyof incomplete observations of a diffusion processrdquo The Annalsof Statistics vol 18 no 3 pp 1158ndash1171 1990
[22] M Uchida ldquoApproximate martingale estimating functions forstochastic differential equations with small noisesrdquo StochasticProcesses and their Applications vol 118 no 9 pp 1706ndash17212008
[23] M Soslashrensen and M Uchida ldquoSmall-diffusion asymptotics fordiscretely sampled stochastic differential equationsrdquo Bernoullivol 9 no 6 pp 1051ndash1069 2003
[24] H Long ldquoParameter estimation for a class of stochastic dif-ferential equations driven by small stable noises from discreteobservationsrdquo Acta Mathematica Scientia B vol 30 no 3 pp645ndash663 2010
Research ArticleOn the Deficiencies of Some Differential-Difference Polynomials
Xiu-Min Zheng1 and Hong Yan Xu2
1 Institute of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China2Department of Informatics and Engineering Jingdezhen Ceramic Institute Jingdezhen 333403 China
Correspondence should be addressed to Xiu-Min Zheng zhengxiumin2008sinacom
Received 1 November 2013 Accepted 4 January 2014 Published 27 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 X-M Zheng and H Y Xu This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
The characteristic functions of differential-difference polynomials are investigated and the result can be viewed as a differential-difference analogue of the classic Valiron-Mokhonrsquoko Theorem in some sense and applied to investigate the deficiencies of somehomogeneous or nonhomogeneous differential-difference polynomials Some special differential-difference polynomials are alsoinvestigated and these results on the value distribution can be viewed as differential-difference analogues of some classic results ofHayman and Yang Examples are given to illustrate our results at the end of this paper
1 Introduction
Throughout this paper we use standard notations in theNevanlinna theory (see eg [1ndash3]) Let 119891(119911) be a meromor-phic function Here and in the following the word ldquomero-morphicrdquo means being meromorphic in the whole complexplane We use normal notations 119898(119903 119891) 119879(119903 119891) 119873(119903 119891)119873(119903 1119891) 120590(119891) 120582(119891) and 120582(1119891) And we also use 120590
2(119891)
to denote the hyperorder of 119891(119911) and 120575(120572 119891) to denote theNevanlinna deficiency of 120572 with respect to 119891(119911) Moreoverwe denote by 119878(119903 119891) any real quantity satisfying 119878(119903 119891) =119900(119879(119903 119891)) as 119903 rarr infin outside of a possible exceptional set offinite logarithmic measure
Recently with some establishments of difference ana-logues of the classic Nevanlinna theory (two typical andmost important ones can be seen in [4ndash6]) there has beena renewed interest in the properties of complex differenceexpressions and meromorphic solutions of complex differ-ence equations (see eg [4ndash17]) By combining complex dif-ferentiates and complex differences we proceed in this way inthis paper
It is well known that the following Valiron-MokhonrsquokoTheorem due to Valiron [18] and A Z Mokhonrsquoko and V DMokhonrsquoko [19] is of essential importance in the theory ofcomplex differential equations and functional equations
Theorem A (see [2 3]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891 (119911)) =
sum119898
119894=0119886119894(119911) 119891(119911)
119894
sum119899
119895=0119887119895(119911) 119891(119911)
119895
(1)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) the characteristic
function of 119877(119911 119891(119911)) satisfies
119879 (119903 119877 (119911 119891 (119911))) = 119889119879 (119903 119891) + 119874 (Ψ (119903)) (2)
where 119889 = max119898 119899 and Ψ(119903) = max119894119895119879(119903 119886
119894) 119879(119903 119887
119895)
Noting that the difference analogue of Theorem A maynot hold we have obtained a result of this type in [16] byadding some additional assumptions as follows
Theorem B (see [16]) Suppose that 119875(119911 119891) is a differencepolynomial of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
1198940119891(119911 + 119888
1)1198941sdot sdot sdot 119891(119911 + 119888
119899)119894119899
(3)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 378151 12 pageshttpdxdoiorg1011552014378151
2 Abstract and Applied Analysis
containing just one monomial of degree 119889(119875) and 119891(119911) is atranscendental meromorphic function of finite order If 119891(119911)also satisfies119873(119903 119891) + 119873(119903 1119891) = 119878(119903 119891) then we have
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (4)
In this paper we consider removing the assumptionldquo119875(119911 119891) contains just one monomial of degree 119889(119875)rdquo in The-orem B and obtain a weaker result which is also generalizedinto differential-difference case The concrete result can beseen in Section 2
Next we recall a classic result concerning Picardrsquos valuesofmeromorphic functions and its derivatives due toHayman[20]
Theorem C (see [20]) Let 119891(119911) be a transcendental entirefunction Then
(a) for 119899 ge 3 and 119886 = 0 Ψ(119911) = 1198911015840(119911) minus 119886(119891(119911))119899 assumesall finite values infinitely often
(b) for 119899 ge 2Φ(119911) = 1198911015840(119911)(119891(119911))119899 assumes all finite valuesexcept possibly zero infinitely often
Corresponding difference analogues ofTheoremC can beseen in [12 17]
Theorem D (see [12 17]) Let 119891(119911) be a transcendental entirefunction of finite order and let 119888 be a nonzero complex constantThen
(a) for 119899 ge 3 and 119886 = 0Ψ1(119911) = 119891(119911+ 119888) minus 119886119891(119911)
119899 assumesall finite complex values infinitely often
(b) for 119899 ge 2 Φ1(119911) = 119891(119911 + 119888)119891(119911)
119899 assumes all finitecomplex values except possibly zero infinitely often
AfterTheoremCmany results have been obtained on thevalue distribution of differential polynomials A typical one isas follows
TheoremE (see [21 22]) Let119891 be a transcendental meromor-phic function with119873(119903 119891) +119873(119903 1119891) = 119878(119903 119891) and let Ψ bea differential polynomial in 119891 of the form
Ψ (119911) = sum119886 (119911) 119891(119911)11989701198911015840
(119911)1198971sdot sdot sdot 119891(119896)
(119911)119897119896 (5)
with no constant term Furthermore assume the degree 119899 ofΨ is greater than one and 119897
0lt 119899 0 le 119897
119894le 119899 for all 119894 = 0 Then
120575(119886 Ψ) lt 1 for all 119886 = 0infin Moreover if all the terms ofΨ havedifferent degrees at least two that is Ψ is nonhomogeneousthen 120575(119886 Ψ) le 1 minus (12119899) for all 119886 =infin
We also consider deficiencies of difference polynomialsof meromorphic functions of finite order in [16] which canbe viewed as difference analogues of Theorem E as well asgeneralizations of Theorem D
In this paper we proceed to investigate deficiencies ofdifferential-difference polynomials of meromorphic func-tions The concrete results can be seen in Section 3
Examples are given in Section 4 to illustrate our results
2 A Differential-Difference Analogue ofValiron-Mokhonrsquoko Theorem
In what follows we will consider differential-difference poly-nomials A differential-difference polynomial is a polynomialin 119891(119911) its shifts its derivatives and derivatives of its shifts(see [14]) that is an expression of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
120582001198911015840
(119911)12058201sdot sdot sdot 119891(119898)
(119911)1205820119898
times 119891(119911 + 1198881)120582101198911015840
(119911 + 1198881)12058211sdot sdot sdot 119891(119898)
(119911 + 1198881)1205821119898
sdot sdot sdot 119891(119911 + 119888119899)12058211989901198911015840
(119911 + 119888119899)1205821198991sdot sdot sdot 119891(119898)
(119911 + 119888119899)120582119899119898
= sum
120582isin119868
119886120582(119911)
119899
prod
119894=0
119898
prod
119895=0
119891(119895)
(119911 + 119888119894)120582119894119895
(6)
where 119868 is a finite set of multi-indices 120582 =
(12058200 120582
0119898 12058210 120582
1119898 120582
1198990 120582
119899119898) and 119888
0(= 0)
and 1198881 119888
119899are distinct complex constants And we assume
that the meromorphic coefficients 119886120582(119911) 120582 isin 119868 of 119875(119911 119891) are
of growth 119878(119903 119891) We denote the degree of the monomialprod119899
119894=0prod119898
119895=0119891(119895)
(119911 + 119888119894)120582119894119895 of 119875(119911 119891) by 119889(120582) = sum119899
119894=0sum119898
119895=0120582119894119895
Then we denote the degree and the lower degree of 119875(119911 119891)by
119889 (119875) = max120582isin119868
119889 (120582) 119889lowast
(119875) = min120582isin119868
119889 (120582) (7)
respectively In particular we call 119875(119911 119891) a homogeneousdifferential-difference polynomial if 119889(119875) = 119889
lowast
(119875) Other-wise 119875(119911 119891) is nonhomogeneous
In the following we assume 119889(119875) ge 1 and 119875(119911 119891) equiv
119875(119911 0)We prove a weaker differential-difference version of the
classic Valiron-MokhonrsquokoTheorem as follows
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic function and 119875(119911 119891) is a differential-difference polyno-mial of the form (6) If 119891(119911) also satisfies 120590
2(119891) lt 1 and
119873(119903 119891) + 119873(119903
1
119891
) = 119878 (119903 119891) (8)
then one has
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891))
le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
(9)
Remark 2 If119875(119911 119891) is a homogeneous differential-differencepolynomial in addition then
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (10)
Remark 3 Especially assumption (8) can be replaced by theassumption ldquomax120582(119891) 120582(1119891) lt 120590(119891)rdquo In fact if 119891(119911)satisfies max120582(119891) 120582(1119891) lt 120590(119891) then 119891(119911) is of regulargrowth and (8) holds consequently
Abstract and Applied Analysis 3
To proveTheorem 1 we need the following lemmas
Lemma 4 (see [6]) Let 119891(119911) be a nonconstant meromorphicfunction 120576 gt 0 and 119888 isin C If 120577 = 120590
2(119891) lt 1 then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120577minus120576
) (11)
for all 119903 outside of a set of finite logarithmic measure
Lemma 5 (see [6]) Let 119879 [0 +infin) rarr [0 +infin) be anondecreasing continuous function and let 119904 isin (0 +infin)If the hyperorder of 119879 is strictly less than one that islim119903rarrinfin
(log2119879(119903) log 119903) = 120577 lt 1 and 120575 isin (0 1 minus 120577) then
119879 (119903 + 119904) = 119879 (119903) + 119900 (
119879 (119903)
119903120575
) (12)
where 119903 runs to infinity outside of a set of finite logarithmicmeasure
It is shown in [23 p66] and [7 Lemma 1] that the ine-quality
(1 + 119900 (1)) 119879 (119903 minus |119888| 119891) le 119879 (119903 119891 (119911 + 119888))
le (1 + 119900 (1)) 119879 (119903 + |119888| 119891)
(13)
holds for 119888 = 0 and 119903 rarr infin And from the proof theabove relation is also true for counting function By combingLemma 5 and these inequalities we immediately deduce thefollowing lemma
Lemma 6 Let 119891(119911) be a nonconstant meromorphic functionof 1205902(119891) lt 1 and let 119888 be a nonzero complex constant Then
one has
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891)
119873 (119903 119891 (119911 + 119888)) = 119873 (119903 119891) + 119878 (119903 119891)
119873(119903
1
119891 (119911 + 119888)
) = 119873(119903
1
119891
) + 119878 (119903 119891)
(14)
Lemma7 Let119891(119911) be a transcendentalmeromorphic functionof 1205902(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then we one has
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891) + 119878 (119903 119891) (15)
Furthermore if 119891(119911) also satisfies
119873(119903 119891) = 119878 (119903 119891) (16)
then one has
119879 (119903 119875 (119911 119891)) le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (17)
Proof For 119894 = 0 1 119899 119895 = 0 1 119898 we define 119892119894119895(119911) =
119891(119895)
(119911 + 119888119894)119891(119911) We also define
119892lowast
119894119895(119911) =
119892119894119895(119911) if 1003816100381610038161003816
1003816119892119894119895(119911)
10038161003816100381610038161003816gt 1
1 if 10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816le 1
119891lowast
(119911) =
119891 (119911) if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816le 1
(18)
Thus
1003816100381610038161003816119875 (119911 119891)
1003816100381610038161003816le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119891 (119911)
1003816100381610038161003816
119889(120582)
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
120582119894119895
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(120582)
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)(
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
1003816100381610038161003816119891lowast
(119911)1003816100381610038161003816)
119889(119875)
(19)
By the definitions of 119891lowast(119911) and 119892lowast119894119895(119911) 119894 = 0 1 119899 119895 =
0 1 119898 we have
119898(119903 119891lowast
) = 119898 (119903 119891)
119898 (119903 119892lowast
119894119895) = 119898 (119903 119892
119894119895) 119894 = 0 119899 119895 = 0 119898
(20)
It follows by (19) and (20) that
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
= 119889 (119875)119898 (119903 119891)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892119894119895) + 119878 (119903 119891)
(21)
Lemmas 4 and 6 and the logarithmic derivative lemma implythat for 119894 = 0 1 119899 and 119895 = 0 1 119898
119898(119903 119892119894119895) = 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911)
)
le 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911 + 119888119894)
) + 119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
= 119878 (119903 119891 (119911 + 119888119894)) + 119878 (119903 119891) = 119878 (119903 119891)
(22)
Then (15) follows by (21) and (22)
4 Abstract and Applied Analysis
It is easy to find that
119873(119903 119875 (119911 119891)) = 119874(119873(119903 119891) +
119899
sum
119894=1
119873(119903 119891 (119911 + 119888119894)))
+ 119878 (119903 119891)
(23)
Then (16) (23) and Lemma 6 yield that
119873(119903 119875 (119911 119891)) = 119878 (119903 119891) (24)
Thus (17) follows by (15) and (24)
Lemma 8 Let 119891(119911) be a transcendental meromorphic func-tion of 120590
2(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then one has
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(25)
Proof Similar to (19) we have
100381610038161003816100381610038161003816100381610038161003816
119875 (119911 119891)
119891(119911)119889(119875)
100381610038161003816100381610038161003816100381610038161003816
le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895 1003816100381610038161003816119892 (119911)
1003816100381610038161003816
119889(119875)minus119889(120582)
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889(120582)
)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889lowast(119875)
(26)
where 119892(119911) = 1119891(119911) and
119892lowast
(119911) =
119892 (119911) if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816le 1
(27)
By the definition of 119892lowast(119911) we have 119898(119903 119892lowast) = 119898(119903 119892) =
119898(119903 1119891) Thus (20) (22) and (26) yield that
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898 (119903 119892lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(28)
that is (25)
Now we can finish the proof of Theorem 1 in the end
Proof of Theorem 1 We deduce from (8) (24) and Lemma 8that
119889 (119875) 119879 (119903 119891)
= 119879 (119903 119891119889(119875)
) le 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119873 (119903 119875 (119911 119891))
+ 119889 (119875)119873(119903
1
119891
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119879 (119903 119875 (119911 119891)) + 119878 (119903 119891)
(29)
that is
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891)) (30)
Then (9) follows by (17) and (30)
3 Deficiencies of SomeDifferential-Difference Polynomials
In the following we assume that 120572(119911)( equiv 0) is ameromorphicfunction of growth 119878(119903 119891)
In this section we will apply Theorem 1 to consider thedeficiencies of general homogeneous or nonhomogeneousdifferential-difference polynomials
Theorem 9 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)
(a) If119875(119911 119891) is a homogeneous differential-difference poly-nomial then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
= 1 120575 (120572 119875 (119911 119891)) = 0
(31)
(b) If 119875(119911 119891) is a nonhomogeneous differential-differencepolynomial with 2119889lowast(119875) gt 119889(119875) then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
ge
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
120575 (120572 119875 (119911 119891)) le 1 minus
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
lt 1
(32)
Thus 119875(119911 119891)minus120572(119911) has infinitely many zeros whether 119875(119911 119891)is homogeneous or nonhomogeneous
Abstract and Applied Analysis 5
Furthermore one considers some differential-differencepolynomials of special forms which are generalizations ofboth differential cases and difference cases that is TheoremsCndashE
Theorem 10 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 V minus 1 of growth 119878(119903 119891) If 119906V gt 119889(119875) 119906 = 1 then
1198761(119911 119891) = 119865 (119891) 119875 (119911 119891) (33)
satisfies
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(34)
Thus 1198761(119911 119891) minus 120572(119911) has infinitely many zeros
When 119865(119891) is of a special form 119891V we can deduce the
following result fromTheorem 9
Theorem 11 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6) If V isin N 1and V + 2119889lowast(119875) gt 119889(119875) then
1198762(119911 119891) = 119891
V119875 (119911 119891) (35)
satisfies 120575(120573 1198762(119911 119891)) lt 1 where 120573 isin C0Thus119876
2(119911 119891)minus
120573 has infinitely many zeros
Remark 12 On the one hand we can also applyTheorem 9 to1198761(119911 119891)with the assumption ldquo2(119889lowast(119875)+119889lowast(119865)) gt 119889(119875)+119906Vrdquo
and obtain the same result as Theorem 10 But our presentassumption ldquo119906V gt 119889(119875)rdquo has no concern with 119889lowast(119875) and119889lowast
(119865) so we think Theorem 10 is better to some extent Onthe other hand we can also apply Theorem 10 to 119876
2(119911 119891)
with the assumption ldquoV gt 119889(119875)rdquo which is stronger thanldquoV + 2119889lowast(119875) gt 119889(119875)rdquo in Theorem 11 showing Theorem 11 isbetter to some extent
Theorem 13 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 Vminus1 of growth 119878(119903 119891) If (119906minus1)119906V(2119906minus1) gt 119889(119875) 119906 = 1then
1198763(119911 119891) = 119865 (119891) + 119875 (119911 119891) (36)
satisfies
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(37)
Thus 1198763(119911 119891) minus 120572(119911) has infinitely many zeros
When 119906 = 1 one can consider some special cases asfollows
Theorem 14 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)(a) If V gt 119889(119875) + 2 ge 3 then
1198764(119911 119891) = 119891
V+ 119875 (119911 119891) (38)
satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(39)
(b) If (Vminus1)V(2Vminus1) gt 119889(119875) V ge 3 then1198764(119911 119891) satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
1
Vminus
2V minus 1V2
119889 (119875)
120575 (120572 1198764(119911 119891)) le
1
V+
2V minus 1V2
119889 (119875) lt 1
(40)
Especially it holds for V = 119889(119875) + 2 = 3(c) If V ge 119889(119875) + 2 ge 3 and 119891 also satisfies 119873(119903 1119891) =119878(119903 119891) then 119876
4(119911 119891) satisfies 120575(120572 119876
4(119911 119891)) lt 1
Especially it holds for V = 119889(119875) + 2 gt 3Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many zeros
If we assume that 119873(119903 1119891) = 119878(119903 119891) in addition thefollowing result follows immediately by Theorem 9
Theorem 15 Suppose that 119891(119911) is a transcendental mero-morphic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891)
is a differential-difference polynomial of the form (6) If2 min119889lowast(119875) V gt max119889(119875) V then 119876
4(119911 119891) satisfies
120575(120572 1198764(119911 119891)) lt 1 Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many
zeros
Remark 16 Noting that when V gt 3 (V minus 1)V(2V minus 1) leV minus 2 hold we see that the assumption ldquoV gt 119889(119875) + 2rdquo inTheorem 14(a) is weaker than the assumption ldquo(V minus 1)V(2V minus1) gt 119889(119875)rdquo in Theorem 14(b) And these assumptions inTheorem 14 have no concernwith119889lowast(119875)) thus they are differ-ent from the assumption ldquo2 min119889lowast(119875) V gt max119889(119875) Vrdquoin Theorem 15
6 Abstract and Applied Analysis
Remark 17 From the proofs behind it is easy to find that
120582 (119875 (119911 119891) minus 120572) = 120590 (119875 (119911 119891)) = 120590 (119891)
120582 (119876119894(119911 119891) minus 120572) = 120590 (119876
119894(119911 119891)) = 120590 (119891) 119894 = 1 3 4
(41)
hold respectively inTheorems 9 10 13 14(a) and (b) and 15Now we give the proofs of Theorems 9ndash15
Proof of Theorem 9 It follows byTheorem 1 that
119878 (119903 119891) = 119878 (119903 119875 (119911 119891)) (42)
We deduce from (8) (24) (25) and (42) that
119873(119903
1
119875 (119911 119891)
)
le 119873(119903
1
119891119889(119875)
) + 119873(119903
119891119889(119901)
119875 (119911 119891)
)
le 119873(119903
1
119891
) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119875 (119911 119891))
(43)
Thus an application of the second main theorem and (24)(42) and (43) imply that
119879 (119903 119875 (119911 119891)) le 119873 (119903 119875 (119911 119891)) + 119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(44)
(a) If 119889(119875) = 119889lowast(119875) then it follows by (44) that
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(45)
by which (31) holds
(b) If 2119889lowast(119875) gt 119889(119875) then we deduce from (30) and (44)that
119879 (119903 119875 (119911 119891)) le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le
119889 (119875) minus 119889lowast
(119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891))
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(46)
that is
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119878 (119903 119875 (119911 119891))
(47)
Since 2119889lowast(119875)minus119889(119875) gt 0 (32) follows immediately by (47)
Proof of Theorem 10 We deduce from (16) (17) and (24) that
119879 (119903 1198761(119911 119891)) le (119906V + 119889 (119875)) 119879 (119903 119891) + 119878 (119903 119891) (48)
119873(119903 1198761(119911 119891)) = 119874 (119873 (119903 119891)) + 119873 (119903 119875 (119911 119891)) + 119878 (119903 119891)
= 119878 (119903 119891)
(49)
hold Next we consider 119873(119903 11198761(119911 119891)) Let 119911
0be a zero of
1198761(119911 119891) and distinguish three cases
(i) 1199110is not a zero of 119865(119891) then 119911
0must be a zero of
119875(119911 119891) and
119906 le 120596(
1
1198761(119911 119891)
1199110) + (119906 minus 1) 120596(
1
119875 (119911 119891)
1199110) (50)
where 120596(119891 1199110) denotes the order of multiplicity of 119911
0or zero
according as 1199110is a pole of 119891(119911) or not
(ii) 1199110is a zero of 119865(119891) but not a pole of 119875(119911 119891) Then
119906 le 120596(
1
1198761(119911 119891)
1199110) (51)
(iii) 1199110is a zero of 119865(119891) and a pole of 119875(119911 119891) Then
119906 le 120596(
1
119865 (119891)
1199110) le 120596(
1
1198761(119911 119891)
1199110) + 120596 (119875 (119911 119891) 119911
0)
(52)
Abstract and Applied Analysis 7
(24) and (50)ndash(52) yield that
119906119873(119903
1
1198761(119911 119891)
) le 119873(119903
1
1198761(119911 119891)
)
+ (119906 minus 1)119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(53)
Then (48) (49) (53) and an application of the second maintheorem to 119876
1(119911 119891) imply that
119879 (119903 1198761(119911 119891))
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
1198761(119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 1198761(119911 119891))
le
1
119906
119873(119903
1
1198761(119911 119891)
) +
119906 minus 1
119906
119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(54)
consequently
119879 (119903 1198761(119911 119891)) le 119873(119903
1
119875 (119911 119891)
)
+
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(55)
Moreover by 119891119889(119875)119865(119891) = 119891119889(119875)
1198761(119911 119891)119875(119911 119891) (16)
(24) (25) andTheorem A we have
(119889 (119875) + 119906V)119898 (119903 119891)
= 119898(119903
119891119889(119875)
1198761(119911 119891)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) minus 119873(119903
119891119889(119875)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898(119903
1
119891
)
+ 119889 (119875) (119873(119903
1
119891
) minus 119873 (119903 119891))
+ 119873 (119903 119875 (119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
= 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898 (119903 119891)
minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(56)
consequently
119906V119898(119903 119891) le 119898 (119903 1198761(119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(57)
On the other hand the evident relation 119906V120596(119891 1199110) le
120596(119865(119891) 1199110) + 119906VsumVminus1
119895=0120596(119886119895 1199110) where the definition of
120596(119891 1199110) is given after (50) results in
119906V119873(119903 119891) le 119873 (119903 119865 (119891)) + 119878 (119903 119891)
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(58)
We deduce from (57) and (58) that
119906V119879 (119903 119891) le 119879 (119903 1198761(119911 119891)) + 119878 (119903 119891) (59)
Then (17) (55) and (59) yield that
119906V119879 (119903 119891)
le 119873(119903
1
119875 (119911 119891)
) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
le 119889 (119875) 119879 (119903 119891) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(60)
that is
(119906 minus 1) (119906V minus 119889 (119875))119906
119879 (119903 119891) le 119873(119903
1
1198761(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(61)
From (48) and (61) we deduce that
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(62)
Proof of Theorem 11 Assume to the contrary that120575(120573 119876
2(119911 119891)) = 1 Denoting
1198762(119911 119891) minus 120573 = 119891
V(119911) 119875 (119911 119891) minus 120573 = 119866 (119911) (63)
we deduce from (16) and (17) that
119873(119903
1
119866
) = 119873(119903
1
1198762(119911 119891) minus 120573
)
= 119878 (119903 1198762(119911 119891)) = 119878 (119903 119891)
(64)
8 Abstract and Applied Analysis
On the other hand (16) and (24) yield that
119873(119903 119866) = 119873 (119903 1198762(119911 119891) minus 120573) = 119878 (119903 119891) (65)
Differentiating both sides of (63) we obtain
119891Vminus1(119911) 119877 (119911 119891) = 119866
1015840
(119911) (66)
where 119877(119911 119891) = V1198911015840(119911)119875(119911 119891) + 119891(119911)1198751015840(119911 119891) Clearly wededuce from (16) and (24) that
119873(119903 119877 (119911 119891)) = 119878 (119903 119891) (67)
Moreover (64) and (65) yield that
119873(119903
1
1198661015840
) le 119873(119903
119866
1198661015840
) + 119873(119903
1
119866
)
le 119879(119903
1198661015840
119866
) + 119873(119903
1
119866
) + 119874 (1)
le 119898(119903
1198661015840
119866
) + 119873 (119903 119866) + 2119873(119903
1
119866
) + 119874 (1)
= 119878 (119903 119866) + 119878 (119903 119891) = 119878 (119903 119891)
(68)
It follows by (66)ndash(68) that
119873(119903
1
119891
) =
1
V minus 1119873(119903
119877 (119911 119891)
1198661015840
) = 119878 (119903 119891) (69)
Then (16) (69) and the fact 2(119889lowast(119875) + V) gt 119889(119875) + V implythat the assumptions of Theorem 9(b) are satisfied ThusTheorem 9(b) yields that 120575(120573 119876
2(119911 119891)) lt 1 a contradiction
Therefore we have 120575(120573 1198762(119911 119891)) lt 1
Proof of Theorem 13 We deduce from (16) (17) and (24) that
119879 (119903 1198763(119911 119891)) le max 119906V 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
= 119906V119879 (119903 119891) + 119878 (119903 119891) (70)
Denote
119867(119911) =
minus119875 (119911 119891) + 120572 (119911)
119865 (119891)
(71)
Now we estimate the poles the zeros and 1-points of119867(119911) accuratelyOn the one handwe see by (71) that the polesof 119867(119911) occur at zeros of 119865(119891) and poles of minus119875(119911 119891) + 120572(119911)which are not simultaneously 1-points of 119867(119911) and thosepoles of119867(119911)which are zeros of 119865(119891) but not simultaneouslyzeros of minus119875(119911 119891) + 120572(119911) also have multiplicities at least 119906 Onthe other handwe also see by (71) that the zeros of119867(119911) occurat zeros of minus119875(119911 119891) + 120572(119911) and poles of 119865(119891) which are notsimultaneously 1-points of 119867(119911) Moreover 1-points of 119867(119911)occur at zeros of 119876
3(119911 119891) minus 120572(119911) and occur at the common
poles zeros of 119865(119891) and minus119875(119911 119891) + 120572(119911) with the samemultiplicities Thus it follows by (16) and (24) that
119873(119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
)
le
1
119906
119873 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(72)
Then (17) (72) and the second main theorem result in
119879 (119903119867) le 119873 (119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
) + 119878 (119903119867)
le
1
119906
119879 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
le
1
119906
119879 (119903119867) + 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(73)
that is
(1 minus
1
119906
)119879 (119903119867) le 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(74)
Moreover Theorem A and (17) imply that
119906V119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119865 (119891)) = 119879(119903minus119875 (119911 119891) + 120572
119867
)
le 119889 (119875) 119879 (119903 119891) + 119879 (119903119867) + 119878 (119903 119891)
(75)
that is
(119906V minus 119889 (119875)) 119879 (119903 119891) le 119879 (119903119867) + 119878 (119903 119891) (76)
Then (74) and (76) yield that
((119906 minus 1) V minus2119906 minus 1
119906
119889 (119875))119879 (119903 119891) le 119873(119903
1
1198763(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(77)
Abstract and Applied Analysis 9
From (70) and (77) we deduce that
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(78)
To prove Theorem 14(c) we also need the followinglemma of one of Tumura-Clunie type theorems
Lemma 18 (see [24]) Let 119891(119911) be a meromorphic functionand suppose that Ψ = 119886
119899119891119899
+ sdot sdot sdot + 1198860has small meromorphic
coefficients 119886119895(119911) 119886119899(119911) equiv 0 in the sense of 119879(119903 119886
119895) = 119878(119903 119891)
Moreover assume that 119873(119903 1Ψ) + 119873(119903 119891) = 119878(119903 119891) ThenΨ = 119886
119899(119891 + (119886
119899minus1119899119886119899))119899
Proof of Theorem 14 (a) We deduce from (16) (17) and (24)that
119879 (119903 1198764(119911 119891)) le V119879 (119903 119891) + 119878 (119903 119891) (79)
Denote
119870 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) + 119875 (119911 119891) minus 120572 (119911) (80)
Differentiating both sides of (80) we obtain
V119891Vminus1(119911) 1198911015840
(119911) + 1198751015840
(119911 119891) minus 1205721015840
(119911)
= 1198701015840
(119911) = (119891V(119911) + 119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
(81)
that is
119891Vminus1(119911) ((V
1198911015840
(119911)
119891 (119911)
minus
1198701015840
(119911)
119870 (119911)
)119891 (119911))
= (119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
minus (1198751015840
(119911 119891) minus 1205721015840
(119911))
= (119875 (119911 119891) minus 120572 (119911)) (
1198701015840
(119911)
119870 (119911)
minus
1198751015840
(119911 119891) minus 1205721015840
(119911)
119875 (119911 119891) minus 120572 (119911)
)
(82)
It follows by (15)ndash(17) (24) (79) and (82) that
119898(119903 119891Vminus1)
le 119898 (119903 119875 (119911 119891) minus 120572) + 119898(119903
1198701015840
119870
)
+ 119898(119903
1198751015840
(119911 119891) minus 1205721015840
119875 (119911 119891) minus 120572
) + 119898(119903
1
(V (1198911015840119891) minus (1198701015840119870)) 119891)
le 119889 (119875)119898 (119903 119891) + 119898(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891)
+ 119873(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891) + 119878 (119903 119870) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1198701015840
119870
) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1
119870
) + 119878 (119903 119891)
(83)
that is
(V minus 119889 (119875) minus 2) 119879 (119903 119891) le 119873(1199031
1198764(119911 119891) minus 120572
) + 119878 (119903 119891)
(84)
From (79) and (84) we deduce that
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(85)
(b) It suffices to note that we may see 119891V as (1198911)V thenTheorem 14(b) follows immediately by Theorem 13
(c) By using a similar reasoning as [13 Theorem 1] wecan rearrange the expression for the differential-differencepolynomial 119875(119911 119891) by collecting together all terms havingthe same total degree and then writing 119875(119911 119891) in the form119875(119911 119891) = sum
119889(119875)
119896=0119887119896(119911)119891119896
(119911) Now each of the coefficients 119887119896(119911)
is a finite sum of products of functions of the form (119891(119895)
(119911 +
119888119894)119891(119911))
120582119894119895= (119891(119895)
(119911+119888119895)119891(119911+119888
119894))120582119894119895(119891(119911+119888
119894)119891(119911))
120582119894119895 with
each such product being multiplied by one of the originalcoefficients 119886
120582(119911) We deduce from the logarithmic derivative
lemma and Lemmas 4 and 6 that 119898(119903 119887119896) = 119878(119903 119891) Clearly
119873(119903 119887119896) = 119878(119903 119891) holds by (8) and Lemma 6Thus 119879(119903 119887
119896) =
119878(119903 119891) Denote
119871 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) +
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) minus 120572 (119911)
(86)
Assume to the contrary that 120575(120572 1198764(119911 119891)) = 1 Thus
Theorem A yields that
119873(119903
1
119871
) = 119873(119903
1
1198764(119911 119891) minus 120572
)
= 119878 (119903 1198764(119911 119891)) = 119878 (119903 119891)
(87)
Then (8) (86) (87) Lemma 18 and the assumption that V ge119889(119875) + 2 imply that 119871(119911) equiv 119891(119911)V that is
119875 (119911 119891) =
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) equiv 120572 (119911) (88)
Noting the fact that 119879(119903 119887119896) = 119878(119903 119891) and 119879(119903 120572) = 119878(119903 119891)
we deduce from Theorem A that (88) is a contradictionTherefore we have 120575(120572 119876
4(119911 119891)) lt 1
10 Abstract and Applied Analysis
4 Examples
Example 1 We consider nonhomogeneous differential-difference polynomials
1198751(119911 119891) = 119891 (119911) 119891
2
(119911 + log 4) minus 411989110158401015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 2) + 119891101584010158402 (119911 + log 3)
1198752(119911 119891) = 3119891
3
(119911) 11989110158402
(119911 + log 4)
minus 21198911015840
(119911) 119891 (119911 + log 3) 119891101584010158403 (119911 + log 2)
+ 1198914
(119911) minus 119891101584010158403
(119911)
1198753(119911 119891) = 119891 (119911) 119891
1015840
(119911 + log 2) 11989110158401015840 (119911 + log 3) minus 6119891101584010158402 (119911)(89)
and a homogeneous differential-difference polynomial
1198754(119911 119891) = 119891
101584010158403
(119911 + log 2) minus 1198911015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 3) minus 119891 (119911) 1198911015840 (119911) 11989110158401015840 (119911) (90)
where 119889(1198751) = 3 gt 2 = 119889
lowast
(1198751) 119889(119875
2) = 5 gt 3 = 119889
lowast
(1198752)
119889(1198753) = 3 gt 2 = 119889
lowast
(1198753) and 119889(119875
4) = 3 = 119889
lowast
(1198754) Clearly the
function 119891(119911) = 119890119911 satisfies (8) and 1205902(119891) = 0 lt 1 Then we
have
119889lowast
(1198751) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198751(119911 119890119911
)) =
2119903
120587
+ 119874 (1)
lt 119889 (1198751) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198752) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198752(119911 119890119911
)) =
4119903
120587
+ 119874 (1)
lt 119889 (1198752) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198753) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198753(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198753) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198754) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198754(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198754) 119879 (119903 119890
119911
) + 119874 (1)
(91)
This example shows that (9) is best possible
Example 2 Consider 119891(119911) = 119890119911 again Then the
homogeneous case 1198754(119911 119891) in Example 1 also illustrates
Theorem 9(a) And the nonhomogeneous differential-difference polynomials 119875
119894(119911 119891) 119894 = 1 2 3 in Example 1
also illustrate Theorem 9(b) where 120575(120572 1198751(119911 119891)) = 0
120575(120572 1198752(119911 119891)) le 14 lt 23 = 1 minus ((2119889
lowast
(1198752) minus 119889(119875
2))119889lowast
(1198752))
and 120575(120572 1198753(119911 119891)) le 13 lt 12 = 1 minus ((2119889
lowast
(1198753) minus
119889(1198753))119889lowast
(1198753)) Next we consider the nonhomogeneous
differential-difference polynomial
1198755(119911 119891) = 119891
1015840
(119911) 119891 (119911 + log 2) minus 1198912 (119911)
+ 1198911015840
(119911 + log 3) minus 3119891 (119911) + 1(92)
where 119889(1198755) = 2 119889
lowast
(1198755) = 0 Clearly 120575(1 119875
5(119911 119891)) =
120575(1 1198902119911
+ 1) = 1 Note that 2119889lowast(1198755) gt 119889(119875
5) fails then this
example shows that the assumption ldquo2119889lowast(119875) gt 119889(119875)rdquo cannotbe omitted inTheorem 9(b)
Example 3 We consider the differential-difference polyno-mials
1198761(119911 119891) = (119891
2
)
2
1198756(119911 119891)
= 1198914
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
1198762(119911 119891) = 119891
2
1198756(119911 119891)
= 1198912
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
(93)
and the function 119891(119911) = sin 119911 On the one hand 119873(119903 119891) =119878(119903 119891) 120590
2(119891) = 0 lt 1 and 119906V
1198761
gt 119889(1198756) and V
1198762
+ 2119889lowast
(1198756) gt
119889(1198756) hold where V
1198761
= V1198762
= 119906 = 2 and 119889(1198756) = 3 gt 2 =
119889lowast
(1198756) On the other hand 120575(120572 119876
1(119911 119891)) le 1 minus (1114) lt
1 minus (114) = 1 minus (119906 minus 1)(119906V1198761
minus 119889(119875))119906(119906V1198761
+ 119889(119875)) lt 1 and120575(120572 119876
2(119911 119891)) lt 1 hold This example shows that Theorems
10 and 11 may holdExample 4 We consider the differential-difference polyno-mials
119876(1)
4(119911 119891) = (119891
2
)
4
+ 1198757(119911 119891) = 119891
8
+ 1198757(119911 119891)
= 1198918
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(2)
4(119911 119891) = 119891
2
+ 1198757(119911 119891)
= 1198912
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(3)
4(119911 119891) = 2119891
3
+ 1198757(119911 119891)
= 21198913
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(4)
4(119911 119891) = 119891
4
+ 1198757(119911 119891)
= 1198914
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
(94)
and the function 119891(119911) = sin 119911 again On the one hand119876(1)
4(119911 119891) satisfies (119906 minus 1)119906V
119876(11)
4
(2119906 minus 1) gt 119889(1198757) and V
119876(12)
4
minus
2 gt (V119876(12)
4
minus 1)V119876(12)
4
(2V119876(12)
4
minus 1) gt 119889(1198757) respectively where
119906 = 4 V119876(11)
4
= 2 V119876(12)
4
= 8 119889(1198757) = 119889
lowast
(1198757) = 3 and
for 119894 = 2 3 4 119876(119894)4(119911 119891) satisfies 2 min119889lowast(119875
7) V119876(119894)
4
gt
max119889(1198757) V119876(119894)
4
where V119876(119894)
4
= 119894 On the other hand120575(120572 119876
(119894)
4(119911 119891)) lt 1 119894 = 1 2 3 4 hold This example shows
Abstract and Applied Analysis 11
that Theorems 13ndash15 may hold Moreover this example alsoshows the assumption ldquo119873(119903 1119891) = 119878(119903 119891)rdquo is not necessaryto Theorems 14(c) and 15 but it is regrettable for us notremoving it in our proofs
Example 5 We consider the differential-difference polyno-mials
1198771(119911 119891) = 119891
2
1198758(119911 119891)
= 1198912
(119911) (1198912
(119911 + 120587) +
1
sin2211991111989110158402
(119911 +
120587
2
))
1198772(119911 119891) = 119891
7
+ 1198759(119911 119891)
= 1198917
(119911) + sin 21199111198911015840 (119911 + 1205872
)1198912
(119911 +
120587
2
)
+ 11989110158402
(119911 +
120587
2
)119891(119911 +
3120587
2
)
+ 119911119891 (119911) 119891 (119911 +
120587
2
)
(95)
and the function 119891(119911) = 119890sin2119911 On the one hand 119877
1(119911 119891)
satisfies V1198771
+2119889lowast
(1198758) gt 119889(119875
8) and 119877
2(119911 119891) satisfies V
1198772
minus2 gt
(V1198772
minus 1)V1198772
(2V1198772
minus 1) gt 119889(1198759) respectively where V
1198771
=
2 and 119889(1198758) = 119889
lowast
(1198758) = 2 and V
1198772
= 7 and 119889(1198759) = 3 On
the other hand 120575(1198902 1198771(119911 119891)) = 120575(119890119911 119877
2(119911 119891)) = 1 hold
showing thatTheorems 11 and 14 fail Noting that the function119891(119911) = 119890
sin2119911 satisfies 1205902(119891) = 1 we know that the assumption
ldquo1205902(119891) lt 1rdquo is essential for Theorems 11 and 14 In fact it is
also essential for our other results in the whole paper but it isunnecessary to give examples one by one
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project was supported by the National Natural Sci-ence Foundation of China (11301233 and 11171119) and theNatural Science Foundation of Jiangxi Province in China(20132BAB211001 and 20132BAB211002) and Sponsored Pro-gram for Cultivating Youths of Outstanding Ability in JiangxiNormal University of China
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Group DordrechtTheNetherlands 2003
[4] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof f(z+120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[5] R G Halburd and R J Korhonen ldquoDifference analogue ofthe Lemma on the Logarithmic Derivative with applicationsto difference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[6] R G Halburd R J Korhonen and K Toghe ldquoHolomor-phic curves with shift-invariant hyper-planepreimagesrdquo Tran-sactions of the American Mathematical Society In press httparxivorgabs09033236
[7] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[8] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[9] Z X Chen ldquoComplex oscillation of meromorphic solutions forthe Pielou logistic equationrdquo Journal of Difference Equations andApplications vol 19 no 11 pp 1795ndash1806 2013
[10] Z X Chen and K H Shon ldquoFixed points of meromorphicsolutions for some difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 496096 7 pages 2013
[11] K Ishizaki and N Yanagihara ldquoWiman-Valiron method fordifference equationsrdquo Nagoya Mathematical Journal vol 175pp 75ndash102 2004
[12] I Laine and C-C Yang ldquoClunie theorems for difference andq-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] I Laine and C C Yang ldquoValue distribution of difference poly-nomialsrdquo Proceedings of the Japan Academy A vol 83 no 8 pp148ndash151 2007
[14] C-C Yang and I Laine ldquoOn analogies between nonlineardifference and differential equationsrdquo Proceedings of the JapanAcademy A vol 86 no 1 pp 10ndash14 2010
[15] R R Zhang and Z B Huang ldquoResults on difference analoguesof Valiron-Mokhonrsquoko theoremrdquoAbstract and Applied Analysisvol 2013 Article ID 273040 6 pages 2013
[16] XMZheng andZXChen ldquoOndeficiencies of somedifferencepolynomialsrdquo Acta Mathematica Sinica vol 54 no 6 pp 983ndash992 2011 (Chinese)
[17] XM Zheng and Z X Chen ldquoOn the value distribution of somedifference polynomialsrdquo Journal of Mathematical Analysis andApplications vol 397 no 2 pp 814ndash821 2013
[18] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Mathematique de France vol 59 pp 17ndash39 1931
[19] A Z Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoSibirskii Matematicheskii Zhurnal vol 15 pp 1305ndash1322 1974(Russian)
[20] W K Hayman ldquoPicard values of meromorphic functions andtheir derivativesrdquo Annals of Mathematics vol 70 no 2 pp 9ndash42 1959
[21] C-C Yang ldquoOn deficiencies of differential polynomialsrdquoMath-ematische Zeitschrift vol 116 no 3 pp 197ndash204 1970
[22] C-C Yang ldquoOn deficiencies of differential polynomials IIrdquoMathematische Zeitschrift vol 125 no 2 pp 107ndash112 1972
[23] A A Golrsquodberg and I V Ostrovskii The Distribution ofValues ofMeromorphic Functions NaukaMoscow Russia 1970(Russian)
12 Abstract and Applied Analysis
[24] E Mues and N Steinmetz ldquoThe theorem of Tumura-Clunie formeromorphic functionsrdquo Journal of the London MathematicalSociety vol 23 no 2 pp 113ndash122 1981
Research ArticleOn Growth of Meromorphic Solutions of Complex FunctionalDifference Equations
Jing Li12 Jianjun Zhang3 and Liangwen Liao1
1 Department of Mathematics Nanjing University Nanjing 210093 China2Nankai University Binhai College Tianjin 300270 China3Mathematics and Information Technology School Jiangsu Second Normal University Nanjing 210013 China
Correspondence should be addressed to Jianjun Zhang zhangjianjun1982163com
Received 29 November 2013 Accepted 13 January 2014 Published 25 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Jing Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The main purpose of this paper is to investigate the growth order of the meromorphic solutions of complex functional differenceequation of the form (sum
120582isin119868120572120582(119911)(prod
119899
]=1119891(119911 + 119888])119897120582]))(sum
120583isin119869120573120583(119911)(prod
119899
]=1119891(119911 + 119888])119898120583]
)) = 119876(119911 119891(119901(119911))) where 119868 = 120582 =
(1198971205821 1198971205822 119897
120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 = (119898
1205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two finite
index sets 119888] (] = 1 2 119899) are distinct complex numbers 120572120582(119911) (120582 isin 119868) and 120573
120583(119911) (120583 isin 119869) are small functions relative to 119891(119911)
and 119876(119911 119906) is a rational function in 119906 with coefficients which are small functions of 119891(119911) 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911]
of degree 119896 ge 1 We also give some examples to show that our results are sharp
1 Introduction and Main Results
Let 119891(119911) be a function meromorphic in the complex planeC We assume that the reader is familiar with the standardnotations and results in Nevanlinnarsquos value distribution the-ory ofmeromorphic functions such as the characteristic func-tion 119879(119903 119891) proximity function 119898(119903 119891) counting function119873(119903 119891) and the first and secondmain theorems (see eg [1ndash4]) We also use 119873(119903 119891) to denote the counting function ofthe poles of 119891(119911) whose every pole is counted only once Thenotations 120588(119891) and 120583(119891) denote the order and the lower orderof119891(119911) respectively 119878(119903 119891) denotes any quantity that satisfiesthe condition 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possiblyoutside an exceptional set of 119903 of finite linear measure Ameromorphic function 119886(119911) is called a small function of 119891(119911)or a small function relative to 119891(119911) if and only if 119879(119903 119886(119911)) =119878(119903 119891)
Recently some papers (see eg [5ndash7]) focusing on com-plex difference and functional difference equations emergedIn 2005 Laine et al [5] firstly considered the growth ofmeromorphic solutions of the complex functional differenceequations by utilizing Nevanlinna theory They obtained thefollowing result
Theorem A Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (1)
where 119869 is a collection of all subsets of 1 2 119899 119888119895rsquos are
distinct complex constants and 119901(119911) is a polynomial of degree119896 ge 2 Moreover we assume that the coefficients120572
119869(119911) are small
functions relative to 119891 and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (2)
where 120572 = log 119899 log 119896
In 2007 Rieppo [6] gave an estimation of growth ofmeromorphic solutions of complex functional equations asfollows
Theorem B Suppose that 119891 is a transcendental meromorphicfunction Let 119876(119911 119891) 119877(119911 119891) be rational functions in 119891
with small meromorphic coefficients relative to 119891 such that0 lt 119902 = deg
119891119876 le 119889 = deg
119891119877 and 119901(119911) = 119901
119896119911119896
+ 119901119896minus1
119911119896minus1
+
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 828746 6 pageshttpdxdoiorg1011552014828746
2 Abstract and Applied Analysis
sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 If 119891 is a solution of the
functional equation
119877 (119911 119891 (119911)) = 119876 (119911 119891 (119901 (119911))) (3)
then 119902119896 le 119889 and for any 120576 0 lt 120576 lt 1 there exist positive realconstants 119870
1and 119870
2such that
1198701(log 119903)120572minus120576 le 119879 (119903 119891) le 119870
2(log 119903)120572+120576 120572 =
log 119889 minus log 119902log 119896
(4)
when 119903 is large enough
Rieppo [6] also considered the growth order ofmeromor-phic solutions of functional equation (3) when 119896 = 1 and gotthe following
Theorem C Suppose that 119891 is a transcendental meromorphicsolution of (3) where 119901(119911) = 119886119911+119887 119886 119887 isin C 119886 = 0 and |119886| = 1Then
120583 (119891) = 120588 (119891) =
log119889 minus log 119902log |119886|
(5)
Two years later Zheng et al [7] extended Theorem A tomore general type and obtained a similar result of TheoremC In fact they got the following two results
TheoremD Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119876 (119911 119891 (119901 (119911))) (6)
where 119869 is a collection of all nonempty subsets of 1 2 119899119888119895(119895 = 1 119899) are distinct complex constants 119901(119911) = 119901
119896119911119896
+
119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a
rational function in 119906 of deg119906119876 = 119902(gt 0) Also suppose that
all the coefficients of (6) are small functions relative to 119891 Then119902119896 le 119899 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (7)
where 120572 = (log 119899 minus log 119902) log 119896
Theorem E Suppose that 119891 is a transcendental meromorphicsolution of (6) where 119869 is a collection of all nonempty subsetsof 1 2 119899 119888
119895(119895 = 1 119899) are distinct complex constants
119901(119911) = 119886119911 + 119887 119886 119887 isin C and 119876(119911 119906) is a rational function in 119906of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of (6)
are small functions relative to 119891(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log 119899minus log |119886|
(8)
(ii) If |119886| gt 1 then we have 119902 le 119899 and
120588 (119891) le
log 119899 minus log 119902log |119886|
(9)
(iii) If |119886| = 1 119902 gt 119899 then we have 120588(119891) = 120583(119891) = infin
In this paper we will consider a more general classof complex functional difference equations We prove thefollowing results which generalize the above related results
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic solution of the functional difference equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119901 (119911))) (10)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 =
120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin N⋃0 ] = 1 2 119899
are two finite index sets 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin
C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a rational function in 119906 ofdeg119906119876 = 119902(gt 0) Also suppose that all the coefficients of (10)
are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (11)
Then 119902119896 le 120590 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (12)
where 120572 = (log120590 minus log 119902) log 119896
Theorem 2 Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119886119911 + 119887))
(13)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 =
(1198981205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two
finite index sets 119886 119887 isin C and 119876(119911 119906) is a rational function in119906 of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of
(10) are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (14)
(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(15)
(ii) If |119886| gt 1 then we have 119902 le 120590 and
120588 (119891) le
log120590 minus log 119902log |119886|
(16)
(iii) If |119886| = 1 and 119902 gt 120590 then we have 120583(119891) = 120588(119891) = infin
Abstract and Applied Analysis 3
Next we will give some examples to show that our resultsare best in some extent
Example 3 Let 1198881= arctan 2 119888
2= minus1205874 Then it is easy to
check that 119891(119911) = tan 119911 solves the following equation
119891(119911 + 1198881)2
119891 (119911 + 1198882)
119891 (119911 + 1198881) + 119891(119911 + 119888
2)2
= (minus4119891(
119911
2
)
8
+ 8119891(
119911
2
)
7
+ 28119891(
119911
2
)
6
minus 56119891(
119911
2
)
5
minus 32119891(
119911
2
)
4
+ 56119891(
119911
2
)
3
+ 28119891(
119911
2
)
2
minus 8119891(
119911
2
) minus 4)
times (3119891(
119911
2
)
8
+ 10119891(
119911
2
)
7
+ 16119891(
119911
2
)
6
+ 122119891(
119911
2
)
5
minus 6119891(
119911
2
)
4
minus122119891(
119911
2
)
3
+16119891(
119911
2
)
2
minus10119891(
119911
2
) + 3)
minus1
(17)
Obviously we have
120583 (119891) = 120588 (119891) = 1 =
log 119902 minus log120590minus log |119886|
(18)
where 119902 = 8 120590 = 4 and 119886 = 12
Example 3 shows that the estimate in Theorem 2(i) issharp
Example 4 It is easy to check that 119891(119911) = tan 119911 satisfies theequation
119891(119911 + (1205873))2
119891 (119911 + (1205876)) minus 119891 (119911 + (1205876))
119891 (119911 + (1205873)) 119891(119911 + (1205876))2
minus 119891 (119911 + (1205873))
=
radic3119891(2119911)2
+ 4119891 (2119911) + radic3
minusradic3119891(2119911)2
+ 4119891 (2119911) minus radic3
(19)
Clearly we have
120583 (119891) = 120588 (119891) = 1 =
log120590 minus log 119902log |119886|
(20)
where 120590 = 4 119902 = 2 and 119886 = 2
Example 4 shows that the estimate in Theorem 2(ii) issharp
Example 5 119891(119911) = tan 119911 satisfies the equation of the form
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
minus(119891(1199112)2
minus 2119891 (1199112) minus 1)
3
8119891 (1199112) (119891(1199112)2
minus 1) (119891(1199112)2
+ 2119891 (1199112) minus 1)
(21)
where 120590 = 4 119902 = 6 and 119886 = 12 120588(119891) = 120583(119891) = 1 gt
log(32) log 2 = (log 119902 minus log120590) minus log |119886|
Example 5 shows that the strict inequality in Theorem 2may occur Therefore we do not have the same estimation asinTheoremC for the growth order ofmeromorphic solutionsof (13)
The following Example shows that the restriction 119902 gt 120590
in case (iii) in Theorem 2 is necessary
Example 6 Meromorphic function 119891(119911) = tan 119911 solves thefollowing equation
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
(119891 (119911) + 1)3
4119891 (119911) (1 minus 119891 (119911))
(22)
where 119886 = 1 and 4 = 120590 gt 119902 = 3 but 120588(119891) = 120583(119891) = 1
Next we give an example to show that case (iii) inTheorem 2 may hold
Example 7 Function 119891(119911) = 119911119890119890119911
satisfies the followingequation
(119911 + log 6) (119911 + log 2)5 [119891(119911 + log 4)4 + 119891 (119911 + log 4)](119911 + log 4) 119891 (119911 + log 6)
=
(119911 + log 4)3119891(119911 + log 2)6 + (119911 + log 2)6
119891 (119911 + log 2)
(23)
where 119886 = 1 and 119902 = 6 gt 5 = 120590 Obviously 120588(119891) = 120583(119891) = infin
2 Main Lemmas
In order to prove our results we need the following lemmas
Lemma 1 (see [4 8]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(24)
such that the meromorphic coefficients 119886119894(119911) 119887119895(119911) satisfy
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 1 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 1 119902
(25)
then one has
119879 (119903 119877 (119911 119891)) = max 119901 119902 sdot 119879 (119903 119891) + 119878 (119903 119891) (26)
From the proof ofTheorem 1 in [9] we have the followingestimate for the Nevanlinna characteristic
Lemma 2 Let 1198911 1198912 119891
119899be distinct meromorphic func-
tions and
119865 (119911) =
119875 (119911)
119876 (119911)
=
sum120582isin119868
120572120582(119911) 119891
1198971205821
1119891
1198971205822
2 119891
119897120582119899
119899
sum120583isin119869
120573120583(119911) 119891
1198981205831
1119891
1198981205832
2 119891
119898120583119899
119899
(27)
4 Abstract and Applied Analysis
Then
119879 (119903 119865 (119911)) le
119899
sum
]=1120590]119879 (119903 119891]) + 119878 (119903 119891) (28)
where 119868 = 120582 = (1198971205821 1198971205822 119897120582119899) | 119897
120582] isin N⋃0 ] =
1 2 119899 and 119869 = 120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin
N⋃0 ] = 1 2 119899 are two finite index sets 120590] =
max120582120583119897120582] 119898120583] (] = 1 2 119899) 120572
120582(119911) = 119900(119879(119903 119891])(120582 isin 119868))
and120573120583(119911) = 119900(119879(119903 119891])(120583 isin 119869)) hold for all ] isin 1 2 119899 and
satisfy 119879(119903 120572120582) = 119878(119903 119891) (120582 isin 119868) and 119879(119903 120573
120583) = 119878(119903 119891) (120583 isin
119869)
Lemma 3 (see [7]) Let 119888 be a complex constant Given 120576 gt 0
and a meromorphic function 119891 one has
119879 (119903 119891 (119911 plusmn 119888)) le (1 + 120576) 119879 (119903 + |119888| 119891) (29)
for all 119903 gt 1199030 where 119903
0is some positive constant
Lemma 4 (see [4]) Let 119892 (0 +infin) rarr R ℎ (0 +infin) rarr R
bemonotone increasing functions such that 119892(119903) le ℎ(119903) outsideof an exceptional set 119864 of finite linear measure Then for any120572 gt 1 there exists 119903
0gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 gt 119903
0
Lemma 5 (see [10]) Let 119891 be a transcendental meromorphicfunction and 119901(119911) = 119886
119896119911119896
+ 119886119896minus1
119911119896minus1
+ sdot sdot sdot + 1198861119911 + 1198860 119886119896
= 0be a nonconstant polynomial of degree 119896 Given 0 lt 120575 lt |119886
119896|
denote 120582 = |119886119896| + 120575 and 120583 = |119886
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C⋃infin one has
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 (119901 (119911))) le 119896119899 (120582119903119896
119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 (119901 (119911)))
le 119873 (120582119903119896
119886 119891) + 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 (119901 (119911))) le (1 + 120576) 119879 (120582119903119896
119891)
(30)
for all 119903 large enough
Lemma 6 (see [11]) Let 120601 [1199030 +infin) rarr (0 +infin) be
positive and bounded in every finite interval and suppose that120601(120583119903119898
) le 119860120601(119903) + 119861 holds for all 119903 large enough where 120583 gt 0119898 gt 1 119860 gt 1 and 119861 are real constants Then
120601 (119903) = 119874 ((log 119903)120572) (31)
where 120572 = log119860 log119898
Lemma 7 (see [6]) Let 120601 (1199030infin) rarr (1infin) where 119903
0ge 1
be a monotone increasing function If for some real constant120572 gt 1 there exists a real number 119870 gt 1 such that 120601(120572119903) ge
119870120601(119903) then
lim119903rarrinfin
log120601 (119903)log 119903
ge
log119870log120572
(32)
Lemma 8 (see [12]) Let 120601 (1infin) rarr (0infin) be a monotoneincreasing function and let 119891 be a nonconstant meromorphic
function If for some real constant 120572 isin (0 1) there exist realconstants 119870
1gt 0 and 119870
2ge 1 such that
119879 (119903 119891) le 1198701120601 (120572119903) + 119870
2119879 (120572119903 119891) + 119878 (120572119903 119891) (33)
then
120588 (119891) le
log1198702
minus log120572+ lim119903rarrinfin
log120601 (119903)log 119903
(34)
3 Proof of Theorems
Proof of Theorem 1 We assume 119891(119911) is a transcendentalmeromorphic solution of (10) Denoting 119862 =
max|1198881| |1198882| |119888
119899| According to Lemmas 1 2 and 3
and the last assertion of Lemma 5 we get that for any 1205761gt 0
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) + 119878 (119903 119891)
le 119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 1205761) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(35)
where 119903 is large enough and120583 = |119901119896|minus120575 for some 0 lt 120575 lt |119901
119896|
Since 119879(119903 + 119862 119891) le 119879(120573119903 119891) holds for 119903 large enough for120573 gt 1 we may assume 119903 to be large enough to satisfy
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120573119903 119891) (36)
outside a possible exceptional set of finite linear measure ByLemma 4 we know that whenever 120574 gt 1
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120574120573119903 119891) (37)
holds for all 119903 large enough Denote 119905 = 120574120573119903 thus theinequality (37) may be written in the form
119879(
120583
(120574120573)119896
119905119896
119891) le
120590 (1 + 1205761)
119902 (1 minus 1205761)
119879 (119905 119891) (38)
By Lemma 6 we have
119879 (119903 119891) = 119874 ((log 119903)1205721) (39)
Abstract and Applied Analysis 5
where
1205721=
log (120590 (1 + 1205761) 119902 (1 minus 120576
1))
log 119896
=
log120590 minus log 119902log 119896
+
log ((1 + 1205761) (1 minus 120576
1))
log 119896
(40)
Denoting now 120572 = (log120590 minus log 119902) log 119896 and 120576 = log((1 +
1205761)(1 minus 120576
1)) log 119896 thus we obtain the required form
Finally we show that 119902119896 le 120590 If 119902119896 gt 120590 then we have120572 lt 1 For sufficiently small 120576 gt 0 we have 120572 + 120576 lt 1 whichcontradicts with the transcendency of 119891 Thus Theorem 1 isproved
Proof of Theorem 2 Suppose 119891(119911) is a transcendental mero-morphic solution of (13) Denoting 119862 = max|119888
1| |1198882|
|119888119899|
(i) 0 lt |119886| lt 1 We may assume that 119902 gt 120590 since the case119902 le 120590 is trivial by the fact that 120583(119891) ge 0 By Lemmas1ndash3 we have for any 120576 gt 0 and 120573 gt 1
119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
le 120590 (1 + 120576) 119879 (120573119903 119891) + 119878 (119903 119891)
(41)
where 119903 is large enoughBy the last assertion of Lemma 5 and (41) we obtain that
for 120583 = |119886| minus 120575 (0 lt 120575 lt |119886| 0 lt 120583 lt 1) the followinginequality
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120573119903 119891) (42)
holds where 119903 is large enough outside of a possible set of finitelinear measure By Lemma 4 we get that for any 120574 gt 1 andsufficiently large 119903
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120574120573119903 119891) (43)
Therefore
119902 (1 minus 120576)
120590 (1 + 120576)
119879 (119903 119891) le 119879(
120574120573
120583
119903 119891) (44)
Since 120573 gt 1 120574 gt 1 0 lt 120583 lt 1 and 119902 gt 120590 we have 120573120574120583 gt 1
and 119902(1 minus 120576)120590(1 + 120576) gt 1 when 120576 is small enough UsingLemma 7 we see that
120583 (119891) ge
log 119902 (1 minus 120576) minus log120590 (1 + 120576)
log 120574120573 minus log 120583 (45)
Letting 120576 rarr 0 120575 rarr 0 120573 rarr 1 and 120574 rarr 1 we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(46)
(ii) |119886| gt 1 By the similar reasoning as is (i) we easilyobtain that
119902 (1 minus 120576) 119879 (120583119903 119891) le 119902119879 (119903 119891 (119901 (119911)))
le 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(47)
for all 119903 large enough We may select sufficiently smallnumbers 120575 gt 0 and 120576 gt 0 such that 120583 = |119886| minus 120575 gt 1 and(1120583) + 120576 lt 1 Thus we have
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891) (48)
namely
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) (49)
where 119903 is large enough possibly outside of a set of finite linearmeasure By Lemma 4 we have for any 1 lt 120574 lt 120583
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (120574119903 119891 (119911)) (50)
that is
119879 (119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879(
120574
120583
119903 119891 (119911)) (51)
holds for all sufficiently large 119903 By Lemma 8 we obtain
120588 (119891) le
log120590 minus log 119902 + log (1 + 120576) minus log (1 minus 120576)
minus log (120574120583) (52)
Letting 120576 rarr 0 120575 rarr 0 and 120574 rarr 1 we have
120588 (119891) le
log120590 minus log 119902log |119886|
(53)
(iii) |119886| = 1 and 119902 gt 120590 The proof of this case is completelysimilar as in the case in (i) In fact we set 120583 = |119886|minus120575 =
1 minus 120575 (0 lt 120575 lt 1 0 lt 120583 lt 1) Similarly we can get
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(54)
Since |119886| = 1 we have 120583(119891) = 120588(119891) = infin
6 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the anonymous ref-erees for their valuable comments and suggestions Theresearch was supported by Colonel-level topics (JSNU-ZY-01) (Jsie2012zd01) and NSF of China (11271179)
References
[1] W Cherry and Z Ye Nevanlinnarsquos Theory of Value DistributionSpringer Monographs in Mathematics Springer Berlin Ger-many 2001
[2] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[3] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Beijing China 1988
[4] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] J Rieppo ldquoOn a class of complex functional equationsrdquoAnnalesAcademiaelig Scientiarum Fennicaelig vol 32 no 1 pp 151ndash170 2007
[7] X-M Zheng Z-X Chen and J Tu ldquoGrowth of meromorphicsolutions of some difference equationsrdquoApplicable Analysis andDiscrete Mathematics vol 4 no 2 pp 309ndash321 2010
[8] A Z Mokhonrsquoko ldquoThe Nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija vol 14 pp 83ndash87 1971 (Russian)
[9] A A Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoAkademija Nauk SSSR vol 15 pp 1305ndash1322 1974
[10] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[11] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[12] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
Research ArticleUnicity of Entire Functions concerning Shifts andDifference Operators
Dan Liu Degui Yang and Mingliang Fang
Institute of Applied Mathematics South China Agricultural University Guangzhou 510642 China
Correspondence should be addressed to Mingliang Fang mlfangscaueducn
Received 29 October 2013 Revised 17 December 2013 Accepted 19 December 2013 Published 3 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Dan Liu et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We prove a unicity theorem of entire functions that share two distinct small functions with their shiftsThe corollary of the theoremconfirms the conjecture posed by Li and Gao (2011)
1 Introduction
Let119891 be a nonconstantmeromorphic function in the complexplane C We will use the standard notations in Nevanlinnatheory of meromorphic functions such as 119879(119903 119891) 119873(119903 119891)and 119898(119903 119891) (see [1 2]) The notation 119878(119903 119891) is defined to beany quantity satisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin pos-sibly outside a set of finite linear measures A meromorphicfunction 119886 is called a small function related to119891 provided that119879(119903 119886) = 119878(119903 119891)
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 be a small function related to both 119891 and 119892 Wesay that 119891 and 119892 share 119886 CM if 119891 minus 119886 and 119892 minus 119886 have thesame zeros with the same multiplicities 119891 and 119892 are said toshare 119886 IM if 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
Let119873(119903 119886) be the counting functions of all common zeroswith the same multiplicities of 119891 minus 119886 and 119892 minus 119886 If
119873(119903
1
119891 minus 119886
) + 119873(119903
1
119892 minus 119886
) minus 2119873 (119903 119886)
= 119878 (119903 119891) + 119878 (119903 119892)
(1)
then we say that 119891 and 119892 share 119886 CM almostFor a nonzero complex constant 119888 isin C we define
difference operators asΔ119888119891(119911) = 119891(119911+119888)minus119891(119911) andΔ119899
119888119891(119911) =
Δ119888(Δ119899minus1
119888119891(119911)) 119899 isin N 119899 gt 2
In 1977 Rubel and Yang [3] proved the following result
Theorem A Let 119891 be a nonconstant entire function If 119891(119911)and 1198911015840(119911) share two distinct finite values CM then 119891(119911) equiv
1198911015840
(119911)
In fact the conclusion still holds if the two CM values arereplaced by two IM values (see Gundersen [4 5] Mues andSteinmetz [6])
Recently a number of articles focused on value dis-tribution in shifts or difference operators of meromorphicfunctions (see [7ndash11]) In particular some papers studied theunicity of meromorphic functions sharing values with theirshifts or difference operators (see [12ndash14]) In 2009 Heit-tokangas et al [12] proved the following result concerningshifts
Theorem B Let 119891 be a nonconstant entire function of finiteorder 119888 isin C If119891(119911) and119891(119911+119888) share two distinct finite valuesCM then 119891(119911) equiv 119891(119911 + 119888)
In 2011 Li and Gao [14] proved the following resultconcerning difference operators
Theorem C Let 119891 be a nonconstant entire function of finiteorder 119888 isin C and let 119899 be a positive integer Suppose that 119891(119911)and Δ119899
119888119891(119911) share two distinct finite values 119886 119887 CM and one of
the following cases is satisfied(i) 119886119887 = 0(ii) 119886119887 = 0 and 120588(119891) notin 119873Then 119891(119911) equiv Δ119899
119888119891(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 380910 5 pageshttpdxdoiorg1011552014380910
2 Abstract and Applied Analysis
In [14] Li andGao conjectured that the restriction 120588(119891) notinN for the case 119886119887 = 0 can be removed In this paper weconfirm their conjecture In fact we prove the followingmoregeneral results
Theorem 1 Let 119891 be a nonconstant entire function of finiteorder let 119899 be a positive integer let 119886(119911) 119887(119911) be two distinctsmall functions related to 119891(119911) let 119898
1 1198982 119898
119899be nonzero
complex numbers and 1198881 1198882 119888
119899distinct finite values and let
119865 (119911) = 1198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119891 (119911 + 119888
119899)
(2)
If 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119865(119911)
Corollary 2 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number let 119899 be apositive integer and let 119886 119887 be two distinct finite values If 119891(119911)and Δ119899
119888119891(119911) share 119886 119887 CM then 119891(119911) equiv Δ119899
119888119891(119911)
Remark 3 Corollary 2 confirms the conjecture of Li and Gaoin [14]
Corollary 4 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number and let 119886(119911)119887(119911) be two distinct small functions related to 119891 If 119891(119911) and119891(119911 + 119888) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119891(119911 + 119888)
2 Some Lemmas
For the proof of Theorem 1 we require the following results
Lemma 5 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903
1
119891
) + 119873 (119903 119891) = 119878 (119903 119891)
119873(119903
1
119892
) + 119873 (119903 119892) = 119878 (119903 119892)
(3)
If 119891(119911) and 119892(119911) share 1 CM almost then either 119891(119911) equiv 119892(119911)
or 119891(119911)119892(119911) equiv 1
Lemma 6 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903 119891) = 119878 (119903 119891) 119873 (119903 119892) = 119878 (119903 119892) (4)
If 119891(119911) and 119892(119911) share 0 and 1 CM almost and
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 119891) + 119879 (119903 119892)
lt
2
3
(5)
where 119868 sub [0infin) is a set of infinitely linear measure then
119891 (119911) =
119886119892 (119911) + 119887
119888119892 (119911) + 119889
(6)
where 119886 119887 119888 and 119889 are constants satisfying 119886119889 minus 119887119888 = 0
Lemma 7 (see [10]) Let 119891 be a nonconstant meromorphicfunction of finite order 119888 isin C Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (7)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
119889119903119903 lt infin
In the following 119878(119903 119891) denotes any function satisfying119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside a set withfinite logarithmic measure
3 Proof of Theorem 1
We prove Theorem 1 by contradiction Suppose that 119891(119911) equiv
119865(119911) Then it follows from 119891(119911) and 119865(119911) being two distinctentire functions that 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) and infinCM By the Nevanlinna second fundamental theorem forthree small functions we have
119879 (119903 119891) le 119873 (119903 119891) + 119873(119903
1
119891 minus 119886
)
+ 119873(119903
1
119891 minus 119887
) + 119878 (119903 119891)
le 119873(119903
1
119865 minus 119886
) + 119873(119903
1
119865 minus 119887
) + 119878 (119903 119891)
le 2119879 (119903 119865) + 119878 (119903 119891)
(8)
Similarly we have 119879(119903 119865) le 2119879(119903 119891) + 119878(119903 119865) Therefore119878(119903 119891) = 119878(119903 119865)
Set
1198911(119911) =
119891 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
1198651(119911) =
119865 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
(9)
Thus 1198911(119911) 1198651(119911) share 0 1 andinfin CM almost
Obviously we have
119879 (119903 1198911) = 119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 1198651) = 119879 (119903 119865) + 119878 (119903 119891)
119878 (119903 119865) = 119878 (119903 1198651) = 119878 (119903 119891
1) = 119878 (119903 119891)
(10)
By Nevanlinnarsquos second fundamental theorem we have
119879 (119903 1198911) le 119873(119903
1
1198911
) + 119873(119903
1
1198911minus 1
) + 119873 (119903 1198911) + 119878 (119903 119891
1)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
Abstract and Applied Analysis 3
le 119873(119903
1
1198651minus 1198911
) + 119878 (119903 119891)
le 119879 (119903 1198651minus 1198911) + 119878 (119903 119891)
le 119879 (119903 119865 minus 119891) + 119878 (119903 119891)
le 119898 (119903 119865 minus 119891) + 119878 (119903 119891)
(11)Since 119865 minus 119891 = 119898
1119891(119911 + 119888
1) + 1198982119891(119911 + 119888
2) + sdot sdot sdot + 119898
119899119891(119911 +
119888119899) minus 119891(119911) = 119891(119911)[119898
1(119891(119911 + 119888
1)119891(119911)) +119898
2(119891(119911 + 119888
2)119891(119911)) +
sdot sdot sdot + 119898119899(119891(119911 + 119888
119899)119891(119911)) minus 1] thus
119898(119903 119865 minus 119891)
le 119898 (119903 119891)
+ 119898(1199031198981
119891 (119911 + 1198881)
119891 (119911)
+ sdot sdot sdot + 119898119899
119891 (119911 + 119888119899)
119891 (119911)
minus 1)
le 119898 (119903 119891) + 119878 (119903 119891)
(12)By (11) we have
119879 (119903 1198911) le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(13)
It follows that119873(119903 0) + 119873 (119903 1) = 119879 (119903 119891
1) + 119878 (119903 119891) (14)
On the other hand byNevanlinna first fundamental theoremwe have
2119879 (119903 1198911) = 119879(119903
1
1198911
) + 119879(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
(15)
So we get
119879 (119903 1198911) le 119898(119903
1
1198911
) + 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119898(119903
1
119891 minus 119886
) + 119898(119903
1
119891 minus 119887
) + 119878 (119903 119891)
(16)
Set1198861(119911) = 119898
1119886 (119911 + 119888
1) + 1198982119886 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119886 (119911 + 119888
119899)
1198871(119911) = 119898
1119887 (119911 + 119888
1) + 1198982119887 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119887 (119911 + 119888
119899)
(17)
If 1198861(119911) equiv 119887
1(119911) we can deduce by (16) that
119879 (119903 1198911) le 119898(119903
1
119891 minus 119886
+
1
119891 minus 119887
) + 119878 (119903 119891)
le 119898(119903
119865 minus 1198861
119891 minus 119886
+
119865 minus 1198871
119891 minus 119887
)
+ 119898(119903
1
119865 minus 1198861
) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(18)
If 1198861(119911) equiv 119887
1(119911) set
119871 (119865) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
119865 11988611198871
1198651015840
1198861015840
11198871015840
1
11986510158401015840
11988610158401015840
111988710158401015840
1
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(19)
Then we have
119898(119903
119865 minus 1198861
119891 minus 119886
) = 119898(119903
119865 minus 1198871
119891 minus 119887
) = 119878 (119903 119891)
119898(119903
119871 (119865)
119865 minus 1198861
) = 119898(119903
119871 (119865)
119865 minus 1198871
) = 119878 (119903 119891)
(20)
It followed from (16) that
119879 (119903 1198911) le 119898(119903
119865 minus 1198861
119891 minus 119886
) + 119898(119903
1
119865 minus 1198861
)
+ 119898(119903
119865 minus 1198871
119891 minus 119887
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
+
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119871 (119865)
) + 119878 (119903 119891)
le 119879 (119903 119871 (119865)) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
4 Abstract and Applied Analysis
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
le 119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 1198911) + 119878 (119903 119891)
(21)
By (18) and (21) we can deduce that
119879 (119903 1198911) = 119879 (119903 119865) + 119878 (119903 119891) = 119879 (119903 119865
1) + 119878 (119903 119891) (22)
It follows from (14) and (22) that
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 1198911) + 119879 (119903 119865
1)
=
1
2
lt
2
3
(23)
By Lemma 6 we have
1198911(119911) =
1198601198651(119911) + 119861
1198621198651(119911) + 119863
(24)
where 119860 119861 119862 and 119863 are complex numbers satisfying 119860119863 minus
119861119862 = 0Now we consider three cases
Case 1 Consider 119873(119903 0) = 119878(119903 1198911) Thus
119873(119903
1
1198911
) + 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (25)
Similarly we have
119873(119903
1
1198651
) + 119873 (119903 1198651) = 119878 (119903 119865
1) = 119878 (119903 119891) (26)
By Lemma 5 we get that either 1198911equiv 1198651or 11989111198651equiv 1
If 1198911equiv 1198651 we can easily deduce that 119891 equiv 119865 which is a
contradiction with our assumptionIf 11989111198651equiv 1 that is
(119891 (119911) minus 119886) (119865 (119911) minus 119886) equiv (119887 minus 119886)2
(27)
then we have
(119891 minus 119886)2
=
(119887 minus 119886)2
(119865 minus 119886) (119891 minus 119886)
(28)
From (28) we have
2119879 (119903 119891) le 119879 (119903 (119891 minus 119886)2
) + 119878 (119903 119891)
= 119879(119903
1
(119887 minus 119886)2
((119865 minus 119886) (119891 minus 119886))
) + 119878 (119903 119891)
le 119879(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
= 119873(119903
119865 minus 119886
119891 minus 119886
) + 119898(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
(119865 minus 1198861) + (119886
1minus 119886)
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
1198861minus 119886
119891 minus 119886
) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
(29)
It follows that 119879(119903 119891) le 119878(119903 119891) a contradiction
Case 2 Consider 119873(119903 1) = 119878(119903 1198911) Using the same argu-
ment as used in Case 1 we deduce that 119879(119903 119891) le 119878(119903 119891) acontradiction
Case 3 Consider 119873(119903 0) = 119878(119903 1198911) 119873(119903 1) = 119878(119903 119891
1) Since
1198911and 119865
1share 0 1 CM almost we deduce from (24) that
1198911(119911) =
(119862 + 119863) 1198651(119911)
1198621198651(119911) + 119863
(30)
If 119862 = 0 then 1198911equiv 1198651 that is 119891 equiv 119865 a contradiction
Hence 119862 = 0 Thus we have
119873(119903
1
1198651+ (119863119862)
) = 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (31)
Obviously 119863119862 = 0 119863119862 = minus 1 Thus by Nevanlinnasecond fundamental theorem and (14) we get
2119879 (119903 1198911) = 2119879 (119903 119865
1) + 119878 (119903 119891
1)
le 119873(119903
1
1198651
) + 119873(119903
1
1198651minus 1
)
+ 119873(119903
1
1198651+ (119863119862)
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119878 (119903 119891)
(32)
It follows that 119879(119903 1198911) le 119878(119903 119891
1) a contradiction Thus
we prove that 119891(119911) equiv 119865(119911) This completes the proof ofTheorem 1
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 5
Acknowledgments
Theauthors thank the referees for careful reading of the paperpointing out a gap in the previous version of this paperand giving many valuable suggestions Research is supportedby the NNSF of China (Grant no 11371149) and NSF ofGuangdong Province China (Grant no S2012010010121)
References
[1] W K Hayman Meromorphic Function Clarendon PressOxford UK 1964
[2] L Yang Value Distribution Theory Springer Berlin Germany1993
[3] L A Rubel and C C Yang Value Shared by an Entire Functionand Its Derivative Lecture Notes in Math Springer BerlinGermany 1977
[4] G G Gundersen ldquoMeromorphic functions that share finitevalues with their derivativerdquo Journal of Mathematical Analysisand Applications vol 75 no 2 pp 441ndash446 1980
[5] G G Gundersen ldquoErrata meromorphic functions that sharefinite values with their derivativerdquo Journal of MathematicalAnalysis and Applications vol 86 no 1 p 307 1982
[6] E Mues and N Steinmetz ldquoMeromorphe Funktionen die mitihrer Ableitung Werte teilenrdquo Manuscripta Mathematica vol29 no 2ndash4 pp 195ndash206 1979
[7] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[8] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoTheRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[9] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[10] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[11] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[12] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[13] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables and Elliptic Equations vol 56 no 1ndash4 pp 81ndash92 2011
[14] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[15] M L Fang ldquoUnicity theorems for meromorphic function andits differential polynomialrdquo Advances in Mathematics vol 24no 3 pp 244ndash249 1995
Research ArticleOn Positive Solutions and Mann Iterative Schemes of a ThirdOrder Difference Equation
Zeqing Liu1 Heng Wu1 Shin Min Kang2 and Young Chel Kwun3
1 Department of Mathematics Liaoning Normal University Dalian Liaoning 116029 China2Department of Mathematics and RINS Gyeongsang National University Jinju 660-701 Republic of Korea3 Department of Mathematics Dong-A University Pusan 614-714 Republic of Korea
Correspondence should be addressed to Young Chel Kwun yckwundauackr
Received 14 October 2013 Accepted 16 December 2013 Published 28 January 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Zeqing Liu et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The existence of uncountably many positive solutions and convergence of the Mann iterative schemes for a third order nonlinearneutral delay difference equation are proved Six examples are given to illustrate the results presented in this paper
1 Introduction and Preliminaries
Recently many researchers studied the oscillation nonoscil-lation and existence of solutions for linear and nonlinearsecond and third order difference equations and systemssee for example [1ndash23] and the references cited therein Bymeans of the Reccati transformation techniques Saker [18]discussed the third order difference equation
Δ3
119909119899+ 119901119899119909119899+1
= 0 forall119899 ge 1198990 (1)
and presented some sufficient conditions which ensure thatall solutions are to be oscillatory or tend to zero Utilizing theSchauder fixed point theorem Yan and Liu [22] proved theexistence of a bounded nonoscillatory solution for the thirdorder difference equation
Δ3
119909119899+ 119891 (119899 119909
119899 119909119899minus120591) = 0 forall119899 ge 119899
0 (2)
Agarwal [2] established the oscillatory and asymptotic prop-erties for the third order nonlinear difference equation
Δ3
119909119899+ 119902119899119891 (119909119899+1) = 0 forall119899 ge 1 (3)
Andruch-Sobiło andMigda [4] studied the third order lineardifference equation of neutral type
Δ3
(119909119899minus 119901119899119909120590119899) plusmn 119902119899119909120591119899= 0 forall119899 ge 119899
0 (4)
and obtained sufficient conditions which ensure that allsolutions of the equation are oscillatory Grace andHamedani[6] discussed the difference equation
Δ3
(119909119899minus 119909119899minus120591) plusmn 119902119899
1003816100381610038161003816119909119899minus120590
1003816100381610038161003816
3 sgn119909119899minus120590
= 0 forall119899 ge 0 (5)
and gave some new criteria for the oscillation of all solutionsand all bounded solutions
Our goal is to discuss solvability and convergence ofthe Mann iterative schemes for the following third ordernonlinear neutral delay difference equation
Δ3
(119909119899+ 119887119899119909119899minus120591) + Δℎ (119899 119909
ℎ1119899
119909ℎ2119899
119909ℎ119896119899
)
+119891 (119899 1199091198911119899
1199091198912119899
119909119891119896119899
) = 119888119899 forall119899 ge 119899
0
(6)
where 120591 119896 1198990isin N 119887
119899119899isinN1198990
119888119899119899isinN1198990
sub R ℎ 119891 isin 119862(N1198990
times
R119896R) ℎ119897119899119899isinN1198990
119891119897119899119899isinN1198990
sube N and
lim119899rarrinfin
ℎ119897119899= lim119899rarrinfin
119891119897119899= +infin 119897 isin 1 2 119896 (7)
By employing the Banach fixed point theorem and somenew techniques we establish the existence of uncountablymany positive solutions of (6) conceive a few Mann iter-ative schemes for approximating these positive solutionsand prove their convergence and the error estimates Sixnontrivial examples are included
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 470181 16 pageshttpdxdoiorg1011552014470181
2 Abstract and Applied Analysis
Throughout this paper we assume that Δ is the forwarddifference operator defined by Δ119909
119899= 119909119899+1
minus 119909119899 R =
(minusinfin +infin) R+ = [0 +infin) N0and N denote the sets of
nonnegative integers and positive integers respectively
N119905= 119899 119899 isin N with 119899 ge 119905 forall119905 isin N
120573 = min 1198990minus 120591 inf ℎ
119897119899 119891119897119899 1 le 119897 le 119896 119899 isin N
1198990
isin N
119867119899= max ℎ2
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
119865119899= max 1198912
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
(8)
and 119897infin
120573represents the Banach space of all real sequences
on N120573with norm
119909 = sup119899isinN120573
1003816100381610038161003816100381610038161003816
119909119899
1198992
1003816100381610038161003816100381610038161003816
lt +infin for each 119909 = 119909119899119899isinN120573
isin 119897infin
120573
119860 (119873119872) = 119909 = 119909119899119899isinN120573
isin 119897infin
120573 119873 le
119909119899
1198992
le 119872 119899 isin N120573
for any 119872 gt 119873 gt 0
(9)
It is easy to see that 119860(119873119872) is a closed and convex subsetof 119897infin120573 By a solution of (6) wemean a sequence 119909
119899119899isinN120573
witha positive integer 119879 ge 119899
0+120591+120573 such that (6) holds for all 119899 ge
119879
Lemma 1 Let 119901119905119905isinN be a nonnegative sequence and 120591 isin N
(i) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199052
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119904=119899+119894120591suminfin
119905=119904119901119905= 0
(ii) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199053
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119906=119899+119894120591suminfin
119904=119906suminfin
119905=119904119901119905= 0
Proof Note that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
(
infin
sum
119905=119899+119894120591
119901119905+
infin
sum
119905=119899+1+119894120591
119901119905+
infin
sum
119905=119899+2+119894120591
119901119905+ sdot sdot sdot )
=
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
(1 + 119905 minus 119899 minus 119894120591) 119901119905le
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
119905119901119905
=
1
1198992
(
infin
sum
119905=119899+120591
119905119901119905+
infin
sum
119905=119899+2120591
119905119901119905+
infin
sum
119905=119899+3120591
119905119901119905+ sdot sdot sdot )
le
1
1198992
infin
sum
119905=119899+120591
(1 +
119905 minus 119899 minus 120591
120591
) 119905119901119905=
1
1198992
infin
sum
119905=119899+120591
119905 minus 119899
120591
119905119901119905
le
1
1198992120591
infin
sum
119905=119899+120591
1199052
119901119905997888rarr 0 as 119899 997888rarr infin
(10)
that is
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905= 0 (11)
As in the proof of (10) we infer that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
(1 + 119905 minus 119906) 119901119905
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
119905119901119905le
1
1198992120591
infin
sum
119905=119899+120591
1199053
119901119905997888rarr 0 as 119899 997888rarr infin
(12)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905= 0 (13)
This completes the proof
2 Uncountably Many Positive Solutions andMann Iterative Schemes
In this section using the Banach fixed point theoremand Mann iterative schemes we establish the existence ofuncountably many positive solutions of (6) prove conver-gence of the Mann iterative schemes relative to these positivesolutions and compute the error estimates between theManniterative schemes and the positive solutions
Theorem 2 Assume that there exist twoconstants 119872 and 119873 with 119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896) minus 119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119875119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896) minus ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119877119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
forall (119899 119906119897 119906119897) isin N1198990
times (R+
0)
2
1 le 119897 le 119896
(14)
1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119876119899
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119882119899
forall (119899 119906119897) isin N1198990
times (R+
0) 1 le 119897 le 119896
(15)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (16)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (17)
119887119899= minus1 eventually (18)
Abstract and Applied Analysis 3
Then one has the following(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
+
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(19)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin 119860(119873119872) of (6)with lim
119899rarrinfin119911119899= +infin and has the following error estimate
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] such thatinfin
sum
119898=0
120572119898= +infin (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Firstly we show that (a) holds Put 119871 isin (119873119872) Itfollows from (16)sim(18) that there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (22)
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119873
(23)
119887119899= minus1 forall119899 ge 119879 (24)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879 119878119871119909119879 120573 le 119899 lt 119879
(25)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In light of (14) (15) (22)(23) and (25) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872)
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
4 Abstract and Applied Analysis
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
lt min 119872 minus 119871 119871 minus 119873
(26)which yield that
119878119871(119860 (119873119872)) sube 119860 (119873119872)
1003817100381710038171003817119878119871119909 minus 1198781198711199101003817100381710038171003817le 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817 forall119909 119910 isin 119860 (119873119872)
(27)
which implies that 119878119871is a contraction in 119860(119873119872) The
Banach fixed point theorem and (27) ensure that 119878119871has a
unique fixed point 119911 = 119911119899119899isinN120573
isin 119860(119873119872) that is
119911119899= 1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879
119911119899minus120591
= (119899 minus 120591)2
119871
+
infin
sum
119894=1
infin
sum
119906=119899+(119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
)
minus 119888119905] forall119899 ge 119879 + 120591
(28)which mean that119911119899minus 119911119899minus120591
= (2119899120591 minus 1205912
) 119871
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
(29)which yields thatΔ (119911119899minus 119911119899minus120591)
= 2120591119871 +
infin
sum
119904=119899
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
Δ2
(119911119899minus 119911119899minus120591)
= minusℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
+
infin
sum
119905=119899
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905] forall119899 ge 119879 + 120591
(30)
which gives that
Δ3
(119911119899minus 119911119899minus120591)
= minusΔℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
minus 119891 (119905 1199111198911119899
1199111198912119899
119911119891119896119899
) + 119888119899 forall119899 ge 119879 + 120591
(31)
which together with (24) implies that 119911 = 119911119899119899isinN120573
is apositive solution of (6) in 119860(119873119872) Note that
119873 le
119911119899
1198992
le 119872 forall119899 isin N120573 (32)
which guarantees that lim119899rarrinfin
119911119899= +infin It follows from (19)
(22) (24) (25) and (27) that for any 119898 isin N0and 119899 ge 119879
1003816100381610038161003816100381610038161003816
119909119898+1119899
1198992
minus
119911119899
1198992
1003816100381610038161003816100381610038161003816
=
1
1198992
1003816100381610038161003816100381610038161003816100381610038161003816
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)] minus 119911
119899
1003816100381610038161003816100381610038161003816100381610038161003816
le (1 minus 120572119898)
1003816100381610038161003816119909119898119899minus 119911119899
1003816100381610038161003816
1198992
+ 120572119898
1003816100381610038161003816119878119871119909119898119899minus 119878119871119911119899
1003816100381610038161003816
1198992
le (1 minus 120572119898)1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817+ 120579120572119898
1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817
le [1 minus (1 minus 120579) 120572119898]1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 119899 ge 119879
(33)
which implies that
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (34)
That is (20) holds Thus Lemma 1 (20) and (21) guaranteethat lim
119898rarrinfin119909119898= 119911
Next we show that (b) holds Let 1198711 1198712
isin
(119873119872) and 1198711=1198712 As in the proof of (a) we deduce
similarly that for each 119888 isin 1 2 there exist constants 120579119888isin
(0 1) and 119879119888ge 1198990+ 120591 + 120573 and a mapping 119878
119871119888
satisfying
Abstract and Applied Analysis 5
(22)sim(27) where 120579 119871 and 119879 are replaced by 120579119888 119871119888 and 119879
119888
respectively and the mapping 119878119871119888
has a fixed point 119911119888 =
119911119888
119899119899isinN120573
isin 119860(119873119872) which is a positive solution of (6) in119860(119873119872) with lim
119899rarrinfin119911119888
119899= +infin It follows that
119911119888
119899= 1198992
119871119888
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911119888
ℎ1119904
119911119888
ℎ2119904
119911119888
ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 119911119888
1198911119905
119911119888
1198912119905
119911119888
119891119896119905
) minus 119888119905]
forall119899 ge 119879119888
(35)
which together with (14) and (20) means that for 119899 ge
max1198791 1198792
100381610038161003816100381610038161003816100381610038161003816
1199111
119899
1198992
minus
1199112
119899
1198992
100381610038161003816100381610038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119911
1
ℎ1119904
1199111
ℎ2119904
1199111
ℎ119896119904
)
minus ℎ (119904 1199112
ℎ1119904
1199112
ℎ2119904
1199112
ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119911
1
1198911119905
1199111
1198912119905
1199111
119891119896119905
) minus 119891 (119905 1199112
1198911119905
1199112
1198912119905
1199112
119891119896119905
)
10038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
10038161199111
ℎ119897119904
minus 1199112
ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
10038161199111
119891119897119905
minus 1199112
119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
max 11987921 1198792
2
infin
sum
119894=1
infin
sum
119906=max1198791 1198792+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816minusmax 120579
1 1205792
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
(36)
which yields that
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817ge
10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
1 +max 1205791 1205792
gt 0 (37)
that is 1199111 =1199112
This completes the proof
Theorem 3 Assume that there exist two constants119872 and 119873with 119872 gt 119873 gt 0 and four nonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) and
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (38)
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (39)
119887119899= 1 eventually (40)
Then one has the following
(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
minus
infin
sum
119894=1
119879+2119894120591minus1
sum
119906=119879+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(41)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
6 Abstract and Applied Analysis
Proof Let 119871 isin (119873119872) It follows from (38)sim(40) that thereexist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (42)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)) lt min 119872 minus 119871 119871 minus 119873
(43)
119887119899= 1 forall119899 ge 119879 (44)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus119888119905] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(45)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Using (14) (15) (42) (43)and (45) we get that for each 119909 = 119909
119899119899isinN120573
119910 = 119910119899119899isinN120573
isin
119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) = 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)])
lt min 119872 minus 119871 119871 minus 119873
(46)
which imply (27) The rest of the proof is similar to the proofof Theorem 2 and is omitted This completes the proof
Theorem 4 Assume that there exist three constants 119887 119872and 119873 with (1 minus 119887)119872 gt 119873 gt 0 and four nonnega-tive sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and0 le 119887119899le 119887 lt 1 eventually (47)
Then one has the following(a) For any 119871 isin (119887119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any
1199090
= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated bythe scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871 minus 119887119899119909119898119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus 119887119879119909119898119879minus120591
minus
infin
sum
119906=119879
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(48)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
Abstract and Applied Analysis 7
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119887119872 + 119873119872) It follows from (38) (39) and(47) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = 119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119887119872 minus119873
0 le 119887119899le 119887 lt 1 forall119899 ge 119879
(49)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus 119887119899119909119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(50)for each 119909 = 119909
119899119899isinN120573
isin 119860(119873119872) In view of (14) (15) and(49) and (50) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 11987910038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le 119887119899
100381610038161003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
(119899 minus 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 minus 120591)2
1198992
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le 1198871003817100381710038171003817119909 minus 119910
1003817100381710038171003817
+
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le [119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817= 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus 119887119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus 119887119872
minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus 119887119872 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minus 119887119872 minusmin 119872 minus 119871 119871 minus 119887119872 minus119873 ge 119873
(51)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 5 Assume that there exist constants 119887 119872 and 119873with (1 + 119887)119872 gt 119873 gt 0 and four nonnegative sequences119875119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14)(15) (38) (39) and
minus1 lt 119887 le 119887119899le 0 eventually (52)
Then one has the following
(a) For any 119871 isin (119873 (1 + 119887)119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for
any 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by(48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
8 Abstract and Applied Analysis
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119873 (1 + 119887)119872) It follows from (38) (39) and(52) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (53)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 119887)119872 minus 119871 119871 minus 119873
(54)
minus1 lt 119887 le 119887119899le 0 forall119899 ge 119879 (55)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by (50) By virtue of
(15) (50) (53) and (55) we infer that for all 119909 = 119909119899119899isinN120573
119910 = 119910
119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le [minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 minus 119887119872
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 minus 119887119872 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 minus 119887119872 +min (1 + 119887)119872 minus 119871 119871 minus 119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minusmin (1 + 119887)119872 minus 119871 119871 minus 119873 ge 119873
(56)
That is (27) holds The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 6 Assume that there exist constants 119887119872 and 119873 with (1 minus 1119887)119872 gt 119873 gt 0 and fournonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882
119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899ge 119887 gt 1 eventually (57)
Then one has the following(a) For any 119871 isin ((1119887)119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any 119909
0=
1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterative sequence119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(58)
Abstract and Applied Analysis 9
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin ((1119887)119872 + 119873119872) It follows from (38) (39)and (57) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 =
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (59)
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus
1
119887
119872 minus119873
(60)
119887119899ge 119887 gt 1 forall119899 ge 119879 (61)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[(119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus 119888119905)] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(62)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In view of (14) (15)and (59)∽(62) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
+
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
+
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus
1
119887
119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
119887
119872
minus
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
119887
119872 minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
gt 119871 minus
1
119887
119872 minusmin 119872 minus 119871 119871 minus
1
119887
119872 minus119873 ge 119873
(63)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 7 Assume that there exist constants 119887 119872and 119873 with (1 + 1119887)119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899le 119887 lt minus1 eventually (64)
10 Abstract and Applied Analysis
Then one has the following
(a) For any 119871 isin (minus(1 + 1119887)119872 minus119873) there exist 120579 isin (0 1)and 119879 ge 119899
0+ 120591 + 120573 such that for any 119909
0= 1199090119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 120572119898 minus 1198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+120572119898 minus 119879
2
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(65)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (minus(1 + 1119887)119872 minus119873) It follows from (38)(39) and (64) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 = minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (66)
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 1
119887
)119872 minus 119871 119871 minus
1
119887
119872 minus119873
(67)
119887119899ge 119887 gt 1 forall119899 ge 119879 (68)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
minus1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(69)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Making use of (15) (66)(68) and (69) we conclude that10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le minus
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
minus
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le minus
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
minus
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le minus119871 minus
119872
119887
minus
1
1198871198992
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
Abstract and Applied Analysis 11
le minus119871 minus
119872
119887
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt minus119871 minus
119872
119887
+min (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 le 119872
119878119871119909119899
1198992
ge minus119871
+
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge minus119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt minus119871 minusmin (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 ge 119873
(70)
which yield (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
3 Examples
In this section we suggest six examples to explain the resultspresented in Section 2
Example 1 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus 119909119899minus120591) + Δ(
sin2119909119899minus3
1198997
) +
1
(1198999+ 21198995+ 1) (1 + 119909
2
1198992)
=
1198992
minus 2119899
1198998+ 1198993+ 1
forall119899 ge 4
(71)
where 120591 isin N is fixed Let 1198990= 4 119896 = 1 and 120573 = min4minus120591 1
and let119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= minus1 119888
119899=
1198992
minus 2119899
1198998+ 1198993+ 1
119891 (119899 119906) =
1
(1198999+ 21198995+ 1) (1 + 119906
2)
ℎ (119899 119906) =
sin21199061198997
1198911119899= 1198992
119865119899= 1198994
ℎ1119899= 119899 minus 3 119867
119899= (119899 minus 3)
2
119875119899=
2119872
1198999
119876119899=
1
1198999
119877119899=
2
1198997
119882119899=
1
1198997
forall (119899 119906) isin N1198990
timesR
(72)
It is easy to see that (14) (15) and (18) are satisfied Note that
1
1198992
infin
sum
119905=119899+120591
1199052max 119877
119905119867119905119882119905
=
1
1198992
infin
sum
119905=119899+120591
1199052max2(119905 minus 3)
2
1199057
1
1199057
=
infin
sum
119905=119899+120591
2(119905 minus 3)2
+ 1
1199055
le
2
1198992
infin
sum
119905=119899+120591
1
1199053
997888rarr 0 as 119899 997888rarr infin
1
1198992
infin
sum
119905=119899+120591
1199053max 119875
119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119905=119899+120591
1199053max2119872
1199055
1
1199059
100381610038161003816100381610038161199052
minus 2119905
10038161003816100381610038161003816
1199058+ 1199053+ 1
le
max 1 2119872
1198992
infin
sum
119905=119899+120591
1
1199052
997888rarr 0 as 119899 997888rarr infin
(73)
which together with Lemma 1 yield that (16) and (17) holdIt follows from Theorem 2 that (71) possesses uncountablymany positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 119899
0+120591+120573 such
that for each 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (19)converges to a positive solution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of(71) with lim
119899rarrinfin119911119899= +infin and has the error estimate (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] satisfying(21)
Example 2 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ 119909119899minus120591) + Δ(
sin211990931198993+1
1198993(1198996+ 2) (1 + 119909
4
21198992minus3
)
)
+
(minus1)119899
1198993
(1199091198992minus119899minus1
+ 119909(119899+1)(119899+2)
)
(11989913+ 1198995+ 1) (1 + 119909
2
1198992minus119899minus1
+ 1199092
(119899+1)(119899+2))
=
1198992
minus ln 1198991198996+ 1198995+ 1
forall119899 ge 5
(74)
where 120591 isin N is fixed Let 1198990= 5 119896 = 2 and 120573 = 5 minus 120591 and let
119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= 1 119888
119899=
1198992
minus ln 1198991198996+ 1198995+ 1
119891 (119899 119906 V) =(minus1)119899
1198993
(119906 + V)(11989913+ 1198995+ 1) (1 + 119906
2+ V2)
12 Abstract and Applied Analysis
ℎ (119899 119906 V) =sin2V
1198993(1198996+ 2) (1 + 119906
4)
1198911119899= 1198992
minus 119899 minus 1
1198912119899= (119899 + 1) (119899 + 2)
119865119899= (119899 + 1)
2
(119899 + 2)2
ℎ1119899= 21198992
minus 3
ℎ2119899= 31198993
+ 1
119867119899= (3119899
3
+ 1)
2
119875119899= 119876119899=
4
11989910
119877119899= 119882119899=
10
1198999
forall (119899 119906 V) isin N1198990
timesR2
(75)
It is clear that (14) (15) and (40) are fulfilled Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
10(31199043
+ 1)
2
1199049
10
1199049
le
160
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
160
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(76)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (77)
Observe that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max4(119905 + 1)2
(119905 + 2)2
11990510
4
11990510
1199052
minus ln 1199051199056+ 1199055+ 1
le
196
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
=
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
119905 minus 119906 + 1
1199054
le
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
196
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(78)
which yields that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (79)
Thus Theorem 3 guarantees that (74) possesses uncount-ably positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 120591 +
1198990+ 120573 such that the Mann iterative sequence 119909
119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (74) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 3 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 + 3 ln 1198992 + 4 ln 119899
119909119899minus120591)
+ Δ(
(minus1)119899 sin (119890minus119899
2|11990951198992minus3|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|11990921198993+1|
)
+
(minus1)119899
1198996(1 + 119909
2
119899minus3)
minus
1
(1198997+ 21198994minus 1) (1 + 119909
2
119899+4)
=
3(minus1)119899
1198992
911989910ln3119899
forall119899 ge 7
(80)
where 120591 isin N is fixed Let 1198990= 7 119896 = 2 119887 = 34 and
120573 = min7 minus 120591 4 and let119872 and119873 be two positive constantswith 119872 gt 4119873 and
119887119899=
1 + 3 ln 1198992 + 4 ln 119899
119888119899=
3(minus1)119899
1198992
911989910ln3119899
119891 (119899 119906 V) =(minus1)119899
1198996(1 + 119906
2)
minus
1
(1198997+ 21198994minus 1) (1 + V2)
ℎ (119899 119906 V) =(minus1)119899 sin (119890minus119899
2|119906|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|V|
1198911119899= 119899 minus 3 119891
2119899= 119899 + 4
119865119899= (119899 + 4)
2
ℎ1119899= 51198992
minus 3
ℎ2119899= 21198993
+ 1 119867119899= (2119899
3
+ 1)
2
119875119899= 119876119899=
3
1198996
119877119899= 119882119899=
2
11989910
forall (119899 119906 V) isin N1198990
timesR2
(81)
It is not difficult to verify that (14) (15) and (47) are fulfilledNote that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
2(21199043
+ 1)
2
11990410
2
11990410
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
18
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(82)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (83)
Abstract and Applied Analysis 13
Observe that1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max3(119905 + 4)2
1199056
3
1199056
100381610038161003816100381610038161003816100381610038161003816
3(minus1)119905
1199052
911990510ln3119905
100381610038161003816100381610038161003816100381610038161003816
le
12
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
le
12
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
12
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(84)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (85)
That is (38) and (39) hold Consequently Theorem 4 impliesthat (80) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((34)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (80) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 4 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 minus 51198993
2 + 61198993
119909119899minus120591) + Δ(
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119909
2
3119899minus7)
)
+
sin (119899211990931198992minus2)
(radic119899 + 14)22
=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
forall119899 ge 9
(86)
where 120591 isin N is fixed Let 1198990= 9 119896 = 1 119887 = minus56 and
120573 = 9 minus 120591 and let 119872 and 119873 be two positive constants with119872 gt 6119873 and
119887119899=
1 minus 51198993
2 + 61198993
119888119899=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
119891 (119899 119906) =
sin (1198992119906)
(radic119899 + 14)22
ℎ (119899 119906) =
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119906
2)
1198911119899= 31198992
minus 2 119865119899= (3119899
2
minus 2)
2
ℎ1119899= 3119899 minus 7
119867119899= (3119899 minus 7)
2
119875119899= 119876119899=
3
1198999
119877119899= 119882119899=
1
1198995
forall (119899 119906) isin N1198990
timesR
(87)
Obviously (14) (15) and (52) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(3119904 minus 7)2
1199045
1
1199045
le
9
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
9
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(88)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (89)
Notice that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max
3(31199052
minus 2)
2
1199059
3
1199059
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905
1199053
+ 51199052
+ 4119905 minus 2
1199059+ 1199058+ 21199055+ 1199053+ 7
100381610038161003816100381610038161003816100381610038161003816
le
27
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
27
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
27
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(90)
which gives that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (91)
That is (38) and (39) hold Thus Theorem 5 shows that(86) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (119873 (16)119872)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that the Mann
iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generatedby (48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
119860(119873119872) of (86) with lim119899rarrinfin
119911119899= +infin and has the error
estimate (20) where 120572119898119898isinN0
is an arbitrary sequencein [0 1] satisfying (21)
14 Abstract and Applied Analysis
Example 5 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ (
120587
2
+ 119899 sin 1119899
) 119909119899minus120591)
+ Δ(
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (11989921199092119899+1
))
)
+
119899 sin (119899119909119899minus2)
2 + (119899 + 5)16
=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
forall119899 ge 3
(92)
where 120591 isin N is fixed Let 1198990= 3 119896 = 1 119887 = 1205872 and
120573 = min3 minus 120591 1 and let119872 and119873 be two positive constantswith (1 minus 2120587)119872 gt 119873 and
119887119899=
120587
2
+ 119899 sin 1119899
119888119899=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
119891 (119899 119906) =
119899 sin (119899119906)2 + (119899 + 5)
16
ℎ (119899 119906) =
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (1198992119906))
1198911119899= 119899 minus 2 119865
119899= (119899 minus 2)
2
ℎ1119899= 2119899 + 1 119867
119899= (2119899 + 1)
2
119875119899= 119876119899=
1
11989914
119877119899= 119882119899=
2
1198999
forall (119899 119906) isin N1198990
timesR
(93)
Clearly (14) (15) and (61) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max2(2119904 + 1)2
1199049
2
1199049
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199047
le
18
1198992
infin
sum
119904=119899
1
1199046
997888rarr 0 as 119899 997888rarr infin
(94)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 2)2
11990514
1
11990514
10038161003816100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1cos3 (1199052 + 1)11990516+ ln 119905
10038161003816100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
11990514
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
11990513
le
1
1198992
infin
sum
119905=119899
1
11990512
997888rarr 0 as 119899 997888rarr infin
(95)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (96)
That is (38) and (39) hold Consequently Theorem 6 impliesthat (92) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((2120587)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (58) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (92) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 6 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119909119899minus120591) + Δ(
cos ((minus1)119899119890119899)
(119899 + 7)6
radic1 +1003816100381610038161003816119909119899minus2
1003816100381610038161003816
)
+
sin (1198992119909119899minus1)
1198999+ 31198995+ 21198992+ 1
=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
forall119899 ge 6
(97)
where 120591 isin N is fixed Let 1198990= 6 119896 = 1 119887 = minus2 and 120573 =
min6 minus 120591 3 and let 119872 and 119873 be two positive constantswith (12)119872 gt 119873 and
119887119899= minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119888119899=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
119891 (119899 119906) =
sin (1198992119906)1198999+ 31198995+ 21198992+ 1
Abstract and Applied Analysis 15
ℎ (119899 119906) =
cos ((minus1)119899119890119899)(119899 + 7)
6
radic1 + |119906|
1198911119899= 119899 minus 1 119865
119899= (119899 minus 1)
2
ℎ1119899= 119899 minus 2 119867
119899= (119899 minus 2)
2
119875119899= 119876119899=
1
1198997
119877119899= 119882119899=
1
1198996
forall (119899 119906) isin N1198990
timesR
(98)
Obviously (14) (15) and (64) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(119904 minus 2)2
1199046
1
1199046
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
1
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(99)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (100)
It is clear that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 1)2
1199057
1
1199057
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1
1199054
+ 41199052
+ 119905 minus 1
11990511+ 61199053+ 7119905 + 2
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
1
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(101)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (102)
That is (38) and (39) hold Consequently Theorem 7 impliesthat (97) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (minus1198722 minus119873)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that
the Mann iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (65) converges to a positive solution 119911 =
119911119899119899isinN120573
isin 119860(119873119872) of (97) with lim119899rarrinfin
119911119899= +infin and
has the error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in[0 1] satisfying (21)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This research was supported by the Science Research Foun-dation of Educational Department of Liaoning Province(L2012380)
References
[1] M H Abu-Risha ldquoOscillation of second-order linear differenceequationsrdquo Applied Mathematics Letters vol 13 no 1 pp 129ndash135 2000
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[3] R P Agarwal and J Henderson ldquoPositive solutions and nonlin-ear eigenvalue problems for third-order difference equationsrdquoComputers amp Mathematics with Applications vol 36 no 10-12pp 347ndash355 1998
[4] A Andruch-Sobiło and M Migda ldquoOn the oscillation ofsolutions of third order linear difference equations of neutraltyperdquoMathematica Bohemica vol 130 no 1 pp 19ndash33 2005
[5] Z Dosla and A Kobza ldquoGlobal asymptotic properties of third-order difference equationsrdquo Computers amp Mathematics withApplications vol 48 no 1-2 pp 191ndash200 2004
[6] S R Grace and G G Hamedani ldquoOn the oscillation of certainneutral difference equationsrdquo Mathematica Bohemica vol 125no 3 pp 307ndash321 2000
[7] J Cheng ldquoExistence of a nonoscillatory solution of a second-order linear neutral difference equationrdquo Applied MathematicsLetters vol 20 no 8 pp 892ndash899 2007
[8] LKongQKong andB Zhang ldquoPositive solutions of boundaryvalue problems for third-order functional difference equationsrdquoComputersampMathematics withApplications vol 44 no 3-4 pp481ndash489 2002
[9] I Y Karaca ldquoDiscrete third-order three-point boundary valueproblemrdquo Journal of Computational and Applied Mathematicsvol 205 no 1 pp 458ndash468 2007
[10] W-T Li and J P Sun ldquoExistence of positive solutions of BVPsfor third-order discrete nonlinear difference systemsrdquo AppliedMathematics and Computation vol 157 no 1 pp 53ndash64 2004
[11] W-T Li and J-P Sun ldquoMultiple positive solutions of BVPs forthird-order discrete difference systemsrdquo Applied Mathematicsand Computation vol 149 no 2 pp 389ndash398 2004
[12] Z Liu M Jia S M Kang and Y C Kwun ldquoBounded positivesolutions for a third order discrete equationrdquo Abstract andApplied Analysis vol 2012 Article ID 237036 12 pages 2012
[13] Z Liu S M Kang and J S Ume ldquoExistence of uncountablymany bounded nonoscillatory solutions and their iterativeapproximations for second order nonlinear neutral delay dif-ference equationsrdquo Applied Mathematics and Computation vol213 no 2 pp 554ndash576 2009
[14] Z Liu Y Xu and S M Kang ldquoBounded oscillation criteriafor certain third order nonlinear difference equations withseveral delays and advancesrdquo Computers amp Mathematics withApplications vol 61 no 4 pp 1145ndash1161 2011
[15] M Migda and J Migda ldquoAsymptotic properties of solutions ofsecond-order neutral difference equationsrdquoNonlinear Analysis
16 Abstract and Applied Analysis
Theory Methods and Applications vol 63 no 5ndash7 pp e789ndashe799 2005
[16] N Parhi ldquoNon-oscillation of solutions of difference equationsof third orderrdquoComputersampMathematics withApplications vol62 no 10 pp 3812ndash3820 2011
[17] N Parhi and A Panda ldquoNonoscillation and oscillation ofsolutions of a class of third order difference equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 1 pp213ndash223 2007
[18] S H Saker ldquoNew oscillation criteria for second-order nonlinearneutral delay difference equationsrdquo Applied Mathematics andComputation vol 142 no 1 pp 99ndash111 2003
[19] S H Saker ldquoOscillation of third-order difference equationsrdquoPortugaliae Mathematica vol 61 no 3 pp 249ndash257 2004
[20] S Stevic ldquoOn a third-order system of difference equationsrdquoApplied Mathematics and Computation vol 218 no 14 pp7649ndash7654 2012
[21] X H Tang ldquoBounded oscillation of second-order delay dif-ference equations of unstable typerdquo Computers amp Mathematicswith Applications vol 44 no 8-9 pp 1147ndash1156 2002
[22] J Yan and B Liu ldquoAsymptotic behavior of a nonlinear delaydifference equationrdquo Applied Mathematics Letters vol 8 no 6pp 1ndash5 1995
[23] Z G Zhang and Q L Li ldquoOscillation theorems for second-order advanced functional difference equationsrdquo Computers ampMathematics with Applications vol 36 no 6 pp 11ndash18 1998
Research ArticleAlgebroid Solutions of Second Order ComplexDifferential Equations
Lingyun Gao1 and Yue Wang2
1 Department of Mathematics Jinan University Guangzhou Guangdong 510632 China2 School of Information Renmin University of China Beijing 100872 China
Correspondence should be addressed to Lingyun Gao tgaolyjnueducn
Received 28 November 2013 Accepted 17 December 2013 Published 2 January 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 L Gao and Y WangThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Using value distribution theory and maximummodulus principle the problem of the algebroid solutions of second order algebraicdifferential equation is investigated Examples show that our results are sharp
1 Introduction and Main Results
We use the standard notations and results of the Nevanlinnatheory of meromorphic or algebroid functions see forexample [1 2]
In this paper we suppose that second order algebraicdifferential equation (3) admit at least one nonconstant ]-valued algebroid solution 119908(119911) in the complex plane Wedenote by 119864 a subset of [0infin) for which 119898(119864) lt infin and by119870 a positive constant where119898(119864) denotes the linearmeasureof 119864 119864 or 119870 does not always mean the same one when theyappear in the following
Let 119886119895119896(119895 = 0 1 119899 119896 = 0 1 119902
119895) be entire func-
tions without common zeroes such that 11988601199020
= 0 We put
119876119895(119911 119908) =
119902119895
sum
119896=0
119886119895119896119908119896
119902119895= deg119876119895
119908
119901 = max 119902119895+ 119895 119895 = 0 1 119899 minus 1
(1)
Some authors had investigated the problem of the exis-tence of algebroid solutions of complex differential equationsand they obtained many results ([2ndash10] etc)
In 1989 Toda [4] considered the existence of algebroidsolutions of algebraic differential equation of the form
119899
sum
119895=0
119876119895(119911 119908) (119908
1015840
)
119895
= 0 (2)
He obtained the following
Theorem A (see [4]) Let 119908(119911) be a nonconstant ]-valuedalgebroid solution of the above differential equation and all 119886
119895119896
are polynomials If 119901 lt 119899 + 119902119899 then 119908(119911) is algebraic
The purpose of this paper is to investigate algebroid solu-tions of the following second order differential equation inthe complex plane with the aid of the Nevanlinna theory andmaximum modulus principle of meromorphic or algebroidfunctions
119899
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
= 0 (3)
where 119876119895(119911 119908) = sum
119902119895
119896=0119886119895119896119908119896 119895 = 0 1 2 119899
We will prove the following two results
Theorem 1 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and all 119886
119895119896are polynomials
If 119901 le 119902119899 then 119908(119911) is algebraic 119901 = max119902
119895 119895 = 0 1 119899 minus
1
Theorem 2 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and the orders of all 119886
119895119896are
finite If 1199020gt max
1le119895le119899minus1119902119895+119895 then the following statements
are equivalent(a) 120575(infin119908) gt 0(b) 1199020= 119902119899+ 119899
(c) infin is a Picard exceptional value of 119908(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 123049 4 pageshttpdxdoiorg1011552014123049
2 Abstract and Applied Analysis
2 Some Lemmas
Lemma 3 (see [2]) Suppose that 119908(119911) 119886119894(119911) (119894 = 1 2 119901)
are meromorphic functions and 119886119901(119911) = 0 Then one has
119898(119903
119901
sum
119894=1
119886119894(119911) 119908119894
) le 119901119898 (119903 119908) +
119901
sum
119894=1
119898(119903 119886119894(119911)) + 119874 (1)
(4)
Examining proof of Lemma 45 presented in [2 pp 192-193] we can verify Lemma 4
Lemma 4 Let 119908(119911) be a transcendental algebroid functionsuch that 119908(119911) has only finite number of poles and let 119908(119911)1199081015840
(119911) and 11990810158401015840(119911) have no poles in |119911| gt 1199030 Then for some
constants 119862119894gt 0 119894 = 1 2 3 and 119903 ge 119903
1ge 1199030it holds
119872(119903 119908) le 1198621+ 1198622119903 + 11986231199032
119872(119903 11990810158401015840
) (5)
where119872(119903 119908) = max|119911|=119903
|119908(119911)|
Lemma 5 (see [11]) The absolute values of roots of equation
119911119899
+ 1198861119911119899minus1
+ sdot sdot sdot + 119886119899= 0 (6)
are bounded by
max 119899 10038161003816100381610038161198861
1003816100381610038161003816 (119899
10038161003816100381610038161198862
1003816100381610038161003816)12
(1198991003816100381610038161003816119886119899
1003816100381610038161003816)1119899
(7)
Lemma 6 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of the differential equation (3) and let 119886
119895119896be a
polynomial If 119901 lt 119899 + 119902119899 then
min 119899 119902119899minus 119901 log+119872(119903 119908)
+max 0 119902119899minus 119901 minus 119899 log+119872(119903 119908)
le +119874 (log 119903) + 119874 (1) (119903 notin 119864)
(8)
where119872(119903 119908) = max|119911|=119903
|119908(119911)| 119870 is a positive constant
Proof We first prove that the poles of 119908 are contained in thezeroes of 119886
119899119902119899
(119911)Suppose that 119911
0is a pole of 119908 of order 120591 and 119911
0is not the
zeroes of 119886119899119902119899
(119911) Then
119908 (119911) = (119911 minus 1199110)minus120591120582
1199081(119911) 119908
1(1199110) = 0infin
11990810158401015840
(119911) = (119911 minus 1199110)minus(120591+2120582)120582
1199082(119911) 119908
2(1199110) = 0infin
(9)
We rewrite differential equation (3) as follows
119886119899119902119899
(11990810158401015840
)
119899
=
119899minus1
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
(10)
It follows from (10) that
119902119899120591 + 119899 (120591 + 2120582) le 119901120591 + (119899 minus 1) (120591 + 2120582) (11)
Noting that 119901 le 119902119899 we have
120591 + 2120582 lt 0 (12)This is a contradiction
This shows that the poles of119908 are contained in the zeroesof 119886119899119902119899
(119911)We rewrite differential equation (3) as follows
119899
sum
119895=0
119876119895(119911 119908)119876
119899(119911 119908)
119899minus119895minus1
(11987611989911990810158401015840
)
119895
= 0 (13)
119872(119903119876111990810158401015840
) ge 119872(119903 119908)119902119899+119872(119903 119908
10158401015840
)
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
ge 119872(119903 119908)119902119899+
119872(119903 119908) minus 1198621minus 1198622119903
11986231199032
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
(14)
For 119895 = 0 1 119899 minus 1 we have10038161003816100381610038161003816119876119895(119911119903 119908)119876
119899(119911119903 119908)119899minus119895minus110038161003816
100381610038161003816le 119870119872(119903 119908)
119902119895+119902119899(119899minus119895minus1)
(15)
Applying Lemma 5 to (13) at 119911 = 119911119903
119872(119903119876111990810158401015840
) le 119870119872(119903 119908)maxℎ
119895 119895=01119899minus1
(16)
where ℎ119895= (119902119895+ 119902119899(119899 minus 119895 minus 1))(119899 minus 119895)
From (14) and (15) we have
119872(119903 119908)119902119899le 119870 119872(119903 119908)
119902119899minus1
+ 1199032
119872(119903 119908)maxℎ
119895 119895=01119899minus1
le 119870 119872(119903 119908)119902119899minus1
+ 1199032
119872(119903 119908)119902119899+((119901minus119902
119899)119899)
(17)Note that
ℎ119895=
119902119895+ 119902119899(119899 minus 119895 minus 1)
119899 minus 119895
lt
119901 + 119902119899(119899 minus 119895 minus 1) minus 119895
119899 minus 119895
= 119902119899+
119901 minus 119902119899
119899
+
119895 (119901 minus 119902119899minus 119899)
119899 (119899 minus 119895)
le 119902119899+
119901 minus 119902119899
119899
119895 = 0 1 119899 minus 1
(18)
Dividing the inequality (17) by119872(119903 119908)
max119902119899minus1119902119899+((119902119899minus119901)119899) we obtain for 119903 notin 119864
119872(119903 119908)min1(119902
119899minus119901)119899
le 11987041 +
1199032
119872(119903 119908)max0(119902
119899minus119901minus119899)119899
(19)
which reduces to our inequality by calculating log+ of theboth sidesmin 119899 119902
119899minus 119901 log+119872(119903 119908)
le minusmax 0 119902119899minus 119901 minus 119899 log+119872(119903 119908) + 119874 (log 119903) + 119874 (1)
(20)Lemma 6 is complete
Abstract and Applied Analysis 3
3 Proof of Theorem 1
First we consider119873(119903 119908)Let 1199110be a pole of 119908 of 120591 Let 119905 be the order of zero of
119886119899119902119899
(119911) at 1199110
(i) When the order of the pole of119876119899(119908)(119908
10158401015840
)119899 is not equal
to that of other terms of the left-hand side of (10) at 1199110 we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 (21)
that is
120591 le
(119905 minus 2119899) 120582
119902119899+ 119899 minus 119901
(22)
(ii)When the order of pole of119876119899(119908)(119908
10158401015840
)119899 is equal to that
of some term 119876119896(119908)(119908
10158401015840
)119899 of the left-hand side of (10) at 119911
0
we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 + 119899 (120591 + 2120582) (23)
that is
120591 le
119905120582
119902119899minus 119901
(24)
Combining cases (i) and (ii) we obtain
119873(119903 119908) le 1198708119873(119903
1
119886119899119902119899
) (25)
where1198708is a positive constant
Secondly by Lemma 6 we obtain
min 119899 119902119899minus 119901119898 (119903 119908)
le 119870[
119901
sum
119894=0
119898(119903 119886119894) +
119902119899minus1
sum
119896=0
119898(119903 119887119896)] + 119874 (log 119903) (119903 notin 119864)
(26)
Combining the inequalities (25) and (26) we have
119879 (119903 119908) = 119874 (log 119903) (119903 notin 119864) (27)
which shows that 119908 is an algebraic solution of (3)This completes the proof of Theorem 1
4 Proof of Theorem 2
(i) (a)rArr (b) Suppose that 120575(infin119908) gt 0 If 1199020gt 119902119899+ 119899 then
we have by (3)
1199081199020= minus
1
11988601199020
119899
sum
119895=1
119876119895(119908)119908
119895
(
11990810158401015840
119908
)
119895
minus
1199020minus1
sum
119896=0
1198860119896119908119896
(28)
Applying Lemma 3 to (28)
1199020119898(119903 119908) le (119902
0minus 1)119898 (119903 119908) +sum
119895119896
119898(119903 119886119895119896)
+ 119870119898(119903
11990810158401015840
119908
) + 119898(119903
1
11988601199020
) + 119874 (1)
(29)
Since 119908(119911) is admissible solution we have
119898(119903 119908) = 119878 (119903 119908) (30)
so that
120575 (119908infin) = 0 (31)
This is a contradiction Thus 1199020le 119902119899+ 119899
If 1199020lt 119902119899+119899 byTheorem 1119908(119911) is nonadmissibleThus
1199020= 119902119899+ 119899 (32)
(ii) (b) rArr (c) Let 1199020= 119902119899+ 119899 Then similar to the proof of
Lemma 6 we obtain that the poles of 119908(119911) are contained inthe set of 119886
119899119902119899
andinfin is a Picard exceptional value of 119908(119911)
(iii) (c) rArr (a) Let infin be a Picard exceptional value of 119908(119911)Then 120575(119908infin) = 1
5 Some Examples
Example 1 The differential equation
]21199082]minus111990810158401015840 minus [((2] + 1)1199084] + 21199082] minus ] + 1)] = 0 (33)
has a transcendental algebroid solution 119908(119911) = (tan 119911)1] Inthis case
119901 = 1199020= 4] gt 1 + 119902
1= 2] minus 1 + 1 = 2] (34)
Remark 7 Example 1 shows that the condition in Theorem 1is sharp
Example 2 Transcendental algebroid function 119908(119911) =
(sin 119911)12 is a 2-valued solution of the following differentialequation
161199086
(11990810158401015840
)
2
+ 21199085
11990810158401015840
minus (1199088
minus
1
2
1199086
+ 21199084
minus
1
2
1199082
+ 1) = 0
(35)
In this case
1199020= 8 119899 = 2 119902
1= 5 119902
2= 6 (36)
By Theorem 2 for transcendental algebroid function 119908(119911) =(sin 119911)12infin is a Picard exceptional value
Remark 8 Example 2 shows that the result in Theorem 2holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project is project Supported by National Natural ScienceFoundation (10471065) of China and NSF of GuangdongProvince (04010474)
4 Abstract and Applied Analysis
References
[1] H Yi and C C YangTheory of the Uniqueness of MeromorphicFunctions Science Press Beijing China 1995 (Chinese)
[2] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Science Press Beijing China 1988
[3] Y Z He and X Z Xiao ldquoAdmissible solutions and ordinarydifferential equationsrdquo Contemporary Mathematics vol 25 pp51ndash61 1983
[4] N Toda ldquoOn algebroid solutions of some algebraic differentialequations in the complex planerdquo Japan Academy A vol 65 no4 pp 94ndash97 1989
[5] T Chen ldquoOne class of ordinary differential equations whichpossess algebroid solutions in the complex domainrdquo ChineseQuarterly Journal of Mathematics vol 6 no 4 pp 45ndash51 1991
[6] K Katajamaki ldquoValue distribution of certain differential poly-nomials of algebroid functionsrdquo Archiv der Mathematik vol 67no 5 pp 422ndash429 1996
[7] L Gao ldquoSome results on admissible algebroid solutions ofcomplex differential equationsrdquo Indian Journal of Pure andApplied Mathematics vol 32 no 7 pp 1041ndash1050 2001
[8] L-Y Gao ldquoOn some generalized higher-order algebraic dif-ferential equations with admissible algebroid solutionsrdquo IndianJournal of Mathematics vol 43 no 2 pp 163ndash175 2001
[9] L Gao ldquoOn the growth of solutions of higher-order algebraicdifferential equationsrdquoActaMathematica Scientia B vol 22 no4 pp 459ndash465 2002
[10] L Y Gao ldquoThe growth of single-valued meromorphic solutionsand finite branch solutionsrdquo Journal of Systems Science andMathematical Sciences vol 24 no 3 pp 303ndash310 2004
[11] T Takagi Lecture on Algebra Kyoritsu Tokyo Japan 1957(Japanese)
Abstract and Applied Analysis
Complex Differences and Difference Equations
Guest Editors Zong-Xuan Chen Kwang Ho Shonand Zhi-Bo Huang
Copyright copy 2014 Hindawi Publishing Corporation All rights reserved
This is a special issue published in ldquoAbstract and Applied Analysisrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited
Editorial Board
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Contents
Complex Differences and Difference Equations Zong-Xuan Chen Kwang Ho Shon and Zhi-Bo HuangVolume 2014 Article ID 124843 1 page
Blow-Up Analysis for a Quasilinear Parabolic Equation with Inner Absorption and NonlinearNeumann Boundary Condition Zhong Bo Fang and Yan ChaiVolume 2014 Article ID 289245 8 pages
Some Properties on Complex Functional Difference Equations Zhi-Bo Huang and Ran-Ran ZhangVolume 2014 Article ID 283895 10 pages
The Regularity of Functions on Dual Split Quaternions in Clifford AnalysisJi Eun Kim and Kwang Ho ShonVolume 2014 Article ID 369430 8 pages
Unicity of Meromorphic Functions Sharing Sets withTheir Linear Difference PolynomialsSheng Li and BaoQin ChenVolume 2014 Article ID 894968 7 pages
A ComparisonTheorem for Oscillation of the Even-Order Nonlinear Neutral Difference EquationQuanxin ZhangVolume 2014 Article ID 492492 5 pages
Difference Equations and Sharing Values Concerning Entire Functions andTheir DifferenceZhiqiang Mao and Huifang LiuVolume 2014 Article ID 584969 6 pages
Admissible Solutions of the Schwarzian Type Difference Equation Baoqin Chen and Sheng LiVolume 2014 Article ID 306360 5 pages
Statistical Inference for Stochastic Differential Equations with Small NoisesLiang Shen and Qingsong XuVolume 2014 Article ID 473681 6 pages
On the Deficiencies of Some Differential-Difference Polynomials Xiu-Min Zheng and Hong Yan XuVolume 2014 Article ID 378151 12 pages
On Growth of Meromorphic Solutions of Complex Functional Difference Equations Jing LiJianjun Zhang and Liangwen LiaoVolume 2014 Article ID 828746 6 pages
Unicity of Entire Functions concerning Shifts and Difference Operators Dan Liu Degui Yangand Mingliang FangVolume 2014 Article ID 380910 5 pages
On Positive Solutions and Mann Iterative Schemes of aThird Order Difference Equation Zeqing LiuHeng Wu Shin Min Kang and Young Chel KwunVolume 2014 Article ID 470181 16 pages
Algebroid Solutions of Second Order Complex Differential Equations Lingyun Gao and Yue WangVolume 2014 Article ID 123049 4 pages
EditorialComplex Differences and Difference Equations
Zong-Xuan Chen1 Kwang Ho Shon2 and Zhi-Bo Huang1
1School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Zong-Xuan Chen chzxvipsinacom
Received 12 August 2014 Accepted 12 August 2014 Published 22 December 2014
Copyright copy 2014 Zong-Xuan Chen et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
In more recent years activity in the area of the complexdifferences and the complex difference equations has fleetlyincreased
This journal has set up a column of this special issue Wewere pleased to invite the interested authors to contributetheir original research papers as well as good expositorypapers to this special issue that will make better improvementon the theory of complex differences and difference equa-tions
In this special issue many good results are obtainedDifference equations are widely applied to mathematical
physics economics and chemistry In this special issue Z-B Huang and R-R Zhang J Li et al and L Gao and YWang investigate the growth a Borel exceptional value ofmeromorphic solutions to different types of higher ordernonliear difference equations respectively B Chen and S Liinvestigate the Schwarzian type difference equation D Liuet al Z Mao and H Liu and S Li and B Chen investigateunicity of meromorphic functions concerning different typesof difference operators Recently many difference analoguesof the classic Nevanlinna theory are obtained
In this special issue X-M Zheng and H Y Xu obtaina differential difference analogue of Valiron-Mohonko theo-rem Related topics with complex difference J E Kim andK H Shon investigate the regularity of functions on dualsplit quaternions in Clifford analysis and the tensor productrepresentation of polynomials of weak type in a DF-spaceQ Zhang and Z Liu et al investigate different types ofreal difference equations respectively L Shen and Q Xuinvestigate stochastic differential equations
This special issue stimulates the continuing efforts to thecomplex differences and the complex difference equations
Zong-XuanChenKwangHo ShonZhi-BoHuang
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 124843 1 pagehttpdxdoiorg1011552014124843
Research ArticleBlow-Up Analysis for a Quasilinear Parabolic Equation withInner Absorption and Nonlinear Neumann Boundary Condition
Zhong Bo Fang and Yan Chai
School of Mathematical Sciences Ocean University of China Qingdao 266100 China
Correspondence should be addressed to Zhong Bo Fang fangzb7777hotmailcom
Received 24 February 2014 Revised 11 April 2014 Accepted 11 April 2014 Published 30 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Z B Fang and Y Chai This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We investigate an initial-boundary value problem for a quasilinear parabolic equation with inner absorption and nonlinearNeumann boundary condition We establish respectively the conditions on nonlinearity to guarantee that 119906(119909 119905) exists globally orblows up at some finite time 119905lowast Moreover an upper bound for 119905lowast is derived Under somewhat more restrictive conditions a lowerbound for 119905lowast is also obtained
1 Introduction
We are concerned with the global existence and blow-upphenomenon for a quasilinear parabolic equation with non-linear inner absorption term
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
minus 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (1)
subjected to the nonlinear Neumann boundary and initialconditions
(|nabla119906|119901
+ 1)
120597119906
120597]= 119892 (119906) (119909 119905) isin 120597Ω times (0 119905
lowast
) (2)
119906 (119909 0) = 1199060(0) ge 0 119909 isin Ω (3)
whereΩ is a bounded star-shaped region of 119877119873 (119873 ge 2) withsmooth boundary 120597Ω ] is the unit outward normal vectoron 120597Ω 119901 ge 0 119905lowast is the blow-up time if blow-up occurs orelse 119905lowast
= +infin the symbol 119894 denotes partial differentiationwith respect to 119909
119894 119894 = 1 2 119873 the repeated index indicates
summation over the index and nabla is gradient operatorMany physical phenomena and biological species theo-
ries such as the concentration of diffusion of some non-Newton fluid through porous medium the density of somebiological species and heat conduction phenomena havebeen formulated as parabolic equation (1) (see [1ndash3]) Thenonlinear Neumann boundary condition (2) can be physi-cally interpreted as the nonlinear radial law (see [4 5])
In the past decades there have been many works dealingwith existence and nonexistence of global solutions blow-upof solutions bounds of blow-up time blow-up rates blow-up sets and asymptotic behavior of solutions to nonlinearparabolic equations see the books [6ndash8] and the surveypapers [9ndash11] Specially we would like to know whether thesolution blows up and at which time when blow-up occursA variety of methods have been used to study the problemabove (see [12]) and in many cases these methods used toshow that solutions blow up often provide an upper boundfor the blow-up time However lower bounds for blow-uptime may be harder to be determined For the study ofthe initial boundary value problem of a parabolic equationwith homogeneous Dirichlet boundary condition see [1314] Payne et al [13] considered the following quasilinearparabolic equation
119906119905= div (120588|nabla119906|2nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (4)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω To get the lower bound for the blow-up time the authorsassumed that 120588 is a positive 1198621 function which satisfies
120588 (119904) + 1199041205881015840
(119904) gt 0 119904 gt 0 (5)
The lower bound for the blow-up time of solution to (4) withRobin boundary condition was obtained in [15] where 120588 is
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 289245 8 pageshttpdxdoiorg1011552014289245
2 Abstract and Applied Analysis
also assumed to satisfy the condition (5) However under thisboundary condition the best constant of Sobolev inequalityused in [13] is no longer applicable They imposed suitableconditions on 119891 and 120588 and determined a lower bound forthe blow-up time if blow-up occurs and determined whenblow-up cannot occur Marras and Vernier Piro [14] studiedthe nonlinear parabolic problem with time dependent coeffi-cients
1198961(119905) div (119892 (|nabla119906|2nabla119906)) + 119896
2(119905) 119891 (119906) = 119896
3(119905) 119906119905
(119909 119905) isin Ω times (0 119905lowast
)
(6)
where Ω is a bounded domain in 119877119873 with smooth boundary
120597Ω Under some conditions on the data and geometry of thespatial domain they obtained upper and lower bounds of theblow-up time Moreover the sufficient conditions for globalexistence of the solution were derived
For the study of the initial boundary value problem ofa parabolic equation with Robin boundary condition werefer to [15ndash19] Li et al [16] investigated the problem of thenonlinear parabolic equation
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
+ 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (7)
where Ω is a bounded domain in 1198773 with smooth boundary
120597ΩThey derived the lower bound for the blow-up timewhenthe blow-up occurs Clearly |nabla119906|119901 + 1 does not satisfy thecondition (5) Enache [17] discussed the quasilinear parabolicproblem
119906119905= (119892 (119906) 119906
119894)119894
+ 119891 (119906) (8)
where Ω is a bounded domain in 119877119873
(119873 ge 2) with smoothboundary 120597Ω By virtue of a first-order differential inequalitytechnique they showed the sufficient conditions to guaranteethat the solution 119906(119909 119905) exists globally or blows up Inaddition a lower bound for the blow-up time when blow-up occurs was also obtained Ding [18] studied the nonlinearparabolic problem
(119887 (119906))119905= nabla sdot (119892 (119906) nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (9)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω They derived conditions on the data which guaranteethe blow-up or the global existence of the solution A lowerbound on blow-up time when blow-up occurs was alsoobtained For the problem of the nonlinear nonlocal porousmedium equation we read the paper of Liu [19]
Recently for the problems with nonlinear Neumannboundary conditions Payne et al [20] studied the semilinearheat equation with inner absorption term
119906119905= Δ119906 minus 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (10)
They established conditions on nonlinearity to guarantee thatthe solution119906(119909 119905) exists for all time 119905 gt 0 or blows up at somefinite time 119905lowast Moreover an upper bound for 119905lowast was derivedUnder somewhat more restrictive conditions a lower bound
for 119905lowast was derivedThereafter they considered the quasilinearparabolic equation
119906119905= nabla sdot (|nabla119906|
119901
nabla119906) (119909 119905) isin Ω times (0 119905lowast
) (11)
and they showed that blow-up occurs at some finite timeunder certain conditions on the nonlinearities and the dataupper and lower bounds for the blow-up time were derivedwhen blow-up occurs see [21] Liu et al The authors [22 23]studied the reaction diffusion problem with nonlocal sourceand inner absorption terms or with local source and gradientabsorption terms Very recently Fang et al [24] consideredlower bounds estimate for the blow-up time to nonlocalproblemwith homogeneousDirichlet orNeumann boundarycondition
Motivated by the above work we intend to study theglobal existence and the blow-up phenomena of problem (1)ndash(3) and the results of the semilinear equations are extended tothe quasilinear equations Unfortunately the techniques usedfor semilinear equation to analysis of blow-up phenomenaare no longer applicable to our problem As a consequenceby using the suitable techniques of differential inequalitieswe establish respectively the conditions on the nonlinearities119891 and 119892 to guarantee that 119906(119909 119905) exists globally or blows upat some finite time If blow-up occurs we derive upper andlower bounds of the blow-up time
The rest of our paper is organized as follows In Section 2we establish conditions on the nonlinearities to guaranteethat 119906(119909 119905) exists globally In Section 3 we show the condi-tions on data forcing the solution 119906(119909 119905) to blow up at somefinite time 119905
lowast and obtain an upper bound for 119905lowast A lower
bound of blow-up time under some assumptions is derivedin Section 4
2 The Global Existence
In this sectionwe establish the conditions on the nonlinearity119891 and nonlinearity 119892 to guarantee that 119906(119909 119905) exists globallyWe state our result as follows
Theorem 1 Assume that the nonnegative functions 119891 and 119892
satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(12)
where 1198961gt 0 119896
2ge 0 119904 gt 1 2119904 lt 119902 + 1 and 119904 minus 1 lt 119901 lt 119902 minus 1
Then the (nonnegative) solution 119906(119909 119905) of problem (1)-(3) doesnot blow up that is 119906(119909 119905) exists for all time 119905 gt 0
Proof Set
Ψ (119905) = int
Ω
1199062
119889119909 (13)
Abstract and Applied Analysis 3
Similar to Theorem 21 in [20] we get
Ψ1015840
(119905) le 21205752
int
Ω
1199062119904
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
+
21198962119873
1205880
int
Ω
119906119904+1
119889119909
minus2int
Ω
|nabla119906|119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 1198681+ 1198682
(14)
where 120575 = 1198962(119904 + 1)1198892120588
0 1205880
= min119909isin120597Ω
(119909 sdot ]) 119889 =
max119909isin120597Ω
|119909| and
1198681le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
times 1198601|Ω|(119902minus119904)(119902+1)
minus 1198602(int
Ω
119906119902+1
119889119909)
(119902minus119904)(119902+1)
(15)
where 1198601= 21205752
120572120576(120572minus1)120572 119860
2= 1198961minus 21205752
(1 minus 120572)120576 120572 = (119902 + 1 minus
2119904)(119902 minus 119904) lt 1 120576 gt 0Next we estimate 119868
2= (211989621198731205880) intΩ
119906119904+1
119889119909minus2 intΩ
|nabla119906|119901+2
119889119909 minus 1198961intΩ
119906119902+1
119889119909 Since
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
= (
119901
2
+ 1)
2
119906119901
|nabla119906|2
(16)
it follows from Holder inequality that
int
Ω
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
119889119909 le (
119901
2
+ 1)
2
(int
Ω
|nabla119906|119901+2
119889119909)
2(119901+2)
times (int
Ω
119906119901+2
119889119909)
119901(119901+2)
(17)
Furthermore we have
int
Ω
119906119901+2
119889119909 le [
(119901 + 2)2
41205821
]
(1199012)+1
int
Ω
|nabla119906|119901+2
119889119909 (18)
which follows from (17) and membrane inequality
1205821int
Ω
1205962
119889119909 le int
Ω
|nabla120596|2
119889119909 (19)
where 1205821is the first eigenvalue in the fixed membrane
problem
Δ120596 + 120582120596 = 0 120596 gt 0 in Ω 120596 = 0 on 120597Ω (20)
Combining 1198682and (18) we have
1198682le
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 2[
41205821
(119901 + 2)2]
(1199012)+1
times int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
=
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 3[
41205821
(119901 + 2)2
]
(1199012)+1
int
Ω
119906119901+2
119889119909
+
[
41205821
(119901 + 2)2]
(1199012)+1
int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 11986821+ 11986822
(21)
Making use of Holder inequality we obtain
int
Ω
119906119904+1
119889119909 le (int
Ω
119906119901+2
119889119909)
(119904+1)(119901+2)
|Ω|(119901minus119904+1)(119901+2)
(22)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119904+1
119889119909)
2(119904+1)
|Ω|(119904minus1)(119904+1)
(23)
Combining (21) (22) with (23) we get
11986821
le (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
(24)
with
1198611=
21198962119873
1205880
1198612= 3[
41205821
(119901 + 2)2]
(1199012)+1
|Ω|minus(119901minus119904+1)2
(25)
Applying Holder inequality we obtain
int
Ω
119906119901+2
119889119909 le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
|Ω|(119902minus119901minus1)(119902+1)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119902+1
119889119909)
2(119902+1)
|Ω|(119902minus1)(119902+1)
(26)
It follows from (26) that
11986822
le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(27)
where
1198621= [
41205821
(119901 + 2)2
]
(1199012)+1
|Ω|(119902minus119901minus1)(119902+1)
1198622= 1198961|Ω|(1minus119902)(119902minus119901minus1)2(119902+1)
(28)
4 Abstract and Applied Analysis
Combining (14) (15) (21) and (24) with (27) we obtain
Ψ1015840
(119905) le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
1198601minus 1198602Ψ(119905)(119902minus119904)2
+ (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
+ (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(29)
with
1198601= 1198601|Ω|(119902minus119904)(119902+1)
1198602= 1198602|Ω|(1minus119902)(119902minus119904)2(119902+1)
(30)
We conclude from (29) that Ψ(119905) is decreasing in each timeinterval on which we obtain
Ψ (119905) ge max(1198601
1198602
)
2(119902minus119904)
(
1198611
1198612
)
2(119901minus119904+1)
(
1198621
1198622
)
2(119902minus119901minus1)
(31)
so that Ψ(119905) remains bounded for all time under theconditions in Theorem 1 This completes the proof ofTheorem 1
3 Blow-Up and Upper Bound of 119905lowast
In this section Ω needs not to be star-shaped We establishthe conditions to assure that the solution of (1)ndash(3) blowsup at finite time 119905lowast and derive an upper bound for 119905lowast Moreprecisely we establish the following result
Theorem 2 Let 119906(119909 119905) be the classical solution of problem (1)-(3) Assume that the nonnegative and integrable functions 119891and 119892 satisfy
120585119891 (120585) le 2 (1 + 120572) 119865 (120585) 120585 ge 0
120585119892 (120585) ge 2 (1 + 120573)119866 (120585) 120585 ge 0
(32)
with
119865 (120585) = int
120585
0
119891 (120578) 119889120578 119866 (120585) = int
120585
0
119892 (120578) 119889120578 (33)
where 120572 ge 0
120573 ge max (119901
2
120572) (34)
Moreover assume that Φ(0) ge 0 with
Φ (119905) = 2int
120597Ω
119866 (119906) 119889119878 minus int
Ω
|nabla119906|2
(1 +
2
119901 + 2
|nabla119906|119901
)119889119909
minus 2int
Ω
119865 (119906) 119889119909
(35)
Then the solution 119906(119909 119905) of problem (1)-(3) blows up at somefinite time 119905lowast lt 119879 with
119879 =
Ψ (0)
2120573 (1 + 120573)Φ (0)
120573 gt 0 (36)
where Ψ(119905) is defined in (13) If 120573 = 0 we have 119879 = infin
Proof We compute
Ψ1015840
(119905) = 2int
Ω
119906119906119905119889119909 = 2int
Ω
119906 [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 2int
120597Ω
119906 (|nabla119906|119901
+ 1)
120597119906
120597]119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
= 2int
120597Ω
119906119892 (119906) 119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
(37)
Making use of the hypotheses stated inTheorem 2 we have
Ψ1015840
(119905) ge 2 (1 + 120573)Φ (119905) (38)
Differentiating (35) we derive
Φ1015840
(119905) = 2int
120597Ω
119892 (119906) 119906119905119889119878 minus int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
minus 2int
Ω
119891 (119906) 119906119905119889119909
(39)
Integrating the identity nabla sdot (119906119905(|nabla119906|119901
+1)nabla119906) = 119906119905nabla sdot ((|nabla119906|
119901
+
1)nabla119906) + (12)(|nabla119906|119901
+ 1)(|nabla119906|2
)119905overΩ we get
int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
= 2int
Ω
nabla sdot (119906119905(|nabla119906|119901
+ 1) nabla119906) 119889119909
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1) nabla119906 sdot ]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
(40)
Substituting (40) into (39) we have
Φ1015840
(119905) = 2int
Ω
1199062
119905119889119909 gt 0 (41)
whichwithΦ(0) gt 0 implyΦ(119905) gt 0 for all 119905 isin (0 119905lowast
) Makinguse of the Schwarz inequality we obtain
2 (1 + 120573)Ψ1015840
Φ le (Ψ1015840
(119905))
2
= 4(int
Ω
119906119906119905119889119909)
2
le 2Ψ (119905)Φ1015840
(119905)
(42)
Abstract and Applied Analysis 5
Multiplying the above inequality by Ψminus2minus120573 we deduce
(ΦΨminus(1+120573)
)
1015840
ge 0 (43)
Arguing as in Theorem 31 in [20] we find
119905lowast
le 119879 =
1
2120573 (1 + 120573)
(Ψ (0))minus120573
=
Ψ (0)
2120573 (1 + 120573)Φ (0)
(44)
valid for 120573 gt 0 If 120573 = 0 we have
Ψ (119905) ge Ψ (0) 1198902119872119905 (45)
valid for 119905 gt 0 implying that 119905lowast = infin This completes theproof of Theorem 2
4 Lower Bounds for 119905lowast
In this section under the assumption that Ω is a star shapeddomain in 119877
3 convex in two orthogonal directions we seek alower bound for the blow-up time 119905lowast Now we state the resultas follows
Theorem 3 Let 119906(119909 119905) be the nonnegative solution of problem(1)-(3) and 119906(119909 119905) blows up at 119905lowast moreover the nonnegativefunctions 119891 and 119892 satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(46)
with 1198961gt 0 119896
2gt 0 119902 gt 1 119904 gt 1 119902 lt 119904 Define
120593 (119905) = int
Ω
119906119899(119904minus1)
119889119909 (47)
where 119899 is a parameter restricted by the condition
119899 gt max 4 2
119904 minus 1
(48)
Then 120593(119905) satisfies inequality
1205931015840
(119905) le Γ (120593) (49)
for some computable function Γ(120593) It follows that 119905lowast is boundedfrom below We have
119905lowast
ge int
infin
120593(0)
119889120578
Γ (120578)
119889120578 (50)
Proof Differentiating (47) and making use of the boundarycondition (2) together with the conditions (46) we have
1205931015840
(119905) = 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119906119905119889119909
= 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
times [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 119899 (119904 minus 1) int
120597Ω
119906119899(119904minus1)minus1
(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119891 (119906) 119889119909
le 1198962119899 (119904 minus 1) int
120597Ω
119906(119899+1)(119904minus1)
119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(51)
Applying inequality (27) in [20] to the first term on the righthand side of (51) we have
int
120597Ω
119906(119899+1)(119904minus1)
119889119878 le
3
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
(119899 + 1) (119904 minus 1) 119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
(52)
6 Abstract and Applied Analysis
Substituting (52) into (51) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(53)
Making use of arithmetic-geometric mean inequality wederive
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909 le
120583
2
int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
+
1
2120583
int
Ω
119906(119899+2)(119904minus1)
119889119909
(54)
for all 120583 gt 0 Choose 120583 gt 0 such that
1198962119899 (119899 + 1) (119904 minus 1)
2
119889120583
21205880
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] = 0 (55)
We rewrite (53) as
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(56)
Using Holder inequality we get
int
Ω
119906119899(119904minus1)
119889119909 le (int
Ω
119906119899(119904minus1)+119902minus1
119889119909)
119899(119904minus1)(119899(119904minus1)+119902minus1)
times |Ω|(119902minus1)(119899(119904minus1)+119902minus1)
(57)
Combining (56) with (57) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
=
31198962119899 (119904 minus 1)
1205880
1198691(119905)
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
1198692(119905)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(58)
where
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
1198692(119905) = int
Ω
119906(119899+2)(119904minus1)
119889119909
120596 (119905) = int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
(59)
Using Sobolev type inequality (A5) derived by Payne et al[21] we obtain
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
le
3
1205880
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 +
(119899 + 1) (119904 minus 1)
3
times(1 +
119889
1205880
)int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
32
(60)
We now make use of Holder inequality to bound the secondintegral on the right hand side of (60) as follows
int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
le (int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909)
(119901+1)(119901+2)
1205961(119901+2)
(61)
with
1205751=
(119899 minus 2) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(62)
Abstract and Applied Analysis 7
Wenote that 1205751lt 1 for 119899 gt (3119901minus2(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Using again Holderrsquosinequality we obtain
int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909
le 1205932(119899+1)(1minus120575
1)3119899
|Ω|1minus(2(119899+1)(1minus120575
1)3119899)
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 le 1205932(119899+1)3119899
|Ω|1minus(2(119899+1)3119899)
(63)
where |Ω| = intΩ
119889119909 is the volume of Ω Substituting (61) and(63) in (60) we obtain the following inequality
1198691(119905) le 119888
11205932(119899+1)3119899
+ 1198882120593(2(119899+1)(1minus120575
1)3119899)((119901+1)(119901+2))
times 1205961(119901+2)
32
le 1198881120593(119899+1)119899
+ 1198882120593((119899+1)(1minus120575
1)119899)((119901+1)(119901+2))
12059632(119901+2)
(64)
where 1198881 1198882are computable positive constants Note that the
last inequality in (64) follows from Holder inequality underthe particular form (119886 + 119887)
32
le radic2(11988632
+ 11988732
) Similarly wecan bound 119869
2and get
1198692(119905) le 119888
3120593(119899+2)119899
+ 1198884120593((119899+2)(1minus120575
2)119899)((119901+1)(119901+2))
12059632(119901+2)
(65)
where 1198883 1198884are computable positive constants
1205752=
(119899 minus 4) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(66)
Wenote that 1205752lt 1 for 119899 gt (3119901minus4(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Inserting (64) and (65)in (58) we arrive at
1205931015840
(119905) le1198891120593(119899+1)119899
+1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
+ 1198893120593(119899+2)119899
+1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(67)
where 120582 = (119901 + 1)(119901 + 2) 1198893and
119889119895(119895 = 1 2 4) are
computable positive constants Next we want to eliminatethe quantity 120596(119905) in inequality (67) By using the followinginequality
120593120572
120596120573
= (120574120596)120573
120593120572(1minus120573)
120574120573(1minus120573)
1minus120573
le 120574120573120596 + (1 minus 120573) 120574120573(120573minus1)
+ (1 minus 120573) 120574120573(120573minus1)
120593120572(1minus120573)
(68)
valid for 0 lt 120573 lt 1 where 120574 is an arbitrary positive constantthen we have
1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
le 1205741120596 (119905) + 119889
2120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
le 1205742120596 (119905) + 119889
4120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(69)
with arbitrary positive constants 1205741 1205742and computable
positive constants 1198892 1198894 Substitute (69) in (67) and choose
the arbitrary (positive) constants 1205741 1205742such that 120574
1+1205742minus119899(119904minus
1)[119899(119904 minus 1) minus 1] = 0 We obtain
1205931015840
(119905) le1198891120593(119899+1)119899
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(70)
We eliminate the last term in (70) by using the followinginequality
120593(119899+1)119899
= 119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(2119899minus1)(119904minus1)((2119899minus1)(119904minus1)+119904minus119902)
times 119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(119904minus119902)((2119899minus1)(119904minus1)+119904minus119902)
le
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
+
119904 minus 119902
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(71)
valid for 119902 lt 119904 and arbitrary119898 gt 0 and choose119898 such that
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
1198891119898 minus 119896
1119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
= 0
(72)
Then (70) can be rewritten as
1205931015840
(119905) le 11988911205933
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(73)
Integrating (73) over [0 119905] we conclude
119905lowast
ge int
infin
120593(0)
119889120578
times (11988911205783
+ 1198892120578(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120578(119899+2)119899
+ 1198894120578(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
)
minus1
(74)
This completes the proof of Theorem 3
8 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Authorsrsquo Contribution
All authors contributed equally to the paper and read andapproved the final paper
Acknowledgments
This work is supported by the Natural Science Foundationof Shandong Province of China (ZR2012AM018) and theFundamental Research Funds for the Central Universities(no 201362032) The authors would like to deeply thank allthe reviewers for their insightful and constructive comments
References
[1] J Bebernes and D Eberly Mathematical Problems from Com-bustion Theory vol 83 of Applied Mathematical SciencesSpringer New York NY USA 1989
[2] C V Pao Nonlinear Parabolic and Elliptic Equations PlenumPress New York NY USA 1992
[3] J L Vazquez The Porous Medium Equations MathematicalTheory Oxford University Press Oxford UK 2007
[4] J Filo ldquoDiffusivity versus absorption through the boundaryrdquoJournal of Differential Equations vol 99 no 2 pp 281ndash305 1992
[5] H A Levine and L E Payne ldquoNonexistence theorems forthe heat equation with nonlinear boundary conditions and forthe porous medium equation backward in timerdquo Journal ofDifferential Equations vol 16 pp 319ndash334 1974
[6] B Straughan Explosive Instabilities in Mechanics SpringerBerlin Germany 1998
[7] A A Samarskii V A Galaktionov S P Kurdyumov and A PMikhailov Blow-Up in Quasilinear Parabolic Equations vol 19of de Gruyter Expositions in Mathematics Walter de GruyterBerlin Germany 1995
[8] PQuittner andP Souplet Superlinear Parabolic Problems Blow-Up Global Existence and Steady States Birkhauser AdvancedTexts Birkhauser Basel Switzerland 2007
[9] C Bandle and H Brunner ldquoBlowup in diffusion equations asurveyrdquo Journal of Computational andAppliedMathematics vol97 no 1-2 pp 3ndash22 1998
[10] V A Galaktionov and J L Vazquez ldquoThe problem of blow-up in nonlinear parabolic equationsrdquo Discrete and ContinuousDynamical Systems vol 8 no 2 pp 399ndash433 2002
[11] H A Levine ldquoThe role of critical exponents in blowup theo-remsrdquo SIAM Review vol 32 no 2 pp 262ndash288 1990
[12] H A Levine ldquoNonexistence of global weak solutions to someproperly and improperly posed problems of mathematicalphysics the method of unbounded Fourier coefficientsrdquoMath-ematische Annalen vol 214 pp 205ndash220 1975
[13] L E Payne G A Philippin and P W Schaefer ldquoBlow-upphenomena for some nonlinear parabolic problemsrdquoNonlinearAnalysis Theory Methods amp Applications vol 69 no 10 pp3495ndash3502 2008
[14] MMarras and S Vernier Piro ldquoOn global existence and boundsfor blow-up time in nonlinear parabolic problems with time
dependent coefficientsrdquo Discrete and Continuous DynamicalSystems vol 2013 pp 535ndash544 2013
[15] Y Li Y Liu and C Lin ldquoBlow-up phenomena for some non-linear parabolic problems under mixed boundary conditionsrdquoNonlinear Analysis Real World Applications vol 11 no 5 pp3815ndash3823 2010
[16] Y Li Y Liu and S Xiao ldquoBlow-up phenomena for some non-linear parabolic problems under Robin boundary conditionsrdquoMathematical and Computer Modelling vol 54 no 11-12 pp3065ndash3069 2011
[17] C Enache ldquoBlow-up phenomena for a class of quasilinearparabolic problems under Robin boundary conditionrdquo AppliedMathematics Letters vol 24 no 3 pp 288ndash292 2011
[18] J Ding ldquoGlobal and blow-up solutions for nonlinear parabolicequations with Robin boundary conditionsrdquo Computers ampMathematics with Applications vol 65 no 11 pp 1808ndash18222013
[19] Y Liu ldquoBlow-up phenomena for the nonlinear nonlocal porousmedium equation under Robin boundary conditionrdquo Comput-ers amp Mathematics with Applications vol 66 no 10 pp 2092ndash2095 2013
[20] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary conditon Irdquo Zeitschrift fur Angewandte Mathematikund Physik vol 61 no 6 pp 999ndash1007 2010
[21] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary condition IIrdquoNonlinear Analysis Theory Methods ampApplications vol 73 no 4 pp 971ndash978 2010
[22] Y Liu ldquoLower bounds for the blow-up time in a non-local reac-tion diffusion problem under nonlinear boundary conditionsrdquoMathematical and ComputerModelling vol 57 no 3-4 pp 926ndash931 2013
[23] Y Liu S Luo and Y Ye ldquoBlow-up phenomena for a parabolicproblem with a gradient nonlinearity under nonlinear bound-ary conditionsrdquo Computers amp Mathematics with Applicationsvol 65 no 8 pp 1194ndash1199 2013
[24] Z B Fang R Yang and Y Chai ldquoLower bounds estimate for theblow-up time of a slow diffusion equation with nonlocal sourceand inner absorptionrdquo Mathematical Problems in Engineeringvol 2014 Article ID 764248 6 pages 2014
Research ArticleSome Properties on Complex Functional Difference Equations
Zhi-Bo Huang12 and Ran-Ran Zhang3
1 School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Physics and Mathematics University of Eastern Finland PO Box 111 80101 Joensuu Finland3Department of Mathematics Guangdong University of Education Guangzhou 510303 China
Correspondence should be addressed to Zhi-Bo Huang huangzhiboscnueducn
Received 15 January 2014 Accepted 13 March 2014 Published 24 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z-B Huang and R-R ZhangThis is an open access article distributed under theCreativeCommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited
We obtain some results on the transcendental meromorphic solutions of complex functional difference equations of the formsum120582isin119868120572120582(119911)(prod
119899
119895=0119891(119911 + 119888
119895)120582119895) = 119877(119911 119891 ∘119901) = ((119886
0(119911) +119886
1(119911)(119891 ∘119901)+ sdot sdot sdot + 119886
119904(119911) (119891 ∘119901)
119904
)(1198870(119911) + 119887
1(119911) (119891 ∘119901)+ sdot sdot sdot + 119887
119905(119911) (119891 ∘119901)
119905
))where 119868 is a finite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119899) are distinct complex constants 119901(119911) is
a polynomial and 120572120582(119911) (120582 isin 119868) 119886
119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative to 119891(119911)
We further investigate the above functional difference equation which has special type if its solution has Borel exceptional zero andpole
1 Introduction and Main Results
In this paper a meromorphic function means meromorphicin the whole complex plane C For a meromorphic function119910(119911) let 120590(119910) be the order of growth and 120583(119910) the lowerorder of119910(119911) Further let 120582(119910) (resp 120582(1119910)) be the exponentof convergence of the zeros (resp poles) of 119910(119911) We alsoassume that the reader is familiar with the fundamentalresults and the standard notations of Nevanlinna theory ofmeromorphic functions (see eg [1]) Given a meromorphicfunction 119910(119911) we call a meromorphic function 119886(119911) a smallfunction relative to 119910(119911) if 119879(119903 119886(119911)) = 119878(119903 119910) = 119900(119879(119903 119910))as 119903 rarr infin possibly outside of an exceptional set of finitelogarithmic measure Moreover if 119877(119911 119910) is rational in 119910(119911)with small functions relative to 119910(119911) as its coefficients we usethe notation 119889 = deg
119910119877(119911 119910) for the degree of 119877(119911 119910) with
respect to119910(119911) Inwhat follows we always assume that119877(119911 119910)is irreducible in 119910(119911)
Meromorphic solutions of complex difference equationshave recently gained increasing interest due to the problemof integrability of difference equations This is related to theactivity concerning Painleve differential equations and theirdiscrete counterparts in the last decades Ablowitz et al [2]considered discrete equations to be delay equations in thecomplex plane This allowed them to analyze these equations
with the methods from complex analysis In regard to relatedpapers concerning a more general class of complex differenceequations we may refer to [3ndash5] These papers mainly dealtwith equations of the form
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119877 (119911 119891) (1)
where 119869 is a collection of all nonempty subsets of1 2 119899 119888
119895(119895 isin 119869) are distinct complex constants 119891(119911) is
a transcendental meromorphic function 120572119869(119911) (119869 isin 119869) are
small functions relative to119891(119911) and119877(119911 119891) is a rational func-tion in 119891(119911) with small meromorphic coefficients Moreoverif the right-hand side of (1) is essentially like the compositefunction 119890 ∘119891 of 119891(119911) and a rational function 119890(119911) Laine et alreversed the order of composition that is they considered thecomposite function 119891 ∘ 119890 of 119891(119911) and a rational function 119890(119911)which resulted in a complex functional difference equationThe following theorem [5 Theorem 28] gives an example
Theorem A (see [5 Theorem 28]) Suppose that 119891(119911) is atranscendental meromorphic solution of equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (2)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 283895 10 pageshttpdxdoiorg1011552014283895
2 Abstract and Applied Analysis
where 119901(119911) is a polynomial of degree 119896 ge 2 Moreover oneassumes that the coefficients 120572
119869(119911) are small functions relative
to 119891(119911) and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (3)
where 120572 = (log 119899)(log 119896)At this point we briefly introduce some notations used in
this paper A difference monomial of a meromorphic function119891(119911) is defined as
119891(119911)1205820119891(119911 + 119888
1)1205821
sdot sdot sdot 119891(119911 + 119888119899)120582119904
=
119904
prod
119895=0
119891(119911 + 119888119895)
120582119895
(4)
where 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119904) are distinct
constants and 120582119895(119895 = 0 1 119904) are natural numbers A
difference polynomial 119867(119911 119891(119911)) of a meromorphic function119891(119911) a finite sum of difference monomials is defined as
119867(119911 119891 (119911)) = sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
) (5)
where 119868 is a finite set of multi-indexes 120582 = (1205820 1205821 120582
119899)
120572120582(119911) (120582 isin 119868) are small functions relative to 119891(119911) The degree
and the weight of the difference polynomial (5) respectively aredefined as
deg119891(119867) = max
120582isin119868
119899
sum
119895=0
120582119895
120581119891(119867) = max
120582isin119868
119899
sum
119895=1
120582119895
(6)
Consequently 120581119891(119867) le deg
119891(119867) For instance the degree and
the weight of the difference polynomial 1198912(119911)119891(119911minus1)119891(119911+1)+119891(119911)119891(119911 + 1)119891(119911 + 2) + 119891
2
(119911 minus 1)119891(119911 + 2) respectively arefour and three Moreover a difference polynomial (5) is said tobe homogeneous with respect to 119891(119911) if the degree sum119899
119895=0120582119895of
each monomial in the sum of (5) is nonzero and the same forall 120582 isin 119868
In the following we proceed to prove generalizations ofTheorem A and investigate some new results for the first timeWe permit more general expressions on both sides of (1)
Theorem 1 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(7)
where119867(119911 119891(119911)) is defined as (5) 119901(119911) = 119889119896119911119896
+ sdot sdot sdot+1198891119911+119889
0
is a polynomial with constant coefficients 119889119896( = 0) 119889
1 1198890
and of the degree 119896 ge 2 and 119886119894(119911) (119894 = 0 1 119904) and
119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative
to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Set 119889 = max119904 119905 If 119896119889 le
(119899 + 1)deg119891(119867) then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (8)
where 120572 = (log(119899 + 1) + log deg119891(119867) minus log119889)(log 119896)
Similar to the proof of Theorem 1 we easily obtain thefollowing result which is a generation of Theorem A
Theorem 2 Let 119888119894isin C (119894 = 1 2 119899) be distinct
constants and 119891(119911) be a transcendental meromorphic solutionof equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895))
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(9)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119896119889 =
119896max119904 119905 le 119899 then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (10)
where 120572 = (log 119899 minus log 119889)(log 119896)
We then proceed to consider the distribution of zeros andpoles of meromorphic solutions of (7) The following resultindicates that solutions having Borel exceptional zeros andpoles appear only in special situations
Theorem 3 Let 1198880= 0 let 119888
119894isin C 0 (119894 = 1 2 119899) be
distinct constants and let 119891(119911) be a finite order transcendentalmeromorphic solution of equation
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(11)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 1
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 If
max120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (12)
then (11) is either of the form119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(13)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (14)
where 120572 isin C 0 is some constant
Abstract and Applied Analysis 3
Example 4 119891(119911) = cos 119911 solves difference equation
4119891(119911)2
119891(119911 + 120587)2
= 119891(2119911)2
+ 2119891 (2119911) + 1 (15)
Here 119901(119911) = 2119911 Clearly 120582(1119891) = 0 lt 1 = 120582(119891) = 120590(119891) Thisexample shows that condition (12) is necessary and cannot bereplaced by
min120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (16)
Moreover we obtain a result parallel toTheorem 54 in [6]for the difference case
Theorem 5 Suppose that the equation119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
=
119888 (119911)
119891(119911)119898 119898 isin N (17)
has a meromorphic solution of finite order where 1198880= 0 119888
119895isin
C 0 (119895 = 1 2 119899) are distinct constants and 119888(119911) is anontrivialmeromorphic function If119891(119911)has only finitelymanypoles then119891(119911) = 119863(119911)119890119864(119911) where119863(119911) is a rational functionand119864(119911) is a polynomial if and only if 119888(119911) = 119866(119911)119890119872(119911) where119866(119911) is a rational function and119872(119911) is a polynomial
Example 6 Difference equation
119891(119911)2
119891 (119911 + 1) 119891 (119911 minus 1) = (
1
1199114(1199112minus 1)
1198906119911
) sdot
1
119891(119911)2
(18)
of the type (17) is solved by 119891(119911) = 119890119911119911 Here 119891(119911) = 119890119911119911and 119888(119911) = 11989061199111199114(1199112 minus 1) satisfy Theorem 5
As an application of Theorem 3 we obtain the following
Theorem 7 Let 119888 isin C 0 and let 119891(119911) be a finite ordertranscendental meromorphic solution of equation
119891 (119911 + 119888) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(19)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 Then 119891(119911) has at most one Borel exceptional value
If the degree 119896 of polynomial 119901(119911) is 1 in Theorem 7 theresult does not hold For example we have the following
Example 8 119891(119911) = tan 119911 solves difference equation
119891 (119911 + 1) =
119891 (119911) + tan 11 minus (tan 1) 119891 (119911)
(20)
of the type (19) Obviously 119891(119911) has two Borel exceptionalvalues plusmn119894
If we remove the assumption max120582(119891) 120582(1119891) lt 120590(119891)used in Theorem 3 we obtain a result similar to Theorem 12in [4]
Theorem 9 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891)
=
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905
(21)
where 119867(119911 119891(119911)) is defined as (5) and 119886119894(119911) (119894 = 0 1 119904)
and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions
relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119889 = max119904 119905 gt
(119899 + 1) deg119891(119867) then 120590(119891) = infin
In fact the following examples show that the assertion ofTheorem 9 does not remain valid identically if 119889 le (119899 +1)deg
119891(119867)
Example 10 119891(119911) = exp119890119911119911 solves the difference equation
(119911 minus 120587119894) (119911 + log 2 minus 120587119894) 119891 (119911) 119891 (119911 minus 120587119894) 119891 (119911 + log 2 minus 120587119894)
+ (119911 + log 8) 119891 (119911 + log 8) =1 + 119911
11
119891(119911)10
1199113119891(119911)
2
(22)
Clearly 119889 = 10 lt (3+1) sdot 3 = (119899+1) deg119891(119867) and 120590(119891) = infin
Example 11 119891(119911) = tan 119911 satisfies the difference equation
119891(119911 +
120587
4
) + 119891(119911 minus
120587
4
) =
4119891 (119911)
1 minus 119891(119911)2
(23)
Obviously 119889 = 2 lt (2+1)times1 = (119899+1) deg119891(119867) and120590(119891) = 1
Example 12 (see [7 pages 103ndash106] and [8 page 8]) Thefollowing difference equation
119891 (119911 + 1) = 120572119891 (119911) (1 minus 119891 (119911)) 120572 = 0 (24)
derives from a well-known discrete logistic model in biologyIt has been proved that all other meromorphic solutions areof infinite order apart from the constant solutions 119891(119911) equiv 0and 119891(119911) = (120572 minus 1)120572 For instance (24) has one-parameterfamilies of entire solutions of infinite order
119891 (119911) =
1
2
(1 minus exp (119860119890119911 log 2)) 119860 isin C 0 120572 = 2
119891 (119911) = sin2 (119861119890119911 log 2) 119861 isin C 0 120572 = 4(25)
Here 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867)
Example 13 119891(119911) = 119911 solves the difference equation
119891 (119911 + 1) =
1 minus 119891(119911)2
minus1199112minus 119911 + 1 + 119891(119911)
2 (26)
We get 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867) and 120590(119891) = 0
If the difference polynomial in the left-hand side of (21)is homogeneous we further obtain the following theorem
4 Abstract and Applied Analysis
Theorem 14 Let 119891(119911) be a transcendental meromorphic solu-tion of (21) where 119867(z 119891(119911)) is defined as (5) and 119886
119894(119911) (119894 =
0 1 119904) and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Suppose
that 119867(119911 119891) is homogeneous and has at least one differencemonomial of type
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
(120582119895isin N
+ 119895 = 0 1 119899) (27)
If 119889 = max119904 119905 gt 3deg119891(119867) then 120590(119891) = infin
2 Proof of Theorem 1
We need some preliminaries to proveTheorem 1
Lemma 15 (see [9 Lemma 4]) Let 119891(119911) be a transcendentalmeromorphic function and let 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot + 1198891119911 +
1198890(119889119896= 0)be a polynomial of degree 119896 Given 0 lt 120575 lt |119889
119896|
denote ] = |119889119896| + 120575 and 120583 = |119889
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C cup infin one has for all 119903 ge 1199030gt 0
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 ∘ 119901) le 119896119899 (]119903119896 119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 ∘ 119901) le 119873(]119903119896 119886 119891)
+ 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 ∘ 119901) le (1 + 120576) 119879 (]119903119896 119891) (28)
Lemma16 (see [10Theorem B16]) Given distinctmeromor-phic functions 119891
1 119891
119899 let 119869 denote the collection of all
nonempty subsets of 1 2 119899 and suppose that 120572119869isin C for
each 119869 isin 119869 Then
119879(119903sum
119869
120572119869(prod
119895isin119869
119891119895)) le
119899
sum
119896=1
119879 (119903 119891119896) + 119874 (1) (29)
By denoting 119891119894+1= 119891(119911 + 119888
119894)120582119894(119894 = 0 1 119899) below it is
an easy exercise to prove the following result from Lemma 16
Lemma 17 Let 119891(119911) be a meromorphic function let 119868 be afinite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) and let 120572
120582(119911)
be small functions relative to 119891(119911) for all 120582 isin 119868 Then thecharacteristic function of the difference polynomial (5) satisfies
119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
))
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(30)
where 119862 = max|1198881| |119888
2| |119888
119899|
Lemma18 (see [11 Lemma5]) Let119892(119903) andℎ(119903) bemonotonenondecreasing functions on [0infin) such that 119892(119903) le ℎ(119903) for all119903 notin 119864 cup [0 1] where 119864 sub (1infin) is a set of finite logarithmicmeasure Let 120572 gt 1 be a given constant Then there exists an1199030= 119903
0(120572) gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 ge 119903
0
Lemma 19 (see [12 Lemma 3]) Let 120595(119903) be a function of119903 (119903 ge 119903
0) positive and bounded in every finite interval
(i) Suppose that 120595(120583119903119898) le 119860120595(119903) + 119861 (119903 ge 1199030) where
120583 (120583 gt 0)119898 (119898 gt 1)119860 (119860 ge 1) and 119861 are constantsThen 120595(119903) = 119874((log 119903)120572) with 120572 = (log119860)(log119898)unless 119860 = 1 and 119861 gt 0 and if 119860 = 1 and 119861 gt 0 thenfor any 120576 gt 0 120595(119903) = 119874((log 119903)120576)
(ii) Suppose that (with the notation of (i)) 120595(120583119903119898) ge119860120595(119903) (119903 ge 119903
0) Then for all sufficiently large values of
119903 120595(119903) ge 119870(log 119903)120572 with 120572 = (log119860)(log119898) for somepositive constant 119870
Proof of Theorem 1 For any 120576 (0 lt 120576 lt 1) we may applyValiron-Mohonrsquoko lemma Lemmas 15 and 17 and (5) and (7)to conclude that119889 (1 minus 120576) 119879 (120583119903
119896
119891)
le 119889119879 (119903 119891 ∘ 119901) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905)
+ 119878 (119903 119891)
= 119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
)) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(31)holds for all sufficiently large 119903 possibly outside of anexceptional set of finite logarithmic measure where 119862 =max|119888
1| |119888
2| |119888
119899| and 120583 is defined as Lemma 15 Now we
may apply Lemma 18 to deal with the exceptional set andconclude that for every 120578 gt 1 there exists an 119903
0gt 0 such
that119889 (1 minus 120576) 119879 (120583119903
119896
119891) le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (120578119903 119891)
(32)holds for all 119903 ge 119903
0 Denote 120596 = 120578119903 Then (32) can be written
in the form
119879(
120583
120578119896
120596119896
119891) le
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
119879 (120596 119891) (33)
Since 119889119896 le (119899 + 1)deg119891(119867) we get ((119899 + 1)deg
119891(119867)(1 +
120576))(119889(1 minus 120576)) gt 1 for all 0 lt 120576 lt 1 Thus we now applyLemma 19(i) to conclude that
119879 (119903 119891) = 119874 ((log 119903)120572+120576)
120572 =
log ((119899 + 1) deg119891(119867) (1 + 120576) 119889 (1 minus 120576))
log 119896
=
log (119899 + 1) + log deg119891(119867) minus log119889
log 119896+ 119900 (1)
(34)
The proof of Theorem 1 is completed
Abstract and Applied Analysis 5
3 Proof of Theorems 3 and 5
We again need some preliminaries
Lemma 20 (see [13 Theorem 15]) Suppose that 119891119895(119911) (119895 =
1 2 119899) (119899 ge 2) are meromorphic functions and 119892119895(119911) (119895 =
1 2 119899) are entire functions satisfying the following condi-tions
(1) sum119899119895=1119891119895(119911)119890
119892119895(119911)
= 0
(2) 119892119895(119911) minus 119892
119896(119911) are not constants for 1 le 119895 lt 119896 le 119899
(3) For 1 le 119895 le 119899 1 le ℎ lt 119896 le 119899
119879 (119903 119891119895) = 119900 119879 (119903 119890
119892ℎminus119892119896) (119903 997888rarr +infin 119903 notin 119864) (35)
where 119864 sub (1 +infin) is of finite linear measure or finitelogarithmic measure
Then 119891119895(119911) equiv 0 (119895 = 1 2 119899)
Lemma21 (see [14Theorem4]) Let119865(119911)119875119899(119911) 119875
0(119911) be
polynomials such that 1198651198751198991198750equiv 0 and then every finite order
transcendental meromorphic solution 119891(119911) of equation
119875119899(119911) 119891 (119911 + 119899) + sdot sdot sdot + 119875
1(119911) 119891 (119911 + 1) + 119875
0(119911) 119891 (119911) = 119865 (119911)
(36)
satisfies 120582(119891) = 120590(119891) ge 1
Remark 22 Replacing 119895 by 119888119895(119895 = 1 2 119899) where 119888
119895(119895 =
1 2 119899) are distinct nonzero complex constants Lemma 21remains valid
Proof of Theorem 3 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed withnonzero poles of 119891(119911) Since max120582(119891) 120582(1119891) lt 120590(119891) thenℎ(119911) = 119911
120591
119902(119911) is an entire function such that
120590 (ℎ) = 120582(
1
119891
) lt 120590 (119891) lt +infin (37)
and 119892(119911) = ℎ(119911)119891(119911) is a transcendental entire function with
119879 (119903 119892) = 119879 (119903 119891) + 119878 (119903 119891) 120590 (119892) = 120590 (119891)
120582 (119892) = 120582 (119891)
(38)
If 119902(119911) is a polynomial we obtain quickly that 120590(ℎ ∘ 119901) =0 lt 120590(119892 ∘ 119901) Otherwise we conclude from the last assertionof Lemma 15 (37) and (38) that
120590 (ℎ ∘ 119901) = 119896120590 (ℎ) = 119896120582(
1
119891
) lt 119896120590 (119892) = 120590 (119892 ∘ 119901) (39)
Therefore
119879 (119903 ℎ ∘ 119901) = 119878 (119903 119892 ∘ 119901) (40)
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (11) we concludethat
(ℎ ∘ 119901)119904minus119905
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
(
119899
prod
119895=0
119892(119911 + 119888119895)
120582119895
)
= (1198860(119911) (ℎ ∘ 119901)
119904
+ 1198861(119911) (ℎ ∘ 119901)
119904minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119886119904(119911) (119892 ∘ 119901)
119904
)
times (1198870(119911) (ℎ ∘ 119901)
119905
+ 1198871(119911) (ℎ ∘ 119901)
119905minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119887119905(119911) (119892 ∘ 119901)
119905
)
minus1
(41)
Obviously it follows from (37)ndash(40) and Lemma 15 that
119879(119903
1
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 (ℎ ∘ 119901)119904minus119905
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 119886119906(119911) (ℎ ∘ 119901)
119904minus119906
) = 119878 (119903 119892 ∘ 119901) 119906 = 0 1 119904
119879 (119903 119887V (119911) (ℎ ∘ 119901)119905minusV) = 119878 (119903 119892 ∘ 119901) V = 0 1 119905
(42)
Denoting119860(119911) = (ℎ ∘ 119901)119904minus119905prod119899119895=0ℎ(119911 + 119888
119895)120582119895 we get from (42)
that119879 (119903 119860) = 119878 (119903 119892 ∘ 119901) (43)
Since zeros and poles are Borel exceptional values of 119891(119911) by(12) we may apply a result due to Whittaker see [15 Satz134] to deduce that 119891(119911) is of regular growth Thus we useLemma 15 and (12) again to get
119879(119903
1198911015840
119891
) = 119873(119903 119891) + 119873(119903
1
119891
) + 119878 (119903 119891) = 119878 (119903 119892 ∘ 119901)
(44)
Similarly if we set 119861(119911) = 119860(119911)(prod119899
119895=0119892(119911 + 119888
119895)120582119895) we
also deduce from the lemma of the logarithmic derivativeLemma 15 (12) (38) and (43) that
119879(119903
1198611015840
119861
) = 119879(119903
1198601015840
119860
+
119899
sum
119895=0
120582119895
1198921015840
(119911 + 119888119895)
119892 (119911 + 119888119895)
) = 119878 (119903 119892 ∘ 119901)
(45)Denoting 119865(119911) = 119892 ∘ 119901
119875 (119911 119865) =
1198860(119911)
119886119904(119911)
(ℎ ∘ 119901)119904
+
1198861(119911)
119886119904(119911)
(ℎ ∘ 119901)119904minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119904
119876 (119911 119865) =
1198870(119911)
119887119905(119911)
(ℎ ∘ 119901)119905
+
1198871(119911)
119887119905(119911)
(ℎ ∘ 119901)119905minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119905
(46)
6 Abstract and Applied Analysis
Therefore we deduce from Lemma 15 and (42) that thecoefficients of 119875(119911 119865) and119876(119911 119865) are small functions relativeto 119865(119911) Thus (41) can be written in the form
119887119905(119911)
119886119904(119911)
119861 (119911) =
119875 (119911 119865)
119876 (119911 119865)
= 119906 (119911 119865) (47)
Denoting
120595 (119911) =
1198651015840
(119911)
119865 (119911)
119880 (119911) =
1199061015840
(119911 119865)
119906 (119911 119865)
(48)
we get 119879(119903 119880) = 119878(119903 119892 ∘ 119901) from (45) and (47) Wealso conclude from the lemma of logarithmic derivativeLemma 15 and (12) that
119879 (119903 120595) = 119879(119903
1198651015840
119865
) = 119898(119903
1198651015840
119865
) + 119873(119903
1198651015840
119865
)
le 119873 (119903 119865) + 119873(119903
1
119865
) + 119878 (119903 119865)
= 119873 (119903 119892 ∘ 119901) + 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(]1199031198961
119892
) + 119878 (119903 119892 ∘ 119901) = 119878 (119903 119892 ∘ 119901)
(49)
where ] is defined as Lemma 15Since
1198751015840
119876 minus 1198751198761015840
1198762
= 1199061015840
= 119880119906 =
119880119875
119876
(50)
we conclude that
1198751015840
119876 minus 1198751198761015840
= 119880119875119876 (51)
Now writing 1198651015840 = 120595119865 in (51) regarding then (51) as analgebraic equation in119865with coefficients of growth 119878(119903 119865) andcomparing the leading coefficients we deduce that
(119904 minus 119905) 120595 = 119880 (52)
By integrating both sides of the last equality above weconclude that
119906 (119911 119865) = 120572119865(119911)119904minus119905
(53)
for some 120572 isin C 0 Therefore by combining therepresentations of 119865 119861 119860 119892 with (53) we conclude that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
(54)
If 119904119905 = 0 we deduce from (11) and (54) that
120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(55)
From this we get that 119877(119911 119891 ∘ 119901) is not irreducible in 119891 ∘ 119901a contradiction Thus 119905 = 0 or 119904 = 0 Therefore we deducefrom (54) that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(56)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (57)
The proof of Theorem 3 is completed
Proof of Theorem 5 Assume first that 119891(119911) = 119863(119911)119890119864(119911)
where 119863(119911) is a rational function and 119864(119911) is a polynomialOne can see from (17) that
119888 (119911) = 119891(119911)119898
119899
prod
119894=0
119891(119911 + 119888119894)120582119894
= [119863(119911)119898
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
] 119890119898119864(119911)+sum
119899
119894=0120582119894119864(119911+119888
119894)
= 119866 (119911) 119890119872(119911)
(58)
where 119866(119911) = 119863(119911)119898prod119899119894=0119863(119911 + 119888
119894)120582119894 is rational and119872(119911) =
119898119864(119911) + sum119899
119894=0120582119894119864(119911 + 119888
119894) is a polynomial
Suppose next that 119888(119911) = 119866(119911)119890119872(119911) where 119866(119911) is arational function and 119872(119911) is a polynomial Since 119891(119911) hasonly finitely many poles we conclude from (17) that
119873(119903
1
119891
) le 119873(119903
1
119891119898
) = 119873(119903
prod119899
119894=0119891(119911 + 119888
119894)120582119894
119888 (119911)
)
= 119874 (log 119903)
(59)
Thus 119891(119911) has only finitely many zeros and poles and119891(119911) = 119863(119911)119890
119864(119911) where119863(119911) is rational and 119864(119911) is an entirefunction In the followingwe only prove119864(119911) is a polynomialNow substituting119891(119911) = 119863(119911)119890119864(119911) and 119888(119911) = 119866(119911)119890119872(119911) into(17) we get
119899
prod
119894=0
119863(119911 + 119888119894)120582119894 exp (120582
119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(60)
(
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
) exp(119899
sum
119894=0
120582119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(61)
Thus we deduce from Lemma 20 that two exponents in (61)cancel each other to a constant 120591 isin C such that
119899
sum
119894=0
120582119894119864 (119911 + 119888
119894) = 119872 (119911) minus 119898119864 (119911) + 120591 (62)
Abstract and Applied Analysis 7
that is119899
sum
119894=1
120582119894119864 (119911 + 119888
119894) + (120582
0+ 119898)119864 (119911) = 119872 (119911) + 120591 (63)
Suppose that 119864(119911) is not a polynomial If 119864(119911) is a transcen-dental entire function of finite order we get from Lemma 21Remark 22 and (63) that 120590(119864) ge 1 Otherwise 119864(119911) is atranscendental entire function of infinite order These bothshow that 120590(119891) = infin contradicting the assumption that119891(119911) is finite order Thus 119864(119911) is a polynomial The proof ofTheorem 5 is completed
4 Proof of Theorem 7
Lemmas 23 and 25 reveal some properties of the maximalmodule of the polynomial in composite function 119891 ∘ 119901 with ameromorphic function 119891(119911) and a polynomial 119901(119911) whichare useful for proving the existence of Borel exceptionalvalue of finite order meromorphic solutions of functionaldifference equation of type (19)
Lemma 23 Let 119892(119911) be a nonconstant entire function of order120590(119892) = 120590 lt infin Suppose that 120572
119895(119911) (119895 = 1 2 119898) are small
meromorphic functions relative to 119892(119911) Then there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 120572119895)
119872 (119903 119892)
997888rarr 0 119895 = 1 2 119898 (64)
hold simultaneously for all 119903 isin 119864 as 119903 rarr infin where the lowerlogarithmic density of set 119864 is defined by
logdens (119864) = lim inf119903rarrinfin
int[1119903]cap119864
(119889119905119905)
log 119903 (65)
Remark 24 The proof of Lemma 23 is similar to the proof ofLemma 24 and Remark 25 in [16] Here we omit it
Lemma 25 Let 119891(119911) be a finite order transcendental mero-morphic function satisfying (12) and 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot+1198891119911+
1198890is a polynomial with constant coefficients 119889
119896( = 0) 119889
1 1198890
and of the degree 119896 ge 1 Suppose that
119867(119911) = 119886119899(119911) (119891 ∘ 119901)
119899
+ 119886119899minus1(119911) (119891 ∘ 119901)
119899minus1
+ sdot sdot sdot + 1198861(119911) (119891 ∘ 119901) + 119886
0(119911)
(66)
is a polynomial in 119891 ∘119901 where 119899 (ge 1) is a positive integer and119886119899(119911) ( equiv 0) 119886
119899minus1(119911) 119886
1(119911) 119886
0(119911) are small meromorphic
functions relative to 119891(119911) Then there exists a set 1198641of lower
logarithmic density 1 such that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (67)
for all 119903 isin 1198641as 119903 rarr infin where 0 lt 120583 lt |119889
119896| Hence119867(119911) equiv
0
Proof of Lemma 25 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed
with the nonzero poles of 119891(119911) Since 119891(119911) satisfies (12) thenℎ(119911) = 119911
120591
119902(119911) is an entire function Thus 119892(119911) = ℎ(119911)119891(119911) isentire and (37) (38) and (40) also hold
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (66) we con-clude that
119867(119911) = 119886119899(119911) sdot
(119892 ∘ 119901)119899
(ℎ ∘ 119901)119899+ 119886
119899minus1(119911) sdot
(119892 ∘ 119901)119899minus1
(ℎ ∘ 119901)119899minus1
+ sdot sdot sdot
+ 1198861(119911) sdot
(119892 ∘ 119901)
(ℎ ∘ 119901)
+ 1198860(119911)
=
119886119899(119911)
(ℎ ∘ 119901)119899(119892 ∘ 119901)
119899
times [1 +
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
(119892 ∘ 119901)minus1
+ sdot sdot sdot
+
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
(119892 ∘ 119901)1minus119899
+
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
(119892 ∘ 119901)minus119899
]
(68)
We note from Lemma 15 and (40) that
119879(119903
119886119899(119911)
(ℎ ∘ 119901)119899) = 119878 (119903 119892 ∘ 119901)
119879(119903
119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911)
) = 119878 (119903 119892 ∘ 119901)
for 119895 = 0 1 119899 minus 1
(69)
Therefore we deduce from Lemma 23 that there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 119886119899(119911) (ℎ ∘ 119901)
119899
)
119872 (119903 119892 ∘ 119901)
997888rarr 0
119872(119903 119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911))
119872 (119903 119892 ∘ 119901)
997888rarr 0
(119895 = 0 1 119899 minus 1)
(70)
Moreover according to the choosing of 119864 in the proofof Lemma 23 we know that 119886
119895(119911)(ℎ ∘ 119901)
119899minus119895
119886119899(119911) for
8 Abstract and Applied Analysis
119895 = 0 1 119899 minus 1 have no zeros and poles for all |119911| = 119903 isin 119864Thus we conclude from (68) and (70) that for any 120576 gt 0
119872(119903119867) ge 119872(119903 119892 ∘ 119901)119899minus120576
times [1 minus
100381610038161003816100381610038161003816100381610038161003816
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus1
minus sdot sdot sdot
minus
1003816100381610038161003816100381610038161003816100381610038161003816
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
1003816100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)1minus119899
minus
100381610038161003816100381610038161003816100381610038161003816
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus119899
]
ge (1 minus 120576)119872(119903 119892 ∘ 119901)119899minus120576
(71)
and so
log+119872(119903119867) ge (119899 minus 32
120576) log+119872(119903 119892 ∘ 119901) (72)
for all |119911| = 119903 isin 119864 and |119892 ∘ 119901(119911)| = 119872(119903 119892 ∘ 119901)Therefore we deduce from Lemma 15 and (38) that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (73)
for all |119911| = 119903 isin 1198641= 119864 cap (119903
0 +infin) where 119903
0gt 0 It is
obvious that 1198641has lower logarithmic density 1 The proof of
Lemma 25 is completed
Proof of Theorem 7 Suppose that 119891(119911) has two finite Borelexceptional values 119886 and 119887 ( = 0 119886) For the case where oneof 119886 and 119887 is infinite we can use a similar method to prove itSet
119892 (119911) =
119891 (119911) minus 119886
119891 (119911) minus 119887
(74)
Then 120590(119892) = 120590(119891) and
120582 (119892) = 120582 (119891 minus 119886) lt 120590 (119892) 120582 (
1
119892
) = 120582 (119891 minus 119887) lt 120590 (119892)
(75)
It follows from (74) that
119891 (119911) =
119886 minus 119887119892 (119911)
1 minus 119892 (119911)
(76)
Now substituting (76) into (19) we conclude that
119892 (119911 + 119888) = (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119886
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
times (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119887
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
minus1
= ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
))
times ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
))
minus1
(77)
Since 1198860(119911) + 119886
1(119911)(119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911)(119891 ∘ 119901)
119904 and 1198870(119911) +
1198871(119911)(119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911)(119891 ∘ 119901)
119905 are irreducible in 119891 ∘ 119901 weconclude that at least one of the following three inequalitiesholds that is
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(78)
Thus we deduce fromTheorem 3 that
119892 (119911 + 119888) = 119888 (119911) (119892 ∘ 119901)119897
(79)
where 119888(119911) is meromorphic function satisfying 119879(119903 119888) =119878(119903 119892) and 119897 isin Z Clearly 119897 = 0 and 119892(119911) is of regular growthfrom (75) see [15 Staz 134] Therefore 120590(119888) lt 120590(119892)
If 119897 ge 1 we conclude from (77) and (79) that
(minus1)119904+119905
119888 (119911)(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)(119892 ∘ 119901)119904+119905+119897
+ sdot sdot sdot + (minus
119904
sum
119894=0
119886119894(119911) 119886
119894
+ 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
) = 0
(80)
Abstract and Applied Analysis 9
Thus we deduce from Lemma 25 that (80) is a contradictionIf 119897 le minus1 we use the same method as above to getanother contradiction Therefore 119891(119911) has at most one Borelexceptional value The proof of Theorem 7 is completed
5 Proof of Theorems 9 and 14
We first recall two lemmas
Lemma 26 (see [17 Lemma 21]) Let 119891(119911) be a nonconstantmeromorphic function 119904 gt 0 120572 lt 1 and 119865 sub R+ the set of all119903 such that
119879 (119903 119891) le 120572119879 (119903 + 119904 119891) (81)
If the logarithmicmeasure of119865 is infinite that isint119865
(119889119905119905) = infinthen 119891(119911) is of infinite order of growth
Lemma 27 (see [18 Corollary 26] and [19 Corollary 22])Let 119891(119911) be a meromorphic function of finite order and let 119888 isinC Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (82)
for all 119903 outside of a possible exceptional set of finite logarithmicmeasure
Proof of Theorem 9 For any 120576 (0 lt 120576 lt (119889 minus (119899 +
1)deg119891(119867))(119889 + (119899 + 1) deg
119891(119867))) we may apply Valiron-
Mohonrsquoko lemma Lemma 17 (5) and (21) to conclude that
119889 (1 minus 120576) 119879 (119903 119891)
le 119889119879 (119903 119891) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905)
= 119879 (119903119867 (119911 119891 (119911))) le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891)
+ 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(83)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote
120572 =
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
(84)
Then 120572 lt 1 since 0 lt 120576 lt (119889 minus (119899 + 1)deg119891(119867))(119889 + (119899 +
1)deg119891(119867)) and 119889 gt (119899 + 1)deg
119891(119867) Thus
119879 (119903 119891) le 120572119879 (119903 + 119862 119891) (85)
holds for all 119903 in a set with infinite logarithmic measureTherefore we deduce from Lemma 26 and (85) that 120590(119891) =infin The proof of Theorem 9 is completed
Proof of Theorem 14 Assume contrary to the assertion that119891(119911) is meromorphic of finite order Taking into account theassumption that119867(119911 119891(119911)) is homogeneous we deduce fromLemma 27 that
119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) = 119878 (119903 119891) (86)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote 119862 = max1le119894le119899|119888119894| Since 119867(119911) is homogeneous
and has at least one difference monomial of typeprod119899119895=0119891(119911 +
119888119895)120582119895 we immediately conclude that by looking at pole
multiplicities summing over |119911| le 119903 and integratinglogarithmically
119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
)
le 120581119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
(87)
Therefore
119879 (119903119867 (119911 119891 (119911)))
= 119898 (119903119867 (119911 119891 (119911))) + 119873 (119903119867 (119911 119891 (119911)))
le 119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119898(119903 119891(119911)deg119891(119867)
)
+ 119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119873(119903 119891(119911)deg119891(119867)
)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
))
+ 119879 (119903 119891(119911)deg119891(119867)
) + 119878 (119903 119891)
le 3deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(88)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure The remainder can be proven by asimilar method in Theorem 9 The proof of Theorem 14 iscompleted
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful to the referees for their helpfulsuggestions to improve this paper The first author alsothanks Professor Ilpo laine and Professor Risto Korhonen for
10 Abstract and Applied Analysis
their valuable suggestion to the present paper Research issupported by National Natural Science Foundation of China(nos 11171119 and 11171121) and Guangdong National NaturalScience Foundation (no S2012040006865)
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[3] BGrammaticos T Tamizhmani A Ramani andKMTamizh-mani ldquoGrowth and integrability in discrete systemsrdquo Journal ofPhysics A vol 34 no 18 pp 3811ndash3821 2001
[4] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of Malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
[7] WGKelley andAC PetersonDifference Equations AcademicPress Boston Mass USA 1991
[8] I Laine and C C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[9] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[10] V I Gromak I Laine and S Shimomura Painleve DifferentialEquations in the Complex Plane vol 28 Walter de GruyterBerlin Germany 2002
[11] G G Gundersen ldquoFinite order solutions of second orderlinear differential equationsrdquo Transactions of the AmericanMathematical Society vol 305 no 1 pp 415ndash429 1988
[12] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[13] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions vol 557 Kluwer Academic Publishers Group Dor-drecht The Netherlands 2003
[14] Z X Chen and K H Shon ldquoOn growth of meromorphicsolutions for linear difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 619296 6 pages 2013
[15] G Jank and L Volkmann Einfuhrung in die Theorie derganzen und meromorphen Funktionen mit Anwendungen aufDifferentialgleichungen Birkhauser Basel Switzerland 1985
[16] J Wang ldquoGrowth and poles of meromorphic solutions of somedifference equationsrdquo Journal of Mathematical Analysis andApplications vol 379 no 1 pp 367ndash377 2011
[17] R G Halburd and R J Korhonen ldquoFinite-order meromorphicsolutions and the discrete Painleve equationsrdquoProceedings of theLondon Mathematical Society vol 94 no 2 pp 443ndash474 2007
[18] Y M Chiang and S J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[19] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
Research ArticleThe Regularity of Functions on Dual Split Quaternionsin Clifford Analysis
Ji Eun Kim and Kwang Ho Shon
Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Kwang Ho Shon khshonpusanackr
Received 28 January 2014 Accepted 2 April 2014 Published 17 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 J E Kim and K H Shon This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
This paper shows some properties of dual split quaternion numbers and expressions of power series in dual split quaternions andprovides differential operators in dual split quaternions and a dual split regular function on Ω sub C2 times C2 that has a dual splitCauchy-Riemann system in dual split quaternions
1 Introduction
Hamilton introduced quaternions extending complex num-bers to higher spatial dimensions in differential geometry (see[1]) A set of quaternions can be represented as
H = 119911 = 1199090+ 1199091119894 + 1199092119895 + 1199093119896 119909119898
isin R 119898 = 0 1 2 3
(1)
where 1198942
= 1198952
= 1198962
= minus1 119894119895119896 = minus1 and R denotes the set ofreal numbers Cockle [2] introduced a set of split quaternionsas
S = 119911 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 119909119898
isin R 119898 = 0 1 2 3
(2)
where 1198902
1= minus1 119890
2
2= 1198902
3= 1 and 119890
111989021198903
= 1 Aset of split quaternions is noncommutative and containszero divisors nilpotent elements and nontrivial idempotents(see [3 4]) Previous studies have examined the geometricand physical applications of split quaternions which arerequired in solving split quaternionic equations (see [5 6])Inoguchi [7] reformulated the Gauss-Codazzi equations informs consistent with the theory of integrable systems in theMinkowski 3-space for split quaternion numbers
A dual quaternion can be represented in a form reflect-ing an ordinary quaternion and a dual symbol Because
dual-quaternion algebra is constructed from real eight-dimensional vector spaces and an ordered pair of quater-nions dual quaternions are used in computer vision appli-cations Kenwright [8] provided the characteristics of dualquaternions and Pennestrı and Stefanelli [9] examined someproperties by using dual quaternions Son [10 11] offeredan extension problem for solutions of partial differentialequations and generalized solutions for the Riesz system Byusing properties of Hamilton operators Kula and Yayli [4]defined dual split quaternions and gave some properties ofthe screw motion in the Minkowski 3-space showing thatHhas a rotation with unit split quaternions in H and a scalarproduct that allows it to be identifiedwith the semi-Euclideanspace for split quaternion numbers
It was shown (see [12 13]) that any complex-valued har-monic function 119891
1in a pseudoconvex domain 119863 of C2 times C2
C being the set of complex numbers has a conjugate function1198912in 119863 such that the quaternion-valued function 119891
1+ 1198912119895 is
hyperholomorphic in119863 and gave a regeneration theorem in aquaternion analysis in view of complex and Clifford analysisIn addition we [14 15] provided a new expression of thequaternionic basis and a regular function on reduced quater-nions by associating hypercomplex numbers 119890
1and 119890
2 We
[16] investigated the existence of hyperconjugate harmonicfunctions of an octonion number system and we [17 18]obtained some regular functions with values in dual quater-nions and researched an extension problem for properties
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 369430 8 pageshttpdxdoiorg1011552014369430
2 Abstract and Applied Analysis
of regular functionswith values in dual quaternions and someapplications for such problems
This paper provides a regular function and some prop-erties of differential operators in dual split quaternions Inadditionwe research some equivalent conditions forCauchy-Riemann systems and expressions of power series in dual splitquaternions from the definition of dual split regular on anopen set Ω sub C2 times C2
2 Preliminaries
A dual number 119860 has the form 119886 + 120576119887 where 119886 and 119887 are realnumbers and 120576 is a dual symbol subject to the rules
120576 = 0 0120576 = 1205760 = 0 1120576 = 1205761 = 120576 1205762
= 0 (3)
and a split quaternion 119902 isin S is an expression of the form
119902 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 (4)
where 119909119898
isin R (119898 = 0 1 2 3) and 119890119903(119903 = 1 2 3) are split
quaternionic units satisfying noncommutativemultiplicationrules (for split quaternions see [1])
1198902
1= minus1 119890
2
2= 1198902
3= 1
11989011198902= minus11989021198901= 1198903 119890
21198903= minus11989031198902= minus1198901
11989031198901= minus11989011198903= 1198902
(5)
Similarly a dual split quaternion 119911 can be written as
D (S) = 119911 | 119911 = 1199010+ 1205761199011 119901119903isin S 119903 = 0 1 (6)
which has elements of the following form
119911 = (1199090+ 11990911198901) + (119909
2+ 11990931198901) 1198902
+ 120576 (1199100+ 11991011198901) + (119910
2+ 11991031198901) 1198902
= (1199110+ 11991111198902) + 120576 (119911
2+ 11991131198902)
= 1199010+ 1205761199011
(7)
where 1199010= 1199110+ 11991111198902and 119901
1= 1199112+ 11991131198902are split quaternion
components 1199110= 1199090+ 11990911198901 1199111= 1199092+ 11990931198901 1199112= 1199100+ 11991011198901
and 1199113= 1199102+ 11991031198901are usual complex numbers and 119909
119898 119910119898
isin
R (119898 = 0 1 2 3) The multiplication of split quaternionicunits with a dual symbol is commutative 120576119890
119903= 119890119903120576 (119903 =
1 2 3) However by properties of split quaternionic unit
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 0 1)
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 2 3)
119890119903119901119896
= 119901119896119890119903 119890119903119901119896= 119901(119896119903)
119890119903
(119903 = 1 2 3 119896 = 0 1)
(8)
where
119901(01)
= 1199110minus 11991111198902= 1199090+ 11990911198901minus 11990921198902minus 11990931198903
119901(02)
= 1199110+ 11991111198902= 1199090minus 11990911198901+ 11990921198902minus 11990931198903
119901(03)
= 1199110minus 11991111198902= 1199090minus 11990911198901minus 11990921198902+ 11990931198903
119901(11)
= 1199112minus 11991131198902= 1199100+ 11991011198901minus 11991021198902minus 11991031198903
119901(12)
= 1199112+ 11991131198902= 1199100minus 11991011198901+ 11991021198902minus 11991031198903
119901(13)
= 1199112minus 11991131198902= 1199100minus 11991011198901minus 11991021198902+ 11991031198903
(9)
with 1199110
= 1199090minus 11990911198901 1199111= 1199092minus 11990931198901 1199112
= 1199100minus 11991011198901 and
1199113= 1199102minus 11991031198901 For instance
11989021199010= 1198902(1199090+ 11990911198901+ 11990921198902+ 11990931198903)
= (1199090minus 11990911198901+ 11990921198902minus 11990931198903) 1198902= 119901(02)
1198902
11989011199011= 1198901(1199100+ 11991011198901+ 11991021198902+ 11991031198903)
= (1199100+ 11991011198901minus 11991021198902minus 11991031198903) 1198901= 119901(11)
1198901
(10)
Because of the properties of the eight-unit equality the addi-tion and subtraction of dual split quaternions are governedby the rules of ordinary algebra Here the symbol 119901
(119896119903)is used
by just enumerating 119903 and 119896 not 119903 times 119896 For example119901(22)
= 1199014and 119901
22= 1199014
For any two elements 119911 = 1199010+ 1205761199011and 119908 = 119902
0+ 1205761199021
of D(S) where 1199020= sum3
119903=0119904119903119890119903and 119902
1= sum3
119903=0119905119903119890119903are split
quaternion components and 119904119903 119905119903isin R (119903 = 0 1 2 3) their
noncommutative product is given by
119911119908 = (1199010+ 1205761199011) (1199020+ 1205761199021) = 11990101199020+ 120576 (119901
01199021+ 11990111199020)
(11)
The conjugation 119911lowast of 119911 and the corresponding modulus 119911119911lowast
inD(S) are defined by
119911lowast
= 119901lowast
0+ 120576119901lowast
1
119911119911lowast
= 119911lowast
119911 = 1199010119901lowast
0+ 120576 (119901
0119901lowast
1+ 1199011119901lowast
0)
= (11991101199110minus 11991111199111) + 2120576 (119911
01199112minus 11991111199113)
=
1
sum
119903=0
(1199092
119903minus 1199092
119903+2) + 120576 (119909
119903119910119903minus 119909119903+2
119910119903+2
)
(12)
where 119901lowast0= 1199110minus 11991111198902and 119901
lowast
1= 1199112minus 11991131198902
Lemma 1 For all 119911 isin D(S) and 119899 isin N = 1 2 3 wehave
119911119899
= 119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0 (13)
Abstract and Applied Analysis 3
Proof If 119899 = 1 then (13) is trivial Now suppose that thisholds for some 119899 isin N Then as desired
119911119899+1
= 119911119911119899
= 119911(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119901119899+1
0+ 120576
119899
sum
119896=1
119901119899minus119896+1
01199011119901119896minus1
0+ 1205761199011119901119899
0
= 119901119899+1
0+ 120576
119899+1
sum
119896=1
119901119899+1minus119896
01199011119901119896minus1
0
(14)
By the principle of mathematical induction (13) holds for all119899 isin N
Let Ω be an open subset of C2 times C2 Then the function119891 Ω rarr D(S) can be expressed as
119891 (119911) = 119891 (1199010 1199011) = 1198910(1199010 1199011) + 1205761198911(1199010 1199011) (15)
where the component functions 119891119903 Ω rarr S (119903 = 0 1) are
split quaternionic-valued functions The component func-tions 119891
119903(119903 = 0 1) are
1198910(1199010 1199011) = 1198910(1199110 1199111 1199112 1199113)
= 1198920(1199110 1199111 1199112 1199113) + 1198921(1199110 1199111 1199112 1199113) 1198902
1198911(1199010 1199011) = 1198911(1199110 1199111 1199112 1199113)
= 1198922(1199110 1199111 1199112 1199113) + 1198923(1199110 1199111 1199112 1199113) 1198902
(16)
where 119892119896
= 1199062119896
+ 1199062119896+1
1198901(119896 = 0 1) and 119892
119896= V2119896minus4
+
V2119896minus3
1198901(119896 = 2 3) are complex-valued functions and 119906
119903and
V119903(119903 = 0 1 2 3) are real-valued functionsNow we let differential operators 119863
1and 119863
2be defined
onD(S) as
1198631= 119863(11)
+ 120576119863(12)
1198632= 119863(21)
+ 120576119863(22)
(17)
Then the conjugate operators119863lowast1and119863
lowast
2are
119863lowast
1= 119863lowast
(11)+ 120576119863lowast
(12) 119863
lowast
2= 119863lowast
(21)+ 120576119863lowast
(22) (18)
where
119863(11)
=
120597
1205971199110
+
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
119863(12)
=
120597
1205971199112
+
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
120597
1205971199102
1198902+
120597
1205971199103
1198903)
119863(21)
=
120597
1205971199110
+
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
1
2
120597
1205971199092
1198902minus
1
2
120597
1205971199093
1198903)
119863(22)
=
120597
1205971199112
+
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
1
2
120597
1205971199102
1198902minus
1
2
120597
1205971199103
1198903)
(19)
119863lowast
(11)=
120597
1205971199110
minus
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
120597
1205971199092
1198902minus
120597
1205971199093
1198903)
119863lowast
(12)=
120597
1205971199112
minus
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
120597
1205971199102
1198902minus
120597
1205971199103
1198903)
119863lowast
(21)=
120597
1205971199110
minus
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
1
2
120597
1205971199092
1198902+
1
2
120597
1205971199093
1198903)
119863lowast
(22)=
120597
1205971199112
minus
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
(20)
act on D(S) These operators are called correspondingCauchy-Riemann operators in D(S) where 120597120597119911
119903and
120597120597119911119903(119903 = 0 1 2 3) are usual differential operators used in
the complex analysis
Remark 2 From the definition of differential operators onD(S)
119863119903119891 = (119863
(1199031)+ 120576119863(1199032)
) (1198910+ 1205761198911)
= 119863(1199031)
1198910+ 120576 (119863
(1199031)1198911+ 119863(1199032)
1198910)
119863lowast
119903119891 = (119863
lowast
(1199031)+ 120576119863lowast
(1199032)) (1198910+ 1205761198911)
= 119863lowast
(1199031)1198910+ 120576 (119863
lowast
(1199031)1198911+ 119863lowast
(1199032)1198910)
(21)
where 119903 = 1 2
Definition 3 LetΩ be an open set inC2 timesC2 A function 119891 =
1198910+ 1205761198911is called an 119871
119903(resp 119877
119903)-regular function (119903 = 1 2)
onΩ if the following two conditions are satisfied
(i) 119891119896(119896 = 0 1) are continuously differential functions
onΩ and
(ii) 119863lowast119903119891(119911) = 0 (resp 119891(119911)119863lowast
119903= 0) onΩ (119903 = 1 2)
In particular the equation 119863lowast
1119891(119911) = 0 of Definition 3 is
equivalent to
119863lowast
(11)1198910= 0 119863
lowast
(12)1198910+ 119863lowast
(11)1198911= 0 (22)
4 Abstract and Applied Analysis
In addition
1205971198920
1205971199110
minus
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1205971198923
1205971199111
minus
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1205971198922
1205971199111
minus
1205971198920
1205971199113
= 0
(23)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1205971199062
1205971199092
minus
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1205971199062
1205971199093
+
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1205971199060
1205971199092
minus
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
minus
1205971199060
1205971199093
+
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
120597V2
1205971199092
minus
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1205971199062
1205971199103
+
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1205971199060
1205971199102
minus
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
120597V0
1205971199092
minus
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
minus
1205971199060
1205971199103
+
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
minus
120597V0
1205971199093
+
120597V1
1205971199092
= 0
(24)
The above systems (23) and (24) are corresponding Cauchy-Riemann systems inD(S) Similarly the equation119863
lowast
2119891(119911) =
0 of Definition 3 is equivalent to
119863lowast
(21)1198910= 0 119863
lowast
(22)1198910+ 119863lowast
(21)1198911= 0 (25)
Then
1205971198920
1205971199110
minus
1
2
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1
2
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1
2
1205971198923
1205971199111
minus
1
2
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1
2
1205971198922
1205971199111
minus
1
2
1205971198920
1205971199113
= 0
(26)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
+
1
2
1205971199062
1205971199093
+
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
+
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
+
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1
2
1205971199062
1205971199103
+
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
+
1
2
120597V2
1205971199093
+
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
+
1
2
1205971199060
1205971199103
+
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
+
1
2
120597V0
1205971199093
+
1
2
120597V1
1205971199092
= 0
(27)
The above systems (26) and (27) are corresponding Cauchy-Riemann systems inD(S)
On the other hand the equation 119891(119911)119863lowast
1= 0 of
Definition 3 is equivalent to
1198910119863lowast
(11)= 0 119891
0119863lowast
(12)= minus1198911119863lowast
(11) (28)
Abstract and Applied Analysis 5
Then
1198920
120597
1205971199110
= 1198921
120597
1205971199111
1198921
120597
1205971199110
= 1198920
120597
1205971199111
1198920
120597
1205971199112
minus 1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+ 1198923
120597
1205971199111
1198921
120597
1205971199112
minus 1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+ 1198922
120597
1205971199111
(29)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(30)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
=
1205971199062
1205971199092
+
1205971199063
1205971199093
1205971199061
1205971199090
+
1205971199060
1205971199091
= minus
1205971199062
1205971199093
+
1205971199063
1205971199092
1205971199062
1205971199090
+
1205971199063
1205971199091
=
1205971199060
1205971199092
minus
1205971199061
1205971199093
1205971199063
1205971199090
minus
1205971199062
1205971199091
=
1205971199060
1205971199093
+
1205971199061
1205971199092
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
= minus
120597V0
1205971199090
+
120597V1
1205971199091
+
120597V2
1205971199092
+
120597V3
1205971199093
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1205971199062
1205971199103
minus
1205971199063
1205971199102
= minus
120597V1
1205971199090
minus
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1205971199060
1205971199102
+
1205971199061
1205971199103
= minus
120597V2
1205971199090
minus
120597V3
1205971199091
+
120597V0
1205971199092
minus
120597V1
1205971199093
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1205971199060
1205971199103
minus
1205971199061
1205971199102
= minus
120597V3
1205971199090
+
120597V2
1205971199091
+
120597V0
1205971199093
+
120597V1
1205971199092
(31)
Similarly the equation 119891(119911)119863lowast
2= 0 of Definition 3 is
equivalent to
1198910119863lowast
(21)= 0 119891
0119863lowast
(22)= minus1198911119863lowast
(21) (32)
Then
1198920
120597
1205971199110
=
1
2
1198921
120597
1205971199111
1198921
120597
1205971199110
=
1
2
1198920
120597
1205971199111
1198920
120597
1205971199112
minus
1
2
1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+
1
2
1198923
120597
1205971199111
1198921
120597
1205971199112
minus
1
2
1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+
1
2
1198922
120597
1205971199111
(33)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(34)
Concretely the system is obtained as follows
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1
2
1205971199062
1205971199093
minus
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
+
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
minus
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
minus
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
minus
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1
2
1205971199062
1205971199103
minus
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
1
2
120597V2
1205971199093
minus
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
+
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1
2
1205971199060
1205971199103
minus
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
minus
120597V2
1205971199091
minus
1
2
120597V0
1205971199093
minus
1
2
120597V1
1205971199092
= 0
(35)
From the systems (24) (27) (31) and (35) the equations119863lowast
119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 = 1 2) are different
Therefore the equations 119863lowast119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 =
1 2) should be distinguished as 119871119903-regular functions (119903 =
1 2) and 119877119903-regular functions (119903 = 1 2) on Ω respectively
Now the properties of the 119871119903-regular function (119903 = 1 2) with
values inD(S) are considered
3 Properties of 119871119903-Regular Functions (119903 = 1 2)
with Values in D(S)
We consider properties of a 119871119903-regular functions (119903 = 1 2)
with values inD(S)
Theorem 4 Let Ω be an open set in C2 times C2 and let 119891 =
1198910+ 1205761198911= (1198920+11989211198902)+ 120576(119892
2+11989231198902) be an 119871
1-regular function
defined on Ω Then
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (36)
6 Abstract and Applied Analysis
Proof By the system (23) we have
1198631119891 = 119863
(11)1198910+ 120576 (119863
(12)1198910+ 119863(11)
1198911)
= (
1205971198920
1205971199110
+
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+ 2
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+ 2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+ 2
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+ 2
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+ 2
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+ 2
1205971198922
1205971199111
) 1198902
= 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(37)
Therefore we obtain
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (38)
Theorem5 LetΩ be an open set inC2timesC2 and119891 = 1198910+1205761198911=
(1198920+11989211198902) + 120576(119892
2+11989231198902) be an 119871
2-regular function defined on
Ω Then
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (39)
Proof By the system (26) we have
1198632119891 = 119863
(21)1198910+ 120576 (119863
(22)1198910+ 119863(21)
1198911)
= (
1205971198920
1205971199110
+
1
2
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1
2
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(40)
Therefore we obtain the following equation
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (41)
Abstract and Applied Analysis 7
Proposition 6 From properties of differential operators thefollowing equations are obtained
119863(1119903)
119901119903minus1
= 2 119863(2119903)
119901119903minus1
= 1
119863lowast
(1119903)119901119903minus1
= minus1 119863lowast
(2119903)119901119903minus1
= 0
119863lowast
(1119903)119901lowast
119903minus1= 2 119863
lowast
(2119903)119901lowast
119903minus1= 1
119863(1199031)
1199011= 119863lowast
(1199031)1199011= 119863(1199031)
119901lowast
1= 119863lowast
(1199031)119901lowast
1
= 119863(1199032)
1199010= 119863lowast
(1199032)1199010= 119863(1199032)
119901lowast
0
= 119863lowast
(1199032)119901lowast
0= 0 (119903 = 1 2)
(42)
Proof By properties of the power of dual split quaternionsand derivatives on D(S) the following derivatives areobtained
119863(11)
1199010=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090+ 11990911198901+ 11990921198902+ 11990931198903) = 2
119863lowast
(22)1199011=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
times (1199100+ 11991011198901+ 11991021198902+ 11991031198903) = 0
119863(11)
119901lowast
0=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090minus 11990911198901minus 11990921198902minus 11990931198903) = minus1
(43)
The other equations are calculated using a similar methodand the above equations are obtained
Theorem 7 LetΩ be an open set in C2 timesC2 and let 119891(119911) be afunction on Ω with values inD(S) Then the power 119911119899 of 119911 inD(S) is not an 119871
1-regular function but an 119871
2-regular function
on Ω where 119899 isin N
Proof From the definition of the 119871119903-regular function (119903 =
1 2) on Ω and Proposition 6 we may consider whether thepower 119911119899 of 119911 in D(S) satisfies the equation 119863
lowast
119903119911119899
= 0 (119903 =
1 2) Since119863lowast(11)
1199010= 2
119863lowast
1119911119899
= (119863lowast
(11)+ 120576119863lowast
(12))(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119863lowast
(11)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(11)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(12)119901119899
0) = 0
(44)
Hence the power 119911119899 of 119911 is not 1198711-regular onΩ On the other
hand from the equations in Proposition 6 we have119863lowast(21)
1199010=
0119863lowast(21)
1199011= 0 and119863
lowast
(22)1199010= 0 Then
119863lowast
2119911119899
= 119863lowast
(21)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(21)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(22)119901119899
0) = 0
(45)
Therefore by the definition of the 119871119903-regular function (119903 =
1 2) onΩ a power 119911119899 of 119911 is 1198712-regular onΩ
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The second author was supported by the Basic ScienceResearch Program through the National Research Founda-tion of Korea (NRF) funded by the Ministry of Science ICTand Future Planning (2013R1A1A2008978)
References
[1] I L Kantor andA S SolodovnikovHypercomplex Numbers AnElementary Introduction to Algebras Springer New York NYUSA 1989
[2] J Cockle ldquoOn systems of algebra involving more than oneimaginaryrdquo PhilosophicalMagazine III vol 35 no 238 pp 434ndash435 1849
[3] M Ozdemir and A A Ergin ldquoRotations with unit timelikequaternions in Minkowski 3-spacerdquo Journal of Geometry andPhysics vol 56 no 2 pp 322ndash336 2006
[4] L Kula and Y Yayli ldquoSplit quaternions and rotations in semiEuclidean spaceE4
2rdquo Journal of the KoreanMathematical Society
vol 44 no 6 pp 1313ndash1327 2007[5] D C Brody and E-M Graefe ldquoOn complexified mechanics
and coquaternionsrdquo Journal of Physics A Mathematical andTheoretical vol 44 no 7 article 072001 2011
[6] I Frenkel and M Libine ldquoSplit quaternionic analysis andseparation of the series for SL(2R) and SL(2C)SL(2R)rdquoAdvances in Mathematics vol 228 no 2 pp 678ndash763 2011
[7] J-i Inoguchi ldquoTimelike surfaces of constant mean curvature inMinkowski 3-spacerdquo Tokyo Journal of Mathematics vol 21 no1 pp 141ndash152 1998
[8] B Kenwright ldquoA beginners guide to dual-quaternions whatthey are how they work and how to use them for 3D characterhierarchiesrdquo in Proceedings of the 20th International Conferenceson Computer Graphics Visualization and Computer Vision pp1ndash10 2012
[9] E Pennestrı and R Stefanelli ldquoLinear algebra and numericalalgorithms using dual numbersrdquo Multibody System Dynamicsvol 18 no 3 pp 323ndash344 2007
[10] L H Son ldquoAn extension problem for solutions of partialdifferential equations in R119899rdquo Complex Variables Theory andApplication vol 15 no 2 pp 87ndash92 1990
[11] L H Son ldquoExtension problem for functions with values in aClifford algebrardquo Archiv der Mathematik vol 55 no 2 pp 146ndash150 1990
[12] J Kajiwara X D Li and K H Shon ldquoRegeneration in complexquaternion and Clifford analysisrdquo in International Colloquiumon Finite or Infinite DimensionalComplex Analysis and ItsApplications vol 2 of Advances in Complex Analysis and ItsApplications no 9 pp 287ndash298 Kluwer Academic PublishersHanoi Vietnam 2004
[13] J Kajiwara X D Li and K H Shon ldquoFunction spaces incomplex and Clifford analysisrdquo in International Colloquium
8 Abstract and Applied Analysis
on Finite Or Infinite Dimensional Complex Analysis and ItsApplications vol 14 of Inhomogeneous Cauchy Riemann Systemof Quaternion andCliffordAnalysis in Ellipsoid pp 127ndash155HueUniversity Hue Vietnam 2006
[14] J E Kim S J Lim and K H Shon ldquoRegular functions withvalues in ternary number system on the complex Cliffordanalysisrdquo Abstract and Applied Analysis vol 2013 Article ID136120 7 pages 2013
[15] J E Kim S J Lim and K H Shon ldquoRegularity of functions onthe reduced quaternion field in Clifford analysisrdquo Abstract andApplied Analysis vol 2014 Article ID 654798 8 pages 2014
[16] S J Lim and K H Shon ldquoHyperholomorphic fucntions andhyperconjugate harmonic functions of octonion variablesrdquoJournal of Inequalities andApplications vol 2013 article 77 2013
[17] S J Lim and K H Shon ldquoDual quaternion functions and itsapplicationsrdquo Journal of Applied Mathematics vol 2013 ArticleID 583813 6 pages 2013
[18] J E Kim S J Lim and K H Shon ldquoTaylor series of functionswith values in dual quaternionrdquo Journal of the Korean Society ofMathematical Education BmdashThePure and AppliedMathematicsvol 20 no 4 pp 251ndash258 2013
Research ArticleUnicity of Meromorphic Functions Sharing Sets withTheir Linear Difference Polynomials
Sheng Li and BaoQin Chen
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to BaoQin Chen chenbaoqin chbq126com
Received 23 January 2014 Accepted 20 March 2014 Published 13 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 S Li and B Chen This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We mainly investigate the unicity of meromorphic functions sharing two or three sets with their linear difference polynomials andprove some results
1 Introduction and Main Results
In this paper we assume the reader is familiar with the fun-damental results and the basic notations of the Nevanlinnatheory of meromorphic functions (see eg [1ndash3]) Let 119891(119911)be meromorphic in the whole plane We use the notation120588(119891) to denote the order of growth of the meromorphicfunction 119891(119911) In addition we denote by 119878(119903 119891) any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin outside of apossible exceptional set of finite logarithmic measure We saythat a meromorphic function 119886(119911) is a small function of 119891(119911)provided that 119879(119903 119886) = 119878(119903 119891) Let 119878(119891) be the set of all smallfunctions of 119891(119911)
For a set 119878 sub 119878(119891) we define the following
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 counting multiplicities
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 ignoring multiplicities
(1)
Let 119891 and 119892 be meromorphic functions If 119864119891(119878) = 119864
119892(119878)
and 119864119891(119878) = 119864
119892(119878) respectively then we say that 119891 and 119892
share a set 119878 CM and IM respectivelyFurthermore let 119888 be a nonzero complex constant We
define the shift of 119891(119911) by 119891(119911 + 119888) and define the differenceoperators of 119891(119911) by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911)
Δ119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911)) 119899 isin N 119899 ge 2
(2)
Theunicity theory ofmeromorphic functions sharing setsis an important topic of the uniqueness theory First of all werecall the following theorem given by Li and Yang in [4]
Theorem A (see [4]) Let 119898 ge 2 and let 119899 gt 2119898 + 6 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0Then for any two nonconstant meromorphic functions 119891 and119892 the conditions 119864
119891(119878) = 119864
119892(119878) and 119864
119891(infin) = 119864
119892(infin)
imply 119891 = 119892
Yi and Lin considered the case 119898 = 1 with the conditionthat two meromorphic functions share three sets and got theresult as follows
Theorem B (see [5]) Let 1198781= 120596 120596
119899
+ 119886120596119899minus1
+ 119887 = 01198782= 0 and 119878
3= infin where 119886 119887 are nonzero constants such
that 120596119899 + 119886120596119899minus1 + 119887 = 0 has no repeated root and 119899(ge 4) is aninteger If for two nonconstant meromorphic functions 119891 and119892 119864119891(119878119895) = 119864
119892(119878119895) for 119895 = 1 2 3 and Θ(infin119891) gt 0 then
119891 equiv 119892
Recently a number of papers have focused on differenceanalogues of the Nevanlinna theory (see eg [6ndash9]) Inparticular there has been an increasing interest in studyingthe uniqueness problems related to meromorphic functionsand their shifts or their difference operators (see eg [10ndash16])
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 894968 7 pageshttpdxdoiorg1011552014894968
2 Abstract and Applied Analysis
In 2010 Zhang considered a meromorphic function 119891(119911)sharing sets with its shift 119891(119911 + 119888) and proved the followingresult
TheoremC (see [16]) Let119898 ge 2 and let 119899 ge 2119898+4with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596119899 + 119886120596119899minus119898 + 119887 = 0 Suppose that119891(119911) is a nonconstant meromorphic function of finite orderThen 119864
119891(119911)(119878) = 119864
119891(119911+119888)(119878) and 119864
119891(119911)(infin) = 119864
119891(119911+119888)(infin)
imply 119891(119911) equiv 119891(119911 + 119888)
For an analogue result in difference operator B Chen andZ Chen proved the following theorem in [10]
Theorem D (see [10]) Let 119898 ge 2 and let 119899 ge 2119898 + 4 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has nomultiple roots Let 119878 = 120596 | 120596119899+119886120596119899minus119898+119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying 119864
119891(119911)(119878) = 119864
Δ119888119891(119878) and 119864
119891(119911)(infin) = 119864
Δ119888119891(infin) If
119873(119903
1
Δ119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (3)
then Δ119888119891 equiv 119891(119911)
It is natural to ask what happens if the shift 119891(119911 + 119888) ordifference operatorΔ
119888119891(119911) is replaced by a general expression
of 119891(119911) such as a linear difference polynomial of 119891(119911)Here a linear difference polynomial of 119891(119911) is an expres-
sion of the form
119871 (119911 119891) = 119887119896(119911) 119891 (119911 + 119888
119896) + sdot sdot sdot + 119887
0(119911) 119891 (119911 + 119888
0) (4)
where 119887119896(119911) equiv 0 119887
0(119911) 119887
119896(119911) are small functions of
119891(119911) 1198880 119888
119896are complex constants and 119896 is a nonnegative
integerIn this paper our aim is to investigate the uniqueness
problems of linear difference polynomials of 119891(119911) In partic-ular we primarily consider the linear difference polynomial119871(119911 119891) which satisfies one of the following conditions
(i) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1
(ii) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0
119873(119903
1
119871 (119911 119891)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(5)
Corresponding to the above question we obtain thefollowing results
Theorem 1 Let 119898 ge 2 and let 119899 ge 2119898 + 4 with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite orderand 119871(119911 119891) is of the form (4) satisfying the condition in (5)If 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) and 119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) then119871(119911 119891) equiv 119891(119911)
Corollary 2 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying the following
119873(119903
1
Δ119896
119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (6)
If 119864119891(119911)
(119878) = 119864Δ119896
119888119891(119911)
(119878) and 119864119891(119911)
(infin) = 119864Δ119896
119888119891(119911)
(infin) thenΔ119896
119888119891 equiv 119891(119911)
With an additional restriction on the order of growth of119891(119911) we prove the following fact
Theorem 3 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersuch that 120588(119891) notin N If 119864
119891(119911)(119878) = 119864
119871(119911119891)(119878) and 119864
119891(119911)(infin) =
119864119871(119911119891)
(infin) then 119871(119911 119891) equiv 119891(119911)
Remark 4 Note that in Theorem 3 we do not assume thatthe linear polynomial 119871(119911 119891) satisfies the condition in (5)In fact since 120588(119891) notin N by (19) we can easily get 120588(119890ℎ(119911)) =deg(ℎ(119911)) lt 120588(119891) which implies 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Thenusing a similar method as in the proof of Theorem 1 we cancomplete the proof of Theorem 3
Now we may ask what happens if the condition119898 ge 2 inTheorem 1 is replaced by a weaker condition containing thecase 119898 = 1 or even 119898 = 0 By considering three sets we getthe following theorem
Theorem 5 Let 119899119898 be nonnegative integers such that 119899 gt 119898Let 119886 and 119887 be nonzero constants such that 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878
1= 120596 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 =1198782= infin and 119878
3= 0 Suppose that 119891(119911) is a nonconstant
meromorphic function of finite order 119871(119911 119891) is of the form (4)satisfying the condition in (5) and 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for
119895 = 1 2 3 Then one has the following(i) If119898 = 0 then 119871(119911 119891) equiv 119905119891(119911) where 119905119899 = 1(ii) If 119899 and119898 are coprime then 119871(119911 119891) equiv 119891(119911)
Remark 6 Taking 119898 = 1 in Theorem 5 we can obtain ananalogue result of Theorem B related to linear differencepolynomials
Furthermore the following result is a corollary ofTheorem 5 related to difference operators
Corollary 7 Let 119899 119898 and 119878119895 119895 = 1 2 3 be given as in
Theorem 5 Suppose that 119891(119911) is a nonconstant meromorphicfunction of finite order satisfying
119873(119903
1
119891 (119911)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (7)
and 119864119891(119911)
(119878119895) = 119864
Δ119896
119888119891(119878119895) for 119895 = 1 2 3 Then one has the
following
(i) If119898 = 0 then Δ119896119888119891 equiv 119905119891(119911) where 119905119899 = 1
(ii) If 119899 and119898 are coprime then Δ119896119888119891 equiv 119891(119911)
Finally we give some examples for our results
Abstract and Applied Analysis 3
Examples In the following let 119892(119911) be an entire functionwith period 1 such that 120588(119892) isin (1infin) N (see [17])
(1) For the case (i) of condition (5) let 1198911(119911) = 119890
21205871198941199111198912(119911) = 119892(119911)119890
2120587119894119911 1198913(119911) = 119890
2120587119894119911
119892(119911) and let119871(119911 119891
119895) = 2119891
119895(119911) minus 119891
119895(119911 + 1) Then for 119895 = 1 2 3
119871(119911 119891119895) = 119891
119895(119911) and the sum of the coefficients of
119871(119911 119891119895) is equal to 1These examples satisfyTheorems
1 and 5 but do not satisfy Theorem D
(2) For the case (ii) of condition (5) let 119891(119911) = 119890119911 log 2119892(119911)and let 119871(119911 119891) = Δ119891(119911) = 119891(119911 + 1) minus 119891(119911) Then119871(119911 119891) = Δ119891(119911) = 119891(119911) the sum of the coefficientsof 119871(119911 119891) equals 0 and
119873(119903
1
Δ119891
) = 119873(119903
1
119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(8)
This example satisfiesTheorems 1 and 5 and Corollar-ies 2 and 7
(3) For Theorem 3 let 119891(119911) = 119890119911 log 3
119892(119911) and let119871(119911 119891) = 119891(119911 + 1) minus 2119891(119911) Then 119871(119911 119891) = 119891(119911)
and the sum of the coefficients of 119871(119911 119891) equals minus1This example satisfies Theorem 3 but does not satisfyTheorem D andTheorems 1 and 5
2 Proof of Theorem 1
We need the following lemmas for the proof of Theorem 1The difference analogue of the logarithmic derivative
lemmawas given byHalburd-Korhonen [7] andChiang-Feng[6] independently We recall the following lemmas
Lemma 8 (see [7]) Let 119891(119911) be a nonconstant meromorphicfunction of finite order 119888 isin C and 120575 lt 1 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 + |119888| 119891)
119903120575
) (9)
for all 119903 outside of a possible exceptional set with finite logarith-mic measure
Lemma 9 (see [8]) Let 119888 isin C let 119899 isin N and let 119891(119911) bea meromorphic function of finite order Then for any smallperiodic function 119886(119911) isin 119878(119891) with period 119888 consider thefollowing
119898(119903
Δ119899
119888119891
119891 (119911) minus 119886 (119911)
) = 119878 (119903 119891) (10)
where the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Let 119891(119911) be a meromorphic function of finite orderNotice that if 119871(119911 119891) ( equiv 0) is of the form (4) such that
1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 then for any given complex con-
stant 119886 119871(119911 119886) = 0 This indicates that 119871(119911 119891) = 119871(119911 119891 minus 119886)
and hence
119898(119903
119871 (119911 119891)
119891 minus 119886
) = 119898(119903
119871 (119911 119891 minus 119886)
119891 minus 119886
)
le
119896
sum
119895=0
119898(119903
119887119895(119911) (119891 (119911 + 119888
119895) minus 119886)
119891 minus 119886
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(11)
With this one can easily prove Lemma 10 below by a similarreasoning as in the proof of the difference analogue of thesecond main theorem of the Nevanlinna theory in [8] byHalburd and Korhonen We omit those details
Lemma 10 Let 119888 isin C let 119891(119911) be a meromorphic function offinite order and let 119871(119911 119891) equiv 0 be of the form (4) such that1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 Let 119902 ge 2 and let 119886
1 119886
119902be distinct
complex constants Then
119898(119903 119891) +
119902
sum
119894=1
119898(119903
1
119891 minus 119886119894
)
le 2119879 (119903 119891) minus 119873lowast
(119903 119891) + 119878 (119903 119891)
(12)
where
119873lowast
(119903 119891) = 2119873 (119903 119891) minus 119873 (119903 119871 (119911 119891)) + 119873(119903
1
119871 (119911 119891)
)
(13)
and the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Remark 11 If the linear difference polynomial 119871(119911 119891) isreplaced by
119871lowast
(119911 119891) = 119887119896(119911) 119891 (119911 + 119896119888)
+ sdot sdot sdot + 1198871(119911) 119891 (119911 + 119888) + 119887
0(119911) 119891 (119911)
(14)
Lemma 10 also holds even if the distinct complex constants1198861 119886
119902are replaced by 119886
1(119911) 119886
119902(119911) which are distinct
meromorphic periodic functions with period 119888 such that 119886119894isin
119878(119891) for all 119894 = 1 119902
The following is the standardValiron-Mohonrsquoko theorem(see Theorem 225 in the book of Laine [2])
Lemma 12 (see [2]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(15)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(16)
4 Abstract and Applied Analysis
The characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (17)
where 119889 = max119901 119902
Proof of Theorem 1 Since 119891(119911) and 119871(119911 119891) shareinfin CM wesee that 119871(119911 119891) equiv 0 and119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) Then byLemma 8 we have
119879 (119903 119871 (119911 119891)) = 119898 (119903 119871 (119911 119891)) + 119873 (119903 119871 (119911 119891))
le 119898(119903
119871 (119911 119891)
119891 (119911)
)
+ 119898 (119903 119891 (119911)) + 119873 (119903 119891 (119911))
le
119896
sum
119894=0
119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
+
119896
sum
119894=0
119898(119903 119887119894(119911)) + 119879 (119903 119891 (119911))
le 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(18)
Since 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) where 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots we know that (119871(119911 119891))119899+119886(119871(119911 119891))119899minus119898+119887 and119891(119911)119899+119886119891(119911)119899minus119898
+ 119887 share 0 CM Then from this and the condition119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) there exists a polynomial ℎ(119911)such that
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
= 119890ℎ(119911)
(19)
Suppose that 119890ℎ(119911) equiv 1 Note that 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots Let 120596
1 120596
119899denote all different roots of the equation
120596119899
+ 119886120596119899minus119898
+ 119887 = 0Next we prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) We know that
119871 (119911 119891) minus 120596119894= 119887119896(119911) (119891 (119911 + 119888
119896) minus 119891 (119911))
+ sdot sdot sdot + 1198870(119911) (119891 (119911 + 119888
0) minus 119891 (119911))
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
(20)
(i) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1 we see that
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 + (119891 (119911) minus 120596119894)
(21)
Then we deduce from this (19) and Lemma 9 that
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
= 119898(119903
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
)
= 119898(119903
(119871 (119911 119891) minus 1205961) sdot sdot sdot (119871 (119911 119891) minus 120596
119899)
(119891 (119911) minus 1205961) sdot sdot sdot (119891 (119911) minus 120596
119899)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(22)
(ii) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 we have
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 minus 120596119894 (23)
From this (19) and Lemma 9 we get
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
=
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
(24)
Applying Lemma 10 to 119891(119911) we get
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
)
le 2119879 (119903 119891 (119911)) minus 119898 (119903 119891 (119911)) minus 2119873 (119903 119891 (119911))
+ 119873 (119903 119871 (119911 119891)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
= 119879 (119903 119891 (119911)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
(25)
Abstract and Applied Analysis 5
Then the assumptions in (5) (24) and (25) yield thefollowing
119879 (119903 119890ℎ(119911)
) le 119879 (119903 119891 (119911))
minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891) = 119878 (119903 119891)
(26)
To sum up we now prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Rewriting(19) we get
(119871 (119911 119891))119899minus119898
[(119871 (119911 119891))119898
+ 119886]
= [119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887 minus 119887119890minusℎ(119911)
] 119890ℎ(119911)
(27)
Denote 119865(119911) = 119891(119911)119899
+ 119886119891(119911)119899minus119898 It follows from
Lemma 12 and119898 gt 0 that
119879 (119903 119865 (119911)) = 119899119879 (119903 119891 (119911)) + 119878 (119903 119891) (28)
Hence 119878(119903 119865) = 119878(119903 119891)By (18) and (27) and applying the second main theorem
for three small target functions we deduce the following
119879 (119903 119865 (119911))
le 119873 (119903 119865 (119911)) + 119873(119903
1
119865 (119911)
)
+ 119873(119903
1
119865 (119911) + 119887 minus 119887119890minus119901(119911)
) + 119878 (119903 119865)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891(119911)119899minus119898
[119891(119911)119898
+ 119886]
)
+ 119873(119903
1
(119871 (119911 119891))119899minus119898
) + 119873(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891 (119911)
) + 119873(119903
1
119891(119911)119898
+ 119886
)
+ 119873(119903
1
119871 (119911 119891)
) + 119879(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119879 (119903 119891 (119911)) + 119879(119903
1
119891 (119911)
) + 119879(119903
1
119891(119911)119898
+ 119886
)
+ 119879(119903
1
119871 (119911 119891)
) + 119898119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (119898 + 2) 119879 (119903 119891 (119911))
+ (119898 + 1) 119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (2119898 + 3) 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(29)
By combining (28) and (29) we have
(119899 minus 2119898 minus 3) 119879 (119903 119891 (119911)) le 119878 (119903 119891) (30)
which contradicts with 119899 ge 2119898 + 4Now we turn to consider the case 119890ℎ(119911) equiv 1 Equation (19)
yields the following
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
equiv 119891(119911)119899
+ 119886119891(119911)119899minus119898
(31)
Set 120593(119911) = 119871(119911 119891)119891(119911) and we have
119891(119911)119898
(120593(119911)119899
minus 1) = minus119886 (120593(119911)119899minus119898
minus 1) (32)
If 120593(119911) is not a constant (32) can be rewritten as
119891(119911)119898
(120593 (119911) minus 1) (120593 (119911) minus 120583) sdot sdot sdot (120593 (119911) minus 120583119899minus1
)
= minus119886 (120593 (119911) minus 1) (120593 (119911) minus ]) sdot sdot sdot (120593 (119911) minus ]119899minus119898minus1) (33)
where 120583 = cos(2120587119899) + 119894 sin(2120587119899) and ] = cos(2120587(119899minus119898)) +119894 sin(2120587(119899 minus 119898))
By the assumption that 119899 and 119899 minus 119898 have no commonfactors we see that 120583 120583119899minus1 ] ]119899minus119898minus1 are differentAssume that 119911
0is a 120583119895-point of 120593(119911) of multiplicity 119906
119895gt 0
where 1 le 119895 le 119899 minus 1 Notice that
minus119886 (120593 (1199110) minus 1) (120593 (119911
0) minus ]) sdot sdot sdot (120593 (119911
0) minus ]119899minus119898minus1) (34)
is a constantThen (33) implies that 1199110is a pole of119891(119911)119898Thus
119906119895ge 119898 This yields the following for 1 le 119895 le 119899 minus 1
119898119873(119903
1
120593 (119911) minus 120583119895
) le 119873(119903
1
120593 (119911) minus 120583119895
)
le 119879 (119903 120593 (119911)) + 119878 (119903 ℎ)
(35)
Then by (35) we get
2 ge
119899minus1
sum
119895=1
Θ(120583119895
120593 (119911)) =
119899minus1
sum
119895=1
1 minus lim119903rarrinfin
119873(119903 1 (120593 (119911) minus 120583119895
))
119879 (119903 120593 (119911))
ge
119899minus1
sum
119895=1
(1 minus
1
119898
) = (119899 minus 1) (1 minus
1
119898
)
(36)
which is impossible with119898 ge 2 and 119899 ge 2119898 + 4Hence 120593(119911) is a constant Since 119891(119911) is a nonconstant
meromorphic function we deduce from (32) that 120593(119911) equiv 1This yields 119871(119911 119891) equiv 119891(119911) which completes the proof ofTheorem 1
3 Proof of Theorem 5
Since 119891(119911) is a nonconstant meromorphic function of finiteorder 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for 119895 = 1 2 3 119878
1= 120596 120596
119899
+
119886120596119899minus119898
+ 119887 = 0 1198782= infin and 119878
3= 0 we have 119871(119911 119891) equiv
0 119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) and 119873(119903 1119871(119911 119891)) = 119873(119903
1119891(119911)) and we also get (18) and (19)
6 Abstract and Applied Analysis
Since 119891(119911) and 119871(119911 119891) share 0 infin CM there exists apolynomial ℎlowast(119911) such that
119871 (119911 119891)
119891 (119911)
= 119890ℎlowast(119911)
(37)
By Lemma 8 we see that
119879(119903 119890ℎlowast(119911)
) = 119898(119903 119890ℎlowast(119911)
) = 119898(119903
119871 (119911 119891)
119891 (119911)
)
le
119896
sum
119895=0
119898(119903
119891 (119911 + 119888119895)
119891 (119911)
)
+
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(38)
As in the proof of Theorem 1 we see that 119879(119903 119890ℎ(119911)) =
119878(119903 119891) still holds in both cases (i) and (ii)Rewriting (19) we have
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
minus 119890ℎ(119911)
119891(119911)119899
minus 119886119890ℎ(119911)
119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(39)
Combining this and (37) we get
(119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
+ 119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(40)
Suppose that 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 If119898 = 0 (40) becomes
(119886 + 1) (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
= 119887 (119890ℎ(119911)
minus 1) (41)
By the condition that 119887 = 0 1198781= 120596 (119886 + 1)120596
119899
+ 119887 = 0 =
implies 119886 = minus 1It follows from (38) (41) and 119879(119903 119890ℎ(119911)) = 119878(119903 119891) that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 119887 (119890ℎ(119911)
minus 1)) = 119878 (119903 119891)
(42)
which is a contradiction since 119899 ge 1If119898 ge 1 it follows from (38) (41) and119879(119903 119890ℎ(119911)) = 119878(119903 119891)
that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 minus119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
) 119891(119911)119899minus119898
+ 119887 (119890ℎ(119911)
minus 1))
le (119899 minus 119898)119879 (119903 119891) + 119878 (119903 119891)
(43)
That is impossible
Therefore 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 Notice that 119886 119887 = 0 Using asimilar method we can prove that 119890(119899minus119898)ℎ
lowast(119911)
minus119890ℎ(119911)
equiv 0Then(40) implies that 119890ℎ(119911) equiv 1
If 119898 = 0 we have 119890119899ℎlowast(119911)
equiv 1 Obviously 119890ℎlowast(119911) is a
constant Set 119905 = 119890ℎlowast(119911) Thus by (37) we get 119871(119911 119891) equiv 119905119891(119911)
where 119905119899 = 1If 119899 and 119898 are coprime 119890119899ℎ
lowast(119911)
equiv 1 and 119890119898ℎlowast(119911)
equiv 1 implythat 119890ℎ
lowast(119911)
equiv 1 Thus by (37) we get 119871(119911 119891) equiv 119891(119911) ThusTheorem 5 is proved
Conflict of Interests
The authors declare that they have no conflict of interests
Authorsrsquo Contribution
Both authors drafted the paper and read and approved thefinal paper
Acknowledgments
The authors are grateful to the editor and referees fortheir valuable suggestions This work was supported by theNNSFC (no 11171119 11301091) the Guangdong Natural Sci-ence Foundation (no S2013040014347) and the Foundationfor Distinguished Young Talents in Higher Education ofGuangdong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 2003
[4] P Li and C-C Yang ldquoSome further results on the unique rangesets of meromorphic functionsrdquo Kodai Mathematical Journalvol 18 no 3 pp 437ndash450 1995
[5] H-X Yi and W-C Lin ldquoUniqueness theorems concerning aquestion of Grossrdquo Proceedings of the Japan Academy Series AMathematical Sciences vol 80 no 7 pp 136ndash140 2004
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[8] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[9] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
Abstract and Applied Analysis 7
[10] B Chen and Z Chen ldquoMeromorphic function sharing twosets with its difference operatorrdquo Bulletin of the MalaysianMathematical Sciences Society Second Series vol 35 no 3 pp765ndash774 2012
[11] B Chen Z Chen and S Li ldquoUniqueness theorems on entirefunctions and their difference operators or shiftsrdquo Abstract andApplied Analysis vol 2012 Article ID 906893 8 pages 2012
[12] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[13] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[14] S Li and B Chen ldquoMeromorphic functions sharing smallfunctions with their linear difference polynomialsrdquoAdvances inDifference Equations vol 2013 article 58 2013
[15] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[16] J Zhang ldquoValue distribution and shared sets of differences ofmeromorphic functionsrdquo Journal of Mathematical Analysis andApplications vol 367 no 2 pp 401ndash408 2010
[17] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleA Comparison Theorem for Oscillation of the Even-OrderNonlinear Neutral Difference Equation
Quanxin Zhang
Department of Mathematics Binzhou University Binzhou Shandong 256603 China
Correspondence should be addressed to Quanxin Zhang 3314744163com
Received 9 December 2013 Accepted 28 March 2014 Published 10 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Quanxin Zhang This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
A comparison theorem on oscillation behavior is firstly established for a class of even-order nonlinear neutral delay differenceequations By using the obtained comparison theorem two oscillation criteria are derived for the class of even-order nonlinearneutral delay difference equations Two examples are given to show the effectiveness of the obtained results
1 Introduction
Recently there have been a lot of research papers in con-nection with the oscillation of solutions of difference equa-tions with or without neutral terms The literature on theoscillation of neutral delay difference equations is growingvery fast and it can be widely applied to the reality Infact neutral delay difference equations arise in modellingof the networks containing lossless transmission lines (asin high speed computers where the lossless transmissionlines are used to interconnect switching circuits) For recentcontributions regarding the theoretical part and providingsystematic treatment of oscillation of solutions of neutraltype difference equations the readers can refer to the recentmonographs by Agarwal [1] Gyori and Ladas [2]
The oscillation behavior of the even-order nonlinearneutral differential equation
[119909 (119905) + 119901 (119905) 119909 (120591 (119905))](119899)
+ 119902 (119905) 119891 [119909 (120590 (119905))] = 0 (1)
has been established by Zhang et al [3] In this paperthe discrete analogue of the above equation is consideredWe consider the even-order nonlinear neutral differenceequation
Δ119898
(119909119899+ 119901119899119909119899minus120591
) + 119902119899119891 (119909119899minus119896
) = 0 (2)
where 119898 ge 2 is an even and 120591 119896 isin N let N denote the setof all natural numbers 119899 isin 119873(119899
0) = 119899
0 1198990+ 1 1198990+ 2
1198990is a nonnegative integer Δ denotes the forward difference
operator defined by Δ119909119899= 119909119899+1
minus 119909119899 Δ119898119909
119899= Δ119898minus1
119909119899+1
minus
Δ119898minus1
119909119899
Throughout this paper the following conditions areassumed to hold
(H1) 119901119899 is a sequence of nonnegative real number 0 le
119901119899
lt 1 and 119902119899 is a sequence of nonnegative real
number with 119902119899 being not eventually identically
equal to zero(H2) 119891 R rarr R (R = (minusinfin +infin)) is a continuous oddfunction and 119909119891(119909) gt 0 for all 119909 = 0
Before deriving themain results the following definitionsare given
Definition 1 By a solution of (2) one means a real sequence119909119899 defined for 119899 ge 119899
0minus120579 (120579 = max120591 119896)which satisfies (2)
for 119899 isin 119873(1198990)
In this paper we restrict our attention to nontrivialsolutions of (2)
Definition 2 A nontrivial solution 119909119899 of (2) is said to
be oscillatory if the terms 119909119899of the sequence are neither
eventually positive nor eventually negative Otherwise it iscalled nonoscillatory
Definition 3 An equation is said to be oscillatory if all itssolutions are oscillatory
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 492492 5 pageshttpdxdoiorg1011552014492492
2 Abstract and Applied Analysis
In 2004 Stavroulakis [4] studied the oscillatory behaviorof all solutions of first-order delay difference equation
119909119899+1
minus 119909119899+ 119901119899119909119899minus119896
= 0 (3)
and established one new oscillation criterion Thandapaniet al [5] studied the oscillatory behavior of all solutions ofsecond-order neutral delay difference equation
Δ2
(119910119899minus 119901119910119899minus119896
) minus 119902119899119891 (119910119899minus119905
) = 0 (4)
and established a number of new oscillation criteria In 2000Zhou et al [6] studied the oscillatory behavior of all solutionsof even-order neutral delay difference equation
Δ119898
(119909119899minus 119901119899119892 (119909119899minus119896
)) minus 119902119899ℎ (119909119899minus119897
) = 0 (5)
and established three new oscillation criteria under certainconditionsThe studies on oscillatory behavior of all solutionsof even-order delay difference equations we recommendreferring to [7ndash10] On the basis of the abovework we studiedthe oscillatory behavior of all solutions of (2) Firstly a com-parison theorem on oscillation behavior is established for (2)The comparison theorem changes the discriminant criteria ofthe oscillation of (2) into the oscillationrsquos discriminant criteriain the first-order nonneutral delay difference equationsThenby using the above comparison theorem we obtain someoscillation criteria for (2) and improve the well-known resultsof Ladas et al [11] Erbe and Zhang [12] and Stavroulakis [4]In particular the results are new when119898 = 2 119901
119899equiv 0
The paper is organized as follows In Section 2 a compar-ison theorem on oscillation behavior is firstly established fora class of even-order nonlinear neutral delay difference equa-tions Then the comparison theorem changes the discrimi-nant of the oscillation in the even-order nonlinear neutraldelay difference equation into the oscillationrsquos discriminantin the first-order nonneutral delay difference equations InSection 3 some oscillation criteria are obtained for the classof even-order nonlinear neutral delay difference equationby using the above comparison theorem In Section 4 twoexamples are given
2 Comparison Theorem
To obtain the comparison theorem in this section we needthe following lemmas which can be founded in [1] see alsoChen [7] andThandapani and Arul [8]
Lemma 4 Let 119906119899 be a sequence of real numbers for 119899 ge 119899
0
Let 119906119899 and Δ
119898
119906119899 be of constant sign where Δ
119898
119906119899is not
identically zero for 119899 ge 1198991 If
119906119899Δ119898
119906119899le 0 (6)
then
(i) there is a natural number 1198992
ge 1198991such that the
sequences Δ119895119906119899 119895 = 1 2 119898 minus 1 are of constant
sign for 119899 ge 1198992
(ii) there exists a number 119897 isin 0 1 119898 minus 1 with(minus1)119898minus119897minus1
= 1 such that
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 0 1 2 119897 119899 ge 119899
2
(minus1)119895minus119897
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 119897 + 1 119898 minus 1 119899 ge 119899
2
(7)
Lemma 5 Observe that under the hypotheses of Lemma 4 if119906119899 is increasing for 119899 ge 119899
0 then there exists a natural number
1198991ge 1198990such that for all 119899 ge 2
119898minus1
1198991
119906119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119906119899 (8)
where 120582119898= 12(119898minus1)
2
Theorem 6 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 If there exists a constant
120582119898= 12(119898minus1)
2
such that the first-order difference equation
Δ119911119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119911119899minus119896
= 0 (9)
is oscillatory then (2) is oscillatory
Proof Suppose that (2) has a nonoscillatory solution 119909119899
Without the loss of generality we assume that 119909119899 is an
eventually positive solution of (2) then there is a naturalnumber 119899
1ge 1198990such that 119909
119899gt 0 119909
119899minus119896gt 0 119909
119899minus120591gt 0 and
119909119899minus119896minus120591
gt 0 for all 119899 ge 1198991 Let
119910119899= 119909119899+ 119901119899119909119899minus120591
(10)
Then from (H1) and (H
2) there exists a natural number 119899
2ge
1198991such that
119910119899gt 0 Δ
119898
119910119899le 0 forall119899 ge 119899
2 (11)
By Lemma 4 there exist an integer 1198993ge 1198992and an integer
119897 (0 le 119897 le 119898) where (119898 + 119897) is an odd integer For all 119899 ge 1198993
we can get
Δ119895
119910119899gt 0 for 119895 = 1 2 119897
(minus1)(119895minus119897)
Δ119895
119910119899gt 0 for 119895 = 119897 + 1 119898 minus 1
(12)
Thus from (12) Δ119910119899
gt 0 and Δ119898minus1
119910119899
gt 0 for 119899 ge 1198993 By
Lemma 5 there exists an integer 1198994ge 1198993 For all 119899 ge 2
119898minus1
1198994
we derive
119910119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119910119899 120582
119898=
1
2(119898minus1)
2 (13)
From (10)
119909119899minus119896
= 119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
(14)
Consequently we have
Δ119898
119910119899+ 119902119899119891 (119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) = 0
for all sufficient large 119899
(15)
Abstract and Applied Analysis 3
Noting that |119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) le 0
for all sufficient large 119899
(16)
By 119910119899ge 119909119899 Δ119910119899gt 0 and 119899 minus 119896 minus 120591 le 119899 minus 119896 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) ge Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
(17)
Therefore we have
Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
le 0
for all sufficient large 119899
(18)
Now by using (13) we have that for 120582119898= 12(119898minus1)
2
119910119899minus119896
ge
120582119898
(119898 minus 1)
(119899 minus 119896)119898minus1
Δ119898minus1
119910119899minus119896
for all sufficient large 119899
(19)
Thus we get
Δ119898
119910119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) Δ119898minus1
119910119899minus119896
le 0
for all sufficient large 119899
(20)
where120582119898= 12(119898minus1)
2
Let119906119899= Δ119898minus1
119910119899 then for large enough
119899 we get
Δ119906119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119906119899minus119896
le 0 (21)
where 120582119898
= 12(119898minus1)
2
Therefore inequality (21) has aneventually positive solution By Lemma 5 in [9] (9) hasan eventually positive solution which contradicts that (9) isoscillatory This completes the proof
3 Applications of the Comparison Theorem
The following lemma is well known (see eg [2 11 12] andthe references therein)
Lemma 7 Let 119902119899 be a sequence of eventually nonnegative
real number and 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt (
119896
1 + 119896
)
1+119896
(22)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894gt 1 (23)
then the first-order difference equation
Δ119909119899+ 119902119899119909119899minus119896
= 0 (24)
is oscillatory
Thus from Theorem 6 and Lemma 7 we can obtain thefollowing results
Theorem 8 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1)(
119896
1 + 119896
)
1+119896
(25)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 2(119898minus1)
2
(119898 minus 1) (26)
then (2) is oscillatory
Proof From (25) and (26) we can obtain
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt (
119896
1 + 119896
)
1+119896
(27)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 (28)
where 120582119898
= 12(119898minus1)
2
By Lemma 7 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According toTheorem 8 we obtain Corollary 9
Corollary 9 Assume that conditions (H1) and (H
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 when 119901
119899equiv 0
119898 = 2 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(
119896
1 + 119896
)
1+119896
(29)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (30)
then the second-order difference equation
Δ2
119909119899+ 119902119899119891 (119909119899minus119896
) = 0 (31)
is oscillatory
The following lemma is given in [4 Theorem 26]
Lemma 10 Let 119902119899 be a sequence of nonnegative real numbers
and 119896 a positive integer Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(32)
4 Abstract and Applied Analysis
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus
1205722
4
(33)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus 120572
119896
(34)
then (24) is oscillatory
Thus fromTheorem 6 and Lemma 10 we can obtain thefollowing results
Theorem 11 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(35)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus
1205722
4
)
(36)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus 120572119896
)
(37)
then (2) is oscillatory
Proof From (36) and (37) we can obtain
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus
1205722
4
(38)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus 120572119896
(39)
where 120582119898
= 12(119898minus1)
2
By Lemma 10 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According to Theorem 11 we can obtain the followingcorollary
Corollary 12 Assume that conditions (1198671) and (119867
2) hold let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
For 119901119899equiv 0119898 = 2 assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(40)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(1 minus
1205722
4
) (41)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (1 minus 120572
119896
) (42)
then (31) is oscillatory
4 Examples
Example 1 Considering the equation
Δ119898
(119909119899+
1
119899
119909119899minus119897
) +
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(43)
where 119899 gt 3119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
1
119899
lt 1 119902119899=
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(44)
where 119902119899 is a positive sequence Then
119899
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
1
119894 minus 2
)
=
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
(45)
Thus
lim sup119899rarrinfin
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
=
6
119890
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1)
(46)
Therefore by Theorem 8 (43) is oscillatory
Example 2 Considering the equation
Δ119898
[119909119899+
119899 minus 1
119899
119909119899minus119897
]
+
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(47)
Abstract and Applied Analysis 5
where 119899 gt 2119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
119899 minus 1
119899
lt 1
119902119899=
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(48)
where 119902119899 is a positive sequence Denote 120572 = 414 then
119899minus1
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
119894 minus 2 minus 1
119894 minus 2
)
=
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
(49)
Thus
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
=
15
16
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1) (1 minus (
4
14
)
2
)
(50)
Therefore by Theorem 11 (47) is oscillatory
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The author sincerely thanks the reviewers for their valuablesuggestions and useful comments that have led to the presentimproved version of the original paper This work wassupported by a Grant from the Natural Science Foundationof Shandong Province of China (no ZR2013AM003) andthe Development Program in Science and Technology ofShandong Province of China (no 2010GWZ20401)
References
[1] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[2] I Gyori and G Ladas Oscillation Theory of Delay DifferentialEquations Oxford Clarendon Press Oxford UK 1991
[3] Q Zhang J Yan and L Gao ldquoOscillation behavior of even-order nonlinear neutral differential equations with variablecoefficientsrdquoComputers andMathematics withApplications vol59 no 1 pp 426ndash430 2010
[4] I P Stavroulakis ldquoOscillation criteria for first order delaydifference equationsrdquo Mediterranean Journal of Mathematicsvol 1 no 2 pp 231ndash240 2004
[5] E Thandapani R Arul and P S Raja ldquoBounded oscillationof second order unstable neutral type difference equationsrdquoJournal of Applied Mathematics and Computing vol 16 no 1-2pp 79ndash90 2004
[6] Z Zhou J Yu and G Lei ldquoOscillations for even-order neu-tral difference equationsrdquo Korean Journal of Computational ampApplied Mathematics vol 7 no 3 pp 601ndash610 2000
[7] S Chen ldquoOscillation criteria for certain even order quasilineardifference equationsrdquo Journal of AppliedMathematics and Com-puting vol 31 no 1-2 pp 495ndash506 2009
[8] EThandapani and R Arul ldquoOscillatory and asymptotic behav-ior of solutions of higher order damped nonlinear differenceequationsrdquoCzechoslovakMathematical Journal vol 49 no 1 pp149ndash161 1999
[9] G Ladas and C Qian ldquoComparison results and linearizedoscillations for higher-order difference equationsrdquo InternationalJournal of Mathematics andMathematical Sciences vol 15 no 1pp 129ndash142 1992
[10] L Yang ldquoOscillation behavior of even-order neutral differenceequations with variable coefficientsrdquo Pure and Applied Mathe-matics vol 24 no 4 pp 796ndash801 2008 (Chinese)
[11] G Ladas Ch G Philos and Y G Sficas ldquoSharp conditions forthe oscillation of delay difference equationsrdquo Journal of AppliedMathematics and Simulation vol 2 no 2 pp 101ndash111 1989
[12] L H Erbe and B G Zhang ldquoOscillation of discrete analogues ofdelay equationsrdquo Differential and Integral Equations vol 2 no3 pp 300ndash309 1989
Research ArticleDifference Equations and Sharing ValuesConcerning Entire Functions and Their Difference
Zhiqiang Mao1 and Huifang Liu2
1 School of Mathematics and Computer Jiangxi Science and Technology Normal University Nanchang 330038 China2 College of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China
Correspondence should be addressed to Huifang Liu liuhuifang73sinacom
Received 19 January 2014 Accepted 22 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z Mao and H Liu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The value distribution of solutions of certain difference equations is investigated As its applications we investigate the differenceanalogue of the Bruck conjecture We obtain some results on entire functions sharing a finite value with their difference operatorsExamples are provided to show that our results are the best possible
1 Introduction and Main Results
In this paper the term meromorphic function will meanbeing meromorphic in the whole complex plane C Itis assumed that the reader is familiar with the standardnotations and the fundamental results of the Nevanlinnatheory see for example [1ndash3] In addition we use notations120590(119891) 120582(119891) to denote the order and the exponent of conver-gence of the sequence of zeros of a meromorphic function 119891respectivelyThe notation 119878(119903 119891) is defined to be any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside aset 119864 of 119903 of finite logarithmic measure
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 isin C We say that 119891 and 119892 share 119886 CM providedthat 119891 minus 119886 and 119892 minus 119886 have the same zeros with the samemultiplicities Similarly we say that 119891 and 119892 share 119886 IMprovided that 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
The famous results in the uniqueness theory of meromor-phic functions are the 5 IM and 4 CM shared values theoremsdue to Nevanlinna [4] It shows that if two nonconstantmeromorphic functions 119891 and 119892 share five different valuesIM or four different values CM then 119891 equiv 119892 or 119891 is a linearfractional transformation of119892 Condition 4CMshared valueshave been improved to 2 CM + 2 IM by Gundersen [5]while the case 1 CM + 3 IM still remains an open problemSpecifically Bruck posed the following conjecture
Conjecture 1 (see [6]) Let 119891 be a nonconstant entire functionsatisfying the hyperorder 120590
2(119891) lt infin where 120590
2(119891) is not a
positive integer If 119891 and 1198911015840 share a finite value 119886 CM then119891 minus 119886 equiv 119888(119891
1015840
minus 119886) for some nonzero constant 119888
In [6] Bruck proved that the conjecture is true providedthat 119886 = 0 or 119873(119903 11198911015840) = 119878(119903 119891) He also gave counter-examples to show that the restriction on the growth of 119891 isnecessary
In recent years as the research on the difference ana-logues of Nevanlinna theory is becoming active lots ofauthors [7ndash11] started to consider the uniqueness of mero-morphic functions sharing values with their shifts or theirdifference operators
Heittokangas et al proved the following result which is ashifted analogue of Bruckrsquos conjecture
Theorem A (see [8]) Let 119891 be a meromorphic function of120590(119891) lt 2 and 120578 a nonzero complex number If119891(119911) and119891(119911+120578)share a finite value 119886 andinfin CM then
119891 (119911 + 120578) minus 119886
119891 (119911) minus 119886
= 120591 (1)
for some constant 120591
In [8] Heittokangas et al gave the example 119891(119911) = 1198901199112
+1
which shows that 120590(119891) lt 2 cannot be relaxed to 120590(119891) le 2
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 584969 6 pageshttpdxdoiorg1011552014584969
2 Abstract and Applied Analysis
For a nonzero complex number 120578 we define differenceoperators as
Δ120578119891 (119911) = 119891 (119911 + 120578) minus 119891 (119911)
Δ119899
120578119891 (119911) = Δ
119899minus1
120578(Δ120578119891 (119911)) 119899 isin N 119899 ge 2
(2)
Regarding the difference analogue of Bruckrsquos conjecture wemention the following results
Theorem B (see [7]) Let 119891 be a finite order transcendentalentire function which has a finite Borel exceptional value 119886 andlet 120578 be a constant such that 119891(119911 + 120578) equiv 119891(119911) If 119891(119911) andΔ120578119891(119911) share 119886 CM then
119886 = 0
119891 (119911 + 120578) minus 119891 (119911)
119891 (119911)
= 119888 (3)
for some nonzero constant 119888
Theorem C (see [11]) Let 119891 be a nonperiodic transcendentalentire function of finite order If 119891(119911) and Δ119899
120578119891(119911) share a
nonzero finite value 119886 CM then 1 le 120590(119891) le 120582(119891 minus 119886) + 1that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (4)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891 minus 119886) and119876(119911) is a polynomial with deg119876 le 120590(119860) + 1
Let 119891 be a nonperiodic transcendental entire function offinite order Theorem B shows that if a nonzero finite value119886 is shared by 119891(119911) and Δ
120578119891(119911) then 120590(119891) = 120582(119891 minus 119886)
It is obvious that the result in Theorem B is sharper thanTheoremC for 119899 = 1 In this paper we continue to investigatethe difference analogue of Bruckrsquos conjecture and obtain thefollowing result
Theorem 2 Let 119891 be a finite order entire function 119899 ge 2 aninteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share a finite value 119886 ( = 0)CM then120582(119891minus119886) = 120590(119891) ge
1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (5)
where 119860(119911) is an entire function with 1 le 120590(119860) = 120582(119891 minus 119886) =120590(119891) and 119876(119911) is a polynomial with deg119876 le 120590(119860)
Remark 3 It is obvious that Theorem 2 is sharper thanTheorem C and a supplement of Theorem B for 119899 ge 2
The discussions in Theorems C and 2 are concerning thecase that shared value 119886 = 0 When 119886 = 0 we obtain thefollowing result
Theorem 4 Let 119891 be a finite order entire function 119899 a positiveinteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share 0 CM then 1 le 120590(119891) le 120582(119891) + 1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
(6)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891) and 119876(119911)is a polynomial with deg119876 le 120590(119860) + 1
It is well known that if a finite order entire function 119891(119911)shares 119886 CM with Δ119899
120578119891(119911) then 119891(119911) satisfies the difference
equation
Δ119899
120578119891 (119911) minus 119886 = 119890
119876(119911)
(119891 (119911) minus 119886) (7)
where119876(119911) is a polynomialHence in order to prove the aboveresults we consider the value distribution of entire solutionsof the difference equation
119886119899(119911) 119891 (119911 + 119899120578) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 120578)
+ (1198860(119911) minus 119890
119876(119911)
) 119891 (119911) = 119861 (119911)
(8)
and obtain the following result
Theorem 5 Let 1198860 119886
119899minus1 119886119899( equiv 0) 119861( equiv 0) be polynomials
and let 119876 be a polynomial with degree 119898(ge 1) Then everyentire solution 119891 of finite order of (8) satisfies 120590(119891) ge 119898 and
(i) if 120590(119891) gt 1 then 120582(119891) = 120590(119891)(ii) if 120590(119891) = 1 then 120582(119891) = 120590(119891) or 119891 has only finitely
many zeros
2 Lemmas
Lemma 6 (see [12]) Let 119879 (0 +infin) rarr (0 +infin) be anondecreasing continuous function 119904 gt 0 120572 lt 1 and let119865 sub R+ be the set of all 119903 such that 119879(119903) le 120572119879(119903 + 119904) If thelogarithmic measure of 119865 is infinite then
lim119903rarrinfin
log119879 (119903)log 119903
= infin (9)
Lemma 7 (see [13]) Let 119891 be a nonconstant meromorphicfunction of finite order 120578 isin C 120575 lt 1 Then
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119900(
119879 (119903 +10038161003816100381610038161205781003816100381610038161003816 119891)
119903120575
) (10)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
(119889119903119903) lt infin
Remark 8 By Lemmas 6 and 7 we know that for a noncon-stant meromorphic function 119891 of finite order
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119878 (119903 119891) (11)
Lemma 9 (see [3]) Let 119891119895(119895 = 1 119899 + 1) and 119892
119895(119895 =
1 119899) be entire functions such that
(i) sum119899119895=1119891119895(119911)119890119892119895(119911)
equiv 119891119899+1(119911)
(ii) the order of 119891119895is less than the order of 119890119892119896 for 1 le 119895 le
119899+1 1 le 119896 le 119899 and furthermore the order of119891119895is less
than the order of 119890119892ℎminus119892119896 for 119899 ge 2 and 1 le 119895 le 119899+1 1 leℎ lt 119896 le 119899
Then 119891119895(119911) equiv 0 (119895 = 1 119899 + 1)
Abstract and Applied Analysis 3
Lemma 10 (see [14]) Let 119891 be a meromorphic function withfinite order 120590(119891) = 120590 lt 1 120578 isin C 0 Then for any given 120576 gt 0and integers 0 le 119895 lt 119896 there exists a set 119864 sub (1infin) of finitelogarithmic measure so that for all |119911| = 119903 notin 119864 ⋃ [0 1] wehave
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119896
120578119891 (119911)
Δ119895
120578119891 (119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|(119896minus119895)(120590minus1)+120576
(12)
Lemma 11 (see [15]) Let 1198860(119911) 119886
119896(119911) be entire functions
with finite order If there exists an integer 119897 (0 le 119897 le 119896) suchthat
120590 (119886119897) gt max0le119895le119896
119895 = 119897
120590 (119886119895) (13)
holds then every meromorphic solution 119891( equiv 0) of thedifference equation
119886119896(119911) 119891 (119911 + 119896) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 1) + 119886
0(119911) 119891 (119911) = 0
(14)
satisfies 120590(119891) ge 120590(119886119897) + 1
3 Proofs of Results
Proof of Theorem 5 Let 119891 be an entire solution of finite orderof (8) By Remark 8 and (8) we get
119879 (119903 119890119876
) = 119879 (119903 119890119876
minus 1198860) + 119878 (119903 119890
119876
)
le
119899
sum
119895=1
119898(119903
119891 (119911 + 119895120578)
119891 (119911)
)
+
119899
sum
119895=0
119898(119903 119886119895) + 119898(119903
119861 (119911)
119891 (119911)
) + 119878 (119903 119890119876
)
le 119879 (119903 119891) + 119878 (119903 119891) + 119878 (119903 119890119876
)
(15)
By (15) we get 120590(119891) ge 119898
Case 1 (120590(119891) gt 1) Suppose that 120582(119891) lt 120590(119891) by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that
120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891) lt 120590 (119891) = deg ℎ
2 (16)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (1198860(119911) minus 119890
119876(119911)
) ℎ1(119911) = 119861 (119911) 119890
minusℎ2(119911)
(17)
If deg ℎ2gt 119898 then by (16) we know that the order of the right
side of (17) is deg ℎ2 and the order of the left side of (17) is less
than deg ℎ2 This is a contradiction Hence deg ℎ
2= 119898 gt 1
Set119876 (119911) = 119887
119898119911119898
+ sdot sdot sdot + 1198870
ℎ2(119911) = 119888
119898119911119898
+ sdot sdot sdot + 1198880
(18)
where 119887119898( = 0) 119887
0 119888119898( = 0) 119888
0are complex numbers
By (17) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 1198860(119911) ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
+ 119861 (119911) 119890minusℎ2(119911)
(19)
Next we discuss the following two subcases
Subcase 1 (119887119898+ 119888119898=0)Then by Lemma 9 (16) and (19) we
get 119861(119911) equiv 0 ℎ1(119911) equiv 0 This is impossible
Subcase 2 (119887119898+ 119888119898= 0) Suppose that
ℎ1(119911) 119890119876(119911)minus119887
119898119911119898
+ 119861 (119911) 119890minusℎ2(119911)minus119887119898119911119898
equiv 0 (20)
Then ℎ1(119911) = minus119861(119911)119890
minusℎ2(119911)minus119876(119911) By 120590(ℎ
1) = 120582(ℎ
1) we
obtain that 119890minusℎ2(119911)minus119876(119911) is a nonzero constant Hence ℎ1(119911) is a
nonzero polynomial By (19) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
= minus1198860(119911) ℎ1(119911)
(21)
Since degℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578) = 119898 minus 1 gt 0 for 119894 = 119895 then
by Lemma 9 and (21) we get
119886119895(119911) ℎ1(119911 + 119895120578) equiv 0 (119895 = 0 1 119899) (22)
This is impossible Hence we have ℎ1(119911)119890119876(119911)minus119887
119898119911119898
+
119861(119911)119890minusℎ2(119911)minus119887119898119911119898
equiv 0 Then from the order considerationwe know that the order of the right side of (19) is 119898 andthe order of the left side of (19) is less than 119898 This is acontradiction Hence 120582(119891) = 120590(119891)
Case 2 (120590(119891) = 1)Then by 120590(119891) ge 119898 we get119898 = 1 Supposethat 119891(119911) has infinitely many zeros and 120582(119891) lt 120590(119891) by theWeierstrass factorization we get
119891 (119911) = ℎ3(119911) 119890120573119911
(23)
where 120573( = 0) is a complex number and ℎ3(119911)( equiv 0) is an
entire function such that
120590 (ℎ3) = 120582 (ℎ
3) = 120582 (119891) lt 1 (24)
Let 119876(119911) = 1198871119911 + 1198870 where 119887
1( = 0) 119887
0are complex numbers
Substituting 119891(119911) = ℎ3(119911)119890120573119911 into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ3(119911 + 119895120578) 119890
120573119895120578
+ 1198860(119911) ℎ3(119911)
= ℎ3(119911) 1198901198871119911+1198870+ 119861 (119911) 119890
minus120573119911
(25)
4 Abstract and Applied Analysis
Note that ℎ3(119911)1198901198870+ 119861(119911) equiv 0 otherwise 119891 has only finitely
many zeros If 1198871+ 120573 = 0 then the order of the right side of
(25) is 1 but the order of the left side of (25) is less than 1Thisis absurd If 119887
1+120573 = 0 then by Lemma 9 (24) and (25) we get
ℎ3(119911) equiv 0 119861(119911) equiv 0 This is impossible Hence 120582(119891) = 120590(119891)
Theorem 5 is thus completely proved
Proof of Theorem 2 Since 119891(119911) and Δ119899120578119891(119911) share 119886 CM and
119891 is of finite order then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119890119876(119911)
(26)
where 119876(119911) is a polynomial with deg119876 le 120590(119891) Now we willtake two steps to complete the proof
Step 1 We prove that 120582(119891 minus 119886) = 120590(119891)Let 119865(119911) = 119891(119911) minus 119886 then
120582 (119865) = 120582 (119891 minus 119886) 120590 (119865) = 120590 (119891) ge deg119876 (27)
and Δ119899120578119891(119911) = Δ
119899
120578119865(119911) = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119865(119911 + 119895120578) By thisand (26) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119890119876(119911)
) 119865 (119911) = 119886 (28)
Next we discuss the following three cases
Case 1 (deg119876 ge 1 and 120590(119865) gt deg119876) Then 120590(119865) gt 1 ByTheorem 5(i) (27) and (28) we get 120582(119891 minus 119886) = 120590(119891)
Case 2 (deg119876 ge 1 and 120590(119865) = deg119876) If 120590(119865) = deg119876 gt 1then byTheorem 5(i) (27) and (28) we get120582(119891minus119886) = 120590(119891) If120590(119865) = deg119876 = 1 then byTheorem 5(ii) and (27) we obtainthat 120582(119891 minus 119886) = 120590(119891) or 119865 has only finitely many zeros
If 119865 has only finitely many zeros set
119865 (119911) = ℎ1(119911) 119890119887119911
(29)
where ℎ1(119911)( equiv 0) is a polynomial and 119887( = 0) is a complex
number then substituting (29) into (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
119887119895120578
+ (minus1)119899
ℎ1(119911)
= ℎ1(119911) 119890119876(119911)
+ 119886119890minus119887119911
(30)
By (30) and Δ119899120578119865(119911) equiv 0 we know that the order of the left
side of (30) is 0 and the order of the right side of (30) is 1unless ℎ
1(119911) = minus119886 and119876(119911) = minus119887119911 In this case take it into the
left side of (30) we have (minus119886)(119890119887120578 minus 1)119899 = 0 Since all 119886 119887 and120578 are not zero it is impossible Hence we get 120582(119891minus119886) = 120590(119891)
Case 3 119876 is a complex constant Then by (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119888) 119865 (119911) = 119886 (31)
where 119888 (= 119890119876 =0) is a complex number Suppose that 120582(119865) lt120590(119865) Let 119865(119911) = ℎ
2(119911)119890ℎ3(119911) where ℎ
2(119911)( equiv 0) is an entire
function and ℎ3(119911) is a polynomial such that
120590 (ℎ2) = 120582 (ℎ
2) = 120582 (119865) lt 120590 (119865) = deg ℎ
3 (32)
Substituting 119865(119911) = ℎ2(119911)119890ℎ3(119911) into (31) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ2(119911 + 119895120578) 119890
ℎ3(119911+119895120578)minusℎ
3(119911)
+ ((minus1)119899
minus 119888) ℎ2(119911) = 119886119890
minusℎ3(119911)
(33)
Since deg(ℎ3(119911 + 119895120578) minus ℎ
3(119911)) = deg ℎ
3(119911) minus 1 (119895 = 1 119899)
by (32) we obtain that the order of the left side of (33) is lessthan deg ℎ
3and the order of the right side of (33) is deg ℎ
3
This is absurd Hence we get 120582(119891 minus 119886) = 120590(119891)
Step 2We prove that 120590(119891) ge 1Suppose that 120590(119891) lt 1 Since 119891(119911) and Δ119899
120578119891(119911) share 119886
CM then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119888 (34)
where 119888 is a nonzero constant Let 119865(119911) = 119891(119911) minus 119886 then by(34) we get
Δ119899
120578119865 (119911) = 119888119865 (119911) + 119886 (35)
Differentiating (35) we get
(Δ119899
120578119865 (119911))
1015840
= 1198881198651015840
(119911) (36)
Note that (Δ119899120578119865(119911))1015840
= Δ119899
120578(1198651015840
(119911)) and 120590(1198651015840) = 120590(119865) = 120590(119891) lt1 So by Lemma 10 and (36) we get
|119888| =
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119899
120578(1198651015840
(119911))
1198651015840(119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|119899(120590(119865)minus1)+120576
997888rarr 0 (37)
This is absurd So 120590(119891) ge 1 Theorem 2 is thus completelyproved
Proof of Theorem 4 Since 119891(119911) and Δ119899120578119891(119911) share 0 CM and
119891 is of finite order then
Δ119899
120578119891 (119911)
119891 (119911)
= 119890119876(119911)
(38)
where119876(119911) is a polynomial ByΔ119899120578119891 = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119891(119911+
119895120578) and (38) we get
119891 (119911 + 119899120578)
+
119899minus1
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119891 (119911 + 119895120578)
+ ((minus1)119899
minus 119890119876(119911)
) 119891 (119911) = 0
(39)
Abstract and Applied Analysis 5
We discuss the following two cases
Case 1 119876 is a polynomial with deg119876 = 119898 ge 1 Then byLemma 11 and (39) we get 120590(119891) ge 119898 + 1 Now we prove120590(119891) le 120582(119891) + 1 Suppose that 120590(119891) gt 120582(119891) + 1 then by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891)
120590 (119891) = deg ℎ2gt 120590 (ℎ
1) + 1
(40)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (39) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (minus1)119899
ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
(41)
If 120590(119891) gt 119898 + 1 then by (40) (41) and deg(ℎ2(119911 + 119895120578) minus
ℎ2(119911)) = deg ℎ
2(119911)minus1 (119895 = 1 119899) we obtain that the order
of the left side of (41) is deg ℎ2minus 1 and the order of the right
side of (41) is less than deg ℎ2minus 1 This is absurd
If 120590(119891) = 119898 + 1 then by (41) we get119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
minus ℎ1(119911) 119890119876(119911)
= (minus1)119899+1
ℎ1(119911)
(42)
Set
ℎ2(119911) = 119889
119898+1119911119898+1
+ sdot sdot sdot + 1198890 (43)
where 119889119898+1( = 0) 119889
0are complex numbers Then
ℎ2(119911 + 119895120578) minus ℎ
2(119911)
= 119889119898+1
(119898 + 1) 119895120578119911119898
+ sdot sdot sdot + 1198891119895120578
(119895 = 1 119899)
(44)
Now we discuss the following two subcases
Subcase 1 deg(119876(119911)minus(ℎ2(119911+119895120578)minusℎ
2(119911))) = 119898 holds for every
119895 isin 1 119899Then by (40) (42) deg(ℎ2(119911+119895120578)minusℎ
2(119911+119894120578)) =
119898 (119895 = 119894) and Lemma 9 we get ℎ1(119911) equiv 0 This is absurd
Subcase 2 There exist some 1198950isin 1 119899 such that
deg(119876(119911)minus(ℎ2(119911+1198950120578)minusℎ2(119911))) le 119898minus1Then by (44) we have
deg(119876(119911) minus (ℎ2(119911 + 119895120578) minus ℎ
2(119911))) = 119898 for 119895 = 119895
0 Merging the
term minusℎ1(119911)119890119876(119911) into ( 119899119895
0) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578)119890ℎ2(119911+1198950120578)minusℎ2(119911)
by (42) we get119899
sum
119895 = 1
119895 =1198950
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 ge 2)
(45)
or
119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 = 1) (46)
where 119860(119911) = (119899
1198950) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578) minus ℎ
1(119911)
119890119876(119911)minus(ℎ
2(119911+1198950120578)minusℎ2(119911)) satisfying 120590(119860) lt 119898 If 119899 ge 2 then
by (40) (45) deg(ℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578)) = 119898 (119895 = 119894) and
Lemma 9 we get ℎ1(119911) equiv 0 This is absurd If 119899 = 1 then by
(46) and ℎ1(119911) equiv 0 we get 119860(119911) equiv 0 By this we know that
the order of the left side of (46) is 119898 and the order of theright side of (46) is less than119898 This is absurd Hence we get120590(119891) le 120582(119891) + 1
Case 2119876 is a complex constantThen by Lemma 10 and (38)we get 120590(119891) ge 1 Now we prove 120590(119891) le 120582(119891) + 1 Supposethat 120590(119891) gt 120582(119891) + 1 If 120590(119891) gt 1 then by the similarargument to that of case 1 we get ℎ
1(119911) equiv 0 This is absurd If
120590(119891) = 1 then by (40) we get 0 le 120582(119891) lt 120590(119891) minus 1 = 0 Since120582(119891) = 0 then 120590(119891) = 120582(119891)+1Theorem 4 is thus completelyproved
4 Some Examples
The following examples show the existence of such entirefunctions which satisfy Theorems 2ndash5 Moreover Example 2shows that the result in Theorem 4 is the best possible
Example 1 Let 120578 = 1 119899 = 2 and 119891(119911) = (119889 + 1)119911 + ((1198892 minus1)1198892
)119886 where 119886( = 0) 119889( = 0 plusmn1) are constants Then 119891(119911)and Δ119899
120578119891(119911) share 119886 CM and 120590(119891) = 120582(119891 minus 119886) = 1
Example 2 Let 120578 = 1 119899 = 2 and 119891(119911) = 119890119911 Then 119891(119911) andΔ119899
120578119891(119911) share 0 CM and 120590(119891) = 1 = 120582(119891) + 1
Example 3 Let 120578 = 1 119899 = 2 and119891(119911) = 119867(119911)119890119911 where119867(119911)is an entire function with period 1 such that 120590(119867) gt 1 and120590(119867) notin N Then 119891(119911) and Δ119899
120578119891(119911) share 0 CM and 120582(119891) =
120582(119867) = 120590(119867) = 120590(119891) gt 1 (Ozawa [16] proved that for any120590 isin [1infin) there exists a period entire function of order 120590)
Example 4 The entire function 119891(119911) = 119911119890minus119911 satisfies the
difference equation
119891 (119911 + 2120578) minus 4119891 (119911 + 120578) + (4 minus 119890119911
) 119891 (119911) = minus119911 (47)
where 120578 = minus log 2 Here120590(119891) = 1 and119891 has only finitelymanyzeros
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11201195 11171119) and the NaturalScience Foundation of Jiangxi China (nos 20122BAB20101220132BAB201008)
6 Abstract and Applied Analysis
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers New York NY USA2003
[4] R Nevanlinna Le Theoreme de Picard-Borel et la Theorie desFonctions Meromorphes Gauthiers-Villars Paris France 1929
[5] G G Gundersen ldquoMeromorphic functions that share fourvaluesrdquo Transactions of the American Mathematical Society vol277 no 2 pp 545ndash567 1983
[6] R Bruck ldquoOn entire functions which share one value CM withtheir first derivativerdquo Results inMathematics vol 30 no 1-2 pp21ndash24 1996
[7] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[8] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[9] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables andElliptic Equations vol 56 pp 81ndash92 2011
[10] K Liu and L-Z Yang ldquoValue distribution of the differenceoperatorrdquo Archiv der Mathematik vol 92 no 3 pp 270ndash2782009
[11] S Li and Z Gao ldquoA note on the Bruck conjecturerdquo Archiv derMathematik vol 95 no 3 pp 257ndash268 2010
[12] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[13] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[14] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[15] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[16] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleAdmissible Solutions of the Schwarzian Type DifferenceEquation
Baoqin Chen and Sheng Li
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to Sheng Li lish lssinacom
Received 14 January 2014 Accepted 20 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 B Chen and S Li This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper is to investigate the Schwarzian type difference equation [(Δ3
119891Δ119891) minus (32) (Δ2
119891Δ119891)
2
]
119896
= 119877 (119911 119891) =
(119875(119911 119891)119876(119911 119891)) where 119877(119911 119891) is a rational function in 119891 with polynomial coefficients 119875(119911 119891) respectively 119876(119911 119891) are twoirreducible polynomials in 119891 of degree 119901 respectively 119902 Relationship between 119901 and 119902 is studied for some special case Denote119889 = max 119901 119902 Let 119891(119911) be an admissible solution of (lowast) such that 120588
2(119891) lt 1 then for 119904 (ge2) distinct complex constants 120572
1 120572
119904
119902 + 2119896sum119904
119895=1120575(120572119895 119891) le 8119896 In particular if119873(119903 119891) = 119878(119903 119891) then 119889 + 2119896sum119904
119895=1120575(120572119895 119891) le 4119896
1 Introduction and Results
Throughout this paper a meromorphic function alwaysmeans being meromorphic in the whole complex plane and119888 always means a nonzero constant For a meromorphicfunction 119891(119911) we define its shift by 119891(119911 + 119888) and define itsdifference operators by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911) Δ
119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911))
119899 isin N 119899 ge 2
(1)
In particular Δ119899119888119891(119911) = Δ
119899
119891(119911) for the case 119888 = 1 We usestandard notations of theNevanlinna theory ofmeromorphicfunctions such as 119879(119903 119891) 119898(119903 119891) and 119873(119903 119891) and as statedin [1ndash3] For a constant 119886 we define theNevanlinna deficiencyby
120575 (119886 119891) = lim inf119903rarrinfin
119898(119903 1 (119891 minus 119886))
119879 (119903 119891)
= 1 minus lim sup119903rarrinfin
119873(119903 1 (119891 minus 119886))
119879 (119903 119891)
(2)
Recently numbers of papers (see eg [4ndash12]) are devotedto considering the complex difference equations and differ-ence analogues of Nevanlinna theory Due to some idea of[13] we consider the admissible solution of the Schwarziantype difference equation
119878119896(119891) = [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(3)
where 119877(119911 119891) is a rational function in 119891 with polynomialcoefficients 119875(119911 119891) respectively119876(119911 119891) are two irreduciblepolynomials in 119891 of degree 119901 respectively 119902 Here and in thefollowing ldquoadmissiblerdquo always means ldquotranscendentalrdquo Andwe denote 119889 = max119901 119902 from now on For the existence ofsolutions of (3) we give some examples below
Examples (1) 119891(119911) = sin 120587119911 + 119911 is an admissible solution ofthe Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus8 [1198912
+ (1 minus 2119911) 119891 + 119911 (119911 minus 1)]
41198912minus 4 (2119911 + 1) 119891 + (2119911 + 1)
2
(4)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 306360 5 pageshttpdxdoiorg1011552014306360
2 Abstract and Applied Analysis
(2) 119891(119911) = (119890119911 ln 2 sin 2120587119911) + 119911 is an admissible solutionof the Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus1198912
+ 2 (119911 + 1) 119891 minus 1199112
minus 2119911
21198912minus 4 (119911 minus 1) 119891 + 2(119911 minus 1)
2 (5)
(3) Let 119891(119911) = 1199112
+ 119911 then 119891(119911) solves the Schwarziantype difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
= minus
3
2 [1198912minus 2 (119911
2minus 1) 119891 + (119911
2minus 1)2
]
(6)
This example shows that (3) may admit polynomial solutions
Considering the relationship between 119901 and 119902 in thoseexamples above we prove the following result
Theorem 1 For the Schwarzian type difference equation (3)with polynomial coefficients note the following
(i) If it admits an admissible solution 119891(119911) such that1205882(119891) lt 1 then
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (7)
In particular if119898(119903 119891) = 119878(119903 119891) then 119901 le 119902(ii) If its coefficients are all constants and it admits a
polynomial solution 119891(119911) with degree 119904 then 119904 ge 2 and119902119904 = 119901119904 + 2119896
Remark 2 From examples (1) and (2) we conjecture that119901 = 119902 inTheorem 1(i) However we cannot prove it currentlyFrom example (3) given before we see that the restriction onthe coefficients in Theorem 1(ii) cannot be omitted
For the Schwarzian differential equation
119878119896(119891) = [
119891101584010158401015840
1198911015840
minus
3
2
(
11989110158401015840
1198911015840
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(8)
where 119877(119911 119891) 119875(119911 119891) and 119876(119911 119891) are as stated beforeIshizaki [13] proved the following result (see also Theorem932 in [2])
Theorem A (see [2 13]) Let 119891(119911) be an admissible solutionof (8) with polynomial coefficients and let 120572
1 120572
119904be 119904 (ge2)
distinct complex constants Then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (9)
For the Schwarzian type difference equation (3) we provethe following result
Theorem 3 Let 119891(119911) be an admissible solution of (3) withpolynomial coefficients such that 120588
2(119891) lt 1 and let 120572
1 120572
119904
be 119904 (ge2) distinct complex constants Then
119902 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 8119896 (10)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (11)
Remark 4 From Theorem 1 under the condition 119873(119903 119891) =119878(119903 119891) in Theorem 3 we have 119889 = 119902 in (11) The behavior ofthe zeros and the poles of 119891(119911) in 119878
119896(119891) is essentially different
from that in the 119878119896(119891) We wonder whether the restriction
119873(119903 119891) = 119878(119903 119891) can be omitted or not
2 Lemmas
The following lemmaplays a very important role in the theoryof complex differential equations and difference equationsIt can be found in Mohonrsquoko [14] and Valiron [15] (see alsoTheorem 225 in the book of Laine and Yang [2])
Lemma 5 (see [14 15]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(12)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(13)
and the characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (14)
where 119889 = max119901 119902
The following two results can be found in [10] In factLemma 6 is a special case of Lemma 83 in [10]
Lemma 6 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891) (15)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 7 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120589minus120576
) = 119878 (119903 119891) (16)
possibly outside of a set of 119903 with finite logarithmic measure
From Lemma 7 we can easily get the following conclu-sion
Abstract and Applied Analysis 3
Lemma 8 Let 119891(119911) be a meromorphic function of hyper order1205882(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
Δ119899
119888119891 (119911)
119891 (119911)
) = 119878 (119903 119891)
119898(119903
Δ119896
119888119891 (119911)
Δ119895
119888119891 (119911)
) = 119878 (119903 119891) 119896 gt 119895
(17)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 9 Let 119891 be an admissible solution of (3) withcoefficients Then using the notation 119876(119911) = 119876(119911 119891(119911))
119902119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (18)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (19)
Proof We use the idea by Ishizaki [13] (see also [2]) to proveLemma 9 It follows from Lemma 8 that
119898(119903 119877) = 119898(119903 [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
)
le 119896119898(119903
Δ3
119891
Δ119891
) + 2119896119898(119903
Δ2
119891
Δ119891
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(20)
From this and Lemma 5 we get
119889119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119877) = 119873 (119903 119877) + 119878 (119903 119891) (21)
and hence
119889119879 (119903 119891) = 119873 (119903 119877) + 119878 (119903 119891) (22)
If 119889 = 119901 gt 119902 since all coefficients of 119875(119911 119891) and 119876(119911 119891)are polynomials there are at the most finitely many poles of119877(119911 119891) neither the poles of 119891(119911) nor the zeros of 119876(119911 119891)Therefore we see that
119873(119903 119877) le (119901 minus 119902)119873 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
le (119901 minus 119902) 119879 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
(23)
We obtain (18) from this and (22) immediatelyIf 119889 = 119902 ge 119901 there are at most finitely many poles of
119877(119911 119891) not the zeros of 119876(119911 119891) then
119873(119903 119877) le 119873(119903
1
119876
) + 119878 (119903 119891) (24)
Now (18) follows from (22) and (24)Notice that if 119873(119903 119891) = 119878(119903 119891) then (24) always holds
This finishes the proof of Lemma 9
3 Proof of Theorem 1
Case 1 Equation (3) admits an admissible solution 119891(119911) suchthat 1205882(119891) lt 1 Since all coefficients of119875(119911 119891) and119876(119911 119891) are
polynomials there are at the most finitely many poles of 119891(119911)that are not the poles of119875(119911 119891) and119876(119911 119891) This implies that
119873(119903 119875) = 119901119873 (119903 119891) + 119878 (119903 119891)
119873 (119903 119876) = 119902119873 (119903 119891) + 119878 (119903 119891)
(25)
From Lemma 5 we get
119879 (119903 119875) = 119901119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 119876) = 119902119879 (119903 119891) + 119878 (119903 119891)
(26)
We can deduce from (3) (25) (26) and Lemma 8 that
119901119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119875)
= 119898 (119903 119875) + 119873 (119903 119875)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891)119876)
+ 119878 (119903 119891)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891))
+ 119898 (119903 119876) + 119878 (119903 119891)
= 119901119873 (119903 119891) + 119879 (119903 119876) minus 119873 (119903 119876)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119879 (119903 119891) minus 119902119873 (119903 119891)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119898 (119903 119891) + 119878 (119903 119891)
(27)
It follows from this that
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (28)
What is more is that if119898(119903 119891) = 119878(119903 119891) then we obtain from(28) that 119901 le 119902
Case 2 The coefficients of (3) are all constants and it admitsa polynomial solution 119891(119911) with degree 119904 Set
119891 (119911) = 119886119904119911119904
+ 119886119904minus1119911119904minus1
+ sdot sdot sdot + 1198861119911 + 1198860 (29)
then
119891 (119911 + 1) = 119886119904119911119904
+ 119887119904minus1119911119904minus1
+ sdot sdot sdot + 1198871119911 + 1198870 (30)
where
119887119904minus119895
= 119886119904119862119895
119904+ 119886119904minus1119862119895minus1
119904minus1+ sdot sdot sdot + 119886
119904minus119895+11198621
119904minus119895+1+ 119886119904minus119895 (31)
From (29) and (30) we obtain that
Δ119891 = 119904119886119904119911119904minus1
+ (119887119904minus2
minus 119886119904minus2) 119911119904minus2
+ sdot sdot sdot + (1198871minus 1198861) 119911 + (119887
0minus 1198860)
(32)
4 Abstract and Applied Analysis
If 119904 = 1 then Δ2119891 = Δ3119891 equiv 0 which yields that 119875(119911 119891) equiv0 That is a contradiction to our assumption Thus 119904 ge 2
If 119904 = 2 thenΔ119891 = 21198862119911+1198862+1198861Δ2119891 = 2119886
2 andΔ3119891 equiv 0
Now from (3) we get
(minus3)119896
119876 (119911 119891) (Δ2
119891)
2119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(33)
Considering degrees of both sides of the equation above wecan see that 119902 = 119901 + 119896
If 119904 ge 3 we can deduce similarly that
Δ2
119891 = 119904 (119904 minus 1) 119886119904119911119904minus2
+ 1198751(119911)
Δ3
119891 = 119904 (119904 minus 1) (119904 minus 2) 119886119904119911119904minus3
+ 1198752(119911)
(34)
where 1198751(119911) 1198752(119911) are polynomials such that deg119875
1le 119904 minus
3 deg1198752le 119904 minus 4
Rewrite (3) as follows
119876 (119911 119891) [2Δ3
119891 sdot Δ119891 minus 3(Δ2
119891)
2
]
119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(35)
From (34) we find that the leading coefficient of 2Δ3119891 sdotΔ119891 minus 3(Δ
2
119891)
2 is
minus1198862
1199041199042
(119904 minus 1) (119904 + 1) = 0 (36)
Considering degrees of both sides of (35) we prove that119902119904 = 119901119904 + 2119896
4 Proof of Theorem 3
Firstly we consider the general case Asmentioned inRemark1 in [13] due to Jank and Volkmann [16] if (3) admits anadmissible solution then there are at most 119878(119903 119891) commonzeros of 119875(119911 119891) and 119876(119911 119891) Since all coefficients of 119876(119911 119891)are polynomials there are at the most finitely many poles of119891 that are the zeros of 119876(119911 119891) Therefore from (3) we have
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119879 (119903 119891 (119911 + 1) minus 119891 (119911)) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119878 (119903 119891)
(37)
Combining this and Lemma 9 applying the second maintheorem we get
119902
2119896
119879 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
119902
2119896
119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
1
2119896
119873(119903
1
119876
) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 4119879 (119903 119891) + 119878 (119903 119891)
(38)
Thus we prove that (10) holdsSecondly we consider the case that 119873(119903 119891) = 119878(119903 119891)
From (3) and Lemma 8 we similarly get that
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119898 (119903 Δ119891) + 119873 (119903 Δ119891) + 119878 (119903 119891)
le 119898(119903
Δ119891
119891
) + 119898 (119903 119891) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
(39)
From this and applying Lemma 9with (19) as arguing beforewe can prove that (11) holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the referees for their valuablesuggestions This work is supported by the NNSFC (nos11226091 and 11301091) the Guangdong Natural ScienceFoundation (no S2013040014347) and the Foundation forDistinguished Young Talents in Higher Education of Guang-dong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic PublishersGroupDordrechtTheNetherlands 2003
[4] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[5] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
Abstract and Applied Analysis 5
[8] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[9] R G Halburd and R J Korhonen ldquoExistence of finite-ordermeromorphic solutions as a detector of integrability in differ-ence equationsrdquo Physica D Nonlinear Phenomena vol 218 no2 pp 191ndash203 2006
[10] R G Halburd R J Korhonen and K Tohge ldquoHolomorphiccurves with shift-invariant hyperplane preimagesrdquo submitted toTransactions of the American Mathematical Society httparxivorgabs09033236
[11] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[12] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] K Ishizaki ldquoAdmissible solutions of the Schwarzian differentialequationrdquoAustralianMathematical SocietyA PureMathematicsand Statistics vol 50 no 2 pp 258ndash278 1991
[14] A Z Mohonrsquoko ldquoThe nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija no 14 pp 83ndash87 1971 (Russian)
[15] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Entomologique de France vol 59 pp 17ndash39 1931
[16] G Jank and L VolkmannMeromorphe Funktionen und Differ-entialgeichungen Birkhauser Verlag Basel Switzerland 1985
Research ArticleStatistical Inference for Stochastic DifferentialEquations with Small Noises
Liang Shen12 and Qingsong Xu1
1 School of Mathematics and Statistics Central South University Changsha Hunan 410075 China2 School of Science Linyi University Linyi Shandong 276005 China
Correspondence should be addressed to Qingsong Xu csuqingsongxu126com
Received 13 November 2013 Accepted 11 February 2014 Published 13 March 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 L Shen and Q Xu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper proposes the least squares method to estimate the drift parameter for the stochastic differential equations driven bysmall noises which is more general than pure jump 120572-stable noises The asymptotic property of this least squares estimator isstudied under some regularity conditions The asymptotic distribution of the estimator is shown to be the convolution of a stabledistribution and a normal distribution which is completely different from the classical cases
1 Introduction
Stochastic differential equations (SDEs) are being extensivelyused as a model to describe some phenomena which aresubject to random influences it has found many applicationsin biology [1] medicine [2] econometrics [3 4] finance[5] geophysics [6] and oceanography [7] Then statisticalinference for these differential equations was of great interestand became a challenging theoretical problem For a morerecent comprehensive discussion we refer to [8 9]
The asymptotic theory of parametric estimation for dif-fusion processes with small white noise based on continuoustime observations is well developed and it has been studiedby many authors (see eg [10ndash14]) There have been manyapplications of small noise in mathematical finance see forexample [15ndash18]
In parametric inference due to the impossibility ofobserving diffusions continuously throughout a time intervalit is more practical and interesting to consider asymptoticestimation for diffusion processes with small noise based ondiscrete observations There are many approaches to driftestimation for discretely observed diffusions (see eg [19ndash23]) Long [24] has started the study on parameter estimationfor a class of stochastic differential equations driven by smallstable noise 119885
119905 119905 ge 0 However there has been no study on
parametric inference for stochastic processes with small Levynoises yet
In this paper we are interested in the study of parameterestimation for the following stochastic differential equationsdriven by more general Levy noise 119871
119905 119905 ge 0 based
on discrete observations We will employ the least squaresmethod to obtain an asymptotically consistent estimator
Let (ΩF F119905ge0
P) be a basic complete filtered prob-ability space satisfying the usual conditions that is thefiltration is continuous on the right and F
0contains all
P-null sets In this paper we consider a class of stochasticdifferential equations as follows
119889119883119905
= 120579119891 (119883119905) 119889119905 + 120576119892 (119883
119905minus
) 119889119871119905 119905 isin [0 1]
119871119905
= 119886119861119905+ 119887119885119905
119883 (0) = 1199090
(1)
where 119891 R rarr R and 119892 R rarr R are known functionsand 119886 119887 are known constants Let 119861
119905 119905 ge 0 be a standard
Brownian motion and let 119885119905 119905 ge 0 be a standard 120572-stable
Levy motion independent of 119861119905 119905 ge 0 with 119885
1sim 119878120572(1 120573 0)
for 120573 isin [0 1] 1 lt 120572 lt 2Let 119883 = 119883
119905 119905 ge 0 be a real-valued stationary process
satisfying the stochastic differential equation (1) and weassume that this process is observed at regularly spaced timepoints 119905
119894= 119894119899 119894 = 1 2 119899 Assume 119883
0
119905is the solution of
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 473681 6 pageshttpdxdoiorg1011552014473681
2 Abstract and Applied Analysis
the underlying ordinary differential equation (ODE) with thetrue value of the drift parameter 120579
0
1198891198830
119905= 1205790119891 (1198830
119905) 119889119905 119883
0
0= 1199090 (2)
Then we get
119883119905119894
minus 119883119905119894minus1
= int
119905119894
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (3)
2 Preliminaries
In this paper we denote 119862 as a generic constant whose valuemay vary from place to place
The following regularity conditions are assumed to hold
(A1) The functions 119891(119909) and 119892(119909) satisfy the Lipschitzconditions that is there exists a constant 119871 gt 0 suchthat
1003816100381610038161003816119891 (119909) minus 119891 (119910)
1003816100381610038161003816+
1003816100381610038161003816119892 (119909) minus 119892 (119910)
1003816100381610038161003816
le 1198711003816100381610038161003816119909 minus 119910
1003816100381610038161003816 119909 119910 isin R
(4)
(A2) There exist constants 119872 gt 0 and 119903 ge 0 satisfying
the growth condition
119892minus2
(119909) le 119872 (1 + |119909|119903
) 119909 isin R (5)
(A3) There exists a positive constant 119873 gt 0 such that
0 lt |119892(119909)| le 119873 lt infin
(A4) For 119862
119903= 2119903minus1
or 1 119903 gt 0
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
le 119862119903(
100381610038161003816100381610038161198830
119905119894minus1
10038161003816100381610038161003816
119903
+
10038161003816100381610038161003816119883119905119894minus1
minus 1198830
119905119894minus1
10038161003816100381610038161003816
119903
) (6)
The LSE of 120579119899120576
is defined as
120579119899120576
= argmin120579
120588119899120576
(120579) (7)
where the contrast function
120588119899120576
(120579) =
119899
sum
119894=1
10038161003816100381610038161003816100381610038161003816100381610038161003816
119883119905119894
minus 119883119905119894minus1
minus 120579119891 (119883119905119894minus1
) Δ119905119894minus1
120576119892 (119883119905119894minus1
)
10038161003816100381610038161003816100381610038161003816100381610038161003816
2
(8)
Then the 120579119899120576
can be represented explicitly as follows
120579119899120576
=
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) (119883119905119894
minus 119883119905119894minus1
)
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
(9)
Based on (3) and (9) there is a special decomposition for 120579119899120576
120579119899120576
=
1205790
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119891 (119883119904) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119887120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119885119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119886120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
= 1205790
+
Φ2
(119899 120576)
Φ1
(119899 120576)
+
Φ3
(119899 120576)
Φ1
(119899 120576)
+
Φ4
(119899 120576)
Φ1
(119899 120576)
(10)
Now we give an explicit expression for 120576minus1
(120579119899120576
minus 1205790) By using
(10) we have
120576minus1
(120579119899120576
minus 1205790) =
120576minus1
Φ2
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ3
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ4
(119899 120576)
Φ1
(119899 120576)
=
Ψ2
(119899 120576)
Φ1
(119899 120576)
+
Ψ3
(119899 120576)
Φ1
(119899 120576)
+
Ψ4
(119899 120576)
Φ1
(119899 120576)
(11)
One of the important tools we will employ is the under-lying lemma (see (35) in the Lemma 32 of [24])
Lemma 1 Under conditions (A1)-(A2) one has
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816
le 120576119890119871|1205790|119905
10038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119883119904minus
) 119889119885119904
10038161003816100381610038161003816100381610038161003816
(12)
sup0le119905le1
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816997888rarr1198750 as 120576 997888rarr 0 (13)
Abstract and Applied Analysis 3
3 Asymptotic Property of the LeastSquares Estimator
Theorem 2 Under the conditions (A1)ndash(A4) as 119899 rarr
infin 120576 rarr 0 119899120576 rarr infin and 119899120576120572(120572minus1)
rarr infin one has
120576minus1
(120579119899120576
minus 1205790)
997904rArr 119886
(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904
119873
+ 119887 (((int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus (int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802)
times (int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
minus1
)
(14)
where 1198801and 119880
2are independent random variables with 120572-
stable distribution 119878120572(1 120573 0) and 119873 is an independent random
variable with standard normal distribution
The theoremwill be proved by establishing several propo-sitionsWewill consider the asymptotic behaviors ofΦ
1(119899 120576)
Ψ119894(119899 120576) 119894 = 2 3 4 respectively
Proposition 3 Under conditions (A1)ndash(A4) and 119899 rarr infin
120576 rarr 0 one has
Φ1
(119899 120576) 997888rarr119875
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904 (15)
Proof Under conditions (A1)ndash(A3) Proposition 3 can be
proved by using condition (A4) (see the proof of Proposition
33 in [24])
Proposition 4 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576 rarr infin one has
Ψ2
(119899 120576) 997888rarr1198750 (16)
Proof For 119905119894minus1
le 119905 le 119905119894 119894 = 1 2 119899
119883119905
= 119883119905119894minus1
+ int
119905
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (17)
It follows that10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le int
119905
119905119894minus1
10038161003816100381610038161205790
1003816100381610038161003816(
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816) 119889119904
+ 120576
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904
100381610038161003816100381610038161003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816119872 int
119905
119905119894minus1
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+ 119899minus1 1003816
1003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+ 119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
(18)
Using Gronwall inequality we get10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872(119905minus119905
119894minus1)
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(19)
which yields
sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872119899
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(20)
thus under conditions (A1) and (A
3)
1003816100381610038161003816Φ2
(119899 120576)1003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816
119899
sum
119894=1
119872 (1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
(119891 (119883119904) minus 119891 (119883
119905119894minus1
)) 119889119904
100381610038161003816100381610038161003816100381610038161003816
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
119899
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
4 Abstract and Applied Analysis
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
1198992
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
2
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890(|1205790|119872)119899
119899
119887120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
119899
119886120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
= Φ21
(119899 120576) + Φ22
(119899 120576) + Φ23
(119899 120576)
(21)
Then
1003816100381610038161003816Ψ2
(119899 120576)1003816100381610038161003816
le 120576minus1
Φ21
(119899 120576) + 120576minus1
Φ22
(119899 120576)
+ 120576minus1
Φ23
(119899 120576)
= Ψ21
(119899 120576) + Ψ22
(119899 120576) + Ψ23
(119899 120576)
(22)
Using (13) in Lemma 1 conditions (A1) and (A
4) we get
Ψ21
(119899 120576) rarr1198750 as 119899 rarr infin 120576 rarr 0 and 119899120576 rarr infin (see (326)
in [24]) By using the same techniques under condition (A2)
we can prove thatΨ2119895
(119899 120576) rarr1198750 119895 = 2 3 as 119899 rarr infin 120576 rarr 0
respectively
Proposition 5 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576120572(120572minus1)
rarr infin one has
Ψ3
(119899 120576)
997904rArr 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802
(23)
Proof Under conditions (A1)ndash(A3) Proposition 5 can be
proved by using condition (A4) (see the proof of Proposition
44 in [24])
Proposition 6 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 one has
Ψ4
(119899 120576) 997904rArr 119886(int
1
0
10038161003816100381610038161003816119892minus2
(1198830
119904)
100381610038161003816100381610038161198912
(1198830
119904) 119889119904)
12
119873 (24)
Proof Note that
Ψ4
(119899 120576) = 119886
119899
sum
119894=1
119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) int
119905119894
119905119894minus1
119892 (1198830
119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
)) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
))
times (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
=
5
sum
119895=1
Ψ4119895
(119899 120576)
(25)
For Ψ41
(119899 120576) let 119884119894
= int
119905119894
119905119894minus1
119892(119883119904minus
)119889119861119904 119895 = 1 119899 Then it is
easy to see that 119884119894
sim 119873(0 int
119905119894
119905119894minus1
1198922
(119883119904minus
)119889119904) and 1198841 119884
119899are
independent normal random variablesIt follows that
Ψ41
(119899 120576)
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) 119884119894
sim 119873 (0 1198862
119899
sum
119894=1
119892minus4
(1198830
119905119894minus1
) 1198912
(1198830
119905119894minus1
) int
119905119894
119905119894minus1
1198922
(119883119904minus
) 119889119904)
997904rArr 119886(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
119873
(26)
as 119899 rarr infin 120576 rarr 0
Abstract and Applied Analysis 5
For Ψ42
(119899 120576) using Markov inequality and Itorsquos isometryproperty for any given 120578 gt 0
Ψ42
(119899 120576)
le
1
120578
E[119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
]
le
119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
))
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119883119904minus
minus 1198830
119904minus
)
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [ sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119904minus
minus 1198830
119904minus
10038161003816100381610038161003816119899minus12
]
(27)
By using (13) Ψ42
(119899 120576) rarr 0 as 119899 rarr infin 120576 rarr 0Applying similar techniques to Ψ
4119895(119899 120576) 119895 = 3 4 5 we
get Ψ4119895
(119899 120576) rarr 0 119895 = 3 4 5 as 119899 rarr infin 120576 rarr 0
Now we can proveTheorem 2
Proof By using Propositions 3 4 5 6 and Slutskyrsquos theoremwe can get the conclusion
4 Example
We consider the following nonlinear SDE driven by generalLevy noises
119889119883119905
= 120579119883119905119889119905 +
120576
1 + 1198832
119905minus
119889119871119905 119905 isin [0 1] 119883
0= 1199090
(28)
where 119891(119909) = 119909 119892(119909) = 1(1 + 1199092
) 1199090and 120576 are known
constants and 120579 = 0 is an unknown parameterFor simplicity let 119909
0gt 0 120576 = 0 we get the ODE
1198891198830
119905= 12057901198830
119905119889119905 119905 isin [0 1] 119883
0
0= 1199090
(29)
and the solution
1198830
119905= 11990901198901205790119905
(30)
Then the asymptotic distribution is
119886(int
1
0
(1 + 1199092
011989021205790119904
)
2
1199092
011989021205790119904
119889119904)
minus12
119873
+ 119887
(int
1
0
(1 + 1199092
011989021205790119904
)
120572
119909120572
01198901205721205790119904
119889119904)
1120572
int
1
0
(1 + 1199092
011989021205790119904)2
1199092
011989021205790119904119889119904
119878120572
(1 120573 0)
(31)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R I Jennrich and P B Bright ldquoFitting systems of linear dif-ferential equations using computer generated exact derivativesrdquoTechnometrics vol 18 no 4 pp 385ndash392 1976
[2] R H Jones ldquoFitting multivariate models to unequally spaceddatardquo in Time Series Analysis of Irregularly Observed Data pp158ndash188 1984
[3] A R Bergstrom Statistical Inference in Continuous TimeEconomic Models vol 99 North-Holland Amsterdam TheNetherlands 1976
[4] A R Bergstrom ldquoThe history of continuous-time econometricmodelsrdquo Econometric Theory vol 4 no 3 pp 365ndash383 1988
[5] F Black and M Scholes ldquoThe pricing of options and corporateliabilitiesrdquoThe Journal of Political Economy vol 81 pp 637ndash6541973
[6] M Arato Linear Stochastic Systems with Constant CoefficientsSpringer Berlin Germany 1982
[7] R J Adler and P Meuller Stochastic Modelling on PhysicalOceanography vol 39 Springer New York NY USA 1996
[8] B P Rao B L P Rao I Statisticien B L P Rao B L P Rao andI Statistician Statistical Inference for di Usion Type ProcessesArnold London UK 1999
[9] Y A Kutoyants Statistical Inference for Ergodic Diffusion Pro-cesses Springer London UK 2004
[10] Yu A Kutoyants Parameter Estimation for Stochastic Processesvol 6 Heldermann Berlin Germany 1984
[11] Yu Kutoyants Identification of Dynamical Systems with SmallNoise Kluwer Academic Dordrecht The Netherlands 1994
[12] N Yoshida ldquoAsymptotic expansions of maximum likelihoodestimators for small diffusions via the theory of Malliavin-Watanaberdquo Probability Theory and Related Fields vol 92 no 3pp 275ndash311 1992
[13] N Yoshida ldquoConditional expansions and their applicationsrdquoStochastic Processes and their Applications vol 107 no 1 pp 53ndash81 2003
[14] M Uchida and N Yoshida ldquoInformation criteria for smalldiffusions via the theory of Malliavin-Watanaberdquo StatisticalInference for Stochastic Processes vol 7 no 1 pp 35ndash67 2004
[15] N Yoshida ldquoAsymptotic expansion of Bayes estimators for smalldiffusionsrdquo Probability Theory and Related Fields vol 95 no 4pp 429ndash450 1993
[16] A Takahashi ldquoAn asymptotic expansion approach to pricingfinancial contingent claimsrdquoAsia-Pacific FinancialMarkets vol6 no 2 pp 115ndash151 1999
6 Abstract and Applied Analysis
[17] N Kunitomo and A Takahashi ldquoThe asymptotic expansionapproach to the valuation of interest rate contingent claimsrdquoMathematical Finance vol 11 no 1 pp 117ndash151 2001
[18] A Takahashi and N Yoshida ldquoAn asymptotic expansionscheme for optimal investment problemsrdquo Statistical Inferencefor Stochastic Processes vol 7 no 2 pp 153ndash188 2004
[19] V Genon-Catalot ldquoMaximum contrast estimation for diffusionprocesses fromdiscrete observationsrdquo Statistics vol 21 no 1 pp99ndash116 1990
[20] A Gloter and M Soslashrensen ldquoEstimation for stochastic differ-ential equations with a small diffusion coefficientrdquo StochasticProcesses and their Applications vol 119 no 3 pp 679ndash6992009
[21] C F Laredo ldquoA sufficient condition for asymptotic sufficiencyof incomplete observations of a diffusion processrdquo The Annalsof Statistics vol 18 no 3 pp 1158ndash1171 1990
[22] M Uchida ldquoApproximate martingale estimating functions forstochastic differential equations with small noisesrdquo StochasticProcesses and their Applications vol 118 no 9 pp 1706ndash17212008
[23] M Soslashrensen and M Uchida ldquoSmall-diffusion asymptotics fordiscretely sampled stochastic differential equationsrdquo Bernoullivol 9 no 6 pp 1051ndash1069 2003
[24] H Long ldquoParameter estimation for a class of stochastic dif-ferential equations driven by small stable noises from discreteobservationsrdquo Acta Mathematica Scientia B vol 30 no 3 pp645ndash663 2010
Research ArticleOn the Deficiencies of Some Differential-Difference Polynomials
Xiu-Min Zheng1 and Hong Yan Xu2
1 Institute of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China2Department of Informatics and Engineering Jingdezhen Ceramic Institute Jingdezhen 333403 China
Correspondence should be addressed to Xiu-Min Zheng zhengxiumin2008sinacom
Received 1 November 2013 Accepted 4 January 2014 Published 27 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 X-M Zheng and H Y Xu This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
The characteristic functions of differential-difference polynomials are investigated and the result can be viewed as a differential-difference analogue of the classic Valiron-Mokhonrsquoko Theorem in some sense and applied to investigate the deficiencies of somehomogeneous or nonhomogeneous differential-difference polynomials Some special differential-difference polynomials are alsoinvestigated and these results on the value distribution can be viewed as differential-difference analogues of some classic results ofHayman and Yang Examples are given to illustrate our results at the end of this paper
1 Introduction
Throughout this paper we use standard notations in theNevanlinna theory (see eg [1ndash3]) Let 119891(119911) be a meromor-phic function Here and in the following the word ldquomero-morphicrdquo means being meromorphic in the whole complexplane We use normal notations 119898(119903 119891) 119879(119903 119891) 119873(119903 119891)119873(119903 1119891) 120590(119891) 120582(119891) and 120582(1119891) And we also use 120590
2(119891)
to denote the hyperorder of 119891(119911) and 120575(120572 119891) to denote theNevanlinna deficiency of 120572 with respect to 119891(119911) Moreoverwe denote by 119878(119903 119891) any real quantity satisfying 119878(119903 119891) =119900(119879(119903 119891)) as 119903 rarr infin outside of a possible exceptional set offinite logarithmic measure
Recently with some establishments of difference ana-logues of the classic Nevanlinna theory (two typical andmost important ones can be seen in [4ndash6]) there has beena renewed interest in the properties of complex differenceexpressions and meromorphic solutions of complex differ-ence equations (see eg [4ndash17]) By combining complex dif-ferentiates and complex differences we proceed in this way inthis paper
It is well known that the following Valiron-MokhonrsquokoTheorem due to Valiron [18] and A Z Mokhonrsquoko and V DMokhonrsquoko [19] is of essential importance in the theory ofcomplex differential equations and functional equations
Theorem A (see [2 3]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891 (119911)) =
sum119898
119894=0119886119894(119911) 119891(119911)
119894
sum119899
119895=0119887119895(119911) 119891(119911)
119895
(1)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) the characteristic
function of 119877(119911 119891(119911)) satisfies
119879 (119903 119877 (119911 119891 (119911))) = 119889119879 (119903 119891) + 119874 (Ψ (119903)) (2)
where 119889 = max119898 119899 and Ψ(119903) = max119894119895119879(119903 119886
119894) 119879(119903 119887
119895)
Noting that the difference analogue of Theorem A maynot hold we have obtained a result of this type in [16] byadding some additional assumptions as follows
Theorem B (see [16]) Suppose that 119875(119911 119891) is a differencepolynomial of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
1198940119891(119911 + 119888
1)1198941sdot sdot sdot 119891(119911 + 119888
119899)119894119899
(3)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 378151 12 pageshttpdxdoiorg1011552014378151
2 Abstract and Applied Analysis
containing just one monomial of degree 119889(119875) and 119891(119911) is atranscendental meromorphic function of finite order If 119891(119911)also satisfies119873(119903 119891) + 119873(119903 1119891) = 119878(119903 119891) then we have
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (4)
In this paper we consider removing the assumptionldquo119875(119911 119891) contains just one monomial of degree 119889(119875)rdquo in The-orem B and obtain a weaker result which is also generalizedinto differential-difference case The concrete result can beseen in Section 2
Next we recall a classic result concerning Picardrsquos valuesofmeromorphic functions and its derivatives due toHayman[20]
Theorem C (see [20]) Let 119891(119911) be a transcendental entirefunction Then
(a) for 119899 ge 3 and 119886 = 0 Ψ(119911) = 1198911015840(119911) minus 119886(119891(119911))119899 assumesall finite values infinitely often
(b) for 119899 ge 2Φ(119911) = 1198911015840(119911)(119891(119911))119899 assumes all finite valuesexcept possibly zero infinitely often
Corresponding difference analogues ofTheoremC can beseen in [12 17]
Theorem D (see [12 17]) Let 119891(119911) be a transcendental entirefunction of finite order and let 119888 be a nonzero complex constantThen
(a) for 119899 ge 3 and 119886 = 0Ψ1(119911) = 119891(119911+ 119888) minus 119886119891(119911)
119899 assumesall finite complex values infinitely often
(b) for 119899 ge 2 Φ1(119911) = 119891(119911 + 119888)119891(119911)
119899 assumes all finitecomplex values except possibly zero infinitely often
AfterTheoremCmany results have been obtained on thevalue distribution of differential polynomials A typical one isas follows
TheoremE (see [21 22]) Let119891 be a transcendental meromor-phic function with119873(119903 119891) +119873(119903 1119891) = 119878(119903 119891) and let Ψ bea differential polynomial in 119891 of the form
Ψ (119911) = sum119886 (119911) 119891(119911)11989701198911015840
(119911)1198971sdot sdot sdot 119891(119896)
(119911)119897119896 (5)
with no constant term Furthermore assume the degree 119899 ofΨ is greater than one and 119897
0lt 119899 0 le 119897
119894le 119899 for all 119894 = 0 Then
120575(119886 Ψ) lt 1 for all 119886 = 0infin Moreover if all the terms ofΨ havedifferent degrees at least two that is Ψ is nonhomogeneousthen 120575(119886 Ψ) le 1 minus (12119899) for all 119886 =infin
We also consider deficiencies of difference polynomialsof meromorphic functions of finite order in [16] which canbe viewed as difference analogues of Theorem E as well asgeneralizations of Theorem D
In this paper we proceed to investigate deficiencies ofdifferential-difference polynomials of meromorphic func-tions The concrete results can be seen in Section 3
Examples are given in Section 4 to illustrate our results
2 A Differential-Difference Analogue ofValiron-Mokhonrsquoko Theorem
In what follows we will consider differential-difference poly-nomials A differential-difference polynomial is a polynomialin 119891(119911) its shifts its derivatives and derivatives of its shifts(see [14]) that is an expression of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
120582001198911015840
(119911)12058201sdot sdot sdot 119891(119898)
(119911)1205820119898
times 119891(119911 + 1198881)120582101198911015840
(119911 + 1198881)12058211sdot sdot sdot 119891(119898)
(119911 + 1198881)1205821119898
sdot sdot sdot 119891(119911 + 119888119899)12058211989901198911015840
(119911 + 119888119899)1205821198991sdot sdot sdot 119891(119898)
(119911 + 119888119899)120582119899119898
= sum
120582isin119868
119886120582(119911)
119899
prod
119894=0
119898
prod
119895=0
119891(119895)
(119911 + 119888119894)120582119894119895
(6)
where 119868 is a finite set of multi-indices 120582 =
(12058200 120582
0119898 12058210 120582
1119898 120582
1198990 120582
119899119898) and 119888
0(= 0)
and 1198881 119888
119899are distinct complex constants And we assume
that the meromorphic coefficients 119886120582(119911) 120582 isin 119868 of 119875(119911 119891) are
of growth 119878(119903 119891) We denote the degree of the monomialprod119899
119894=0prod119898
119895=0119891(119895)
(119911 + 119888119894)120582119894119895 of 119875(119911 119891) by 119889(120582) = sum119899
119894=0sum119898
119895=0120582119894119895
Then we denote the degree and the lower degree of 119875(119911 119891)by
119889 (119875) = max120582isin119868
119889 (120582) 119889lowast
(119875) = min120582isin119868
119889 (120582) (7)
respectively In particular we call 119875(119911 119891) a homogeneousdifferential-difference polynomial if 119889(119875) = 119889
lowast
(119875) Other-wise 119875(119911 119891) is nonhomogeneous
In the following we assume 119889(119875) ge 1 and 119875(119911 119891) equiv
119875(119911 0)We prove a weaker differential-difference version of the
classic Valiron-MokhonrsquokoTheorem as follows
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic function and 119875(119911 119891) is a differential-difference polyno-mial of the form (6) If 119891(119911) also satisfies 120590
2(119891) lt 1 and
119873(119903 119891) + 119873(119903
1
119891
) = 119878 (119903 119891) (8)
then one has
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891))
le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
(9)
Remark 2 If119875(119911 119891) is a homogeneous differential-differencepolynomial in addition then
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (10)
Remark 3 Especially assumption (8) can be replaced by theassumption ldquomax120582(119891) 120582(1119891) lt 120590(119891)rdquo In fact if 119891(119911)satisfies max120582(119891) 120582(1119891) lt 120590(119891) then 119891(119911) is of regulargrowth and (8) holds consequently
Abstract and Applied Analysis 3
To proveTheorem 1 we need the following lemmas
Lemma 4 (see [6]) Let 119891(119911) be a nonconstant meromorphicfunction 120576 gt 0 and 119888 isin C If 120577 = 120590
2(119891) lt 1 then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120577minus120576
) (11)
for all 119903 outside of a set of finite logarithmic measure
Lemma 5 (see [6]) Let 119879 [0 +infin) rarr [0 +infin) be anondecreasing continuous function and let 119904 isin (0 +infin)If the hyperorder of 119879 is strictly less than one that islim119903rarrinfin
(log2119879(119903) log 119903) = 120577 lt 1 and 120575 isin (0 1 minus 120577) then
119879 (119903 + 119904) = 119879 (119903) + 119900 (
119879 (119903)
119903120575
) (12)
where 119903 runs to infinity outside of a set of finite logarithmicmeasure
It is shown in [23 p66] and [7 Lemma 1] that the ine-quality
(1 + 119900 (1)) 119879 (119903 minus |119888| 119891) le 119879 (119903 119891 (119911 + 119888))
le (1 + 119900 (1)) 119879 (119903 + |119888| 119891)
(13)
holds for 119888 = 0 and 119903 rarr infin And from the proof theabove relation is also true for counting function By combingLemma 5 and these inequalities we immediately deduce thefollowing lemma
Lemma 6 Let 119891(119911) be a nonconstant meromorphic functionof 1205902(119891) lt 1 and let 119888 be a nonzero complex constant Then
one has
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891)
119873 (119903 119891 (119911 + 119888)) = 119873 (119903 119891) + 119878 (119903 119891)
119873(119903
1
119891 (119911 + 119888)
) = 119873(119903
1
119891
) + 119878 (119903 119891)
(14)
Lemma7 Let119891(119911) be a transcendentalmeromorphic functionof 1205902(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then we one has
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891) + 119878 (119903 119891) (15)
Furthermore if 119891(119911) also satisfies
119873(119903 119891) = 119878 (119903 119891) (16)
then one has
119879 (119903 119875 (119911 119891)) le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (17)
Proof For 119894 = 0 1 119899 119895 = 0 1 119898 we define 119892119894119895(119911) =
119891(119895)
(119911 + 119888119894)119891(119911) We also define
119892lowast
119894119895(119911) =
119892119894119895(119911) if 1003816100381610038161003816
1003816119892119894119895(119911)
10038161003816100381610038161003816gt 1
1 if 10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816le 1
119891lowast
(119911) =
119891 (119911) if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816le 1
(18)
Thus
1003816100381610038161003816119875 (119911 119891)
1003816100381610038161003816le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119891 (119911)
1003816100381610038161003816
119889(120582)
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
120582119894119895
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(120582)
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)(
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
1003816100381610038161003816119891lowast
(119911)1003816100381610038161003816)
119889(119875)
(19)
By the definitions of 119891lowast(119911) and 119892lowast119894119895(119911) 119894 = 0 1 119899 119895 =
0 1 119898 we have
119898(119903 119891lowast
) = 119898 (119903 119891)
119898 (119903 119892lowast
119894119895) = 119898 (119903 119892
119894119895) 119894 = 0 119899 119895 = 0 119898
(20)
It follows by (19) and (20) that
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
= 119889 (119875)119898 (119903 119891)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892119894119895) + 119878 (119903 119891)
(21)
Lemmas 4 and 6 and the logarithmic derivative lemma implythat for 119894 = 0 1 119899 and 119895 = 0 1 119898
119898(119903 119892119894119895) = 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911)
)
le 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911 + 119888119894)
) + 119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
= 119878 (119903 119891 (119911 + 119888119894)) + 119878 (119903 119891) = 119878 (119903 119891)
(22)
Then (15) follows by (21) and (22)
4 Abstract and Applied Analysis
It is easy to find that
119873(119903 119875 (119911 119891)) = 119874(119873(119903 119891) +
119899
sum
119894=1
119873(119903 119891 (119911 + 119888119894)))
+ 119878 (119903 119891)
(23)
Then (16) (23) and Lemma 6 yield that
119873(119903 119875 (119911 119891)) = 119878 (119903 119891) (24)
Thus (17) follows by (15) and (24)
Lemma 8 Let 119891(119911) be a transcendental meromorphic func-tion of 120590
2(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then one has
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(25)
Proof Similar to (19) we have
100381610038161003816100381610038161003816100381610038161003816
119875 (119911 119891)
119891(119911)119889(119875)
100381610038161003816100381610038161003816100381610038161003816
le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895 1003816100381610038161003816119892 (119911)
1003816100381610038161003816
119889(119875)minus119889(120582)
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889(120582)
)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889lowast(119875)
(26)
where 119892(119911) = 1119891(119911) and
119892lowast
(119911) =
119892 (119911) if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816le 1
(27)
By the definition of 119892lowast(119911) we have 119898(119903 119892lowast) = 119898(119903 119892) =
119898(119903 1119891) Thus (20) (22) and (26) yield that
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898 (119903 119892lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(28)
that is (25)
Now we can finish the proof of Theorem 1 in the end
Proof of Theorem 1 We deduce from (8) (24) and Lemma 8that
119889 (119875) 119879 (119903 119891)
= 119879 (119903 119891119889(119875)
) le 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119873 (119903 119875 (119911 119891))
+ 119889 (119875)119873(119903
1
119891
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119879 (119903 119875 (119911 119891)) + 119878 (119903 119891)
(29)
that is
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891)) (30)
Then (9) follows by (17) and (30)
3 Deficiencies of SomeDifferential-Difference Polynomials
In the following we assume that 120572(119911)( equiv 0) is ameromorphicfunction of growth 119878(119903 119891)
In this section we will apply Theorem 1 to consider thedeficiencies of general homogeneous or nonhomogeneousdifferential-difference polynomials
Theorem 9 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)
(a) If119875(119911 119891) is a homogeneous differential-difference poly-nomial then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
= 1 120575 (120572 119875 (119911 119891)) = 0
(31)
(b) If 119875(119911 119891) is a nonhomogeneous differential-differencepolynomial with 2119889lowast(119875) gt 119889(119875) then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
ge
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
120575 (120572 119875 (119911 119891)) le 1 minus
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
lt 1
(32)
Thus 119875(119911 119891)minus120572(119911) has infinitely many zeros whether 119875(119911 119891)is homogeneous or nonhomogeneous
Abstract and Applied Analysis 5
Furthermore one considers some differential-differencepolynomials of special forms which are generalizations ofboth differential cases and difference cases that is TheoremsCndashE
Theorem 10 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 V minus 1 of growth 119878(119903 119891) If 119906V gt 119889(119875) 119906 = 1 then
1198761(119911 119891) = 119865 (119891) 119875 (119911 119891) (33)
satisfies
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(34)
Thus 1198761(119911 119891) minus 120572(119911) has infinitely many zeros
When 119865(119891) is of a special form 119891V we can deduce the
following result fromTheorem 9
Theorem 11 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6) If V isin N 1and V + 2119889lowast(119875) gt 119889(119875) then
1198762(119911 119891) = 119891
V119875 (119911 119891) (35)
satisfies 120575(120573 1198762(119911 119891)) lt 1 where 120573 isin C0Thus119876
2(119911 119891)minus
120573 has infinitely many zeros
Remark 12 On the one hand we can also applyTheorem 9 to1198761(119911 119891)with the assumption ldquo2(119889lowast(119875)+119889lowast(119865)) gt 119889(119875)+119906Vrdquo
and obtain the same result as Theorem 10 But our presentassumption ldquo119906V gt 119889(119875)rdquo has no concern with 119889lowast(119875) and119889lowast
(119865) so we think Theorem 10 is better to some extent Onthe other hand we can also apply Theorem 10 to 119876
2(119911 119891)
with the assumption ldquoV gt 119889(119875)rdquo which is stronger thanldquoV + 2119889lowast(119875) gt 119889(119875)rdquo in Theorem 11 showing Theorem 11 isbetter to some extent
Theorem 13 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 Vminus1 of growth 119878(119903 119891) If (119906minus1)119906V(2119906minus1) gt 119889(119875) 119906 = 1then
1198763(119911 119891) = 119865 (119891) + 119875 (119911 119891) (36)
satisfies
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(37)
Thus 1198763(119911 119891) minus 120572(119911) has infinitely many zeros
When 119906 = 1 one can consider some special cases asfollows
Theorem 14 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)(a) If V gt 119889(119875) + 2 ge 3 then
1198764(119911 119891) = 119891
V+ 119875 (119911 119891) (38)
satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(39)
(b) If (Vminus1)V(2Vminus1) gt 119889(119875) V ge 3 then1198764(119911 119891) satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
1
Vminus
2V minus 1V2
119889 (119875)
120575 (120572 1198764(119911 119891)) le
1
V+
2V minus 1V2
119889 (119875) lt 1
(40)
Especially it holds for V = 119889(119875) + 2 = 3(c) If V ge 119889(119875) + 2 ge 3 and 119891 also satisfies 119873(119903 1119891) =119878(119903 119891) then 119876
4(119911 119891) satisfies 120575(120572 119876
4(119911 119891)) lt 1
Especially it holds for V = 119889(119875) + 2 gt 3Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many zeros
If we assume that 119873(119903 1119891) = 119878(119903 119891) in addition thefollowing result follows immediately by Theorem 9
Theorem 15 Suppose that 119891(119911) is a transcendental mero-morphic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891)
is a differential-difference polynomial of the form (6) If2 min119889lowast(119875) V gt max119889(119875) V then 119876
4(119911 119891) satisfies
120575(120572 1198764(119911 119891)) lt 1 Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many
zeros
Remark 16 Noting that when V gt 3 (V minus 1)V(2V minus 1) leV minus 2 hold we see that the assumption ldquoV gt 119889(119875) + 2rdquo inTheorem 14(a) is weaker than the assumption ldquo(V minus 1)V(2V minus1) gt 119889(119875)rdquo in Theorem 14(b) And these assumptions inTheorem 14 have no concernwith119889lowast(119875)) thus they are differ-ent from the assumption ldquo2 min119889lowast(119875) V gt max119889(119875) Vrdquoin Theorem 15
6 Abstract and Applied Analysis
Remark 17 From the proofs behind it is easy to find that
120582 (119875 (119911 119891) minus 120572) = 120590 (119875 (119911 119891)) = 120590 (119891)
120582 (119876119894(119911 119891) minus 120572) = 120590 (119876
119894(119911 119891)) = 120590 (119891) 119894 = 1 3 4
(41)
hold respectively inTheorems 9 10 13 14(a) and (b) and 15Now we give the proofs of Theorems 9ndash15
Proof of Theorem 9 It follows byTheorem 1 that
119878 (119903 119891) = 119878 (119903 119875 (119911 119891)) (42)
We deduce from (8) (24) (25) and (42) that
119873(119903
1
119875 (119911 119891)
)
le 119873(119903
1
119891119889(119875)
) + 119873(119903
119891119889(119901)
119875 (119911 119891)
)
le 119873(119903
1
119891
) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119875 (119911 119891))
(43)
Thus an application of the second main theorem and (24)(42) and (43) imply that
119879 (119903 119875 (119911 119891)) le 119873 (119903 119875 (119911 119891)) + 119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(44)
(a) If 119889(119875) = 119889lowast(119875) then it follows by (44) that
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(45)
by which (31) holds
(b) If 2119889lowast(119875) gt 119889(119875) then we deduce from (30) and (44)that
119879 (119903 119875 (119911 119891)) le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le
119889 (119875) minus 119889lowast
(119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891))
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(46)
that is
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119878 (119903 119875 (119911 119891))
(47)
Since 2119889lowast(119875)minus119889(119875) gt 0 (32) follows immediately by (47)
Proof of Theorem 10 We deduce from (16) (17) and (24) that
119879 (119903 1198761(119911 119891)) le (119906V + 119889 (119875)) 119879 (119903 119891) + 119878 (119903 119891) (48)
119873(119903 1198761(119911 119891)) = 119874 (119873 (119903 119891)) + 119873 (119903 119875 (119911 119891)) + 119878 (119903 119891)
= 119878 (119903 119891)
(49)
hold Next we consider 119873(119903 11198761(119911 119891)) Let 119911
0be a zero of
1198761(119911 119891) and distinguish three cases
(i) 1199110is not a zero of 119865(119891) then 119911
0must be a zero of
119875(119911 119891) and
119906 le 120596(
1
1198761(119911 119891)
1199110) + (119906 minus 1) 120596(
1
119875 (119911 119891)
1199110) (50)
where 120596(119891 1199110) denotes the order of multiplicity of 119911
0or zero
according as 1199110is a pole of 119891(119911) or not
(ii) 1199110is a zero of 119865(119891) but not a pole of 119875(119911 119891) Then
119906 le 120596(
1
1198761(119911 119891)
1199110) (51)
(iii) 1199110is a zero of 119865(119891) and a pole of 119875(119911 119891) Then
119906 le 120596(
1
119865 (119891)
1199110) le 120596(
1
1198761(119911 119891)
1199110) + 120596 (119875 (119911 119891) 119911
0)
(52)
Abstract and Applied Analysis 7
(24) and (50)ndash(52) yield that
119906119873(119903
1
1198761(119911 119891)
) le 119873(119903
1
1198761(119911 119891)
)
+ (119906 minus 1)119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(53)
Then (48) (49) (53) and an application of the second maintheorem to 119876
1(119911 119891) imply that
119879 (119903 1198761(119911 119891))
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
1198761(119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 1198761(119911 119891))
le
1
119906
119873(119903
1
1198761(119911 119891)
) +
119906 minus 1
119906
119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(54)
consequently
119879 (119903 1198761(119911 119891)) le 119873(119903
1
119875 (119911 119891)
)
+
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(55)
Moreover by 119891119889(119875)119865(119891) = 119891119889(119875)
1198761(119911 119891)119875(119911 119891) (16)
(24) (25) andTheorem A we have
(119889 (119875) + 119906V)119898 (119903 119891)
= 119898(119903
119891119889(119875)
1198761(119911 119891)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) minus 119873(119903
119891119889(119875)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898(119903
1
119891
)
+ 119889 (119875) (119873(119903
1
119891
) minus 119873 (119903 119891))
+ 119873 (119903 119875 (119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
= 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898 (119903 119891)
minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(56)
consequently
119906V119898(119903 119891) le 119898 (119903 1198761(119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(57)
On the other hand the evident relation 119906V120596(119891 1199110) le
120596(119865(119891) 1199110) + 119906VsumVminus1
119895=0120596(119886119895 1199110) where the definition of
120596(119891 1199110) is given after (50) results in
119906V119873(119903 119891) le 119873 (119903 119865 (119891)) + 119878 (119903 119891)
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(58)
We deduce from (57) and (58) that
119906V119879 (119903 119891) le 119879 (119903 1198761(119911 119891)) + 119878 (119903 119891) (59)
Then (17) (55) and (59) yield that
119906V119879 (119903 119891)
le 119873(119903
1
119875 (119911 119891)
) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
le 119889 (119875) 119879 (119903 119891) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(60)
that is
(119906 minus 1) (119906V minus 119889 (119875))119906
119879 (119903 119891) le 119873(119903
1
1198761(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(61)
From (48) and (61) we deduce that
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(62)
Proof of Theorem 11 Assume to the contrary that120575(120573 119876
2(119911 119891)) = 1 Denoting
1198762(119911 119891) minus 120573 = 119891
V(119911) 119875 (119911 119891) minus 120573 = 119866 (119911) (63)
we deduce from (16) and (17) that
119873(119903
1
119866
) = 119873(119903
1
1198762(119911 119891) minus 120573
)
= 119878 (119903 1198762(119911 119891)) = 119878 (119903 119891)
(64)
8 Abstract and Applied Analysis
On the other hand (16) and (24) yield that
119873(119903 119866) = 119873 (119903 1198762(119911 119891) minus 120573) = 119878 (119903 119891) (65)
Differentiating both sides of (63) we obtain
119891Vminus1(119911) 119877 (119911 119891) = 119866
1015840
(119911) (66)
where 119877(119911 119891) = V1198911015840(119911)119875(119911 119891) + 119891(119911)1198751015840(119911 119891) Clearly wededuce from (16) and (24) that
119873(119903 119877 (119911 119891)) = 119878 (119903 119891) (67)
Moreover (64) and (65) yield that
119873(119903
1
1198661015840
) le 119873(119903
119866
1198661015840
) + 119873(119903
1
119866
)
le 119879(119903
1198661015840
119866
) + 119873(119903
1
119866
) + 119874 (1)
le 119898(119903
1198661015840
119866
) + 119873 (119903 119866) + 2119873(119903
1
119866
) + 119874 (1)
= 119878 (119903 119866) + 119878 (119903 119891) = 119878 (119903 119891)
(68)
It follows by (66)ndash(68) that
119873(119903
1
119891
) =
1
V minus 1119873(119903
119877 (119911 119891)
1198661015840
) = 119878 (119903 119891) (69)
Then (16) (69) and the fact 2(119889lowast(119875) + V) gt 119889(119875) + V implythat the assumptions of Theorem 9(b) are satisfied ThusTheorem 9(b) yields that 120575(120573 119876
2(119911 119891)) lt 1 a contradiction
Therefore we have 120575(120573 1198762(119911 119891)) lt 1
Proof of Theorem 13 We deduce from (16) (17) and (24) that
119879 (119903 1198763(119911 119891)) le max 119906V 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
= 119906V119879 (119903 119891) + 119878 (119903 119891) (70)
Denote
119867(119911) =
minus119875 (119911 119891) + 120572 (119911)
119865 (119891)
(71)
Now we estimate the poles the zeros and 1-points of119867(119911) accuratelyOn the one handwe see by (71) that the polesof 119867(119911) occur at zeros of 119865(119891) and poles of minus119875(119911 119891) + 120572(119911)which are not simultaneously 1-points of 119867(119911) and thosepoles of119867(119911)which are zeros of 119865(119891) but not simultaneouslyzeros of minus119875(119911 119891) + 120572(119911) also have multiplicities at least 119906 Onthe other handwe also see by (71) that the zeros of119867(119911) occurat zeros of minus119875(119911 119891) + 120572(119911) and poles of 119865(119891) which are notsimultaneously 1-points of 119867(119911) Moreover 1-points of 119867(119911)occur at zeros of 119876
3(119911 119891) minus 120572(119911) and occur at the common
poles zeros of 119865(119891) and minus119875(119911 119891) + 120572(119911) with the samemultiplicities Thus it follows by (16) and (24) that
119873(119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
)
le
1
119906
119873 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(72)
Then (17) (72) and the second main theorem result in
119879 (119903119867) le 119873 (119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
) + 119878 (119903119867)
le
1
119906
119879 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
le
1
119906
119879 (119903119867) + 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(73)
that is
(1 minus
1
119906
)119879 (119903119867) le 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(74)
Moreover Theorem A and (17) imply that
119906V119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119865 (119891)) = 119879(119903minus119875 (119911 119891) + 120572
119867
)
le 119889 (119875) 119879 (119903 119891) + 119879 (119903119867) + 119878 (119903 119891)
(75)
that is
(119906V minus 119889 (119875)) 119879 (119903 119891) le 119879 (119903119867) + 119878 (119903 119891) (76)
Then (74) and (76) yield that
((119906 minus 1) V minus2119906 minus 1
119906
119889 (119875))119879 (119903 119891) le 119873(119903
1
1198763(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(77)
Abstract and Applied Analysis 9
From (70) and (77) we deduce that
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(78)
To prove Theorem 14(c) we also need the followinglemma of one of Tumura-Clunie type theorems
Lemma 18 (see [24]) Let 119891(119911) be a meromorphic functionand suppose that Ψ = 119886
119899119891119899
+ sdot sdot sdot + 1198860has small meromorphic
coefficients 119886119895(119911) 119886119899(119911) equiv 0 in the sense of 119879(119903 119886
119895) = 119878(119903 119891)
Moreover assume that 119873(119903 1Ψ) + 119873(119903 119891) = 119878(119903 119891) ThenΨ = 119886
119899(119891 + (119886
119899minus1119899119886119899))119899
Proof of Theorem 14 (a) We deduce from (16) (17) and (24)that
119879 (119903 1198764(119911 119891)) le V119879 (119903 119891) + 119878 (119903 119891) (79)
Denote
119870 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) + 119875 (119911 119891) minus 120572 (119911) (80)
Differentiating both sides of (80) we obtain
V119891Vminus1(119911) 1198911015840
(119911) + 1198751015840
(119911 119891) minus 1205721015840
(119911)
= 1198701015840
(119911) = (119891V(119911) + 119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
(81)
that is
119891Vminus1(119911) ((V
1198911015840
(119911)
119891 (119911)
minus
1198701015840
(119911)
119870 (119911)
)119891 (119911))
= (119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
minus (1198751015840
(119911 119891) minus 1205721015840
(119911))
= (119875 (119911 119891) minus 120572 (119911)) (
1198701015840
(119911)
119870 (119911)
minus
1198751015840
(119911 119891) minus 1205721015840
(119911)
119875 (119911 119891) minus 120572 (119911)
)
(82)
It follows by (15)ndash(17) (24) (79) and (82) that
119898(119903 119891Vminus1)
le 119898 (119903 119875 (119911 119891) minus 120572) + 119898(119903
1198701015840
119870
)
+ 119898(119903
1198751015840
(119911 119891) minus 1205721015840
119875 (119911 119891) minus 120572
) + 119898(119903
1
(V (1198911015840119891) minus (1198701015840119870)) 119891)
le 119889 (119875)119898 (119903 119891) + 119898(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891)
+ 119873(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891) + 119878 (119903 119870) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1198701015840
119870
) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1
119870
) + 119878 (119903 119891)
(83)
that is
(V minus 119889 (119875) minus 2) 119879 (119903 119891) le 119873(1199031
1198764(119911 119891) minus 120572
) + 119878 (119903 119891)
(84)
From (79) and (84) we deduce that
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(85)
(b) It suffices to note that we may see 119891V as (1198911)V thenTheorem 14(b) follows immediately by Theorem 13
(c) By using a similar reasoning as [13 Theorem 1] wecan rearrange the expression for the differential-differencepolynomial 119875(119911 119891) by collecting together all terms havingthe same total degree and then writing 119875(119911 119891) in the form119875(119911 119891) = sum
119889(119875)
119896=0119887119896(119911)119891119896
(119911) Now each of the coefficients 119887119896(119911)
is a finite sum of products of functions of the form (119891(119895)
(119911 +
119888119894)119891(119911))
120582119894119895= (119891(119895)
(119911+119888119895)119891(119911+119888
119894))120582119894119895(119891(119911+119888
119894)119891(119911))
120582119894119895 with
each such product being multiplied by one of the originalcoefficients 119886
120582(119911) We deduce from the logarithmic derivative
lemma and Lemmas 4 and 6 that 119898(119903 119887119896) = 119878(119903 119891) Clearly
119873(119903 119887119896) = 119878(119903 119891) holds by (8) and Lemma 6Thus 119879(119903 119887
119896) =
119878(119903 119891) Denote
119871 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) +
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) minus 120572 (119911)
(86)
Assume to the contrary that 120575(120572 1198764(119911 119891)) = 1 Thus
Theorem A yields that
119873(119903
1
119871
) = 119873(119903
1
1198764(119911 119891) minus 120572
)
= 119878 (119903 1198764(119911 119891)) = 119878 (119903 119891)
(87)
Then (8) (86) (87) Lemma 18 and the assumption that V ge119889(119875) + 2 imply that 119871(119911) equiv 119891(119911)V that is
119875 (119911 119891) =
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) equiv 120572 (119911) (88)
Noting the fact that 119879(119903 119887119896) = 119878(119903 119891) and 119879(119903 120572) = 119878(119903 119891)
we deduce from Theorem A that (88) is a contradictionTherefore we have 120575(120572 119876
4(119911 119891)) lt 1
10 Abstract and Applied Analysis
4 Examples
Example 1 We consider nonhomogeneous differential-difference polynomials
1198751(119911 119891) = 119891 (119911) 119891
2
(119911 + log 4) minus 411989110158401015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 2) + 119891101584010158402 (119911 + log 3)
1198752(119911 119891) = 3119891
3
(119911) 11989110158402
(119911 + log 4)
minus 21198911015840
(119911) 119891 (119911 + log 3) 119891101584010158403 (119911 + log 2)
+ 1198914
(119911) minus 119891101584010158403
(119911)
1198753(119911 119891) = 119891 (119911) 119891
1015840
(119911 + log 2) 11989110158401015840 (119911 + log 3) minus 6119891101584010158402 (119911)(89)
and a homogeneous differential-difference polynomial
1198754(119911 119891) = 119891
101584010158403
(119911 + log 2) minus 1198911015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 3) minus 119891 (119911) 1198911015840 (119911) 11989110158401015840 (119911) (90)
where 119889(1198751) = 3 gt 2 = 119889
lowast
(1198751) 119889(119875
2) = 5 gt 3 = 119889
lowast
(1198752)
119889(1198753) = 3 gt 2 = 119889
lowast
(1198753) and 119889(119875
4) = 3 = 119889
lowast
(1198754) Clearly the
function 119891(119911) = 119890119911 satisfies (8) and 1205902(119891) = 0 lt 1 Then we
have
119889lowast
(1198751) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198751(119911 119890119911
)) =
2119903
120587
+ 119874 (1)
lt 119889 (1198751) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198752) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198752(119911 119890119911
)) =
4119903
120587
+ 119874 (1)
lt 119889 (1198752) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198753) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198753(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198753) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198754) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198754(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198754) 119879 (119903 119890
119911
) + 119874 (1)
(91)
This example shows that (9) is best possible
Example 2 Consider 119891(119911) = 119890119911 again Then the
homogeneous case 1198754(119911 119891) in Example 1 also illustrates
Theorem 9(a) And the nonhomogeneous differential-difference polynomials 119875
119894(119911 119891) 119894 = 1 2 3 in Example 1
also illustrate Theorem 9(b) where 120575(120572 1198751(119911 119891)) = 0
120575(120572 1198752(119911 119891)) le 14 lt 23 = 1 minus ((2119889
lowast
(1198752) minus 119889(119875
2))119889lowast
(1198752))
and 120575(120572 1198753(119911 119891)) le 13 lt 12 = 1 minus ((2119889
lowast
(1198753) minus
119889(1198753))119889lowast
(1198753)) Next we consider the nonhomogeneous
differential-difference polynomial
1198755(119911 119891) = 119891
1015840
(119911) 119891 (119911 + log 2) minus 1198912 (119911)
+ 1198911015840
(119911 + log 3) minus 3119891 (119911) + 1(92)
where 119889(1198755) = 2 119889
lowast
(1198755) = 0 Clearly 120575(1 119875
5(119911 119891)) =
120575(1 1198902119911
+ 1) = 1 Note that 2119889lowast(1198755) gt 119889(119875
5) fails then this
example shows that the assumption ldquo2119889lowast(119875) gt 119889(119875)rdquo cannotbe omitted inTheorem 9(b)
Example 3 We consider the differential-difference polyno-mials
1198761(119911 119891) = (119891
2
)
2
1198756(119911 119891)
= 1198914
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
1198762(119911 119891) = 119891
2
1198756(119911 119891)
= 1198912
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
(93)
and the function 119891(119911) = sin 119911 On the one hand 119873(119903 119891) =119878(119903 119891) 120590
2(119891) = 0 lt 1 and 119906V
1198761
gt 119889(1198756) and V
1198762
+ 2119889lowast
(1198756) gt
119889(1198756) hold where V
1198761
= V1198762
= 119906 = 2 and 119889(1198756) = 3 gt 2 =
119889lowast
(1198756) On the other hand 120575(120572 119876
1(119911 119891)) le 1 minus (1114) lt
1 minus (114) = 1 minus (119906 minus 1)(119906V1198761
minus 119889(119875))119906(119906V1198761
+ 119889(119875)) lt 1 and120575(120572 119876
2(119911 119891)) lt 1 hold This example shows that Theorems
10 and 11 may holdExample 4 We consider the differential-difference polyno-mials
119876(1)
4(119911 119891) = (119891
2
)
4
+ 1198757(119911 119891) = 119891
8
+ 1198757(119911 119891)
= 1198918
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(2)
4(119911 119891) = 119891
2
+ 1198757(119911 119891)
= 1198912
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(3)
4(119911 119891) = 2119891
3
+ 1198757(119911 119891)
= 21198913
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(4)
4(119911 119891) = 119891
4
+ 1198757(119911 119891)
= 1198914
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
(94)
and the function 119891(119911) = sin 119911 again On the one hand119876(1)
4(119911 119891) satisfies (119906 minus 1)119906V
119876(11)
4
(2119906 minus 1) gt 119889(1198757) and V
119876(12)
4
minus
2 gt (V119876(12)
4
minus 1)V119876(12)
4
(2V119876(12)
4
minus 1) gt 119889(1198757) respectively where
119906 = 4 V119876(11)
4
= 2 V119876(12)
4
= 8 119889(1198757) = 119889
lowast
(1198757) = 3 and
for 119894 = 2 3 4 119876(119894)4(119911 119891) satisfies 2 min119889lowast(119875
7) V119876(119894)
4
gt
max119889(1198757) V119876(119894)
4
where V119876(119894)
4
= 119894 On the other hand120575(120572 119876
(119894)
4(119911 119891)) lt 1 119894 = 1 2 3 4 hold This example shows
Abstract and Applied Analysis 11
that Theorems 13ndash15 may hold Moreover this example alsoshows the assumption ldquo119873(119903 1119891) = 119878(119903 119891)rdquo is not necessaryto Theorems 14(c) and 15 but it is regrettable for us notremoving it in our proofs
Example 5 We consider the differential-difference polyno-mials
1198771(119911 119891) = 119891
2
1198758(119911 119891)
= 1198912
(119911) (1198912
(119911 + 120587) +
1
sin2211991111989110158402
(119911 +
120587
2
))
1198772(119911 119891) = 119891
7
+ 1198759(119911 119891)
= 1198917
(119911) + sin 21199111198911015840 (119911 + 1205872
)1198912
(119911 +
120587
2
)
+ 11989110158402
(119911 +
120587
2
)119891(119911 +
3120587
2
)
+ 119911119891 (119911) 119891 (119911 +
120587
2
)
(95)
and the function 119891(119911) = 119890sin2119911 On the one hand 119877
1(119911 119891)
satisfies V1198771
+2119889lowast
(1198758) gt 119889(119875
8) and 119877
2(119911 119891) satisfies V
1198772
minus2 gt
(V1198772
minus 1)V1198772
(2V1198772
minus 1) gt 119889(1198759) respectively where V
1198771
=
2 and 119889(1198758) = 119889
lowast
(1198758) = 2 and V
1198772
= 7 and 119889(1198759) = 3 On
the other hand 120575(1198902 1198771(119911 119891)) = 120575(119890119911 119877
2(119911 119891)) = 1 hold
showing thatTheorems 11 and 14 fail Noting that the function119891(119911) = 119890
sin2119911 satisfies 1205902(119891) = 1 we know that the assumption
ldquo1205902(119891) lt 1rdquo is essential for Theorems 11 and 14 In fact it is
also essential for our other results in the whole paper but it isunnecessary to give examples one by one
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project was supported by the National Natural Sci-ence Foundation of China (11301233 and 11171119) and theNatural Science Foundation of Jiangxi Province in China(20132BAB211001 and 20132BAB211002) and Sponsored Pro-gram for Cultivating Youths of Outstanding Ability in JiangxiNormal University of China
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Group DordrechtTheNetherlands 2003
[4] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof f(z+120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[5] R G Halburd and R J Korhonen ldquoDifference analogue ofthe Lemma on the Logarithmic Derivative with applicationsto difference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[6] R G Halburd R J Korhonen and K Toghe ldquoHolomor-phic curves with shift-invariant hyper-planepreimagesrdquo Tran-sactions of the American Mathematical Society In press httparxivorgabs09033236
[7] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[8] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[9] Z X Chen ldquoComplex oscillation of meromorphic solutions forthe Pielou logistic equationrdquo Journal of Difference Equations andApplications vol 19 no 11 pp 1795ndash1806 2013
[10] Z X Chen and K H Shon ldquoFixed points of meromorphicsolutions for some difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 496096 7 pages 2013
[11] K Ishizaki and N Yanagihara ldquoWiman-Valiron method fordifference equationsrdquo Nagoya Mathematical Journal vol 175pp 75ndash102 2004
[12] I Laine and C-C Yang ldquoClunie theorems for difference andq-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] I Laine and C C Yang ldquoValue distribution of difference poly-nomialsrdquo Proceedings of the Japan Academy A vol 83 no 8 pp148ndash151 2007
[14] C-C Yang and I Laine ldquoOn analogies between nonlineardifference and differential equationsrdquo Proceedings of the JapanAcademy A vol 86 no 1 pp 10ndash14 2010
[15] R R Zhang and Z B Huang ldquoResults on difference analoguesof Valiron-Mokhonrsquoko theoremrdquoAbstract and Applied Analysisvol 2013 Article ID 273040 6 pages 2013
[16] XMZheng andZXChen ldquoOndeficiencies of somedifferencepolynomialsrdquo Acta Mathematica Sinica vol 54 no 6 pp 983ndash992 2011 (Chinese)
[17] XM Zheng and Z X Chen ldquoOn the value distribution of somedifference polynomialsrdquo Journal of Mathematical Analysis andApplications vol 397 no 2 pp 814ndash821 2013
[18] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Mathematique de France vol 59 pp 17ndash39 1931
[19] A Z Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoSibirskii Matematicheskii Zhurnal vol 15 pp 1305ndash1322 1974(Russian)
[20] W K Hayman ldquoPicard values of meromorphic functions andtheir derivativesrdquo Annals of Mathematics vol 70 no 2 pp 9ndash42 1959
[21] C-C Yang ldquoOn deficiencies of differential polynomialsrdquoMath-ematische Zeitschrift vol 116 no 3 pp 197ndash204 1970
[22] C-C Yang ldquoOn deficiencies of differential polynomials IIrdquoMathematische Zeitschrift vol 125 no 2 pp 107ndash112 1972
[23] A A Golrsquodberg and I V Ostrovskii The Distribution ofValues ofMeromorphic Functions NaukaMoscow Russia 1970(Russian)
12 Abstract and Applied Analysis
[24] E Mues and N Steinmetz ldquoThe theorem of Tumura-Clunie formeromorphic functionsrdquo Journal of the London MathematicalSociety vol 23 no 2 pp 113ndash122 1981
Research ArticleOn Growth of Meromorphic Solutions of Complex FunctionalDifference Equations
Jing Li12 Jianjun Zhang3 and Liangwen Liao1
1 Department of Mathematics Nanjing University Nanjing 210093 China2Nankai University Binhai College Tianjin 300270 China3Mathematics and Information Technology School Jiangsu Second Normal University Nanjing 210013 China
Correspondence should be addressed to Jianjun Zhang zhangjianjun1982163com
Received 29 November 2013 Accepted 13 January 2014 Published 25 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Jing Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The main purpose of this paper is to investigate the growth order of the meromorphic solutions of complex functional differenceequation of the form (sum
120582isin119868120572120582(119911)(prod
119899
]=1119891(119911 + 119888])119897120582]))(sum
120583isin119869120573120583(119911)(prod
119899
]=1119891(119911 + 119888])119898120583]
)) = 119876(119911 119891(119901(119911))) where 119868 = 120582 =
(1198971205821 1198971205822 119897
120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 = (119898
1205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two finite
index sets 119888] (] = 1 2 119899) are distinct complex numbers 120572120582(119911) (120582 isin 119868) and 120573
120583(119911) (120583 isin 119869) are small functions relative to 119891(119911)
and 119876(119911 119906) is a rational function in 119906 with coefficients which are small functions of 119891(119911) 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911]
of degree 119896 ge 1 We also give some examples to show that our results are sharp
1 Introduction and Main Results
Let 119891(119911) be a function meromorphic in the complex planeC We assume that the reader is familiar with the standardnotations and results in Nevanlinnarsquos value distribution the-ory ofmeromorphic functions such as the characteristic func-tion 119879(119903 119891) proximity function 119898(119903 119891) counting function119873(119903 119891) and the first and secondmain theorems (see eg [1ndash4]) We also use 119873(119903 119891) to denote the counting function ofthe poles of 119891(119911) whose every pole is counted only once Thenotations 120588(119891) and 120583(119891) denote the order and the lower orderof119891(119911) respectively 119878(119903 119891) denotes any quantity that satisfiesthe condition 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possiblyoutside an exceptional set of 119903 of finite linear measure Ameromorphic function 119886(119911) is called a small function of 119891(119911)or a small function relative to 119891(119911) if and only if 119879(119903 119886(119911)) =119878(119903 119891)
Recently some papers (see eg [5ndash7]) focusing on com-plex difference and functional difference equations emergedIn 2005 Laine et al [5] firstly considered the growth ofmeromorphic solutions of the complex functional differenceequations by utilizing Nevanlinna theory They obtained thefollowing result
Theorem A Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (1)
where 119869 is a collection of all subsets of 1 2 119899 119888119895rsquos are
distinct complex constants and 119901(119911) is a polynomial of degree119896 ge 2 Moreover we assume that the coefficients120572
119869(119911) are small
functions relative to 119891 and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (2)
where 120572 = log 119899 log 119896
In 2007 Rieppo [6] gave an estimation of growth ofmeromorphic solutions of complex functional equations asfollows
Theorem B Suppose that 119891 is a transcendental meromorphicfunction Let 119876(119911 119891) 119877(119911 119891) be rational functions in 119891
with small meromorphic coefficients relative to 119891 such that0 lt 119902 = deg
119891119876 le 119889 = deg
119891119877 and 119901(119911) = 119901
119896119911119896
+ 119901119896minus1
119911119896minus1
+
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 828746 6 pageshttpdxdoiorg1011552014828746
2 Abstract and Applied Analysis
sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 If 119891 is a solution of the
functional equation
119877 (119911 119891 (119911)) = 119876 (119911 119891 (119901 (119911))) (3)
then 119902119896 le 119889 and for any 120576 0 lt 120576 lt 1 there exist positive realconstants 119870
1and 119870
2such that
1198701(log 119903)120572minus120576 le 119879 (119903 119891) le 119870
2(log 119903)120572+120576 120572 =
log 119889 minus log 119902log 119896
(4)
when 119903 is large enough
Rieppo [6] also considered the growth order ofmeromor-phic solutions of functional equation (3) when 119896 = 1 and gotthe following
Theorem C Suppose that 119891 is a transcendental meromorphicsolution of (3) where 119901(119911) = 119886119911+119887 119886 119887 isin C 119886 = 0 and |119886| = 1Then
120583 (119891) = 120588 (119891) =
log119889 minus log 119902log |119886|
(5)
Two years later Zheng et al [7] extended Theorem A tomore general type and obtained a similar result of TheoremC In fact they got the following two results
TheoremD Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119876 (119911 119891 (119901 (119911))) (6)
where 119869 is a collection of all nonempty subsets of 1 2 119899119888119895(119895 = 1 119899) are distinct complex constants 119901(119911) = 119901
119896119911119896
+
119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a
rational function in 119906 of deg119906119876 = 119902(gt 0) Also suppose that
all the coefficients of (6) are small functions relative to 119891 Then119902119896 le 119899 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (7)
where 120572 = (log 119899 minus log 119902) log 119896
Theorem E Suppose that 119891 is a transcendental meromorphicsolution of (6) where 119869 is a collection of all nonempty subsetsof 1 2 119899 119888
119895(119895 = 1 119899) are distinct complex constants
119901(119911) = 119886119911 + 119887 119886 119887 isin C and 119876(119911 119906) is a rational function in 119906of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of (6)
are small functions relative to 119891(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log 119899minus log |119886|
(8)
(ii) If |119886| gt 1 then we have 119902 le 119899 and
120588 (119891) le
log 119899 minus log 119902log |119886|
(9)
(iii) If |119886| = 1 119902 gt 119899 then we have 120588(119891) = 120583(119891) = infin
In this paper we will consider a more general classof complex functional difference equations We prove thefollowing results which generalize the above related results
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic solution of the functional difference equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119901 (119911))) (10)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 =
120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin N⋃0 ] = 1 2 119899
are two finite index sets 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin
C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a rational function in 119906 ofdeg119906119876 = 119902(gt 0) Also suppose that all the coefficients of (10)
are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (11)
Then 119902119896 le 120590 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (12)
where 120572 = (log120590 minus log 119902) log 119896
Theorem 2 Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119886119911 + 119887))
(13)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 =
(1198981205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two
finite index sets 119886 119887 isin C and 119876(119911 119906) is a rational function in119906 of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of
(10) are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (14)
(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(15)
(ii) If |119886| gt 1 then we have 119902 le 120590 and
120588 (119891) le
log120590 minus log 119902log |119886|
(16)
(iii) If |119886| = 1 and 119902 gt 120590 then we have 120583(119891) = 120588(119891) = infin
Abstract and Applied Analysis 3
Next we will give some examples to show that our resultsare best in some extent
Example 3 Let 1198881= arctan 2 119888
2= minus1205874 Then it is easy to
check that 119891(119911) = tan 119911 solves the following equation
119891(119911 + 1198881)2
119891 (119911 + 1198882)
119891 (119911 + 1198881) + 119891(119911 + 119888
2)2
= (minus4119891(
119911
2
)
8
+ 8119891(
119911
2
)
7
+ 28119891(
119911
2
)
6
minus 56119891(
119911
2
)
5
minus 32119891(
119911
2
)
4
+ 56119891(
119911
2
)
3
+ 28119891(
119911
2
)
2
minus 8119891(
119911
2
) minus 4)
times (3119891(
119911
2
)
8
+ 10119891(
119911
2
)
7
+ 16119891(
119911
2
)
6
+ 122119891(
119911
2
)
5
minus 6119891(
119911
2
)
4
minus122119891(
119911
2
)
3
+16119891(
119911
2
)
2
minus10119891(
119911
2
) + 3)
minus1
(17)
Obviously we have
120583 (119891) = 120588 (119891) = 1 =
log 119902 minus log120590minus log |119886|
(18)
where 119902 = 8 120590 = 4 and 119886 = 12
Example 3 shows that the estimate in Theorem 2(i) issharp
Example 4 It is easy to check that 119891(119911) = tan 119911 satisfies theequation
119891(119911 + (1205873))2
119891 (119911 + (1205876)) minus 119891 (119911 + (1205876))
119891 (119911 + (1205873)) 119891(119911 + (1205876))2
minus 119891 (119911 + (1205873))
=
radic3119891(2119911)2
+ 4119891 (2119911) + radic3
minusradic3119891(2119911)2
+ 4119891 (2119911) minus radic3
(19)
Clearly we have
120583 (119891) = 120588 (119891) = 1 =
log120590 minus log 119902log |119886|
(20)
where 120590 = 4 119902 = 2 and 119886 = 2
Example 4 shows that the estimate in Theorem 2(ii) issharp
Example 5 119891(119911) = tan 119911 satisfies the equation of the form
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
minus(119891(1199112)2
minus 2119891 (1199112) minus 1)
3
8119891 (1199112) (119891(1199112)2
minus 1) (119891(1199112)2
+ 2119891 (1199112) minus 1)
(21)
where 120590 = 4 119902 = 6 and 119886 = 12 120588(119891) = 120583(119891) = 1 gt
log(32) log 2 = (log 119902 minus log120590) minus log |119886|
Example 5 shows that the strict inequality in Theorem 2may occur Therefore we do not have the same estimation asinTheoremC for the growth order ofmeromorphic solutionsof (13)
The following Example shows that the restriction 119902 gt 120590
in case (iii) in Theorem 2 is necessary
Example 6 Meromorphic function 119891(119911) = tan 119911 solves thefollowing equation
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
(119891 (119911) + 1)3
4119891 (119911) (1 minus 119891 (119911))
(22)
where 119886 = 1 and 4 = 120590 gt 119902 = 3 but 120588(119891) = 120583(119891) = 1
Next we give an example to show that case (iii) inTheorem 2 may hold
Example 7 Function 119891(119911) = 119911119890119890119911
satisfies the followingequation
(119911 + log 6) (119911 + log 2)5 [119891(119911 + log 4)4 + 119891 (119911 + log 4)](119911 + log 4) 119891 (119911 + log 6)
=
(119911 + log 4)3119891(119911 + log 2)6 + (119911 + log 2)6
119891 (119911 + log 2)
(23)
where 119886 = 1 and 119902 = 6 gt 5 = 120590 Obviously 120588(119891) = 120583(119891) = infin
2 Main Lemmas
In order to prove our results we need the following lemmas
Lemma 1 (see [4 8]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(24)
such that the meromorphic coefficients 119886119894(119911) 119887119895(119911) satisfy
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 1 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 1 119902
(25)
then one has
119879 (119903 119877 (119911 119891)) = max 119901 119902 sdot 119879 (119903 119891) + 119878 (119903 119891) (26)
From the proof ofTheorem 1 in [9] we have the followingestimate for the Nevanlinna characteristic
Lemma 2 Let 1198911 1198912 119891
119899be distinct meromorphic func-
tions and
119865 (119911) =
119875 (119911)
119876 (119911)
=
sum120582isin119868
120572120582(119911) 119891
1198971205821
1119891
1198971205822
2 119891
119897120582119899
119899
sum120583isin119869
120573120583(119911) 119891
1198981205831
1119891
1198981205832
2 119891
119898120583119899
119899
(27)
4 Abstract and Applied Analysis
Then
119879 (119903 119865 (119911)) le
119899
sum
]=1120590]119879 (119903 119891]) + 119878 (119903 119891) (28)
where 119868 = 120582 = (1198971205821 1198971205822 119897120582119899) | 119897
120582] isin N⋃0 ] =
1 2 119899 and 119869 = 120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin
N⋃0 ] = 1 2 119899 are two finite index sets 120590] =
max120582120583119897120582] 119898120583] (] = 1 2 119899) 120572
120582(119911) = 119900(119879(119903 119891])(120582 isin 119868))
and120573120583(119911) = 119900(119879(119903 119891])(120583 isin 119869)) hold for all ] isin 1 2 119899 and
satisfy 119879(119903 120572120582) = 119878(119903 119891) (120582 isin 119868) and 119879(119903 120573
120583) = 119878(119903 119891) (120583 isin
119869)
Lemma 3 (see [7]) Let 119888 be a complex constant Given 120576 gt 0
and a meromorphic function 119891 one has
119879 (119903 119891 (119911 plusmn 119888)) le (1 + 120576) 119879 (119903 + |119888| 119891) (29)
for all 119903 gt 1199030 where 119903
0is some positive constant
Lemma 4 (see [4]) Let 119892 (0 +infin) rarr R ℎ (0 +infin) rarr R
bemonotone increasing functions such that 119892(119903) le ℎ(119903) outsideof an exceptional set 119864 of finite linear measure Then for any120572 gt 1 there exists 119903
0gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 gt 119903
0
Lemma 5 (see [10]) Let 119891 be a transcendental meromorphicfunction and 119901(119911) = 119886
119896119911119896
+ 119886119896minus1
119911119896minus1
+ sdot sdot sdot + 1198861119911 + 1198860 119886119896
= 0be a nonconstant polynomial of degree 119896 Given 0 lt 120575 lt |119886
119896|
denote 120582 = |119886119896| + 120575 and 120583 = |119886
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C⋃infin one has
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 (119901 (119911))) le 119896119899 (120582119903119896
119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 (119901 (119911)))
le 119873 (120582119903119896
119886 119891) + 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 (119901 (119911))) le (1 + 120576) 119879 (120582119903119896
119891)
(30)
for all 119903 large enough
Lemma 6 (see [11]) Let 120601 [1199030 +infin) rarr (0 +infin) be
positive and bounded in every finite interval and suppose that120601(120583119903119898
) le 119860120601(119903) + 119861 holds for all 119903 large enough where 120583 gt 0119898 gt 1 119860 gt 1 and 119861 are real constants Then
120601 (119903) = 119874 ((log 119903)120572) (31)
where 120572 = log119860 log119898
Lemma 7 (see [6]) Let 120601 (1199030infin) rarr (1infin) where 119903
0ge 1
be a monotone increasing function If for some real constant120572 gt 1 there exists a real number 119870 gt 1 such that 120601(120572119903) ge
119870120601(119903) then
lim119903rarrinfin
log120601 (119903)log 119903
ge
log119870log120572
(32)
Lemma 8 (see [12]) Let 120601 (1infin) rarr (0infin) be a monotoneincreasing function and let 119891 be a nonconstant meromorphic
function If for some real constant 120572 isin (0 1) there exist realconstants 119870
1gt 0 and 119870
2ge 1 such that
119879 (119903 119891) le 1198701120601 (120572119903) + 119870
2119879 (120572119903 119891) + 119878 (120572119903 119891) (33)
then
120588 (119891) le
log1198702
minus log120572+ lim119903rarrinfin
log120601 (119903)log 119903
(34)
3 Proof of Theorems
Proof of Theorem 1 We assume 119891(119911) is a transcendentalmeromorphic solution of (10) Denoting 119862 =
max|1198881| |1198882| |119888
119899| According to Lemmas 1 2 and 3
and the last assertion of Lemma 5 we get that for any 1205761gt 0
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) + 119878 (119903 119891)
le 119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 1205761) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(35)
where 119903 is large enough and120583 = |119901119896|minus120575 for some 0 lt 120575 lt |119901
119896|
Since 119879(119903 + 119862 119891) le 119879(120573119903 119891) holds for 119903 large enough for120573 gt 1 we may assume 119903 to be large enough to satisfy
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120573119903 119891) (36)
outside a possible exceptional set of finite linear measure ByLemma 4 we know that whenever 120574 gt 1
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120574120573119903 119891) (37)
holds for all 119903 large enough Denote 119905 = 120574120573119903 thus theinequality (37) may be written in the form
119879(
120583
(120574120573)119896
119905119896
119891) le
120590 (1 + 1205761)
119902 (1 minus 1205761)
119879 (119905 119891) (38)
By Lemma 6 we have
119879 (119903 119891) = 119874 ((log 119903)1205721) (39)
Abstract and Applied Analysis 5
where
1205721=
log (120590 (1 + 1205761) 119902 (1 minus 120576
1))
log 119896
=
log120590 minus log 119902log 119896
+
log ((1 + 1205761) (1 minus 120576
1))
log 119896
(40)
Denoting now 120572 = (log120590 minus log 119902) log 119896 and 120576 = log((1 +
1205761)(1 minus 120576
1)) log 119896 thus we obtain the required form
Finally we show that 119902119896 le 120590 If 119902119896 gt 120590 then we have120572 lt 1 For sufficiently small 120576 gt 0 we have 120572 + 120576 lt 1 whichcontradicts with the transcendency of 119891 Thus Theorem 1 isproved
Proof of Theorem 2 Suppose 119891(119911) is a transcendental mero-morphic solution of (13) Denoting 119862 = max|119888
1| |1198882|
|119888119899|
(i) 0 lt |119886| lt 1 We may assume that 119902 gt 120590 since the case119902 le 120590 is trivial by the fact that 120583(119891) ge 0 By Lemmas1ndash3 we have for any 120576 gt 0 and 120573 gt 1
119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
le 120590 (1 + 120576) 119879 (120573119903 119891) + 119878 (119903 119891)
(41)
where 119903 is large enoughBy the last assertion of Lemma 5 and (41) we obtain that
for 120583 = |119886| minus 120575 (0 lt 120575 lt |119886| 0 lt 120583 lt 1) the followinginequality
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120573119903 119891) (42)
holds where 119903 is large enough outside of a possible set of finitelinear measure By Lemma 4 we get that for any 120574 gt 1 andsufficiently large 119903
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120574120573119903 119891) (43)
Therefore
119902 (1 minus 120576)
120590 (1 + 120576)
119879 (119903 119891) le 119879(
120574120573
120583
119903 119891) (44)
Since 120573 gt 1 120574 gt 1 0 lt 120583 lt 1 and 119902 gt 120590 we have 120573120574120583 gt 1
and 119902(1 minus 120576)120590(1 + 120576) gt 1 when 120576 is small enough UsingLemma 7 we see that
120583 (119891) ge
log 119902 (1 minus 120576) minus log120590 (1 + 120576)
log 120574120573 minus log 120583 (45)
Letting 120576 rarr 0 120575 rarr 0 120573 rarr 1 and 120574 rarr 1 we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(46)
(ii) |119886| gt 1 By the similar reasoning as is (i) we easilyobtain that
119902 (1 minus 120576) 119879 (120583119903 119891) le 119902119879 (119903 119891 (119901 (119911)))
le 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(47)
for all 119903 large enough We may select sufficiently smallnumbers 120575 gt 0 and 120576 gt 0 such that 120583 = |119886| minus 120575 gt 1 and(1120583) + 120576 lt 1 Thus we have
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891) (48)
namely
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) (49)
where 119903 is large enough possibly outside of a set of finite linearmeasure By Lemma 4 we have for any 1 lt 120574 lt 120583
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (120574119903 119891 (119911)) (50)
that is
119879 (119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879(
120574
120583
119903 119891 (119911)) (51)
holds for all sufficiently large 119903 By Lemma 8 we obtain
120588 (119891) le
log120590 minus log 119902 + log (1 + 120576) minus log (1 minus 120576)
minus log (120574120583) (52)
Letting 120576 rarr 0 120575 rarr 0 and 120574 rarr 1 we have
120588 (119891) le
log120590 minus log 119902log |119886|
(53)
(iii) |119886| = 1 and 119902 gt 120590 The proof of this case is completelysimilar as in the case in (i) In fact we set 120583 = |119886|minus120575 =
1 minus 120575 (0 lt 120575 lt 1 0 lt 120583 lt 1) Similarly we can get
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(54)
Since |119886| = 1 we have 120583(119891) = 120588(119891) = infin
6 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the anonymous ref-erees for their valuable comments and suggestions Theresearch was supported by Colonel-level topics (JSNU-ZY-01) (Jsie2012zd01) and NSF of China (11271179)
References
[1] W Cherry and Z Ye Nevanlinnarsquos Theory of Value DistributionSpringer Monographs in Mathematics Springer Berlin Ger-many 2001
[2] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[3] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Beijing China 1988
[4] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] J Rieppo ldquoOn a class of complex functional equationsrdquoAnnalesAcademiaelig Scientiarum Fennicaelig vol 32 no 1 pp 151ndash170 2007
[7] X-M Zheng Z-X Chen and J Tu ldquoGrowth of meromorphicsolutions of some difference equationsrdquoApplicable Analysis andDiscrete Mathematics vol 4 no 2 pp 309ndash321 2010
[8] A Z Mokhonrsquoko ldquoThe Nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija vol 14 pp 83ndash87 1971 (Russian)
[9] A A Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoAkademija Nauk SSSR vol 15 pp 1305ndash1322 1974
[10] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[11] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[12] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
Research ArticleUnicity of Entire Functions concerning Shifts andDifference Operators
Dan Liu Degui Yang and Mingliang Fang
Institute of Applied Mathematics South China Agricultural University Guangzhou 510642 China
Correspondence should be addressed to Mingliang Fang mlfangscaueducn
Received 29 October 2013 Revised 17 December 2013 Accepted 19 December 2013 Published 3 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Dan Liu et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We prove a unicity theorem of entire functions that share two distinct small functions with their shiftsThe corollary of the theoremconfirms the conjecture posed by Li and Gao (2011)
1 Introduction
Let119891 be a nonconstantmeromorphic function in the complexplane C We will use the standard notations in Nevanlinnatheory of meromorphic functions such as 119879(119903 119891) 119873(119903 119891)and 119898(119903 119891) (see [1 2]) The notation 119878(119903 119891) is defined to beany quantity satisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin pos-sibly outside a set of finite linear measures A meromorphicfunction 119886 is called a small function related to119891 provided that119879(119903 119886) = 119878(119903 119891)
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 be a small function related to both 119891 and 119892 Wesay that 119891 and 119892 share 119886 CM if 119891 minus 119886 and 119892 minus 119886 have thesame zeros with the same multiplicities 119891 and 119892 are said toshare 119886 IM if 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
Let119873(119903 119886) be the counting functions of all common zeroswith the same multiplicities of 119891 minus 119886 and 119892 minus 119886 If
119873(119903
1
119891 minus 119886
) + 119873(119903
1
119892 minus 119886
) minus 2119873 (119903 119886)
= 119878 (119903 119891) + 119878 (119903 119892)
(1)
then we say that 119891 and 119892 share 119886 CM almostFor a nonzero complex constant 119888 isin C we define
difference operators asΔ119888119891(119911) = 119891(119911+119888)minus119891(119911) andΔ119899
119888119891(119911) =
Δ119888(Δ119899minus1
119888119891(119911)) 119899 isin N 119899 gt 2
In 1977 Rubel and Yang [3] proved the following result
Theorem A Let 119891 be a nonconstant entire function If 119891(119911)and 1198911015840(119911) share two distinct finite values CM then 119891(119911) equiv
1198911015840
(119911)
In fact the conclusion still holds if the two CM values arereplaced by two IM values (see Gundersen [4 5] Mues andSteinmetz [6])
Recently a number of articles focused on value dis-tribution in shifts or difference operators of meromorphicfunctions (see [7ndash11]) In particular some papers studied theunicity of meromorphic functions sharing values with theirshifts or difference operators (see [12ndash14]) In 2009 Heit-tokangas et al [12] proved the following result concerningshifts
Theorem B Let 119891 be a nonconstant entire function of finiteorder 119888 isin C If119891(119911) and119891(119911+119888) share two distinct finite valuesCM then 119891(119911) equiv 119891(119911 + 119888)
In 2011 Li and Gao [14] proved the following resultconcerning difference operators
Theorem C Let 119891 be a nonconstant entire function of finiteorder 119888 isin C and let 119899 be a positive integer Suppose that 119891(119911)and Δ119899
119888119891(119911) share two distinct finite values 119886 119887 CM and one of
the following cases is satisfied(i) 119886119887 = 0(ii) 119886119887 = 0 and 120588(119891) notin 119873Then 119891(119911) equiv Δ119899
119888119891(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 380910 5 pageshttpdxdoiorg1011552014380910
2 Abstract and Applied Analysis
In [14] Li andGao conjectured that the restriction 120588(119891) notinN for the case 119886119887 = 0 can be removed In this paper weconfirm their conjecture In fact we prove the followingmoregeneral results
Theorem 1 Let 119891 be a nonconstant entire function of finiteorder let 119899 be a positive integer let 119886(119911) 119887(119911) be two distinctsmall functions related to 119891(119911) let 119898
1 1198982 119898
119899be nonzero
complex numbers and 1198881 1198882 119888
119899distinct finite values and let
119865 (119911) = 1198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119891 (119911 + 119888
119899)
(2)
If 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119865(119911)
Corollary 2 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number let 119899 be apositive integer and let 119886 119887 be two distinct finite values If 119891(119911)and Δ119899
119888119891(119911) share 119886 119887 CM then 119891(119911) equiv Δ119899
119888119891(119911)
Remark 3 Corollary 2 confirms the conjecture of Li and Gaoin [14]
Corollary 4 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number and let 119886(119911)119887(119911) be two distinct small functions related to 119891 If 119891(119911) and119891(119911 + 119888) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119891(119911 + 119888)
2 Some Lemmas
For the proof of Theorem 1 we require the following results
Lemma 5 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903
1
119891
) + 119873 (119903 119891) = 119878 (119903 119891)
119873(119903
1
119892
) + 119873 (119903 119892) = 119878 (119903 119892)
(3)
If 119891(119911) and 119892(119911) share 1 CM almost then either 119891(119911) equiv 119892(119911)
or 119891(119911)119892(119911) equiv 1
Lemma 6 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903 119891) = 119878 (119903 119891) 119873 (119903 119892) = 119878 (119903 119892) (4)
If 119891(119911) and 119892(119911) share 0 and 1 CM almost and
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 119891) + 119879 (119903 119892)
lt
2
3
(5)
where 119868 sub [0infin) is a set of infinitely linear measure then
119891 (119911) =
119886119892 (119911) + 119887
119888119892 (119911) + 119889
(6)
where 119886 119887 119888 and 119889 are constants satisfying 119886119889 minus 119887119888 = 0
Lemma 7 (see [10]) Let 119891 be a nonconstant meromorphicfunction of finite order 119888 isin C Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (7)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
119889119903119903 lt infin
In the following 119878(119903 119891) denotes any function satisfying119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside a set withfinite logarithmic measure
3 Proof of Theorem 1
We prove Theorem 1 by contradiction Suppose that 119891(119911) equiv
119865(119911) Then it follows from 119891(119911) and 119865(119911) being two distinctentire functions that 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) and infinCM By the Nevanlinna second fundamental theorem forthree small functions we have
119879 (119903 119891) le 119873 (119903 119891) + 119873(119903
1
119891 minus 119886
)
+ 119873(119903
1
119891 minus 119887
) + 119878 (119903 119891)
le 119873(119903
1
119865 minus 119886
) + 119873(119903
1
119865 minus 119887
) + 119878 (119903 119891)
le 2119879 (119903 119865) + 119878 (119903 119891)
(8)
Similarly we have 119879(119903 119865) le 2119879(119903 119891) + 119878(119903 119865) Therefore119878(119903 119891) = 119878(119903 119865)
Set
1198911(119911) =
119891 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
1198651(119911) =
119865 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
(9)
Thus 1198911(119911) 1198651(119911) share 0 1 andinfin CM almost
Obviously we have
119879 (119903 1198911) = 119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 1198651) = 119879 (119903 119865) + 119878 (119903 119891)
119878 (119903 119865) = 119878 (119903 1198651) = 119878 (119903 119891
1) = 119878 (119903 119891)
(10)
By Nevanlinnarsquos second fundamental theorem we have
119879 (119903 1198911) le 119873(119903
1
1198911
) + 119873(119903
1
1198911minus 1
) + 119873 (119903 1198911) + 119878 (119903 119891
1)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
Abstract and Applied Analysis 3
le 119873(119903
1
1198651minus 1198911
) + 119878 (119903 119891)
le 119879 (119903 1198651minus 1198911) + 119878 (119903 119891)
le 119879 (119903 119865 minus 119891) + 119878 (119903 119891)
le 119898 (119903 119865 minus 119891) + 119878 (119903 119891)
(11)Since 119865 minus 119891 = 119898
1119891(119911 + 119888
1) + 1198982119891(119911 + 119888
2) + sdot sdot sdot + 119898
119899119891(119911 +
119888119899) minus 119891(119911) = 119891(119911)[119898
1(119891(119911 + 119888
1)119891(119911)) +119898
2(119891(119911 + 119888
2)119891(119911)) +
sdot sdot sdot + 119898119899(119891(119911 + 119888
119899)119891(119911)) minus 1] thus
119898(119903 119865 minus 119891)
le 119898 (119903 119891)
+ 119898(1199031198981
119891 (119911 + 1198881)
119891 (119911)
+ sdot sdot sdot + 119898119899
119891 (119911 + 119888119899)
119891 (119911)
minus 1)
le 119898 (119903 119891) + 119878 (119903 119891)
(12)By (11) we have
119879 (119903 1198911) le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(13)
It follows that119873(119903 0) + 119873 (119903 1) = 119879 (119903 119891
1) + 119878 (119903 119891) (14)
On the other hand byNevanlinna first fundamental theoremwe have
2119879 (119903 1198911) = 119879(119903
1
1198911
) + 119879(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
(15)
So we get
119879 (119903 1198911) le 119898(119903
1
1198911
) + 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119898(119903
1
119891 minus 119886
) + 119898(119903
1
119891 minus 119887
) + 119878 (119903 119891)
(16)
Set1198861(119911) = 119898
1119886 (119911 + 119888
1) + 1198982119886 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119886 (119911 + 119888
119899)
1198871(119911) = 119898
1119887 (119911 + 119888
1) + 1198982119887 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119887 (119911 + 119888
119899)
(17)
If 1198861(119911) equiv 119887
1(119911) we can deduce by (16) that
119879 (119903 1198911) le 119898(119903
1
119891 minus 119886
+
1
119891 minus 119887
) + 119878 (119903 119891)
le 119898(119903
119865 minus 1198861
119891 minus 119886
+
119865 minus 1198871
119891 minus 119887
)
+ 119898(119903
1
119865 minus 1198861
) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(18)
If 1198861(119911) equiv 119887
1(119911) set
119871 (119865) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
119865 11988611198871
1198651015840
1198861015840
11198871015840
1
11986510158401015840
11988610158401015840
111988710158401015840
1
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(19)
Then we have
119898(119903
119865 minus 1198861
119891 minus 119886
) = 119898(119903
119865 minus 1198871
119891 minus 119887
) = 119878 (119903 119891)
119898(119903
119871 (119865)
119865 minus 1198861
) = 119898(119903
119871 (119865)
119865 minus 1198871
) = 119878 (119903 119891)
(20)
It followed from (16) that
119879 (119903 1198911) le 119898(119903
119865 minus 1198861
119891 minus 119886
) + 119898(119903
1
119865 minus 1198861
)
+ 119898(119903
119865 minus 1198871
119891 minus 119887
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
+
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119871 (119865)
) + 119878 (119903 119891)
le 119879 (119903 119871 (119865)) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
4 Abstract and Applied Analysis
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
le 119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 1198911) + 119878 (119903 119891)
(21)
By (18) and (21) we can deduce that
119879 (119903 1198911) = 119879 (119903 119865) + 119878 (119903 119891) = 119879 (119903 119865
1) + 119878 (119903 119891) (22)
It follows from (14) and (22) that
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 1198911) + 119879 (119903 119865
1)
=
1
2
lt
2
3
(23)
By Lemma 6 we have
1198911(119911) =
1198601198651(119911) + 119861
1198621198651(119911) + 119863
(24)
where 119860 119861 119862 and 119863 are complex numbers satisfying 119860119863 minus
119861119862 = 0Now we consider three cases
Case 1 Consider 119873(119903 0) = 119878(119903 1198911) Thus
119873(119903
1
1198911
) + 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (25)
Similarly we have
119873(119903
1
1198651
) + 119873 (119903 1198651) = 119878 (119903 119865
1) = 119878 (119903 119891) (26)
By Lemma 5 we get that either 1198911equiv 1198651or 11989111198651equiv 1
If 1198911equiv 1198651 we can easily deduce that 119891 equiv 119865 which is a
contradiction with our assumptionIf 11989111198651equiv 1 that is
(119891 (119911) minus 119886) (119865 (119911) minus 119886) equiv (119887 minus 119886)2
(27)
then we have
(119891 minus 119886)2
=
(119887 minus 119886)2
(119865 minus 119886) (119891 minus 119886)
(28)
From (28) we have
2119879 (119903 119891) le 119879 (119903 (119891 minus 119886)2
) + 119878 (119903 119891)
= 119879(119903
1
(119887 minus 119886)2
((119865 minus 119886) (119891 minus 119886))
) + 119878 (119903 119891)
le 119879(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
= 119873(119903
119865 minus 119886
119891 minus 119886
) + 119898(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
(119865 minus 1198861) + (119886
1minus 119886)
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
1198861minus 119886
119891 minus 119886
) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
(29)
It follows that 119879(119903 119891) le 119878(119903 119891) a contradiction
Case 2 Consider 119873(119903 1) = 119878(119903 1198911) Using the same argu-
ment as used in Case 1 we deduce that 119879(119903 119891) le 119878(119903 119891) acontradiction
Case 3 Consider 119873(119903 0) = 119878(119903 1198911) 119873(119903 1) = 119878(119903 119891
1) Since
1198911and 119865
1share 0 1 CM almost we deduce from (24) that
1198911(119911) =
(119862 + 119863) 1198651(119911)
1198621198651(119911) + 119863
(30)
If 119862 = 0 then 1198911equiv 1198651 that is 119891 equiv 119865 a contradiction
Hence 119862 = 0 Thus we have
119873(119903
1
1198651+ (119863119862)
) = 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (31)
Obviously 119863119862 = 0 119863119862 = minus 1 Thus by Nevanlinnasecond fundamental theorem and (14) we get
2119879 (119903 1198911) = 2119879 (119903 119865
1) + 119878 (119903 119891
1)
le 119873(119903
1
1198651
) + 119873(119903
1
1198651minus 1
)
+ 119873(119903
1
1198651+ (119863119862)
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119878 (119903 119891)
(32)
It follows that 119879(119903 1198911) le 119878(119903 119891
1) a contradiction Thus
we prove that 119891(119911) equiv 119865(119911) This completes the proof ofTheorem 1
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 5
Acknowledgments
Theauthors thank the referees for careful reading of the paperpointing out a gap in the previous version of this paperand giving many valuable suggestions Research is supportedby the NNSF of China (Grant no 11371149) and NSF ofGuangdong Province China (Grant no S2012010010121)
References
[1] W K Hayman Meromorphic Function Clarendon PressOxford UK 1964
[2] L Yang Value Distribution Theory Springer Berlin Germany1993
[3] L A Rubel and C C Yang Value Shared by an Entire Functionand Its Derivative Lecture Notes in Math Springer BerlinGermany 1977
[4] G G Gundersen ldquoMeromorphic functions that share finitevalues with their derivativerdquo Journal of Mathematical Analysisand Applications vol 75 no 2 pp 441ndash446 1980
[5] G G Gundersen ldquoErrata meromorphic functions that sharefinite values with their derivativerdquo Journal of MathematicalAnalysis and Applications vol 86 no 1 p 307 1982
[6] E Mues and N Steinmetz ldquoMeromorphe Funktionen die mitihrer Ableitung Werte teilenrdquo Manuscripta Mathematica vol29 no 2ndash4 pp 195ndash206 1979
[7] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[8] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoTheRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[9] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[10] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[11] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[12] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[13] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables and Elliptic Equations vol 56 no 1ndash4 pp 81ndash92 2011
[14] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[15] M L Fang ldquoUnicity theorems for meromorphic function andits differential polynomialrdquo Advances in Mathematics vol 24no 3 pp 244ndash249 1995
Research ArticleOn Positive Solutions and Mann Iterative Schemes of a ThirdOrder Difference Equation
Zeqing Liu1 Heng Wu1 Shin Min Kang2 and Young Chel Kwun3
1 Department of Mathematics Liaoning Normal University Dalian Liaoning 116029 China2Department of Mathematics and RINS Gyeongsang National University Jinju 660-701 Republic of Korea3 Department of Mathematics Dong-A University Pusan 614-714 Republic of Korea
Correspondence should be addressed to Young Chel Kwun yckwundauackr
Received 14 October 2013 Accepted 16 December 2013 Published 28 January 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Zeqing Liu et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The existence of uncountably many positive solutions and convergence of the Mann iterative schemes for a third order nonlinearneutral delay difference equation are proved Six examples are given to illustrate the results presented in this paper
1 Introduction and Preliminaries
Recently many researchers studied the oscillation nonoscil-lation and existence of solutions for linear and nonlinearsecond and third order difference equations and systemssee for example [1ndash23] and the references cited therein Bymeans of the Reccati transformation techniques Saker [18]discussed the third order difference equation
Δ3
119909119899+ 119901119899119909119899+1
= 0 forall119899 ge 1198990 (1)
and presented some sufficient conditions which ensure thatall solutions are to be oscillatory or tend to zero Utilizing theSchauder fixed point theorem Yan and Liu [22] proved theexistence of a bounded nonoscillatory solution for the thirdorder difference equation
Δ3
119909119899+ 119891 (119899 119909
119899 119909119899minus120591) = 0 forall119899 ge 119899
0 (2)
Agarwal [2] established the oscillatory and asymptotic prop-erties for the third order nonlinear difference equation
Δ3
119909119899+ 119902119899119891 (119909119899+1) = 0 forall119899 ge 1 (3)
Andruch-Sobiło andMigda [4] studied the third order lineardifference equation of neutral type
Δ3
(119909119899minus 119901119899119909120590119899) plusmn 119902119899119909120591119899= 0 forall119899 ge 119899
0 (4)
and obtained sufficient conditions which ensure that allsolutions of the equation are oscillatory Grace andHamedani[6] discussed the difference equation
Δ3
(119909119899minus 119909119899minus120591) plusmn 119902119899
1003816100381610038161003816119909119899minus120590
1003816100381610038161003816
3 sgn119909119899minus120590
= 0 forall119899 ge 0 (5)
and gave some new criteria for the oscillation of all solutionsand all bounded solutions
Our goal is to discuss solvability and convergence ofthe Mann iterative schemes for the following third ordernonlinear neutral delay difference equation
Δ3
(119909119899+ 119887119899119909119899minus120591) + Δℎ (119899 119909
ℎ1119899
119909ℎ2119899
119909ℎ119896119899
)
+119891 (119899 1199091198911119899
1199091198912119899
119909119891119896119899
) = 119888119899 forall119899 ge 119899
0
(6)
where 120591 119896 1198990isin N 119887
119899119899isinN1198990
119888119899119899isinN1198990
sub R ℎ 119891 isin 119862(N1198990
times
R119896R) ℎ119897119899119899isinN1198990
119891119897119899119899isinN1198990
sube N and
lim119899rarrinfin
ℎ119897119899= lim119899rarrinfin
119891119897119899= +infin 119897 isin 1 2 119896 (7)
By employing the Banach fixed point theorem and somenew techniques we establish the existence of uncountablymany positive solutions of (6) conceive a few Mann iter-ative schemes for approximating these positive solutionsand prove their convergence and the error estimates Sixnontrivial examples are included
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 470181 16 pageshttpdxdoiorg1011552014470181
2 Abstract and Applied Analysis
Throughout this paper we assume that Δ is the forwarddifference operator defined by Δ119909
119899= 119909119899+1
minus 119909119899 R =
(minusinfin +infin) R+ = [0 +infin) N0and N denote the sets of
nonnegative integers and positive integers respectively
N119905= 119899 119899 isin N with 119899 ge 119905 forall119905 isin N
120573 = min 1198990minus 120591 inf ℎ
119897119899 119891119897119899 1 le 119897 le 119896 119899 isin N
1198990
isin N
119867119899= max ℎ2
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
119865119899= max 1198912
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
(8)
and 119897infin
120573represents the Banach space of all real sequences
on N120573with norm
119909 = sup119899isinN120573
1003816100381610038161003816100381610038161003816
119909119899
1198992
1003816100381610038161003816100381610038161003816
lt +infin for each 119909 = 119909119899119899isinN120573
isin 119897infin
120573
119860 (119873119872) = 119909 = 119909119899119899isinN120573
isin 119897infin
120573 119873 le
119909119899
1198992
le 119872 119899 isin N120573
for any 119872 gt 119873 gt 0
(9)
It is easy to see that 119860(119873119872) is a closed and convex subsetof 119897infin120573 By a solution of (6) wemean a sequence 119909
119899119899isinN120573
witha positive integer 119879 ge 119899
0+120591+120573 such that (6) holds for all 119899 ge
119879
Lemma 1 Let 119901119905119905isinN be a nonnegative sequence and 120591 isin N
(i) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199052
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119904=119899+119894120591suminfin
119905=119904119901119905= 0
(ii) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199053
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119906=119899+119894120591suminfin
119904=119906suminfin
119905=119904119901119905= 0
Proof Note that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
(
infin
sum
119905=119899+119894120591
119901119905+
infin
sum
119905=119899+1+119894120591
119901119905+
infin
sum
119905=119899+2+119894120591
119901119905+ sdot sdot sdot )
=
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
(1 + 119905 minus 119899 minus 119894120591) 119901119905le
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
119905119901119905
=
1
1198992
(
infin
sum
119905=119899+120591
119905119901119905+
infin
sum
119905=119899+2120591
119905119901119905+
infin
sum
119905=119899+3120591
119905119901119905+ sdot sdot sdot )
le
1
1198992
infin
sum
119905=119899+120591
(1 +
119905 minus 119899 minus 120591
120591
) 119905119901119905=
1
1198992
infin
sum
119905=119899+120591
119905 minus 119899
120591
119905119901119905
le
1
1198992120591
infin
sum
119905=119899+120591
1199052
119901119905997888rarr 0 as 119899 997888rarr infin
(10)
that is
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905= 0 (11)
As in the proof of (10) we infer that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
(1 + 119905 minus 119906) 119901119905
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
119905119901119905le
1
1198992120591
infin
sum
119905=119899+120591
1199053
119901119905997888rarr 0 as 119899 997888rarr infin
(12)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905= 0 (13)
This completes the proof
2 Uncountably Many Positive Solutions andMann Iterative Schemes
In this section using the Banach fixed point theoremand Mann iterative schemes we establish the existence ofuncountably many positive solutions of (6) prove conver-gence of the Mann iterative schemes relative to these positivesolutions and compute the error estimates between theManniterative schemes and the positive solutions
Theorem 2 Assume that there exist twoconstants 119872 and 119873 with 119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896) minus 119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119875119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896) minus ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119877119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
forall (119899 119906119897 119906119897) isin N1198990
times (R+
0)
2
1 le 119897 le 119896
(14)
1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119876119899
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119882119899
forall (119899 119906119897) isin N1198990
times (R+
0) 1 le 119897 le 119896
(15)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (16)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (17)
119887119899= minus1 eventually (18)
Abstract and Applied Analysis 3
Then one has the following(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
+
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(19)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin 119860(119873119872) of (6)with lim
119899rarrinfin119911119899= +infin and has the following error estimate
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] such thatinfin
sum
119898=0
120572119898= +infin (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Firstly we show that (a) holds Put 119871 isin (119873119872) Itfollows from (16)sim(18) that there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (22)
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119873
(23)
119887119899= minus1 forall119899 ge 119879 (24)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879 119878119871119909119879 120573 le 119899 lt 119879
(25)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In light of (14) (15) (22)(23) and (25) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872)
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
4 Abstract and Applied Analysis
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
lt min 119872 minus 119871 119871 minus 119873
(26)which yield that
119878119871(119860 (119873119872)) sube 119860 (119873119872)
1003817100381710038171003817119878119871119909 minus 1198781198711199101003817100381710038171003817le 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817 forall119909 119910 isin 119860 (119873119872)
(27)
which implies that 119878119871is a contraction in 119860(119873119872) The
Banach fixed point theorem and (27) ensure that 119878119871has a
unique fixed point 119911 = 119911119899119899isinN120573
isin 119860(119873119872) that is
119911119899= 1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879
119911119899minus120591
= (119899 minus 120591)2
119871
+
infin
sum
119894=1
infin
sum
119906=119899+(119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
)
minus 119888119905] forall119899 ge 119879 + 120591
(28)which mean that119911119899minus 119911119899minus120591
= (2119899120591 minus 1205912
) 119871
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
(29)which yields thatΔ (119911119899minus 119911119899minus120591)
= 2120591119871 +
infin
sum
119904=119899
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
Δ2
(119911119899minus 119911119899minus120591)
= minusℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
+
infin
sum
119905=119899
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905] forall119899 ge 119879 + 120591
(30)
which gives that
Δ3
(119911119899minus 119911119899minus120591)
= minusΔℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
minus 119891 (119905 1199111198911119899
1199111198912119899
119911119891119896119899
) + 119888119899 forall119899 ge 119879 + 120591
(31)
which together with (24) implies that 119911 = 119911119899119899isinN120573
is apositive solution of (6) in 119860(119873119872) Note that
119873 le
119911119899
1198992
le 119872 forall119899 isin N120573 (32)
which guarantees that lim119899rarrinfin
119911119899= +infin It follows from (19)
(22) (24) (25) and (27) that for any 119898 isin N0and 119899 ge 119879
1003816100381610038161003816100381610038161003816
119909119898+1119899
1198992
minus
119911119899
1198992
1003816100381610038161003816100381610038161003816
=
1
1198992
1003816100381610038161003816100381610038161003816100381610038161003816
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)] minus 119911
119899
1003816100381610038161003816100381610038161003816100381610038161003816
le (1 minus 120572119898)
1003816100381610038161003816119909119898119899minus 119911119899
1003816100381610038161003816
1198992
+ 120572119898
1003816100381610038161003816119878119871119909119898119899minus 119878119871119911119899
1003816100381610038161003816
1198992
le (1 minus 120572119898)1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817+ 120579120572119898
1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817
le [1 minus (1 minus 120579) 120572119898]1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 119899 ge 119879
(33)
which implies that
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (34)
That is (20) holds Thus Lemma 1 (20) and (21) guaranteethat lim
119898rarrinfin119909119898= 119911
Next we show that (b) holds Let 1198711 1198712
isin
(119873119872) and 1198711=1198712 As in the proof of (a) we deduce
similarly that for each 119888 isin 1 2 there exist constants 120579119888isin
(0 1) and 119879119888ge 1198990+ 120591 + 120573 and a mapping 119878
119871119888
satisfying
Abstract and Applied Analysis 5
(22)sim(27) where 120579 119871 and 119879 are replaced by 120579119888 119871119888 and 119879
119888
respectively and the mapping 119878119871119888
has a fixed point 119911119888 =
119911119888
119899119899isinN120573
isin 119860(119873119872) which is a positive solution of (6) in119860(119873119872) with lim
119899rarrinfin119911119888
119899= +infin It follows that
119911119888
119899= 1198992
119871119888
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911119888
ℎ1119904
119911119888
ℎ2119904
119911119888
ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 119911119888
1198911119905
119911119888
1198912119905
119911119888
119891119896119905
) minus 119888119905]
forall119899 ge 119879119888
(35)
which together with (14) and (20) means that for 119899 ge
max1198791 1198792
100381610038161003816100381610038161003816100381610038161003816
1199111
119899
1198992
minus
1199112
119899
1198992
100381610038161003816100381610038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119911
1
ℎ1119904
1199111
ℎ2119904
1199111
ℎ119896119904
)
minus ℎ (119904 1199112
ℎ1119904
1199112
ℎ2119904
1199112
ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119911
1
1198911119905
1199111
1198912119905
1199111
119891119896119905
) minus 119891 (119905 1199112
1198911119905
1199112
1198912119905
1199112
119891119896119905
)
10038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
10038161199111
ℎ119897119904
minus 1199112
ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
10038161199111
119891119897119905
minus 1199112
119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
max 11987921 1198792
2
infin
sum
119894=1
infin
sum
119906=max1198791 1198792+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816minusmax 120579
1 1205792
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
(36)
which yields that
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817ge
10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
1 +max 1205791 1205792
gt 0 (37)
that is 1199111 =1199112
This completes the proof
Theorem 3 Assume that there exist two constants119872 and 119873with 119872 gt 119873 gt 0 and four nonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) and
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (38)
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (39)
119887119899= 1 eventually (40)
Then one has the following
(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
minus
infin
sum
119894=1
119879+2119894120591minus1
sum
119906=119879+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(41)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
6 Abstract and Applied Analysis
Proof Let 119871 isin (119873119872) It follows from (38)sim(40) that thereexist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (42)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)) lt min 119872 minus 119871 119871 minus 119873
(43)
119887119899= 1 forall119899 ge 119879 (44)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus119888119905] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(45)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Using (14) (15) (42) (43)and (45) we get that for each 119909 = 119909
119899119899isinN120573
119910 = 119910119899119899isinN120573
isin
119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) = 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)])
lt min 119872 minus 119871 119871 minus 119873
(46)
which imply (27) The rest of the proof is similar to the proofof Theorem 2 and is omitted This completes the proof
Theorem 4 Assume that there exist three constants 119887 119872and 119873 with (1 minus 119887)119872 gt 119873 gt 0 and four nonnega-tive sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and0 le 119887119899le 119887 lt 1 eventually (47)
Then one has the following(a) For any 119871 isin (119887119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any
1199090
= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated bythe scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871 minus 119887119899119909119898119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus 119887119879119909119898119879minus120591
minus
infin
sum
119906=119879
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(48)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
Abstract and Applied Analysis 7
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119887119872 + 119873119872) It follows from (38) (39) and(47) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = 119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119887119872 minus119873
0 le 119887119899le 119887 lt 1 forall119899 ge 119879
(49)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus 119887119899119909119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(50)for each 119909 = 119909
119899119899isinN120573
isin 119860(119873119872) In view of (14) (15) and(49) and (50) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 11987910038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le 119887119899
100381610038161003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
(119899 minus 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 minus 120591)2
1198992
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le 1198871003817100381710038171003817119909 minus 119910
1003817100381710038171003817
+
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le [119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817= 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus 119887119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus 119887119872
minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus 119887119872 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minus 119887119872 minusmin 119872 minus 119871 119871 minus 119887119872 minus119873 ge 119873
(51)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 5 Assume that there exist constants 119887 119872 and 119873with (1 + 119887)119872 gt 119873 gt 0 and four nonnegative sequences119875119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14)(15) (38) (39) and
minus1 lt 119887 le 119887119899le 0 eventually (52)
Then one has the following
(a) For any 119871 isin (119873 (1 + 119887)119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for
any 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by(48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
8 Abstract and Applied Analysis
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119873 (1 + 119887)119872) It follows from (38) (39) and(52) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (53)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 119887)119872 minus 119871 119871 minus 119873
(54)
minus1 lt 119887 le 119887119899le 0 forall119899 ge 119879 (55)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by (50) By virtue of
(15) (50) (53) and (55) we infer that for all 119909 = 119909119899119899isinN120573
119910 = 119910
119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le [minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 minus 119887119872
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 minus 119887119872 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 minus 119887119872 +min (1 + 119887)119872 minus 119871 119871 minus 119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minusmin (1 + 119887)119872 minus 119871 119871 minus 119873 ge 119873
(56)
That is (27) holds The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 6 Assume that there exist constants 119887119872 and 119873 with (1 minus 1119887)119872 gt 119873 gt 0 and fournonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882
119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899ge 119887 gt 1 eventually (57)
Then one has the following(a) For any 119871 isin ((1119887)119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any 119909
0=
1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterative sequence119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(58)
Abstract and Applied Analysis 9
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin ((1119887)119872 + 119873119872) It follows from (38) (39)and (57) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 =
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (59)
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus
1
119887
119872 minus119873
(60)
119887119899ge 119887 gt 1 forall119899 ge 119879 (61)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[(119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus 119888119905)] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(62)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In view of (14) (15)and (59)∽(62) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
+
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
+
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus
1
119887
119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
119887
119872
minus
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
119887
119872 minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
gt 119871 minus
1
119887
119872 minusmin 119872 minus 119871 119871 minus
1
119887
119872 minus119873 ge 119873
(63)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 7 Assume that there exist constants 119887 119872and 119873 with (1 + 1119887)119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899le 119887 lt minus1 eventually (64)
10 Abstract and Applied Analysis
Then one has the following
(a) For any 119871 isin (minus(1 + 1119887)119872 minus119873) there exist 120579 isin (0 1)and 119879 ge 119899
0+ 120591 + 120573 such that for any 119909
0= 1199090119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 120572119898 minus 1198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+120572119898 minus 119879
2
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(65)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (minus(1 + 1119887)119872 minus119873) It follows from (38)(39) and (64) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 = minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (66)
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 1
119887
)119872 minus 119871 119871 minus
1
119887
119872 minus119873
(67)
119887119899ge 119887 gt 1 forall119899 ge 119879 (68)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
minus1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(69)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Making use of (15) (66)(68) and (69) we conclude that10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le minus
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
minus
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le minus
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
minus
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le minus119871 minus
119872
119887
minus
1
1198871198992
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
Abstract and Applied Analysis 11
le minus119871 minus
119872
119887
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt minus119871 minus
119872
119887
+min (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 le 119872
119878119871119909119899
1198992
ge minus119871
+
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge minus119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt minus119871 minusmin (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 ge 119873
(70)
which yield (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
3 Examples
In this section we suggest six examples to explain the resultspresented in Section 2
Example 1 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus 119909119899minus120591) + Δ(
sin2119909119899minus3
1198997
) +
1
(1198999+ 21198995+ 1) (1 + 119909
2
1198992)
=
1198992
minus 2119899
1198998+ 1198993+ 1
forall119899 ge 4
(71)
where 120591 isin N is fixed Let 1198990= 4 119896 = 1 and 120573 = min4minus120591 1
and let119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= minus1 119888
119899=
1198992
minus 2119899
1198998+ 1198993+ 1
119891 (119899 119906) =
1
(1198999+ 21198995+ 1) (1 + 119906
2)
ℎ (119899 119906) =
sin21199061198997
1198911119899= 1198992
119865119899= 1198994
ℎ1119899= 119899 minus 3 119867
119899= (119899 minus 3)
2
119875119899=
2119872
1198999
119876119899=
1
1198999
119877119899=
2
1198997
119882119899=
1
1198997
forall (119899 119906) isin N1198990
timesR
(72)
It is easy to see that (14) (15) and (18) are satisfied Note that
1
1198992
infin
sum
119905=119899+120591
1199052max 119877
119905119867119905119882119905
=
1
1198992
infin
sum
119905=119899+120591
1199052max2(119905 minus 3)
2
1199057
1
1199057
=
infin
sum
119905=119899+120591
2(119905 minus 3)2
+ 1
1199055
le
2
1198992
infin
sum
119905=119899+120591
1
1199053
997888rarr 0 as 119899 997888rarr infin
1
1198992
infin
sum
119905=119899+120591
1199053max 119875
119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119905=119899+120591
1199053max2119872
1199055
1
1199059
100381610038161003816100381610038161199052
minus 2119905
10038161003816100381610038161003816
1199058+ 1199053+ 1
le
max 1 2119872
1198992
infin
sum
119905=119899+120591
1
1199052
997888rarr 0 as 119899 997888rarr infin
(73)
which together with Lemma 1 yield that (16) and (17) holdIt follows from Theorem 2 that (71) possesses uncountablymany positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 119899
0+120591+120573 such
that for each 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (19)converges to a positive solution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of(71) with lim
119899rarrinfin119911119899= +infin and has the error estimate (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] satisfying(21)
Example 2 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ 119909119899minus120591) + Δ(
sin211990931198993+1
1198993(1198996+ 2) (1 + 119909
4
21198992minus3
)
)
+
(minus1)119899
1198993
(1199091198992minus119899minus1
+ 119909(119899+1)(119899+2)
)
(11989913+ 1198995+ 1) (1 + 119909
2
1198992minus119899minus1
+ 1199092
(119899+1)(119899+2))
=
1198992
minus ln 1198991198996+ 1198995+ 1
forall119899 ge 5
(74)
where 120591 isin N is fixed Let 1198990= 5 119896 = 2 and 120573 = 5 minus 120591 and let
119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= 1 119888
119899=
1198992
minus ln 1198991198996+ 1198995+ 1
119891 (119899 119906 V) =(minus1)119899
1198993
(119906 + V)(11989913+ 1198995+ 1) (1 + 119906
2+ V2)
12 Abstract and Applied Analysis
ℎ (119899 119906 V) =sin2V
1198993(1198996+ 2) (1 + 119906
4)
1198911119899= 1198992
minus 119899 minus 1
1198912119899= (119899 + 1) (119899 + 2)
119865119899= (119899 + 1)
2
(119899 + 2)2
ℎ1119899= 21198992
minus 3
ℎ2119899= 31198993
+ 1
119867119899= (3119899
3
+ 1)
2
119875119899= 119876119899=
4
11989910
119877119899= 119882119899=
10
1198999
forall (119899 119906 V) isin N1198990
timesR2
(75)
It is clear that (14) (15) and (40) are fulfilled Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
10(31199043
+ 1)
2
1199049
10
1199049
le
160
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
160
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(76)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (77)
Observe that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max4(119905 + 1)2
(119905 + 2)2
11990510
4
11990510
1199052
minus ln 1199051199056+ 1199055+ 1
le
196
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
=
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
119905 minus 119906 + 1
1199054
le
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
196
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(78)
which yields that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (79)
Thus Theorem 3 guarantees that (74) possesses uncount-ably positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 120591 +
1198990+ 120573 such that the Mann iterative sequence 119909
119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (74) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 3 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 + 3 ln 1198992 + 4 ln 119899
119909119899minus120591)
+ Δ(
(minus1)119899 sin (119890minus119899
2|11990951198992minus3|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|11990921198993+1|
)
+
(minus1)119899
1198996(1 + 119909
2
119899minus3)
minus
1
(1198997+ 21198994minus 1) (1 + 119909
2
119899+4)
=
3(minus1)119899
1198992
911989910ln3119899
forall119899 ge 7
(80)
where 120591 isin N is fixed Let 1198990= 7 119896 = 2 119887 = 34 and
120573 = min7 minus 120591 4 and let119872 and119873 be two positive constantswith 119872 gt 4119873 and
119887119899=
1 + 3 ln 1198992 + 4 ln 119899
119888119899=
3(minus1)119899
1198992
911989910ln3119899
119891 (119899 119906 V) =(minus1)119899
1198996(1 + 119906
2)
minus
1
(1198997+ 21198994minus 1) (1 + V2)
ℎ (119899 119906 V) =(minus1)119899 sin (119890minus119899
2|119906|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|V|
1198911119899= 119899 minus 3 119891
2119899= 119899 + 4
119865119899= (119899 + 4)
2
ℎ1119899= 51198992
minus 3
ℎ2119899= 21198993
+ 1 119867119899= (2119899
3
+ 1)
2
119875119899= 119876119899=
3
1198996
119877119899= 119882119899=
2
11989910
forall (119899 119906 V) isin N1198990
timesR2
(81)
It is not difficult to verify that (14) (15) and (47) are fulfilledNote that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
2(21199043
+ 1)
2
11990410
2
11990410
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
18
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(82)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (83)
Abstract and Applied Analysis 13
Observe that1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max3(119905 + 4)2
1199056
3
1199056
100381610038161003816100381610038161003816100381610038161003816
3(minus1)119905
1199052
911990510ln3119905
100381610038161003816100381610038161003816100381610038161003816
le
12
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
le
12
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
12
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(84)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (85)
That is (38) and (39) hold Consequently Theorem 4 impliesthat (80) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((34)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (80) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 4 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 minus 51198993
2 + 61198993
119909119899minus120591) + Δ(
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119909
2
3119899minus7)
)
+
sin (119899211990931198992minus2)
(radic119899 + 14)22
=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
forall119899 ge 9
(86)
where 120591 isin N is fixed Let 1198990= 9 119896 = 1 119887 = minus56 and
120573 = 9 minus 120591 and let 119872 and 119873 be two positive constants with119872 gt 6119873 and
119887119899=
1 minus 51198993
2 + 61198993
119888119899=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
119891 (119899 119906) =
sin (1198992119906)
(radic119899 + 14)22
ℎ (119899 119906) =
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119906
2)
1198911119899= 31198992
minus 2 119865119899= (3119899
2
minus 2)
2
ℎ1119899= 3119899 minus 7
119867119899= (3119899 minus 7)
2
119875119899= 119876119899=
3
1198999
119877119899= 119882119899=
1
1198995
forall (119899 119906) isin N1198990
timesR
(87)
Obviously (14) (15) and (52) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(3119904 minus 7)2
1199045
1
1199045
le
9
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
9
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(88)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (89)
Notice that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max
3(31199052
minus 2)
2
1199059
3
1199059
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905
1199053
+ 51199052
+ 4119905 minus 2
1199059+ 1199058+ 21199055+ 1199053+ 7
100381610038161003816100381610038161003816100381610038161003816
le
27
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
27
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
27
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(90)
which gives that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (91)
That is (38) and (39) hold Thus Theorem 5 shows that(86) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (119873 (16)119872)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that the Mann
iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generatedby (48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
119860(119873119872) of (86) with lim119899rarrinfin
119911119899= +infin and has the error
estimate (20) where 120572119898119898isinN0
is an arbitrary sequencein [0 1] satisfying (21)
14 Abstract and Applied Analysis
Example 5 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ (
120587
2
+ 119899 sin 1119899
) 119909119899minus120591)
+ Δ(
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (11989921199092119899+1
))
)
+
119899 sin (119899119909119899minus2)
2 + (119899 + 5)16
=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
forall119899 ge 3
(92)
where 120591 isin N is fixed Let 1198990= 3 119896 = 1 119887 = 1205872 and
120573 = min3 minus 120591 1 and let119872 and119873 be two positive constantswith (1 minus 2120587)119872 gt 119873 and
119887119899=
120587
2
+ 119899 sin 1119899
119888119899=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
119891 (119899 119906) =
119899 sin (119899119906)2 + (119899 + 5)
16
ℎ (119899 119906) =
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (1198992119906))
1198911119899= 119899 minus 2 119865
119899= (119899 minus 2)
2
ℎ1119899= 2119899 + 1 119867
119899= (2119899 + 1)
2
119875119899= 119876119899=
1
11989914
119877119899= 119882119899=
2
1198999
forall (119899 119906) isin N1198990
timesR
(93)
Clearly (14) (15) and (61) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max2(2119904 + 1)2
1199049
2
1199049
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199047
le
18
1198992
infin
sum
119904=119899
1
1199046
997888rarr 0 as 119899 997888rarr infin
(94)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 2)2
11990514
1
11990514
10038161003816100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1cos3 (1199052 + 1)11990516+ ln 119905
10038161003816100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
11990514
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
11990513
le
1
1198992
infin
sum
119905=119899
1
11990512
997888rarr 0 as 119899 997888rarr infin
(95)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (96)
That is (38) and (39) hold Consequently Theorem 6 impliesthat (92) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((2120587)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (58) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (92) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 6 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119909119899minus120591) + Δ(
cos ((minus1)119899119890119899)
(119899 + 7)6
radic1 +1003816100381610038161003816119909119899minus2
1003816100381610038161003816
)
+
sin (1198992119909119899minus1)
1198999+ 31198995+ 21198992+ 1
=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
forall119899 ge 6
(97)
where 120591 isin N is fixed Let 1198990= 6 119896 = 1 119887 = minus2 and 120573 =
min6 minus 120591 3 and let 119872 and 119873 be two positive constantswith (12)119872 gt 119873 and
119887119899= minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119888119899=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
119891 (119899 119906) =
sin (1198992119906)1198999+ 31198995+ 21198992+ 1
Abstract and Applied Analysis 15
ℎ (119899 119906) =
cos ((minus1)119899119890119899)(119899 + 7)
6
radic1 + |119906|
1198911119899= 119899 minus 1 119865
119899= (119899 minus 1)
2
ℎ1119899= 119899 minus 2 119867
119899= (119899 minus 2)
2
119875119899= 119876119899=
1
1198997
119877119899= 119882119899=
1
1198996
forall (119899 119906) isin N1198990
timesR
(98)
Obviously (14) (15) and (64) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(119904 minus 2)2
1199046
1
1199046
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
1
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(99)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (100)
It is clear that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 1)2
1199057
1
1199057
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1
1199054
+ 41199052
+ 119905 minus 1
11990511+ 61199053+ 7119905 + 2
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
1
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(101)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (102)
That is (38) and (39) hold Consequently Theorem 7 impliesthat (97) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (minus1198722 minus119873)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that
the Mann iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (65) converges to a positive solution 119911 =
119911119899119899isinN120573
isin 119860(119873119872) of (97) with lim119899rarrinfin
119911119899= +infin and
has the error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in[0 1] satisfying (21)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This research was supported by the Science Research Foun-dation of Educational Department of Liaoning Province(L2012380)
References
[1] M H Abu-Risha ldquoOscillation of second-order linear differenceequationsrdquo Applied Mathematics Letters vol 13 no 1 pp 129ndash135 2000
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[3] R P Agarwal and J Henderson ldquoPositive solutions and nonlin-ear eigenvalue problems for third-order difference equationsrdquoComputers amp Mathematics with Applications vol 36 no 10-12pp 347ndash355 1998
[4] A Andruch-Sobiło and M Migda ldquoOn the oscillation ofsolutions of third order linear difference equations of neutraltyperdquoMathematica Bohemica vol 130 no 1 pp 19ndash33 2005
[5] Z Dosla and A Kobza ldquoGlobal asymptotic properties of third-order difference equationsrdquo Computers amp Mathematics withApplications vol 48 no 1-2 pp 191ndash200 2004
[6] S R Grace and G G Hamedani ldquoOn the oscillation of certainneutral difference equationsrdquo Mathematica Bohemica vol 125no 3 pp 307ndash321 2000
[7] J Cheng ldquoExistence of a nonoscillatory solution of a second-order linear neutral difference equationrdquo Applied MathematicsLetters vol 20 no 8 pp 892ndash899 2007
[8] LKongQKong andB Zhang ldquoPositive solutions of boundaryvalue problems for third-order functional difference equationsrdquoComputersampMathematics withApplications vol 44 no 3-4 pp481ndash489 2002
[9] I Y Karaca ldquoDiscrete third-order three-point boundary valueproblemrdquo Journal of Computational and Applied Mathematicsvol 205 no 1 pp 458ndash468 2007
[10] W-T Li and J P Sun ldquoExistence of positive solutions of BVPsfor third-order discrete nonlinear difference systemsrdquo AppliedMathematics and Computation vol 157 no 1 pp 53ndash64 2004
[11] W-T Li and J-P Sun ldquoMultiple positive solutions of BVPs forthird-order discrete difference systemsrdquo Applied Mathematicsand Computation vol 149 no 2 pp 389ndash398 2004
[12] Z Liu M Jia S M Kang and Y C Kwun ldquoBounded positivesolutions for a third order discrete equationrdquo Abstract andApplied Analysis vol 2012 Article ID 237036 12 pages 2012
[13] Z Liu S M Kang and J S Ume ldquoExistence of uncountablymany bounded nonoscillatory solutions and their iterativeapproximations for second order nonlinear neutral delay dif-ference equationsrdquo Applied Mathematics and Computation vol213 no 2 pp 554ndash576 2009
[14] Z Liu Y Xu and S M Kang ldquoBounded oscillation criteriafor certain third order nonlinear difference equations withseveral delays and advancesrdquo Computers amp Mathematics withApplications vol 61 no 4 pp 1145ndash1161 2011
[15] M Migda and J Migda ldquoAsymptotic properties of solutions ofsecond-order neutral difference equationsrdquoNonlinear Analysis
16 Abstract and Applied Analysis
Theory Methods and Applications vol 63 no 5ndash7 pp e789ndashe799 2005
[16] N Parhi ldquoNon-oscillation of solutions of difference equationsof third orderrdquoComputersampMathematics withApplications vol62 no 10 pp 3812ndash3820 2011
[17] N Parhi and A Panda ldquoNonoscillation and oscillation ofsolutions of a class of third order difference equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 1 pp213ndash223 2007
[18] S H Saker ldquoNew oscillation criteria for second-order nonlinearneutral delay difference equationsrdquo Applied Mathematics andComputation vol 142 no 1 pp 99ndash111 2003
[19] S H Saker ldquoOscillation of third-order difference equationsrdquoPortugaliae Mathematica vol 61 no 3 pp 249ndash257 2004
[20] S Stevic ldquoOn a third-order system of difference equationsrdquoApplied Mathematics and Computation vol 218 no 14 pp7649ndash7654 2012
[21] X H Tang ldquoBounded oscillation of second-order delay dif-ference equations of unstable typerdquo Computers amp Mathematicswith Applications vol 44 no 8-9 pp 1147ndash1156 2002
[22] J Yan and B Liu ldquoAsymptotic behavior of a nonlinear delaydifference equationrdquo Applied Mathematics Letters vol 8 no 6pp 1ndash5 1995
[23] Z G Zhang and Q L Li ldquoOscillation theorems for second-order advanced functional difference equationsrdquo Computers ampMathematics with Applications vol 36 no 6 pp 11ndash18 1998
Research ArticleAlgebroid Solutions of Second Order ComplexDifferential Equations
Lingyun Gao1 and Yue Wang2
1 Department of Mathematics Jinan University Guangzhou Guangdong 510632 China2 School of Information Renmin University of China Beijing 100872 China
Correspondence should be addressed to Lingyun Gao tgaolyjnueducn
Received 28 November 2013 Accepted 17 December 2013 Published 2 January 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 L Gao and Y WangThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Using value distribution theory and maximummodulus principle the problem of the algebroid solutions of second order algebraicdifferential equation is investigated Examples show that our results are sharp
1 Introduction and Main Results
We use the standard notations and results of the Nevanlinnatheory of meromorphic or algebroid functions see forexample [1 2]
In this paper we suppose that second order algebraicdifferential equation (3) admit at least one nonconstant ]-valued algebroid solution 119908(119911) in the complex plane Wedenote by 119864 a subset of [0infin) for which 119898(119864) lt infin and by119870 a positive constant where119898(119864) denotes the linearmeasureof 119864 119864 or 119870 does not always mean the same one when theyappear in the following
Let 119886119895119896(119895 = 0 1 119899 119896 = 0 1 119902
119895) be entire func-
tions without common zeroes such that 11988601199020
= 0 We put
119876119895(119911 119908) =
119902119895
sum
119896=0
119886119895119896119908119896
119902119895= deg119876119895
119908
119901 = max 119902119895+ 119895 119895 = 0 1 119899 minus 1
(1)
Some authors had investigated the problem of the exis-tence of algebroid solutions of complex differential equationsand they obtained many results ([2ndash10] etc)
In 1989 Toda [4] considered the existence of algebroidsolutions of algebraic differential equation of the form
119899
sum
119895=0
119876119895(119911 119908) (119908
1015840
)
119895
= 0 (2)
He obtained the following
Theorem A (see [4]) Let 119908(119911) be a nonconstant ]-valuedalgebroid solution of the above differential equation and all 119886
119895119896
are polynomials If 119901 lt 119899 + 119902119899 then 119908(119911) is algebraic
The purpose of this paper is to investigate algebroid solu-tions of the following second order differential equation inthe complex plane with the aid of the Nevanlinna theory andmaximum modulus principle of meromorphic or algebroidfunctions
119899
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
= 0 (3)
where 119876119895(119911 119908) = sum
119902119895
119896=0119886119895119896119908119896 119895 = 0 1 2 119899
We will prove the following two results
Theorem 1 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and all 119886
119895119896are polynomials
If 119901 le 119902119899 then 119908(119911) is algebraic 119901 = max119902
119895 119895 = 0 1 119899 minus
1
Theorem 2 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and the orders of all 119886
119895119896are
finite If 1199020gt max
1le119895le119899minus1119902119895+119895 then the following statements
are equivalent(a) 120575(infin119908) gt 0(b) 1199020= 119902119899+ 119899
(c) infin is a Picard exceptional value of 119908(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 123049 4 pageshttpdxdoiorg1011552014123049
2 Abstract and Applied Analysis
2 Some Lemmas
Lemma 3 (see [2]) Suppose that 119908(119911) 119886119894(119911) (119894 = 1 2 119901)
are meromorphic functions and 119886119901(119911) = 0 Then one has
119898(119903
119901
sum
119894=1
119886119894(119911) 119908119894
) le 119901119898 (119903 119908) +
119901
sum
119894=1
119898(119903 119886119894(119911)) + 119874 (1)
(4)
Examining proof of Lemma 45 presented in [2 pp 192-193] we can verify Lemma 4
Lemma 4 Let 119908(119911) be a transcendental algebroid functionsuch that 119908(119911) has only finite number of poles and let 119908(119911)1199081015840
(119911) and 11990810158401015840(119911) have no poles in |119911| gt 1199030 Then for some
constants 119862119894gt 0 119894 = 1 2 3 and 119903 ge 119903
1ge 1199030it holds
119872(119903 119908) le 1198621+ 1198622119903 + 11986231199032
119872(119903 11990810158401015840
) (5)
where119872(119903 119908) = max|119911|=119903
|119908(119911)|
Lemma 5 (see [11]) The absolute values of roots of equation
119911119899
+ 1198861119911119899minus1
+ sdot sdot sdot + 119886119899= 0 (6)
are bounded by
max 119899 10038161003816100381610038161198861
1003816100381610038161003816 (119899
10038161003816100381610038161198862
1003816100381610038161003816)12
(1198991003816100381610038161003816119886119899
1003816100381610038161003816)1119899
(7)
Lemma 6 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of the differential equation (3) and let 119886
119895119896be a
polynomial If 119901 lt 119899 + 119902119899 then
min 119899 119902119899minus 119901 log+119872(119903 119908)
+max 0 119902119899minus 119901 minus 119899 log+119872(119903 119908)
le +119874 (log 119903) + 119874 (1) (119903 notin 119864)
(8)
where119872(119903 119908) = max|119911|=119903
|119908(119911)| 119870 is a positive constant
Proof We first prove that the poles of 119908 are contained in thezeroes of 119886
119899119902119899
(119911)Suppose that 119911
0is a pole of 119908 of order 120591 and 119911
0is not the
zeroes of 119886119899119902119899
(119911) Then
119908 (119911) = (119911 minus 1199110)minus120591120582
1199081(119911) 119908
1(1199110) = 0infin
11990810158401015840
(119911) = (119911 minus 1199110)minus(120591+2120582)120582
1199082(119911) 119908
2(1199110) = 0infin
(9)
We rewrite differential equation (3) as follows
119886119899119902119899
(11990810158401015840
)
119899
=
119899minus1
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
(10)
It follows from (10) that
119902119899120591 + 119899 (120591 + 2120582) le 119901120591 + (119899 minus 1) (120591 + 2120582) (11)
Noting that 119901 le 119902119899 we have
120591 + 2120582 lt 0 (12)This is a contradiction
This shows that the poles of119908 are contained in the zeroesof 119886119899119902119899
(119911)We rewrite differential equation (3) as follows
119899
sum
119895=0
119876119895(119911 119908)119876
119899(119911 119908)
119899minus119895minus1
(11987611989911990810158401015840
)
119895
= 0 (13)
119872(119903119876111990810158401015840
) ge 119872(119903 119908)119902119899+119872(119903 119908
10158401015840
)
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
ge 119872(119903 119908)119902119899+
119872(119903 119908) minus 1198621minus 1198622119903
11986231199032
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
(14)
For 119895 = 0 1 119899 minus 1 we have10038161003816100381610038161003816119876119895(119911119903 119908)119876
119899(119911119903 119908)119899minus119895minus110038161003816
100381610038161003816le 119870119872(119903 119908)
119902119895+119902119899(119899minus119895minus1)
(15)
Applying Lemma 5 to (13) at 119911 = 119911119903
119872(119903119876111990810158401015840
) le 119870119872(119903 119908)maxℎ
119895 119895=01119899minus1
(16)
where ℎ119895= (119902119895+ 119902119899(119899 minus 119895 minus 1))(119899 minus 119895)
From (14) and (15) we have
119872(119903 119908)119902119899le 119870 119872(119903 119908)
119902119899minus1
+ 1199032
119872(119903 119908)maxℎ
119895 119895=01119899minus1
le 119870 119872(119903 119908)119902119899minus1
+ 1199032
119872(119903 119908)119902119899+((119901minus119902
119899)119899)
(17)Note that
ℎ119895=
119902119895+ 119902119899(119899 minus 119895 minus 1)
119899 minus 119895
lt
119901 + 119902119899(119899 minus 119895 minus 1) minus 119895
119899 minus 119895
= 119902119899+
119901 minus 119902119899
119899
+
119895 (119901 minus 119902119899minus 119899)
119899 (119899 minus 119895)
le 119902119899+
119901 minus 119902119899
119899
119895 = 0 1 119899 minus 1
(18)
Dividing the inequality (17) by119872(119903 119908)
max119902119899minus1119902119899+((119902119899minus119901)119899) we obtain for 119903 notin 119864
119872(119903 119908)min1(119902
119899minus119901)119899
le 11987041 +
1199032
119872(119903 119908)max0(119902
119899minus119901minus119899)119899
(19)
which reduces to our inequality by calculating log+ of theboth sidesmin 119899 119902
119899minus 119901 log+119872(119903 119908)
le minusmax 0 119902119899minus 119901 minus 119899 log+119872(119903 119908) + 119874 (log 119903) + 119874 (1)
(20)Lemma 6 is complete
Abstract and Applied Analysis 3
3 Proof of Theorem 1
First we consider119873(119903 119908)Let 1199110be a pole of 119908 of 120591 Let 119905 be the order of zero of
119886119899119902119899
(119911) at 1199110
(i) When the order of the pole of119876119899(119908)(119908
10158401015840
)119899 is not equal
to that of other terms of the left-hand side of (10) at 1199110 we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 (21)
that is
120591 le
(119905 minus 2119899) 120582
119902119899+ 119899 minus 119901
(22)
(ii)When the order of pole of119876119899(119908)(119908
10158401015840
)119899 is equal to that
of some term 119876119896(119908)(119908
10158401015840
)119899 of the left-hand side of (10) at 119911
0
we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 + 119899 (120591 + 2120582) (23)
that is
120591 le
119905120582
119902119899minus 119901
(24)
Combining cases (i) and (ii) we obtain
119873(119903 119908) le 1198708119873(119903
1
119886119899119902119899
) (25)
where1198708is a positive constant
Secondly by Lemma 6 we obtain
min 119899 119902119899minus 119901119898 (119903 119908)
le 119870[
119901
sum
119894=0
119898(119903 119886119894) +
119902119899minus1
sum
119896=0
119898(119903 119887119896)] + 119874 (log 119903) (119903 notin 119864)
(26)
Combining the inequalities (25) and (26) we have
119879 (119903 119908) = 119874 (log 119903) (119903 notin 119864) (27)
which shows that 119908 is an algebraic solution of (3)This completes the proof of Theorem 1
4 Proof of Theorem 2
(i) (a)rArr (b) Suppose that 120575(infin119908) gt 0 If 1199020gt 119902119899+ 119899 then
we have by (3)
1199081199020= minus
1
11988601199020
119899
sum
119895=1
119876119895(119908)119908
119895
(
11990810158401015840
119908
)
119895
minus
1199020minus1
sum
119896=0
1198860119896119908119896
(28)
Applying Lemma 3 to (28)
1199020119898(119903 119908) le (119902
0minus 1)119898 (119903 119908) +sum
119895119896
119898(119903 119886119895119896)
+ 119870119898(119903
11990810158401015840
119908
) + 119898(119903
1
11988601199020
) + 119874 (1)
(29)
Since 119908(119911) is admissible solution we have
119898(119903 119908) = 119878 (119903 119908) (30)
so that
120575 (119908infin) = 0 (31)
This is a contradiction Thus 1199020le 119902119899+ 119899
If 1199020lt 119902119899+119899 byTheorem 1119908(119911) is nonadmissibleThus
1199020= 119902119899+ 119899 (32)
(ii) (b) rArr (c) Let 1199020= 119902119899+ 119899 Then similar to the proof of
Lemma 6 we obtain that the poles of 119908(119911) are contained inthe set of 119886
119899119902119899
andinfin is a Picard exceptional value of 119908(119911)
(iii) (c) rArr (a) Let infin be a Picard exceptional value of 119908(119911)Then 120575(119908infin) = 1
5 Some Examples
Example 1 The differential equation
]21199082]minus111990810158401015840 minus [((2] + 1)1199084] + 21199082] minus ] + 1)] = 0 (33)
has a transcendental algebroid solution 119908(119911) = (tan 119911)1] Inthis case
119901 = 1199020= 4] gt 1 + 119902
1= 2] minus 1 + 1 = 2] (34)
Remark 7 Example 1 shows that the condition in Theorem 1is sharp
Example 2 Transcendental algebroid function 119908(119911) =
(sin 119911)12 is a 2-valued solution of the following differentialequation
161199086
(11990810158401015840
)
2
+ 21199085
11990810158401015840
minus (1199088
minus
1
2
1199086
+ 21199084
minus
1
2
1199082
+ 1) = 0
(35)
In this case
1199020= 8 119899 = 2 119902
1= 5 119902
2= 6 (36)
By Theorem 2 for transcendental algebroid function 119908(119911) =(sin 119911)12infin is a Picard exceptional value
Remark 8 Example 2 shows that the result in Theorem 2holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project is project Supported by National Natural ScienceFoundation (10471065) of China and NSF of GuangdongProvince (04010474)
4 Abstract and Applied Analysis
References
[1] H Yi and C C YangTheory of the Uniqueness of MeromorphicFunctions Science Press Beijing China 1995 (Chinese)
[2] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Science Press Beijing China 1988
[3] Y Z He and X Z Xiao ldquoAdmissible solutions and ordinarydifferential equationsrdquo Contemporary Mathematics vol 25 pp51ndash61 1983
[4] N Toda ldquoOn algebroid solutions of some algebraic differentialequations in the complex planerdquo Japan Academy A vol 65 no4 pp 94ndash97 1989
[5] T Chen ldquoOne class of ordinary differential equations whichpossess algebroid solutions in the complex domainrdquo ChineseQuarterly Journal of Mathematics vol 6 no 4 pp 45ndash51 1991
[6] K Katajamaki ldquoValue distribution of certain differential poly-nomials of algebroid functionsrdquo Archiv der Mathematik vol 67no 5 pp 422ndash429 1996
[7] L Gao ldquoSome results on admissible algebroid solutions ofcomplex differential equationsrdquo Indian Journal of Pure andApplied Mathematics vol 32 no 7 pp 1041ndash1050 2001
[8] L-Y Gao ldquoOn some generalized higher-order algebraic dif-ferential equations with admissible algebroid solutionsrdquo IndianJournal of Mathematics vol 43 no 2 pp 163ndash175 2001
[9] L Gao ldquoOn the growth of solutions of higher-order algebraicdifferential equationsrdquoActaMathematica Scientia B vol 22 no4 pp 459ndash465 2002
[10] L Y Gao ldquoThe growth of single-valued meromorphic solutionsand finite branch solutionsrdquo Journal of Systems Science andMathematical Sciences vol 24 no 3 pp 303ndash310 2004
[11] T Takagi Lecture on Algebra Kyoritsu Tokyo Japan 1957(Japanese)
Copyright copy 2014 Hindawi Publishing Corporation All rights reserved
This is a special issue published in ldquoAbstract and Applied Analysisrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited
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Contents
Complex Differences and Difference Equations Zong-Xuan Chen Kwang Ho Shon and Zhi-Bo HuangVolume 2014 Article ID 124843 1 page
Blow-Up Analysis for a Quasilinear Parabolic Equation with Inner Absorption and NonlinearNeumann Boundary Condition Zhong Bo Fang and Yan ChaiVolume 2014 Article ID 289245 8 pages
Some Properties on Complex Functional Difference Equations Zhi-Bo Huang and Ran-Ran ZhangVolume 2014 Article ID 283895 10 pages
The Regularity of Functions on Dual Split Quaternions in Clifford AnalysisJi Eun Kim and Kwang Ho ShonVolume 2014 Article ID 369430 8 pages
Unicity of Meromorphic Functions Sharing Sets withTheir Linear Difference PolynomialsSheng Li and BaoQin ChenVolume 2014 Article ID 894968 7 pages
A ComparisonTheorem for Oscillation of the Even-Order Nonlinear Neutral Difference EquationQuanxin ZhangVolume 2014 Article ID 492492 5 pages
Difference Equations and Sharing Values Concerning Entire Functions andTheir DifferenceZhiqiang Mao and Huifang LiuVolume 2014 Article ID 584969 6 pages
Admissible Solutions of the Schwarzian Type Difference Equation Baoqin Chen and Sheng LiVolume 2014 Article ID 306360 5 pages
Statistical Inference for Stochastic Differential Equations with Small NoisesLiang Shen and Qingsong XuVolume 2014 Article ID 473681 6 pages
On the Deficiencies of Some Differential-Difference Polynomials Xiu-Min Zheng and Hong Yan XuVolume 2014 Article ID 378151 12 pages
On Growth of Meromorphic Solutions of Complex Functional Difference Equations Jing LiJianjun Zhang and Liangwen LiaoVolume 2014 Article ID 828746 6 pages
Unicity of Entire Functions concerning Shifts and Difference Operators Dan Liu Degui Yangand Mingliang FangVolume 2014 Article ID 380910 5 pages
On Positive Solutions and Mann Iterative Schemes of aThird Order Difference Equation Zeqing LiuHeng Wu Shin Min Kang and Young Chel KwunVolume 2014 Article ID 470181 16 pages
Algebroid Solutions of Second Order Complex Differential Equations Lingyun Gao and Yue WangVolume 2014 Article ID 123049 4 pages
EditorialComplex Differences and Difference Equations
Zong-Xuan Chen1 Kwang Ho Shon2 and Zhi-Bo Huang1
1School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Zong-Xuan Chen chzxvipsinacom
Received 12 August 2014 Accepted 12 August 2014 Published 22 December 2014
Copyright copy 2014 Zong-Xuan Chen et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
In more recent years activity in the area of the complexdifferences and the complex difference equations has fleetlyincreased
This journal has set up a column of this special issue Wewere pleased to invite the interested authors to contributetheir original research papers as well as good expositorypapers to this special issue that will make better improvementon the theory of complex differences and difference equa-tions
In this special issue many good results are obtainedDifference equations are widely applied to mathematical
physics economics and chemistry In this special issue Z-B Huang and R-R Zhang J Li et al and L Gao and YWang investigate the growth a Borel exceptional value ofmeromorphic solutions to different types of higher ordernonliear difference equations respectively B Chen and S Liinvestigate the Schwarzian type difference equation D Liuet al Z Mao and H Liu and S Li and B Chen investigateunicity of meromorphic functions concerning different typesof difference operators Recently many difference analoguesof the classic Nevanlinna theory are obtained
In this special issue X-M Zheng and H Y Xu obtaina differential difference analogue of Valiron-Mohonko theo-rem Related topics with complex difference J E Kim andK H Shon investigate the regularity of functions on dualsplit quaternions in Clifford analysis and the tensor productrepresentation of polynomials of weak type in a DF-spaceQ Zhang and Z Liu et al investigate different types ofreal difference equations respectively L Shen and Q Xuinvestigate stochastic differential equations
This special issue stimulates the continuing efforts to thecomplex differences and the complex difference equations
Zong-XuanChenKwangHo ShonZhi-BoHuang
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 124843 1 pagehttpdxdoiorg1011552014124843
Research ArticleBlow-Up Analysis for a Quasilinear Parabolic Equation withInner Absorption and Nonlinear Neumann Boundary Condition
Zhong Bo Fang and Yan Chai
School of Mathematical Sciences Ocean University of China Qingdao 266100 China
Correspondence should be addressed to Zhong Bo Fang fangzb7777hotmailcom
Received 24 February 2014 Revised 11 April 2014 Accepted 11 April 2014 Published 30 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Z B Fang and Y Chai This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We investigate an initial-boundary value problem for a quasilinear parabolic equation with inner absorption and nonlinearNeumann boundary condition We establish respectively the conditions on nonlinearity to guarantee that 119906(119909 119905) exists globally orblows up at some finite time 119905lowast Moreover an upper bound for 119905lowast is derived Under somewhat more restrictive conditions a lowerbound for 119905lowast is also obtained
1 Introduction
We are concerned with the global existence and blow-upphenomenon for a quasilinear parabolic equation with non-linear inner absorption term
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
minus 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (1)
subjected to the nonlinear Neumann boundary and initialconditions
(|nabla119906|119901
+ 1)
120597119906
120597]= 119892 (119906) (119909 119905) isin 120597Ω times (0 119905
lowast
) (2)
119906 (119909 0) = 1199060(0) ge 0 119909 isin Ω (3)
whereΩ is a bounded star-shaped region of 119877119873 (119873 ge 2) withsmooth boundary 120597Ω ] is the unit outward normal vectoron 120597Ω 119901 ge 0 119905lowast is the blow-up time if blow-up occurs orelse 119905lowast
= +infin the symbol 119894 denotes partial differentiationwith respect to 119909
119894 119894 = 1 2 119873 the repeated index indicates
summation over the index and nabla is gradient operatorMany physical phenomena and biological species theo-
ries such as the concentration of diffusion of some non-Newton fluid through porous medium the density of somebiological species and heat conduction phenomena havebeen formulated as parabolic equation (1) (see [1ndash3]) Thenonlinear Neumann boundary condition (2) can be physi-cally interpreted as the nonlinear radial law (see [4 5])
In the past decades there have been many works dealingwith existence and nonexistence of global solutions blow-upof solutions bounds of blow-up time blow-up rates blow-up sets and asymptotic behavior of solutions to nonlinearparabolic equations see the books [6ndash8] and the surveypapers [9ndash11] Specially we would like to know whether thesolution blows up and at which time when blow-up occursA variety of methods have been used to study the problemabove (see [12]) and in many cases these methods used toshow that solutions blow up often provide an upper boundfor the blow-up time However lower bounds for blow-uptime may be harder to be determined For the study ofthe initial boundary value problem of a parabolic equationwith homogeneous Dirichlet boundary condition see [1314] Payne et al [13] considered the following quasilinearparabolic equation
119906119905= div (120588|nabla119906|2nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (4)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω To get the lower bound for the blow-up time the authorsassumed that 120588 is a positive 1198621 function which satisfies
120588 (119904) + 1199041205881015840
(119904) gt 0 119904 gt 0 (5)
The lower bound for the blow-up time of solution to (4) withRobin boundary condition was obtained in [15] where 120588 is
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 289245 8 pageshttpdxdoiorg1011552014289245
2 Abstract and Applied Analysis
also assumed to satisfy the condition (5) However under thisboundary condition the best constant of Sobolev inequalityused in [13] is no longer applicable They imposed suitableconditions on 119891 and 120588 and determined a lower bound forthe blow-up time if blow-up occurs and determined whenblow-up cannot occur Marras and Vernier Piro [14] studiedthe nonlinear parabolic problem with time dependent coeffi-cients
1198961(119905) div (119892 (|nabla119906|2nabla119906)) + 119896
2(119905) 119891 (119906) = 119896
3(119905) 119906119905
(119909 119905) isin Ω times (0 119905lowast
)
(6)
where Ω is a bounded domain in 119877119873 with smooth boundary
120597Ω Under some conditions on the data and geometry of thespatial domain they obtained upper and lower bounds of theblow-up time Moreover the sufficient conditions for globalexistence of the solution were derived
For the study of the initial boundary value problem ofa parabolic equation with Robin boundary condition werefer to [15ndash19] Li et al [16] investigated the problem of thenonlinear parabolic equation
119906119905= [(|nabla119906|
119901
+ 1) 119906119894]119894
+ 119891 (119906) (119909 119905) isin Ω times (0 119905lowast
) (7)
where Ω is a bounded domain in 1198773 with smooth boundary
120597ΩThey derived the lower bound for the blow-up timewhenthe blow-up occurs Clearly |nabla119906|119901 + 1 does not satisfy thecondition (5) Enache [17] discussed the quasilinear parabolicproblem
119906119905= (119892 (119906) 119906
119894)119894
+ 119891 (119906) (8)
where Ω is a bounded domain in 119877119873
(119873 ge 2) with smoothboundary 120597Ω By virtue of a first-order differential inequalitytechnique they showed the sufficient conditions to guaranteethat the solution 119906(119909 119905) exists globally or blows up Inaddition a lower bound for the blow-up time when blow-up occurs was also obtained Ding [18] studied the nonlinearparabolic problem
(119887 (119906))119905= nabla sdot (119892 (119906) nabla119906) + 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (9)
where Ω is a bounded domain in 1198773 with smooth boundary
120597Ω They derived conditions on the data which guaranteethe blow-up or the global existence of the solution A lowerbound on blow-up time when blow-up occurs was alsoobtained For the problem of the nonlinear nonlocal porousmedium equation we read the paper of Liu [19]
Recently for the problems with nonlinear Neumannboundary conditions Payne et al [20] studied the semilinearheat equation with inner absorption term
119906119905= Δ119906 minus 119891 (119906) (119909 119905) isin Ω times (0 119905
lowast
) (10)
They established conditions on nonlinearity to guarantee thatthe solution119906(119909 119905) exists for all time 119905 gt 0 or blows up at somefinite time 119905lowast Moreover an upper bound for 119905lowast was derivedUnder somewhat more restrictive conditions a lower bound
for 119905lowast was derivedThereafter they considered the quasilinearparabolic equation
119906119905= nabla sdot (|nabla119906|
119901
nabla119906) (119909 119905) isin Ω times (0 119905lowast
) (11)
and they showed that blow-up occurs at some finite timeunder certain conditions on the nonlinearities and the dataupper and lower bounds for the blow-up time were derivedwhen blow-up occurs see [21] Liu et al The authors [22 23]studied the reaction diffusion problem with nonlocal sourceand inner absorption terms or with local source and gradientabsorption terms Very recently Fang et al [24] consideredlower bounds estimate for the blow-up time to nonlocalproblemwith homogeneousDirichlet orNeumann boundarycondition
Motivated by the above work we intend to study theglobal existence and the blow-up phenomena of problem (1)ndash(3) and the results of the semilinear equations are extended tothe quasilinear equations Unfortunately the techniques usedfor semilinear equation to analysis of blow-up phenomenaare no longer applicable to our problem As a consequenceby using the suitable techniques of differential inequalitieswe establish respectively the conditions on the nonlinearities119891 and 119892 to guarantee that 119906(119909 119905) exists globally or blows upat some finite time If blow-up occurs we derive upper andlower bounds of the blow-up time
The rest of our paper is organized as follows In Section 2we establish conditions on the nonlinearities to guaranteethat 119906(119909 119905) exists globally In Section 3 we show the condi-tions on data forcing the solution 119906(119909 119905) to blow up at somefinite time 119905
lowast and obtain an upper bound for 119905lowast A lower
bound of blow-up time under some assumptions is derivedin Section 4
2 The Global Existence
In this sectionwe establish the conditions on the nonlinearity119891 and nonlinearity 119892 to guarantee that 119906(119909 119905) exists globallyWe state our result as follows
Theorem 1 Assume that the nonnegative functions 119891 and 119892
satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(12)
where 1198961gt 0 119896
2ge 0 119904 gt 1 2119904 lt 119902 + 1 and 119904 minus 1 lt 119901 lt 119902 minus 1
Then the (nonnegative) solution 119906(119909 119905) of problem (1)-(3) doesnot blow up that is 119906(119909 119905) exists for all time 119905 gt 0
Proof Set
Ψ (119905) = int
Ω
1199062
119889119909 (13)
Abstract and Applied Analysis 3
Similar to Theorem 21 in [20] we get
Ψ1015840
(119905) le 21205752
int
Ω
1199062119904
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
+
21198962119873
1205880
int
Ω
119906119904+1
119889119909
minus2int
Ω
|nabla119906|119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 1198681+ 1198682
(14)
where 120575 = 1198962(119904 + 1)1198892120588
0 1205880
= min119909isin120597Ω
(119909 sdot ]) 119889 =
max119909isin120597Ω
|119909| and
1198681le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
times 1198601|Ω|(119902minus119904)(119902+1)
minus 1198602(int
Ω
119906119902+1
119889119909)
(119902minus119904)(119902+1)
(15)
where 1198601= 21205752
120572120576(120572minus1)120572 119860
2= 1198961minus 21205752
(1 minus 120572)120576 120572 = (119902 + 1 minus
2119904)(119902 minus 119904) lt 1 120576 gt 0Next we estimate 119868
2= (211989621198731205880) intΩ
119906119904+1
119889119909minus2 intΩ
|nabla119906|119901+2
119889119909 minus 1198961intΩ
119906119902+1
119889119909 Since
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
= (
119901
2
+ 1)
2
119906119901
|nabla119906|2
(16)
it follows from Holder inequality that
int
Ω
10038161003816100381610038161003816nabla119906(1199012)+1
10038161003816100381610038161003816
2
119889119909 le (
119901
2
+ 1)
2
(int
Ω
|nabla119906|119901+2
119889119909)
2(119901+2)
times (int
Ω
119906119901+2
119889119909)
119901(119901+2)
(17)
Furthermore we have
int
Ω
119906119901+2
119889119909 le [
(119901 + 2)2
41205821
]
(1199012)+1
int
Ω
|nabla119906|119901+2
119889119909 (18)
which follows from (17) and membrane inequality
1205821int
Ω
1205962
119889119909 le int
Ω
|nabla120596|2
119889119909 (19)
where 1205821is the first eigenvalue in the fixed membrane
problem
Δ120596 + 120582120596 = 0 120596 gt 0 in Ω 120596 = 0 on 120597Ω (20)
Combining 1198682and (18) we have
1198682le
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 2[
41205821
(119901 + 2)2]
(1199012)+1
times int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
=
21198962119873
1205880
int
Ω
119906119904+1
119889119909 minus 3[
41205821
(119901 + 2)2
]
(1199012)+1
int
Ω
119906119901+2
119889119909
+
[
41205821
(119901 + 2)2]
(1199012)+1
int
Ω
119906119901+2
119889119909 minus 1198961int
Ω
119906119902+1
119889119909
= 11986821+ 11986822
(21)
Making use of Holder inequality we obtain
int
Ω
119906119904+1
119889119909 le (int
Ω
119906119901+2
119889119909)
(119904+1)(119901+2)
|Ω|(119901minus119904+1)(119901+2)
(22)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119904+1
119889119909)
2(119904+1)
|Ω|(119904minus1)(119904+1)
(23)
Combining (21) (22) with (23) we get
11986821
le (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
(24)
with
1198611=
21198962119873
1205880
1198612= 3[
41205821
(119901 + 2)2]
(1199012)+1
|Ω|minus(119901minus119904+1)2
(25)
Applying Holder inequality we obtain
int
Ω
119906119901+2
119889119909 le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
|Ω|(119902minus119901minus1)(119902+1)
Ψ (119905) = int
Ω
1199062
119889119909 le (int
Ω
119906119902+1
119889119909)
2(119902+1)
|Ω|(119902minus1)(119902+1)
(26)
It follows from (26) that
11986822
le (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(27)
where
1198621= [
41205821
(119901 + 2)2
]
(1199012)+1
|Ω|(119902minus119901minus1)(119902+1)
1198622= 1198961|Ω|(1minus119902)(119902minus119901minus1)2(119902+1)
(28)
4 Abstract and Applied Analysis
Combining (14) (15) (21) and (24) with (27) we obtain
Ψ1015840
(119905) le (int
Ω
119906119902+1
119889119909)
(119904+1)(119902+1)
1198601minus 1198602Ψ(119905)(119902minus119904)2
+ (int
Ω
119906119904+1
119889119909) 1198611minus 1198612Ψ(119901minus119904+1)2
+ (int
Ω
119906119902+1
119889119909)
(119901+2)(119902+1)
1198621minus 1198622Ψ(119902minus119901minus1)2
(29)
with
1198601= 1198601|Ω|(119902minus119904)(119902+1)
1198602= 1198602|Ω|(1minus119902)(119902minus119904)2(119902+1)
(30)
We conclude from (29) that Ψ(119905) is decreasing in each timeinterval on which we obtain
Ψ (119905) ge max(1198601
1198602
)
2(119902minus119904)
(
1198611
1198612
)
2(119901minus119904+1)
(
1198621
1198622
)
2(119902minus119901minus1)
(31)
so that Ψ(119905) remains bounded for all time under theconditions in Theorem 1 This completes the proof ofTheorem 1
3 Blow-Up and Upper Bound of 119905lowast
In this section Ω needs not to be star-shaped We establishthe conditions to assure that the solution of (1)ndash(3) blowsup at finite time 119905lowast and derive an upper bound for 119905lowast Moreprecisely we establish the following result
Theorem 2 Let 119906(119909 119905) be the classical solution of problem (1)-(3) Assume that the nonnegative and integrable functions 119891and 119892 satisfy
120585119891 (120585) le 2 (1 + 120572) 119865 (120585) 120585 ge 0
120585119892 (120585) ge 2 (1 + 120573)119866 (120585) 120585 ge 0
(32)
with
119865 (120585) = int
120585
0
119891 (120578) 119889120578 119866 (120585) = int
120585
0
119892 (120578) 119889120578 (33)
where 120572 ge 0
120573 ge max (119901
2
120572) (34)
Moreover assume that Φ(0) ge 0 with
Φ (119905) = 2int
120597Ω
119866 (119906) 119889119878 minus int
Ω
|nabla119906|2
(1 +
2
119901 + 2
|nabla119906|119901
)119889119909
minus 2int
Ω
119865 (119906) 119889119909
(35)
Then the solution 119906(119909 119905) of problem (1)-(3) blows up at somefinite time 119905lowast lt 119879 with
119879 =
Ψ (0)
2120573 (1 + 120573)Φ (0)
120573 gt 0 (36)
where Ψ(119905) is defined in (13) If 120573 = 0 we have 119879 = infin
Proof We compute
Ψ1015840
(119905) = 2int
Ω
119906119906119905119889119909 = 2int
Ω
119906 [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 2int
120597Ω
119906 (|nabla119906|119901
+ 1)
120597119906
120597]119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
= 2int
120597Ω
119906119892 (119906) 119889119878 minus 2int
Ω
(|nabla119906|119901
+ 1) |nabla119906|2
119889119909
minus 2int
Ω
119906119891 (119906) 119889119909
(37)
Making use of the hypotheses stated inTheorem 2 we have
Ψ1015840
(119905) ge 2 (1 + 120573)Φ (119905) (38)
Differentiating (35) we derive
Φ1015840
(119905) = 2int
120597Ω
119892 (119906) 119906119905119889119878 minus int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
minus 2int
Ω
119891 (119906) 119906119905119889119909
(39)
Integrating the identity nabla sdot (119906119905(|nabla119906|119901
+1)nabla119906) = 119906119905nabla sdot ((|nabla119906|
119901
+
1)nabla119906) + (12)(|nabla119906|119901
+ 1)(|nabla119906|2
)119905overΩ we get
int
Ω
(|nabla119906|119901
+ 1) (|nabla119906|2
)119905
119889119909
= 2int
Ω
nabla sdot (119906119905(|nabla119906|119901
+ 1) nabla119906) 119889119909
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1) nabla119906 sdot ]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
= 2int
120597Ω
119906119905(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 2int
Ω
119906119905nabla sdot ((|nabla119906|
119901
+ 1) nabla119906) 119889119909
(40)
Substituting (40) into (39) we have
Φ1015840
(119905) = 2int
Ω
1199062
119905119889119909 gt 0 (41)
whichwithΦ(0) gt 0 implyΦ(119905) gt 0 for all 119905 isin (0 119905lowast
) Makinguse of the Schwarz inequality we obtain
2 (1 + 120573)Ψ1015840
Φ le (Ψ1015840
(119905))
2
= 4(int
Ω
119906119906119905119889119909)
2
le 2Ψ (119905)Φ1015840
(119905)
(42)
Abstract and Applied Analysis 5
Multiplying the above inequality by Ψminus2minus120573 we deduce
(ΦΨminus(1+120573)
)
1015840
ge 0 (43)
Arguing as in Theorem 31 in [20] we find
119905lowast
le 119879 =
1
2120573 (1 + 120573)
(Ψ (0))minus120573
=
Ψ (0)
2120573 (1 + 120573)Φ (0)
(44)
valid for 120573 gt 0 If 120573 = 0 we have
Ψ (119905) ge Ψ (0) 1198902119872119905 (45)
valid for 119905 gt 0 implying that 119905lowast = infin This completes theproof of Theorem 2
4 Lower Bounds for 119905lowast
In this section under the assumption that Ω is a star shapeddomain in 119877
3 convex in two orthogonal directions we seek alower bound for the blow-up time 119905lowast Now we state the resultas follows
Theorem 3 Let 119906(119909 119905) be the nonnegative solution of problem(1)-(3) and 119906(119909 119905) blows up at 119905lowast moreover the nonnegativefunctions 119891 and 119892 satisfy
119891 (120585) ge 1198961120585119902
120585 ge 0
119892 (120585) le 1198962120585119904
120585 ge 0
(46)
with 1198961gt 0 119896
2gt 0 119902 gt 1 119904 gt 1 119902 lt 119904 Define
120593 (119905) = int
Ω
119906119899(119904minus1)
119889119909 (47)
where 119899 is a parameter restricted by the condition
119899 gt max 4 2
119904 minus 1
(48)
Then 120593(119905) satisfies inequality
1205931015840
(119905) le Γ (120593) (49)
for some computable function Γ(120593) It follows that 119905lowast is boundedfrom below We have
119905lowast
ge int
infin
120593(0)
119889120578
Γ (120578)
119889120578 (50)
Proof Differentiating (47) and making use of the boundarycondition (2) together with the conditions (46) we have
1205931015840
(119905) = 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119906119905119889119909
= 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
times [((|nabla119906|119901
+ 1) 119906119894)119894
minus 119891 (119906)] 119889119909
= 119899 (119904 minus 1) int
120597Ω
119906119899(119904minus1)minus1
(|nabla119906|119901
+ 1)
120597119906
120597]119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 119899 (119904 minus 1) int
Ω
119906119899(119904minus1)minus1
119891 (119906) 119889119909
le 1198962119899 (119904 minus 1) int
120597Ω
119906(119899+1)(119904minus1)
119889119878
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(51)
Applying inequality (27) in [20] to the first term on the righthand side of (51) we have
int
120597Ω
119906(119899+1)(119904minus1)
119889119878 le
3
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
(119899 + 1) (119904 minus 1) 119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
(52)
6 Abstract and Applied Analysis
Substituting (52) into (51) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
1205880
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(53)
Making use of arithmetic-geometric mean inequality wederive
int
Ω
119906(119899+1)(119904minus1)minus1
|nabla119906| 119889119909 le
120583
2
int
Ω
119906119899(119904minus1)minus2
|nabla119906|2
119889119909
+
1
2120583
int
Ω
119906(119899+2)(119904minus1)
119889119909
(54)
for all 120583 gt 0 Choose 120583 gt 0 such that
1198962119899 (119899 + 1) (119904 minus 1)
2
119889120583
21205880
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] = 0 (55)
We rewrite (53) as
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) int
Ω
119906119899(119904minus1)+119902minus1
119889119909
(56)
Using Holder inequality we get
int
Ω
119906119899(119904minus1)
119889119909 le (int
Ω
119906119899(119904minus1)+119902minus1
119889119909)
119899(119904minus1)(119899(119904minus1)+119902minus1)
times |Ω|(119902minus1)(119899(119904minus1)+119902minus1)
(57)
Combining (56) with (57) we obtain
1205931015840
(119905) le
31198962119899 (119904 minus 1)
1205880
int
Ω
119906(119899+1)(119904minus1)
119889119909
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
int
Ω
119906(119899+2)(119904minus1)
119889119909
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1]
times int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
=
31198962119899 (119904 minus 1)
1205880
1198691(119905)
+
1198962119899 (119899 + 1) (119904 minus 1)
2
119889
21205831205880
1198692(119905)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(58)
where
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
1198692(119905) = int
Ω
119906(119899+2)(119904minus1)
119889119909
120596 (119905) = int
Ω
119906119899(119904minus1)minus2
|nabla119906|119901+2
119889119909
(59)
Using Sobolev type inequality (A5) derived by Payne et al[21] we obtain
1198691(119905) = int
Ω
119906(119899+1)(119904minus1)
119889119909
le
3
1205880
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 +
(119899 + 1) (119904 minus 1)
3
times(1 +
119889
1205880
)int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
32
(60)
We now make use of Holder inequality to bound the secondintegral on the right hand side of (60) as follows
int
Ω
119906(23)(119899+1)(119904minus1)minus1
|nabla119906| 119889119909
le (int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909)
(119901+1)(119901+2)
1205961(119901+2)
(61)
with
1205751=
(119899 minus 2) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(62)
Abstract and Applied Analysis 7
Wenote that 1205751lt 1 for 119899 gt (3119901minus2(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Using again Holderrsquosinequality we obtain
int
Ω
119906(23)(119899+1)(119904minus1)(1minus120575
1)
119889119909
le 1205932(119899+1)(1minus120575
1)3119899
|Ω|1minus(2(119899+1)(1minus120575
1)3119899)
int
Ω
119906(23)(119899+1)(119904minus1)
119889119909 le 1205932(119899+1)3119899
|Ω|1minus(2(119899+1)3119899)
(63)
where |Ω| = intΩ
119889119909 is the volume of Ω Substituting (61) and(63) in (60) we obtain the following inequality
1198691(119905) le 119888
11205932(119899+1)3119899
+ 1198882120593(2(119899+1)(1minus120575
1)3119899)((119901+1)(119901+2))
times 1205961(119901+2)
32
le 1198881120593(119899+1)119899
+ 1198882120593((119899+1)(1minus120575
1)119899)((119901+1)(119901+2))
12059632(119901+2)
(64)
where 1198881 1198882are computable positive constants Note that the
last inequality in (64) follows from Holder inequality underthe particular form (119886 + 119887)
32
le radic2(11988632
+ 11988732
) Similarly wecan bound 119869
2and get
1198692(119905) le 119888
3120593(119899+2)119899
+ 1198884120593((119899+2)(1minus120575
2)119899)((119901+1)(119901+2))
12059632(119901+2)
(65)
where 1198883 1198884are computable positive constants
1205752=
(119899 minus 4) (119904 minus 1) + 3119901
2 (119899 + 1) (119904 minus 1) (119901 + 1)
(66)
Wenote that 1205752lt 1 for 119899 gt (3119901minus4(119904minus1)(119901+2))(119904minus1)(2119901+1)
an inequality satisfied in view of (48) Inserting (64) and (65)in (58) we arrive at
1205931015840
(119905) le1198891120593(119899+1)119899
+1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
+ 1198893120593(119899+2)119899
+1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
minus 119899 (119904 minus 1) [119899 (119904 minus 1) minus 1] 120596 (119905)
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(67)
where 120582 = (119901 + 1)(119901 + 2) 1198893and
119889119895(119895 = 1 2 4) are
computable positive constants Next we want to eliminatethe quantity 120596(119905) in inequality (67) By using the followinginequality
120593120572
120596120573
= (120574120596)120573
120593120572(1minus120573)
120574120573(1minus120573)
1minus120573
le 120574120573120596 + (1 minus 120573) 120574120573(120573minus1)
+ (1 minus 120573) 120574120573(120573minus1)
120593120572(1minus120573)
(68)
valid for 0 lt 120573 lt 1 where 120574 is an arbitrary positive constantthen we have
1198892120593((119899+1)(1minus120575
1)119899)120582
12059632(119901+2)
le 1205741120596 (119905) + 119889
2120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
1198894120593((119899+2)(1minus120575
2)119899)120582
12059632(119901+2)
le 1205742120596 (119905) + 119889
4120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(69)
with arbitrary positive constants 1205741 1205742and computable
positive constants 1198892 1198894 Substitute (69) in (67) and choose
the arbitrary (positive) constants 1205741 1205742such that 120574
1+1205742minus119899(119904minus
1)[119899(119904 minus 1) minus 1] = 0 We obtain
1205931015840
(119905) le1198891120593(119899+1)119899
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
minus 1198961119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(70)
We eliminate the last term in (70) by using the followinginequality
120593(119899+1)119899
= 119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
(2119899minus1)(119904minus1)((2119899minus1)(119904minus1)+119904minus119902)
times 119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(119904minus119902)((2119899minus1)(119904minus1)+119904minus119902)
le
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898120593(119899(119904minus1)+119902minus1)119899(119904minus1)
+
119904 minus 119902
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
119898(2119899minus1)(1minus119904)(119904minus119902)
1205933
(71)
valid for 119902 lt 119904 and arbitrary119898 gt 0 and choose119898 such that
(2119899 minus 1) (119904 minus 1)
(2119899 minus 1) (119904 minus 1) + 119904 minus 119902
1198891119898 minus 119896
1119899 (119904 minus 1) |Ω|
(1minus119902)119899(119904minus1)
= 0
(72)
Then (70) can be rewritten as
1205931015840
(119905) le 11988911205933
+ 1198892120593(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120593(119899+2)119899
+ 1198894120593(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
(73)
Integrating (73) over [0 119905] we conclude
119905lowast
ge int
infin
120593(0)
119889120578
times (11988911205783
+ 1198892120578(2(119899+1)(1minus120575
1)(119901+2)119899(2119901+1))120582
+ 1198893120578(119899+2)119899
+ 1198894120578(2(119899+2)(1minus120575
2)(119901+2)119899(2119901+1))120582
)
minus1
(74)
This completes the proof of Theorem 3
8 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Authorsrsquo Contribution
All authors contributed equally to the paper and read andapproved the final paper
Acknowledgments
This work is supported by the Natural Science Foundationof Shandong Province of China (ZR2012AM018) and theFundamental Research Funds for the Central Universities(no 201362032) The authors would like to deeply thank allthe reviewers for their insightful and constructive comments
References
[1] J Bebernes and D Eberly Mathematical Problems from Com-bustion Theory vol 83 of Applied Mathematical SciencesSpringer New York NY USA 1989
[2] C V Pao Nonlinear Parabolic and Elliptic Equations PlenumPress New York NY USA 1992
[3] J L Vazquez The Porous Medium Equations MathematicalTheory Oxford University Press Oxford UK 2007
[4] J Filo ldquoDiffusivity versus absorption through the boundaryrdquoJournal of Differential Equations vol 99 no 2 pp 281ndash305 1992
[5] H A Levine and L E Payne ldquoNonexistence theorems forthe heat equation with nonlinear boundary conditions and forthe porous medium equation backward in timerdquo Journal ofDifferential Equations vol 16 pp 319ndash334 1974
[6] B Straughan Explosive Instabilities in Mechanics SpringerBerlin Germany 1998
[7] A A Samarskii V A Galaktionov S P Kurdyumov and A PMikhailov Blow-Up in Quasilinear Parabolic Equations vol 19of de Gruyter Expositions in Mathematics Walter de GruyterBerlin Germany 1995
[8] PQuittner andP Souplet Superlinear Parabolic Problems Blow-Up Global Existence and Steady States Birkhauser AdvancedTexts Birkhauser Basel Switzerland 2007
[9] C Bandle and H Brunner ldquoBlowup in diffusion equations asurveyrdquo Journal of Computational andAppliedMathematics vol97 no 1-2 pp 3ndash22 1998
[10] V A Galaktionov and J L Vazquez ldquoThe problem of blow-up in nonlinear parabolic equationsrdquo Discrete and ContinuousDynamical Systems vol 8 no 2 pp 399ndash433 2002
[11] H A Levine ldquoThe role of critical exponents in blowup theo-remsrdquo SIAM Review vol 32 no 2 pp 262ndash288 1990
[12] H A Levine ldquoNonexistence of global weak solutions to someproperly and improperly posed problems of mathematicalphysics the method of unbounded Fourier coefficientsrdquoMath-ematische Annalen vol 214 pp 205ndash220 1975
[13] L E Payne G A Philippin and P W Schaefer ldquoBlow-upphenomena for some nonlinear parabolic problemsrdquoNonlinearAnalysis Theory Methods amp Applications vol 69 no 10 pp3495ndash3502 2008
[14] MMarras and S Vernier Piro ldquoOn global existence and boundsfor blow-up time in nonlinear parabolic problems with time
dependent coefficientsrdquo Discrete and Continuous DynamicalSystems vol 2013 pp 535ndash544 2013
[15] Y Li Y Liu and C Lin ldquoBlow-up phenomena for some non-linear parabolic problems under mixed boundary conditionsrdquoNonlinear Analysis Real World Applications vol 11 no 5 pp3815ndash3823 2010
[16] Y Li Y Liu and S Xiao ldquoBlow-up phenomena for some non-linear parabolic problems under Robin boundary conditionsrdquoMathematical and Computer Modelling vol 54 no 11-12 pp3065ndash3069 2011
[17] C Enache ldquoBlow-up phenomena for a class of quasilinearparabolic problems under Robin boundary conditionrdquo AppliedMathematics Letters vol 24 no 3 pp 288ndash292 2011
[18] J Ding ldquoGlobal and blow-up solutions for nonlinear parabolicequations with Robin boundary conditionsrdquo Computers ampMathematics with Applications vol 65 no 11 pp 1808ndash18222013
[19] Y Liu ldquoBlow-up phenomena for the nonlinear nonlocal porousmedium equation under Robin boundary conditionrdquo Comput-ers amp Mathematics with Applications vol 66 no 10 pp 2092ndash2095 2013
[20] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary conditon Irdquo Zeitschrift fur Angewandte Mathematikund Physik vol 61 no 6 pp 999ndash1007 2010
[21] L E Payne G A Philippin and S Vernier Piro ldquoBlow-up phenomena for a semilinear heat equation with nonlinearboundary condition IIrdquoNonlinear Analysis Theory Methods ampApplications vol 73 no 4 pp 971ndash978 2010
[22] Y Liu ldquoLower bounds for the blow-up time in a non-local reac-tion diffusion problem under nonlinear boundary conditionsrdquoMathematical and ComputerModelling vol 57 no 3-4 pp 926ndash931 2013
[23] Y Liu S Luo and Y Ye ldquoBlow-up phenomena for a parabolicproblem with a gradient nonlinearity under nonlinear bound-ary conditionsrdquo Computers amp Mathematics with Applicationsvol 65 no 8 pp 1194ndash1199 2013
[24] Z B Fang R Yang and Y Chai ldquoLower bounds estimate for theblow-up time of a slow diffusion equation with nonlocal sourceand inner absorptionrdquo Mathematical Problems in Engineeringvol 2014 Article ID 764248 6 pages 2014
Research ArticleSome Properties on Complex Functional Difference Equations
Zhi-Bo Huang12 and Ran-Ran Zhang3
1 School of Mathematical Sciences South China Normal University Guangzhou 510631 China2Department of Physics and Mathematics University of Eastern Finland PO Box 111 80101 Joensuu Finland3Department of Mathematics Guangdong University of Education Guangzhou 510303 China
Correspondence should be addressed to Zhi-Bo Huang huangzhiboscnueducn
Received 15 January 2014 Accepted 13 March 2014 Published 24 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z-B Huang and R-R ZhangThis is an open access article distributed under theCreativeCommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited
We obtain some results on the transcendental meromorphic solutions of complex functional difference equations of the formsum120582isin119868120572120582(119911)(prod
119899
119895=0119891(119911 + 119888
119895)120582119895) = 119877(119911 119891 ∘119901) = ((119886
0(119911) +119886
1(119911)(119891 ∘119901)+ sdot sdot sdot + 119886
119904(119911) (119891 ∘119901)
119904
)(1198870(119911) + 119887
1(119911) (119891 ∘119901)+ sdot sdot sdot + 119887
119905(119911) (119891 ∘119901)
119905
))where 119868 is a finite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119899) are distinct complex constants 119901(119911) is
a polynomial and 120572120582(119911) (120582 isin 119868) 119886
119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative to 119891(119911)
We further investigate the above functional difference equation which has special type if its solution has Borel exceptional zero andpole
1 Introduction and Main Results
In this paper a meromorphic function means meromorphicin the whole complex plane C For a meromorphic function119910(119911) let 120590(119910) be the order of growth and 120583(119910) the lowerorder of119910(119911) Further let 120582(119910) (resp 120582(1119910)) be the exponentof convergence of the zeros (resp poles) of 119910(119911) We alsoassume that the reader is familiar with the fundamentalresults and the standard notations of Nevanlinna theory ofmeromorphic functions (see eg [1]) Given a meromorphicfunction 119910(119911) we call a meromorphic function 119886(119911) a smallfunction relative to 119910(119911) if 119879(119903 119886(119911)) = 119878(119903 119910) = 119900(119879(119903 119910))as 119903 rarr infin possibly outside of an exceptional set of finitelogarithmic measure Moreover if 119877(119911 119910) is rational in 119910(119911)with small functions relative to 119910(119911) as its coefficients we usethe notation 119889 = deg
119910119877(119911 119910) for the degree of 119877(119911 119910) with
respect to119910(119911) Inwhat follows we always assume that119877(119911 119910)is irreducible in 119910(119911)
Meromorphic solutions of complex difference equationshave recently gained increasing interest due to the problemof integrability of difference equations This is related to theactivity concerning Painleve differential equations and theirdiscrete counterparts in the last decades Ablowitz et al [2]considered discrete equations to be delay equations in thecomplex plane This allowed them to analyze these equations
with the methods from complex analysis In regard to relatedpapers concerning a more general class of complex differenceequations we may refer to [3ndash5] These papers mainly dealtwith equations of the form
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119877 (119911 119891) (1)
where 119869 is a collection of all nonempty subsets of1 2 119899 119888
119895(119895 isin 119869) are distinct complex constants 119891(119911) is
a transcendental meromorphic function 120572119869(119911) (119869 isin 119869) are
small functions relative to119891(119911) and119877(119911 119891) is a rational func-tion in 119891(119911) with small meromorphic coefficients Moreoverif the right-hand side of (1) is essentially like the compositefunction 119890 ∘119891 of 119891(119911) and a rational function 119890(119911) Laine et alreversed the order of composition that is they considered thecomposite function 119891 ∘ 119890 of 119891(119911) and a rational function 119890(119911)which resulted in a complex functional difference equationThe following theorem [5 Theorem 28] gives an example
Theorem A (see [5 Theorem 28]) Suppose that 119891(119911) is atranscendental meromorphic solution of equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (2)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 283895 10 pageshttpdxdoiorg1011552014283895
2 Abstract and Applied Analysis
where 119901(119911) is a polynomial of degree 119896 ge 2 Moreover oneassumes that the coefficients 120572
119869(119911) are small functions relative
to 119891(119911) and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (3)
where 120572 = (log 119899)(log 119896)At this point we briefly introduce some notations used in
this paper A difference monomial of a meromorphic function119891(119911) is defined as
119891(119911)1205820119891(119911 + 119888
1)1205821
sdot sdot sdot 119891(119911 + 119888119899)120582119904
=
119904
prod
119895=0
119891(119911 + 119888119895)
120582119895
(4)
where 1198880= 0 119888
119895isin C 0 (119895 = 1 2 119904) are distinct
constants and 120582119895(119895 = 0 1 119904) are natural numbers A
difference polynomial 119867(119911 119891(119911)) of a meromorphic function119891(119911) a finite sum of difference monomials is defined as
119867(119911 119891 (119911)) = sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
) (5)
where 119868 is a finite set of multi-indexes 120582 = (1205820 1205821 120582
119899)
120572120582(119911) (120582 isin 119868) are small functions relative to 119891(119911) The degree
and the weight of the difference polynomial (5) respectively aredefined as
deg119891(119867) = max
120582isin119868
119899
sum
119895=0
120582119895
120581119891(119867) = max
120582isin119868
119899
sum
119895=1
120582119895
(6)
Consequently 120581119891(119867) le deg
119891(119867) For instance the degree and
the weight of the difference polynomial 1198912(119911)119891(119911minus1)119891(119911+1)+119891(119911)119891(119911 + 1)119891(119911 + 2) + 119891
2
(119911 minus 1)119891(119911 + 2) respectively arefour and three Moreover a difference polynomial (5) is said tobe homogeneous with respect to 119891(119911) if the degree sum119899
119895=0120582119895of
each monomial in the sum of (5) is nonzero and the same forall 120582 isin 119868
In the following we proceed to prove generalizations ofTheorem A and investigate some new results for the first timeWe permit more general expressions on both sides of (1)
Theorem 1 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(7)
where119867(119911 119891(119911)) is defined as (5) 119901(119911) = 119889119896119911119896
+ sdot sdot sdot+1198891119911+119889
0
is a polynomial with constant coefficients 119889119896( = 0) 119889
1 1198890
and of the degree 119896 ge 2 and 119886119894(119911) (119894 = 0 1 119904) and
119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions relative
to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Set 119889 = max119904 119905 If 119896119889 le
(119899 + 1)deg119891(119867) then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (8)
where 120572 = (log(119899 + 1) + log deg119891(119867) minus log119889)(log 119896)
Similar to the proof of Theorem 1 we easily obtain thefollowing result which is a generation of Theorem A
Theorem 2 Let 119888119894isin C (119894 = 1 2 119899) be distinct
constants and 119891(119911) be a transcendental meromorphic solutionof equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895))
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(9)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119896119889 =
119896max119904 119905 le 119899 then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (10)
where 120572 = (log 119899 minus log 119889)(log 119896)
We then proceed to consider the distribution of zeros andpoles of meromorphic solutions of (7) The following resultindicates that solutions having Borel exceptional zeros andpoles appear only in special situations
Theorem 3 Let 1198880= 0 let 119888
119894isin C 0 (119894 = 1 2 119899) be
distinct constants and let 119891(119911) be a finite order transcendentalmeromorphic solution of equation
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(11)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 1
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 If
max120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (12)
then (11) is either of the form119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(13)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (14)
where 120572 isin C 0 is some constant
Abstract and Applied Analysis 3
Example 4 119891(119911) = cos 119911 solves difference equation
4119891(119911)2
119891(119911 + 120587)2
= 119891(2119911)2
+ 2119891 (2119911) + 1 (15)
Here 119901(119911) = 2119911 Clearly 120582(1119891) = 0 lt 1 = 120582(119891) = 120590(119891) Thisexample shows that condition (12) is necessary and cannot bereplaced by
min120582 (119891) 120582 ( 1119891
) lt 120590 (119891) (16)
Moreover we obtain a result parallel toTheorem 54 in [6]for the difference case
Theorem 5 Suppose that the equation119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
=
119888 (119911)
119891(119911)119898 119898 isin N (17)
has a meromorphic solution of finite order where 1198880= 0 119888
119895isin
C 0 (119895 = 1 2 119899) are distinct constants and 119888(119911) is anontrivialmeromorphic function If119891(119911)has only finitelymanypoles then119891(119911) = 119863(119911)119890119864(119911) where119863(119911) is a rational functionand119864(119911) is a polynomial if and only if 119888(119911) = 119866(119911)119890119872(119911) where119866(119911) is a rational function and119872(119911) is a polynomial
Example 6 Difference equation
119891(119911)2
119891 (119911 + 1) 119891 (119911 minus 1) = (
1
1199114(1199112minus 1)
1198906119911
) sdot
1
119891(119911)2
(18)
of the type (17) is solved by 119891(119911) = 119890119911119911 Here 119891(119911) = 119890119911119911and 119888(119911) = 11989061199111199114(1199112 minus 1) satisfy Theorem 5
As an application of Theorem 3 we obtain the following
Theorem 7 Let 119888 isin C 0 and let 119891(119911) be a finite ordertranscendental meromorphic solution of equation
119891 (119911 + 119888) = 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(19)
where 119901(119911) = 119889119896119911119896
+ sdot sdot sdot + 1198891119911 + 119889
0is a polynomial with
constant coefficients 119889119896( = 0) 119889
1 1198890and of the degree 119896 ge 2
and 119886119894(119911) (119894 = 0 1 119904) and 119887
119895(119911) (119895 = 0 1 119905) are small
meromorphic functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv
0 Then 119891(119911) has at most one Borel exceptional value
If the degree 119896 of polynomial 119901(119911) is 1 in Theorem 7 theresult does not hold For example we have the following
Example 8 119891(119911) = tan 119911 solves difference equation
119891 (119911 + 1) =
119891 (119911) + tan 11 minus (tan 1) 119891 (119911)
(20)
of the type (19) Obviously 119891(119911) has two Borel exceptionalvalues plusmn119894
If we remove the assumption max120582(119891) 120582(1119891) lt 120590(119891)used in Theorem 3 we obtain a result similar to Theorem 12in [4]
Theorem 9 Let 119891(119911) be a transcendental meromorphic solu-tion of equation
119867(119911 119891 (119911)) = 119877 (119911 119891)
=
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905
(21)
where 119867(119911 119891(119911)) is defined as (5) and 119886119894(119911) (119894 = 0 1 119904)
and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic functions
relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 If 119889 = max119904 119905 gt
(119899 + 1) deg119891(119867) then 120590(119891) = infin
In fact the following examples show that the assertion ofTheorem 9 does not remain valid identically if 119889 le (119899 +1)deg
119891(119867)
Example 10 119891(119911) = exp119890119911119911 solves the difference equation
(119911 minus 120587119894) (119911 + log 2 minus 120587119894) 119891 (119911) 119891 (119911 minus 120587119894) 119891 (119911 + log 2 minus 120587119894)
+ (119911 + log 8) 119891 (119911 + log 8) =1 + 119911
11
119891(119911)10
1199113119891(119911)
2
(22)
Clearly 119889 = 10 lt (3+1) sdot 3 = (119899+1) deg119891(119867) and 120590(119891) = infin
Example 11 119891(119911) = tan 119911 satisfies the difference equation
119891(119911 +
120587
4
) + 119891(119911 minus
120587
4
) =
4119891 (119911)
1 minus 119891(119911)2
(23)
Obviously 119889 = 2 lt (2+1)times1 = (119899+1) deg119891(119867) and120590(119891) = 1
Example 12 (see [7 pages 103ndash106] and [8 page 8]) Thefollowing difference equation
119891 (119911 + 1) = 120572119891 (119911) (1 minus 119891 (119911)) 120572 = 0 (24)
derives from a well-known discrete logistic model in biologyIt has been proved that all other meromorphic solutions areof infinite order apart from the constant solutions 119891(119911) equiv 0and 119891(119911) = (120572 minus 1)120572 For instance (24) has one-parameterfamilies of entire solutions of infinite order
119891 (119911) =
1
2
(1 minus exp (119860119890119911 log 2)) 119860 isin C 0 120572 = 2
119891 (119911) = sin2 (119861119890119911 log 2) 119861 isin C 0 120572 = 4(25)
Here 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867)
Example 13 119891(119911) = 119911 solves the difference equation
119891 (119911 + 1) =
1 minus 119891(119911)2
minus1199112minus 119911 + 1 + 119891(119911)
2 (26)
We get 119889 = 2 = (1 + 1) times 1 = (119899 + 1)deg119891(119867) and 120590(119891) = 0
If the difference polynomial in the left-hand side of (21)is homogeneous we further obtain the following theorem
4 Abstract and Applied Analysis
Theorem 14 Let 119891(119911) be a transcendental meromorphic solu-tion of (21) where 119867(z 119891(119911)) is defined as (5) and 119886
119894(119911) (119894 =
0 1 119904) and 119887119895(119911) (119895 = 0 1 119905) are small meromorphic
functions relative to 119891(119911) such that 119886119904(119911)119887
119905(119911) equiv 0 Suppose
that 119867(119911 119891) is homogeneous and has at least one differencemonomial of type
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
(120582119895isin N
+ 119895 = 0 1 119899) (27)
If 119889 = max119904 119905 gt 3deg119891(119867) then 120590(119891) = infin
2 Proof of Theorem 1
We need some preliminaries to proveTheorem 1
Lemma 15 (see [9 Lemma 4]) Let 119891(119911) be a transcendentalmeromorphic function and let 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot + 1198891119911 +
1198890(119889119896= 0)be a polynomial of degree 119896 Given 0 lt 120575 lt |119889
119896|
denote ] = |119889119896| + 120575 and 120583 = |119889
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C cup infin one has for all 119903 ge 1199030gt 0
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 ∘ 119901) le 119896119899 (]119903119896 119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 ∘ 119901) le 119873(]119903119896 119886 119891)
+ 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 ∘ 119901) le (1 + 120576) 119879 (]119903119896 119891) (28)
Lemma16 (see [10Theorem B16]) Given distinctmeromor-phic functions 119891
1 119891
119899 let 119869 denote the collection of all
nonempty subsets of 1 2 119899 and suppose that 120572119869isin C for
each 119869 isin 119869 Then
119879(119903sum
119869
120572119869(prod
119895isin119869
119891119895)) le
119899
sum
119896=1
119879 (119903 119891119896) + 119874 (1) (29)
By denoting 119891119894+1= 119891(119911 + 119888
119894)120582119894(119894 = 0 1 119899) below it is
an easy exercise to prove the following result from Lemma 16
Lemma 17 Let 119891(119911) be a meromorphic function let 119868 be afinite set of multi-indexes 120582 = (120582
0 1205821 120582
119899) and let 120572
120582(119911)
be small functions relative to 119891(119911) for all 120582 isin 119868 Then thecharacteristic function of the difference polynomial (5) satisfies
119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
))
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(30)
where 119862 = max|1198881| |119888
2| |119888
119899|
Lemma18 (see [11 Lemma5]) Let119892(119903) andℎ(119903) bemonotonenondecreasing functions on [0infin) such that 119892(119903) le ℎ(119903) for all119903 notin 119864 cup [0 1] where 119864 sub (1infin) is a set of finite logarithmicmeasure Let 120572 gt 1 be a given constant Then there exists an1199030= 119903
0(120572) gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 ge 119903
0
Lemma 19 (see [12 Lemma 3]) Let 120595(119903) be a function of119903 (119903 ge 119903
0) positive and bounded in every finite interval
(i) Suppose that 120595(120583119903119898) le 119860120595(119903) + 119861 (119903 ge 1199030) where
120583 (120583 gt 0)119898 (119898 gt 1)119860 (119860 ge 1) and 119861 are constantsThen 120595(119903) = 119874((log 119903)120572) with 120572 = (log119860)(log119898)unless 119860 = 1 and 119861 gt 0 and if 119860 = 1 and 119861 gt 0 thenfor any 120576 gt 0 120595(119903) = 119874((log 119903)120576)
(ii) Suppose that (with the notation of (i)) 120595(120583119903119898) ge119860120595(119903) (119903 ge 119903
0) Then for all sufficiently large values of
119903 120595(119903) ge 119870(log 119903)120572 with 120572 = (log119860)(log119898) for somepositive constant 119870
Proof of Theorem 1 For any 120576 (0 lt 120576 lt 1) we may applyValiron-Mohonrsquoko lemma Lemmas 15 and 17 and (5) and (7)to conclude that119889 (1 minus 120576) 119879 (120583119903
119896
119891)
le 119889119879 (119903 119891 ∘ 119901) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905)
+ 119878 (119903 119891)
= 119879(119903 sum
120582isin119868
120572120582(119911)(
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
)) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(31)holds for all sufficiently large 119903 possibly outside of anexceptional set of finite logarithmic measure where 119862 =max|119888
1| |119888
2| |119888
119899| and 120583 is defined as Lemma 15 Now we
may apply Lemma 18 to deal with the exceptional set andconclude that for every 120578 gt 1 there exists an 119903
0gt 0 such
that119889 (1 minus 120576) 119879 (120583119903
119896
119891) le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (120578119903 119891)
(32)holds for all 119903 ge 119903
0 Denote 120596 = 120578119903 Then (32) can be written
in the form
119879(
120583
120578119896
120596119896
119891) le
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
119879 (120596 119891) (33)
Since 119889119896 le (119899 + 1)deg119891(119867) we get ((119899 + 1)deg
119891(119867)(1 +
120576))(119889(1 minus 120576)) gt 1 for all 0 lt 120576 lt 1 Thus we now applyLemma 19(i) to conclude that
119879 (119903 119891) = 119874 ((log 119903)120572+120576)
120572 =
log ((119899 + 1) deg119891(119867) (1 + 120576) 119889 (1 minus 120576))
log 119896
=
log (119899 + 1) + log deg119891(119867) minus log119889
log 119896+ 119900 (1)
(34)
The proof of Theorem 1 is completed
Abstract and Applied Analysis 5
3 Proof of Theorems 3 and 5
We again need some preliminaries
Lemma 20 (see [13 Theorem 15]) Suppose that 119891119895(119911) (119895 =
1 2 119899) (119899 ge 2) are meromorphic functions and 119892119895(119911) (119895 =
1 2 119899) are entire functions satisfying the following condi-tions
(1) sum119899119895=1119891119895(119911)119890
119892119895(119911)
= 0
(2) 119892119895(119911) minus 119892
119896(119911) are not constants for 1 le 119895 lt 119896 le 119899
(3) For 1 le 119895 le 119899 1 le ℎ lt 119896 le 119899
119879 (119903 119891119895) = 119900 119879 (119903 119890
119892ℎminus119892119896) (119903 997888rarr +infin 119903 notin 119864) (35)
where 119864 sub (1 +infin) is of finite linear measure or finitelogarithmic measure
Then 119891119895(119911) equiv 0 (119895 = 1 2 119899)
Lemma21 (see [14Theorem4]) Let119865(119911)119875119899(119911) 119875
0(119911) be
polynomials such that 1198651198751198991198750equiv 0 and then every finite order
transcendental meromorphic solution 119891(119911) of equation
119875119899(119911) 119891 (119911 + 119899) + sdot sdot sdot + 119875
1(119911) 119891 (119911 + 1) + 119875
0(119911) 119891 (119911) = 119865 (119911)
(36)
satisfies 120582(119891) = 120590(119891) ge 1
Remark 22 Replacing 119895 by 119888119895(119895 = 1 2 119899) where 119888
119895(119895 =
1 2 119899) are distinct nonzero complex constants Lemma 21remains valid
Proof of Theorem 3 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed withnonzero poles of 119891(119911) Since max120582(119891) 120582(1119891) lt 120590(119891) thenℎ(119911) = 119911
120591
119902(119911) is an entire function such that
120590 (ℎ) = 120582(
1
119891
) lt 120590 (119891) lt +infin (37)
and 119892(119911) = ℎ(119911)119891(119911) is a transcendental entire function with
119879 (119903 119892) = 119879 (119903 119891) + 119878 (119903 119891) 120590 (119892) = 120590 (119891)
120582 (119892) = 120582 (119891)
(38)
If 119902(119911) is a polynomial we obtain quickly that 120590(ℎ ∘ 119901) =0 lt 120590(119892 ∘ 119901) Otherwise we conclude from the last assertionof Lemma 15 (37) and (38) that
120590 (ℎ ∘ 119901) = 119896120590 (ℎ) = 119896120582(
1
119891
) lt 119896120590 (119892) = 120590 (119892 ∘ 119901) (39)
Therefore
119879 (119903 ℎ ∘ 119901) = 119878 (119903 119892 ∘ 119901) (40)
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (11) we concludethat
(ℎ ∘ 119901)119904minus119905
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
(
119899
prod
119895=0
119892(119911 + 119888119895)
120582119895
)
= (1198860(119911) (ℎ ∘ 119901)
119904
+ 1198861(119911) (ℎ ∘ 119901)
119904minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119886119904(119911) (119892 ∘ 119901)
119904
)
times (1198870(119911) (ℎ ∘ 119901)
119905
+ 1198871(119911) (ℎ ∘ 119901)
119905minus1
(119892 ∘ 119901)
+ sdot sdot sdot + 119887119905(119911) (119892 ∘ 119901)
119905
)
minus1
(41)
Obviously it follows from (37)ndash(40) and Lemma 15 that
119879(119903
1
prod119899
119895=0ℎ(119911 + 119888
119895)
120582119895
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 (ℎ ∘ 119901)119904minus119905
) = 119878 (119903 119892 ∘ 119901)
119879 (119903 119886119906(119911) (ℎ ∘ 119901)
119904minus119906
) = 119878 (119903 119892 ∘ 119901) 119906 = 0 1 119904
119879 (119903 119887V (119911) (ℎ ∘ 119901)119905minusV) = 119878 (119903 119892 ∘ 119901) V = 0 1 119905
(42)
Denoting119860(119911) = (ℎ ∘ 119901)119904minus119905prod119899119895=0ℎ(119911 + 119888
119895)120582119895 we get from (42)
that119879 (119903 119860) = 119878 (119903 119892 ∘ 119901) (43)
Since zeros and poles are Borel exceptional values of 119891(119911) by(12) we may apply a result due to Whittaker see [15 Satz134] to deduce that 119891(119911) is of regular growth Thus we useLemma 15 and (12) again to get
119879(119903
1198911015840
119891
) = 119873(119903 119891) + 119873(119903
1
119891
) + 119878 (119903 119891) = 119878 (119903 119892 ∘ 119901)
(44)
Similarly if we set 119861(119911) = 119860(119911)(prod119899
119895=0119892(119911 + 119888
119895)120582119895) we
also deduce from the lemma of the logarithmic derivativeLemma 15 (12) (38) and (43) that
119879(119903
1198611015840
119861
) = 119879(119903
1198601015840
119860
+
119899
sum
119895=0
120582119895
1198921015840
(119911 + 119888119895)
119892 (119911 + 119888119895)
) = 119878 (119903 119892 ∘ 119901)
(45)Denoting 119865(119911) = 119892 ∘ 119901
119875 (119911 119865) =
1198860(119911)
119886119904(119911)
(ℎ ∘ 119901)119904
+
1198861(119911)
119886119904(119911)
(ℎ ∘ 119901)119904minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119904
119876 (119911 119865) =
1198870(119911)
119887119905(119911)
(ℎ ∘ 119901)119905
+
1198871(119911)
119887119905(119911)
(ℎ ∘ 119901)119905minus1
119865 (119911) + sdot sdot sdot + 119865(119911)119905
(46)
6 Abstract and Applied Analysis
Therefore we deduce from Lemma 15 and (42) that thecoefficients of 119875(119911 119865) and119876(119911 119865) are small functions relativeto 119865(119911) Thus (41) can be written in the form
119887119905(119911)
119886119904(119911)
119861 (119911) =
119875 (119911 119865)
119876 (119911 119865)
= 119906 (119911 119865) (47)
Denoting
120595 (119911) =
1198651015840
(119911)
119865 (119911)
119880 (119911) =
1199061015840
(119911 119865)
119906 (119911 119865)
(48)
we get 119879(119903 119880) = 119878(119903 119892 ∘ 119901) from (45) and (47) Wealso conclude from the lemma of logarithmic derivativeLemma 15 and (12) that
119879 (119903 120595) = 119879(119903
1198651015840
119865
) = 119898(119903
1198651015840
119865
) + 119873(119903
1198651015840
119865
)
le 119873 (119903 119865) + 119873(119903
1
119865
) + 119878 (119903 119865)
= 119873 (119903 119892 ∘ 119901) + 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(119903
1
119892 ∘ 119901
) + 119878 (119903 119892 ∘ 119901)
le 119873(]1199031198961
119892
) + 119878 (119903 119892 ∘ 119901) = 119878 (119903 119892 ∘ 119901)
(49)
where ] is defined as Lemma 15Since
1198751015840
119876 minus 1198751198761015840
1198762
= 1199061015840
= 119880119906 =
119880119875
119876
(50)
we conclude that
1198751015840
119876 minus 1198751198761015840
= 119880119875119876 (51)
Now writing 1198651015840 = 120595119865 in (51) regarding then (51) as analgebraic equation in119865with coefficients of growth 119878(119903 119865) andcomparing the leading coefficients we deduce that
(119904 minus 119905) 120595 = 119880 (52)
By integrating both sides of the last equality above weconclude that
119906 (119911 119865) = 120572119865(119911)119904minus119905
(53)
for some 120572 isin C 0 Therefore by combining therepresentations of 119865 119861 119860 119892 with (53) we conclude that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
(54)
If 119904119905 = 0 we deduce from (11) and (54) that
120572
119886119904(119911)
119887119905(119911)
(119891 ∘ 119901)119904minus119905
= 119877 (119911 119891 ∘ 119901)
=
1198860(119911) + 119886
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911) (119891 ∘ 119901)
119904
1198870(119911) + 119887
1(119911) (119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911) (119891 ∘ 119901)
119905
(55)
From this we get that 119877(119911 119891 ∘ 119901) is not irreducible in 119891 ∘ 119901a contradiction Thus 119905 = 0 or 119904 = 0 Therefore we deducefrom (54) that
119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
119886119904(119911)
1198870(119911)
(119891 ∘ 119901)119904
(56)
or119899
prod
119895=0
119891(119911 + 119888119895)
120582119895
= 120572
1198860(119911)
119887119905(119911)
1
(119891 ∘ 119901)119905 (57)
The proof of Theorem 3 is completed
Proof of Theorem 5 Assume first that 119891(119911) = 119863(119911)119890119864(119911)
where 119863(119911) is a rational function and 119864(119911) is a polynomialOne can see from (17) that
119888 (119911) = 119891(119911)119898
119899
prod
119894=0
119891(119911 + 119888119894)120582119894
= [119863(119911)119898
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
] 119890119898119864(119911)+sum
119899
119894=0120582119894119864(119911+119888
119894)
= 119866 (119911) 119890119872(119911)
(58)
where 119866(119911) = 119863(119911)119898prod119899119894=0119863(119911 + 119888
119894)120582119894 is rational and119872(119911) =
119898119864(119911) + sum119899
119894=0120582119894119864(119911 + 119888
119894) is a polynomial
Suppose next that 119888(119911) = 119866(119911)119890119872(119911) where 119866(119911) is arational function and 119872(119911) is a polynomial Since 119891(119911) hasonly finitely many poles we conclude from (17) that
119873(119903
1
119891
) le 119873(119903
1
119891119898
) = 119873(119903
prod119899
119894=0119891(119911 + 119888
119894)120582119894
119888 (119911)
)
= 119874 (log 119903)
(59)
Thus 119891(119911) has only finitely many zeros and poles and119891(119911) = 119863(119911)119890
119864(119911) where119863(119911) is rational and 119864(119911) is an entirefunction In the followingwe only prove119864(119911) is a polynomialNow substituting119891(119911) = 119863(119911)119890119864(119911) and 119888(119911) = 119866(119911)119890119872(119911) into(17) we get
119899
prod
119894=0
119863(119911 + 119888119894)120582119894 exp (120582
119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(60)
(
119899
prod
119894=0
119863(119911 + 119888119894)120582119894
) exp(119899
sum
119894=0
120582119894119864 (119911 + 119888
119894))
=
119866 (119911)
119863(119911)119898exp (119872 (119911) minus 119898119864 (119911))
(61)
Thus we deduce from Lemma 20 that two exponents in (61)cancel each other to a constant 120591 isin C such that
119899
sum
119894=0
120582119894119864 (119911 + 119888
119894) = 119872 (119911) minus 119898119864 (119911) + 120591 (62)
Abstract and Applied Analysis 7
that is119899
sum
119894=1
120582119894119864 (119911 + 119888
119894) + (120582
0+ 119898)119864 (119911) = 119872 (119911) + 120591 (63)
Suppose that 119864(119911) is not a polynomial If 119864(119911) is a transcen-dental entire function of finite order we get from Lemma 21Remark 22 and (63) that 120590(119864) ge 1 Otherwise 119864(119911) is atranscendental entire function of infinite order These bothshow that 120590(119891) = infin contradicting the assumption that119891(119911) is finite order Thus 119864(119911) is a polynomial The proof ofTheorem 5 is completed
4 Proof of Theorem 7
Lemmas 23 and 25 reveal some properties of the maximalmodule of the polynomial in composite function 119891 ∘ 119901 with ameromorphic function 119891(119911) and a polynomial 119901(119911) whichare useful for proving the existence of Borel exceptionalvalue of finite order meromorphic solutions of functionaldifference equation of type (19)
Lemma 23 Let 119892(119911) be a nonconstant entire function of order120590(119892) = 120590 lt infin Suppose that 120572
119895(119911) (119895 = 1 2 119898) are small
meromorphic functions relative to 119892(119911) Then there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 120572119895)
119872 (119903 119892)
997888rarr 0 119895 = 1 2 119898 (64)
hold simultaneously for all 119903 isin 119864 as 119903 rarr infin where the lowerlogarithmic density of set 119864 is defined by
logdens (119864) = lim inf119903rarrinfin
int[1119903]cap119864
(119889119905119905)
log 119903 (65)
Remark 24 The proof of Lemma 23 is similar to the proof ofLemma 24 and Remark 25 in [16] Here we omit it
Lemma 25 Let 119891(119911) be a finite order transcendental mero-morphic function satisfying (12) and 119901(119911) = 119889
119896119911119896
+ sdot sdot sdot+1198891119911+
1198890is a polynomial with constant coefficients 119889
119896( = 0) 119889
1 1198890
and of the degree 119896 ge 1 Suppose that
119867(119911) = 119886119899(119911) (119891 ∘ 119901)
119899
+ 119886119899minus1(119911) (119891 ∘ 119901)
119899minus1
+ sdot sdot sdot + 1198861(119911) (119891 ∘ 119901) + 119886
0(119911)
(66)
is a polynomial in 119891 ∘119901 where 119899 (ge 1) is a positive integer and119886119899(119911) ( equiv 0) 119886
119899minus1(119911) 119886
1(119911) 119886
0(119911) are small meromorphic
functions relative to 119891(119911) Then there exists a set 1198641of lower
logarithmic density 1 such that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (67)
for all 119903 isin 1198641as 119903 rarr infin where 0 lt 120583 lt |119889
119896| Hence119867(119911) equiv
0
Proof of Lemma 25 Let 120591 be the multiplicity of pole of 119891(119911)at the origin and let 119902(119911) be a canonical product formed
with the nonzero poles of 119891(119911) Since 119891(119911) satisfies (12) thenℎ(119911) = 119911
120591
119902(119911) is an entire function Thus 119892(119911) = ℎ(119911)119891(119911) isentire and (37) (38) and (40) also hold
Now substituting 119891(119911) = 119892(119911)ℎ(119911) into (66) we con-clude that
119867(119911) = 119886119899(119911) sdot
(119892 ∘ 119901)119899
(ℎ ∘ 119901)119899+ 119886
119899minus1(119911) sdot
(119892 ∘ 119901)119899minus1
(ℎ ∘ 119901)119899minus1
+ sdot sdot sdot
+ 1198861(119911) sdot
(119892 ∘ 119901)
(ℎ ∘ 119901)
+ 1198860(119911)
=
119886119899(119911)
(ℎ ∘ 119901)119899(119892 ∘ 119901)
119899
times [1 +
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
(119892 ∘ 119901)minus1
+ sdot sdot sdot
+
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
(119892 ∘ 119901)1minus119899
+
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
(119892 ∘ 119901)minus119899
]
(68)
We note from Lemma 15 and (40) that
119879(119903
119886119899(119911)
(ℎ ∘ 119901)119899) = 119878 (119903 119892 ∘ 119901)
119879(119903
119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911)
) = 119878 (119903 119892 ∘ 119901)
for 119895 = 0 1 119899 minus 1
(69)
Therefore we deduce from Lemma 23 that there exists a set119864 sub (1infin) of lower logarithmic density 1 such that
119872(119903 119886119899(119911) (ℎ ∘ 119901)
119899
)
119872 (119903 119892 ∘ 119901)
997888rarr 0
119872(119903 119886119895(119911) (ℎ ∘ 119901)
119899minus119895
119886119899(119911))
119872 (119903 119892 ∘ 119901)
997888rarr 0
(119895 = 0 1 119899 minus 1)
(70)
Moreover according to the choosing of 119864 in the proofof Lemma 23 we know that 119886
119895(119911)(ℎ ∘ 119901)
119899minus119895
119886119899(119911) for
8 Abstract and Applied Analysis
119895 = 0 1 119899 minus 1 have no zeros and poles for all |119911| = 119903 isin 119864Thus we conclude from (68) and (70) that for any 120576 gt 0
119872(119903119867) ge 119872(119903 119892 ∘ 119901)119899minus120576
times [1 minus
100381610038161003816100381610038161003816100381610038161003816
119886119899minus1(119911) (ℎ ∘ 119901)
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus1
minus sdot sdot sdot
minus
1003816100381610038161003816100381610038161003816100381610038161003816
1198861(119911) (ℎ ∘ 119901)
119899minus1
119886119899(119911)
1003816100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)1minus119899
minus
100381610038161003816100381610038161003816100381610038161003816
1198860(119911) (ℎ ∘ 119901)
119899
119886119899(119911)
100381610038161003816100381610038161003816100381610038161003816
119872(119903 119892 ∘ 119901)minus119899
]
ge (1 minus 120576)119872(119903 119892 ∘ 119901)119899minus120576
(71)
and so
log+119872(119903119867) ge (119899 minus 32
120576) log+119872(119903 119892 ∘ 119901) (72)
for all |119911| = 119903 isin 119864 and |119892 ∘ 119901(119911)| = 119872(119903 119892 ∘ 119901)Therefore we deduce from Lemma 15 and (38) that
log+119872(119903119867) ge (119899 minus 2120576) 119879 (120583119903119896 119891) (73)
for all |119911| = 119903 isin 1198641= 119864 cap (119903
0 +infin) where 119903
0gt 0 It is
obvious that 1198641has lower logarithmic density 1 The proof of
Lemma 25 is completed
Proof of Theorem 7 Suppose that 119891(119911) has two finite Borelexceptional values 119886 and 119887 ( = 0 119886) For the case where oneof 119886 and 119887 is infinite we can use a similar method to prove itSet
119892 (119911) =
119891 (119911) minus 119886
119891 (119911) minus 119887
(74)
Then 120590(119892) = 120590(119891) and
120582 (119892) = 120582 (119891 minus 119886) lt 120590 (119892) 120582 (
1
119892
) = 120582 (119891 minus 119887) lt 120590 (119892)
(75)
It follows from (74) that
119891 (119911) =
119886 minus 119887119892 (119911)
1 minus 119892 (119911)
(76)
Now substituting (76) into (19) we conclude that
119892 (119911 + 119888) = (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119886
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
times (
119904
sum
119894=0
119886119894(119911) (119886 minus 119887119892 ∘ 119901)
119894
(1 minus 119892 ∘ 119901)119904+119905minus119894
minus119887
119905
sum
119895=0
119887119895(119911) (119886 minus 119887119892 ∘ 119901)
119895
(1 minus 119892 ∘ 119901)119904+119905minus119895
)
minus1
= ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
))
times ((minus119892 ∘ 119901)119904+119905
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
+ sdot sdot sdot + (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
))
minus1
(77)
Since 1198860(119911) + 119886
1(119911)(119891 ∘ 119901) + sdot sdot sdot + 119886
119904(119911)(119891 ∘ 119901)
119904 and 1198870(119911) +
1198871(119911)(119891 ∘ 119901) + sdot sdot sdot + 119887
119905(119911)(119891 ∘ 119901)
119905 are irreducible in 119891 ∘ 119901 weconclude that at least one of the following three inequalitiesholds that is
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
) sdot (
119904
sum
119894=0
119886119894(119911) 119886
119894
minus 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
)
equiv 0
(78)
Thus we deduce fromTheorem 3 that
119892 (119911 + 119888) = 119888 (119911) (119892 ∘ 119901)119897
(79)
where 119888(119911) is meromorphic function satisfying 119879(119903 119888) =119878(119903 119892) and 119897 isin Z Clearly 119897 = 0 and 119892(119911) is of regular growthfrom (75) see [15 Staz 134] Therefore 120590(119888) lt 120590(119892)
If 119897 ge 1 we conclude from (77) and (79) that
(minus1)119904+119905
119888 (119911)(
119904
sum
119894=0
119886119894(119911) 119887
119894
minus 119887
119905
sum
119895=0
119887119895(119911) 119887
119895
)(119892 ∘ 119901)119904+119905+119897
+ sdot sdot sdot + (minus
119904
sum
119894=0
119886119894(119911) 119886
119894
+ 119886
119905
sum
119895=0
119887119895(119911) 119886
119895
) = 0
(80)
Abstract and Applied Analysis 9
Thus we deduce from Lemma 25 that (80) is a contradictionIf 119897 le minus1 we use the same method as above to getanother contradiction Therefore 119891(119911) has at most one Borelexceptional value The proof of Theorem 7 is completed
5 Proof of Theorems 9 and 14
We first recall two lemmas
Lemma 26 (see [17 Lemma 21]) Let 119891(119911) be a nonconstantmeromorphic function 119904 gt 0 120572 lt 1 and 119865 sub R+ the set of all119903 such that
119879 (119903 119891) le 120572119879 (119903 + 119904 119891) (81)
If the logarithmicmeasure of119865 is infinite that isint119865
(119889119905119905) = infinthen 119891(119911) is of infinite order of growth
Lemma 27 (see [18 Corollary 26] and [19 Corollary 22])Let 119891(119911) be a meromorphic function of finite order and let 119888 isinC Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (82)
for all 119903 outside of a possible exceptional set of finite logarithmicmeasure
Proof of Theorem 9 For any 120576 (0 lt 120576 lt (119889 minus (119899 +
1)deg119891(119867))(119889 + (119899 + 1) deg
119891(119867))) we may apply Valiron-
Mohonrsquoko lemma Lemma 17 (5) and (21) to conclude that
119889 (1 minus 120576) 119879 (119903 119891)
le 119889119879 (119903 119891) + 119878 (119903 119891)
= 119879(119903
1198860(119911) + 119886
1(119911) 119891 (119911) + sdot sdot sdot + 119886
119904(119911) 119891(119911)
119904
1198870(119911) + 119887
1(119911) 119891 (119911) + sdot sdot sdot + 119887
119905(119911) 119891(119911)
119905)
= 119879 (119903119867 (119911 119891 (119911))) le (119899 + 1) deg119891(119867) 119879 (119903 + 119862 119891)
+ 119878 (119903 119891)
le (119899 + 1) deg119891(119867) (1 + 120576) 119879 (119903 + 119862 119891)
(83)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote
120572 =
(119899 + 1) deg119891(119867) (1 + 120576)
119889 (1 minus 120576)
(84)
Then 120572 lt 1 since 0 lt 120576 lt (119889 minus (119899 + 1)deg119891(119867))(119889 + (119899 +
1)deg119891(119867)) and 119889 gt (119899 + 1)deg
119891(119867) Thus
119879 (119903 119891) le 120572119879 (119903 + 119862 119891) (85)
holds for all 119903 in a set with infinite logarithmic measureTherefore we deduce from Lemma 26 and (85) that 120590(119891) =infin The proof of Theorem 9 is completed
Proof of Theorem 14 Assume contrary to the assertion that119891(119911) is meromorphic of finite order Taking into account theassumption that119867(119911 119891(119911)) is homogeneous we deduce fromLemma 27 that
119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) = 119878 (119903 119891) (86)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure
Denote 119862 = max1le119894le119899|119888119894| Since 119867(119911) is homogeneous
and has at least one difference monomial of typeprod119899119895=0119891(119911 +
119888119895)120582119895 we immediately conclude that by looking at pole
multiplicities summing over |119911| le 119903 and integratinglogarithmically
119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
)
le 120581119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
)) + 119878 (119903 119891)
(87)
Therefore
119879 (119903119867 (119911 119891 (119911)))
= 119898 (119903119867 (119911 119891 (119911))) + 119873 (119903119867 (119911 119891 (119911)))
le 119898(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119898(119903 119891(119911)deg119891(119867)
)
+ 119873(119903
119867 (119911 119891 (119911))
119891(119911)deg119891(119867)
) + 119873(119903 119891(119911)deg119891(119867)
)
le deg119891(119867) (119873 (119903 + 119862 119891) + 119873(119903
1
119891
))
+ 119879 (119903 119891(119911)deg119891(119867)
) + 119878 (119903 119891)
le 3deg119891(119867) 119879 (119903 + 119862 119891) + 119878 (119903 119891)
(88)
for all 119903 outside of a possible exceptional set of finitelogarithmic measure The remainder can be proven by asimilar method in Theorem 9 The proof of Theorem 14 iscompleted
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful to the referees for their helpfulsuggestions to improve this paper The first author alsothanks Professor Ilpo laine and Professor Risto Korhonen for
10 Abstract and Applied Analysis
their valuable suggestion to the present paper Research issupported by National Natural Science Foundation of China(nos 11171119 and 11171121) and Guangdong National NaturalScience Foundation (no S2012040006865)
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[3] BGrammaticos T Tamizhmani A Ramani andKMTamizh-mani ldquoGrowth and integrability in discrete systemsrdquo Journal ofPhysics A vol 34 no 18 pp 3811ndash3821 2001
[4] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of Malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
[7] WGKelley andAC PetersonDifference Equations AcademicPress Boston Mass USA 1991
[8] I Laine and C C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[9] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[10] V I Gromak I Laine and S Shimomura Painleve DifferentialEquations in the Complex Plane vol 28 Walter de GruyterBerlin Germany 2002
[11] G G Gundersen ldquoFinite order solutions of second orderlinear differential equationsrdquo Transactions of the AmericanMathematical Society vol 305 no 1 pp 415ndash429 1988
[12] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[13] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions vol 557 Kluwer Academic Publishers Group Dor-drecht The Netherlands 2003
[14] Z X Chen and K H Shon ldquoOn growth of meromorphicsolutions for linear difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 619296 6 pages 2013
[15] G Jank and L Volkmann Einfuhrung in die Theorie derganzen und meromorphen Funktionen mit Anwendungen aufDifferentialgleichungen Birkhauser Basel Switzerland 1985
[16] J Wang ldquoGrowth and poles of meromorphic solutions of somedifference equationsrdquo Journal of Mathematical Analysis andApplications vol 379 no 1 pp 367ndash377 2011
[17] R G Halburd and R J Korhonen ldquoFinite-order meromorphicsolutions and the discrete Painleve equationsrdquoProceedings of theLondon Mathematical Society vol 94 no 2 pp 443ndash474 2007
[18] Y M Chiang and S J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[19] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
Research ArticleThe Regularity of Functions on Dual Split Quaternionsin Clifford Analysis
Ji Eun Kim and Kwang Ho Shon
Department of Mathematics Pusan National University Busan 609-735 Republic of Korea
Correspondence should be addressed to Kwang Ho Shon khshonpusanackr
Received 28 January 2014 Accepted 2 April 2014 Published 17 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 J E Kim and K H Shon This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
This paper shows some properties of dual split quaternion numbers and expressions of power series in dual split quaternions andprovides differential operators in dual split quaternions and a dual split regular function on Ω sub C2 times C2 that has a dual splitCauchy-Riemann system in dual split quaternions
1 Introduction
Hamilton introduced quaternions extending complex num-bers to higher spatial dimensions in differential geometry (see[1]) A set of quaternions can be represented as
H = 119911 = 1199090+ 1199091119894 + 1199092119895 + 1199093119896 119909119898
isin R 119898 = 0 1 2 3
(1)
where 1198942
= 1198952
= 1198962
= minus1 119894119895119896 = minus1 and R denotes the set ofreal numbers Cockle [2] introduced a set of split quaternionsas
S = 119911 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 119909119898
isin R 119898 = 0 1 2 3
(2)
where 1198902
1= minus1 119890
2
2= 1198902
3= 1 and 119890
111989021198903
= 1 Aset of split quaternions is noncommutative and containszero divisors nilpotent elements and nontrivial idempotents(see [3 4]) Previous studies have examined the geometricand physical applications of split quaternions which arerequired in solving split quaternionic equations (see [5 6])Inoguchi [7] reformulated the Gauss-Codazzi equations informs consistent with the theory of integrable systems in theMinkowski 3-space for split quaternion numbers
A dual quaternion can be represented in a form reflect-ing an ordinary quaternion and a dual symbol Because
dual-quaternion algebra is constructed from real eight-dimensional vector spaces and an ordered pair of quater-nions dual quaternions are used in computer vision appli-cations Kenwright [8] provided the characteristics of dualquaternions and Pennestrı and Stefanelli [9] examined someproperties by using dual quaternions Son [10 11] offeredan extension problem for solutions of partial differentialequations and generalized solutions for the Riesz system Byusing properties of Hamilton operators Kula and Yayli [4]defined dual split quaternions and gave some properties ofthe screw motion in the Minkowski 3-space showing thatHhas a rotation with unit split quaternions in H and a scalarproduct that allows it to be identifiedwith the semi-Euclideanspace for split quaternion numbers
It was shown (see [12 13]) that any complex-valued har-monic function 119891
1in a pseudoconvex domain 119863 of C2 times C2
C being the set of complex numbers has a conjugate function1198912in 119863 such that the quaternion-valued function 119891
1+ 1198912119895 is
hyperholomorphic in119863 and gave a regeneration theorem in aquaternion analysis in view of complex and Clifford analysisIn addition we [14 15] provided a new expression of thequaternionic basis and a regular function on reduced quater-nions by associating hypercomplex numbers 119890
1and 119890
2 We
[16] investigated the existence of hyperconjugate harmonicfunctions of an octonion number system and we [17 18]obtained some regular functions with values in dual quater-nions and researched an extension problem for properties
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 369430 8 pageshttpdxdoiorg1011552014369430
2 Abstract and Applied Analysis
of regular functionswith values in dual quaternions and someapplications for such problems
This paper provides a regular function and some prop-erties of differential operators in dual split quaternions Inadditionwe research some equivalent conditions forCauchy-Riemann systems and expressions of power series in dual splitquaternions from the definition of dual split regular on anopen set Ω sub C2 times C2
2 Preliminaries
A dual number 119860 has the form 119886 + 120576119887 where 119886 and 119887 are realnumbers and 120576 is a dual symbol subject to the rules
120576 = 0 0120576 = 1205760 = 0 1120576 = 1205761 = 120576 1205762
= 0 (3)
and a split quaternion 119902 isin S is an expression of the form
119902 = 1199090+ 11990911198901+ 11990921198902+ 11990931198903 (4)
where 119909119898
isin R (119898 = 0 1 2 3) and 119890119903(119903 = 1 2 3) are split
quaternionic units satisfying noncommutativemultiplicationrules (for split quaternions see [1])
1198902
1= minus1 119890
2
2= 1198902
3= 1
11989011198902= minus11989021198901= 1198903 119890
21198903= minus11989031198902= minus1198901
11989031198901= minus11989011198903= 1198902
(5)
Similarly a dual split quaternion 119911 can be written as
D (S) = 119911 | 119911 = 1199010+ 1205761199011 119901119903isin S 119903 = 0 1 (6)
which has elements of the following form
119911 = (1199090+ 11990911198901) + (119909
2+ 11990931198901) 1198902
+ 120576 (1199100+ 11991011198901) + (119910
2+ 11991031198901) 1198902
= (1199110+ 11991111198902) + 120576 (119911
2+ 11991131198902)
= 1199010+ 1205761199011
(7)
where 1199010= 1199110+ 11991111198902and 119901
1= 1199112+ 11991131198902are split quaternion
components 1199110= 1199090+ 11990911198901 1199111= 1199092+ 11990931198901 1199112= 1199100+ 11991011198901
and 1199113= 1199102+ 11991031198901are usual complex numbers and 119909
119898 119910119898
isin
R (119898 = 0 1 2 3) The multiplication of split quaternionicunits with a dual symbol is commutative 120576119890
119903= 119890119903120576 (119903 =
1 2 3) However by properties of split quaternionic unit
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 0 1)
119911119896119890119903= 119890119903119911119896
(119896 = 0 1 2 3 119903 = 2 3)
119890119903119901119896
= 119901119896119890119903 119890119903119901119896= 119901(119896119903)
119890119903
(119903 = 1 2 3 119896 = 0 1)
(8)
where
119901(01)
= 1199110minus 11991111198902= 1199090+ 11990911198901minus 11990921198902minus 11990931198903
119901(02)
= 1199110+ 11991111198902= 1199090minus 11990911198901+ 11990921198902minus 11990931198903
119901(03)
= 1199110minus 11991111198902= 1199090minus 11990911198901minus 11990921198902+ 11990931198903
119901(11)
= 1199112minus 11991131198902= 1199100+ 11991011198901minus 11991021198902minus 11991031198903
119901(12)
= 1199112+ 11991131198902= 1199100minus 11991011198901+ 11991021198902minus 11991031198903
119901(13)
= 1199112minus 11991131198902= 1199100minus 11991011198901minus 11991021198902+ 11991031198903
(9)
with 1199110
= 1199090minus 11990911198901 1199111= 1199092minus 11990931198901 1199112
= 1199100minus 11991011198901 and
1199113= 1199102minus 11991031198901 For instance
11989021199010= 1198902(1199090+ 11990911198901+ 11990921198902+ 11990931198903)
= (1199090minus 11990911198901+ 11990921198902minus 11990931198903) 1198902= 119901(02)
1198902
11989011199011= 1198901(1199100+ 11991011198901+ 11991021198902+ 11991031198903)
= (1199100+ 11991011198901minus 11991021198902minus 11991031198903) 1198901= 119901(11)
1198901
(10)
Because of the properties of the eight-unit equality the addi-tion and subtraction of dual split quaternions are governedby the rules of ordinary algebra Here the symbol 119901
(119896119903)is used
by just enumerating 119903 and 119896 not 119903 times 119896 For example119901(22)
= 1199014and 119901
22= 1199014
For any two elements 119911 = 1199010+ 1205761199011and 119908 = 119902
0+ 1205761199021
of D(S) where 1199020= sum3
119903=0119904119903119890119903and 119902
1= sum3
119903=0119905119903119890119903are split
quaternion components and 119904119903 119905119903isin R (119903 = 0 1 2 3) their
noncommutative product is given by
119911119908 = (1199010+ 1205761199011) (1199020+ 1205761199021) = 11990101199020+ 120576 (119901
01199021+ 11990111199020)
(11)
The conjugation 119911lowast of 119911 and the corresponding modulus 119911119911lowast
inD(S) are defined by
119911lowast
= 119901lowast
0+ 120576119901lowast
1
119911119911lowast
= 119911lowast
119911 = 1199010119901lowast
0+ 120576 (119901
0119901lowast
1+ 1199011119901lowast
0)
= (11991101199110minus 11991111199111) + 2120576 (119911
01199112minus 11991111199113)
=
1
sum
119903=0
(1199092
119903minus 1199092
119903+2) + 120576 (119909
119903119910119903minus 119909119903+2
119910119903+2
)
(12)
where 119901lowast0= 1199110minus 11991111198902and 119901
lowast
1= 1199112minus 11991131198902
Lemma 1 For all 119911 isin D(S) and 119899 isin N = 1 2 3 wehave
119911119899
= 119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0 (13)
Abstract and Applied Analysis 3
Proof If 119899 = 1 then (13) is trivial Now suppose that thisholds for some 119899 isin N Then as desired
119911119899+1
= 119911119911119899
= 119911(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119901119899+1
0+ 120576
119899
sum
119896=1
119901119899minus119896+1
01199011119901119896minus1
0+ 1205761199011119901119899
0
= 119901119899+1
0+ 120576
119899+1
sum
119896=1
119901119899+1minus119896
01199011119901119896minus1
0
(14)
By the principle of mathematical induction (13) holds for all119899 isin N
Let Ω be an open subset of C2 times C2 Then the function119891 Ω rarr D(S) can be expressed as
119891 (119911) = 119891 (1199010 1199011) = 1198910(1199010 1199011) + 1205761198911(1199010 1199011) (15)
where the component functions 119891119903 Ω rarr S (119903 = 0 1) are
split quaternionic-valued functions The component func-tions 119891
119903(119903 = 0 1) are
1198910(1199010 1199011) = 1198910(1199110 1199111 1199112 1199113)
= 1198920(1199110 1199111 1199112 1199113) + 1198921(1199110 1199111 1199112 1199113) 1198902
1198911(1199010 1199011) = 1198911(1199110 1199111 1199112 1199113)
= 1198922(1199110 1199111 1199112 1199113) + 1198923(1199110 1199111 1199112 1199113) 1198902
(16)
where 119892119896
= 1199062119896
+ 1199062119896+1
1198901(119896 = 0 1) and 119892
119896= V2119896minus4
+
V2119896minus3
1198901(119896 = 2 3) are complex-valued functions and 119906
119903and
V119903(119903 = 0 1 2 3) are real-valued functionsNow we let differential operators 119863
1and 119863
2be defined
onD(S) as
1198631= 119863(11)
+ 120576119863(12)
1198632= 119863(21)
+ 120576119863(22)
(17)
Then the conjugate operators119863lowast1and119863
lowast
2are
119863lowast
1= 119863lowast
(11)+ 120576119863lowast
(12) 119863
lowast
2= 119863lowast
(21)+ 120576119863lowast
(22) (18)
where
119863(11)
=
120597
1205971199110
+
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
119863(12)
=
120597
1205971199112
+
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
120597
1205971199102
1198902+
120597
1205971199103
1198903)
119863(21)
=
120597
1205971199110
+
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
1
2
120597
1205971199092
1198902minus
1
2
120597
1205971199093
1198903)
119863(22)
=
120597
1205971199112
+
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
minus
120597
1205971199101
1198901+
1
2
120597
1205971199102
1198902minus
1
2
120597
1205971199103
1198903)
(19)
119863lowast
(11)=
120597
1205971199110
minus
120597
1205971199111
1198902=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
120597
1205971199092
1198902minus
120597
1205971199093
1198903)
119863lowast
(12)=
120597
1205971199112
minus
120597
1205971199113
1198902=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
120597
1205971199102
1198902minus
120597
1205971199103
1198903)
119863lowast
(21)=
120597
1205971199110
minus
1
2
120597
1205971199111
1198902
=
1
2
(
120597
1205971199090
+
120597
1205971199091
1198901minus
1
2
120597
1205971199092
1198902+
1
2
120597
1205971199093
1198903)
119863lowast
(22)=
120597
1205971199112
minus
1
2
120597
1205971199113
1198902
=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
(20)
act on D(S) These operators are called correspondingCauchy-Riemann operators in D(S) where 120597120597119911
119903and
120597120597119911119903(119903 = 0 1 2 3) are usual differential operators used in
the complex analysis
Remark 2 From the definition of differential operators onD(S)
119863119903119891 = (119863
(1199031)+ 120576119863(1199032)
) (1198910+ 1205761198911)
= 119863(1199031)
1198910+ 120576 (119863
(1199031)1198911+ 119863(1199032)
1198910)
119863lowast
119903119891 = (119863
lowast
(1199031)+ 120576119863lowast
(1199032)) (1198910+ 1205761198911)
= 119863lowast
(1199031)1198910+ 120576 (119863
lowast
(1199031)1198911+ 119863lowast
(1199032)1198910)
(21)
where 119903 = 1 2
Definition 3 LetΩ be an open set inC2 timesC2 A function 119891 =
1198910+ 1205761198911is called an 119871
119903(resp 119877
119903)-regular function (119903 = 1 2)
onΩ if the following two conditions are satisfied
(i) 119891119896(119896 = 0 1) are continuously differential functions
onΩ and
(ii) 119863lowast119903119891(119911) = 0 (resp 119891(119911)119863lowast
119903= 0) onΩ (119903 = 1 2)
In particular the equation 119863lowast
1119891(119911) = 0 of Definition 3 is
equivalent to
119863lowast
(11)1198910= 0 119863
lowast
(12)1198910+ 119863lowast
(11)1198911= 0 (22)
4 Abstract and Applied Analysis
In addition
1205971198920
1205971199110
minus
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1205971198923
1205971199111
minus
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1205971198922
1205971199111
minus
1205971198920
1205971199113
= 0
(23)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1205971199062
1205971199092
minus
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1205971199062
1205971199093
+
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1205971199060
1205971199092
minus
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
minus
1205971199060
1205971199093
+
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
120597V2
1205971199092
minus
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1205971199062
1205971199103
+
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1205971199060
1205971199102
minus
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
120597V0
1205971199092
minus
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
minus
1205971199060
1205971199103
+
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
minus
120597V0
1205971199093
+
120597V1
1205971199092
= 0
(24)
The above systems (23) and (24) are corresponding Cauchy-Riemann systems inD(S) Similarly the equation119863
lowast
2119891(119911) =
0 of Definition 3 is equivalent to
119863lowast
(21)1198910= 0 119863
lowast
(22)1198910+ 119863lowast
(21)1198911= 0 (25)
Then
1205971198920
1205971199110
minus
1
2
1205971198921
1205971199111
= 0
1205971198921
1205971199110
minus
1
2
1205971198920
1205971199111
= 0
1205971198922
1205971199110
+
1205971198920
1205971199112
minus
1
2
1205971198923
1205971199111
minus
1
2
1205971198921
1205971199113
= 0
1205971198923
1205971199110
+
1205971198921
1205971199112
minus
1
2
1205971198922
1205971199111
minus
1
2
1205971198920
1205971199113
= 0
(26)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
+
1
2
1205971199062
1205971199093
+
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
minus
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
+
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
+
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
+
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1
2
1205971199062
1205971199103
+
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
+
1
2
120597V2
1205971199093
+
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
minus
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
minus
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
+
1205971199062
1205971199101
+
1
2
1205971199060
1205971199103
+
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
+
120597V2
1205971199091
+
1
2
120597V0
1205971199093
+
1
2
120597V1
1205971199092
= 0
(27)
The above systems (26) and (27) are corresponding Cauchy-Riemann systems inD(S)
On the other hand the equation 119891(119911)119863lowast
1= 0 of
Definition 3 is equivalent to
1198910119863lowast
(11)= 0 119891
0119863lowast
(12)= minus1198911119863lowast
(11) (28)
Abstract and Applied Analysis 5
Then
1198920
120597
1205971199110
= 1198921
120597
1205971199111
1198921
120597
1205971199110
= 1198920
120597
1205971199111
1198920
120597
1205971199112
minus 1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+ 1198923
120597
1205971199111
1198921
120597
1205971199112
minus 1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+ 1198922
120597
1205971199111
(29)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(30)
Concretely the following system is obtained
1205971199060
1205971199090
minus
1205971199061
1205971199091
=
1205971199062
1205971199092
+
1205971199063
1205971199093
1205971199061
1205971199090
+
1205971199060
1205971199091
= minus
1205971199062
1205971199093
+
1205971199063
1205971199092
1205971199062
1205971199090
+
1205971199063
1205971199091
=
1205971199060
1205971199092
minus
1205971199061
1205971199093
1205971199063
1205971199090
minus
1205971199062
1205971199091
=
1205971199060
1205971199093
+
1205971199061
1205971199092
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1205971199062
1205971199102
minus
1205971199063
1205971199103
= minus
120597V0
1205971199090
+
120597V1
1205971199091
+
120597V2
1205971199092
+
120597V3
1205971199093
1205971199061
1205971199100
+
1205971199060
1205971199101
+
1205971199062
1205971199103
minus
1205971199063
1205971199102
= minus
120597V1
1205971199090
minus
120597V0
1205971199091
minus
120597V2
1205971199093
+
120597V3
1205971199092
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1205971199060
1205971199102
+
1205971199061
1205971199103
= minus
120597V2
1205971199090
minus
120597V3
1205971199091
+
120597V0
1205971199092
minus
120597V1
1205971199093
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1205971199060
1205971199103
minus
1205971199061
1205971199102
= minus
120597V3
1205971199090
+
120597V2
1205971199091
+
120597V0
1205971199093
+
120597V1
1205971199092
(31)
Similarly the equation 119891(119911)119863lowast
2= 0 of Definition 3 is
equivalent to
1198910119863lowast
(21)= 0 119891
0119863lowast
(22)= minus1198911119863lowast
(21) (32)
Then
1198920
120597
1205971199110
=
1
2
1198921
120597
1205971199111
1198921
120597
1205971199110
=
1
2
1198920
120597
1205971199111
1198920
120597
1205971199112
minus
1
2
1198921
120597
1205971199113
= minus 1198922
120597
1205971199110
+
1
2
1198923
120597
1205971199111
1198921
120597
1205971199112
minus
1
2
1198920
120597
1205971199113
= minus 1198923
120597
1205971199110
+
1
2
1198922
120597
1205971199111
(33)
where
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
119892119896
120597
120597119911119898
=
120597119892119896
120597119911119898
(119896119898 = 0 1 2 3)
(34)
Concretely the system is obtained as follows
1205971199060
1205971199090
minus
1205971199061
1205971199091
minus
1
2
1205971199062
1205971199092
+
1
2
1205971199063
1205971199093
= 0
1205971199061
1205971199090
+
1205971199060
1205971199091
minus
1
2
1205971199062
1205971199093
minus
1
2
1205971199063
1205971199092
= 0
1205971199062
1205971199090
+
1205971199063
1205971199091
minus
1
2
1205971199060
1205971199092
minus
1
2
1205971199061
1205971199093
= 0
1205971199063
1205971199090
minus
1205971199062
1205971199091
+
1
2
1205971199060
1205971199093
minus
1
2
1205971199061
1205971199092
= 0
1205971199060
1205971199100
minus
1205971199061
1205971199101
minus
1
2
1205971199062
1205971199102
+
1
2
1205971199063
1205971199103
+
120597V0
1205971199090
minus
120597V1
1205971199091
minus
1
2
120597V2
1205971199092
+
1
2
120597V3
1205971199093
= 0
1205971199061
1205971199100
+
1205971199060
1205971199101
minus
1
2
1205971199062
1205971199103
minus
1
2
1205971199063
1205971199102
+
120597V1
1205971199090
+
120597V0
1205971199091
minus
1
2
120597V2
1205971199093
minus
1
2
120597V3
1205971199092
= 0
1205971199062
1205971199100
+
1205971199063
1205971199101
minus
1
2
1205971199060
1205971199102
+
1
2
1205971199061
1205971199103
+
120597V2
1205971199090
+
120597V3
1205971199091
minus
1
2
120597V0
1205971199092
+
1
2
120597V1
1205971199093
= 0
1205971199063
1205971199100
minus
1205971199062
1205971199101
minus
1
2
1205971199060
1205971199103
minus
1
2
1205971199061
1205971199102
+
120597V3
1205971199090
minus
120597V2
1205971199091
minus
1
2
120597V0
1205971199093
minus
1
2
120597V1
1205971199092
= 0
(35)
From the systems (24) (27) (31) and (35) the equations119863lowast
119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 = 1 2) are different
Therefore the equations 119863lowast119903119891(119911) = 0 and 119891(119911)119863
lowast
119903= 0 (119903 =
1 2) should be distinguished as 119871119903-regular functions (119903 =
1 2) and 119877119903-regular functions (119903 = 1 2) on Ω respectively
Now the properties of the 119871119903-regular function (119903 = 1 2) with
values inD(S) are considered
3 Properties of 119871119903-Regular Functions (119903 = 1 2)
with Values in D(S)
We consider properties of a 119871119903-regular functions (119903 = 1 2)
with values inD(S)
Theorem 4 Let Ω be an open set in C2 times C2 and let 119891 =
1198910+ 1205761198911= (1198920+11989211198902)+ 120576(119892
2+11989231198902) be an 119871
1-regular function
defined on Ω Then
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (36)
6 Abstract and Applied Analysis
Proof By the system (23) we have
1198631119891 = 119863
(11)1198910+ 120576 (119863
(12)1198910+ 119863(11)
1198911)
= (
1205971198920
1205971199110
+
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+ 2
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+ 2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+ 2
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+ 2
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+ 2
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+ 2
1205971198922
1205971199111
) 1198902
= 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(37)
Therefore we obtain
1198631119891 = 2(
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (38)
Theorem5 LetΩ be an open set inC2timesC2 and119891 = 1198910+1205761198911=
(1198920+11989211198902) + 120576(119892
2+11989231198902) be an 119871
2-regular function defined on
Ω Then
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (39)
Proof By the system (26) we have
1198632119891 = 119863
(21)1198910+ 120576 (119863
(22)1198910+ 119863(21)
1198911)
= (
1205971198920
1205971199110
+
1
2
1205971198921
1205971199111
) + (
1205971198921
1205971199110
+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971198920
1205971199110
+
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1
2
1205971198921
1205971199111
)
+ (
1205971198921
1205971199110
+
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1
2
1205971198920
1205971199111
) 1198902
+ 120576(
1205971198920
1205971199112
+
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1
2
1205971198921
1205971199113
+
1205971198922
1205971199110
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1
2
1205971198923
1205971199111
)
+ 120576(
1205971198921
1205971199112
+
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1
2
1205971198920
1205971199113
+
1205971198923
1205971199110
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1
2
1205971198922
1205971199111
) 1198902
= (
1205971199061
1205971199091
minus
1205971199060
1205971199091
1198901+
1205971198921
1205971199111
)
+ (
1205971199063
1205971199091
minus
1205971199062
1205971199091
1198901+
1205971198920
1205971199111
) 1198902
+ 120576(
1205971199061
1205971199101
minus
1205971199060
1205971199101
1198901+
1205971198921
1205971199113
+
120597V1
1205971199091
minus
120597V0
1205971199091
1198901+
1205971198923
1205971199111
)
+ 120576(
1205971199063
1205971199101
minus
1205971199062
1205971199101
1198901+
1205971198920
1205971199113
+
120597V3
1205971199091
minus
120597V2
1205971199091
1198901+
1205971198922
1205971199111
) 1198902
= (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891
(40)
Therefore we obtain the following equation
1198632119891 = (
120597
1205971199111
+ 120576
120597
1205971199113
) 1198902minus (
120597
1205971199091
+ 120576
120597
1205971199101
) 1198901119891 (41)
Abstract and Applied Analysis 7
Proposition 6 From properties of differential operators thefollowing equations are obtained
119863(1119903)
119901119903minus1
= 2 119863(2119903)
119901119903minus1
= 1
119863lowast
(1119903)119901119903minus1
= minus1 119863lowast
(2119903)119901119903minus1
= 0
119863lowast
(1119903)119901lowast
119903minus1= 2 119863
lowast
(2119903)119901lowast
119903minus1= 1
119863(1199031)
1199011= 119863lowast
(1199031)1199011= 119863(1199031)
119901lowast
1= 119863lowast
(1199031)119901lowast
1
= 119863(1199032)
1199010= 119863lowast
(1199032)1199010= 119863(1199032)
119901lowast
0
= 119863lowast
(1199032)119901lowast
0= 0 (119903 = 1 2)
(42)
Proof By properties of the power of dual split quaternionsand derivatives on D(S) the following derivatives areobtained
119863(11)
1199010=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090+ 11990911198901+ 11990921198902+ 11990931198903) = 2
119863lowast
(22)1199011=
1
2
(
120597
1205971199100
+
120597
1205971199101
1198901minus
1
2
120597
1205971199102
1198902+
1
2
120597
1205971199103
1198903)
times (1199100+ 11991011198901+ 11991021198902+ 11991031198903) = 0
119863(11)
119901lowast
0=
1
2
(
120597
1205971199090
minus
120597
1205971199091
1198901+
120597
1205971199092
1198902+
120597
1205971199093
1198903)
times (1199090minus 11990911198901minus 11990921198902minus 11990931198903) = minus1
(43)
The other equations are calculated using a similar methodand the above equations are obtained
Theorem 7 LetΩ be an open set in C2 timesC2 and let 119891(119911) be afunction on Ω with values inD(S) Then the power 119911119899 of 119911 inD(S) is not an 119871
1-regular function but an 119871
2-regular function
on Ω where 119899 isin N
Proof From the definition of the 119871119903-regular function (119903 =
1 2) on Ω and Proposition 6 we may consider whether thepower 119911119899 of 119911 in D(S) satisfies the equation 119863
lowast
119903119911119899
= 0 (119903 =
1 2) Since119863lowast(11)
1199010= 2
119863lowast
1119911119899
= (119863lowast
(11)+ 120576119863lowast
(12))(119901119899
0+ 120576
119899
sum
119896=1
119901119899minus119896
01199011119901119896minus1
0)
= 119863lowast
(11)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(11)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(12)119901119899
0) = 0
(44)
Hence the power 119911119899 of 119911 is not 1198711-regular onΩ On the other
hand from the equations in Proposition 6 we have119863lowast(21)
1199010=
0119863lowast(21)
1199011= 0 and119863
lowast
(22)1199010= 0 Then
119863lowast
2119911119899
= 119863lowast
(21)119901119899
0+ 120576(
119899
sum
119896=1
119863lowast
(21)119901119899minus119896
01199011119901119896minus1
0+ 119863lowast
(22)119901119899
0) = 0
(45)
Therefore by the definition of the 119871119903-regular function (119903 =
1 2) onΩ a power 119911119899 of 119911 is 1198712-regular onΩ
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The second author was supported by the Basic ScienceResearch Program through the National Research Founda-tion of Korea (NRF) funded by the Ministry of Science ICTand Future Planning (2013R1A1A2008978)
References
[1] I L Kantor andA S SolodovnikovHypercomplex Numbers AnElementary Introduction to Algebras Springer New York NYUSA 1989
[2] J Cockle ldquoOn systems of algebra involving more than oneimaginaryrdquo PhilosophicalMagazine III vol 35 no 238 pp 434ndash435 1849
[3] M Ozdemir and A A Ergin ldquoRotations with unit timelikequaternions in Minkowski 3-spacerdquo Journal of Geometry andPhysics vol 56 no 2 pp 322ndash336 2006
[4] L Kula and Y Yayli ldquoSplit quaternions and rotations in semiEuclidean spaceE4
2rdquo Journal of the KoreanMathematical Society
vol 44 no 6 pp 1313ndash1327 2007[5] D C Brody and E-M Graefe ldquoOn complexified mechanics
and coquaternionsrdquo Journal of Physics A Mathematical andTheoretical vol 44 no 7 article 072001 2011
[6] I Frenkel and M Libine ldquoSplit quaternionic analysis andseparation of the series for SL(2R) and SL(2C)SL(2R)rdquoAdvances in Mathematics vol 228 no 2 pp 678ndash763 2011
[7] J-i Inoguchi ldquoTimelike surfaces of constant mean curvature inMinkowski 3-spacerdquo Tokyo Journal of Mathematics vol 21 no1 pp 141ndash152 1998
[8] B Kenwright ldquoA beginners guide to dual-quaternions whatthey are how they work and how to use them for 3D characterhierarchiesrdquo in Proceedings of the 20th International Conferenceson Computer Graphics Visualization and Computer Vision pp1ndash10 2012
[9] E Pennestrı and R Stefanelli ldquoLinear algebra and numericalalgorithms using dual numbersrdquo Multibody System Dynamicsvol 18 no 3 pp 323ndash344 2007
[10] L H Son ldquoAn extension problem for solutions of partialdifferential equations in R119899rdquo Complex Variables Theory andApplication vol 15 no 2 pp 87ndash92 1990
[11] L H Son ldquoExtension problem for functions with values in aClifford algebrardquo Archiv der Mathematik vol 55 no 2 pp 146ndash150 1990
[12] J Kajiwara X D Li and K H Shon ldquoRegeneration in complexquaternion and Clifford analysisrdquo in International Colloquiumon Finite or Infinite DimensionalComplex Analysis and ItsApplications vol 2 of Advances in Complex Analysis and ItsApplications no 9 pp 287ndash298 Kluwer Academic PublishersHanoi Vietnam 2004
[13] J Kajiwara X D Li and K H Shon ldquoFunction spaces incomplex and Clifford analysisrdquo in International Colloquium
8 Abstract and Applied Analysis
on Finite Or Infinite Dimensional Complex Analysis and ItsApplications vol 14 of Inhomogeneous Cauchy Riemann Systemof Quaternion andCliffordAnalysis in Ellipsoid pp 127ndash155HueUniversity Hue Vietnam 2006
[14] J E Kim S J Lim and K H Shon ldquoRegular functions withvalues in ternary number system on the complex Cliffordanalysisrdquo Abstract and Applied Analysis vol 2013 Article ID136120 7 pages 2013
[15] J E Kim S J Lim and K H Shon ldquoRegularity of functions onthe reduced quaternion field in Clifford analysisrdquo Abstract andApplied Analysis vol 2014 Article ID 654798 8 pages 2014
[16] S J Lim and K H Shon ldquoHyperholomorphic fucntions andhyperconjugate harmonic functions of octonion variablesrdquoJournal of Inequalities andApplications vol 2013 article 77 2013
[17] S J Lim and K H Shon ldquoDual quaternion functions and itsapplicationsrdquo Journal of Applied Mathematics vol 2013 ArticleID 583813 6 pages 2013
[18] J E Kim S J Lim and K H Shon ldquoTaylor series of functionswith values in dual quaternionrdquo Journal of the Korean Society ofMathematical Education BmdashThePure and AppliedMathematicsvol 20 no 4 pp 251ndash258 2013
Research ArticleUnicity of Meromorphic Functions Sharing Sets withTheir Linear Difference Polynomials
Sheng Li and BaoQin Chen
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to BaoQin Chen chenbaoqin chbq126com
Received 23 January 2014 Accepted 20 March 2014 Published 13 April 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 S Li and B Chen This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We mainly investigate the unicity of meromorphic functions sharing two or three sets with their linear difference polynomials andprove some results
1 Introduction and Main Results
In this paper we assume the reader is familiar with the fun-damental results and the basic notations of the Nevanlinnatheory of meromorphic functions (see eg [1ndash3]) Let 119891(119911)be meromorphic in the whole plane We use the notation120588(119891) to denote the order of growth of the meromorphicfunction 119891(119911) In addition we denote by 119878(119903 119891) any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin outside of apossible exceptional set of finite logarithmic measure We saythat a meromorphic function 119886(119911) is a small function of 119891(119911)provided that 119879(119903 119886) = 119878(119903 119891) Let 119878(119891) be the set of all smallfunctions of 119891(119911)
For a set 119878 sub 119878(119891) we define the following
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 counting multiplicities
119864119891(119878) = ⋃
119886isin119878
119911 | 119891 (119911) minus 119886 (119911) = 0 ignoring multiplicities
(1)
Let 119891 and 119892 be meromorphic functions If 119864119891(119878) = 119864
119892(119878)
and 119864119891(119878) = 119864
119892(119878) respectively then we say that 119891 and 119892
share a set 119878 CM and IM respectivelyFurthermore let 119888 be a nonzero complex constant We
define the shift of 119891(119911) by 119891(119911 + 119888) and define the differenceoperators of 119891(119911) by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911)
Δ119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911)) 119899 isin N 119899 ge 2
(2)
Theunicity theory ofmeromorphic functions sharing setsis an important topic of the uniqueness theory First of all werecall the following theorem given by Li and Yang in [4]
Theorem A (see [4]) Let 119898 ge 2 and let 119899 gt 2119898 + 6 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0Then for any two nonconstant meromorphic functions 119891 and119892 the conditions 119864
119891(119878) = 119864
119892(119878) and 119864
119891(infin) = 119864
119892(infin)
imply 119891 = 119892
Yi and Lin considered the case 119898 = 1 with the conditionthat two meromorphic functions share three sets and got theresult as follows
Theorem B (see [5]) Let 1198781= 120596 120596
119899
+ 119886120596119899minus1
+ 119887 = 01198782= 0 and 119878
3= infin where 119886 119887 are nonzero constants such
that 120596119899 + 119886120596119899minus1 + 119887 = 0 has no repeated root and 119899(ge 4) is aninteger If for two nonconstant meromorphic functions 119891 and119892 119864119891(119878119895) = 119864
119892(119878119895) for 119895 = 1 2 3 and Θ(infin119891) gt 0 then
119891 equiv 119892
Recently a number of papers have focused on differenceanalogues of the Nevanlinna theory (see eg [6ndash9]) Inparticular there has been an increasing interest in studyingthe uniqueness problems related to meromorphic functionsand their shifts or their difference operators (see eg [10ndash16])
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 894968 7 pageshttpdxdoiorg1011552014894968
2 Abstract and Applied Analysis
In 2010 Zhang considered a meromorphic function 119891(119911)sharing sets with its shift 119891(119911 + 119888) and proved the followingresult
TheoremC (see [16]) Let119898 ge 2 and let 119899 ge 2119898+4with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596119899 + 119886120596119899minus119898 + 119887 = 0 Suppose that119891(119911) is a nonconstant meromorphic function of finite orderThen 119864
119891(119911)(119878) = 119864
119891(119911+119888)(119878) and 119864
119891(119911)(infin) = 119864
119891(119911+119888)(infin)
imply 119891(119911) equiv 119891(119911 + 119888)
For an analogue result in difference operator B Chen andZ Chen proved the following theorem in [10]
Theorem D (see [10]) Let 119898 ge 2 and let 119899 ge 2119898 + 4 with119899 and 119899 minus 119898 having no common factors Let 119886 and 119887 be twononzero constants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0has nomultiple roots Let 119878 = 120596 | 120596119899+119886120596119899minus119898+119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying 119864
119891(119911)(119878) = 119864
Δ119888119891(119878) and 119864
119891(119911)(infin) = 119864
Δ119888119891(infin) If
119873(119903
1
Δ119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (3)
then Δ119888119891 equiv 119891(119911)
It is natural to ask what happens if the shift 119891(119911 + 119888) ordifference operatorΔ
119888119891(119911) is replaced by a general expression
of 119891(119911) such as a linear difference polynomial of 119891(119911)Here a linear difference polynomial of 119891(119911) is an expres-
sion of the form
119871 (119911 119891) = 119887119896(119911) 119891 (119911 + 119888
119896) + sdot sdot sdot + 119887
0(119911) 119891 (119911 + 119888
0) (4)
where 119887119896(119911) equiv 0 119887
0(119911) 119887
119896(119911) are small functions of
119891(119911) 1198880 119888
119896are complex constants and 119896 is a nonnegative
integerIn this paper our aim is to investigate the uniqueness
problems of linear difference polynomials of 119891(119911) In partic-ular we primarily consider the linear difference polynomial119871(119911 119891) which satisfies one of the following conditions
(i) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1
(ii) 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0
119873(119903
1
119871 (119911 119891)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(5)
Corresponding to the above question we obtain thefollowing results
Theorem 1 Let 119898 ge 2 and let 119899 ge 2119898 + 4 with 119899 and119899 minus 119898 having no common factors Let 119886 and 119887 be two nonzeroconstants such that the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has nomultiple roots Let 119878 = 120596 | 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 Supposethat119891(119911) is a nonconstantmeromorphic function of finite orderand 119871(119911 119891) is of the form (4) satisfying the condition in (5)If 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) and 119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) then119871(119911 119891) equiv 119891(119911)
Corollary 2 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersatisfying the following
119873(119903
1
Δ119896
119888119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (6)
If 119864119891(119911)
(119878) = 119864Δ119896
119888119891(119911)
(119878) and 119864119891(119911)
(infin) = 119864Δ119896
119888119891(119911)
(infin) thenΔ119896
119888119891 equiv 119891(119911)
With an additional restriction on the order of growth of119891(119911) we prove the following fact
Theorem 3 Let 119899119898 and 119878 be given as inTheorem 1 Supposethat119891(119911) is a nonconstantmeromorphic function of finite ordersuch that 120588(119891) notin N If 119864
119891(119911)(119878) = 119864
119871(119911119891)(119878) and 119864
119891(119911)(infin) =
119864119871(119911119891)
(infin) then 119871(119911 119891) equiv 119891(119911)
Remark 4 Note that in Theorem 3 we do not assume thatthe linear polynomial 119871(119911 119891) satisfies the condition in (5)In fact since 120588(119891) notin N by (19) we can easily get 120588(119890ℎ(119911)) =deg(ℎ(119911)) lt 120588(119891) which implies 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Thenusing a similar method as in the proof of Theorem 1 we cancomplete the proof of Theorem 3
Now we may ask what happens if the condition119898 ge 2 inTheorem 1 is replaced by a weaker condition containing thecase 119898 = 1 or even 119898 = 0 By considering three sets we getthe following theorem
Theorem 5 Let 119899119898 be nonnegative integers such that 119899 gt 119898Let 119886 and 119887 be nonzero constants such that 120596119899 + 119886120596119899minus119898 + 119887 = 0has no multiple roots Let 119878
1= 120596 120596
119899
+ 119886120596119899minus119898
+ 119887 = 0 =1198782= infin and 119878
3= 0 Suppose that 119891(119911) is a nonconstant
meromorphic function of finite order 119871(119911 119891) is of the form (4)satisfying the condition in (5) and 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for
119895 = 1 2 3 Then one has the following(i) If119898 = 0 then 119871(119911 119891) equiv 119905119891(119911) where 119905119899 = 1(ii) If 119899 and119898 are coprime then 119871(119911 119891) equiv 119891(119911)
Remark 6 Taking 119898 = 1 in Theorem 5 we can obtain ananalogue result of Theorem B related to linear differencepolynomials
Furthermore the following result is a corollary ofTheorem 5 related to difference operators
Corollary 7 Let 119899 119898 and 119878119895 119895 = 1 2 3 be given as in
Theorem 5 Suppose that 119891(119911) is a nonconstant meromorphicfunction of finite order satisfying
119873(119903
1
119891 (119911)
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891) (7)
and 119864119891(119911)
(119878119895) = 119864
Δ119896
119888119891(119878119895) for 119895 = 1 2 3 Then one has the
following
(i) If119898 = 0 then Δ119896119888119891 equiv 119905119891(119911) where 119905119899 = 1
(ii) If 119899 and119898 are coprime then Δ119896119888119891 equiv 119891(119911)
Finally we give some examples for our results
Abstract and Applied Analysis 3
Examples In the following let 119892(119911) be an entire functionwith period 1 such that 120588(119892) isin (1infin) N (see [17])
(1) For the case (i) of condition (5) let 1198911(119911) = 119890
21205871198941199111198912(119911) = 119892(119911)119890
2120587119894119911 1198913(119911) = 119890
2120587119894119911
119892(119911) and let119871(119911 119891
119895) = 2119891
119895(119911) minus 119891
119895(119911 + 1) Then for 119895 = 1 2 3
119871(119911 119891119895) = 119891
119895(119911) and the sum of the coefficients of
119871(119911 119891119895) is equal to 1These examples satisfyTheorems
1 and 5 but do not satisfy Theorem D
(2) For the case (ii) of condition (5) let 119891(119911) = 119890119911 log 2119892(119911)and let 119871(119911 119891) = Δ119891(119911) = 119891(119911 + 1) minus 119891(119911) Then119871(119911 119891) = Δ119891(119911) = 119891(119911) the sum of the coefficientsof 119871(119911 119891) equals 0 and
119873(119903
1
Δ119891
) = 119873(119903
1
119891
) = 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(8)
This example satisfiesTheorems 1 and 5 and Corollar-ies 2 and 7
(3) For Theorem 3 let 119891(119911) = 119890119911 log 3
119892(119911) and let119871(119911 119891) = 119891(119911 + 1) minus 2119891(119911) Then 119871(119911 119891) = 119891(119911)
and the sum of the coefficients of 119871(119911 119891) equals minus1This example satisfies Theorem 3 but does not satisfyTheorem D andTheorems 1 and 5
2 Proof of Theorem 1
We need the following lemmas for the proof of Theorem 1The difference analogue of the logarithmic derivative
lemmawas given byHalburd-Korhonen [7] andChiang-Feng[6] independently We recall the following lemmas
Lemma 8 (see [7]) Let 119891(119911) be a nonconstant meromorphicfunction of finite order 119888 isin C and 120575 lt 1 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 + |119888| 119891)
119903120575
) (9)
for all 119903 outside of a possible exceptional set with finite logarith-mic measure
Lemma 9 (see [8]) Let 119888 isin C let 119899 isin N and let 119891(119911) bea meromorphic function of finite order Then for any smallperiodic function 119886(119911) isin 119878(119891) with period 119888 consider thefollowing
119898(119903
Δ119899
119888119891
119891 (119911) minus 119886 (119911)
) = 119878 (119903 119891) (10)
where the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Let 119891(119911) be a meromorphic function of finite orderNotice that if 119871(119911 119891) ( equiv 0) is of the form (4) such that
1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 then for any given complex con-
stant 119886 119871(119911 119886) = 0 This indicates that 119871(119911 119891) = 119871(119911 119891 minus 119886)
and hence
119898(119903
119871 (119911 119891)
119891 minus 119886
) = 119898(119903
119871 (119911 119891 minus 119886)
119891 minus 119886
)
le
119896
sum
119895=0
119898(119903
119887119895(119911) (119891 (119911 + 119888
119895) minus 119886)
119891 minus 119886
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(11)
With this one can easily prove Lemma 10 below by a similarreasoning as in the proof of the difference analogue of thesecond main theorem of the Nevanlinna theory in [8] byHalburd and Korhonen We omit those details
Lemma 10 Let 119888 isin C let 119891(119911) be a meromorphic function offinite order and let 119871(119911 119891) equiv 0 be of the form (4) such that1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 Let 119902 ge 2 and let 119886
1 119886
119902be distinct
complex constants Then
119898(119903 119891) +
119902
sum
119894=1
119898(119903
1
119891 minus 119886119894
)
le 2119879 (119903 119891) minus 119873lowast
(119903 119891) + 119878 (119903 119891)
(12)
where
119873lowast
(119903 119891) = 2119873 (119903 119891) minus 119873 (119903 119871 (119911 119891)) + 119873(119903
1
119871 (119911 119891)
)
(13)
and the exceptional set associated with 119878(119903 119891) is of at mostfinite logarithmic measure
Remark 11 If the linear difference polynomial 119871(119911 119891) isreplaced by
119871lowast
(119911 119891) = 119887119896(119911) 119891 (119911 + 119896119888)
+ sdot sdot sdot + 1198871(119911) 119891 (119911 + 119888) + 119887
0(119911) 119891 (119911)
(14)
Lemma 10 also holds even if the distinct complex constants1198861 119886
119902are replaced by 119886
1(119911) 119886
119902(119911) which are distinct
meromorphic periodic functions with period 119888 such that 119886119894isin
119878(119891) for all 119894 = 1 119902
The following is the standardValiron-Mohonrsquoko theorem(see Theorem 225 in the book of Laine [2])
Lemma 12 (see [2]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(15)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(16)
4 Abstract and Applied Analysis
The characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (17)
where 119889 = max119901 119902
Proof of Theorem 1 Since 119891(119911) and 119871(119911 119891) shareinfin CM wesee that 119871(119911 119891) equiv 0 and119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) Then byLemma 8 we have
119879 (119903 119871 (119911 119891)) = 119898 (119903 119871 (119911 119891)) + 119873 (119903 119871 (119911 119891))
le 119898(119903
119871 (119911 119891)
119891 (119911)
)
+ 119898 (119903 119891 (119911)) + 119873 (119903 119891 (119911))
le
119896
sum
119894=0
119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
+
119896
sum
119894=0
119898(119903 119887119894(119911)) + 119879 (119903 119891 (119911))
le 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(18)
Since 119864119891(119911)
(119878) = 119864119871(119911119891)
(119878) where 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots we know that (119871(119911 119891))119899+119886(119871(119911 119891))119899minus119898+119887 and119891(119911)119899+119886119891(119911)119899minus119898
+ 119887 share 0 CM Then from this and the condition119864119891(119911)
(infin) = 119864119871(119911119891)
(infin) there exists a polynomial ℎ(119911)such that
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
= 119890ℎ(119911)
(19)
Suppose that 119890ℎ(119911) equiv 1 Note that 119878 = 120596 | 120596119899
+ 119886120596119899minus119898
+
119887 = 0 and the equation 120596119899 + 119886120596119899minus119898 + 119887 = 0 has no multipleroots Let 120596
1 120596
119899denote all different roots of the equation
120596119899
+ 119886120596119899minus119898
+ 119887 = 0Next we prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) We know that
119871 (119911 119891) minus 120596119894= 119887119896(119911) (119891 (119911 + 119888
119896) minus 119891 (119911))
+ sdot sdot sdot + 1198870(119911) (119891 (119911 + 119888
0) minus 119891 (119911))
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891
+ (119887119896(119911) + sdot sdot sdot + 119887
0(119911)) 119891 (119911) minus 120596
119894
(20)
(i) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 1 we see that
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 + (119891 (119911) minus 120596119894)
(21)
Then we deduce from this (19) and Lemma 9 that
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
= 119898(119903
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
+ 119887
119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887
)
= 119898(119903
(119871 (119911 119891) minus 1205961) sdot sdot sdot (119871 (119911 119891) minus 120596
119899)
(119891 (119911) minus 1205961) sdot sdot sdot (119891 (119911) minus 120596
119899)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(22)
(ii) If 1198870(119911) + sdot sdot sdot + 119887
119896(119911) equiv 0 we have
119871 (119911 119891) minus 120596119894= 119887119896(119911) Δ119888119896
119891 + sdot sdot sdot + 1198870(119911) Δ1198880
119891 minus 120596119894 (23)
From this (19) and Lemma 9 we get
119879 (119903 119890ℎ(119911)
) = 119898 (119903 119890ℎ(119911)
)
le
119899
sum
119894=1
119898(119903
119871 (119911 119891) minus 120596119894
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
le
119899
sum
119894=1
119896
sum
119895=0
119898(119903
Δ119888119895
119891
119891 (119911) minus 120596119894
)
+
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
=
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
) + 119878 (119903 119891)
(24)
Applying Lemma 10 to 119891(119911) we get
119899
sum
119894=1
119898(119903
1
119891 (119911) minus 120596119894
)
le 2119879 (119903 119891 (119911)) minus 119898 (119903 119891 (119911)) minus 2119873 (119903 119891 (119911))
+ 119873 (119903 119871 (119911 119891)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
= 119879 (119903 119891 (119911)) minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891)
(25)
Abstract and Applied Analysis 5
Then the assumptions in (5) (24) and (25) yield thefollowing
119879 (119903 119890ℎ(119911)
) le 119879 (119903 119891 (119911))
minus 119873(119903
1
119871 (119911 119891)
) + 119878 (119903 119891) = 119878 (119903 119891)
(26)
To sum up we now prove that 119879(119903 119890ℎ(119911)) = 119878(119903 119891) Rewriting(19) we get
(119871 (119911 119891))119899minus119898
[(119871 (119911 119891))119898
+ 119886]
= [119891(119911)119899
+ 119886119891(119911)119899minus119898
+ 119887 minus 119887119890minusℎ(119911)
] 119890ℎ(119911)
(27)
Denote 119865(119911) = 119891(119911)119899
+ 119886119891(119911)119899minus119898 It follows from
Lemma 12 and119898 gt 0 that
119879 (119903 119865 (119911)) = 119899119879 (119903 119891 (119911)) + 119878 (119903 119891) (28)
Hence 119878(119903 119865) = 119878(119903 119891)By (18) and (27) and applying the second main theorem
for three small target functions we deduce the following
119879 (119903 119865 (119911))
le 119873 (119903 119865 (119911)) + 119873(119903
1
119865 (119911)
)
+ 119873(119903
1
119865 (119911) + 119887 minus 119887119890minus119901(119911)
) + 119878 (119903 119865)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891(119911)119899minus119898
[119891(119911)119898
+ 119886]
)
+ 119873(119903
1
(119871 (119911 119891))119899minus119898
) + 119873(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119873 (119903 119891 (119911)) + 119873(119903
1
119891 (119911)
) + 119873(119903
1
119891(119911)119898
+ 119886
)
+ 119873(119903
1
119871 (119911 119891)
) + 119879(119903
1
(119871 (119911 119891))119898
+ 119886
)
+ 119878 (119903 119891)
le 119879 (119903 119891 (119911)) + 119879(119903
1
119891 (119911)
) + 119879(119903
1
119891(119911)119898
+ 119886
)
+ 119879(119903
1
119871 (119911 119891)
) + 119898119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (119898 + 2) 119879 (119903 119891 (119911))
+ (119898 + 1) 119879 (119903 119871 (119911 119891)) + 119878 (119903 119891)
le (2119898 + 3) 119879 (119903 119891 (119911)) + 119878 (119903 119891)
(29)
By combining (28) and (29) we have
(119899 minus 2119898 minus 3) 119879 (119903 119891 (119911)) le 119878 (119903 119891) (30)
which contradicts with 119899 ge 2119898 + 4Now we turn to consider the case 119890ℎ(119911) equiv 1 Equation (19)
yields the following
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
equiv 119891(119911)119899
+ 119886119891(119911)119899minus119898
(31)
Set 120593(119911) = 119871(119911 119891)119891(119911) and we have
119891(119911)119898
(120593(119911)119899
minus 1) = minus119886 (120593(119911)119899minus119898
minus 1) (32)
If 120593(119911) is not a constant (32) can be rewritten as
119891(119911)119898
(120593 (119911) minus 1) (120593 (119911) minus 120583) sdot sdot sdot (120593 (119911) minus 120583119899minus1
)
= minus119886 (120593 (119911) minus 1) (120593 (119911) minus ]) sdot sdot sdot (120593 (119911) minus ]119899minus119898minus1) (33)
where 120583 = cos(2120587119899) + 119894 sin(2120587119899) and ] = cos(2120587(119899minus119898)) +119894 sin(2120587(119899 minus 119898))
By the assumption that 119899 and 119899 minus 119898 have no commonfactors we see that 120583 120583119899minus1 ] ]119899minus119898minus1 are differentAssume that 119911
0is a 120583119895-point of 120593(119911) of multiplicity 119906
119895gt 0
where 1 le 119895 le 119899 minus 1 Notice that
minus119886 (120593 (1199110) minus 1) (120593 (119911
0) minus ]) sdot sdot sdot (120593 (119911
0) minus ]119899minus119898minus1) (34)
is a constantThen (33) implies that 1199110is a pole of119891(119911)119898Thus
119906119895ge 119898 This yields the following for 1 le 119895 le 119899 minus 1
119898119873(119903
1
120593 (119911) minus 120583119895
) le 119873(119903
1
120593 (119911) minus 120583119895
)
le 119879 (119903 120593 (119911)) + 119878 (119903 ℎ)
(35)
Then by (35) we get
2 ge
119899minus1
sum
119895=1
Θ(120583119895
120593 (119911)) =
119899minus1
sum
119895=1
1 minus lim119903rarrinfin
119873(119903 1 (120593 (119911) minus 120583119895
))
119879 (119903 120593 (119911))
ge
119899minus1
sum
119895=1
(1 minus
1
119898
) = (119899 minus 1) (1 minus
1
119898
)
(36)
which is impossible with119898 ge 2 and 119899 ge 2119898 + 4Hence 120593(119911) is a constant Since 119891(119911) is a nonconstant
meromorphic function we deduce from (32) that 120593(119911) equiv 1This yields 119871(119911 119891) equiv 119891(119911) which completes the proof ofTheorem 1
3 Proof of Theorem 5
Since 119891(119911) is a nonconstant meromorphic function of finiteorder 119864
119891(119911)(119878119895) = 119864
119871(119911119891)(119878119895) for 119895 = 1 2 3 119878
1= 120596 120596
119899
+
119886120596119899minus119898
+ 119887 = 0 1198782= infin and 119878
3= 0 we have 119871(119911 119891) equiv
0 119873(119903 119871(119911 119891)) = 119873(119903 119891(119911)) and 119873(119903 1119871(119911 119891)) = 119873(119903
1119891(119911)) and we also get (18) and (19)
6 Abstract and Applied Analysis
Since 119891(119911) and 119871(119911 119891) share 0 infin CM there exists apolynomial ℎlowast(119911) such that
119871 (119911 119891)
119891 (119911)
= 119890ℎlowast(119911)
(37)
By Lemma 8 we see that
119879(119903 119890ℎlowast(119911)
) = 119898(119903 119890ℎlowast(119911)
) = 119898(119903
119871 (119911 119891)
119891 (119911)
)
le
119896
sum
119895=0
119898(119903
119891 (119911 + 119888119895)
119891 (119911)
)
+
119896
sum
119895=0
119898(119903 119887119895(119911)) + 119878 (119903 119891)
= 119878 (119903 119891)
(38)
As in the proof of Theorem 1 we see that 119879(119903 119890ℎ(119911)) =
119878(119903 119891) still holds in both cases (i) and (ii)Rewriting (19) we have
(119871 (119911 119891))119899
+ 119886(119871 (119911 119891))119899minus119898
minus 119890ℎ(119911)
119891(119911)119899
minus 119886119890ℎ(119911)
119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(39)
Combining this and (37) we get
(119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
+ 119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899minus119898
= 119887 (119890ℎ(119911)
minus 1)
(40)
Suppose that 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 If119898 = 0 (40) becomes
(119886 + 1) (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
= 119887 (119890ℎ(119911)
minus 1) (41)
By the condition that 119887 = 0 1198781= 120596 (119886 + 1)120596
119899
+ 119887 = 0 =
implies 119886 = minus 1It follows from (38) (41) and 119879(119903 119890ℎ(119911)) = 119878(119903 119891) that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 119887 (119890ℎ(119911)
minus 1)) = 119878 (119903 119891)
(42)
which is a contradiction since 119899 ge 1If119898 ge 1 it follows from (38) (41) and119879(119903 119890ℎ(119911)) = 119878(119903 119891)
that
119899119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 (119890119899ℎlowast(119911)
minus 119890ℎ(119911)
)119891(119911)119899
)
= 119879 (119903 minus119886 (119890(119899minus119898)ℎ
lowast(119911)
minus 119890ℎ(119911)
) 119891(119911)119899minus119898
+ 119887 (119890ℎ(119911)
minus 1))
le (119899 minus 119898)119879 (119903 119891) + 119878 (119903 119891)
(43)
That is impossible
Therefore 119890119899ℎlowast(119911)
minus 119890ℎ(119911)
equiv 0 Notice that 119886 119887 = 0 Using asimilar method we can prove that 119890(119899minus119898)ℎ
lowast(119911)
minus119890ℎ(119911)
equiv 0Then(40) implies that 119890ℎ(119911) equiv 1
If 119898 = 0 we have 119890119899ℎlowast(119911)
equiv 1 Obviously 119890ℎlowast(119911) is a
constant Set 119905 = 119890ℎlowast(119911) Thus by (37) we get 119871(119911 119891) equiv 119905119891(119911)
where 119905119899 = 1If 119899 and 119898 are coprime 119890119899ℎ
lowast(119911)
equiv 1 and 119890119898ℎlowast(119911)
equiv 1 implythat 119890ℎ
lowast(119911)
equiv 1 Thus by (37) we get 119871(119911 119891) equiv 119891(119911) ThusTheorem 5 is proved
Conflict of Interests
The authors declare that they have no conflict of interests
Authorsrsquo Contribution
Both authors drafted the paper and read and approved thefinal paper
Acknowledgments
The authors are grateful to the editor and referees fortheir valuable suggestions This work was supported by theNNSFC (no 11171119 11301091) the Guangdong Natural Sci-ence Foundation (no S2013040014347) and the Foundationfor Distinguished Young Talents in Higher Education ofGuangdong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 2003
[4] P Li and C-C Yang ldquoSome further results on the unique rangesets of meromorphic functionsrdquo Kodai Mathematical Journalvol 18 no 3 pp 437ndash450 1995
[5] H-X Yi and W-C Lin ldquoUniqueness theorems concerning aquestion of Grossrdquo Proceedings of the Japan Academy Series AMathematical Sciences vol 80 no 7 pp 136ndash140 2004
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[8] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[9] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
Abstract and Applied Analysis 7
[10] B Chen and Z Chen ldquoMeromorphic function sharing twosets with its difference operatorrdquo Bulletin of the MalaysianMathematical Sciences Society Second Series vol 35 no 3 pp765ndash774 2012
[11] B Chen Z Chen and S Li ldquoUniqueness theorems on entirefunctions and their difference operators or shiftsrdquo Abstract andApplied Analysis vol 2012 Article ID 906893 8 pages 2012
[12] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[13] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[14] S Li and B Chen ldquoMeromorphic functions sharing smallfunctions with their linear difference polynomialsrdquoAdvances inDifference Equations vol 2013 article 58 2013
[15] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[16] J Zhang ldquoValue distribution and shared sets of differences ofmeromorphic functionsrdquo Journal of Mathematical Analysis andApplications vol 367 no 2 pp 401ndash408 2010
[17] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleA Comparison Theorem for Oscillation of the Even-OrderNonlinear Neutral Difference Equation
Quanxin Zhang
Department of Mathematics Binzhou University Binzhou Shandong 256603 China
Correspondence should be addressed to Quanxin Zhang 3314744163com
Received 9 December 2013 Accepted 28 March 2014 Published 10 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Quanxin Zhang This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
A comparison theorem on oscillation behavior is firstly established for a class of even-order nonlinear neutral delay differenceequations By using the obtained comparison theorem two oscillation criteria are derived for the class of even-order nonlinearneutral delay difference equations Two examples are given to show the effectiveness of the obtained results
1 Introduction
Recently there have been a lot of research papers in con-nection with the oscillation of solutions of difference equa-tions with or without neutral terms The literature on theoscillation of neutral delay difference equations is growingvery fast and it can be widely applied to the reality Infact neutral delay difference equations arise in modellingof the networks containing lossless transmission lines (asin high speed computers where the lossless transmissionlines are used to interconnect switching circuits) For recentcontributions regarding the theoretical part and providingsystematic treatment of oscillation of solutions of neutraltype difference equations the readers can refer to the recentmonographs by Agarwal [1] Gyori and Ladas [2]
The oscillation behavior of the even-order nonlinearneutral differential equation
[119909 (119905) + 119901 (119905) 119909 (120591 (119905))](119899)
+ 119902 (119905) 119891 [119909 (120590 (119905))] = 0 (1)
has been established by Zhang et al [3] In this paperthe discrete analogue of the above equation is consideredWe consider the even-order nonlinear neutral differenceequation
Δ119898
(119909119899+ 119901119899119909119899minus120591
) + 119902119899119891 (119909119899minus119896
) = 0 (2)
where 119898 ge 2 is an even and 120591 119896 isin N let N denote the setof all natural numbers 119899 isin 119873(119899
0) = 119899
0 1198990+ 1 1198990+ 2
1198990is a nonnegative integer Δ denotes the forward difference
operator defined by Δ119909119899= 119909119899+1
minus 119909119899 Δ119898119909
119899= Δ119898minus1
119909119899+1
minus
Δ119898minus1
119909119899
Throughout this paper the following conditions areassumed to hold
(H1) 119901119899 is a sequence of nonnegative real number 0 le
119901119899
lt 1 and 119902119899 is a sequence of nonnegative real
number with 119902119899 being not eventually identically
equal to zero(H2) 119891 R rarr R (R = (minusinfin +infin)) is a continuous oddfunction and 119909119891(119909) gt 0 for all 119909 = 0
Before deriving themain results the following definitionsare given
Definition 1 By a solution of (2) one means a real sequence119909119899 defined for 119899 ge 119899
0minus120579 (120579 = max120591 119896)which satisfies (2)
for 119899 isin 119873(1198990)
In this paper we restrict our attention to nontrivialsolutions of (2)
Definition 2 A nontrivial solution 119909119899 of (2) is said to
be oscillatory if the terms 119909119899of the sequence are neither
eventually positive nor eventually negative Otherwise it iscalled nonoscillatory
Definition 3 An equation is said to be oscillatory if all itssolutions are oscillatory
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 492492 5 pageshttpdxdoiorg1011552014492492
2 Abstract and Applied Analysis
In 2004 Stavroulakis [4] studied the oscillatory behaviorof all solutions of first-order delay difference equation
119909119899+1
minus 119909119899+ 119901119899119909119899minus119896
= 0 (3)
and established one new oscillation criterion Thandapaniet al [5] studied the oscillatory behavior of all solutions ofsecond-order neutral delay difference equation
Δ2
(119910119899minus 119901119910119899minus119896
) minus 119902119899119891 (119910119899minus119905
) = 0 (4)
and established a number of new oscillation criteria In 2000Zhou et al [6] studied the oscillatory behavior of all solutionsof even-order neutral delay difference equation
Δ119898
(119909119899minus 119901119899119892 (119909119899minus119896
)) minus 119902119899ℎ (119909119899minus119897
) = 0 (5)
and established three new oscillation criteria under certainconditionsThe studies on oscillatory behavior of all solutionsof even-order delay difference equations we recommendreferring to [7ndash10] On the basis of the abovework we studiedthe oscillatory behavior of all solutions of (2) Firstly a com-parison theorem on oscillation behavior is established for (2)The comparison theorem changes the discriminant criteria ofthe oscillation of (2) into the oscillationrsquos discriminant criteriain the first-order nonneutral delay difference equationsThenby using the above comparison theorem we obtain someoscillation criteria for (2) and improve the well-known resultsof Ladas et al [11] Erbe and Zhang [12] and Stavroulakis [4]In particular the results are new when119898 = 2 119901
119899equiv 0
The paper is organized as follows In Section 2 a compar-ison theorem on oscillation behavior is firstly established fora class of even-order nonlinear neutral delay difference equa-tions Then the comparison theorem changes the discrimi-nant of the oscillation in the even-order nonlinear neutraldelay difference equation into the oscillationrsquos discriminantin the first-order nonneutral delay difference equations InSection 3 some oscillation criteria are obtained for the classof even-order nonlinear neutral delay difference equationby using the above comparison theorem In Section 4 twoexamples are given
2 Comparison Theorem
To obtain the comparison theorem in this section we needthe following lemmas which can be founded in [1] see alsoChen [7] andThandapani and Arul [8]
Lemma 4 Let 119906119899 be a sequence of real numbers for 119899 ge 119899
0
Let 119906119899 and Δ
119898
119906119899 be of constant sign where Δ
119898
119906119899is not
identically zero for 119899 ge 1198991 If
119906119899Δ119898
119906119899le 0 (6)
then
(i) there is a natural number 1198992
ge 1198991such that the
sequences Δ119895119906119899 119895 = 1 2 119898 minus 1 are of constant
sign for 119899 ge 1198992
(ii) there exists a number 119897 isin 0 1 119898 minus 1 with(minus1)119898minus119897minus1
= 1 such that
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 0 1 2 119897 119899 ge 119899
2
(minus1)119895minus119897
119906119899Δ119895
119906119899gt 0 119891119900119903 119895 = 119897 + 1 119898 minus 1 119899 ge 119899
2
(7)
Lemma 5 Observe that under the hypotheses of Lemma 4 if119906119899 is increasing for 119899 ge 119899
0 then there exists a natural number
1198991ge 1198990such that for all 119899 ge 2
119898minus1
1198991
119906119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119906119899 (8)
where 120582119898= 12(119898minus1)
2
Theorem 6 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 If there exists a constant
120582119898= 12(119898minus1)
2
such that the first-order difference equation
Δ119911119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119911119899minus119896
= 0 (9)
is oscillatory then (2) is oscillatory
Proof Suppose that (2) has a nonoscillatory solution 119909119899
Without the loss of generality we assume that 119909119899 is an
eventually positive solution of (2) then there is a naturalnumber 119899
1ge 1198990such that 119909
119899gt 0 119909
119899minus119896gt 0 119909
119899minus120591gt 0 and
119909119899minus119896minus120591
gt 0 for all 119899 ge 1198991 Let
119910119899= 119909119899+ 119901119899119909119899minus120591
(10)
Then from (H1) and (H
2) there exists a natural number 119899
2ge
1198991such that
119910119899gt 0 Δ
119898
119910119899le 0 forall119899 ge 119899
2 (11)
By Lemma 4 there exist an integer 1198993ge 1198992and an integer
119897 (0 le 119897 le 119898) where (119898 + 119897) is an odd integer For all 119899 ge 1198993
we can get
Δ119895
119910119899gt 0 for 119895 = 1 2 119897
(minus1)(119895minus119897)
Δ119895
119910119899gt 0 for 119895 = 119897 + 1 119898 minus 1
(12)
Thus from (12) Δ119910119899
gt 0 and Δ119898minus1
119910119899
gt 0 for 119899 ge 1198993 By
Lemma 5 there exists an integer 1198994ge 1198993 For all 119899 ge 2
119898minus1
1198994
we derive
119910119899ge
120582119898
(119898 minus 1)
119899119898minus1
Δ119898minus1
119910119899 120582
119898=
1
2(119898minus1)
2 (13)
From (10)
119909119899minus119896
= 119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
(14)
Consequently we have
Δ119898
119910119899+ 119902119899119891 (119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) = 0
for all sufficient large 119899
(15)
Abstract and Applied Analysis 3
Noting that |119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) le 0
for all sufficient large 119899
(16)
By 119910119899ge 119909119899 Δ119910119899gt 0 and 119899 minus 119896 minus 120591 le 119899 minus 119896 we obtain
Δ119898
119910119899+ 119902119899(119910119899minus119896
minus 119901119899minus119896
119909119899minus119896minus120591
) ge Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
(17)
Therefore we have
Δ119898
119910119899+ 119902119899(1 minus 119901
119899minus119896) 119910119899minus119896
le 0
for all sufficient large 119899
(18)
Now by using (13) we have that for 120582119898= 12(119898minus1)
2
119910119899minus119896
ge
120582119898
(119898 minus 1)
(119899 minus 119896)119898minus1
Δ119898minus1
119910119899minus119896
for all sufficient large 119899
(19)
Thus we get
Δ119898
119910119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) Δ119898minus1
119910119899minus119896
le 0
for all sufficient large 119899
(20)
where120582119898= 12(119898minus1)
2
Let119906119899= Δ119898minus1
119910119899 then for large enough
119899 we get
Δ119906119899+
120582119898
(119898 minus 1)
119902119899(119899 minus 119896)
119898minus1
(1 minus 119901119899minus119896
) 119906119899minus119896
le 0 (21)
where 120582119898
= 12(119898minus1)
2
Therefore inequality (21) has aneventually positive solution By Lemma 5 in [9] (9) hasan eventually positive solution which contradicts that (9) isoscillatory This completes the proof
3 Applications of the Comparison Theorem
The following lemma is well known (see eg [2 11 12] andthe references therein)
Lemma 7 Let 119902119899 be a sequence of eventually nonnegative
real number and 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt (
119896
1 + 119896
)
1+119896
(22)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894gt 1 (23)
then the first-order difference equation
Δ119909119899+ 119902119899119909119899minus119896
= 0 (24)
is oscillatory
Thus from Theorem 6 and Lemma 7 we can obtain thefollowing results
Theorem 8 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1)(
119896
1 + 119896
)
1+119896
(25)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 2(119898minus1)
2
(119898 minus 1) (26)
then (2) is oscillatory
Proof From (25) and (26) we can obtain
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt (
119896
1 + 119896
)
1+119896
(27)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 (28)
where 120582119898
= 12(119898minus1)
2
By Lemma 7 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According toTheorem 8 we obtain Corollary 9
Corollary 9 Assume that conditions (H1) and (H
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 For 119896 ge 1 when 119901
119899equiv 0
119898 = 2 if either
lim inf119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(
119896
1 + 119896
)
1+119896
(29)
or
lim sup119899rarrinfin
119899
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (30)
then the second-order difference equation
Δ2
119909119899+ 119902119899119891 (119909119899minus119896
) = 0 (31)
is oscillatory
The following lemma is given in [4 Theorem 26]
Lemma 10 Let 119902119899 be a sequence of nonnegative real numbers
and 119896 a positive integer Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(32)
4 Abstract and Applied Analysis
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus
1205722
4
(33)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894gt 1 minus 120572
119896
(34)
then (24) is oscillatory
Thus fromTheorem 6 and Lemma 10 we can obtain thefollowing results
Theorem 11 Assume that conditions (1198671) and (119867
2) hold Let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
Assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(35)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus
1205722
4
)
(36)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
)
gt 2(119898minus1)
2
(119898 minus 1) (1 minus 120572119896
)
(37)
then (2) is oscillatory
Proof From (36) and (37) we can obtain
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus
1205722
4
(38)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
120582119898
(119898 minus 1)
119902119894(119894 minus 119896)
119898minus1
(1 minus 119901119894minus119896
) gt 1 minus 120572119896
(39)
where 120582119898
= 12(119898minus1)
2
By Lemma 10 we know (9) isoscillatoryThen similar to the proof ofTheorem 6 the resultsfollow immediately This completes the proof
According to Theorem 11 we can obtain the followingcorollary
Corollary 12 Assume that conditions (1198671) and (119867
2) hold let
|119891(119909)| ge |119909| for all |119909| ge 1199090gt 0 and let 119896 be a positive integer
For 119901119899equiv 0119898 = 2 assume that
0 lt 120572 le (
119896
1 + 119896
)
1+119896
(40)
if either
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2(1 minus
1205722
4
) (41)
or
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus119896
119902119894(119894 minus 119896) gt 2 (1 minus 120572
119896
) (42)
then (31) is oscillatory
4 Examples
Example 1 Considering the equation
Δ119898
(119909119899+
1
119899
119909119899minus119897
) +
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(43)
where 119899 gt 3119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
1
119899
lt 1 119902119899=
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890 (119899 minus 3) (119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(44)
where 119902119899 is a positive sequence Then
119899
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
1
119894 minus 2
)
=
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
(45)
Thus
lim sup119899rarrinfin
119899
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [2 + ((minus1)119899
119899)]
119890
=
6
119890
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1)
(46)
Therefore by Theorem 8 (43) is oscillatory
Example 2 Considering the equation
Δ119898
[119909119899+
119899 minus 1
119899
119909119899minus119897
]
+
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
times 119909119899minus2
ln (119890 + 1199092
119899minus2) = 0
(47)
Abstract and Applied Analysis 5
where 119899 gt 2119898 is an even and 119897 is a positive integer then wehave
0 lt 119901119899=
119899 minus 1
119899
lt 1
119902119899=
2(119898minus1)
2
(119898 minus 1) [(1532) + ((minus1)119899
119899)]
(119899 minus 2)119898minus2
119891 (119909) = 119909 ln (119890 + 1199092
) 119896 = 2
(48)
where 119902119899 is a positive sequence Denote 120572 = 414 then
119899minus1
sum
119894=119899minus2
119902119894(119894 minus 2)
119898minus1
(1 minus
119894 minus 2 minus 1
119894 minus 2
)
=
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
(49)
Thus
lim sup119899rarrinfin
119899minus1
sum
119894=119899minus2
2(119898minus1)
2
(119898 minus 1) [
15
32
+
(minus1)119899
119899
]
=
15
16
2(119898minus1)
2
(119898 minus 1) gt 2(119898minus1)
2
(119898 minus 1) (1 minus (
4
14
)
2
)
(50)
Therefore by Theorem 11 (47) is oscillatory
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The author sincerely thanks the reviewers for their valuablesuggestions and useful comments that have led to the presentimproved version of the original paper This work wassupported by a Grant from the Natural Science Foundationof Shandong Province of China (no ZR2013AM003) andthe Development Program in Science and Technology ofShandong Province of China (no 2010GWZ20401)
References
[1] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[2] I Gyori and G Ladas Oscillation Theory of Delay DifferentialEquations Oxford Clarendon Press Oxford UK 1991
[3] Q Zhang J Yan and L Gao ldquoOscillation behavior of even-order nonlinear neutral differential equations with variablecoefficientsrdquoComputers andMathematics withApplications vol59 no 1 pp 426ndash430 2010
[4] I P Stavroulakis ldquoOscillation criteria for first order delaydifference equationsrdquo Mediterranean Journal of Mathematicsvol 1 no 2 pp 231ndash240 2004
[5] E Thandapani R Arul and P S Raja ldquoBounded oscillationof second order unstable neutral type difference equationsrdquoJournal of Applied Mathematics and Computing vol 16 no 1-2pp 79ndash90 2004
[6] Z Zhou J Yu and G Lei ldquoOscillations for even-order neu-tral difference equationsrdquo Korean Journal of Computational ampApplied Mathematics vol 7 no 3 pp 601ndash610 2000
[7] S Chen ldquoOscillation criteria for certain even order quasilineardifference equationsrdquo Journal of AppliedMathematics and Com-puting vol 31 no 1-2 pp 495ndash506 2009
[8] EThandapani and R Arul ldquoOscillatory and asymptotic behav-ior of solutions of higher order damped nonlinear differenceequationsrdquoCzechoslovakMathematical Journal vol 49 no 1 pp149ndash161 1999
[9] G Ladas and C Qian ldquoComparison results and linearizedoscillations for higher-order difference equationsrdquo InternationalJournal of Mathematics andMathematical Sciences vol 15 no 1pp 129ndash142 1992
[10] L Yang ldquoOscillation behavior of even-order neutral differenceequations with variable coefficientsrdquo Pure and Applied Mathe-matics vol 24 no 4 pp 796ndash801 2008 (Chinese)
[11] G Ladas Ch G Philos and Y G Sficas ldquoSharp conditions forthe oscillation of delay difference equationsrdquo Journal of AppliedMathematics and Simulation vol 2 no 2 pp 101ndash111 1989
[12] L H Erbe and B G Zhang ldquoOscillation of discrete analogues ofdelay equationsrdquo Differential and Integral Equations vol 2 no3 pp 300ndash309 1989
Research ArticleDifference Equations and Sharing ValuesConcerning Entire Functions and Their Difference
Zhiqiang Mao1 and Huifang Liu2
1 School of Mathematics and Computer Jiangxi Science and Technology Normal University Nanchang 330038 China2 College of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China
Correspondence should be addressed to Huifang Liu liuhuifang73sinacom
Received 19 January 2014 Accepted 22 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Z Mao and H Liu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The value distribution of solutions of certain difference equations is investigated As its applications we investigate the differenceanalogue of the Bruck conjecture We obtain some results on entire functions sharing a finite value with their difference operatorsExamples are provided to show that our results are the best possible
1 Introduction and Main Results
In this paper the term meromorphic function will meanbeing meromorphic in the whole complex plane C Itis assumed that the reader is familiar with the standardnotations and the fundamental results of the Nevanlinnatheory see for example [1ndash3] In addition we use notations120590(119891) 120582(119891) to denote the order and the exponent of conver-gence of the sequence of zeros of a meromorphic function 119891respectivelyThe notation 119878(119903 119891) is defined to be any quantitysatisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside aset 119864 of 119903 of finite logarithmic measure
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 isin C We say that 119891 and 119892 share 119886 CM providedthat 119891 minus 119886 and 119892 minus 119886 have the same zeros with the samemultiplicities Similarly we say that 119891 and 119892 share 119886 IMprovided that 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
The famous results in the uniqueness theory of meromor-phic functions are the 5 IM and 4 CM shared values theoremsdue to Nevanlinna [4] It shows that if two nonconstantmeromorphic functions 119891 and 119892 share five different valuesIM or four different values CM then 119891 equiv 119892 or 119891 is a linearfractional transformation of119892 Condition 4CMshared valueshave been improved to 2 CM + 2 IM by Gundersen [5]while the case 1 CM + 3 IM still remains an open problemSpecifically Bruck posed the following conjecture
Conjecture 1 (see [6]) Let 119891 be a nonconstant entire functionsatisfying the hyperorder 120590
2(119891) lt infin where 120590
2(119891) is not a
positive integer If 119891 and 1198911015840 share a finite value 119886 CM then119891 minus 119886 equiv 119888(119891
1015840
minus 119886) for some nonzero constant 119888
In [6] Bruck proved that the conjecture is true providedthat 119886 = 0 or 119873(119903 11198911015840) = 119878(119903 119891) He also gave counter-examples to show that the restriction on the growth of 119891 isnecessary
In recent years as the research on the difference ana-logues of Nevanlinna theory is becoming active lots ofauthors [7ndash11] started to consider the uniqueness of mero-morphic functions sharing values with their shifts or theirdifference operators
Heittokangas et al proved the following result which is ashifted analogue of Bruckrsquos conjecture
Theorem A (see [8]) Let 119891 be a meromorphic function of120590(119891) lt 2 and 120578 a nonzero complex number If119891(119911) and119891(119911+120578)share a finite value 119886 andinfin CM then
119891 (119911 + 120578) minus 119886
119891 (119911) minus 119886
= 120591 (1)
for some constant 120591
In [8] Heittokangas et al gave the example 119891(119911) = 1198901199112
+1
which shows that 120590(119891) lt 2 cannot be relaxed to 120590(119891) le 2
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 584969 6 pageshttpdxdoiorg1011552014584969
2 Abstract and Applied Analysis
For a nonzero complex number 120578 we define differenceoperators as
Δ120578119891 (119911) = 119891 (119911 + 120578) minus 119891 (119911)
Δ119899
120578119891 (119911) = Δ
119899minus1
120578(Δ120578119891 (119911)) 119899 isin N 119899 ge 2
(2)
Regarding the difference analogue of Bruckrsquos conjecture wemention the following results
Theorem B (see [7]) Let 119891 be a finite order transcendentalentire function which has a finite Borel exceptional value 119886 andlet 120578 be a constant such that 119891(119911 + 120578) equiv 119891(119911) If 119891(119911) andΔ120578119891(119911) share 119886 CM then
119886 = 0
119891 (119911 + 120578) minus 119891 (119911)
119891 (119911)
= 119888 (3)
for some nonzero constant 119888
Theorem C (see [11]) Let 119891 be a nonperiodic transcendentalentire function of finite order If 119891(119911) and Δ119899
120578119891(119911) share a
nonzero finite value 119886 CM then 1 le 120590(119891) le 120582(119891 minus 119886) + 1that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (4)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891 minus 119886) and119876(119911) is a polynomial with deg119876 le 120590(119860) + 1
Let 119891 be a nonperiodic transcendental entire function offinite order Theorem B shows that if a nonzero finite value119886 is shared by 119891(119911) and Δ
120578119891(119911) then 120590(119891) = 120582(119891 minus 119886)
It is obvious that the result in Theorem B is sharper thanTheoremC for 119899 = 1 In this paper we continue to investigatethe difference analogue of Bruckrsquos conjecture and obtain thefollowing result
Theorem 2 Let 119891 be a finite order entire function 119899 ge 2 aninteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share a finite value 119886 ( = 0)CM then120582(119891minus119886) = 120590(119891) ge
1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
+ 119886 (5)
where 119860(119911) is an entire function with 1 le 120590(119860) = 120582(119891 minus 119886) =120590(119891) and 119876(119911) is a polynomial with deg119876 le 120590(119860)
Remark 3 It is obvious that Theorem 2 is sharper thanTheorem C and a supplement of Theorem B for 119899 ge 2
The discussions in Theorems C and 2 are concerning thecase that shared value 119886 = 0 When 119886 = 0 we obtain thefollowing result
Theorem 4 Let 119891 be a finite order entire function 119899 a positiveinteger and 120578 a constant such that Δ119899
120578119891(119911) equiv 0 If 119891(119911) and
Δ119899
120578119891(119911) share 0 CM then 1 le 120590(119891) le 120582(119891) + 1 that is
119891 (119911) = 119860 (119911) 119890119876(119911)
(6)
where 119860(119911) is an entire function with 120590(119860) = 120582(119891) and 119876(119911)is a polynomial with deg119876 le 120590(119860) + 1
It is well known that if a finite order entire function 119891(119911)shares 119886 CM with Δ119899
120578119891(119911) then 119891(119911) satisfies the difference
equation
Δ119899
120578119891 (119911) minus 119886 = 119890
119876(119911)
(119891 (119911) minus 119886) (7)
where119876(119911) is a polynomialHence in order to prove the aboveresults we consider the value distribution of entire solutionsof the difference equation
119886119899(119911) 119891 (119911 + 119899120578) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 120578)
+ (1198860(119911) minus 119890
119876(119911)
) 119891 (119911) = 119861 (119911)
(8)
and obtain the following result
Theorem 5 Let 1198860 119886
119899minus1 119886119899( equiv 0) 119861( equiv 0) be polynomials
and let 119876 be a polynomial with degree 119898(ge 1) Then everyentire solution 119891 of finite order of (8) satisfies 120590(119891) ge 119898 and
(i) if 120590(119891) gt 1 then 120582(119891) = 120590(119891)(ii) if 120590(119891) = 1 then 120582(119891) = 120590(119891) or 119891 has only finitely
many zeros
2 Lemmas
Lemma 6 (see [12]) Let 119879 (0 +infin) rarr (0 +infin) be anondecreasing continuous function 119904 gt 0 120572 lt 1 and let119865 sub R+ be the set of all 119903 such that 119879(119903) le 120572119879(119903 + 119904) If thelogarithmic measure of 119865 is infinite then
lim119903rarrinfin
log119879 (119903)log 119903
= infin (9)
Lemma 7 (see [13]) Let 119891 be a nonconstant meromorphicfunction of finite order 120578 isin C 120575 lt 1 Then
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119900(
119879 (119903 +10038161003816100381610038161205781003816100381610038161003816 119891)
119903120575
) (10)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
(119889119903119903) lt infin
Remark 8 By Lemmas 6 and 7 we know that for a noncon-stant meromorphic function 119891 of finite order
119898(119903
119891 (119911 + 120578)
119891 (119911)
) = 119878 (119903 119891) (11)
Lemma 9 (see [3]) Let 119891119895(119895 = 1 119899 + 1) and 119892
119895(119895 =
1 119899) be entire functions such that
(i) sum119899119895=1119891119895(119911)119890119892119895(119911)
equiv 119891119899+1(119911)
(ii) the order of 119891119895is less than the order of 119890119892119896 for 1 le 119895 le
119899+1 1 le 119896 le 119899 and furthermore the order of119891119895is less
than the order of 119890119892ℎminus119892119896 for 119899 ge 2 and 1 le 119895 le 119899+1 1 leℎ lt 119896 le 119899
Then 119891119895(119911) equiv 0 (119895 = 1 119899 + 1)
Abstract and Applied Analysis 3
Lemma 10 (see [14]) Let 119891 be a meromorphic function withfinite order 120590(119891) = 120590 lt 1 120578 isin C 0 Then for any given 120576 gt 0and integers 0 le 119895 lt 119896 there exists a set 119864 sub (1infin) of finitelogarithmic measure so that for all |119911| = 119903 notin 119864 ⋃ [0 1] wehave
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119896
120578119891 (119911)
Δ119895
120578119891 (119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|(119896minus119895)(120590minus1)+120576
(12)
Lemma 11 (see [15]) Let 1198860(119911) 119886
119896(119911) be entire functions
with finite order If there exists an integer 119897 (0 le 119897 le 119896) suchthat
120590 (119886119897) gt max0le119895le119896
119895 = 119897
120590 (119886119895) (13)
holds then every meromorphic solution 119891( equiv 0) of thedifference equation
119886119896(119911) 119891 (119911 + 119896) + sdot sdot sdot + 119886
1(119911) 119891 (119911 + 1) + 119886
0(119911) 119891 (119911) = 0
(14)
satisfies 120590(119891) ge 120590(119886119897) + 1
3 Proofs of Results
Proof of Theorem 5 Let 119891 be an entire solution of finite orderof (8) By Remark 8 and (8) we get
119879 (119903 119890119876
) = 119879 (119903 119890119876
minus 1198860) + 119878 (119903 119890
119876
)
le
119899
sum
119895=1
119898(119903
119891 (119911 + 119895120578)
119891 (119911)
)
+
119899
sum
119895=0
119898(119903 119886119895) + 119898(119903
119861 (119911)
119891 (119911)
) + 119878 (119903 119890119876
)
le 119879 (119903 119891) + 119878 (119903 119891) + 119878 (119903 119890119876
)
(15)
By (15) we get 120590(119891) ge 119898
Case 1 (120590(119891) gt 1) Suppose that 120582(119891) lt 120590(119891) by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that
120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891) lt 120590 (119891) = deg ℎ
2 (16)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (1198860(119911) minus 119890
119876(119911)
) ℎ1(119911) = 119861 (119911) 119890
minusℎ2(119911)
(17)
If deg ℎ2gt 119898 then by (16) we know that the order of the right
side of (17) is deg ℎ2 and the order of the left side of (17) is less
than deg ℎ2 This is a contradiction Hence deg ℎ
2= 119898 gt 1
Set119876 (119911) = 119887
119898119911119898
+ sdot sdot sdot + 1198870
ℎ2(119911) = 119888
119898119911119898
+ sdot sdot sdot + 1198880
(18)
where 119887119898( = 0) 119887
0 119888119898( = 0) 119888
0are complex numbers
By (17) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 1198860(119911) ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
+ 119861 (119911) 119890minusℎ2(119911)
(19)
Next we discuss the following two subcases
Subcase 1 (119887119898+ 119888119898=0)Then by Lemma 9 (16) and (19) we
get 119861(119911) equiv 0 ℎ1(119911) equiv 0 This is impossible
Subcase 2 (119887119898+ 119888119898= 0) Suppose that
ℎ1(119911) 119890119876(119911)minus119887
119898119911119898
+ 119861 (119911) 119890minusℎ2(119911)minus119887119898119911119898
equiv 0 (20)
Then ℎ1(119911) = minus119861(119911)119890
minusℎ2(119911)minus119876(119911) By 120590(ℎ
1) = 120582(ℎ
1) we
obtain that 119890minusℎ2(119911)minus119876(119911) is a nonzero constant Hence ℎ1(119911) is a
nonzero polynomial By (19) we get119899
sum
119895=1
119886119895(119911) ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
= minus1198860(119911) ℎ1(119911)
(21)
Since degℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578) = 119898 minus 1 gt 0 for 119894 = 119895 then
by Lemma 9 and (21) we get
119886119895(119911) ℎ1(119911 + 119895120578) equiv 0 (119895 = 0 1 119899) (22)
This is impossible Hence we have ℎ1(119911)119890119876(119911)minus119887
119898119911119898
+
119861(119911)119890minusℎ2(119911)minus119887119898119911119898
equiv 0 Then from the order considerationwe know that the order of the right side of (19) is 119898 andthe order of the left side of (19) is less than 119898 This is acontradiction Hence 120582(119891) = 120590(119891)
Case 2 (120590(119891) = 1)Then by 120590(119891) ge 119898 we get119898 = 1 Supposethat 119891(119911) has infinitely many zeros and 120582(119891) lt 120590(119891) by theWeierstrass factorization we get
119891 (119911) = ℎ3(119911) 119890120573119911
(23)
where 120573( = 0) is a complex number and ℎ3(119911)( equiv 0) is an
entire function such that
120590 (ℎ3) = 120582 (ℎ
3) = 120582 (119891) lt 1 (24)
Let 119876(119911) = 1198871119911 + 1198870 where 119887
1( = 0) 119887
0are complex numbers
Substituting 119891(119911) = ℎ3(119911)119890120573119911 into (8) we get
119899
sum
119895=1
119886119895(119911) ℎ3(119911 + 119895120578) 119890
120573119895120578
+ 1198860(119911) ℎ3(119911)
= ℎ3(119911) 1198901198871119911+1198870+ 119861 (119911) 119890
minus120573119911
(25)
4 Abstract and Applied Analysis
Note that ℎ3(119911)1198901198870+ 119861(119911) equiv 0 otherwise 119891 has only finitely
many zeros If 1198871+ 120573 = 0 then the order of the right side of
(25) is 1 but the order of the left side of (25) is less than 1Thisis absurd If 119887
1+120573 = 0 then by Lemma 9 (24) and (25) we get
ℎ3(119911) equiv 0 119861(119911) equiv 0 This is impossible Hence 120582(119891) = 120590(119891)
Theorem 5 is thus completely proved
Proof of Theorem 2 Since 119891(119911) and Δ119899120578119891(119911) share 119886 CM and
119891 is of finite order then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119890119876(119911)
(26)
where 119876(119911) is a polynomial with deg119876 le 120590(119891) Now we willtake two steps to complete the proof
Step 1 We prove that 120582(119891 minus 119886) = 120590(119891)Let 119865(119911) = 119891(119911) minus 119886 then
120582 (119865) = 120582 (119891 minus 119886) 120590 (119865) = 120590 (119891) ge deg119876 (27)
and Δ119899120578119891(119911) = Δ
119899
120578119865(119911) = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119865(119911 + 119895120578) By thisand (26) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119890119876(119911)
) 119865 (119911) = 119886 (28)
Next we discuss the following three cases
Case 1 (deg119876 ge 1 and 120590(119865) gt deg119876) Then 120590(119865) gt 1 ByTheorem 5(i) (27) and (28) we get 120582(119891 minus 119886) = 120590(119891)
Case 2 (deg119876 ge 1 and 120590(119865) = deg119876) If 120590(119865) = deg119876 gt 1then byTheorem 5(i) (27) and (28) we get120582(119891minus119886) = 120590(119891) If120590(119865) = deg119876 = 1 then byTheorem 5(ii) and (27) we obtainthat 120582(119891 minus 119886) = 120590(119891) or 119865 has only finitely many zeros
If 119865 has only finitely many zeros set
119865 (119911) = ℎ1(119911) 119890119887119911
(29)
where ℎ1(119911)( equiv 0) is a polynomial and 119887( = 0) is a complex
number then substituting (29) into (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
119887119895120578
+ (minus1)119899
ℎ1(119911)
= ℎ1(119911) 119890119876(119911)
+ 119886119890minus119887119911
(30)
By (30) and Δ119899120578119865(119911) equiv 0 we know that the order of the left
side of (30) is 0 and the order of the right side of (30) is 1unless ℎ
1(119911) = minus119886 and119876(119911) = minus119887119911 In this case take it into the
left side of (30) we have (minus119886)(119890119887120578 minus 1)119899 = 0 Since all 119886 119887 and120578 are not zero it is impossible Hence we get 120582(119891minus119886) = 120590(119891)
Case 3 119876 is a complex constant Then by (28) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119865 (119911 + 119895120578) + ((minus1)119899
minus 119888) 119865 (119911) = 119886 (31)
where 119888 (= 119890119876 =0) is a complex number Suppose that 120582(119865) lt120590(119865) Let 119865(119911) = ℎ
2(119911)119890ℎ3(119911) where ℎ
2(119911)( equiv 0) is an entire
function and ℎ3(119911) is a polynomial such that
120590 (ℎ2) = 120582 (ℎ
2) = 120582 (119865) lt 120590 (119865) = deg ℎ
3 (32)
Substituting 119865(119911) = ℎ2(119911)119890ℎ3(119911) into (31) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ2(119911 + 119895120578) 119890
ℎ3(119911+119895120578)minusℎ
3(119911)
+ ((minus1)119899
minus 119888) ℎ2(119911) = 119886119890
minusℎ3(119911)
(33)
Since deg(ℎ3(119911 + 119895120578) minus ℎ
3(119911)) = deg ℎ
3(119911) minus 1 (119895 = 1 119899)
by (32) we obtain that the order of the left side of (33) is lessthan deg ℎ
3and the order of the right side of (33) is deg ℎ
3
This is absurd Hence we get 120582(119891 minus 119886) = 120590(119891)
Step 2We prove that 120590(119891) ge 1Suppose that 120590(119891) lt 1 Since 119891(119911) and Δ119899
120578119891(119911) share 119886
CM then
Δ119899
120578119891 (119911) minus 119886
119891 (119911) minus 119886
= 119888 (34)
where 119888 is a nonzero constant Let 119865(119911) = 119891(119911) minus 119886 then by(34) we get
Δ119899
120578119865 (119911) = 119888119865 (119911) + 119886 (35)
Differentiating (35) we get
(Δ119899
120578119865 (119911))
1015840
= 1198881198651015840
(119911) (36)
Note that (Δ119899120578119865(119911))1015840
= Δ119899
120578(1198651015840
(119911)) and 120590(1198651015840) = 120590(119865) = 120590(119891) lt1 So by Lemma 10 and (36) we get
|119888| =
10038161003816100381610038161003816100381610038161003816100381610038161003816
Δ119899
120578(1198651015840
(119911))
1198651015840(119911)
10038161003816100381610038161003816100381610038161003816100381610038161003816
le |119911|119899(120590(119865)minus1)+120576
997888rarr 0 (37)
This is absurd So 120590(119891) ge 1 Theorem 2 is thus completelyproved
Proof of Theorem 4 Since 119891(119911) and Δ119899120578119891(119911) share 0 CM and
119891 is of finite order then
Δ119899
120578119891 (119911)
119891 (119911)
= 119890119876(119911)
(38)
where119876(119911) is a polynomial ByΔ119899120578119891 = sum
119899
119895=0(119899
119895 ) (minus1)119899minus119895
119891(119911+
119895120578) and (38) we get
119891 (119911 + 119899120578)
+
119899minus1
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
119891 (119911 + 119895120578)
+ ((minus1)119899
minus 119890119876(119911)
) 119891 (119911) = 0
(39)
Abstract and Applied Analysis 5
We discuss the following two cases
Case 1 119876 is a polynomial with deg119876 = 119898 ge 1 Then byLemma 11 and (39) we get 120590(119891) ge 119898 + 1 Now we prove120590(119891) le 120582(119891) + 1 Suppose that 120590(119891) gt 120582(119891) + 1 then by theWeierstrass factorization we get 119891(119911) = ℎ
1(119911)119890ℎ2(119911) where
ℎ1(119911)( equiv 0) is an entire function and ℎ
2(119911) is a polynomial
such that120590 (ℎ1) = 120582 (ℎ
1) = 120582 (119891)
120590 (119891) = deg ℎ2gt 120590 (ℎ
1) + 1
(40)
Substituting 119891(119911) = ℎ1(119911)119890ℎ2(119911) into (39) we get
119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ (minus1)119899
ℎ1(119911) = ℎ
1(119911) 119890119876(119911)
(41)
If 120590(119891) gt 119898 + 1 then by (40) (41) and deg(ℎ2(119911 + 119895120578) minus
ℎ2(119911)) = deg ℎ
2(119911)minus1 (119895 = 1 119899) we obtain that the order
of the left side of (41) is deg ℎ2minus 1 and the order of the right
side of (41) is less than deg ℎ2minus 1 This is absurd
If 120590(119891) = 119898 + 1 then by (41) we get119899
sum
119895=1
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
minus ℎ1(119911) 119890119876(119911)
= (minus1)119899+1
ℎ1(119911)
(42)
Set
ℎ2(119911) = 119889
119898+1119911119898+1
+ sdot sdot sdot + 1198890 (43)
where 119889119898+1( = 0) 119889
0are complex numbers Then
ℎ2(119911 + 119895120578) minus ℎ
2(119911)
= 119889119898+1
(119898 + 1) 119895120578119911119898
+ sdot sdot sdot + 1198891119895120578
(119895 = 1 119899)
(44)
Now we discuss the following two subcases
Subcase 1 deg(119876(119911)minus(ℎ2(119911+119895120578)minusℎ
2(119911))) = 119898 holds for every
119895 isin 1 119899Then by (40) (42) deg(ℎ2(119911+119895120578)minusℎ
2(119911+119894120578)) =
119898 (119895 = 119894) and Lemma 9 we get ℎ1(119911) equiv 0 This is absurd
Subcase 2 There exist some 1198950isin 1 119899 such that
deg(119876(119911)minus(ℎ2(119911+1198950120578)minusℎ2(119911))) le 119898minus1Then by (44) we have
deg(119876(119911) minus (ℎ2(119911 + 119895120578) minus ℎ
2(119911))) = 119898 for 119895 = 119895
0 Merging the
term minusℎ1(119911)119890119876(119911) into ( 119899119895
0) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578)119890ℎ2(119911+1198950120578)minusℎ2(119911)
by (42) we get119899
sum
119895 = 1
119895 =1198950
(
119899
119895) (minus1)
119899minus119895
ℎ1(119911 + 119895120578) 119890
ℎ2(119911+119895120578)minusℎ
2(119911)
+ 119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 ge 2)
(45)
or
119860 (119911) 119890ℎ2(119911+1198950120578)minusℎ2(119911)
= (minus1)119899+1
ℎ1(119911) (119899 = 1) (46)
where 119860(119911) = (119899
1198950) (minus1)
119899minus1198950ℎ1(119911 + 119895
0120578) minus ℎ
1(119911)
119890119876(119911)minus(ℎ
2(119911+1198950120578)minusℎ2(119911)) satisfying 120590(119860) lt 119898 If 119899 ge 2 then
by (40) (45) deg(ℎ2(119911 + 119895120578) minus ℎ
2(119911 + 119894120578)) = 119898 (119895 = 119894) and
Lemma 9 we get ℎ1(119911) equiv 0 This is absurd If 119899 = 1 then by
(46) and ℎ1(119911) equiv 0 we get 119860(119911) equiv 0 By this we know that
the order of the left side of (46) is 119898 and the order of theright side of (46) is less than119898 This is absurd Hence we get120590(119891) le 120582(119891) + 1
Case 2119876 is a complex constantThen by Lemma 10 and (38)we get 120590(119891) ge 1 Now we prove 120590(119891) le 120582(119891) + 1 Supposethat 120590(119891) gt 120582(119891) + 1 If 120590(119891) gt 1 then by the similarargument to that of case 1 we get ℎ
1(119911) equiv 0 This is absurd If
120590(119891) = 1 then by (40) we get 0 le 120582(119891) lt 120590(119891) minus 1 = 0 Since120582(119891) = 0 then 120590(119891) = 120582(119891)+1Theorem 4 is thus completelyproved
4 Some Examples
The following examples show the existence of such entirefunctions which satisfy Theorems 2ndash5 Moreover Example 2shows that the result in Theorem 4 is the best possible
Example 1 Let 120578 = 1 119899 = 2 and 119891(119911) = (119889 + 1)119911 + ((1198892 minus1)1198892
)119886 where 119886( = 0) 119889( = 0 plusmn1) are constants Then 119891(119911)and Δ119899
120578119891(119911) share 119886 CM and 120590(119891) = 120582(119891 minus 119886) = 1
Example 2 Let 120578 = 1 119899 = 2 and 119891(119911) = 119890119911 Then 119891(119911) andΔ119899
120578119891(119911) share 0 CM and 120590(119891) = 1 = 120582(119891) + 1
Example 3 Let 120578 = 1 119899 = 2 and119891(119911) = 119867(119911)119890119911 where119867(119911)is an entire function with period 1 such that 120590(119867) gt 1 and120590(119867) notin N Then 119891(119911) and Δ119899
120578119891(119911) share 0 CM and 120582(119891) =
120582(119867) = 120590(119867) = 120590(119891) gt 1 (Ozawa [16] proved that for any120590 isin [1infin) there exists a period entire function of order 120590)
Example 4 The entire function 119891(119911) = 119911119890minus119911 satisfies the
difference equation
119891 (119911 + 2120578) minus 4119891 (119911 + 120578) + (4 minus 119890119911
) 119891 (119911) = minus119911 (47)
where 120578 = minus log 2 Here120590(119891) = 1 and119891 has only finitelymanyzeros
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11201195 11171119) and the NaturalScience Foundation of Jiangxi China (nos 20122BAB20101220132BAB201008)
6 Abstract and Applied Analysis
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers New York NY USA2003
[4] R Nevanlinna Le Theoreme de Picard-Borel et la Theorie desFonctions Meromorphes Gauthiers-Villars Paris France 1929
[5] G G Gundersen ldquoMeromorphic functions that share fourvaluesrdquo Transactions of the American Mathematical Society vol277 no 2 pp 545ndash567 1983
[6] R Bruck ldquoOn entire functions which share one value CM withtheir first derivativerdquo Results inMathematics vol 30 no 1-2 pp21ndash24 1996
[7] Z-X Chen and H-X Yi ldquoOn sharing values of meromorphicfunctions and their differencesrdquo Results in Mathematics vol 63no 1-2 pp 557ndash565 2013
[8] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[9] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables andElliptic Equations vol 56 pp 81ndash92 2011
[10] K Liu and L-Z Yang ldquoValue distribution of the differenceoperatorrdquo Archiv der Mathematik vol 92 no 3 pp 270ndash2782009
[11] S Li and Z Gao ldquoA note on the Bruck conjecturerdquo Archiv derMathematik vol 95 no 3 pp 257ndash268 2010
[12] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[13] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[14] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[15] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[16] M Ozawa ldquoOn the existence of prime periodic entire func-tionsrdquo Kodai Mathematical Seminar Reports vol 29 no 3 pp308ndash321 1978
Research ArticleAdmissible Solutions of the Schwarzian Type DifferenceEquation
Baoqin Chen and Sheng Li
College of Science Guangdong Ocean University Zhanjiang 524088 China
Correspondence should be addressed to Sheng Li lish lssinacom
Received 14 January 2014 Accepted 20 March 2014 Published 7 April 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 B Chen and S Li This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper is to investigate the Schwarzian type difference equation [(Δ3
119891Δ119891) minus (32) (Δ2
119891Δ119891)
2
]
119896
= 119877 (119911 119891) =
(119875(119911 119891)119876(119911 119891)) where 119877(119911 119891) is a rational function in 119891 with polynomial coefficients 119875(119911 119891) respectively 119876(119911 119891) are twoirreducible polynomials in 119891 of degree 119901 respectively 119902 Relationship between 119901 and 119902 is studied for some special case Denote119889 = max 119901 119902 Let 119891(119911) be an admissible solution of (lowast) such that 120588
2(119891) lt 1 then for 119904 (ge2) distinct complex constants 120572
1 120572
119904
119902 + 2119896sum119904
119895=1120575(120572119895 119891) le 8119896 In particular if119873(119903 119891) = 119878(119903 119891) then 119889 + 2119896sum119904
119895=1120575(120572119895 119891) le 4119896
1 Introduction and Results
Throughout this paper a meromorphic function alwaysmeans being meromorphic in the whole complex plane and119888 always means a nonzero constant For a meromorphicfunction 119891(119911) we define its shift by 119891(119911 + 119888) and define itsdifference operators by
Δ119888119891 (119911) = 119891 (119911 + 119888) minus 119891 (119911) Δ
119899
119888119891 (119911) = Δ
119899minus1
119888(Δ119888119891 (119911))
119899 isin N 119899 ge 2
(1)
In particular Δ119899119888119891(119911) = Δ
119899
119891(119911) for the case 119888 = 1 We usestandard notations of theNevanlinna theory ofmeromorphicfunctions such as 119879(119903 119891) 119898(119903 119891) and 119873(119903 119891) and as statedin [1ndash3] For a constant 119886 we define theNevanlinna deficiencyby
120575 (119886 119891) = lim inf119903rarrinfin
119898(119903 1 (119891 minus 119886))
119879 (119903 119891)
= 1 minus lim sup119903rarrinfin
119873(119903 1 (119891 minus 119886))
119879 (119903 119891)
(2)
Recently numbers of papers (see eg [4ndash12]) are devotedto considering the complex difference equations and differ-ence analogues of Nevanlinna theory Due to some idea of[13] we consider the admissible solution of the Schwarziantype difference equation
119878119896(119891) = [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(3)
where 119877(119911 119891) is a rational function in 119891 with polynomialcoefficients 119875(119911 119891) respectively119876(119911 119891) are two irreduciblepolynomials in 119891 of degree 119901 respectively 119902 Here and in thefollowing ldquoadmissiblerdquo always means ldquotranscendentalrdquo Andwe denote 119889 = max119901 119902 from now on For the existence ofsolutions of (3) we give some examples below
Examples (1) 119891(119911) = sin 120587119911 + 119911 is an admissible solution ofthe Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus8 [1198912
+ (1 minus 2119911) 119891 + 119911 (119911 minus 1)]
41198912minus 4 (2119911 + 1) 119891 + (2119911 + 1)
2
(4)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 306360 5 pageshttpdxdoiorg1011552014306360
2 Abstract and Applied Analysis
(2) 119891(119911) = (119890119911 ln 2 sin 2120587119911) + 119911 is an admissible solutionof the Schwarzian type difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
=
minus1198912
+ 2 (119911 + 1) 119891 minus 1199112
minus 2119911
21198912minus 4 (119911 minus 1) 119891 + 2(119911 minus 1)
2 (5)
(3) Let 119891(119911) = 1199112
+ 119911 then 119891(119911) solves the Schwarziantype difference equation
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
= minus
3
2 [1198912minus 2 (119911
2minus 1) 119891 + (119911
2minus 1)2
]
(6)
This example shows that (3) may admit polynomial solutions
Considering the relationship between 119901 and 119902 in thoseexamples above we prove the following result
Theorem 1 For the Schwarzian type difference equation (3)with polynomial coefficients note the following
(i) If it admits an admissible solution 119891(119911) such that1205882(119891) lt 1 then
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (7)
In particular if119898(119903 119891) = 119878(119903 119891) then 119901 le 119902(ii) If its coefficients are all constants and it admits a
polynomial solution 119891(119911) with degree 119904 then 119904 ge 2 and119902119904 = 119901119904 + 2119896
Remark 2 From examples (1) and (2) we conjecture that119901 = 119902 inTheorem 1(i) However we cannot prove it currentlyFrom example (3) given before we see that the restriction onthe coefficients in Theorem 1(ii) cannot be omitted
For the Schwarzian differential equation
119878119896(119891) = [
119891101584010158401015840
1198911015840
minus
3
2
(
11989110158401015840
1198911015840
)
2
]
119896
= 119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
(8)
where 119877(119911 119891) 119875(119911 119891) and 119876(119911 119891) are as stated beforeIshizaki [13] proved the following result (see also Theorem932 in [2])
Theorem A (see [2 13]) Let 119891(119911) be an admissible solutionof (8) with polynomial coefficients and let 120572
1 120572
119904be 119904 (ge2)
distinct complex constants Then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (9)
For the Schwarzian type difference equation (3) we provethe following result
Theorem 3 Let 119891(119911) be an admissible solution of (3) withpolynomial coefficients such that 120588
2(119891) lt 1 and let 120572
1 120572
119904
be 119904 (ge2) distinct complex constants Then
119902 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 8119896 (10)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889 + 2119896
119904
sum
119895=1
120575 (120572119895 119891) le 4119896 (11)
Remark 4 From Theorem 1 under the condition 119873(119903 119891) =119878(119903 119891) in Theorem 3 we have 119889 = 119902 in (11) The behavior ofthe zeros and the poles of 119891(119911) in 119878
119896(119891) is essentially different
from that in the 119878119896(119891) We wonder whether the restriction
119873(119903 119891) = 119878(119903 119891) can be omitted or not
2 Lemmas
The following lemmaplays a very important role in the theoryof complex differential equations and difference equationsIt can be found in Mohonrsquoko [14] and Valiron [15] (see alsoTheorem 225 in the book of Laine and Yang [2])
Lemma 5 (see [14 15]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(12)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) such that
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 119902
(13)
and the characteristic function of 119877(119911 119891) satisfies
119879 (119903 119877 (119911 119891)) = 119889119879 (119903 119891) + 119878 (119903 119891) (14)
where 119889 = max119901 119902
The following two results can be found in [10] In factLemma 6 is a special case of Lemma 83 in [10]
Lemma 6 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891) (15)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 7 (see [10]) Let 119891(119911) be a meromorphic function ofhyper order 120588
2(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120589minus120576
) = 119878 (119903 119891) (16)
possibly outside of a set of 119903 with finite logarithmic measure
From Lemma 7 we can easily get the following conclu-sion
Abstract and Applied Analysis 3
Lemma 8 Let 119891(119911) be a meromorphic function of hyper order1205882(119891) = 120589 lt 1 119888 isin C and 120576 gt 0 Then
119898(119903
Δ119899
119888119891 (119911)
119891 (119911)
) = 119878 (119903 119891)
119898(119903
Δ119896
119888119891 (119911)
Δ119895
119888119891 (119911)
) = 119878 (119903 119891) 119896 gt 119895
(17)
possibly outside of a set of 119903 with finite logarithmic measure
Lemma 9 Let 119891 be an admissible solution of (3) withcoefficients Then using the notation 119876(119911) = 119876(119911 119891(119911))
119902119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (18)
In particular if119873(119903 119891) = 119878(119903 119891) then
119889119879 (119903 119891) + 119878 (119903 119891) le 119873(119903
1
119876
) (19)
Proof We use the idea by Ishizaki [13] (see also [2]) to proveLemma 9 It follows from Lemma 8 that
119898(119903 119877) = 119898(119903 [
Δ3
119891
Δ119891
minus
3
2
(
Δ2
119891
Δ119891
)
2
]
119896
)
le 119896119898(119903
Δ3
119891
Δ119891
) + 2119896119898(119903
Δ2
119891
Δ119891
)
+ 119878 (119903 119891) = 119878 (119903 119891)
(20)
From this and Lemma 5 we get
119889119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119877) = 119873 (119903 119877) + 119878 (119903 119891) (21)
and hence
119889119879 (119903 119891) = 119873 (119903 119877) + 119878 (119903 119891) (22)
If 119889 = 119901 gt 119902 since all coefficients of 119875(119911 119891) and 119876(119911 119891)are polynomials there are at the most finitely many poles of119877(119911 119891) neither the poles of 119891(119911) nor the zeros of 119876(119911 119891)Therefore we see that
119873(119903 119877) le (119901 minus 119902)119873 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
le (119901 minus 119902) 119879 (119903 119891) + 119873(119903
1
119876
) + 119878 (119903 119891)
(23)
We obtain (18) from this and (22) immediatelyIf 119889 = 119902 ge 119901 there are at most finitely many poles of
119877(119911 119891) not the zeros of 119876(119911 119891) then
119873(119903 119877) le 119873(119903
1
119876
) + 119878 (119903 119891) (24)
Now (18) follows from (22) and (24)Notice that if 119873(119903 119891) = 119878(119903 119891) then (24) always holds
This finishes the proof of Lemma 9
3 Proof of Theorem 1
Case 1 Equation (3) admits an admissible solution 119891(119911) suchthat 1205882(119891) lt 1 Since all coefficients of119875(119911 119891) and119876(119911 119891) are
polynomials there are at the most finitely many poles of 119891(119911)that are not the poles of119875(119911 119891) and119876(119911 119891) This implies that
119873(119903 119875) = 119901119873 (119903 119891) + 119878 (119903 119891)
119873 (119903 119876) = 119902119873 (119903 119891) + 119878 (119903 119891)
(25)
From Lemma 5 we get
119879 (119903 119875) = 119901119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 119876) = 119902119879 (119903 119891) + 119878 (119903 119891)
(26)
We can deduce from (3) (25) (26) and Lemma 8 that
119901119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119875)
= 119898 (119903 119875) + 119873 (119903 119875)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891)119876)
+ 119878 (119903 119891)
le 119901119873 (119903 119891) + 119898 (119903 119878119896(119891))
+ 119898 (119903 119876) + 119878 (119903 119891)
= 119901119873 (119903 119891) + 119879 (119903 119876) minus 119873 (119903 119876)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119879 (119903 119891) minus 119902119873 (119903 119891)
+ 119878 (119903 119891)
= 119901119873 (119903 119891) + 119902119898 (119903 119891) + 119878 (119903 119891)
(27)
It follows from this that
119901119898 (119903 119891) le 119902119898 (119903 119891) + 119878 (119903 119891) (28)
What is more is that if119898(119903 119891) = 119878(119903 119891) then we obtain from(28) that 119901 le 119902
Case 2 The coefficients of (3) are all constants and it admitsa polynomial solution 119891(119911) with degree 119904 Set
119891 (119911) = 119886119904119911119904
+ 119886119904minus1119911119904minus1
+ sdot sdot sdot + 1198861119911 + 1198860 (29)
then
119891 (119911 + 1) = 119886119904119911119904
+ 119887119904minus1119911119904minus1
+ sdot sdot sdot + 1198871119911 + 1198870 (30)
where
119887119904minus119895
= 119886119904119862119895
119904+ 119886119904minus1119862119895minus1
119904minus1+ sdot sdot sdot + 119886
119904minus119895+11198621
119904minus119895+1+ 119886119904minus119895 (31)
From (29) and (30) we obtain that
Δ119891 = 119904119886119904119911119904minus1
+ (119887119904minus2
minus 119886119904minus2) 119911119904minus2
+ sdot sdot sdot + (1198871minus 1198861) 119911 + (119887
0minus 1198860)
(32)
4 Abstract and Applied Analysis
If 119904 = 1 then Δ2119891 = Δ3119891 equiv 0 which yields that 119875(119911 119891) equiv0 That is a contradiction to our assumption Thus 119904 ge 2
If 119904 = 2 thenΔ119891 = 21198862119911+1198862+1198861Δ2119891 = 2119886
2 andΔ3119891 equiv 0
Now from (3) we get
(minus3)119896
119876 (119911 119891) (Δ2
119891)
2119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(33)
Considering degrees of both sides of the equation above wecan see that 119902 = 119901 + 119896
If 119904 ge 3 we can deduce similarly that
Δ2
119891 = 119904 (119904 minus 1) 119886119904119911119904minus2
+ 1198751(119911)
Δ3
119891 = 119904 (119904 minus 1) (119904 minus 2) 119886119904119911119904minus3
+ 1198752(119911)
(34)
where 1198751(119911) 1198752(119911) are polynomials such that deg119875
1le 119904 minus
3 deg1198752le 119904 minus 4
Rewrite (3) as follows
119876 (119911 119891) [2Δ3
119891 sdot Δ119891 minus 3(Δ2
119891)
2
]
119896
= 2119896
119875 (119911 119891) (Δ119891)2119896
(35)
From (34) we find that the leading coefficient of 2Δ3119891 sdotΔ119891 minus 3(Δ
2
119891)
2 is
minus1198862
1199041199042
(119904 minus 1) (119904 + 1) = 0 (36)
Considering degrees of both sides of (35) we prove that119902119904 = 119901119904 + 2119896
4 Proof of Theorem 3
Firstly we consider the general case Asmentioned inRemark1 in [13] due to Jank and Volkmann [16] if (3) admits anadmissible solution then there are at most 119878(119903 119891) commonzeros of 119875(119911 119891) and 119876(119911 119891) Since all coefficients of 119876(119911 119891)are polynomials there are at the most finitely many poles of119891 that are the zeros of 119876(119911 119891) Therefore from (3) we have
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119879 (119903 119891 (119911 + 1) minus 119891 (119911)) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119878 (119903 119891)
(37)
Combining this and Lemma 9 applying the second maintheorem we get
119902
2119896
119879 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
119902
2119896
119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
)
le
1
2119896
119873(119903
1
119876
) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 2119879 (119903 119891) + 119898 (119903 119891) +
119904
sum
119895=1
119898(119903
1
119891 minus 120572119895
) + 119878 (119903 119891)
le 4119879 (119903 119891) + 119878 (119903 119891)
(38)
Thus we prove that (10) holdsSecondly we consider the case that 119873(119903 119891) = 119878(119903 119891)
From (3) and Lemma 8 we similarly get that
1
2119896
119873(119903
1
119876
) le 119873(119903
1
Δ119891
) + 119878 (119903 119891) le 119879 (119903 Δ119891) + 119878 (119903 119891)
= 119898 (119903 Δ119891) + 119873 (119903 Δ119891) + 119878 (119903 119891)
le 119898(119903
Δ119891
119891
) + 119898 (119903 119891) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
(39)
From this and applying Lemma 9with (19) as arguing beforewe can prove that (11) holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the referees for their valuablesuggestions This work is supported by the NNSFC (nos11226091 and 11301091) the Guangdong Natural ScienceFoundation (no S2013040014347) and the Foundation forDistinguished Young Talents in Higher Education of Guang-dong (no 2013LYM 0037)
References
[1] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C-C Yang and H-X Yi Uniqueness Theory of MeromorphicFunctions vol 557 ofMathematics and Its Applications KluwerAcademic PublishersGroupDordrechtTheNetherlands 2003
[4] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[5] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[6] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[7] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
Abstract and Applied Analysis 5
[8] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[9] R G Halburd and R J Korhonen ldquoExistence of finite-ordermeromorphic solutions as a detector of integrability in differ-ence equationsrdquo Physica D Nonlinear Phenomena vol 218 no2 pp 191ndash203 2006
[10] R G Halburd R J Korhonen and K Tohge ldquoHolomorphiccurves with shift-invariant hyperplane preimagesrdquo submitted toTransactions of the American Mathematical Society httparxivorgabs09033236
[11] J Heittokangas R Korhonen I Laine J Rieppo and KTohge ldquoComplex difference equations of malmquist typerdquoComputational Methods and Function Theory vol 1 no 1 pp27ndash39 2001
[12] I Laine and C-C Yang ldquoClunie theorems for difference and119902-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] K Ishizaki ldquoAdmissible solutions of the Schwarzian differentialequationrdquoAustralianMathematical SocietyA PureMathematicsand Statistics vol 50 no 2 pp 258ndash278 1991
[14] A Z Mohonrsquoko ldquoThe nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija no 14 pp 83ndash87 1971 (Russian)
[15] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Entomologique de France vol 59 pp 17ndash39 1931
[16] G Jank and L VolkmannMeromorphe Funktionen und Differ-entialgeichungen Birkhauser Verlag Basel Switzerland 1985
Research ArticleStatistical Inference for Stochastic DifferentialEquations with Small Noises
Liang Shen12 and Qingsong Xu1
1 School of Mathematics and Statistics Central South University Changsha Hunan 410075 China2 School of Science Linyi University Linyi Shandong 276005 China
Correspondence should be addressed to Qingsong Xu csuqingsongxu126com
Received 13 November 2013 Accepted 11 February 2014 Published 13 March 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 L Shen and Q Xu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper proposes the least squares method to estimate the drift parameter for the stochastic differential equations driven bysmall noises which is more general than pure jump 120572-stable noises The asymptotic property of this least squares estimator isstudied under some regularity conditions The asymptotic distribution of the estimator is shown to be the convolution of a stabledistribution and a normal distribution which is completely different from the classical cases
1 Introduction
Stochastic differential equations (SDEs) are being extensivelyused as a model to describe some phenomena which aresubject to random influences it has found many applicationsin biology [1] medicine [2] econometrics [3 4] finance[5] geophysics [6] and oceanography [7] Then statisticalinference for these differential equations was of great interestand became a challenging theoretical problem For a morerecent comprehensive discussion we refer to [8 9]
The asymptotic theory of parametric estimation for dif-fusion processes with small white noise based on continuoustime observations is well developed and it has been studiedby many authors (see eg [10ndash14]) There have been manyapplications of small noise in mathematical finance see forexample [15ndash18]
In parametric inference due to the impossibility ofobserving diffusions continuously throughout a time intervalit is more practical and interesting to consider asymptoticestimation for diffusion processes with small noise based ondiscrete observations There are many approaches to driftestimation for discretely observed diffusions (see eg [19ndash23]) Long [24] has started the study on parameter estimationfor a class of stochastic differential equations driven by smallstable noise 119885
119905 119905 ge 0 However there has been no study on
parametric inference for stochastic processes with small Levynoises yet
In this paper we are interested in the study of parameterestimation for the following stochastic differential equationsdriven by more general Levy noise 119871
119905 119905 ge 0 based
on discrete observations We will employ the least squaresmethod to obtain an asymptotically consistent estimator
Let (ΩF F119905ge0
P) be a basic complete filtered prob-ability space satisfying the usual conditions that is thefiltration is continuous on the right and F
0contains all
P-null sets In this paper we consider a class of stochasticdifferential equations as follows
119889119883119905
= 120579119891 (119883119905) 119889119905 + 120576119892 (119883
119905minus
) 119889119871119905 119905 isin [0 1]
119871119905
= 119886119861119905+ 119887119885119905
119883 (0) = 1199090
(1)
where 119891 R rarr R and 119892 R rarr R are known functionsand 119886 119887 are known constants Let 119861
119905 119905 ge 0 be a standard
Brownian motion and let 119885119905 119905 ge 0 be a standard 120572-stable
Levy motion independent of 119861119905 119905 ge 0 with 119885
1sim 119878120572(1 120573 0)
for 120573 isin [0 1] 1 lt 120572 lt 2Let 119883 = 119883
119905 119905 ge 0 be a real-valued stationary process
satisfying the stochastic differential equation (1) and weassume that this process is observed at regularly spaced timepoints 119905
119894= 119894119899 119894 = 1 2 119899 Assume 119883
0
119905is the solution of
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 473681 6 pageshttpdxdoiorg1011552014473681
2 Abstract and Applied Analysis
the underlying ordinary differential equation (ODE) with thetrue value of the drift parameter 120579
0
1198891198830
119905= 1205790119891 (1198830
119905) 119889119905 119883
0
0= 1199090 (2)
Then we get
119883119905119894
minus 119883119905119894minus1
= int
119905119894
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (3)
2 Preliminaries
In this paper we denote 119862 as a generic constant whose valuemay vary from place to place
The following regularity conditions are assumed to hold
(A1) The functions 119891(119909) and 119892(119909) satisfy the Lipschitzconditions that is there exists a constant 119871 gt 0 suchthat
1003816100381610038161003816119891 (119909) minus 119891 (119910)
1003816100381610038161003816+
1003816100381610038161003816119892 (119909) minus 119892 (119910)
1003816100381610038161003816
le 1198711003816100381610038161003816119909 minus 119910
1003816100381610038161003816 119909 119910 isin R
(4)
(A2) There exist constants 119872 gt 0 and 119903 ge 0 satisfying
the growth condition
119892minus2
(119909) le 119872 (1 + |119909|119903
) 119909 isin R (5)
(A3) There exists a positive constant 119873 gt 0 such that
0 lt |119892(119909)| le 119873 lt infin
(A4) For 119862
119903= 2119903minus1
or 1 119903 gt 0
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
le 119862119903(
100381610038161003816100381610038161198830
119905119894minus1
10038161003816100381610038161003816
119903
+
10038161003816100381610038161003816119883119905119894minus1
minus 1198830
119905119894minus1
10038161003816100381610038161003816
119903
) (6)
The LSE of 120579119899120576
is defined as
120579119899120576
= argmin120579
120588119899120576
(120579) (7)
where the contrast function
120588119899120576
(120579) =
119899
sum
119894=1
10038161003816100381610038161003816100381610038161003816100381610038161003816
119883119905119894
minus 119883119905119894minus1
minus 120579119891 (119883119905119894minus1
) Δ119905119894minus1
120576119892 (119883119905119894minus1
)
10038161003816100381610038161003816100381610038161003816100381610038161003816
2
(8)
Then the 120579119899120576
can be represented explicitly as follows
120579119899120576
=
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) (119883119905119894
minus 119883119905119894minus1
)
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
(9)
Based on (3) and (9) there is a special decomposition for 120579119899120576
120579119899120576
=
1205790
sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119891 (119883119904) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119871119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
=1205790+
1205790sum119899
119894=1119892minus2
(119883119905119894minus1
)119891 (119883119905119894minus1
)int
119905119894
119905119894minus1
(119891 (119883119904)minus119891 (119883
119905119894minus1
)) 119889119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119887120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119885119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
+
119886120576 sum119899
119894=1119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
119899minus1
sum119899
119894=1119892minus2
(119883119905119894minus1
) 1198912
(119883119905119894minus1
)
= 1205790
+
Φ2
(119899 120576)
Φ1
(119899 120576)
+
Φ3
(119899 120576)
Φ1
(119899 120576)
+
Φ4
(119899 120576)
Φ1
(119899 120576)
(10)
Now we give an explicit expression for 120576minus1
(120579119899120576
minus 1205790) By using
(10) we have
120576minus1
(120579119899120576
minus 1205790) =
120576minus1
Φ2
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ3
(119899 120576)
Φ1
(119899 120576)
+
120576minus1
Φ4
(119899 120576)
Φ1
(119899 120576)
=
Ψ2
(119899 120576)
Φ1
(119899 120576)
+
Ψ3
(119899 120576)
Φ1
(119899 120576)
+
Ψ4
(119899 120576)
Φ1
(119899 120576)
(11)
One of the important tools we will employ is the under-lying lemma (see (35) in the Lemma 32 of [24])
Lemma 1 Under conditions (A1)-(A2) one has
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816
le 120576119890119871|1205790|119905
10038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119883119904minus
) 119889119885119904
10038161003816100381610038161003816100381610038161003816
(12)
sup0le119905le1
10038161003816100381610038161003816119883119905minus 1198830
119905
10038161003816100381610038161003816997888rarr1198750 as 120576 997888rarr 0 (13)
Abstract and Applied Analysis 3
3 Asymptotic Property of the LeastSquares Estimator
Theorem 2 Under the conditions (A1)ndash(A4) as 119899 rarr
infin 120576 rarr 0 119899120576 rarr infin and 119899120576120572(120572minus1)
rarr infin one has
120576minus1
(120579119899120576
minus 1205790)
997904rArr 119886
(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904
119873
+ 119887 (((int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus (int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802)
times (int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
minus1
)
(14)
where 1198801and 119880
2are independent random variables with 120572-
stable distribution 119878120572(1 120573 0) and 119873 is an independent random
variable with standard normal distribution
The theoremwill be proved by establishing several propo-sitionsWewill consider the asymptotic behaviors ofΦ
1(119899 120576)
Ψ119894(119899 120576) 119894 = 2 3 4 respectively
Proposition 3 Under conditions (A1)ndash(A4) and 119899 rarr infin
120576 rarr 0 one has
Φ1
(119899 120576) 997888rarr119875
int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904 (15)
Proof Under conditions (A1)ndash(A3) Proposition 3 can be
proved by using condition (A4) (see the proof of Proposition
33 in [24])
Proposition 4 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576 rarr infin one has
Ψ2
(119899 120576) 997888rarr1198750 (16)
Proof For 119905119894minus1
le 119905 le 119905119894 119894 = 1 2 119899
119883119905
= 119883119905119894minus1
+ int
119905
119905119894minus1
1205790119891 (119883119904) 119889119904 + 120576 int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904 (17)
It follows that10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le int
119905
119905119894minus1
10038161003816100381610038161205790
1003816100381610038161003816(
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816) 119889119904
+ 120576
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119871119904
100381610038161003816100381610038161003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816119872 int
119905
119905119894minus1
10038161003816100381610038161003816119891 (119883119904) minus 119891 (119883
119905119894minus1
)
10038161003816100381610038161003816+ 119899minus1 1003816
1003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+ 119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
(18)
Using Gronwall inequality we get10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872(119905minus119905
119894minus1)
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(19)
which yields
sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
le 119890|1205790|119872119899
[
10038161003816100381610038161205790
1003816100381610038161003816
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
119899
+ 119886120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
+119887120576 sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
]
(20)
thus under conditions (A1) and (A
3)
1003816100381610038161003816Φ2
(119899 120576)1003816100381610038161003816
le10038161003816100381610038161205790
1003816100381610038161003816
119899
sum
119894=1
119872 (1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
(119891 (119883119904) minus 119891 (119883
119905119894minus1
)) 119889119904
100381610038161003816100381610038161003816100381610038161003816
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
119899
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119905minus 119883119905119894minus1
10038161003816100381610038161003816
4 Abstract and Applied Analysis
le
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
1198992
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
2
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890(|1205790|119872)119899
119899
119887120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119885119904
100381610038161003816100381610038161003816100381610038161003816
+
11987211987010038161003816100381610038161205790
1003816100381610038161003816
2
119890|1205790|119872119899
119899
119886120576
119899
sum
119894=1
(1 +
10038161003816100381610038161003816119883119905119894minus1
10038161003816100381610038161003816
119903
)
10038161003816100381610038161003816119891 (119883119905119894minus1
)
10038161003816100381610038161003816
times sup119905119894minus1le119905le119905119894
100381610038161003816100381610038161003816100381610038161003816
int
119905
119905119894minus1
119892 (119883119904minus
) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
= Φ21
(119899 120576) + Φ22
(119899 120576) + Φ23
(119899 120576)
(21)
Then
1003816100381610038161003816Ψ2
(119899 120576)1003816100381610038161003816
le 120576minus1
Φ21
(119899 120576) + 120576minus1
Φ22
(119899 120576)
+ 120576minus1
Φ23
(119899 120576)
= Ψ21
(119899 120576) + Ψ22
(119899 120576) + Ψ23
(119899 120576)
(22)
Using (13) in Lemma 1 conditions (A1) and (A
4) we get
Ψ21
(119899 120576) rarr1198750 as 119899 rarr infin 120576 rarr 0 and 119899120576 rarr infin (see (326)
in [24]) By using the same techniques under condition (A2)
we can prove thatΨ2119895
(119899 120576) rarr1198750 119895 = 2 3 as 119899 rarr infin 120576 rarr 0
respectively
Proposition 5 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 and 119899120576120572(120572minus1)
rarr infin one has
Ψ3
(119899 120576)
997904rArr 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
+
119889119904)
1120572
1198801
minus 119887(int
1
0
10038161003816100381610038161003816119892 (1198830
119904)
10038161003816100381610038161003816
minus2120572
(119891 (1198830
119904) 119892 (119883
0
119904))
120572
minus
119889119904)
1120572
1198802
(23)
Proof Under conditions (A1)ndash(A3) Proposition 5 can be
proved by using condition (A4) (see the proof of Proposition
44 in [24])
Proposition 6 Under conditions (A1)ndash(A4) as 119899 rarr infin
120576 rarr 0 one has
Ψ4
(119899 120576) 997904rArr 119886(int
1
0
10038161003816100381610038161003816119892minus2
(1198830
119904)
100381610038161003816100381610038161198912
(1198830
119904) 119889119904)
12
119873 (24)
Proof Note that
Ψ4
(119899 120576) = 119886
119899
sum
119894=1
119892minus2
(119883119905119894minus1
) 119891 (119883119905119894minus1
) int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) int
119905119894
119905119894minus1
119892 (1198830
119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
+ 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
)) 119891 (1198830
119905119894minus1
)
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
+ 119886
119899
sum
119894=1
(119892minus2
(119883119905119894minus1
) minus 119892minus2
(1198830
119905119894minus1
))
times (119891 (119883119905119894minus1
) minus 119891 (1198830
119905119894minus1
))
times int
119905119894
119905119894minus1
119892 (119883119904minus
) 119889119861119904
=
5
sum
119895=1
Ψ4119895
(119899 120576)
(25)
For Ψ41
(119899 120576) let 119884119894
= int
119905119894
119905119894minus1
119892(119883119904minus
)119889119861119904 119895 = 1 119899 Then it is
easy to see that 119884119894
sim 119873(0 int
119905119894
119905119894minus1
1198922
(119883119904minus
)119889119904) and 1198841 119884
119899are
independent normal random variablesIt follows that
Ψ41
(119899 120576)
= 119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
) 119891 (1198830
119905119894minus1
) 119884119894
sim 119873 (0 1198862
119899
sum
119894=1
119892minus4
(1198830
119905119894minus1
) 1198912
(1198830
119905119894minus1
) int
119905119894
119905119894minus1
1198922
(119883119904minus
) 119889119904)
997904rArr 119886(int
1
0
119892minus2
(1198830
119904) 1198912
(1198830
119904) 119889119904)
12
119873
(26)
as 119899 rarr infin 120576 rarr 0
Abstract and Applied Analysis 5
For Ψ42
(119899 120576) using Markov inequality and Itorsquos isometryproperty for any given 120578 gt 0
Ψ42
(119899 120576)
le
1
120578
E[119886
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times
100381610038161003816100381610038161003816100381610038161003816
int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
)) 119889119861119904
100381610038161003816100381610038161003816100381610038161003816
]
le
119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119892 (119883119904minus
) minus 119892 (1198830
119904minus
))
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [int
119905119894
119905119894minus1
(119883119904minus
minus 1198830
119904minus
)
2
119889119904]
12
le
119871119886
120578
119899
sum
119894=1
119892minus2
(1198830
119905119894minus1
)
10038161003816100381610038161003816119891 (1198830
119905119894minus1
)
10038161003816100381610038161003816
times [ sup119905119894minus1le119905le119905119894
10038161003816100381610038161003816119883119904minus
minus 1198830
119904minus
10038161003816100381610038161003816119899minus12
]
(27)
By using (13) Ψ42
(119899 120576) rarr 0 as 119899 rarr infin 120576 rarr 0Applying similar techniques to Ψ
4119895(119899 120576) 119895 = 3 4 5 we
get Ψ4119895
(119899 120576) rarr 0 119895 = 3 4 5 as 119899 rarr infin 120576 rarr 0
Now we can proveTheorem 2
Proof By using Propositions 3 4 5 6 and Slutskyrsquos theoremwe can get the conclusion
4 Example
We consider the following nonlinear SDE driven by generalLevy noises
119889119883119905
= 120579119883119905119889119905 +
120576
1 + 1198832
119905minus
119889119871119905 119905 isin [0 1] 119883
0= 1199090
(28)
where 119891(119909) = 119909 119892(119909) = 1(1 + 1199092
) 1199090and 120576 are known
constants and 120579 = 0 is an unknown parameterFor simplicity let 119909
0gt 0 120576 = 0 we get the ODE
1198891198830
119905= 12057901198830
119905119889119905 119905 isin [0 1] 119883
0
0= 1199090
(29)
and the solution
1198830
119905= 11990901198901205790119905
(30)
Then the asymptotic distribution is
119886(int
1
0
(1 + 1199092
011989021205790119904
)
2
1199092
011989021205790119904
119889119904)
minus12
119873
+ 119887
(int
1
0
(1 + 1199092
011989021205790119904
)
120572
119909120572
01198901205721205790119904
119889119904)
1120572
int
1
0
(1 + 1199092
011989021205790119904)2
1199092
011989021205790119904119889119904
119878120572
(1 120573 0)
(31)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R I Jennrich and P B Bright ldquoFitting systems of linear dif-ferential equations using computer generated exact derivativesrdquoTechnometrics vol 18 no 4 pp 385ndash392 1976
[2] R H Jones ldquoFitting multivariate models to unequally spaceddatardquo in Time Series Analysis of Irregularly Observed Data pp158ndash188 1984
[3] A R Bergstrom Statistical Inference in Continuous TimeEconomic Models vol 99 North-Holland Amsterdam TheNetherlands 1976
[4] A R Bergstrom ldquoThe history of continuous-time econometricmodelsrdquo Econometric Theory vol 4 no 3 pp 365ndash383 1988
[5] F Black and M Scholes ldquoThe pricing of options and corporateliabilitiesrdquoThe Journal of Political Economy vol 81 pp 637ndash6541973
[6] M Arato Linear Stochastic Systems with Constant CoefficientsSpringer Berlin Germany 1982
[7] R J Adler and P Meuller Stochastic Modelling on PhysicalOceanography vol 39 Springer New York NY USA 1996
[8] B P Rao B L P Rao I Statisticien B L P Rao B L P Rao andI Statistician Statistical Inference for di Usion Type ProcessesArnold London UK 1999
[9] Y A Kutoyants Statistical Inference for Ergodic Diffusion Pro-cesses Springer London UK 2004
[10] Yu A Kutoyants Parameter Estimation for Stochastic Processesvol 6 Heldermann Berlin Germany 1984
[11] Yu Kutoyants Identification of Dynamical Systems with SmallNoise Kluwer Academic Dordrecht The Netherlands 1994
[12] N Yoshida ldquoAsymptotic expansions of maximum likelihoodestimators for small diffusions via the theory of Malliavin-Watanaberdquo Probability Theory and Related Fields vol 92 no 3pp 275ndash311 1992
[13] N Yoshida ldquoConditional expansions and their applicationsrdquoStochastic Processes and their Applications vol 107 no 1 pp 53ndash81 2003
[14] M Uchida and N Yoshida ldquoInformation criteria for smalldiffusions via the theory of Malliavin-Watanaberdquo StatisticalInference for Stochastic Processes vol 7 no 1 pp 35ndash67 2004
[15] N Yoshida ldquoAsymptotic expansion of Bayes estimators for smalldiffusionsrdquo Probability Theory and Related Fields vol 95 no 4pp 429ndash450 1993
[16] A Takahashi ldquoAn asymptotic expansion approach to pricingfinancial contingent claimsrdquoAsia-Pacific FinancialMarkets vol6 no 2 pp 115ndash151 1999
6 Abstract and Applied Analysis
[17] N Kunitomo and A Takahashi ldquoThe asymptotic expansionapproach to the valuation of interest rate contingent claimsrdquoMathematical Finance vol 11 no 1 pp 117ndash151 2001
[18] A Takahashi and N Yoshida ldquoAn asymptotic expansionscheme for optimal investment problemsrdquo Statistical Inferencefor Stochastic Processes vol 7 no 2 pp 153ndash188 2004
[19] V Genon-Catalot ldquoMaximum contrast estimation for diffusionprocesses fromdiscrete observationsrdquo Statistics vol 21 no 1 pp99ndash116 1990
[20] A Gloter and M Soslashrensen ldquoEstimation for stochastic differ-ential equations with a small diffusion coefficientrdquo StochasticProcesses and their Applications vol 119 no 3 pp 679ndash6992009
[21] C F Laredo ldquoA sufficient condition for asymptotic sufficiencyof incomplete observations of a diffusion processrdquo The Annalsof Statistics vol 18 no 3 pp 1158ndash1171 1990
[22] M Uchida ldquoApproximate martingale estimating functions forstochastic differential equations with small noisesrdquo StochasticProcesses and their Applications vol 118 no 9 pp 1706ndash17212008
[23] M Soslashrensen and M Uchida ldquoSmall-diffusion asymptotics fordiscretely sampled stochastic differential equationsrdquo Bernoullivol 9 no 6 pp 1051ndash1069 2003
[24] H Long ldquoParameter estimation for a class of stochastic dif-ferential equations driven by small stable noises from discreteobservationsrdquo Acta Mathematica Scientia B vol 30 no 3 pp645ndash663 2010
Research ArticleOn the Deficiencies of Some Differential-Difference Polynomials
Xiu-Min Zheng1 and Hong Yan Xu2
1 Institute of Mathematics and Information Science Jiangxi Normal University Nanchang 330022 China2Department of Informatics and Engineering Jingdezhen Ceramic Institute Jingdezhen 333403 China
Correspondence should be addressed to Xiu-Min Zheng zhengxiumin2008sinacom
Received 1 November 2013 Accepted 4 January 2014 Published 27 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 X-M Zheng and H Y Xu This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
The characteristic functions of differential-difference polynomials are investigated and the result can be viewed as a differential-difference analogue of the classic Valiron-Mokhonrsquoko Theorem in some sense and applied to investigate the deficiencies of somehomogeneous or nonhomogeneous differential-difference polynomials Some special differential-difference polynomials are alsoinvestigated and these results on the value distribution can be viewed as differential-difference analogues of some classic results ofHayman and Yang Examples are given to illustrate our results at the end of this paper
1 Introduction
Throughout this paper we use standard notations in theNevanlinna theory (see eg [1ndash3]) Let 119891(119911) be a meromor-phic function Here and in the following the word ldquomero-morphicrdquo means being meromorphic in the whole complexplane We use normal notations 119898(119903 119891) 119879(119903 119891) 119873(119903 119891)119873(119903 1119891) 120590(119891) 120582(119891) and 120582(1119891) And we also use 120590
2(119891)
to denote the hyperorder of 119891(119911) and 120575(120572 119891) to denote theNevanlinna deficiency of 120572 with respect to 119891(119911) Moreoverwe denote by 119878(119903 119891) any real quantity satisfying 119878(119903 119891) =119900(119879(119903 119891)) as 119903 rarr infin outside of a possible exceptional set offinite logarithmic measure
Recently with some establishments of difference ana-logues of the classic Nevanlinna theory (two typical andmost important ones can be seen in [4ndash6]) there has beena renewed interest in the properties of complex differenceexpressions and meromorphic solutions of complex differ-ence equations (see eg [4ndash17]) By combining complex dif-ferentiates and complex differences we proceed in this way inthis paper
It is well known that the following Valiron-MokhonrsquokoTheorem due to Valiron [18] and A Z Mokhonrsquoko and V DMokhonrsquoko [19] is of essential importance in the theory ofcomplex differential equations and functional equations
Theorem A (see [2 3]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891 (119911)) =
sum119898
119894=0119886119894(119911) 119891(119911)
119894
sum119899
119895=0119887119895(119911) 119891(119911)
119895
(1)
with meromorphic coefficients 119886119894(119911) 119887119895(119911) the characteristic
function of 119877(119911 119891(119911)) satisfies
119879 (119903 119877 (119911 119891 (119911))) = 119889119879 (119903 119891) + 119874 (Ψ (119903)) (2)
where 119889 = max119898 119899 and Ψ(119903) = max119894119895119879(119903 119886
119894) 119879(119903 119887
119895)
Noting that the difference analogue of Theorem A maynot hold we have obtained a result of this type in [16] byadding some additional assumptions as follows
Theorem B (see [16]) Suppose that 119875(119911 119891) is a differencepolynomial of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
1198940119891(119911 + 119888
1)1198941sdot sdot sdot 119891(119911 + 119888
119899)119894119899
(3)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 378151 12 pageshttpdxdoiorg1011552014378151
2 Abstract and Applied Analysis
containing just one monomial of degree 119889(119875) and 119891(119911) is atranscendental meromorphic function of finite order If 119891(119911)also satisfies119873(119903 119891) + 119873(119903 1119891) = 119878(119903 119891) then we have
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (4)
In this paper we consider removing the assumptionldquo119875(119911 119891) contains just one monomial of degree 119889(119875)rdquo in The-orem B and obtain a weaker result which is also generalizedinto differential-difference case The concrete result can beseen in Section 2
Next we recall a classic result concerning Picardrsquos valuesofmeromorphic functions and its derivatives due toHayman[20]
Theorem C (see [20]) Let 119891(119911) be a transcendental entirefunction Then
(a) for 119899 ge 3 and 119886 = 0 Ψ(119911) = 1198911015840(119911) minus 119886(119891(119911))119899 assumesall finite values infinitely often
(b) for 119899 ge 2Φ(119911) = 1198911015840(119911)(119891(119911))119899 assumes all finite valuesexcept possibly zero infinitely often
Corresponding difference analogues ofTheoremC can beseen in [12 17]
Theorem D (see [12 17]) Let 119891(119911) be a transcendental entirefunction of finite order and let 119888 be a nonzero complex constantThen
(a) for 119899 ge 3 and 119886 = 0Ψ1(119911) = 119891(119911+ 119888) minus 119886119891(119911)
119899 assumesall finite complex values infinitely often
(b) for 119899 ge 2 Φ1(119911) = 119891(119911 + 119888)119891(119911)
119899 assumes all finitecomplex values except possibly zero infinitely often
AfterTheoremCmany results have been obtained on thevalue distribution of differential polynomials A typical one isas follows
TheoremE (see [21 22]) Let119891 be a transcendental meromor-phic function with119873(119903 119891) +119873(119903 1119891) = 119878(119903 119891) and let Ψ bea differential polynomial in 119891 of the form
Ψ (119911) = sum119886 (119911) 119891(119911)11989701198911015840
(119911)1198971sdot sdot sdot 119891(119896)
(119911)119897119896 (5)
with no constant term Furthermore assume the degree 119899 ofΨ is greater than one and 119897
0lt 119899 0 le 119897
119894le 119899 for all 119894 = 0 Then
120575(119886 Ψ) lt 1 for all 119886 = 0infin Moreover if all the terms ofΨ havedifferent degrees at least two that is Ψ is nonhomogeneousthen 120575(119886 Ψ) le 1 minus (12119899) for all 119886 =infin
We also consider deficiencies of difference polynomialsof meromorphic functions of finite order in [16] which canbe viewed as difference analogues of Theorem E as well asgeneralizations of Theorem D
In this paper we proceed to investigate deficiencies ofdifferential-difference polynomials of meromorphic func-tions The concrete results can be seen in Section 3
Examples are given in Section 4 to illustrate our results
2 A Differential-Difference Analogue ofValiron-Mokhonrsquoko Theorem
In what follows we will consider differential-difference poly-nomials A differential-difference polynomial is a polynomialin 119891(119911) its shifts its derivatives and derivatives of its shifts(see [14]) that is an expression of the form
119875 (119911 119891) = sum
120582isin119868
119886120582(119911) 119891(119911)
120582001198911015840
(119911)12058201sdot sdot sdot 119891(119898)
(119911)1205820119898
times 119891(119911 + 1198881)120582101198911015840
(119911 + 1198881)12058211sdot sdot sdot 119891(119898)
(119911 + 1198881)1205821119898
sdot sdot sdot 119891(119911 + 119888119899)12058211989901198911015840
(119911 + 119888119899)1205821198991sdot sdot sdot 119891(119898)
(119911 + 119888119899)120582119899119898
= sum
120582isin119868
119886120582(119911)
119899
prod
119894=0
119898
prod
119895=0
119891(119895)
(119911 + 119888119894)120582119894119895
(6)
where 119868 is a finite set of multi-indices 120582 =
(12058200 120582
0119898 12058210 120582
1119898 120582
1198990 120582
119899119898) and 119888
0(= 0)
and 1198881 119888
119899are distinct complex constants And we assume
that the meromorphic coefficients 119886120582(119911) 120582 isin 119868 of 119875(119911 119891) are
of growth 119878(119903 119891) We denote the degree of the monomialprod119899
119894=0prod119898
119895=0119891(119895)
(119911 + 119888119894)120582119894119895 of 119875(119911 119891) by 119889(120582) = sum119899
119894=0sum119898
119895=0120582119894119895
Then we denote the degree and the lower degree of 119875(119911 119891)by
119889 (119875) = max120582isin119868
119889 (120582) 119889lowast
(119875) = min120582isin119868
119889 (120582) (7)
respectively In particular we call 119875(119911 119891) a homogeneousdifferential-difference polynomial if 119889(119875) = 119889
lowast
(119875) Other-wise 119875(119911 119891) is nonhomogeneous
In the following we assume 119889(119875) ge 1 and 119875(119911 119891) equiv
119875(119911 0)We prove a weaker differential-difference version of the
classic Valiron-MokhonrsquokoTheorem as follows
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic function and 119875(119911 119891) is a differential-difference polyno-mial of the form (6) If 119891(119911) also satisfies 120590
2(119891) lt 1 and
119873(119903 119891) + 119873(119903
1
119891
) = 119878 (119903 119891) (8)
then one has
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891))
le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
(9)
Remark 2 If119875(119911 119891) is a homogeneous differential-differencepolynomial in addition then
119879 (119903 119875 (119911 119891)) = 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (10)
Remark 3 Especially assumption (8) can be replaced by theassumption ldquomax120582(119891) 120582(1119891) lt 120590(119891)rdquo In fact if 119891(119911)satisfies max120582(119891) 120582(1119891) lt 120590(119891) then 119891(119911) is of regulargrowth and (8) holds consequently
Abstract and Applied Analysis 3
To proveTheorem 1 we need the following lemmas
Lemma 4 (see [6]) Let 119891(119911) be a nonconstant meromorphicfunction 120576 gt 0 and 119888 isin C If 120577 = 120590
2(119891) lt 1 then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119900(
119879 (119903 119891)
1199031minus120577minus120576
) (11)
for all 119903 outside of a set of finite logarithmic measure
Lemma 5 (see [6]) Let 119879 [0 +infin) rarr [0 +infin) be anondecreasing continuous function and let 119904 isin (0 +infin)If the hyperorder of 119879 is strictly less than one that islim119903rarrinfin
(log2119879(119903) log 119903) = 120577 lt 1 and 120575 isin (0 1 minus 120577) then
119879 (119903 + 119904) = 119879 (119903) + 119900 (
119879 (119903)
119903120575
) (12)
where 119903 runs to infinity outside of a set of finite logarithmicmeasure
It is shown in [23 p66] and [7 Lemma 1] that the ine-quality
(1 + 119900 (1)) 119879 (119903 minus |119888| 119891) le 119879 (119903 119891 (119911 + 119888))
le (1 + 119900 (1)) 119879 (119903 + |119888| 119891)
(13)
holds for 119888 = 0 and 119903 rarr infin And from the proof theabove relation is also true for counting function By combingLemma 5 and these inequalities we immediately deduce thefollowing lemma
Lemma 6 Let 119891(119911) be a nonconstant meromorphic functionof 1205902(119891) lt 1 and let 119888 be a nonzero complex constant Then
one has
119879 (119903 119891 (119911 + 119888)) = 119879 (119903 119891) + 119878 (119903 119891)
119873 (119903 119891 (119911 + 119888)) = 119873 (119903 119891) + 119878 (119903 119891)
119873(119903
1
119891 (119911 + 119888)
) = 119873(119903
1
119891
) + 119878 (119903 119891)
(14)
Lemma7 Let119891(119911) be a transcendentalmeromorphic functionof 1205902(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then we one has
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891) + 119878 (119903 119891) (15)
Furthermore if 119891(119911) also satisfies
119873(119903 119891) = 119878 (119903 119891) (16)
then one has
119879 (119903 119875 (119911 119891)) le 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891) (17)
Proof For 119894 = 0 1 119899 119895 = 0 1 119898 we define 119892119894119895(119911) =
119891(119895)
(119911 + 119888119894)119891(119911) We also define
119892lowast
119894119895(119911) =
119892119894119895(119911) if 1003816100381610038161003816
1003816119892119894119895(119911)
10038161003816100381610038161003816gt 1
1 if 10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816le 1
119891lowast
(119911) =
119891 (119911) if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119891 (119911)
1003816100381610038161003816le 1
(18)
Thus
1003816100381610038161003816119875 (119911 119891)
1003816100381610038161003816le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119891 (119911)
1003816100381610038161003816
119889(120582)
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
120582119894119895
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(120582)
) |119891lowast
(119911)|119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)(
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
1003816100381610038161003816119891lowast
(119911)1003816100381610038161003816)
119889(119875)
(19)
By the definitions of 119891lowast(119911) and 119892lowast119894119895(119911) 119894 = 0 1 119899 119895 =
0 1 119898 we have
119898(119903 119891lowast
) = 119898 (119903 119891)
119898 (119903 119892lowast
119894119895) = 119898 (119903 119892
119894119895) 119894 = 0 119899 119895 = 0 119898
(20)
It follows by (19) and (20) that
119898(119903 119875 (119911 119891)) le 119889 (119875)119898 (119903 119891lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
= 119889 (119875)119898 (119903 119891)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892119894119895) + 119878 (119903 119891)
(21)
Lemmas 4 and 6 and the logarithmic derivative lemma implythat for 119894 = 0 1 119899 and 119895 = 0 1 119898
119898(119903 119892119894119895) = 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911)
)
le 119898(119903
119891(119895)
(119911 + 119888119894)
119891 (119911 + 119888119894)
) + 119898(119903
119891 (119911 + 119888119894)
119891 (119911)
)
= 119878 (119903 119891 (119911 + 119888119894)) + 119878 (119903 119891) = 119878 (119903 119891)
(22)
Then (15) follows by (21) and (22)
4 Abstract and Applied Analysis
It is easy to find that
119873(119903 119875 (119911 119891)) = 119874(119873(119903 119891) +
119899
sum
119894=1
119873(119903 119891 (119911 + 119888119894)))
+ 119878 (119903 119891)
(23)
Then (16) (23) and Lemma 6 yield that
119873(119903 119875 (119911 119891)) = 119878 (119903 119891) (24)
Thus (17) follows by (15) and (24)
Lemma 8 Let 119891(119911) be a transcendental meromorphic func-tion of 120590
2(119891) lt 1 and let 119875(119911 119891) be a differential-difference
polynomial of the form (6) then one has
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(25)
Proof Similar to (19) we have
100381610038161003816100381610038161003816100381610038161003816
119875 (119911 119891)
119891(119911)119889(119875)
100381610038161003816100381610038161003816100381610038161003816
le sum
120582isin119868
(1003816100381610038161003816119886120582(119911)1003816100381610038161003816
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892119894119895(119911)
10038161003816100381610038161003816
120582119894119895 1003816100381610038161003816119892 (119911)
1003816100381610038161003816
119889(119875)minus119889(120582)
)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816
1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889(120582)
)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)
le (sum
120582isin119868
1003816100381610038161003816119886120582(119911)1003816100381610038161003816)
times
119899
prod
119894=0
119898
prod
119895=0
10038161003816100381610038161003816119892lowast
119894119895(119911)
10038161003816100381610038161003816
119889(119875)1003816100381610038161003816119892lowast
(119911)1003816100381610038161003816
119889(119875)minus119889lowast(119875)
(26)
where 119892(119911) = 1119891(119911) and
119892lowast
(119911) =
119892 (119911) if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816gt 1
1 if 1003816100381610038161003816119892 (119911)
1003816100381610038161003816le 1
(27)
By the definition of 119892lowast(119911) we have 119898(119903 119892lowast) = 119898(119903 119892) =
119898(119903 1119891) Thus (20) (22) and (26) yield that
119898(119903
119875 (119911 119891)
119891119889(119875)
) le (119889 (119875) minus 119889lowast
(119875))119898 (119903 119892lowast
)
+ 119889 (119875)
119899
sum
119894=0
119898
sum
119895=0
119898(119903 119892lowast
119894119895) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
(28)
that is (25)
Now we can finish the proof of Theorem 1 in the end
Proof of Theorem 1 We deduce from (8) (24) and Lemma 8that
119889 (119875) 119879 (119903 119891)
= 119879 (119903 119891119889(119875)
) le 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119873 (119903 119875 (119911 119891))
+ 119889 (119875)119873(119903
1
119891
) + 119879 (119903 119875 (119911 119891)) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119879 (119903 119875 (119911 119891)) + 119878 (119903 119891)
(29)
that is
119889lowast
(119875) 119879 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119875 (119911 119891)) (30)
Then (9) follows by (17) and (30)
3 Deficiencies of SomeDifferential-Difference Polynomials
In the following we assume that 120572(119911)( equiv 0) is ameromorphicfunction of growth 119878(119903 119891)
In this section we will apply Theorem 1 to consider thedeficiencies of general homogeneous or nonhomogeneousdifferential-difference polynomials
Theorem 9 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)
(a) If119875(119911 119891) is a homogeneous differential-difference poly-nomial then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
= 1 120575 (120572 119875 (119911 119891)) = 0
(31)
(b) If 119875(119911 119891) is a nonhomogeneous differential-differencepolynomial with 2119889lowast(119875) gt 119889(119875) then one has
lim119903rarrinfin
119873(119903 1 (119875 (119911 119891) minus 120572))
119879 (119903 119875 (119911 119891))
ge
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
120575 (120572 119875 (119911 119891)) le 1 minus
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
lt 1
(32)
Thus 119875(119911 119891)minus120572(119911) has infinitely many zeros whether 119875(119911 119891)is homogeneous or nonhomogeneous
Abstract and Applied Analysis 5
Furthermore one considers some differential-differencepolynomials of special forms which are generalizations ofboth differential cases and difference cases that is TheoremsCndashE
Theorem 10 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 V minus 1 of growth 119878(119903 119891) If 119906V gt 119889(119875) 119906 = 1 then
1198761(119911 119891) = 119865 (119891) 119875 (119911 119891) (33)
satisfies
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(34)
Thus 1198761(119911 119891) minus 120572(119911) has infinitely many zeros
When 119865(119891) is of a special form 119891V we can deduce the
following result fromTheorem 9
Theorem 11 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6) If V isin N 1and V + 2119889lowast(119875) gt 119889(119875) then
1198762(119911 119891) = 119891
V119875 (119911 119891) (35)
satisfies 120575(120573 1198762(119911 119891)) lt 1 where 120573 isin C0Thus119876
2(119911 119891)minus
120573 has infinitely many zeros
Remark 12 On the one hand we can also applyTheorem 9 to1198761(119911 119891)with the assumption ldquo2(119889lowast(119875)+119889lowast(119865)) gt 119889(119875)+119906Vrdquo
and obtain the same result as Theorem 10 But our presentassumption ldquo119906V gt 119889(119875)rdquo has no concern with 119889lowast(119875) and119889lowast
(119865) so we think Theorem 10 is better to some extent Onthe other hand we can also apply Theorem 10 to 119876
2(119911 119891)
with the assumption ldquoV gt 119889(119875)rdquo which is stronger thanldquoV + 2119889lowast(119875) gt 119889(119875)rdquo in Theorem 11 showing Theorem 11 isbetter to some extent
Theorem 13 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) 119875(119911 119891) is a
differential-difference polynomial of the form (6) and 119865(119891) =(119891
V+ 119886Vminus1(119911)119891
Vminus1+ sdot sdot sdot + 119886
1(119911)119891 + 119886
0(119911))119906 119906 V isin N is a
polynomial of 119891(119911) with meromorphic coefficients 119886119894(119911) 119894 =
0 Vminus1 of growth 119878(119903 119891) If (119906minus1)119906V(2119906minus1) gt 119889(119875) 119906 = 1then
1198763(119911 119891) = 119865 (119891) + 119875 (119911 119891) (36)
satisfies
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(37)
Thus 1198763(119911 119891) minus 120572(119911) has infinitely many zeros
When 119906 = 1 one can consider some special cases asfollows
Theorem 14 Suppose that 119891(119911) is a transcendental meromor-phic function satisfying 120590
2(119891) lt 1 and (16) and 119875(119911 119891) is a
differential-difference polynomial of the form (6)(a) If V gt 119889(119875) + 2 ge 3 then
1198764(119911 119891) = 119891
V+ 119875 (119911 119891) (38)
satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(39)
(b) If (Vminus1)V(2Vminus1) gt 119889(119875) V ge 3 then1198764(119911 119891) satisfies
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
1
Vminus
2V minus 1V2
119889 (119875)
120575 (120572 1198764(119911 119891)) le
1
V+
2V minus 1V2
119889 (119875) lt 1
(40)
Especially it holds for V = 119889(119875) + 2 = 3(c) If V ge 119889(119875) + 2 ge 3 and 119891 also satisfies 119873(119903 1119891) =119878(119903 119891) then 119876
4(119911 119891) satisfies 120575(120572 119876
4(119911 119891)) lt 1
Especially it holds for V = 119889(119875) + 2 gt 3Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many zeros
If we assume that 119873(119903 1119891) = 119878(119903 119891) in addition thefollowing result follows immediately by Theorem 9
Theorem 15 Suppose that 119891(119911) is a transcendental mero-morphic function satisfying 120590
2(119891) lt 1 and (8) and 119875(119911 119891)
is a differential-difference polynomial of the form (6) If2 min119889lowast(119875) V gt max119889(119875) V then 119876
4(119911 119891) satisfies
120575(120572 1198764(119911 119891)) lt 1 Thus 119876
4(119911 119891) minus 120572(119911) has infinitely many
zeros
Remark 16 Noting that when V gt 3 (V minus 1)V(2V minus 1) leV minus 2 hold we see that the assumption ldquoV gt 119889(119875) + 2rdquo inTheorem 14(a) is weaker than the assumption ldquo(V minus 1)V(2V minus1) gt 119889(119875)rdquo in Theorem 14(b) And these assumptions inTheorem 14 have no concernwith119889lowast(119875)) thus they are differ-ent from the assumption ldquo2 min119889lowast(119875) V gt max119889(119875) Vrdquoin Theorem 15
6 Abstract and Applied Analysis
Remark 17 From the proofs behind it is easy to find that
120582 (119875 (119911 119891) minus 120572) = 120590 (119875 (119911 119891)) = 120590 (119891)
120582 (119876119894(119911 119891) minus 120572) = 120590 (119876
119894(119911 119891)) = 120590 (119891) 119894 = 1 3 4
(41)
hold respectively inTheorems 9 10 13 14(a) and (b) and 15Now we give the proofs of Theorems 9ndash15
Proof of Theorem 9 It follows byTheorem 1 that
119878 (119903 119891) = 119878 (119903 119875 (119911 119891)) (42)
We deduce from (8) (24) (25) and (42) that
119873(119903
1
119875 (119911 119891)
)
le 119873(119903
1
119891119889(119875)
) + 119873(119903
119891119889(119901)
119875 (119911 119891)
)
le 119873(119903
1
119891
) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) + 119874 (1)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119891)
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
) + 119878 (119903 119875 (119911 119891))
(43)
Thus an application of the second main theorem and (24)(42) and (43) imply that
119879 (119903 119875 (119911 119891)) le 119873 (119903 119875 (119911 119891)) + 119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le (119889 (119875) minus 119889lowast
(119875))119898(119903
1
119891
)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(44)
(a) If 119889(119875) = 119889lowast(119875) then it follows by (44) that
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(45)
by which (31) holds
(b) If 2119889lowast(119875) gt 119889(119875) then we deduce from (30) and (44)that
119879 (119903 119875 (119911 119891)) le (119889 (119875) minus 119889lowast
(119875)) 119879 (119903 119891)
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
le
119889 (119875) minus 119889lowast
(119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891))
+ 119873(119903
1
119875 (119911 119891) minus 120572
) + 119878 (119903 119875 (119911 119891))
(46)
that is
2119889lowast
(119875) minus 119889 (119875)
119889lowast(119875)
119879 (119903 119875 (119911 119891)) le 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119878 (119903 119875 (119911 119891))
(47)
Since 2119889lowast(119875)minus119889(119875) gt 0 (32) follows immediately by (47)
Proof of Theorem 10 We deduce from (16) (17) and (24) that
119879 (119903 1198761(119911 119891)) le (119906V + 119889 (119875)) 119879 (119903 119891) + 119878 (119903 119891) (48)
119873(119903 1198761(119911 119891)) = 119874 (119873 (119903 119891)) + 119873 (119903 119875 (119911 119891)) + 119878 (119903 119891)
= 119878 (119903 119891)
(49)
hold Next we consider 119873(119903 11198761(119911 119891)) Let 119911
0be a zero of
1198761(119911 119891) and distinguish three cases
(i) 1199110is not a zero of 119865(119891) then 119911
0must be a zero of
119875(119911 119891) and
119906 le 120596(
1
1198761(119911 119891)
1199110) + (119906 minus 1) 120596(
1
119875 (119911 119891)
1199110) (50)
where 120596(119891 1199110) denotes the order of multiplicity of 119911
0or zero
according as 1199110is a pole of 119891(119911) or not
(ii) 1199110is a zero of 119865(119891) but not a pole of 119875(119911 119891) Then
119906 le 120596(
1
1198761(119911 119891)
1199110) (51)
(iii) 1199110is a zero of 119865(119891) and a pole of 119875(119911 119891) Then
119906 le 120596(
1
119865 (119891)
1199110) le 120596(
1
1198761(119911 119891)
1199110) + 120596 (119875 (119911 119891) 119911
0)
(52)
Abstract and Applied Analysis 7
(24) and (50)ndash(52) yield that
119906119873(119903
1
1198761(119911 119891)
) le 119873(119903
1
1198761(119911 119891)
)
+ (119906 minus 1)119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(53)
Then (48) (49) (53) and an application of the second maintheorem to 119876
1(119911 119891) imply that
119879 (119903 1198761(119911 119891))
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
1198761(119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 1198761(119911 119891))
le
1
119906
119873(119903
1
1198761(119911 119891)
) +
119906 minus 1
119906
119873(119903
1
119875 (119911 119891)
)
+ 119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(54)
consequently
119879 (119903 1198761(119911 119891)) le 119873(119903
1
119875 (119911 119891)
)
+
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(55)
Moreover by 119891119889(119875)119865(119891) = 119891119889(119875)
1198761(119911 119891)119875(119911 119891) (16)
(24) (25) andTheorem A we have
(119889 (119875) + 119906V)119898 (119903 119891)
= 119898(119903
119891119889(119875)
1198761(119911 119891)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119898(119903
119875 (119911 119891)
119891119889(119875)
)
+ 119873(119903
119875 (119911 119891)
119891119889(119875)
) minus 119873(119903
119891119889(119875)
119875 (119911 119891)
) + 119878 (119903 119891)
le 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898(119903
1
119891
)
+ 119889 (119875) (119873(119903
1
119891
) minus 119873 (119903 119891))
+ 119873 (119903 119875 (119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
= 119898 (119903 1198761(119911 119891)) + 119889 (119875)119898 (119903 119891)
minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(56)
consequently
119906V119898(119903 119891) le 119898 (119903 1198761(119911 119891)) minus 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(57)
On the other hand the evident relation 119906V120596(119891 1199110) le
120596(119865(119891) 1199110) + 119906VsumVminus1
119895=0120596(119886119895 1199110) where the definition of
120596(119891 1199110) is given after (50) results in
119906V119873(119903 119891) le 119873 (119903 119865 (119891)) + 119878 (119903 119891)
le 119873 (119903 1198761(119911 119891)) + 119873(119903
1
119875 (119911 119891)
) + 119878 (119903 119891)
(58)
We deduce from (57) and (58) that
119906V119879 (119903 119891) le 119879 (119903 1198761(119911 119891)) + 119878 (119903 119891) (59)
Then (17) (55) and (59) yield that
119906V119879 (119903 119891)
le 119873(119903
1
119875 (119911 119891)
) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
le 119889 (119875) 119879 (119903 119891) +
119906
119906 minus 1
119873(119903
1
1198761(119911 119891) minus 120572
) + 119878 (119903 119891)
(60)
that is
(119906 minus 1) (119906V minus 119889 (119875))119906
119879 (119903 119891) le 119873(119903
1
1198761(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(61)
From (48) and (61) we deduce that
lim119903rarrinfin
119873(119903 1 (1198761(119911 119891) minus 120572))
119879 (119903 1198761(119911 119891))
ge
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
120575 (120572 1198761(119911 119891)) le 1 minus
(119906 minus 1) (119906V minus 119889 (119875))119906 (119906V + 119889 (119875))
lt 1
(62)
Proof of Theorem 11 Assume to the contrary that120575(120573 119876
2(119911 119891)) = 1 Denoting
1198762(119911 119891) minus 120573 = 119891
V(119911) 119875 (119911 119891) minus 120573 = 119866 (119911) (63)
we deduce from (16) and (17) that
119873(119903
1
119866
) = 119873(119903
1
1198762(119911 119891) minus 120573
)
= 119878 (119903 1198762(119911 119891)) = 119878 (119903 119891)
(64)
8 Abstract and Applied Analysis
On the other hand (16) and (24) yield that
119873(119903 119866) = 119873 (119903 1198762(119911 119891) minus 120573) = 119878 (119903 119891) (65)
Differentiating both sides of (63) we obtain
119891Vminus1(119911) 119877 (119911 119891) = 119866
1015840
(119911) (66)
where 119877(119911 119891) = V1198911015840(119911)119875(119911 119891) + 119891(119911)1198751015840(119911 119891) Clearly wededuce from (16) and (24) that
119873(119903 119877 (119911 119891)) = 119878 (119903 119891) (67)
Moreover (64) and (65) yield that
119873(119903
1
1198661015840
) le 119873(119903
119866
1198661015840
) + 119873(119903
1
119866
)
le 119879(119903
1198661015840
119866
) + 119873(119903
1
119866
) + 119874 (1)
le 119898(119903
1198661015840
119866
) + 119873 (119903 119866) + 2119873(119903
1
119866
) + 119874 (1)
= 119878 (119903 119866) + 119878 (119903 119891) = 119878 (119903 119891)
(68)
It follows by (66)ndash(68) that
119873(119903
1
119891
) =
1
V minus 1119873(119903
119877 (119911 119891)
1198661015840
) = 119878 (119903 119891) (69)
Then (16) (69) and the fact 2(119889lowast(119875) + V) gt 119889(119875) + V implythat the assumptions of Theorem 9(b) are satisfied ThusTheorem 9(b) yields that 120575(120573 119876
2(119911 119891)) lt 1 a contradiction
Therefore we have 120575(120573 1198762(119911 119891)) lt 1
Proof of Theorem 13 We deduce from (16) (17) and (24) that
119879 (119903 1198763(119911 119891)) le max 119906V 119889 (119875) 119879 (119903 119891) + 119878 (119903 119891)
= 119906V119879 (119903 119891) + 119878 (119903 119891) (70)
Denote
119867(119911) =
minus119875 (119911 119891) + 120572 (119911)
119865 (119891)
(71)
Now we estimate the poles the zeros and 1-points of119867(119911) accuratelyOn the one handwe see by (71) that the polesof 119867(119911) occur at zeros of 119865(119891) and poles of minus119875(119911 119891) + 120572(119911)which are not simultaneously 1-points of 119867(119911) and thosepoles of119867(119911)which are zeros of 119865(119891) but not simultaneouslyzeros of minus119875(119911 119891) + 120572(119911) also have multiplicities at least 119906 Onthe other handwe also see by (71) that the zeros of119867(119911) occurat zeros of minus119875(119911 119891) + 120572(119911) and poles of 119865(119891) which are notsimultaneously 1-points of 119867(119911) Moreover 1-points of 119867(119911)occur at zeros of 119876
3(119911 119891) minus 120572(119911) and occur at the common
poles zeros of 119865(119891) and minus119875(119911 119891) + 120572(119911) with the samemultiplicities Thus it follows by (16) and (24) that
119873(119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
)
le
1
119906
119873 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(72)
Then (17) (72) and the second main theorem result in
119879 (119903119867) le 119873 (119903119867) + 119873(119903
1
119867
) + 119873(119903
1
119867 minus 1
) + 119878 (119903119867)
le
1
119906
119879 (119903119867) + 119873(119903
1
119875 (119911 119891) minus 120572
)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
le
1
119906
119879 (119903119867) + 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(73)
that is
(1 minus
1
119906
)119879 (119903119867) le 119889 (119875) 119879 (119903 119891)
+ 119873(119903
1
1198763(119911 119891) minus 120572
) + 119878 (119903 119891)
(74)
Moreover Theorem A and (17) imply that
119906V119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 119865 (119891)) = 119879(119903minus119875 (119911 119891) + 120572
119867
)
le 119889 (119875) 119879 (119903 119891) + 119879 (119903119867) + 119878 (119903 119891)
(75)
that is
(119906V minus 119889 (119875)) 119879 (119903 119891) le 119879 (119903119867) + 119878 (119903 119891) (76)
Then (74) and (76) yield that
((119906 minus 1) V minus2119906 minus 1
119906
119889 (119875))119879 (119903 119891) le 119873(119903
1
1198763(119911 119891) minus 120572
)
+ 119878 (119903 119891)
(77)
Abstract and Applied Analysis 9
From (70) and (77) we deduce that
lim119903rarrinfin
119873(119903 1 (1198763(119911 119891) minus 120572))
119879 (119903 1198763(119911 119891))
ge 1 minus
1
119906
minus
2119906 minus 1
1199062V
119889 (119875)
120575 (120572 1198763(119911 119891)) le
1
119906
+
2119906 minus 1
1199062V
119889 (119875) lt 1
(78)
To prove Theorem 14(c) we also need the followinglemma of one of Tumura-Clunie type theorems
Lemma 18 (see [24]) Let 119891(119911) be a meromorphic functionand suppose that Ψ = 119886
119899119891119899
+ sdot sdot sdot + 1198860has small meromorphic
coefficients 119886119895(119911) 119886119899(119911) equiv 0 in the sense of 119879(119903 119886
119895) = 119878(119903 119891)
Moreover assume that 119873(119903 1Ψ) + 119873(119903 119891) = 119878(119903 119891) ThenΨ = 119886
119899(119891 + (119886
119899minus1119899119886119899))119899
Proof of Theorem 14 (a) We deduce from (16) (17) and (24)that
119879 (119903 1198764(119911 119891)) le V119879 (119903 119891) + 119878 (119903 119891) (79)
Denote
119870 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) + 119875 (119911 119891) minus 120572 (119911) (80)
Differentiating both sides of (80) we obtain
V119891Vminus1(119911) 1198911015840
(119911) + 1198751015840
(119911 119891) minus 1205721015840
(119911)
= 1198701015840
(119911) = (119891V(119911) + 119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
(81)
that is
119891Vminus1(119911) ((V
1198911015840
(119911)
119891 (119911)
minus
1198701015840
(119911)
119870 (119911)
)119891 (119911))
= (119875 (119911 119891) minus 120572 (119911))
1198701015840
(119911)
119870 (119911)
minus (1198751015840
(119911 119891) minus 1205721015840
(119911))
= (119875 (119911 119891) minus 120572 (119911)) (
1198701015840
(119911)
119870 (119911)
minus
1198751015840
(119911 119891) minus 1205721015840
(119911)
119875 (119911 119891) minus 120572 (119911)
)
(82)
It follows by (15)ndash(17) (24) (79) and (82) that
119898(119903 119891Vminus1)
le 119898 (119903 119875 (119911 119891) minus 120572) + 119898(119903
1198701015840
119870
)
+ 119898(119903
1198751015840
(119911 119891) minus 1205721015840
119875 (119911 119891) minus 120572
) + 119898(119903
1
(V (1198911015840119891) minus (1198701015840119870)) 119891)
le 119889 (119875)119898 (119903 119891) + 119898(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891)
+ 119873(119903 (V1198911015840
119891
minus
1198701015840
119870
)119891) + 119878 (119903 119870) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1198701015840
119870
) + 119878 (119903 119891)
le (119889 (119875) + 1)119898 (119903 119891) + 119873(119903
1
119870
) + 119878 (119903 119891)
(83)
that is
(V minus 119889 (119875) minus 2) 119879 (119903 119891) le 119873(1199031
1198764(119911 119891) minus 120572
) + 119878 (119903 119891)
(84)
From (79) and (84) we deduce that
lim119903rarrinfin
119873(119903 1 (1198764(119911 119891) minus 120572))
119879 (119903 1198764(119911 119891))
ge 1 minus
119889 (119875) + 2
V
120575 (120572 1198764(119911 119891)) le
119889 (119875) + 2
Vlt 1
(85)
(b) It suffices to note that we may see 119891V as (1198911)V thenTheorem 14(b) follows immediately by Theorem 13
(c) By using a similar reasoning as [13 Theorem 1] wecan rearrange the expression for the differential-differencepolynomial 119875(119911 119891) by collecting together all terms havingthe same total degree and then writing 119875(119911 119891) in the form119875(119911 119891) = sum
119889(119875)
119896=0119887119896(119911)119891119896
(119911) Now each of the coefficients 119887119896(119911)
is a finite sum of products of functions of the form (119891(119895)
(119911 +
119888119894)119891(119911))
120582119894119895= (119891(119895)
(119911+119888119895)119891(119911+119888
119894))120582119894119895(119891(119911+119888
119894)119891(119911))
120582119894119895 with
each such product being multiplied by one of the originalcoefficients 119886
120582(119911) We deduce from the logarithmic derivative
lemma and Lemmas 4 and 6 that 119898(119903 119887119896) = 119878(119903 119891) Clearly
119873(119903 119887119896) = 119878(119903 119891) holds by (8) and Lemma 6Thus 119879(119903 119887
119896) =
119878(119903 119891) Denote
119871 (119911) = 1198764(119911 119891) minus 120572 (119911) = 119891
V(119911) +
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) minus 120572 (119911)
(86)
Assume to the contrary that 120575(120572 1198764(119911 119891)) = 1 Thus
Theorem A yields that
119873(119903
1
119871
) = 119873(119903
1
1198764(119911 119891) minus 120572
)
= 119878 (119903 1198764(119911 119891)) = 119878 (119903 119891)
(87)
Then (8) (86) (87) Lemma 18 and the assumption that V ge119889(119875) + 2 imply that 119871(119911) equiv 119891(119911)V that is
119875 (119911 119891) =
119889(119875)
sum
119896=0
119887119896(119911) 119891119896
(119911) equiv 120572 (119911) (88)
Noting the fact that 119879(119903 119887119896) = 119878(119903 119891) and 119879(119903 120572) = 119878(119903 119891)
we deduce from Theorem A that (88) is a contradictionTherefore we have 120575(120572 119876
4(119911 119891)) lt 1
10 Abstract and Applied Analysis
4 Examples
Example 1 We consider nonhomogeneous differential-difference polynomials
1198751(119911 119891) = 119891 (119911) 119891
2
(119911 + log 4) minus 411989110158401015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 2) + 119891101584010158402 (119911 + log 3)
1198752(119911 119891) = 3119891
3
(119911) 11989110158402
(119911 + log 4)
minus 21198911015840
(119911) 119891 (119911 + log 3) 119891101584010158403 (119911 + log 2)
+ 1198914
(119911) minus 119891101584010158403
(119911)
1198753(119911 119891) = 119891 (119911) 119891
1015840
(119911 + log 2) 11989110158401015840 (119911 + log 3) minus 6119891101584010158402 (119911)(89)
and a homogeneous differential-difference polynomial
1198754(119911 119891) = 119891
101584010158403
(119911 + log 2) minus 1198911015840 (119911) 119891 (119911 + log 2)
times 1198911015840
(119911 + log 3) minus 119891 (119911) 1198911015840 (119911) 11989110158401015840 (119911) (90)
where 119889(1198751) = 3 gt 2 = 119889
lowast
(1198751) 119889(119875
2) = 5 gt 3 = 119889
lowast
(1198752)
119889(1198753) = 3 gt 2 = 119889
lowast
(1198753) and 119889(119875
4) = 3 = 119889
lowast
(1198754) Clearly the
function 119891(119911) = 119890119911 satisfies (8) and 1205902(119891) = 0 lt 1 Then we
have
119889lowast
(1198751) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198751(119911 119890119911
)) =
2119903
120587
+ 119874 (1)
lt 119889 (1198751) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198752) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198752(119911 119890119911
)) =
4119903
120587
+ 119874 (1)
lt 119889 (1198752) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198753) 119879 (119903 119890
119911
) + 119874 (1) lt 119879 (119903 1198753(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198753) 119879 (119903 119890
119911
) + 119874 (1)
119889lowast
(1198754) 119879 (119903 119890
119911
) + 119874 (1) = 119879 (119903 1198754(119911 119890119911
)) =
3119903
120587
+ 119874 (1)
= 119889 (1198754) 119879 (119903 119890
119911
) + 119874 (1)
(91)
This example shows that (9) is best possible
Example 2 Consider 119891(119911) = 119890119911 again Then the
homogeneous case 1198754(119911 119891) in Example 1 also illustrates
Theorem 9(a) And the nonhomogeneous differential-difference polynomials 119875
119894(119911 119891) 119894 = 1 2 3 in Example 1
also illustrate Theorem 9(b) where 120575(120572 1198751(119911 119891)) = 0
120575(120572 1198752(119911 119891)) le 14 lt 23 = 1 minus ((2119889
lowast
(1198752) minus 119889(119875
2))119889lowast
(1198752))
and 120575(120572 1198753(119911 119891)) le 13 lt 12 = 1 minus ((2119889
lowast
(1198753) minus
119889(1198753))119889lowast
(1198753)) Next we consider the nonhomogeneous
differential-difference polynomial
1198755(119911 119891) = 119891
1015840
(119911) 119891 (119911 + log 2) minus 1198912 (119911)
+ 1198911015840
(119911 + log 3) minus 3119891 (119911) + 1(92)
where 119889(1198755) = 2 119889
lowast
(1198755) = 0 Clearly 120575(1 119875
5(119911 119891)) =
120575(1 1198902119911
+ 1) = 1 Note that 2119889lowast(1198755) gt 119889(119875
5) fails then this
example shows that the assumption ldquo2119889lowast(119875) gt 119889(119875)rdquo cannotbe omitted inTheorem 9(b)
Example 3 We consider the differential-difference polyno-mials
1198761(119911 119891) = (119891
2
)
2
1198756(119911 119891)
= 1198914
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
1198762(119911 119891) = 119891
2
1198756(119911 119891)
= 1198912
(119911) (1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
+1198912
(119911 + 120587) )
(93)
and the function 119891(119911) = sin 119911 On the one hand 119873(119903 119891) =119878(119903 119891) 120590
2(119891) = 0 lt 1 and 119906V
1198761
gt 119889(1198756) and V
1198762
+ 2119889lowast
(1198756) gt
119889(1198756) hold where V
1198761
= V1198762
= 119906 = 2 and 119889(1198756) = 3 gt 2 =
119889lowast
(1198756) On the other hand 120575(120572 119876
1(119911 119891)) le 1 minus (1114) lt
1 minus (114) = 1 minus (119906 minus 1)(119906V1198761
minus 119889(119875))119906(119906V1198761
+ 119889(119875)) lt 1 and120575(120572 119876
2(119911 119891)) lt 1 hold This example shows that Theorems
10 and 11 may holdExample 4 We consider the differential-difference polyno-mials
119876(1)
4(119911 119891) = (119891
2
)
4
+ 1198757(119911 119891) = 119891
8
+ 1198757(119911 119891)
= 1198918
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(2)
4(119911 119891) = 119891
2
+ 1198757(119911 119891)
= 1198912
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(3)
4(119911 119891) = 2119891
3
+ 1198757(119911 119891)
= 21198913
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
119876(4)
4(119911 119891) = 119891
4
+ 1198757(119911 119891)
= 1198914
(119911) + 1198911015840
(119911 +
120587
2
)119891 (119911 + 120587) 11989110158401015840
(119911 + 2120587)
(94)
and the function 119891(119911) = sin 119911 again On the one hand119876(1)
4(119911 119891) satisfies (119906 minus 1)119906V
119876(11)
4
(2119906 minus 1) gt 119889(1198757) and V
119876(12)
4
minus
2 gt (V119876(12)
4
minus 1)V119876(12)
4
(2V119876(12)
4
minus 1) gt 119889(1198757) respectively where
119906 = 4 V119876(11)
4
= 2 V119876(12)
4
= 8 119889(1198757) = 119889
lowast
(1198757) = 3 and
for 119894 = 2 3 4 119876(119894)4(119911 119891) satisfies 2 min119889lowast(119875
7) V119876(119894)
4
gt
max119889(1198757) V119876(119894)
4
where V119876(119894)
4
= 119894 On the other hand120575(120572 119876
(119894)
4(119911 119891)) lt 1 119894 = 1 2 3 4 hold This example shows
Abstract and Applied Analysis 11
that Theorems 13ndash15 may hold Moreover this example alsoshows the assumption ldquo119873(119903 1119891) = 119878(119903 119891)rdquo is not necessaryto Theorems 14(c) and 15 but it is regrettable for us notremoving it in our proofs
Example 5 We consider the differential-difference polyno-mials
1198771(119911 119891) = 119891
2
1198758(119911 119891)
= 1198912
(119911) (1198912
(119911 + 120587) +
1
sin2211991111989110158402
(119911 +
120587
2
))
1198772(119911 119891) = 119891
7
+ 1198759(119911 119891)
= 1198917
(119911) + sin 21199111198911015840 (119911 + 1205872
)1198912
(119911 +
120587
2
)
+ 11989110158402
(119911 +
120587
2
)119891(119911 +
3120587
2
)
+ 119911119891 (119911) 119891 (119911 +
120587
2
)
(95)
and the function 119891(119911) = 119890sin2119911 On the one hand 119877
1(119911 119891)
satisfies V1198771
+2119889lowast
(1198758) gt 119889(119875
8) and 119877
2(119911 119891) satisfies V
1198772
minus2 gt
(V1198772
minus 1)V1198772
(2V1198772
minus 1) gt 119889(1198759) respectively where V
1198771
=
2 and 119889(1198758) = 119889
lowast
(1198758) = 2 and V
1198772
= 7 and 119889(1198759) = 3 On
the other hand 120575(1198902 1198771(119911 119891)) = 120575(119890119911 119877
2(119911 119891)) = 1 hold
showing thatTheorems 11 and 14 fail Noting that the function119891(119911) = 119890
sin2119911 satisfies 1205902(119891) = 1 we know that the assumption
ldquo1205902(119891) lt 1rdquo is essential for Theorems 11 and 14 In fact it is
also essential for our other results in the whole paper but it isunnecessary to give examples one by one
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project was supported by the National Natural Sci-ence Foundation of China (11301233 and 11171119) and theNatural Science Foundation of Jiangxi Province in China(20132BAB211001 and 20132BAB211002) and Sponsored Pro-gram for Cultivating Youths of Outstanding Ability in JiangxiNormal University of China
References
[1] W K Hayman Meromorphic Functions Clarendon PressOxford UK 1964
[2] I LaineNevanlinnaTheory andComplexDifferential EquationsWalter de Gruyter Berlin Germany 1993
[3] C C Yang and H X Yi Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Group DordrechtTheNetherlands 2003
[4] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof f(z+120578) and difference equations in the complex planerdquoRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[5] R G Halburd and R J Korhonen ldquoDifference analogue ofthe Lemma on the Logarithmic Derivative with applicationsto difference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[6] R G Halburd R J Korhonen and K Toghe ldquoHolomor-phic curves with shift-invariant hyper-planepreimagesrdquo Tran-sactions of the American Mathematical Society In press httparxivorgabs09033236
[7] M J Ablowitz R Halburd and B Herbst ldquoOn the extensionof the Painleve property to difference equationsrdquo Nonlinearityvol 13 no 3 pp 889ndash905 2000
[8] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[9] Z X Chen ldquoComplex oscillation of meromorphic solutions forthe Pielou logistic equationrdquo Journal of Difference Equations andApplications vol 19 no 11 pp 1795ndash1806 2013
[10] Z X Chen and K H Shon ldquoFixed points of meromorphicsolutions for some difference equationsrdquo Abstract and AppliedAnalysis vol 2013 Article ID 496096 7 pages 2013
[11] K Ishizaki and N Yanagihara ldquoWiman-Valiron method fordifference equationsrdquo Nagoya Mathematical Journal vol 175pp 75ndash102 2004
[12] I Laine and C-C Yang ldquoClunie theorems for difference andq-difference polynomialsrdquo Journal of the London MathematicalSociety vol 76 no 3 pp 556ndash566 2007
[13] I Laine and C C Yang ldquoValue distribution of difference poly-nomialsrdquo Proceedings of the Japan Academy A vol 83 no 8 pp148ndash151 2007
[14] C-C Yang and I Laine ldquoOn analogies between nonlineardifference and differential equationsrdquo Proceedings of the JapanAcademy A vol 86 no 1 pp 10ndash14 2010
[15] R R Zhang and Z B Huang ldquoResults on difference analoguesof Valiron-Mokhonrsquoko theoremrdquoAbstract and Applied Analysisvol 2013 Article ID 273040 6 pages 2013
[16] XMZheng andZXChen ldquoOndeficiencies of somedifferencepolynomialsrdquo Acta Mathematica Sinica vol 54 no 6 pp 983ndash992 2011 (Chinese)
[17] XM Zheng and Z X Chen ldquoOn the value distribution of somedifference polynomialsrdquo Journal of Mathematical Analysis andApplications vol 397 no 2 pp 814ndash821 2013
[18] G Valiron ldquoSur la derivee des fonctions algebroidesrdquo Bulletinde la Societe Mathematique de France vol 59 pp 17ndash39 1931
[19] A Z Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoSibirskii Matematicheskii Zhurnal vol 15 pp 1305ndash1322 1974(Russian)
[20] W K Hayman ldquoPicard values of meromorphic functions andtheir derivativesrdquo Annals of Mathematics vol 70 no 2 pp 9ndash42 1959
[21] C-C Yang ldquoOn deficiencies of differential polynomialsrdquoMath-ematische Zeitschrift vol 116 no 3 pp 197ndash204 1970
[22] C-C Yang ldquoOn deficiencies of differential polynomials IIrdquoMathematische Zeitschrift vol 125 no 2 pp 107ndash112 1972
[23] A A Golrsquodberg and I V Ostrovskii The Distribution ofValues ofMeromorphic Functions NaukaMoscow Russia 1970(Russian)
12 Abstract and Applied Analysis
[24] E Mues and N Steinmetz ldquoThe theorem of Tumura-Clunie formeromorphic functionsrdquo Journal of the London MathematicalSociety vol 23 no 2 pp 113ndash122 1981
Research ArticleOn Growth of Meromorphic Solutions of Complex FunctionalDifference Equations
Jing Li12 Jianjun Zhang3 and Liangwen Liao1
1 Department of Mathematics Nanjing University Nanjing 210093 China2Nankai University Binhai College Tianjin 300270 China3Mathematics and Information Technology School Jiangsu Second Normal University Nanjing 210013 China
Correspondence should be addressed to Jianjun Zhang zhangjianjun1982163com
Received 29 November 2013 Accepted 13 January 2014 Published 25 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Jing Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The main purpose of this paper is to investigate the growth order of the meromorphic solutions of complex functional differenceequation of the form (sum
120582isin119868120572120582(119911)(prod
119899
]=1119891(119911 + 119888])119897120582]))(sum
120583isin119869120573120583(119911)(prod
119899
]=1119891(119911 + 119888])119898120583]
)) = 119876(119911 119891(119901(119911))) where 119868 = 120582 =
(1198971205821 1198971205822 119897
120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 = (119898
1205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two finite
index sets 119888] (] = 1 2 119899) are distinct complex numbers 120572120582(119911) (120582 isin 119868) and 120573
120583(119911) (120583 isin 119869) are small functions relative to 119891(119911)
and 119876(119911 119906) is a rational function in 119906 with coefficients which are small functions of 119891(119911) 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911]
of degree 119896 ge 1 We also give some examples to show that our results are sharp
1 Introduction and Main Results
Let 119891(119911) be a function meromorphic in the complex planeC We assume that the reader is familiar with the standardnotations and results in Nevanlinnarsquos value distribution the-ory ofmeromorphic functions such as the characteristic func-tion 119879(119903 119891) proximity function 119898(119903 119891) counting function119873(119903 119891) and the first and secondmain theorems (see eg [1ndash4]) We also use 119873(119903 119891) to denote the counting function ofthe poles of 119891(119911) whose every pole is counted only once Thenotations 120588(119891) and 120583(119891) denote the order and the lower orderof119891(119911) respectively 119878(119903 119891) denotes any quantity that satisfiesthe condition 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possiblyoutside an exceptional set of 119903 of finite linear measure Ameromorphic function 119886(119911) is called a small function of 119891(119911)or a small function relative to 119891(119911) if and only if 119879(119903 119886(119911)) =119878(119903 119891)
Recently some papers (see eg [5ndash7]) focusing on com-plex difference and functional difference equations emergedIn 2005 Laine et al [5] firstly considered the growth ofmeromorphic solutions of the complex functional differenceequations by utilizing Nevanlinna theory They obtained thefollowing result
Theorem A Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119891 (119901 (119911)) (1)
where 119869 is a collection of all subsets of 1 2 119899 119888119895rsquos are
distinct complex constants and 119901(119911) is a polynomial of degree119896 ge 2 Moreover we assume that the coefficients120572
119869(119911) are small
functions relative to 119891 and that 119899 ge 119896 Then
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (2)
where 120572 = log 119899 log 119896
In 2007 Rieppo [6] gave an estimation of growth ofmeromorphic solutions of complex functional equations asfollows
Theorem B Suppose that 119891 is a transcendental meromorphicfunction Let 119876(119911 119891) 119877(119911 119891) be rational functions in 119891
with small meromorphic coefficients relative to 119891 such that0 lt 119902 = deg
119891119876 le 119889 = deg
119891119877 and 119901(119911) = 119901
119896119911119896
+ 119901119896minus1
119911119896minus1
+
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 828746 6 pageshttpdxdoiorg1011552014828746
2 Abstract and Applied Analysis
sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 If 119891 is a solution of the
functional equation
119877 (119911 119891 (119911)) = 119876 (119911 119891 (119901 (119911))) (3)
then 119902119896 le 119889 and for any 120576 0 lt 120576 lt 1 there exist positive realconstants 119870
1and 119870
2such that
1198701(log 119903)120572minus120576 le 119879 (119903 119891) le 119870
2(log 119903)120572+120576 120572 =
log 119889 minus log 119902log 119896
(4)
when 119903 is large enough
Rieppo [6] also considered the growth order ofmeromor-phic solutions of functional equation (3) when 119896 = 1 and gotthe following
Theorem C Suppose that 119891 is a transcendental meromorphicsolution of (3) where 119901(119911) = 119886119911+119887 119886 119887 isin C 119886 = 0 and |119886| = 1Then
120583 (119891) = 120588 (119891) =
log119889 minus log 119902log |119886|
(5)
Two years later Zheng et al [7] extended Theorem A tomore general type and obtained a similar result of TheoremC In fact they got the following two results
TheoremD Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum
119869
120572119869(119911)(prod
119895isin119869
119891 (119911 + 119888119895)) = 119876 (119911 119891 (119901 (119911))) (6)
where 119869 is a collection of all nonempty subsets of 1 2 119899119888119895(119895 = 1 119899) are distinct complex constants 119901(119911) = 119901
119896119911119896
+
119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a
rational function in 119906 of deg119906119876 = 119902(gt 0) Also suppose that
all the coefficients of (6) are small functions relative to 119891 Then119902119896 le 119899 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (7)
where 120572 = (log 119899 minus log 119902) log 119896
Theorem E Suppose that 119891 is a transcendental meromorphicsolution of (6) where 119869 is a collection of all nonempty subsetsof 1 2 119899 119888
119895(119895 = 1 119899) are distinct complex constants
119901(119911) = 119886119911 + 119887 119886 119887 isin C and 119876(119911 119906) is a rational function in 119906of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of (6)
are small functions relative to 119891(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log 119899minus log |119886|
(8)
(ii) If |119886| gt 1 then we have 119902 le 119899 and
120588 (119891) le
log 119899 minus log 119902log |119886|
(9)
(iii) If |119886| = 1 119902 gt 119899 then we have 120588(119891) = 120583(119891) = infin
In this paper we will consider a more general classof complex functional difference equations We prove thefollowing results which generalize the above related results
Theorem 1 Suppose that 119891(119911) is a transcendental meromor-phic solution of the functional difference equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119901 (119911))) (10)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 =
120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin N⋃0 ] = 1 2 119899
are two finite index sets 119901(119911) = 119901119896119911119896
+ 119901119896minus1
119911119896minus1
+ sdot sdot sdot + 1199010isin
C[119911] of degree 119896 gt 1 and 119876(119911 119906) is a rational function in 119906 ofdeg119906119876 = 119902(gt 0) Also suppose that all the coefficients of (10)
are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (11)
Then 119902119896 le 120590 and
119879 (119903 119891) = 119874 ((log 119903)120572+120576) (12)
where 120572 = (log120590 minus log 119902) log 119896
Theorem 2 Suppose that 119891 is a transcendental meromorphicsolution of the equation
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
= 119876 (119911 119891 (119886119911 + 119887))
(13)
where 119888] (] = 1 119899) are distinct complex constants 119868 = 120582 =
(1198971205821 1198971205822 119897120582119899) | 119897120582] isin N⋃0 ] = 1 2 119899 and 119869 = 120583 =
(1198981205831 1198981205832 119898
120583119899) | 119898120583] isin N⋃0 ] = 1 2 119899 are two
finite index sets 119886 119887 isin C and 119876(119911 119906) is a rational function in119906 of deg
119906119876 = 119902(gt 0) Also suppose that all the coefficients of
(10) are small functions relative to 119891 Denoting
120590] = max120582120583
119897120582] 119898120583] (] = 1 2 119899) 120590 =
119899
sum
]=1120590] (14)
(i) If 0 lt |119886| lt 1 then we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(15)
(ii) If |119886| gt 1 then we have 119902 le 120590 and
120588 (119891) le
log120590 minus log 119902log |119886|
(16)
(iii) If |119886| = 1 and 119902 gt 120590 then we have 120583(119891) = 120588(119891) = infin
Abstract and Applied Analysis 3
Next we will give some examples to show that our resultsare best in some extent
Example 3 Let 1198881= arctan 2 119888
2= minus1205874 Then it is easy to
check that 119891(119911) = tan 119911 solves the following equation
119891(119911 + 1198881)2
119891 (119911 + 1198882)
119891 (119911 + 1198881) + 119891(119911 + 119888
2)2
= (minus4119891(
119911
2
)
8
+ 8119891(
119911
2
)
7
+ 28119891(
119911
2
)
6
minus 56119891(
119911
2
)
5
minus 32119891(
119911
2
)
4
+ 56119891(
119911
2
)
3
+ 28119891(
119911
2
)
2
minus 8119891(
119911
2
) minus 4)
times (3119891(
119911
2
)
8
+ 10119891(
119911
2
)
7
+ 16119891(
119911
2
)
6
+ 122119891(
119911
2
)
5
minus 6119891(
119911
2
)
4
minus122119891(
119911
2
)
3
+16119891(
119911
2
)
2
minus10119891(
119911
2
) + 3)
minus1
(17)
Obviously we have
120583 (119891) = 120588 (119891) = 1 =
log 119902 minus log120590minus log |119886|
(18)
where 119902 = 8 120590 = 4 and 119886 = 12
Example 3 shows that the estimate in Theorem 2(i) issharp
Example 4 It is easy to check that 119891(119911) = tan 119911 satisfies theequation
119891(119911 + (1205873))2
119891 (119911 + (1205876)) minus 119891 (119911 + (1205876))
119891 (119911 + (1205873)) 119891(119911 + (1205876))2
minus 119891 (119911 + (1205873))
=
radic3119891(2119911)2
+ 4119891 (2119911) + radic3
minusradic3119891(2119911)2
+ 4119891 (2119911) minus radic3
(19)
Clearly we have
120583 (119891) = 120588 (119891) = 1 =
log120590 minus log 119902log |119886|
(20)
where 120590 = 4 119902 = 2 and 119886 = 2
Example 4 shows that the estimate in Theorem 2(ii) issharp
Example 5 119891(119911) = tan 119911 satisfies the equation of the form
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
minus(119891(1199112)2
minus 2119891 (1199112) minus 1)
3
8119891 (1199112) (119891(1199112)2
minus 1) (119891(1199112)2
+ 2119891 (1199112) minus 1)
(21)
where 120590 = 4 119902 = 6 and 119886 = 12 120588(119891) = 120583(119891) = 1 gt
log(32) log 2 = (log 119902 minus log120590) minus log |119886|
Example 5 shows that the strict inequality in Theorem 2may occur Therefore we do not have the same estimation asinTheoremC for the growth order ofmeromorphic solutionsof (13)
The following Example shows that the restriction 119902 gt 120590
in case (iii) in Theorem 2 is necessary
Example 6 Meromorphic function 119891(119911) = tan 119911 solves thefollowing equation
119891(119911 + (1205874))2
119891 (119911 + (1205874)) + 119891(119911 minus (1205874))2
=
(119891 (119911) + 1)3
4119891 (119911) (1 minus 119891 (119911))
(22)
where 119886 = 1 and 4 = 120590 gt 119902 = 3 but 120588(119891) = 120583(119891) = 1
Next we give an example to show that case (iii) inTheorem 2 may hold
Example 7 Function 119891(119911) = 119911119890119890119911
satisfies the followingequation
(119911 + log 6) (119911 + log 2)5 [119891(119911 + log 4)4 + 119891 (119911 + log 4)](119911 + log 4) 119891 (119911 + log 6)
=
(119911 + log 4)3119891(119911 + log 2)6 + (119911 + log 2)6
119891 (119911 + log 2)
(23)
where 119886 = 1 and 119902 = 6 gt 5 = 120590 Obviously 120588(119891) = 120583(119891) = infin
2 Main Lemmas
In order to prove our results we need the following lemmas
Lemma 1 (see [4 8]) Let 119891(119911) be a meromorphic functionThen for all irreducible rational functions in 119891
119877 (119911 119891) =
119875 (119911 119891)
119876 (119911 119891)
=
sum119901
119894=0119886119894(119911) 119891119894
sum119902
119895=0119887119895(119911) 119891119895
(24)
such that the meromorphic coefficients 119886119894(119911) 119887119895(119911) satisfy
119879 (119903 119886119894) = 119878 (119903 119891) 119894 = 0 1 119901
119879 (119903 119887119895) = 119878 (119903 119891) 119895 = 0 1 119902
(25)
then one has
119879 (119903 119877 (119911 119891)) = max 119901 119902 sdot 119879 (119903 119891) + 119878 (119903 119891) (26)
From the proof ofTheorem 1 in [9] we have the followingestimate for the Nevanlinna characteristic
Lemma 2 Let 1198911 1198912 119891
119899be distinct meromorphic func-
tions and
119865 (119911) =
119875 (119911)
119876 (119911)
=
sum120582isin119868
120572120582(119911) 119891
1198971205821
1119891
1198971205822
2 119891
119897120582119899
119899
sum120583isin119869
120573120583(119911) 119891
1198981205831
1119891
1198981205832
2 119891
119898120583119899
119899
(27)
4 Abstract and Applied Analysis
Then
119879 (119903 119865 (119911)) le
119899
sum
]=1120590]119879 (119903 119891]) + 119878 (119903 119891) (28)
where 119868 = 120582 = (1198971205821 1198971205822 119897120582119899) | 119897
120582] isin N⋃0 ] =
1 2 119899 and 119869 = 120583 = (1198981205831 1198981205832 119898
120583119899) | 119898
120583] isin
N⋃0 ] = 1 2 119899 are two finite index sets 120590] =
max120582120583119897120582] 119898120583] (] = 1 2 119899) 120572
120582(119911) = 119900(119879(119903 119891])(120582 isin 119868))
and120573120583(119911) = 119900(119879(119903 119891])(120583 isin 119869)) hold for all ] isin 1 2 119899 and
satisfy 119879(119903 120572120582) = 119878(119903 119891) (120582 isin 119868) and 119879(119903 120573
120583) = 119878(119903 119891) (120583 isin
119869)
Lemma 3 (see [7]) Let 119888 be a complex constant Given 120576 gt 0
and a meromorphic function 119891 one has
119879 (119903 119891 (119911 plusmn 119888)) le (1 + 120576) 119879 (119903 + |119888| 119891) (29)
for all 119903 gt 1199030 where 119903
0is some positive constant
Lemma 4 (see [4]) Let 119892 (0 +infin) rarr R ℎ (0 +infin) rarr R
bemonotone increasing functions such that 119892(119903) le ℎ(119903) outsideof an exceptional set 119864 of finite linear measure Then for any120572 gt 1 there exists 119903
0gt 0 such that 119892(119903) le ℎ(120572119903) for all 119903 gt 119903
0
Lemma 5 (see [10]) Let 119891 be a transcendental meromorphicfunction and 119901(119911) = 119886
119896119911119896
+ 119886119896minus1
119911119896minus1
+ sdot sdot sdot + 1198861119911 + 1198860 119886119896
= 0be a nonconstant polynomial of degree 119896 Given 0 lt 120575 lt |119886
119896|
denote 120582 = |119886119896| + 120575 and 120583 = |119886
119896| minus 120575 Then given 120576 gt 0 and
119886 isin C⋃infin one has
119896119899 (120583119903119896
119886 119891) le 119899 (119903 119886 119891 (119901 (119911))) le 119896119899 (120582119903119896
119886 119891)
119873 (120583119903119896
119886 119891) + 119874 (log 119903) le 119873 (119903 119886 119891 (119901 (119911)))
le 119873 (120582119903119896
119886 119891) + 119874 (log 119903)
(1 minus 120576) 119879 (120583119903119896
119891) le 119879 (119903 119891 (119901 (119911))) le (1 + 120576) 119879 (120582119903119896
119891)
(30)
for all 119903 large enough
Lemma 6 (see [11]) Let 120601 [1199030 +infin) rarr (0 +infin) be
positive and bounded in every finite interval and suppose that120601(120583119903119898
) le 119860120601(119903) + 119861 holds for all 119903 large enough where 120583 gt 0119898 gt 1 119860 gt 1 and 119861 are real constants Then
120601 (119903) = 119874 ((log 119903)120572) (31)
where 120572 = log119860 log119898
Lemma 7 (see [6]) Let 120601 (1199030infin) rarr (1infin) where 119903
0ge 1
be a monotone increasing function If for some real constant120572 gt 1 there exists a real number 119870 gt 1 such that 120601(120572119903) ge
119870120601(119903) then
lim119903rarrinfin
log120601 (119903)log 119903
ge
log119870log120572
(32)
Lemma 8 (see [12]) Let 120601 (1infin) rarr (0infin) be a monotoneincreasing function and let 119891 be a nonconstant meromorphic
function If for some real constant 120572 isin (0 1) there exist realconstants 119870
1gt 0 and 119870
2ge 1 such that
119879 (119903 119891) le 1198701120601 (120572119903) + 119870
2119879 (120572119903 119891) + 119878 (120572119903 119891) (33)
then
120588 (119891) le
log1198702
minus log120572+ lim119903rarrinfin
log120601 (119903)log 119903
(34)
3 Proof of Theorems
Proof of Theorem 1 We assume 119891(119911) is a transcendentalmeromorphic solution of (10) Denoting 119862 =
max|1198881| |1198882| |119888
119899| According to Lemmas 1 2 and 3
and the last assertion of Lemma 5 we get that for any 1205761gt 0
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) + 119878 (119903 119891)
le 119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576
1) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 1205761) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(35)
where 119903 is large enough and120583 = |119901119896|minus120575 for some 0 lt 120575 lt |119901
119896|
Since 119879(119903 + 119862 119891) le 119879(120573119903 119891) holds for 119903 large enough for120573 gt 1 we may assume 119903 to be large enough to satisfy
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120573119903 119891) (36)
outside a possible exceptional set of finite linear measure ByLemma 4 we know that whenever 120574 gt 1
119902 (1 minus 1205761) 119879 (120583119903
119896
119891) le 120590 (1 + 1205761) 119879 (120574120573119903 119891) (37)
holds for all 119903 large enough Denote 119905 = 120574120573119903 thus theinequality (37) may be written in the form
119879(
120583
(120574120573)119896
119905119896
119891) le
120590 (1 + 1205761)
119902 (1 minus 1205761)
119879 (119905 119891) (38)
By Lemma 6 we have
119879 (119903 119891) = 119874 ((log 119903)1205721) (39)
Abstract and Applied Analysis 5
where
1205721=
log (120590 (1 + 1205761) 119902 (1 minus 120576
1))
log 119896
=
log120590 minus log 119902log 119896
+
log ((1 + 1205761) (1 minus 120576
1))
log 119896
(40)
Denoting now 120572 = (log120590 minus log 119902) log 119896 and 120576 = log((1 +
1205761)(1 minus 120576
1)) log 119896 thus we obtain the required form
Finally we show that 119902119896 le 120590 If 119902119896 gt 120590 then we have120572 lt 1 For sufficiently small 120576 gt 0 we have 120572 + 120576 lt 1 whichcontradicts with the transcendency of 119891 Thus Theorem 1 isproved
Proof of Theorem 2 Suppose 119891(119911) is a transcendental mero-morphic solution of (13) Denoting 119862 = max|119888
1| |1198882|
|119888119899|
(i) 0 lt |119886| lt 1 We may assume that 119902 gt 120590 since the case119902 le 120590 is trivial by the fact that 120583(119891) ge 0 By Lemmas1ndash3 we have for any 120576 gt 0 and 120573 gt 1
119902119879 (119903 119891 (119901 (119911))) + 119878 (119903 119891)
= 119879 (119903 119876 (119911 119891 (119901 (119911))))
= 119879(119903
sum120582isin119868
120572120582(119911) (prod
119899
]=1119891(119911 + 119888])119897120582])
sum120583isin119869
120573120583(119911) (prod
119899
]=1119891(119911 + 119888])119898120583])
)
le
119899
sum
]=1120590]119879 (119903 119891 (119911 + 119888])) + 119878 (119903 119891)
le
119899
sum
]=1120590] (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= (
119899
sum
]=1120590]) (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
= 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
le 120590 (1 + 120576) 119879 (120573119903 119891) + 119878 (119903 119891)
(41)
where 119903 is large enoughBy the last assertion of Lemma 5 and (41) we obtain that
for 120583 = |119886| minus 120575 (0 lt 120575 lt |119886| 0 lt 120583 lt 1) the followinginequality
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120573119903 119891) (42)
holds where 119903 is large enough outside of a possible set of finitelinear measure By Lemma 4 we get that for any 120574 gt 1 andsufficiently large 119903
119902 (1 minus 120576) 119879 (120583119903 119891) le 120590 (1 + 120576) 119879 (120574120573119903 119891) (43)
Therefore
119902 (1 minus 120576)
120590 (1 + 120576)
119879 (119903 119891) le 119879(
120574120573
120583
119903 119891) (44)
Since 120573 gt 1 120574 gt 1 0 lt 120583 lt 1 and 119902 gt 120590 we have 120573120574120583 gt 1
and 119902(1 minus 120576)120590(1 + 120576) gt 1 when 120576 is small enough UsingLemma 7 we see that
120583 (119891) ge
log 119902 (1 minus 120576) minus log120590 (1 + 120576)
log 120574120573 minus log 120583 (45)
Letting 120576 rarr 0 120575 rarr 0 120573 rarr 1 and 120574 rarr 1 we have
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(46)
(ii) |119886| gt 1 By the similar reasoning as is (i) we easilyobtain that
119902 (1 minus 120576) 119879 (120583119903 119891) le 119902119879 (119903 119891 (119901 (119911)))
le 120590 (1 + 120576) 119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891)
(47)
for all 119903 large enough We may select sufficiently smallnumbers 120575 gt 0 and 120576 gt 0 such that 120583 = |119886| minus 120575 gt 1 and(1120583) + 120576 lt 1 Thus we have
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) + 119878 (119903 119891) (48)
namely
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (119903 + 119862 119891 (119911)) (49)
where 119903 is large enough possibly outside of a set of finite linearmeasure By Lemma 4 we have for any 1 lt 120574 lt 120583
119879 (120583119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879 (120574119903 119891 (119911)) (50)
that is
119879 (119903 119891) le
120590 (1 + 120576)
119902 (1 minus 120576)
119879(
120574
120583
119903 119891 (119911)) (51)
holds for all sufficiently large 119903 By Lemma 8 we obtain
120588 (119891) le
log120590 minus log 119902 + log (1 + 120576) minus log (1 minus 120576)
minus log (120574120583) (52)
Letting 120576 rarr 0 120575 rarr 0 and 120574 rarr 1 we have
120588 (119891) le
log120590 minus log 119902log |119886|
(53)
(iii) |119886| = 1 and 119902 gt 120590 The proof of this case is completelysimilar as in the case in (i) In fact we set 120583 = |119886|minus120575 =
1 minus 120575 (0 lt 120575 lt 1 0 lt 120583 lt 1) Similarly we can get
120583 (119891) ge
log 119902 minus log120590minus log |119886|
(54)
Since |119886| = 1 we have 120583(119891) = 120588(119891) = infin
6 Abstract and Applied Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors would like to thank the anonymous ref-erees for their valuable comments and suggestions Theresearch was supported by Colonel-level topics (JSNU-ZY-01) (Jsie2012zd01) and NSF of China (11271179)
References
[1] W Cherry and Z Ye Nevanlinnarsquos Theory of Value DistributionSpringer Monographs in Mathematics Springer Berlin Ger-many 2001
[2] W K Hayman Meromorphic Functions Oxford MathematicalMonographs Clarendon Press Oxford UK 1964
[3] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Beijing China 1988
[4] I LaineNevanlinnaTheory andComplexDifferential Equationsvol 15 of de Gruyter Studies in Mathematics Walter de GruyterBerlin Germany 1993
[5] I Laine J Rieppo and H Silvennoinen ldquoRemarks on complexdifference equationsrdquo Computational Methods and FunctionTheory vol 5 no 1 pp 77ndash88 2005
[6] J Rieppo ldquoOn a class of complex functional equationsrdquoAnnalesAcademiaelig Scientiarum Fennicaelig vol 32 no 1 pp 151ndash170 2007
[7] X-M Zheng Z-X Chen and J Tu ldquoGrowth of meromorphicsolutions of some difference equationsrdquoApplicable Analysis andDiscrete Mathematics vol 4 no 2 pp 309ndash321 2010
[8] A Z Mokhonrsquoko ldquoThe Nevanlinna characteristics of certainmeromorphic functionsrdquo Teorija Funkciı Funkcionalrsquonyı Analizi ih Prilozenija vol 14 pp 83ndash87 1971 (Russian)
[9] A A Mokhonrsquoko and V D Mokhonrsquoko ldquoEstimates of theNevanlinna characteristics of certain classes of meromorphicfunctions and their applications to differential equationsrdquoAkademija Nauk SSSR vol 15 pp 1305ndash1322 1974
[10] R Goldstein ldquoSome results on factorisation of meromorphicfunctionsrdquo Journal of the London Mathematical Society vol 4pp 357ndash364 1971
[11] R Goldstein ldquoOn meromorphic solutions of certain functionalequationsrdquo Aequationes Mathematicae vol 18 no 1-2 pp 112ndash157 1978
[12] G G Gundersen J Heittokangas I Laine J Rieppo andD Yang ldquoMeromorphic solutions of generalized Schroderequationsrdquo Aequationes Mathematicae vol 63 no 1-2 pp 110ndash135 2002
Research ArticleUnicity of Entire Functions concerning Shifts andDifference Operators
Dan Liu Degui Yang and Mingliang Fang
Institute of Applied Mathematics South China Agricultural University Guangzhou 510642 China
Correspondence should be addressed to Mingliang Fang mlfangscaueducn
Received 29 October 2013 Revised 17 December 2013 Accepted 19 December 2013 Published 3 February 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 Dan Liu et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We prove a unicity theorem of entire functions that share two distinct small functions with their shiftsThe corollary of the theoremconfirms the conjecture posed by Li and Gao (2011)
1 Introduction
Let119891 be a nonconstantmeromorphic function in the complexplane C We will use the standard notations in Nevanlinnatheory of meromorphic functions such as 119879(119903 119891) 119873(119903 119891)and 119898(119903 119891) (see [1 2]) The notation 119878(119903 119891) is defined to beany quantity satisfying 119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin pos-sibly outside a set of finite linear measures A meromorphicfunction 119886 is called a small function related to119891 provided that119879(119903 119886) = 119878(119903 119891)
Let 119891 and 119892 be two nonconstant meromorphic functionsand let 119886 be a small function related to both 119891 and 119892 Wesay that 119891 and 119892 share 119886 CM if 119891 minus 119886 and 119892 minus 119886 have thesame zeros with the same multiplicities 119891 and 119892 are said toshare 119886 IM if 119891 minus 119886 and 119892 minus 119886 have the same zeros ignoringmultiplicities
Let119873(119903 119886) be the counting functions of all common zeroswith the same multiplicities of 119891 minus 119886 and 119892 minus 119886 If
119873(119903
1
119891 minus 119886
) + 119873(119903
1
119892 minus 119886
) minus 2119873 (119903 119886)
= 119878 (119903 119891) + 119878 (119903 119892)
(1)
then we say that 119891 and 119892 share 119886 CM almostFor a nonzero complex constant 119888 isin C we define
difference operators asΔ119888119891(119911) = 119891(119911+119888)minus119891(119911) andΔ119899
119888119891(119911) =
Δ119888(Δ119899minus1
119888119891(119911)) 119899 isin N 119899 gt 2
In 1977 Rubel and Yang [3] proved the following result
Theorem A Let 119891 be a nonconstant entire function If 119891(119911)and 1198911015840(119911) share two distinct finite values CM then 119891(119911) equiv
1198911015840
(119911)
In fact the conclusion still holds if the two CM values arereplaced by two IM values (see Gundersen [4 5] Mues andSteinmetz [6])
Recently a number of articles focused on value dis-tribution in shifts or difference operators of meromorphicfunctions (see [7ndash11]) In particular some papers studied theunicity of meromorphic functions sharing values with theirshifts or difference operators (see [12ndash14]) In 2009 Heit-tokangas et al [12] proved the following result concerningshifts
Theorem B Let 119891 be a nonconstant entire function of finiteorder 119888 isin C If119891(119911) and119891(119911+119888) share two distinct finite valuesCM then 119891(119911) equiv 119891(119911 + 119888)
In 2011 Li and Gao [14] proved the following resultconcerning difference operators
Theorem C Let 119891 be a nonconstant entire function of finiteorder 119888 isin C and let 119899 be a positive integer Suppose that 119891(119911)and Δ119899
119888119891(119911) share two distinct finite values 119886 119887 CM and one of
the following cases is satisfied(i) 119886119887 = 0(ii) 119886119887 = 0 and 120588(119891) notin 119873Then 119891(119911) equiv Δ119899
119888119891(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 380910 5 pageshttpdxdoiorg1011552014380910
2 Abstract and Applied Analysis
In [14] Li andGao conjectured that the restriction 120588(119891) notinN for the case 119886119887 = 0 can be removed In this paper weconfirm their conjecture In fact we prove the followingmoregeneral results
Theorem 1 Let 119891 be a nonconstant entire function of finiteorder let 119899 be a positive integer let 119886(119911) 119887(119911) be two distinctsmall functions related to 119891(119911) let 119898
1 1198982 119898
119899be nonzero
complex numbers and 1198881 1198882 119888
119899distinct finite values and let
119865 (119911) = 1198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119891 (119911 + 119888
119899)
(2)
If 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119865(119911)
Corollary 2 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number let 119899 be apositive integer and let 119886 119887 be two distinct finite values If 119891(119911)and Δ119899
119888119891(119911) share 119886 119887 CM then 119891(119911) equiv Δ119899
119888119891(119911)
Remark 3 Corollary 2 confirms the conjecture of Li and Gaoin [14]
Corollary 4 Let 119891 be a nonconstant entire function of finiteorder let 119888 be a nonzero finite complex number and let 119886(119911)119887(119911) be two distinct small functions related to 119891 If 119891(119911) and119891(119911 + 119888) share 119886(119911) 119887(119911) CM then 119891(119911) equiv 119891(119911 + 119888)
2 Some Lemmas
For the proof of Theorem 1 we require the following results
Lemma 5 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903
1
119891
) + 119873 (119903 119891) = 119878 (119903 119891)
119873(119903
1
119892
) + 119873 (119903 119892) = 119878 (119903 119892)
(3)
If 119891(119911) and 119892(119911) share 1 CM almost then either 119891(119911) equiv 119892(119911)
or 119891(119911)119892(119911) equiv 1
Lemma 6 (see [15]) Let 119891 and 119892 be two nonconstant mero-morphic functions satisfying
119873(119903 119891) = 119878 (119903 119891) 119873 (119903 119892) = 119878 (119903 119892) (4)
If 119891(119911) and 119892(119911) share 0 and 1 CM almost and
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 119891) + 119879 (119903 119892)
lt
2
3
(5)
where 119868 sub [0infin) is a set of infinitely linear measure then
119891 (119911) =
119886119892 (119911) + 119887
119888119892 (119911) + 119889
(6)
where 119886 119887 119888 and 119889 are constants satisfying 119886119889 minus 119887119888 = 0
Lemma 7 (see [10]) Let 119891 be a nonconstant meromorphicfunction of finite order 119888 isin C Then
119898(119903
119891 (119911 + 119888)
119891 (119911)
) = 119878 (119903 119891) (7)
for all 119903 outside a possible exceptional set 119864 with finite loga-rithmic measure int
119864
119889119903119903 lt infin
In the following 119878(119903 119891) denotes any function satisfying119878(119903 119891) = 119900(119879(119903 119891)) as 119903 rarr infin possibly outside a set withfinite logarithmic measure
3 Proof of Theorem 1
We prove Theorem 1 by contradiction Suppose that 119891(119911) equiv
119865(119911) Then it follows from 119891(119911) and 119865(119911) being two distinctentire functions that 119891(119911) and 119865(119911) share 119886(119911) 119887(119911) and infinCM By the Nevanlinna second fundamental theorem forthree small functions we have
119879 (119903 119891) le 119873 (119903 119891) + 119873(119903
1
119891 minus 119886
)
+ 119873(119903
1
119891 minus 119887
) + 119878 (119903 119891)
le 119873(119903
1
119865 minus 119886
) + 119873(119903
1
119865 minus 119887
) + 119878 (119903 119891)
le 2119879 (119903 119865) + 119878 (119903 119891)
(8)
Similarly we have 119879(119903 119865) le 2119879(119903 119891) + 119878(119903 119865) Therefore119878(119903 119891) = 119878(119903 119865)
Set
1198911(119911) =
119891 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
1198651(119911) =
119865 (119911) minus 119886 (119911)
119887 (119911) minus 119886 (119911)
(9)
Thus 1198911(119911) 1198651(119911) share 0 1 andinfin CM almost
Obviously we have
119879 (119903 1198911) = 119879 (119903 119891) + 119878 (119903 119891)
119879 (119903 1198651) = 119879 (119903 119865) + 119878 (119903 119891)
119878 (119903 119865) = 119878 (119903 1198651) = 119878 (119903 119891
1) = 119878 (119903 119891)
(10)
By Nevanlinnarsquos second fundamental theorem we have
119879 (119903 1198911) le 119873(119903
1
1198911
) + 119873(119903
1
1198911minus 1
) + 119873 (119903 1198911) + 119878 (119903 119891
1)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
Abstract and Applied Analysis 3
le 119873(119903
1
1198651minus 1198911
) + 119878 (119903 119891)
le 119879 (119903 1198651minus 1198911) + 119878 (119903 119891)
le 119879 (119903 119865 minus 119891) + 119878 (119903 119891)
le 119898 (119903 119865 minus 119891) + 119878 (119903 119891)
(11)Since 119865 minus 119891 = 119898
1119891(119911 + 119888
1) + 1198982119891(119911 + 119888
2) + sdot sdot sdot + 119898
119899119891(119911 +
119888119899) minus 119891(119911) = 119891(119911)[119898
1(119891(119911 + 119888
1)119891(119911)) +119898
2(119891(119911 + 119888
2)119891(119911)) +
sdot sdot sdot + 119898119899(119891(119911 + 119888
119899)119891(119911)) minus 1] thus
119898(119903 119865 minus 119891)
le 119898 (119903 119891)
+ 119898(1199031198981
119891 (119911 + 1198881)
119891 (119911)
+ sdot sdot sdot + 119898119899
119891 (119911 + 119888119899)
119891 (119911)
minus 1)
le 119898 (119903 119891) + 119878 (119903 119891)
(12)By (11) we have
119879 (119903 1198911) le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(13)
It follows that119873(119903 0) + 119873 (119903 1) = 119879 (119903 119891
1) + 119878 (119903 119891) (14)
On the other hand byNevanlinna first fundamental theoremwe have
2119879 (119903 1198911) = 119879(119903
1
1198911
) + 119879(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119898(119903
1
1198911
)
+ 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
(15)
So we get
119879 (119903 1198911) le 119898(119903
1
1198911
) + 119898(119903
1
1198911minus 1
) + 119878 (119903 119891)
le 119898(119903
1
119891 minus 119886
) + 119898(119903
1
119891 minus 119887
) + 119878 (119903 119891)
(16)
Set1198861(119911) = 119898
1119886 (119911 + 119888
1) + 1198982119886 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119886 (119911 + 119888
119899)
1198871(119911) = 119898
1119887 (119911 + 119888
1) + 1198982119887 (119911 + 119888
2) + sdot sdot sdot + 119898
119899119887 (119911 + 119888
119899)
(17)
If 1198861(119911) equiv 119887
1(119911) we can deduce by (16) that
119879 (119903 1198911) le 119898(119903
1
119891 minus 119886
+
1
119891 minus 119887
) + 119878 (119903 119891)
le 119898(119903
119865 minus 1198861
119891 minus 119886
+
119865 minus 1198871
119891 minus 119887
)
+ 119898(119903
1
119865 minus 1198861
) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
= 119879 (119903 1198911) + 119878 (119903 119891)
(18)
If 1198861(119911) equiv 119887
1(119911) set
119871 (119865) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
119865 11988611198871
1198651015840
1198861015840
11198871015840
1
11986510158401015840
11988610158401015840
111988710158401015840
1
100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(19)
Then we have
119898(119903
119865 minus 1198861
119891 minus 119886
) = 119898(119903
119865 minus 1198871
119891 minus 119887
) = 119878 (119903 119891)
119898(119903
119871 (119865)
119865 minus 1198861
) = 119898(119903
119871 (119865)
119865 minus 1198871
) = 119878 (119903 119891)
(20)
It followed from (16) that
119879 (119903 1198911) le 119898(119903
119865 minus 1198861
119891 minus 119886
) + 119898(119903
1
119865 minus 1198861
)
+ 119898(119903
119865 minus 1198871
119891 minus 119887
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
) + 119898(119903
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119865 minus 1198861
+
1
119865 minus 1198871
) + 119878 (119903 119891)
le 119898(119903
1
119871 (119865)
) + 119878 (119903 119891)
le 119879 (119903 119871 (119865)) + 119878 (119903 119891)
le 119879 (119903 119865) + 119878 (119903 119891)
4 Abstract and Applied Analysis
le 119879 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
= 119898 (1199031198981119891 (119911 + 119888
1) + 1198982119891 (119911 + 119888
2)
+ sdot sdot sdot + 119898119899119891 (119911 + 119888
119899)) + 119878 (119903 119891)
le 119898 (119903 119891) + 119878 (119903 119891)
le 119879 (119903 119891) + 119878 (119903 119891) = 119879 (119903 1198911) + 119878 (119903 119891)
(21)
By (18) and (21) we can deduce that
119879 (119903 1198911) = 119879 (119903 119865) + 119878 (119903 119891) = 119879 (119903 119865
1) + 119878 (119903 119891) (22)
It follows from (14) and (22) that
lim119903rarrinfin
119903isin119868
119873(119903 0) + 119873 (119903 1)
119879 (119903 1198911) + 119879 (119903 119865
1)
=
1
2
lt
2
3
(23)
By Lemma 6 we have
1198911(119911) =
1198601198651(119911) + 119861
1198621198651(119911) + 119863
(24)
where 119860 119861 119862 and 119863 are complex numbers satisfying 119860119863 minus
119861119862 = 0Now we consider three cases
Case 1 Consider 119873(119903 0) = 119878(119903 1198911) Thus
119873(119903
1
1198911
) + 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (25)
Similarly we have
119873(119903
1
1198651
) + 119873 (119903 1198651) = 119878 (119903 119865
1) = 119878 (119903 119891) (26)
By Lemma 5 we get that either 1198911equiv 1198651or 11989111198651equiv 1
If 1198911equiv 1198651 we can easily deduce that 119891 equiv 119865 which is a
contradiction with our assumptionIf 11989111198651equiv 1 that is
(119891 (119911) minus 119886) (119865 (119911) minus 119886) equiv (119887 minus 119886)2
(27)
then we have
(119891 minus 119886)2
=
(119887 minus 119886)2
(119865 minus 119886) (119891 minus 119886)
(28)
From (28) we have
2119879 (119903 119891) le 119879 (119903 (119891 minus 119886)2
) + 119878 (119903 119891)
= 119879(119903
1
(119887 minus 119886)2
((119865 minus 119886) (119891 minus 119886))
) + 119878 (119903 119891)
le 119879(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
= 119873(119903
119865 minus 119886
119891 minus 119886
) + 119898(119903
119865 minus 119886
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
(119865 minus 1198861) + (119886
1minus 119886)
119891 minus 119886
) + 119878 (119903 119891)
le 119898(119903
1198861minus 119886
119891 minus 119886
) + 119878 (119903 119891) le 119879 (119903 119891) + 119878 (119903 119891)
(29)
It follows that 119879(119903 119891) le 119878(119903 119891) a contradiction
Case 2 Consider 119873(119903 1) = 119878(119903 1198911) Using the same argu-
ment as used in Case 1 we deduce that 119879(119903 119891) le 119878(119903 119891) acontradiction
Case 3 Consider 119873(119903 0) = 119878(119903 1198911) 119873(119903 1) = 119878(119903 119891
1) Since
1198911and 119865
1share 0 1 CM almost we deduce from (24) that
1198911(119911) =
(119862 + 119863) 1198651(119911)
1198621198651(119911) + 119863
(30)
If 119862 = 0 then 1198911equiv 1198651 that is 119891 equiv 119865 a contradiction
Hence 119862 = 0 Thus we have
119873(119903
1
1198651+ (119863119862)
) = 119873 (119903 1198911) = 119878 (119903 119891
1) = 119878 (119903 119891) (31)
Obviously 119863119862 = 0 119863119862 = minus 1 Thus by Nevanlinnasecond fundamental theorem and (14) we get
2119879 (119903 1198911) = 2119879 (119903 119865
1) + 119878 (119903 119891
1)
le 119873(119903
1
1198651
) + 119873(119903
1
1198651minus 1
)
+ 119873(119903
1
1198651+ (119863119862)
) + 119878 (119903 119891)
le 119873 (119903 0) + 119873 (119903 1) + 119878 (119903 119891)
le 119879 (119903 1198911) + 119878 (119903 119891)
(32)
It follows that 119879(119903 1198911) le 119878(119903 119891
1) a contradiction Thus
we prove that 119891(119911) equiv 119865(119911) This completes the proof ofTheorem 1
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 5
Acknowledgments
Theauthors thank the referees for careful reading of the paperpointing out a gap in the previous version of this paperand giving many valuable suggestions Research is supportedby the NNSF of China (Grant no 11371149) and NSF ofGuangdong Province China (Grant no S2012010010121)
References
[1] W K Hayman Meromorphic Function Clarendon PressOxford UK 1964
[2] L Yang Value Distribution Theory Springer Berlin Germany1993
[3] L A Rubel and C C Yang Value Shared by an Entire Functionand Its Derivative Lecture Notes in Math Springer BerlinGermany 1977
[4] G G Gundersen ldquoMeromorphic functions that share finitevalues with their derivativerdquo Journal of Mathematical Analysisand Applications vol 75 no 2 pp 441ndash446 1980
[5] G G Gundersen ldquoErrata meromorphic functions that sharefinite values with their derivativerdquo Journal of MathematicalAnalysis and Applications vol 86 no 1 p 307 1982
[6] E Mues and N Steinmetz ldquoMeromorphe Funktionen die mitihrer Ableitung Werte teilenrdquo Manuscripta Mathematica vol29 no 2ndash4 pp 195ndash206 1979
[7] W Bergweiler and J K Langley ldquoZeros of differences of mero-morphic functionsrdquoMathematical Proceedings of the CambridgePhilosophical Society vol 142 no 1 pp 133ndash147 2007
[8] Y-M Chiang and S-J Feng ldquoOn the Nevanlinna characteristicof 119891(119911 + 120578) and difference equations in the complex planerdquoTheRamanujan Journal vol 16 no 1 pp 105ndash129 2008
[9] Y-M Chiang and S-J Feng ldquoOn the growth of logarithmicdifferences difference quotients and logarithmic derivatives ofmeromorphic functionsrdquo Transactions of the American Mathe-matical Society vol 361 no 7 pp 3767ndash3791 2009
[10] R G Halburd and R J Korhonen ldquoNevanlinna theory for thedifference operatorrdquo Annales Academiaelig Scientiarum FennicaeligMathematica vol 31 no 2 pp 463ndash478 2006
[11] R G Halburd and R J Korhonen ldquoDifference analogue ofthe lemma on the logarithmic derivative with applications todifference equationsrdquo Journal of Mathematical Analysis andApplications vol 314 no 2 pp 477ndash487 2006
[12] J Heittokangas R Korhonen I Laine J Rieppo and J ZhangldquoValue sharing results for shifts of meromorphic functions andsufficient conditions for periodicityrdquo Journal of MathematicalAnalysis and Applications vol 355 no 1 pp 352ndash363 2009
[13] J Heittokangas R Korhonen I Laine and J Rieppo ldquoUnique-ness of meromorphic functions sharing values with their shiftsrdquoComplexVariables and Elliptic Equations vol 56 no 1ndash4 pp 81ndash92 2011
[14] S Li and Z Gao ldquoEntire functions sharing one or two finitevalues CM with their shifts or difference operatorsrdquo Archiv derMathematik vol 97 no 5 pp 475ndash483 2011
[15] M L Fang ldquoUnicity theorems for meromorphic function andits differential polynomialrdquo Advances in Mathematics vol 24no 3 pp 244ndash249 1995
Research ArticleOn Positive Solutions and Mann Iterative Schemes of a ThirdOrder Difference Equation
Zeqing Liu1 Heng Wu1 Shin Min Kang2 and Young Chel Kwun3
1 Department of Mathematics Liaoning Normal University Dalian Liaoning 116029 China2Department of Mathematics and RINS Gyeongsang National University Jinju 660-701 Republic of Korea3 Department of Mathematics Dong-A University Pusan 614-714 Republic of Korea
Correspondence should be addressed to Young Chel Kwun yckwundauackr
Received 14 October 2013 Accepted 16 December 2013 Published 28 January 2014
Academic Editor Zhi-Bo Huang
Copyright copy 2014 Zeqing Liu et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The existence of uncountably many positive solutions and convergence of the Mann iterative schemes for a third order nonlinearneutral delay difference equation are proved Six examples are given to illustrate the results presented in this paper
1 Introduction and Preliminaries
Recently many researchers studied the oscillation nonoscil-lation and existence of solutions for linear and nonlinearsecond and third order difference equations and systemssee for example [1ndash23] and the references cited therein Bymeans of the Reccati transformation techniques Saker [18]discussed the third order difference equation
Δ3
119909119899+ 119901119899119909119899+1
= 0 forall119899 ge 1198990 (1)
and presented some sufficient conditions which ensure thatall solutions are to be oscillatory or tend to zero Utilizing theSchauder fixed point theorem Yan and Liu [22] proved theexistence of a bounded nonoscillatory solution for the thirdorder difference equation
Δ3
119909119899+ 119891 (119899 119909
119899 119909119899minus120591) = 0 forall119899 ge 119899
0 (2)
Agarwal [2] established the oscillatory and asymptotic prop-erties for the third order nonlinear difference equation
Δ3
119909119899+ 119902119899119891 (119909119899+1) = 0 forall119899 ge 1 (3)
Andruch-Sobiło andMigda [4] studied the third order lineardifference equation of neutral type
Δ3
(119909119899minus 119901119899119909120590119899) plusmn 119902119899119909120591119899= 0 forall119899 ge 119899
0 (4)
and obtained sufficient conditions which ensure that allsolutions of the equation are oscillatory Grace andHamedani[6] discussed the difference equation
Δ3
(119909119899minus 119909119899minus120591) plusmn 119902119899
1003816100381610038161003816119909119899minus120590
1003816100381610038161003816
3 sgn119909119899minus120590
= 0 forall119899 ge 0 (5)
and gave some new criteria for the oscillation of all solutionsand all bounded solutions
Our goal is to discuss solvability and convergence ofthe Mann iterative schemes for the following third ordernonlinear neutral delay difference equation
Δ3
(119909119899+ 119887119899119909119899minus120591) + Δℎ (119899 119909
ℎ1119899
119909ℎ2119899
119909ℎ119896119899
)
+119891 (119899 1199091198911119899
1199091198912119899
119909119891119896119899
) = 119888119899 forall119899 ge 119899
0
(6)
where 120591 119896 1198990isin N 119887
119899119899isinN1198990
119888119899119899isinN1198990
sub R ℎ 119891 isin 119862(N1198990
times
R119896R) ℎ119897119899119899isinN1198990
119891119897119899119899isinN1198990
sube N and
lim119899rarrinfin
ℎ119897119899= lim119899rarrinfin
119891119897119899= +infin 119897 isin 1 2 119896 (7)
By employing the Banach fixed point theorem and somenew techniques we establish the existence of uncountablymany positive solutions of (6) conceive a few Mann iter-ative schemes for approximating these positive solutionsand prove their convergence and the error estimates Sixnontrivial examples are included
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 470181 16 pageshttpdxdoiorg1011552014470181
2 Abstract and Applied Analysis
Throughout this paper we assume that Δ is the forwarddifference operator defined by Δ119909
119899= 119909119899+1
minus 119909119899 R =
(minusinfin +infin) R+ = [0 +infin) N0and N denote the sets of
nonnegative integers and positive integers respectively
N119905= 119899 119899 isin N with 119899 ge 119905 forall119905 isin N
120573 = min 1198990minus 120591 inf ℎ
119897119899 119891119897119899 1 le 119897 le 119896 119899 isin N
1198990
isin N
119867119899= max ℎ2
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
119865119899= max 1198912
119897119899 119897 isin 1 2 119896 forall119899 isin N
1198990
(8)
and 119897infin
120573represents the Banach space of all real sequences
on N120573with norm
119909 = sup119899isinN120573
1003816100381610038161003816100381610038161003816
119909119899
1198992
1003816100381610038161003816100381610038161003816
lt +infin for each 119909 = 119909119899119899isinN120573
isin 119897infin
120573
119860 (119873119872) = 119909 = 119909119899119899isinN120573
isin 119897infin
120573 119873 le
119909119899
1198992
le 119872 119899 isin N120573
for any 119872 gt 119873 gt 0
(9)
It is easy to see that 119860(119873119872) is a closed and convex subsetof 119897infin120573 By a solution of (6) wemean a sequence 119909
119899119899isinN120573
witha positive integer 119879 ge 119899
0+120591+120573 such that (6) holds for all 119899 ge
119879
Lemma 1 Let 119901119905119905isinN be a nonnegative sequence and 120591 isin N
(i) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199052
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119904=119899+119894120591suminfin
119905=119904119901119905= 0
(ii) If lim119899rarrinfin
(11198992
) suminfin
119905=119899+1205911199053
119901119905
= 0
then lim119899rarrinfin
(11198992
) suminfin
119894=1suminfin
119906=119899+119894120591suminfin
119904=119906suminfin
119905=119904119901119905= 0
Proof Note that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
(
infin
sum
119905=119899+119894120591
119901119905+
infin
sum
119905=119899+1+119894120591
119901119905+
infin
sum
119905=119899+2+119894120591
119901119905+ sdot sdot sdot )
=
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
(1 + 119905 minus 119899 minus 119894120591) 119901119905le
1
1198992
infin
sum
119894=1
infin
sum
119905=119899+119894120591
119905119901119905
=
1
1198992
(
infin
sum
119905=119899+120591
119905119901119905+
infin
sum
119905=119899+2120591
119905119901119905+
infin
sum
119905=119899+3120591
119905119901119905+ sdot sdot sdot )
le
1
1198992
infin
sum
119905=119899+120591
(1 +
119905 minus 119899 minus 120591
120591
) 119905119901119905=
1
1198992
infin
sum
119905=119899+120591
119905 minus 119899
120591
119905119901119905
le
1
1198992120591
infin
sum
119905=119899+120591
1199052
119901119905997888rarr 0 as 119899 997888rarr infin
(10)
that is
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119904=119899+119894120591
infin
sum
119905=119904
119901119905= 0 (11)
As in the proof of (10) we infer that
0 le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905
=
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
(1 + 119905 minus 119906) 119901119905
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119905=119906
119905119901119905le
1
1198992120591
infin
sum
119905=119899+120591
1199053
119901119905997888rarr 0 as 119899 997888rarr infin
(12)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
119901119905= 0 (13)
This completes the proof
2 Uncountably Many Positive Solutions andMann Iterative Schemes
In this section using the Banach fixed point theoremand Mann iterative schemes we establish the existence ofuncountably many positive solutions of (6) prove conver-gence of the Mann iterative schemes relative to these positivesolutions and compute the error estimates between theManniterative schemes and the positive solutions
Theorem 2 Assume that there exist twoconstants 119872 and 119873 with 119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896) minus 119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119875119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896) minus ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816
le 119877119899max 100381610038161003816
1003816119906119897minus 119906119897
1003816100381610038161003816 1 le 119897 le 119896
forall (119899 119906119897 119906119897) isin N1198990
times (R+
0)
2
1 le 119897 le 119896
(14)
1003816100381610038161003816119891 (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119876119899
1003816100381610038161003816ℎ (119899 119906
1 1199062 119906
119896)1003816100381610038161003816le 119882119899
forall (119899 119906119897) isin N1198990
times (R+
0) 1 le 119897 le 119896
(15)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (16)
lim119899rarrinfin
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (17)
119887119899= minus1 eventually (18)
Abstract and Applied Analysis 3
Then one has the following(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
+
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(19)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin 119860(119873119872) of (6)with lim
119899rarrinfin119911119899= +infin and has the following error estimate
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] such thatinfin
sum
119898=0
120572119898= +infin (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Firstly we show that (a) holds Put 119871 isin (119873119872) Itfollows from (16)sim(18) that there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (22)
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119873
(23)
119887119899= minus1 forall119899 ge 119879 (24)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879 119878119871119909119879 120573 le 119899 lt 119879
(25)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In light of (14) (15) (22)(23) and (25) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872)
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
4 Abstract and Applied Analysis
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119894=1
infin
sum
119906=119879+119894120591
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
lt min 119872 minus 119871 119871 minus 119873
(26)which yield that
119878119871(119860 (119873119872)) sube 119860 (119873119872)
1003817100381710038171003817119878119871119909 minus 1198781198711199101003817100381710038171003817le 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817 forall119909 119910 isin 119860 (119873119872)
(27)
which implies that 119878119871is a contraction in 119860(119873119872) The
Banach fixed point theorem and (27) ensure that 119878119871has a
unique fixed point 119911 = 119911119899119899isinN120573
isin 119860(119873119872) that is
119911119899= 1198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879
119911119899minus120591
= (119899 minus 120591)2
119871
+
infin
sum
119894=1
infin
sum
119906=119899+(119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
)
minus 119888119905] forall119899 ge 119879 + 120591
(28)which mean that119911119899minus 119911119899minus120591
= (2119899120591 minus 1205912
) 119871
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
(29)which yields thatΔ (119911119899minus 119911119899minus120591)
= 2120591119871 +
infin
sum
119904=119899
ℎ (119904 119911ℎ1119904
119911ℎ2119904
119911ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905]
forall119899 ge 119879 + 120591
Δ2
(119911119899minus 119911119899minus120591)
= minusℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
+
infin
sum
119905=119899
[119891 (119905 1199111198911119905
1199111198912119905
119911119891119896119905
) minus 119888119905] forall119899 ge 119879 + 120591
(30)
which gives that
Δ3
(119911119899minus 119911119899minus120591)
= minusΔℎ (119899 119911ℎ1119899
119911ℎ2119899
119911ℎ119896119899
)
minus 119891 (119905 1199111198911119899
1199111198912119899
119911119891119896119899
) + 119888119899 forall119899 ge 119879 + 120591
(31)
which together with (24) implies that 119911 = 119911119899119899isinN120573
is apositive solution of (6) in 119860(119873119872) Note that
119873 le
119911119899
1198992
le 119872 forall119899 isin N120573 (32)
which guarantees that lim119899rarrinfin
119911119899= +infin It follows from (19)
(22) (24) (25) and (27) that for any 119898 isin N0and 119899 ge 119879
1003816100381610038161003816100381610038161003816
119909119898+1119899
1198992
minus
119911119899
1198992
1003816100381610038161003816100381610038161003816
=
1
1198992
1003816100381610038161003816100381610038161003816100381610038161003816
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)] minus 119911
119899
1003816100381610038161003816100381610038161003816100381610038161003816
le (1 minus 120572119898)
1003816100381610038161003816119909119898119899minus 119911119899
1003816100381610038161003816
1198992
+ 120572119898
1003816100381610038161003816119878119871119909119898119899minus 119878119871119911119899
1003816100381610038161003816
1198992
le (1 minus 120572119898)1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817+ 120579120572119898
1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817
le [1 minus (1 minus 120579) 120572119898]1003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 119899 ge 119879
(33)
which implies that
1003817100381710038171003817119909119898+1
minus 1199111003817100381710038171003817le 119890minus(1minus120579)sum
119898
119894=01205721198941003817100381710038171003817119909119898minus 119911
1003817100381710038171003817 forall119898 isin N
0 (34)
That is (20) holds Thus Lemma 1 (20) and (21) guaranteethat lim
119898rarrinfin119909119898= 119911
Next we show that (b) holds Let 1198711 1198712
isin
(119873119872) and 1198711=1198712 As in the proof of (a) we deduce
similarly that for each 119888 isin 1 2 there exist constants 120579119888isin
(0 1) and 119879119888ge 1198990+ 120591 + 120573 and a mapping 119878
119871119888
satisfying
Abstract and Applied Analysis 5
(22)sim(27) where 120579 119871 and 119879 are replaced by 120579119888 119871119888 and 119879
119888
respectively and the mapping 119878119871119888
has a fixed point 119911119888 =
119911119888
119899119899isinN120573
isin 119860(119873119872) which is a positive solution of (6) in119860(119873119872) with lim
119899rarrinfin119911119888
119899= +infin It follows that
119911119888
119899= 1198992
119871119888
+
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
ℎ (119904 119911119888
ℎ1119904
119911119888
ℎ2119904
119911119888
ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 119911119888
1198911119905
119911119888
1198912119905
119911119888
119891119896119905
) minus 119888119905]
forall119899 ge 119879119888
(35)
which together with (14) and (20) means that for 119899 ge
max1198791 1198792
100381610038161003816100381610038161003816100381610038161003816
1199111
119899
1198992
minus
1199112
119899
1198992
100381610038161003816100381610038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119911
1
ℎ1119904
1199111
ℎ2119904
1199111
ℎ119896119904
)
minus ℎ (119904 1199112
ℎ1119904
1199112
ℎ2119904
1199112
ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119911
1
1198911119905
1199111
1198912119905
1199111
119891119896119905
) minus 119891 (119905 1199112
1198911119905
1199112
1198912119905
1199112
119891119896119905
)
10038161003816100381610038161003816
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
1
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
10038161199111
ℎ119897119904
minus 1199112
ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
10038161199111
119891119897119905
minus 1199112
119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
1198992
infin
sum
119894=1
infin
sum
119906=119899+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
minus
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
max 11987921 1198792
2
infin
sum
119894=1
infin
sum
119906=max1198791 1198792+119894120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
ge10038161003816100381610038161198711minus 1198712
1003816100381610038161003816minusmax 120579
1 1205792
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817
(36)
which yields that
100381710038171003817100381710038171199111
minus 119911210038171003817100381710038171003817ge
10038161003816100381610038161198711minus 1198712
1003816100381610038161003816
1 +max 1205791 1205792
gt 0 (37)
that is 1199111 =1199112
This completes the proof
Theorem 3 Assume that there exist two constants119872 and 119873with 119872 gt 119873 gt 0 and four nonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) and
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (38)
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (39)
119887119899= 1 eventually (40)
Then one has the following
(a) For any 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge
1198990+ 120591 + 120573 such that for each 119909
0= 119909
0119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871
minus
infin
sum
119894=1
119879+2119894120591minus1
sum
119906=119879+(2119894minus1)120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
) minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(41)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
6 Abstract and Applied Analysis
Proof Let 119871 isin (119873119872) It follows from (38)sim(40) that thereexist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 =
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (42)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)) lt min 119872 minus 119871 119871 minus 119873
(43)
119887119899= 1 forall119899 ge 119879 (44)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871
minus
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus119888119905] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(45)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Using (14) (15) (42) (43)and (45) we get that for each 119909 = 119909
119899119899isinN120573
119910 = 119910119899119899isinN120573
isin
119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
le
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) = 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus 119871
10038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119894=1
119899+2119894120591minus1
sum
119906=119899+(2119894minus1)120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)])
lt min 119872 minus 119871 119871 minus 119873
(46)
which imply (27) The rest of the proof is similar to the proofof Theorem 2 and is omitted This completes the proof
Theorem 4 Assume that there exist three constants 119887 119872and 119873 with (1 minus 119887)119872 gt 119873 gt 0 and four nonnega-tive sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and0 le 119887119899le 119887 lt 1 eventually (47)
Then one has the following(a) For any 119871 isin (119887119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any
1199090
= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated bythe scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+1205721198981198992
119871 minus 119887119899119909119898119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus 119887119879119909119898119879minus120591
minus
infin
sum
119906=119879
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(48)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
Abstract and Applied Analysis 7
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119887119872 + 119873119872) It follows from (38) (39) and(47) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = 119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus 119887119872 minus119873
0 le 119887119899le 119887 lt 1 forall119899 ge 119879
(49)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus 119887119899119909119899minus120591
minus
infin
sum
119906=119899
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(50)for each 119909 = 119909
119899119899isinN120573
isin 119860(119873119872) In view of (14) (15) and(49) and (50) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 11987910038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le 119887119899
100381610038161003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
(119899 minus 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 minus 120591)2
1198992
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le 1198871003817100381710038171003817119909 minus 119910
1003817100381710038171003817
+
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max ℎ2
119897119904 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1198912
119897119905 1 le 119897 le 119896]
le [119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817= 120579
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus 119887119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus 119887119872
minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus 119887119872 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minus 119887119872 minusmin 119872 minus 119871 119871 minus 119887119872 minus119873 ge 119873
(51)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 5 Assume that there exist constants 119887 119872 and 119873with (1 + 119887)119872 gt 119873 gt 0 and four nonnegative sequences119875119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14)(15) (38) (39) and
minus1 lt 119887 le 119887119899le 0 eventually (52)
Then one has the following
(a) For any 119871 isin (119873 (1 + 119887)119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for
any 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by(48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
8 Abstract and Applied Analysis
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (119873 (1 + 119887)119872) It follows from (38) (39) and(52) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 satisfying
120579 = minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905) (53)
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 119887)119872 minus 119871 119871 minus 119873
(54)
minus1 lt 119887 le 119887119899le 0 forall119899 ge 119879 (55)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by (50) By virtue of
(15) (50) (53) and (55) we infer that for all 119909 = 119909119899119899isinN120573
119910 = 119910
119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le 119887119899
1003816100381610038161003816100381610038161003816
119909119899minus120591
minus 119910119899minus120591
1198992
1003816100381610038161003816100381610038161003816
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le [minus119887 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 minus 119887119872
+
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 minus 119887119872 +
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 minus 119887119872 +min (1 + 119887)119872 minus 119871 119871 minus 119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt 119871 minusmin (1 + 119887)119872 minus 119871 119871 minus 119873 ge 119873
(56)
That is (27) holds The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 6 Assume that there exist constants 119887119872 and 119873 with (1 minus 1119887)119872 gt 119873 gt 0 and fournonnegative sequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882
119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899ge 119887 gt 1 eventually (57)
Then one has the following(a) For any 119871 isin ((1119887)119872 + 119873119872) there exist 120579 isin
(0 1) and 119879 ge 1198990+ 120591 + 120573 such that for any 119909
0=
1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterative sequence119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 1205721198981198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+1205721198981198792
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(58)
Abstract and Applied Analysis 9
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin ((1119887)119872 + 119873119872) It follows from (38) (39)and (57) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 =
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (59)
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min 119872 minus 119871 119871 minus
1
119887
119872 minus119873
(60)
119887119899ge 119887 gt 1 forall119899 ge 119879 (61)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[(119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
)
minus 119888119905)] 119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(62)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) In view of (14) (15)and (59)∽(62) we obtain that for each 119909 = 119909
119899119899isinN120573
119910 =
119910119899119899isinN120573
isin 119860(119873119872) and 119899 ge 119879
10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
+
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
+
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le 119871 +
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816
+1003816100381610038161003816119888119905
1003816100381610038161003816]
le 119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt 119871 +min 119872 minus 119871 119871 minus
1
119887
119872 minus119873 le 119872
119878119871119909119899
1198992
ge 119871 minus
1
119887
119872
minus
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge 119871 minus
1
119887
119872 minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
gt 119871 minus
1
119887
119872 minusmin 119872 minus 119871 119871 minus
1
119887
119872 minus119873 ge 119873
(63)
which imply (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
Theorem 7 Assume that there exist constants 119887 119872and 119873 with (1 + 1119887)119872 gt 119873 gt 0 and four nonnegativesequences 119875
119899119899isinN1198990
119876119899119899isinN1198990
119877119899119899isinN1198990
and 119882119899119899isinN1198990
satisfying (14) (15) (38) (39) and
119887119899le 119887 lt minus1 eventually (64)
10 Abstract and Applied Analysis
Then one has the following
(a) For any 119871 isin (minus(1 + 1119887)119872 minus119873) there exist 120579 isin (0 1)and 119879 ge 119899
0+ 120591 + 120573 such that for any 119909
0= 1199090119899119899isinN120573
isin
119860(119873119872) the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by the scheme
119909119898+1119899
=
(1 minus 120572119898) 119909119898119899
+ 120572119898 minus 1198992
119871 minus
119909119898119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
119899 ge 119879 119898 ge 0
(1 minus 120572119898) 119909119898119879
+120572119898 minus 119879
2
119871 minus
119909119898119879+120591
119887119879+120591
minus
infin
sum
119906=119879+120591
infin
sum
119904=119906
[ℎ (119904 119909119898ℎ1119904
119909119898ℎ2119904
119909119898ℎ119896119904
)
minus
infin
sum
119905=119904
(119891 (119905 1199091198981198911119905
1199091198981198912119905
119909119898119891119896119905
)
minus 119888119905)]
120573 le 119899 lt 119879 119898 ge 0
(65)
converges to a positive solution 119911 = 119911119899119899isinN120573
isin
119860(119873119872) of (6) with lim119899rarrinfin
119911119899= +infin and has the
error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in [0 1] satisfying (21)
(b) Equation (6) possesses uncountablymany positive solu-tions in 119860(119873119872)
Proof Put 119871 isin (minus(1 + 1119887)119872 minus119873) It follows from (38)(39) and (64) that there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 +
120573 satisfying
120579 = minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)] (66)
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt min (1 + 1
119887
)119872 minus 119871 119871 minus
1
119887
119872 minus119873
(67)
119887119899ge 119887 gt 1 forall119899 ge 119879 (68)
Define a mapping 119878119871 119860(119873119872) rarr 119897
infin
120573by
119878119871119909119899
=
minus1198992
119871 minus
119909119899+120591
119887119899+120591
minus
1
119887119899+120591
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
ℎ (119904 119909ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus
infin
sum
119905=119904
[119891 (119905 1199091198911119905
1199091198912119905
119909119891119896119905
) minus 119888119905]
119899 ge 119879
119878119871119909119879 120573 le 119899 lt 119879
(69)
for each 119909 = 119909119899119899isinN120573
isin 119860(119873119872) Making use of (15) (66)(68) and (69) we conclude that10038161003816100381610038161003816100381610038161003816
119878119871119909119899
1198992
minus
119878119871119910119899
1198992
10038161003816100381610038161003816100381610038161003816
le minus
1
119887119899+120591
1003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
1198992
1003816100381610038161003816100381610038161003816
minus
1
119887119899+1205911198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
[
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
minus ℎ (119904 119910ℎ1119904
119910ℎ2119904
119910ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
minus 119891 (119905 1199101198911119905
1199101198912119905
119910119891119896119905
)
10038161003816100381610038161003816]
le minus
1
119887119899+120591
100381610038161003816100381610038161003816100381610038161003816
119909119899+120591
minus 119910119899+120591
(119899 + 120591)2
100381610038161003816100381610038161003816100381610038161003816
(119899 + 120591)2
1198992
minus
1
119887119899+1205911198992
infin
sum
119906=119899
infin
sum
119904=119906
[119877119904max 1003816100381610038161003816
1003816119909ℎ119897119904
minus 119910ℎ119897119904
10038161003816100381610038161003816 1 le 119897 le 119896
+
infin
sum
119905=119904
119875119905max 1003816100381610038161003816
1003816119909119891119897119905
minus 119910119891119897119905
10038161003816100381610038161003816 1 le 119897 le 119896]
le minus
1
119887
[(1 +
120591
119879
)
2
+
1
1198792
infin
sum
119906=119879
infin
sum
119904=119906
(119877119904119867119904+
infin
sum
119905=119904
119875119905119865119905)]
1003817100381710038171003817119909 minus 119910
1003817100381710038171003817
= 1205791003817100381710038171003817119909 minus 119910
1003817100381710038171003817
119878119871119909119899
1198992
le minus119871 minus
119872
119887
minus
1
1198871198992
times
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
Abstract and Applied Analysis 11
le minus119871 minus
119872
119887
minus
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
(119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816))
lt minus119871 minus
119872
119887
+min (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 le 119872
119878119871119909119899
1198992
ge minus119871
+
1
1198871198992
infin
sum
119906=119899+120591
infin
sum
119904=119906
10038161003816100381610038161003816ℎ (119904 119909
ℎ1119904
119909ℎ2119904
119909ℎ119896119904
)
10038161003816100381610038161003816
+
infin
sum
119905=119904
[
10038161003816100381610038161003816119891 (119905 119909
1198911119905
1199091198912119905
119909119891119896119905
)
10038161003816100381610038161003816+1003816100381610038161003816119888119905
1003816100381610038161003816]
ge minus119871 +
1
1198871198792
infin
sum
119906=119879
infin
sum
119904=119906
[119882119904+
infin
sum
119905=119904
(119876119905+1003816100381610038161003816119888119905
1003816100381610038161003816)]
gt minus119871 minusmin (1 + 1
119887
)119872 + 119871 minus119871 minus 119873 ge 119873
(70)
which yield (27) The rest of the proof is similar to that ofTheorem 2 and is omitted This completes the proof
3 Examples
In this section we suggest six examples to explain the resultspresented in Section 2
Example 1 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus 119909119899minus120591) + Δ(
sin2119909119899minus3
1198997
) +
1
(1198999+ 21198995+ 1) (1 + 119909
2
1198992)
=
1198992
minus 2119899
1198998+ 1198993+ 1
forall119899 ge 4
(71)
where 120591 isin N is fixed Let 1198990= 4 119896 = 1 and 120573 = min4minus120591 1
and let119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= minus1 119888
119899=
1198992
minus 2119899
1198998+ 1198993+ 1
119891 (119899 119906) =
1
(1198999+ 21198995+ 1) (1 + 119906
2)
ℎ (119899 119906) =
sin21199061198997
1198911119899= 1198992
119865119899= 1198994
ℎ1119899= 119899 minus 3 119867
119899= (119899 minus 3)
2
119875119899=
2119872
1198999
119876119899=
1
1198999
119877119899=
2
1198997
119882119899=
1
1198997
forall (119899 119906) isin N1198990
timesR
(72)
It is easy to see that (14) (15) and (18) are satisfied Note that
1
1198992
infin
sum
119905=119899+120591
1199052max 119877
119905119867119905119882119905
=
1
1198992
infin
sum
119905=119899+120591
1199052max2(119905 minus 3)
2
1199057
1
1199057
=
infin
sum
119905=119899+120591
2(119905 minus 3)2
+ 1
1199055
le
2
1198992
infin
sum
119905=119899+120591
1
1199053
997888rarr 0 as 119899 997888rarr infin
1
1198992
infin
sum
119905=119899+120591
1199053max 119875
119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119905=119899+120591
1199053max2119872
1199055
1
1199059
100381610038161003816100381610038161199052
minus 2119905
10038161003816100381610038161003816
1199058+ 1199053+ 1
le
max 1 2119872
1198992
infin
sum
119905=119899+120591
1
1199052
997888rarr 0 as 119899 997888rarr infin
(73)
which together with Lemma 1 yield that (16) and (17) holdIt follows from Theorem 2 that (71) possesses uncountablymany positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 119899
0+120591+120573 such
that for each 1199090= 1199090119899119899isinN120573
isin 119860(119873119872) the Mann iterativesequence 119909
119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (19)converges to a positive solution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of(71) with lim
119899rarrinfin119911119899= +infin and has the error estimate (20)
where 120572119898119898isinN0
is an arbitrary sequence in [0 1] satisfying(21)
Example 2 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ 119909119899minus120591) + Δ(
sin211990931198993+1
1198993(1198996+ 2) (1 + 119909
4
21198992minus3
)
)
+
(minus1)119899
1198993
(1199091198992minus119899minus1
+ 119909(119899+1)(119899+2)
)
(11989913+ 1198995+ 1) (1 + 119909
2
1198992minus119899minus1
+ 1199092
(119899+1)(119899+2))
=
1198992
minus ln 1198991198996+ 1198995+ 1
forall119899 ge 5
(74)
where 120591 isin N is fixed Let 1198990= 5 119896 = 2 and 120573 = 5 minus 120591 and let
119872 and 119873 be two positive constants with 119872 gt 119873 and
119887119899= 1 119888
119899=
1198992
minus ln 1198991198996+ 1198995+ 1
119891 (119899 119906 V) =(minus1)119899
1198993
(119906 + V)(11989913+ 1198995+ 1) (1 + 119906
2+ V2)
12 Abstract and Applied Analysis
ℎ (119899 119906 V) =sin2V
1198993(1198996+ 2) (1 + 119906
4)
1198911119899= 1198992
minus 119899 minus 1
1198912119899= (119899 + 1) (119899 + 2)
119865119899= (119899 + 1)
2
(119899 + 2)2
ℎ1119899= 21198992
minus 3
ℎ2119899= 31198993
+ 1
119867119899= (3119899
3
+ 1)
2
119875119899= 119876119899=
4
11989910
119877119899= 119882119899=
10
1198999
forall (119899 119906 V) isin N1198990
timesR2
(75)
It is clear that (14) (15) and (40) are fulfilled Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
10(31199043
+ 1)
2
1199049
10
1199049
le
160
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
160
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(76)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (77)
Observe that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max4(119905 + 1)2
(119905 + 2)2
11990510
4
11990510
1199052
minus ln 1199051199056+ 1199055+ 1
le
196
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
=
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
119905 minus 119906 + 1
1199054
le
196
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
196
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(78)
which yields that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (79)
Thus Theorem 3 guarantees that (74) possesses uncount-ably positive solutions in 119860(119873119872) On the other hand forany 119871 isin (119873119872) there exist 120579 isin (0 1) and 119879 ge 120591 +
1198990+ 120573 such that the Mann iterative sequence 119909
119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (74) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 3 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 + 3 ln 1198992 + 4 ln 119899
119909119899minus120591)
+ Δ(
(minus1)119899 sin (119890minus119899
2|11990951198992minus3|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|11990921198993+1|
)
+
(minus1)119899
1198996(1 + 119909
2
119899minus3)
minus
1
(1198997+ 21198994minus 1) (1 + 119909
2
119899+4)
=
3(minus1)119899
1198992
911989910ln3119899
forall119899 ge 7
(80)
where 120591 isin N is fixed Let 1198990= 7 119896 = 2 119887 = 34 and
120573 = min7 minus 120591 4 and let119872 and119873 be two positive constantswith 119872 gt 4119873 and
119887119899=
1 + 3 ln 1198992 + 4 ln 119899
119888119899=
3(minus1)119899
1198992
911989910ln3119899
119891 (119899 119906 V) =(minus1)119899
1198996(1 + 119906
2)
minus
1
(1198997+ 21198994minus 1) (1 + V2)
ℎ (119899 119906 V) =(minus1)119899 sin (119890minus119899
2|119906|
)
11989915minus radic119899 + 3
+
1198992
+ (minus1)119899(119899+1)2
(11989912+ 611989910+ 7) 119890
|V|
1198911119899= 119899 minus 3 119891
2119899= 119899 + 4
119865119899= (119899 + 4)
2
ℎ1119899= 51198992
minus 3
ℎ2119899= 21198993
+ 1 119867119899= (2119899
3
+ 1)
2
119875119899= 119876119899=
3
1198996
119877119899= 119882119899=
2
11989910
forall (119899 119906 V) isin N1198990
timesR2
(81)
It is not difficult to verify that (14) (15) and (47) are fulfilledNote that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max
2(21199043
+ 1)
2
11990410
2
11990410
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
18
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(82)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (83)
Abstract and Applied Analysis 13
Observe that1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max3(119905 + 4)2
1199056
3
1199056
100381610038161003816100381610038161003816100381610038161003816
3(minus1)119905
1199052
911990510ln3119905
100381610038161003816100381610038161003816100381610038161003816
le
12
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199054
le
12
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199053
le
12
1198992
infin
sum
119905=119899
1
1199052
997888rarr 0 as 119899 997888rarr infin
(84)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (85)
That is (38) and (39) hold Consequently Theorem 4 impliesthat (80) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((34)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (41) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (80) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 4 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+
1 minus 51198993
2 + 61198993
119909119899minus120591) + Δ(
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119909
2
3119899minus7)
)
+
sin (119899211990931198992minus2)
(radic119899 + 14)22
=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
forall119899 ge 9
(86)
where 120591 isin N is fixed Let 1198990= 9 119896 = 1 119887 = minus56 and
120573 = 9 minus 120591 and let 119872 and 119873 be two positive constants with119872 gt 6119873 and
119887119899=
1 minus 51198993
2 + 61198993
119888119899=
(minus1)119899
1198993
+ 51198992
+ 4119899 minus 2
1198999+ 1198998+ 21198995+ 1198993+ 7
119891 (119899 119906) =
sin (1198992119906)
(radic119899 + 14)22
ℎ (119899 119906) =
21198992
+ 119899 minus 1
(1198998+ 31198996+ 2) (1 + 119906
2)
1198911119899= 31198992
minus 2 119865119899= (3119899
2
minus 2)
2
ℎ1119899= 3119899 minus 7
119867119899= (3119899 minus 7)
2
119875119899= 119876119899=
3
1198999
119877119899= 119882119899=
1
1198995
forall (119899 119906) isin N1198990
timesR
(87)
Obviously (14) (15) and (52) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(3119904 minus 7)2
1199045
1
1199045
le
9
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199043
le
9
1198992
infin
sum
119904=119899
1
1199042
997888rarr 0 as 119899 997888rarr infin
(88)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (89)
Notice that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max
3(31199052
minus 2)
2
1199059
3
1199059
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905
1199053
+ 51199052
+ 4119905 minus 2
1199059+ 1199058+ 21199055+ 1199053+ 7
100381610038161003816100381610038161003816100381610038161003816
le
27
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
27
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
27
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(90)
which gives that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (91)
That is (38) and (39) hold Thus Theorem 5 shows that(86) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (119873 (16)119872)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that the Mann
iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generatedby (48) converges to a positive solution 119911 = 119911
119899119899isinN120573
isin
119860(119873119872) of (86) with lim119899rarrinfin
119911119899= +infin and has the error
estimate (20) where 120572119898119898isinN0
is an arbitrary sequencein [0 1] satisfying (21)
14 Abstract and Applied Analysis
Example 5 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899+ (
120587
2
+ 119899 sin 1119899
) 119909119899minus120591)
+ Δ(
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (11989921199092119899+1
))
)
+
119899 sin (119899119909119899minus2)
2 + (119899 + 5)16
=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
forall119899 ge 3
(92)
where 120591 isin N is fixed Let 1198990= 3 119896 = 1 119887 = 1205872 and
120573 = min3 minus 120591 1 and let119872 and119873 be two positive constantswith (1 minus 2120587)119872 gt 119873 and
119887119899=
120587
2
+ 119899 sin 1119899
119888119899=
(minus1)119899minus1cos3 (1198992 + 1)11989916+ ln 119899
119891 (119899 119906) =
119899 sin (119899119906)2 + (119899 + 5)
16
ℎ (119899 119906) =
(minus1)119899(119899+1)2
(119899 + 4)8
(119899 + 5)3
(1 + cos (1198992119906))
1198911119899= 119899 minus 2 119865
119899= (119899 minus 2)
2
ℎ1119899= 2119899 + 1 119867
119899= (2119899 + 1)
2
119875119899= 119876119899=
1
11989914
119877119899= 119882119899=
2
1198999
forall (119899 119906) isin N1198990
timesR
(93)
Clearly (14) (15) and (61) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max2(2119904 + 1)2
1199049
2
1199049
le
18
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199047
le
18
1198992
infin
sum
119904=119899
1
1199046
997888rarr 0 as 119899 997888rarr infin
(94)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 2)2
11990514
1
11990514
10038161003816100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1cos3 (1199052 + 1)11990516+ ln 119905
10038161003816100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
11990514
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
11990513
le
1
1198992
infin
sum
119905=119899
1
11990512
997888rarr 0 as 119899 997888rarr infin
(95)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (96)
That is (38) and (39) hold Consequently Theorem 6 impliesthat (92) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin ((2120587)119872 +
119873119872) there exist 120579 isin (0 1) and 119879 ge 1198990+ 120591 +
120573 such that the Mann iterative sequence 119909119898119898isinN0
=
119909119898119899119899isinN120573
119898isinN0
generated by (58) converges to a positivesolution 119911 = 119911
119899119899isinN120573
isin 119860(119873119872) of (92) with lim119899rarrinfin
119911119899=
+infin and has the error estimate (20) where 120572119898119898isinN0
is anarbitrary sequence in [0 1] satisfying (21)
Example 6 Consider the third order nonlinear neutral delaydifference equation
Δ3
(119909119899minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119909119899minus120591) + Δ(
cos ((minus1)119899119890119899)
(119899 + 7)6
radic1 +1003816100381610038161003816119909119899minus2
1003816100381610038161003816
)
+
sin (1198992119909119899minus1)
1198999+ 31198995+ 21198992+ 1
=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
forall119899 ge 6
(97)
where 120591 isin N is fixed Let 1198990= 6 119896 = 1 119887 = minus2 and 120573 =
min6 minus 120591 3 and let 119872 and 119873 be two positive constantswith (12)119872 gt 119873 and
119887119899= minus
21198995
+ 91198992
minus 1
1198995+ 31198992+ 2
119888119899=
(minus1)119899minus1
1198994
+ 41198992
+ 119899 minus 1
11989911+ 61198993+ 7119899 + 2
119891 (119899 119906) =
sin (1198992119906)1198999+ 31198995+ 21198992+ 1
Abstract and Applied Analysis 15
ℎ (119899 119906) =
cos ((minus1)119899119890119899)(119899 + 7)
6
radic1 + |119906|
1198911119899= 119899 minus 1 119865
119899= (119899 minus 1)
2
ℎ1119899= 119899 minus 2 119867
119899= (119899 minus 2)
2
119875119899= 119876119899=
1
1198997
119877119899= 119882119899=
1
1198996
forall (119899 119906) isin N1198990
timesR
(98)
Obviously (14) (15) and (64) are satisfied Note that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max(119904 minus 2)2
1199046
1
1199046
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
1
1199044
le
1
1198992
infin
sum
119904=119899
1
1199043
997888rarr 0 as 119899 997888rarr infin
(99)
which means that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
max 119877119904119867119904119882119904 = 0 (100)
It is clear that
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816
=
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max(119905 minus 1)2
1199057
1
1199057
100381610038161003816100381610038161003816100381610038161003816
(minus1)119905minus1
1199054
+ 41199052
+ 119905 minus 1
11990511+ 61199053+ 7119905 + 2
100381610038161003816100381610038161003816100381610038161003816
le
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
1
1199055
le
1
1198992
infin
sum
119906=119899
infin
sum
119905=119906
1
1199054
le
1
1198992
infin
sum
119905=119899
1
1199053
997888rarr 0 as 119899 997888rarr infin
(101)
which implies that
lim119899rarrinfin
1
1198992
infin
sum
119906=119899
infin
sum
119904=119906
infin
sum
119905=119904
max 119875119905119865119905 1198761199051003816100381610038161003816119888119905
1003816100381610038161003816 = 0 (102)
That is (38) and (39) hold Consequently Theorem 7 impliesthat (97) possesses uncountably many positive solutionsin 119860(119873119872) On the other hand for any 119871 isin (minus1198722 minus119873)there exist 120579 isin (0 1) and 119879 ge 119899
0+ 120591 + 120573 such that
the Mann iterative sequence 119909119898119898isinN0
= 119909119898119899119899isinN120573
119898isinN0
generated by (65) converges to a positive solution 119911 =
119911119899119899isinN120573
isin 119860(119873119872) of (97) with lim119899rarrinfin
119911119899= +infin and
has the error estimate (20) where 120572119898119898isinN0
is an arbitrarysequence in[0 1] satisfying (21)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This research was supported by the Science Research Foun-dation of Educational Department of Liaoning Province(L2012380)
References
[1] M H Abu-Risha ldquoOscillation of second-order linear differenceequationsrdquo Applied Mathematics Letters vol 13 no 1 pp 129ndash135 2000
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker New York NY USA 2nd edition 2000
[3] R P Agarwal and J Henderson ldquoPositive solutions and nonlin-ear eigenvalue problems for third-order difference equationsrdquoComputers amp Mathematics with Applications vol 36 no 10-12pp 347ndash355 1998
[4] A Andruch-Sobiło and M Migda ldquoOn the oscillation ofsolutions of third order linear difference equations of neutraltyperdquoMathematica Bohemica vol 130 no 1 pp 19ndash33 2005
[5] Z Dosla and A Kobza ldquoGlobal asymptotic properties of third-order difference equationsrdquo Computers amp Mathematics withApplications vol 48 no 1-2 pp 191ndash200 2004
[6] S R Grace and G G Hamedani ldquoOn the oscillation of certainneutral difference equationsrdquo Mathematica Bohemica vol 125no 3 pp 307ndash321 2000
[7] J Cheng ldquoExistence of a nonoscillatory solution of a second-order linear neutral difference equationrdquo Applied MathematicsLetters vol 20 no 8 pp 892ndash899 2007
[8] LKongQKong andB Zhang ldquoPositive solutions of boundaryvalue problems for third-order functional difference equationsrdquoComputersampMathematics withApplications vol 44 no 3-4 pp481ndash489 2002
[9] I Y Karaca ldquoDiscrete third-order three-point boundary valueproblemrdquo Journal of Computational and Applied Mathematicsvol 205 no 1 pp 458ndash468 2007
[10] W-T Li and J P Sun ldquoExistence of positive solutions of BVPsfor third-order discrete nonlinear difference systemsrdquo AppliedMathematics and Computation vol 157 no 1 pp 53ndash64 2004
[11] W-T Li and J-P Sun ldquoMultiple positive solutions of BVPs forthird-order discrete difference systemsrdquo Applied Mathematicsand Computation vol 149 no 2 pp 389ndash398 2004
[12] Z Liu M Jia S M Kang and Y C Kwun ldquoBounded positivesolutions for a third order discrete equationrdquo Abstract andApplied Analysis vol 2012 Article ID 237036 12 pages 2012
[13] Z Liu S M Kang and J S Ume ldquoExistence of uncountablymany bounded nonoscillatory solutions and their iterativeapproximations for second order nonlinear neutral delay dif-ference equationsrdquo Applied Mathematics and Computation vol213 no 2 pp 554ndash576 2009
[14] Z Liu Y Xu and S M Kang ldquoBounded oscillation criteriafor certain third order nonlinear difference equations withseveral delays and advancesrdquo Computers amp Mathematics withApplications vol 61 no 4 pp 1145ndash1161 2011
[15] M Migda and J Migda ldquoAsymptotic properties of solutions ofsecond-order neutral difference equationsrdquoNonlinear Analysis
16 Abstract and Applied Analysis
Theory Methods and Applications vol 63 no 5ndash7 pp e789ndashe799 2005
[16] N Parhi ldquoNon-oscillation of solutions of difference equationsof third orderrdquoComputersampMathematics withApplications vol62 no 10 pp 3812ndash3820 2011
[17] N Parhi and A Panda ldquoNonoscillation and oscillation ofsolutions of a class of third order difference equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 1 pp213ndash223 2007
[18] S H Saker ldquoNew oscillation criteria for second-order nonlinearneutral delay difference equationsrdquo Applied Mathematics andComputation vol 142 no 1 pp 99ndash111 2003
[19] S H Saker ldquoOscillation of third-order difference equationsrdquoPortugaliae Mathematica vol 61 no 3 pp 249ndash257 2004
[20] S Stevic ldquoOn a third-order system of difference equationsrdquoApplied Mathematics and Computation vol 218 no 14 pp7649ndash7654 2012
[21] X H Tang ldquoBounded oscillation of second-order delay dif-ference equations of unstable typerdquo Computers amp Mathematicswith Applications vol 44 no 8-9 pp 1147ndash1156 2002
[22] J Yan and B Liu ldquoAsymptotic behavior of a nonlinear delaydifference equationrdquo Applied Mathematics Letters vol 8 no 6pp 1ndash5 1995
[23] Z G Zhang and Q L Li ldquoOscillation theorems for second-order advanced functional difference equationsrdquo Computers ampMathematics with Applications vol 36 no 6 pp 11ndash18 1998
Research ArticleAlgebroid Solutions of Second Order ComplexDifferential Equations
Lingyun Gao1 and Yue Wang2
1 Department of Mathematics Jinan University Guangzhou Guangdong 510632 China2 School of Information Renmin University of China Beijing 100872 China
Correspondence should be addressed to Lingyun Gao tgaolyjnueducn
Received 28 November 2013 Accepted 17 December 2013 Published 2 January 2014
Academic Editor Zong-Xuan Chen
Copyright copy 2014 L Gao and Y WangThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Using value distribution theory and maximummodulus principle the problem of the algebroid solutions of second order algebraicdifferential equation is investigated Examples show that our results are sharp
1 Introduction and Main Results
We use the standard notations and results of the Nevanlinnatheory of meromorphic or algebroid functions see forexample [1 2]
In this paper we suppose that second order algebraicdifferential equation (3) admit at least one nonconstant ]-valued algebroid solution 119908(119911) in the complex plane Wedenote by 119864 a subset of [0infin) for which 119898(119864) lt infin and by119870 a positive constant where119898(119864) denotes the linearmeasureof 119864 119864 or 119870 does not always mean the same one when theyappear in the following
Let 119886119895119896(119895 = 0 1 119899 119896 = 0 1 119902
119895) be entire func-
tions without common zeroes such that 11988601199020
= 0 We put
119876119895(119911 119908) =
119902119895
sum
119896=0
119886119895119896119908119896
119902119895= deg119876119895
119908
119901 = max 119902119895+ 119895 119895 = 0 1 119899 minus 1
(1)
Some authors had investigated the problem of the exis-tence of algebroid solutions of complex differential equationsand they obtained many results ([2ndash10] etc)
In 1989 Toda [4] considered the existence of algebroidsolutions of algebraic differential equation of the form
119899
sum
119895=0
119876119895(119911 119908) (119908
1015840
)
119895
= 0 (2)
He obtained the following
Theorem A (see [4]) Let 119908(119911) be a nonconstant ]-valuedalgebroid solution of the above differential equation and all 119886
119895119896
are polynomials If 119901 lt 119899 + 119902119899 then 119908(119911) is algebraic
The purpose of this paper is to investigate algebroid solu-tions of the following second order differential equation inthe complex plane with the aid of the Nevanlinna theory andmaximum modulus principle of meromorphic or algebroidfunctions
119899
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
= 0 (3)
where 119876119895(119911 119908) = sum
119902119895
119896=0119886119895119896119908119896 119895 = 0 1 2 119899
We will prove the following two results
Theorem 1 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and all 119886
119895119896are polynomials
If 119901 le 119902119899 then 119908(119911) is algebraic 119901 = max119902
119895 119895 = 0 1 119899 minus
1
Theorem 2 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of differential equation (3) and the orders of all 119886
119895119896are
finite If 1199020gt max
1le119895le119899minus1119902119895+119895 then the following statements
are equivalent(a) 120575(infin119908) gt 0(b) 1199020= 119902119899+ 119899
(c) infin is a Picard exceptional value of 119908(119911)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 123049 4 pageshttpdxdoiorg1011552014123049
2 Abstract and Applied Analysis
2 Some Lemmas
Lemma 3 (see [2]) Suppose that 119908(119911) 119886119894(119911) (119894 = 1 2 119901)
are meromorphic functions and 119886119901(119911) = 0 Then one has
119898(119903
119901
sum
119894=1
119886119894(119911) 119908119894
) le 119901119898 (119903 119908) +
119901
sum
119894=1
119898(119903 119886119894(119911)) + 119874 (1)
(4)
Examining proof of Lemma 45 presented in [2 pp 192-193] we can verify Lemma 4
Lemma 4 Let 119908(119911) be a transcendental algebroid functionsuch that 119908(119911) has only finite number of poles and let 119908(119911)1199081015840
(119911) and 11990810158401015840(119911) have no poles in |119911| gt 1199030 Then for some
constants 119862119894gt 0 119894 = 1 2 3 and 119903 ge 119903
1ge 1199030it holds
119872(119903 119908) le 1198621+ 1198622119903 + 11986231199032
119872(119903 11990810158401015840
) (5)
where119872(119903 119908) = max|119911|=119903
|119908(119911)|
Lemma 5 (see [11]) The absolute values of roots of equation
119911119899
+ 1198861119911119899minus1
+ sdot sdot sdot + 119886119899= 0 (6)
are bounded by
max 119899 10038161003816100381610038161198861
1003816100381610038161003816 (119899
10038161003816100381610038161198862
1003816100381610038161003816)12
(1198991003816100381610038161003816119886119899
1003816100381610038161003816)1119899
(7)
Lemma 6 Let 119908(119911) be a nonconstant ]-valued algebroidsolution of the differential equation (3) and let 119886
119895119896be a
polynomial If 119901 lt 119899 + 119902119899 then
min 119899 119902119899minus 119901 log+119872(119903 119908)
+max 0 119902119899minus 119901 minus 119899 log+119872(119903 119908)
le +119874 (log 119903) + 119874 (1) (119903 notin 119864)
(8)
where119872(119903 119908) = max|119911|=119903
|119908(119911)| 119870 is a positive constant
Proof We first prove that the poles of 119908 are contained in thezeroes of 119886
119899119902119899
(119911)Suppose that 119911
0is a pole of 119908 of order 120591 and 119911
0is not the
zeroes of 119886119899119902119899
(119911) Then
119908 (119911) = (119911 minus 1199110)minus120591120582
1199081(119911) 119908
1(1199110) = 0infin
11990810158401015840
(119911) = (119911 minus 1199110)minus(120591+2120582)120582
1199082(119911) 119908
2(1199110) = 0infin
(9)
We rewrite differential equation (3) as follows
119886119899119902119899
(11990810158401015840
)
119899
=
119899minus1
sum
119895=0
119876119895(119911 119908) (119908
10158401015840
)
119895
(10)
It follows from (10) that
119902119899120591 + 119899 (120591 + 2120582) le 119901120591 + (119899 minus 1) (120591 + 2120582) (11)
Noting that 119901 le 119902119899 we have
120591 + 2120582 lt 0 (12)This is a contradiction
This shows that the poles of119908 are contained in the zeroesof 119886119899119902119899
(119911)We rewrite differential equation (3) as follows
119899
sum
119895=0
119876119895(119911 119908)119876
119899(119911 119908)
119899minus119895minus1
(11987611989911990810158401015840
)
119895
= 0 (13)
119872(119903119876111990810158401015840
) ge 119872(119903 119908)119902119899+119872(119903 119908
10158401015840
)
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
ge 119872(119903 119908)119902119899+
119872(119903 119908) minus 1198621minus 1198622119903
11986231199032
minus
119902119899minus1
sum
119896=0
119872(119903 119886119899119896)119872(119903 119908)
119896
(14)
For 119895 = 0 1 119899 minus 1 we have10038161003816100381610038161003816119876119895(119911119903 119908)119876
119899(119911119903 119908)119899minus119895minus110038161003816
100381610038161003816le 119870119872(119903 119908)
119902119895+119902119899(119899minus119895minus1)
(15)
Applying Lemma 5 to (13) at 119911 = 119911119903
119872(119903119876111990810158401015840
) le 119870119872(119903 119908)maxℎ
119895 119895=01119899minus1
(16)
where ℎ119895= (119902119895+ 119902119899(119899 minus 119895 minus 1))(119899 minus 119895)
From (14) and (15) we have
119872(119903 119908)119902119899le 119870 119872(119903 119908)
119902119899minus1
+ 1199032
119872(119903 119908)maxℎ
119895 119895=01119899minus1
le 119870 119872(119903 119908)119902119899minus1
+ 1199032
119872(119903 119908)119902119899+((119901minus119902
119899)119899)
(17)Note that
ℎ119895=
119902119895+ 119902119899(119899 minus 119895 minus 1)
119899 minus 119895
lt
119901 + 119902119899(119899 minus 119895 minus 1) minus 119895
119899 minus 119895
= 119902119899+
119901 minus 119902119899
119899
+
119895 (119901 minus 119902119899minus 119899)
119899 (119899 minus 119895)
le 119902119899+
119901 minus 119902119899
119899
119895 = 0 1 119899 minus 1
(18)
Dividing the inequality (17) by119872(119903 119908)
max119902119899minus1119902119899+((119902119899minus119901)119899) we obtain for 119903 notin 119864
119872(119903 119908)min1(119902
119899minus119901)119899
le 11987041 +
1199032
119872(119903 119908)max0(119902
119899minus119901minus119899)119899
(19)
which reduces to our inequality by calculating log+ of theboth sidesmin 119899 119902
119899minus 119901 log+119872(119903 119908)
le minusmax 0 119902119899minus 119901 minus 119899 log+119872(119903 119908) + 119874 (log 119903) + 119874 (1)
(20)Lemma 6 is complete
Abstract and Applied Analysis 3
3 Proof of Theorem 1
First we consider119873(119903 119908)Let 1199110be a pole of 119908 of 120591 Let 119905 be the order of zero of
119886119899119902119899
(119911) at 1199110
(i) When the order of the pole of119876119899(119908)(119908
10158401015840
)119899 is not equal
to that of other terms of the left-hand side of (10) at 1199110 we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 (21)
that is
120591 le
(119905 minus 2119899) 120582
119902119899+ 119899 minus 119901
(22)
(ii)When the order of pole of119876119899(119908)(119908
10158401015840
)119899 is equal to that
of some term 119876119896(119908)(119908
10158401015840
)119899 of the left-hand side of (10) at 119911
0
we get
119902119899120591 + 119899 (120591 + 2120582) minus 119905120582 le 119901120591 + 119899 (120591 + 2120582) (23)
that is
120591 le
119905120582
119902119899minus 119901
(24)
Combining cases (i) and (ii) we obtain
119873(119903 119908) le 1198708119873(119903
1
119886119899119902119899
) (25)
where1198708is a positive constant
Secondly by Lemma 6 we obtain
min 119899 119902119899minus 119901119898 (119903 119908)
le 119870[
119901
sum
119894=0
119898(119903 119886119894) +
119902119899minus1
sum
119896=0
119898(119903 119887119896)] + 119874 (log 119903) (119903 notin 119864)
(26)
Combining the inequalities (25) and (26) we have
119879 (119903 119908) = 119874 (log 119903) (119903 notin 119864) (27)
which shows that 119908 is an algebraic solution of (3)This completes the proof of Theorem 1
4 Proof of Theorem 2
(i) (a)rArr (b) Suppose that 120575(infin119908) gt 0 If 1199020gt 119902119899+ 119899 then
we have by (3)
1199081199020= minus
1
11988601199020
119899
sum
119895=1
119876119895(119908)119908
119895
(
11990810158401015840
119908
)
119895
minus
1199020minus1
sum
119896=0
1198860119896119908119896
(28)
Applying Lemma 3 to (28)
1199020119898(119903 119908) le (119902
0minus 1)119898 (119903 119908) +sum
119895119896
119898(119903 119886119895119896)
+ 119870119898(119903
11990810158401015840
119908
) + 119898(119903
1
11988601199020
) + 119874 (1)
(29)
Since 119908(119911) is admissible solution we have
119898(119903 119908) = 119878 (119903 119908) (30)
so that
120575 (119908infin) = 0 (31)
This is a contradiction Thus 1199020le 119902119899+ 119899
If 1199020lt 119902119899+119899 byTheorem 1119908(119911) is nonadmissibleThus
1199020= 119902119899+ 119899 (32)
(ii) (b) rArr (c) Let 1199020= 119902119899+ 119899 Then similar to the proof of
Lemma 6 we obtain that the poles of 119908(119911) are contained inthe set of 119886
119899119902119899
andinfin is a Picard exceptional value of 119908(119911)
(iii) (c) rArr (a) Let infin be a Picard exceptional value of 119908(119911)Then 120575(119908infin) = 1
5 Some Examples
Example 1 The differential equation
]21199082]minus111990810158401015840 minus [((2] + 1)1199084] + 21199082] minus ] + 1)] = 0 (33)
has a transcendental algebroid solution 119908(119911) = (tan 119911)1] Inthis case
119901 = 1199020= 4] gt 1 + 119902
1= 2] minus 1 + 1 = 2] (34)
Remark 7 Example 1 shows that the condition in Theorem 1is sharp
Example 2 Transcendental algebroid function 119908(119911) =
(sin 119911)12 is a 2-valued solution of the following differentialequation
161199086
(11990810158401015840
)
2
+ 21199085
11990810158401015840
minus (1199088
minus
1
2
1199086
+ 21199084
minus
1
2
1199082
+ 1) = 0
(35)
In this case
1199020= 8 119899 = 2 119902
1= 5 119902
2= 6 (36)
By Theorem 2 for transcendental algebroid function 119908(119911) =(sin 119911)12infin is a Picard exceptional value
Remark 8 Example 2 shows that the result in Theorem 2holds
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This project is project Supported by National Natural ScienceFoundation (10471065) of China and NSF of GuangdongProvince (04010474)
4 Abstract and Applied Analysis
References
[1] H Yi and C C YangTheory of the Uniqueness of MeromorphicFunctions Science Press Beijing China 1995 (Chinese)
[2] Y Z He and X Z Xiao Algebroid Functions and OrdinaryDifferential Equations Science Press Beijing China 1988
[3] Y Z He and X Z Xiao ldquoAdmissible solutions and ordinarydifferential equationsrdquo Contemporary Mathematics vol 25 pp51ndash61 1983
[4] N Toda ldquoOn algebroid solutions of some algebraic differentialequations in the complex planerdquo Japan Academy A vol 65 no4 pp 94ndash97 1989
[5] T Chen ldquoOne class of ordinary differential equations whichpossess algebroid solutions in the complex domainrdquo ChineseQuarterly Journal of Mathematics vol 6 no 4 pp 45ndash51 1991
[6] K Katajamaki ldquoValue distribution of certain differential poly-nomials of algebroid functionsrdquo Archiv der Mathematik vol 67no 5 pp 422ndash429 1996
[7] L Gao ldquoSome results on admissible algebroid solutions ofcomplex differential equationsrdquo Indian Journal of Pure andApplied Mathematics vol 32 no 7 pp 1041ndash1050 2001
[8] L-Y Gao ldquoOn some generalized higher-order algebraic dif-ferential equations with admissible algebroid solutionsrdquo IndianJournal of Mathematics vol 43 no 2 pp 163ndash175 2001
[9] L Gao ldquoOn the growth of solutions of higher-order algebraicdifferential equationsrdquoActaMathematica Scientia B vol 22 no4 pp 459ndash465 2002
[10] L Y Gao ldquoThe growth of single-valued meromorphic solutionsand finite branch solutionsrdquo Journal of Systems Science andMathematical Sciences vol 24 no 3 pp 303ndash310 2004
[11] T Takagi Lecture on Algebra Kyoritsu Tokyo Japan 1957(Japanese)
