Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 735360 12 pageshttpdxdoiorg1011552013735360

Research ArticleNew Oscillation Criteria for Second-Order ForcedQuasilinear Functional Differential Equations

Mervan PašiT

Department of Mathematics Faculty of Electrical Engineering and Computing University of Zagreb 10000 Zagreb Croatia

Correspondence should be addressed to Mervan Pasic mervanpasicgmailcom

Received 29 April 2013 Accepted 17 July 2013

Academic Editor Milan Tvrdy

Copyright copy 2013 Mervan PasicThis 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 establish some new interval oscillation criteria for a general class of second-order forced quasilinear functional differentialequations with 120601-Laplacian operator and mixed nonlinearities It especially includes the linear the one-dimensional p-Laplacianand the prescribedmean curvature quasilinear differential operators It continues some recently published results on the oscillationsof the second-order functional differential equations including functional arguments of delay advanced or delay-advanced typesThe nonlinear terms are of superlinear or supersublinear (mixed) types Consequences and examples are shown to illustrate thenovelty and simplicity of our oscillation criteria

1 Introduction

We study the oscillation of the following three kinds ofsecond-order forced quasilinear functional differential equa-tions of delay advanced and delay-advanced types

(119903 (119905) 120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (119909 (ℎ

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (ℎ119894 (119905))1003816100381610038161003816120572119894 sgn119909 (ℎ

119894(119905)) = 119890 (119905)

(1)

where 119905 ge 1199050gt 0 and 119903(119905) 119903

119894(119905) 119902119894(119905) and 119890(119905) are continuous

functions on [1199050infin) and 119909 = 119909(119905) 119909 isin 1198622((119905

0infin)R) is a

classic solution of (1) A continuous function 119909(119905) is said tobe nonoscillatory if there is a 119879 gt 119905

0such that 119909(119905) = 0 on

[119879infin) Otherwise 119909(119905) is said to be oscillatory Equation (1)is called oscillatory if all of its classic solutions are oscillatory

In the delay case ℎ119894(119905) = 120591

119894(119905) le 119905 119894 isin 1 119899 and

119903(119905) is nondecreasing in the advanced case ℎ119894(119905) = 120590

119894(119905) ge

119905 119894 isin 1 119899 and 119903(119905) is nonincreasing in the delay-advanced case ℎ

119894(119905) = 120591

119894(119905) le 119905 119894 isin 1 119898 ℎ

119894(119905) =

120590119894(119905) ge 119905 119894 isin 119898 + 1 119899 and 119903(119905) equiv const gt 0

The exponents 120572119894 satisfy a superlinear or a supersublinear

(mixed) condition On the function 120601 R rarr R whichappears in the first termof (1) we impose such conditions thatthe following three main classes of second-order differentialoperators are especially included the linear (119903(119905)1199091015840)1015840 the one-dimensional 119901-Laplacian (119903(119905)|1199091015840|119901minus11199091015840)

1015840

and the prescribedmean curvature operator (119903(119905)1199091015840(1 + 11990910158402)

minus12

)1015840

The function119891(119906) satisfies a usual growth condition and the coefficients119903119894(119905) and 119902

119894(119905) are positive only on some intervals where 119890(119905)

changes the signRecently Bai and Liu [1] have studied the oscillation of

second-order delay differential equation

(119903 (119905) 1199091015840

(119905))1015840

+119899

sum119894=1

119903119894(119905) 119909 (119905 minus 120591

119894)

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (119905 minus 120591119894)1003816100381610038161003816120572119894 sgn119909 (119905 minus 120591

119894) = 119890 (119905) 119905 ge 119905

0

(2)

2 Abstract and Applied Analysis

where 120591119894

ge 0 In Murugadass et al [2] authors havestudied the oscillation of the second-order quasilinear delaydifferential equation

(119903 (119905) (1199091015840

(119905))119901

)1015840

+ 119902 (119905) 119909119901

(119905 minus 120591)

+119899

sum119894=1

119902119894(119905) |119909 (119905 minus 120591)|

120572119894 sgn119909 (119905 minus 120591) = 119890 (119905) 119905 ge 1199050

(3)

where 119901 and 120572119894 are ratio of odd positive integers Later in

Hassan et al [3] authors consider the oscillation of the half-linear functional differential equation

(119903 (119905) (1199091015840

(119905))119901

)1015840

+ 1199030(119905) 119909119901(ℎ0(119905))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (ℎ119894 (119905))1003816100381610038161003816120572119894 sgn119909 (ℎ

119894(119905)) = 119890 (119905) 119905 ge 119905

0

(4)

where the function ℎ119894

= ℎ119894(119905) is positive continuous

functions with lim119905rarrinfin

ℎ119894(119905) = infin Moreover in [1ndash3] some

doubts concerning the proof of the main result of [4] areresolved The well-known variational technique that uses thegeneralized Philosrsquo results based on the so-called119867-functionhas been used in [1 Theorem 22] [2 Theorems 23 and 26]and [3 Theorems 25 26 and 27] (see also [5ndash11] and refer-ences therein) Since the second-order quasilinear differentialoperator (119903(119905)120601(1199091015840(119905)))

1015840 usually causes some difficulties inmany problems the application of previous method on (1) isstated here as an open problem In contrast to the precedingwe use a combination of the Riccati classic transformationa blow-up argument and a comparison pointwise principlerecently established in [12 13] but for differential equationswithout functional arguments It seems that our criteria areslightly simpler to be verified which is discussed on someexamples given in the next section

Among some recently published papers on the oscillationof quasilinear functional differential equations with bothdelay and advanced terms we point out an oscillationcriterion by Zafer [5] obtained for equation

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1199021(119905) |119909 (120591 (119905))|

120573minus1119909 (120591 (119905))

+ 1199022(119905) |119909 (120590 (119905))|

120574minus1119909 (120590 (119905)) = 119890 (119905) 119905 ge 0

(5)

where 119901 gt 0 120573 ge 119901 and 120574 ge 119901 On an interesting examplein [5] the author established the oscillations provided at leastone of constants appearing in the coefficients is sufficientlylarge (see also [6] for 119901 = 1) which is presented here in thenext section as a particular case of our main results

On the properties of some classes of the one-dimensionalquasilinear differential equations with 120601-Laplacian operatorswe refer reader to [14ndash20] and the references therein Aboutthe applications of second-order functional differential equa-tions in the mathematical description of certain phenomenain physics technics and biology (oscillation in a vacuumtube interaction of an oscillator with an energy sourcecoupled oscillators in electronics chemistry and ecologyrelativistic motion of a mass in a central field ship course

stabilization moving of the tip of a growing plant etc) wesuggest reading Kolmanovskii and Myshkis book [21]

2 Main Results and Examples

Let 120601 = 120601(V) be a function which appears in the first term of(1) such that

120601 isin 1198621

(RR) 120601 is odd and increasing function on R

(6)

120601 (V) V ge 1003816100381610038161003816120601 (V)1003816100381610038161003816(119901+1)119901

forallV isin R and some 119901 gt 0 (7)

These two assumptions are fulfilled respectively inthe linear case (119903(119905)1199091015840(119905))

1015840 where 120601(V) equiv V and 119901 =1 in the one-dimensional 119901-Laplacian quasilinear case(119903(119905)|1199091015840(119905)|

119901minus1

1199091015840(119905))1015840

where 120601(V) equiv |V|119901minus1V and 119901 gt 0 andin the mean prescribed curvature quasilinear case

[119903 (119905) 1199091015840

(119905) (1 + 11990910158402

(119905))minus12

]1015840

(8)

where 120601(V) = V(1 + V2)minus12 and 119901 = 1 The importance of theprescribed mean curvature quasilinear differential operatorslies in the capillarity type problems in fluid mechanics flux-limited diffusion phenomena and prescribedmean curvatureproblems see for instance [14 22 23]

The function 119891 = 119891(119906) which appears in the second termof (1) satisfies

119891 (119906) is odd function on R

119891 (119906)

119906119901ge 119870 gt 0 forall119906 gt 0 and some 119870 isin R

(9)

where 119901 gt 0 is from assumption (7)In the first two theorems below the exponents 120572

119894 satisfy

1205721ge sdot sdot sdot ge 120572

119898gt 119901 gt 120572

119898+1ge sdot sdot sdot ge 120572

119899gt 0 119898 isin N

120572119894gt 120572119894+1 119894 isin 1 119899 minus 1

there exists (119899 + 1)-tuple (1205780 1205781 120578

119899)

such that 0 lt 120578119894lt 1

119899

sum119894=1

120578119894lt 1 120578

0= 1 minus

119899

sum119894=1

120578119894

119899

sum119894=1

120572119894120578119894= 119901

(10)

where 119901 is from assumption (7) For instance if 119899 = 2 1205721=

52 119901 = 1 and 1205722= 12 then 120578

0= 1205781= 1205782= 13 On

assumption (10) see for instance [9 Lemma 1]Unlike recently published oscillation criteria for the linear

and half-linear second-order forced functional differential

Abstract and Applied Analysis 3

equations our oscillation criterion is only based on thefollowing elementary integral inequality

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905 ge 1

119895 isin 1 2

(11)

where 1198861lt 1198871le 1198862lt 1198872 119901 is from (7) 120587

119901= (119901(119901 +

1))120587 sin(119901120587(119901 + 1)) 120582119895

gt 0 and functions 119876119895(119905) and

119877119895(119905) are explicitly expressed by the coefficients of (1) just like

it is done in the next three main results

Theorem 1 (delay equation) One assumes (6) (7) (9) and(10) Let 119903(119905) be a nondecreasing positive function on [119905

0infin)

Let ℎ119894(119905) = 120591

119894(119905) le 119905 on [119905

0infin) and lim

119905rarrinfin120591119894(119905) = infin 119894 isin

1 2 119899 Let for every 119879 ge 1199050there exist 119886

1 1198871 1198862 1198872 119879 le

1198861lt 1198871le 120591min(1198862) le 119886

2lt 1198872such that

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [120591min (1198861) 1198871] cup [120591min (1198862) 1198872] (12)

119890 (119905) le 0 119900119899 [120591min (1198861) 1198871]

119890 (119905) ge 0 119900119899 [120591min (1198862) 1198872] (13)

where 120591min(119905) = min1205911(119905) 1205912(119905) 120591

119899(119905) Equation (1) is

oscillatory provided there are two real parameters 1205821 1205822gt 0

such that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

(14)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

120572119894120578119894

(15)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

120578119894appearing respectively in (7) (9) and (10)

As we can see in (13) the forcing term 119890(119905) can bean oscillatory function The main property of any intervaloscillation criterion (see [8]) is that the coefficients of theconsidered equation do not satisfy some conditions on thewhole [119905

0infin) than only on intervals [119886

1 1198871] cup [119886

2 1198872] The

main consequence of Theorem 1 is the following oscillationcriterion for (1) with 119903(119905) equiv 1 and 119903

119894(119905) equiv 0 119894 isin 1 2 119899

Corollary 2 One assumes (6) (7) (10) (12) and (13) Thenequation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905)) = 119890 (119905) (16)

is oscillatory provided

119901119901(119901+1)

2120587119901

int119887119895

119886119895

119876119895(119905) 119889119905 ge ( max

119905isin[119886119895 119887119895]119876119895(119905))

119901(119901+1)

gt 0

119895 isin 1 2

(17)

where 119876119895(119905) is defined in (15)

In particular for 119901 = 1 previous corollary takes thefollowing form

Corollary 3 Let (10) hold with 119901 = 1 120591119894(119905) le 119905 on [119905

0infin)

and lim119905rarrinfin

120591119894(119905) = infin 119894 isin 1 2 119899 Let for every 119879 ge

1199050there exist 119886

1 1198871 1198862 1198872 119879 le 119886

1lt 1198871le 120591min(1198862) le 119886

2lt

1198872such that assumptions (12) and (13) hold with 119903

119894(119905) equiv 0

Equation

11990910158401015840

(119905) +119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905)) = 119890 (119905) (18)

is oscillatory provided

1

120587int119887119895

119886119895

119876119895(119905) 119889119905 ge radic max

119905isin[119886119895 119887119895]119876119895(119905) gt 0 119895 isin 1 2 (19)

where 119876119895(119905) is defined in (15)

In the next examples and remarks we discuss the applica-tion of oscillation criteria from Corollary 3 and [1 Theorem22] on the following equation

11990910158401015840+ 1198981sin (119905) 1003816100381610038161003816119909 (1205911 (119905))

10038161003816100381610038161205721 sgn119909 (120591

1(119905))

+ 1198982cos (119905) 1003816100381610038161003816119909 (1205912 (119905))

10038161003816100381610038161205722 sgn119909 (120591

2(119905)) = minus119890

0cos (2119905)

(20)

where 1205721gt 1 gt 120572

2gt 0 120591

1(119905) = 120591

2(119905) = 119905 minus 1205878 and 119898

1 1198982

and 1198900are positive constants

Example 4 As a consequence of Corollary 3 we show inthis example that (20) is oscillatory provided the constants1198981 1198982 and 119890

0satisfy the following simple inequality

119868119895

120587radic1205781198901205780

01198981205781

11198981205782

2ge 1 119895 isin 1 2 (21)

where 120578119894satisfy (10) 120578 = prod

2

119894=0120578minus120578119894

119894 and 119868

1 1198682gt 0 are two real

numbers defined by

1198681= int1205874

1205878

119882(119905)119905 minus 1205878

119905119889119905

1198682= int1205872

31205878

119882(119905)119905 minus 31205878

119905 minus 1205874119889119905

119882 (119905) = |cos 2119905|1205780 |sin 119905|1205781 |cos 119905|1205782

(22)

Indeed let 120587119901= 1205872 120591min(119905) = 120591

1(119905) = 120591

2(119905) = 119905 minus

1205878 [1198861 1198871] = [1205878 + 2119896120587 1205874 + 2119896120587][119886

2 1198872] = [31205878 +

4 Abstract and Applied Analysis

2119896120587 1205872 + 2119896120587] 119896 isin N 1199021(119905) = sin(119905) 119902

2(119905) = cos(119905)

and 119890(119905) = minus1198900cos(2119905) Firstly it is elementary to check that

the required inequalities (12) and (13) are satisfied Next weprove that required inequality (19) immediately follows fromassumption (21) On the first hand we estimate from abovethe term on the right hand side in (19) using that120572

11205781+12057221205782=

1

119876119895(119905) = 120578119890

1205780

01198981205781

11198981205782

2119882(119905)

119905 minus 119886119895

119905 minus 119886119895+ 1205878

le 1205781198901205780

01198981205781

11198981205782

2

119905 isin [119886119895 119887119895]

(23)

that is

0 lt radic max119905isin[119886119895 119887119895]

119876119895(119905) le radic120578119890

1205780

01198981205781

11198981205782

2 119895 isin 1 2 (24)

On the other hand we calculate the integral on the lefthand side in (19)

int1198871

1198861

1198761(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205874+2119896120587

1205878+2119896120587

119882(119905)119905 minus (1205878 + 2119896120587)

119905 minus (1205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198681

int1198872

1198862

1198762(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205872+2119896120587

31205878+2119896120587

119882(119905)119905 minus (31205878 + 2119896120587)

119905 minus (31205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198682

(25)

Now from previous two equalities and inequalities (21)and (24) we conclude

1

120587int119887119895

119886119895

119876119895(119905) 119889119905 =

119868119895

1205871205781198901205780

01198981205781

11198981205782

2

ge radic1205781198901205780

01198981205781

11198981205782

2ge radic max119905isin[119886119895 119887119895]

119876119895(119905) gt 0

(26)

Thus assumption (21) proves desired inequality (19) andthus Corollary 2 verifies that (20) is oscillatory provided (21)holds

Remark 5 Oneway to obtain the inequality (21) is to supposefor instance that at least one of the constants 119898

1 1198982

and 1198900is large enough

Remark 6 We can consider the slightly more general equa-tion than (20)

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) 1003816100381610038161003816119909 (119905 minus 120591

1)10038161003816100381610038161205721 sgn119909 (119905 minus 120591

1)

+ 1198982cos (119905) 1003816100381610038161003816119909 (119905 minus 120591

2)10038161003816100381610038161205722 sgn119909 (119905 minus 120591

2) = minus119890

0cos (2119905)

(27)

where 1205911= 1205912= 1205878 and 120601 = 120601(V) satisfy (6) and (7)

and 1205721gt 119901 gt 120572

2gt 0 In a similar way as in Example 4

one can show that (27) is oscillatory provided at least one ofthe constants 119898

1 1198982 and 119890

0is large enough

Example 7 In this example we present an application of[1 Theorem 22] for getting another condition on theconstants 119898

1 1198982 and 119890

0for oscillation of (20) Let 120591

1=

1205912= 1205878

[1198861 1198881] = [

120587

8+ 2119896120587

3120587

16+ 2119896120587]

[1198881 1198871] = [

3120587

16+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

3120587

8+ 2119896120587

7120587

16+ 2119896120587]

[1198882 1198872] = [

7120587

16+ 2119896120587

120587

2+ 2119896120587]

(28)

1199021(119905) = 119898

1sin(119905) 119902

2(119905) = 119898

2cos(119905) and 119890(119905) = minus119890

0cos(2119905)

It is clear that 119902119894(119905) ge 0 on [119886

1minus 120591119894 1198871] cup [119886

2minus 120591119894 1198872] 119894 =

1 2 119890(119905) le 0 on [1198861minus 120591119894 1198871] and 119890(119905) ge 0 on [119886

2minus

120591119894 1198872] 119894 = 1 2 Therefore we may apply [1 Theorem 22] on

(20) provided the following inequality holds

1

119867119895(119888119895 119886119895)int119888119895

119886119895

(119876119895(119905)119867119895(119905 119886119895) minus 1) 119889119905

+1

119867119895(119887119895 119888119895)int119887119895

119888119895

(119876119895(119905)119867119895(119887119895 119905) minus 1) 119889119905 gt 0

119895 isin 1 2

(29)

Now if we put 1198671(119905 119904) = 119867

2(119905 119904) = (119905 minus 119904)2 and ℎ

1198951

(119905 119904) = ℎ1198952(119905 119904) = 1 for 119905 gt 119904 then previous inequality takes

the following concrete form

int312058716+2119896120587

1205878+2119896120587

1198761(119905) (119905 minus

120587

8minus 2119896120587)

2

119889119905

+ int1205874+2119896120587

312058716+2119896120587

1198761(119905) (119905 minus

120587

4minus 2119896120587)

2

119889119905 gt120587

8

int712058716+2119896120587

31205878+2119896120587

1198762(119905) (119905 minus

3120587

8minus 2119896120587)

2

119889119905

+ int1205872+2119896120587

712058716+2119896120587

1198762(119905) (119905 minus

120587

2minus 2119896120587)

2

119889119905 gt120587

8

(30)

Similarly as in Example 4 we can calculate previousintegrals and conclude by [1 Theorem 22] that (20) isoscillatory provided the following inequalities hold

1205781198901205780

01198981205781

11198981205782

2(11986811+ 11986812) gt

120587

8

1205781198901205780

01198981205781

11198981205782

2(11986821+ 11986822) gt

120587

8

(31)

Abstract and Applied Analysis 5

where

11986811= int312058716

1205878

119882(119905)(119905 minus 1205878)3

119905119889119905

11986812= int1205874

312058716

119882(119905)(119905 minus 1205878) (119905 minus 1205874)2

119905119889119905

11986821= int712058716

31205878

119882(119905)(119905 minus 31205878)3

119905 minus 1205874119889119905

11986822= int1205872

712058716

119882(119905)(119905 minus 31205878) (119905 minus 1205872)2

119905 minus 1205874119889119905

(32)

and119882(119905) = | cos 2119905|1205780 | sin 119905|1205781 | cos 119905|1205782

We leave to the reader to compare and verify whichone of the two inequalities (21) and (31) is simpler to beverified Also similar to the previous example it is possibleto get corresponding inequalities analogously to (31) for othercriteria published in the papers cited in the references

Remark 8 In [1 Section 3] the authors give an application of[1 Theorem 22] to (20) where 120591

1(119905) = 119905 minus 120591

1 1205912(119905) = 119905 minus

1205912 1205911= 1205878 and 120591

2= 1205874 However the following choice

(see [1 Section 3])

[1198861 1198881] = [2119896120587

120587

8+ 2119896120587]

[1198881 1198871] = [

120587

8+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

120587

4+ 2119896120587

3120587

8+ 2119896120587]

[1198882 1198872] = [

3120587

8+ 2119896120587

120587

2+ 2119896120587]

(33)

is not correct since the desired conditions 119902119894(119905) ge 0 on [119886

1minus

120591119894 1198871]cup[1198862minus120591119894 1198872] 119894 = 1 2 are not fulfilled Hence we suggest

reader to use the intervals proposed in the previous example

Open Question 9 The well-known variational techniquebased on the generalized Philosrsquo 119867-function has been usedin [1ndash11] to obtain some oscillation criteria for (1) where thesecond-order quasilinear differential operator (119903(119905)120601(1199091015840(119905)))1015840

is linear or half-linear Is it possible to use this technique inthe case of any function 120601(V) satisfying general conditions (6)and (7)

Theorem 10 (advanced equation) Under assumptions (6)(7) (9) and (10) let 119903(119905) be a nonincreasing positive functionon [1199050infin) and ℎ

119894(119905) = 120590

119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 2 119899

Let for every 119879 ge 1199050there exist numbers 119886

1 1198871 1198862 1198872 such that

119879 le 1198861lt 1198871le 120590max(1198871) le 119886

2lt 1198872and

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [1198861 120590max (1198871)] cup [119886

2 120590max (1198872)]

119890 (119905) le 0 119900119899 [1198861 120590max (1198871)]

119890 (119905) ge 0 119900119899 [1198862 120590max (1198872)]

(34)

where 120590max(119905) = max1205901(119905) 120590

119899(119905) Equation (1) is oscilla-

tory provided there are two real parameters 1205821 1205822gt 0 such

that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(35)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(119887119895)minus120590119894(119905)

120590119894(119887119895) minus 119905

)

120572119894120578119894

(36)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

and 120578119894appearing respectively in (7) (9) and (10)

Now analogously with (16) we consider the oscillation ofthe advanced equation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(37)

As a consequence of Theorem 10 we have the followingcriterion

Corollary 11 One assumes (6) (7) (10) and (34) Then (37)is oscillatory provided condition (17) is satisfied where 119876

119895(119905) is

defined in (36)

As the third case we consider second-order functionaldifferential equations with both delay and advanced argu-ments

Theorem 12 (delay-advanced equation) One assumes (6)(7) and (9) 120572

119894gt 119901 for 119894 isin 1 2 119899 and 119898 isin N 1 lt

119898 lt 119899 Let 119903(119905) equiv 119888119900119899119904119905 gt 0 on [1199050infin) ℎ

119894(119905) = 120591

119894(119905) le

119905 for 119894 isin 1 119898 and ℎ119894(119905) = 120590

119894(119905) ge 119905 for 119894 isin 119898 +

1 119899 on [1199050infin) Let 119903

119894(119905) 119902119894(119905) and 119890(119905) satisfy (12)-(13)

for 119894 isin 1 119898 and (34) for 119894 isin 119898 + 1 119899 Equation (1)is oscillatory provided there are two real parameters 120582

1 1205822gt

0 such that (11) is fulfilled where

119877119895(119905) = 119870

119898

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+ 119870119899

sum119894=119898+1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(38)

6 Abstract and Applied Analysis

119876119895(119905) =

119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(39)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870 appear-

ing respectively in (7) and (9)

Analogously with (16) and (37) we consider the oscilla-tion of delay-advanced equation

(120601 (1199091015840

(119905)))1015840

+119898

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905))

+119899

sum119894=119898+1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(40)

As a consequence of Theorem 12 we derive the followingoscillation criterion for (40)

Corollary 13 Let all assumptions of Theorem 12 hold withrespect to 119903

119894(119905) equiv 0 and 119891(119906) equiv 0 Equation (40) is oscillatory

provided condition (17) is satisfied where 119876119895(119905) is defined in

(39)

In Zafer [5] (see also [6]) the author has illustratedits main oscillation result on the following example of thesecond-order functional differential equations with delay andadvanced arguments Since |119909|120572minus1119909 = |119909|120572 sgn119909 it can berewritten in the form

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(41)

where 1198981 1198982 and 119890

0are positive constants 120591(119905) = 119905 minus 1205875

120590(119905) = 119905 + 12058720 and 1205721 1205722gt 0 It has been proved that

previous equation is oscillatory provided at least one of theconstants 119898

1 1198982 and 119890

0is large enough Here according

to Corollary 13 we can repeat this interesting conclusion forslightly complicated equation but with 120572

1 1205722gt 1

Example 14 We consider the following 120601-Laplacian func-tional differential equation with delay and advanced argu-ments

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(42)

where 120601 = 120601(V) satisfies (6) (7) 120591(119905) = 119905 minus 1205910 120590(119905) = 119905 + 120590

0

and 1205910 1205900isin [0 1205874) We claim that the previous equation

is oscillatory provided at least one of the constants 1198981 1198982

and 1198900is large enough Indeed it is enough to show that

condition (17) is fulfilled by using Corollary 13 It can be donein a similarway as in Example 4with (20) andCorollary 3Weleave it to the reader

3 Auxiliary Results QualitativeProperties of Concave-Like Functions

Let 119886 lt 119887 be two arbitrary real numbers It is known that if119909(119905) is a concave and smooth enough function on (119886 119887) thatis if 11990910158401015840(119905) le 0 for all 119905 isin (119886 119887) then 119909(119905) minus 119909(119904) ge 1199091015840(119905)(119905 minus 119904)for all 119904 119905 isin (119886 119887) 119904 lt 119905 Moreover if 119909(119905) gt 0 for all 119905 isin (119886 119887)then from previous inequality we obtain

1199091015840 (119905)

119909 (119905)le

1

119905 minus 119904forall119904 119905 isin (119886 119887) 119904 lt 119905 (43)

However often we do not have any information aboutthe sign of the second-order linear differential operator in(1) except only in a particular case when 119903(119905) equiv 1 and120601(V) equiv V Hence the main goal of this section is to findsome sufficient conditions on 119903(119905) and 120601(V) such that theassumption ldquo(119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)rdquo implies thedesired inequality (43) It is done in the following results

Lemma 15 Let 120601 = 120601(V) satisfy (6) Let 0 lt 119903(119904) le 119903(119905) forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862([119886 119887)R) such that 119909(119905) gt 0 for all 119905 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (119886 119887)

119905ℎ1198901198991199091015840 (119905)

119909 (119905)le

1

119905 minus 119886forall119905 isin (119886 119887)

(44)

Proof From assumption (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)it follows that

119903 (119905) 120601 (1199091015840

(119905)) le 119903 (120585) 120601 (1199091015840

(120585)) forall120585 119905 isin (119886 119887) 120585 le 119905

(45)

To the end of this proof let 119905 isin (119886 119887) be fixed Thanks to(6) if 120601(1199091015840(119905)) le 0 then 1199091015840(119905) le 0 too because 120601 is odd andincreasing In this case since 119903(119904) gt 0 and 119909(119904) gt 0 on (119886 119887)it implies that

1199091015840 (119905)

119909 (119905)le 0 lt

1

119905 minus 119886forall119905 isin (119886 119887) (46)

Thus it remains to show (44) for the case of 120601(1199091015840(119905)) ge 0From 0 lt 119903(120585) le 119903(119905) for all 120585 isin (119886 119887) 120585 le 119905 we obtain119903(119905)119903(120585) ge 1 which together with 120601(1199091015840(119905)) ge 0 and (45) gives

120601 (1199091015840

(119905)) le119903 (119905)

119903 (120585)120601 (1199091015840

(119905)) le 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 le 119905

(47)

Abstract and Applied Analysis 7

Acting on (47) with 120601minus1 and using that 120601minus1 is an increas-ing function because of (6) we obtain

1199091015840

(120585) ge 1199091015840

(119905) forall120585 isin (119886 119887) 120585 le 119905 (48)

Next since 119909(119905) gt 0 on (119886 119887) and 119909 isin 119862([119886 119887)R) wehave 119909(119886) ge 0 From the mean-value theorem for 119909(119905) on(119886 119905) we get a 120585 isin (119886 119905) depending on 119886 and 119905 such that119909(119905) minus 119909(119886) = 1199091015840(120585)(119905 minus 119886) and since 119909(119886) ge 0 from (48)we obtain

119909 (119905) ge 1199091015840

(120585) (119905 minus 119886) ge 1199091015840

(119905) (119905 minus 119886) (49)

which proves the desired inequality in (44)

In the advanced case of (1) that is when ℎ119894(119905) = 120590

119894(119905) ge 119905

we have the analogous result to Lemma 15

Lemma 16 Let 120601 = 120601(V) satisfy (6) Let 119903(119904) ge 119903(119905) gt 0 forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862((119886 119887]R) such that 119909(119904) gt 0 for all 119904 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (119886 119887)

119905ℎ1198901198991199091015840 (119904)

119909 (119904)ge minus

1

119887 minus 119904forall119904 isin (119886 119887)

(50)

Proof From assumption (119903(119904)120601(1199091015840(119904)))1015840 le 0 for all 119904 isin (119886 119887)we have

119903 (119904) 120601 (1199091015840

(119904)) ge 119903 (120585) 120601 (1199091015840

(120585)) forall119904 120585 isin (119886 119887) 119904 le 120585

(51)

To the end of this proof let 119904 isin (119886 119887) be fixed If120601(1199091015840(119904)) ge0 then 1199091015840(119904) ge 0 because 120601 is odd and increasing and sinceby assumption 119909(119905) gt 0 for all 119905 isin (119886 119887) we have that

1199091015840 (119904)

119909 (119904)ge 0 ge minus

1

119887 minus 119904forall119904 isin (119886 119887) (52)

which proves (50) in this case Let now 120601(1199091015840(119904)) le 0 Itimplies that 1199091015840(119904) le 0 and 120601(1199091015840(120585)) le 0 because of (51) and119903(119904) gt 0 for all 119904 isin (119886 119887) Hence from 119903(120585) le 119903(119904) for all120585 isin (119886 119887) 120585 ge 119904 from (51) and 120601(1199091015840(120585)) le 0 we especiallyconclude that

120601 (1199091015840

(119904)) ge119903 (120585)

119903 (119904)120601 (1199091015840

(120585)) ge 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 ge 119904

(53)

Acting on (53) with 120601minus1 we obtain

1199091015840

(119904) = 120601minus1(120601 (1199091015840

(119904))) ge 120601minus1(120601 (1199091015840

(120585))) = 1199091015840

(120585)

forall120585 isin (119886 119887) 120585 ge 119904(54)

By the Lagrangersquos mean-value theorem there exists a 120585 isin(119904 119887) depending on 119904 119887 such that 119909(119887) minus 119909(119904) = 1199091015840(120585)(119887 minus 119904)Since 119909(119904) gt 0 for all 119904 isin (119886 119887) and 119909 isin 119862((119886 119887]R) from(54) we obtain

minus119909 (119904) le 119909 (119887) minus 119909 (119904) = 1199091015840

(120585) (119887 minus 119904) le 1199091015840

(119904) (119887 minus 119904) (55)

which proves the desired inequality

Statement (44) will be frequently used in the followingform

Corollary 17 Let 120601 = 120601(V) satisfy (6) 120591(119905) le 119905 and 0 lt119903(119904) le 119903(119905) for all 119904 119905 119905

0le 119904 le 119905 Let 119886

1 1198871be two arbitrary

real numbers such that 1199050le 120591(119886

1) le 119886

1lt 1198871 Then for any

function 119909 isin 1198622((120591(1198861) 1198871)R) cap 119862([120591(119886

1) 1198871)R) such that

119909(119905) gt 0 for all 119905 isin (120591(1198861) 1198871) the following statement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (120591 (1198861) 1198871)

119905ℎ119890119899119909 (120591 (119905))

119909 (119905)ge120591 (119905) minus 120591 (119886

1)

119905 minus 120591 (1198861)

forall119905 isin (1198861 1198871)

(56)

Proof By assumptions of this corollary we have 120591(119905) le 119905119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))

1015840

le 0 for all 119905 isin (120591(1198861) 1198871) In

particular from statement (44) applied on (119886 119887) = (120591(1198861) 1198871)

we get

1199091015840 (119905)

119909 (119905)le

1

119905 minus 120591 (1198861)

forall119905 isin (120591 (1198861) 1198871) (57)

Integrating this inequality over the interval [120591(119905) 119905] for all119905 isin (1198861 1198871) we obtain

119909 (119905)

119909 (120591 (119905))le

119905 minus 120591 (1198861)

120591 (119905) minus 120591 (1198861)

forall119905 isin (1198861 1198871) (58)

which proves the desired inequality in (56)

Statement (50) will appear in the following form

Corollary 18 Let 1198861lt 1198871be two arbitrary real numbers Let

120601 = 120601(V) satisfy (6) 120590(119905) ge 119905 119903(119905) gt 0 and 119903(119904) ge 119903(119905) forall 119904 119905 119904 le 119905 Then for any function 119909 isin 1198622((119886

1 120590(1198871))R) cap

119862((1198861 120590(1198871)]R) such that 119909(119905) gt 0 for all 119905 isin (119886

1 120590(1198871)) the

following statement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (1198861 120590 (1198871))

119905ℎ119890119899119909 (120590 (119904))

119909 (119904)ge120590 (1198871) minus 120590 (119904)

120590 (1198871) minus 119904

forall119904 isin (1198861 1198871)

(59)

Proof Analogously with the proof of Corollary 17 we justneed to use the second inequality in (50) on (119886 119887) =(1198861 120590(1198871)) and integrate it over the interval [119904 120590(119904)] for all

119904 isin (1198861 1198871)

Next by Corollary 17 and the arithmetic-geometric meaninequality

119899

sum119894=0

120578119894119906119894ge119899

prod119894=0

119906120578119894

119894 119906119894ge 0 120578

119894gt 0 (60)

we can prove the following proposition

Proposition 19 (with delay arguments) Let 0 lt 119903(119904) le 119903(119905)for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

where 120591119894

ge 0 In Murugadass et al [2] authors havestudied the oscillation of the second-order quasilinear delaydifferential equation

(119903 (119905) (1199091015840

(119905))119901

)1015840

+ 119902 (119905) 119909119901

(119905 minus 120591)

+119899

sum119894=1

119902119894(119905) |119909 (119905 minus 120591)|

120572119894 sgn119909 (119905 minus 120591) = 119890 (119905) 119905 ge 1199050

(3)

where 119901 and 120572119894 are ratio of odd positive integers Later in

Hassan et al [3] authors consider the oscillation of the half-linear functional differential equation

(119903 (119905) (1199091015840

(119905))119901

)1015840

+ 1199030(119905) 119909119901(ℎ0(119905))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (ℎ119894 (119905))1003816100381610038161003816120572119894 sgn119909 (ℎ

119894(119905)) = 119890 (119905) 119905 ge 119905

0

(4)

where the function ℎ119894

= ℎ119894(119905) is positive continuous

functions with lim119905rarrinfin

ℎ119894(119905) = infin Moreover in [1ndash3] some

doubts concerning the proof of the main result of [4] areresolved The well-known variational technique that uses thegeneralized Philosrsquo results based on the so-called119867-functionhas been used in [1 Theorem 22] [2 Theorems 23 and 26]and [3 Theorems 25 26 and 27] (see also [5ndash11] and refer-ences therein) Since the second-order quasilinear differentialoperator (119903(119905)120601(1199091015840(119905)))

1015840 usually causes some difficulties inmany problems the application of previous method on (1) isstated here as an open problem In contrast to the precedingwe use a combination of the Riccati classic transformationa blow-up argument and a comparison pointwise principlerecently established in [12 13] but for differential equationswithout functional arguments It seems that our criteria areslightly simpler to be verified which is discussed on someexamples given in the next section

Among some recently published papers on the oscillationof quasilinear functional differential equations with bothdelay and advanced terms we point out an oscillationcriterion by Zafer [5] obtained for equation

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1199021(119905) |119909 (120591 (119905))|

120573minus1119909 (120591 (119905))

+ 1199022(119905) |119909 (120590 (119905))|

120574minus1119909 (120590 (119905)) = 119890 (119905) 119905 ge 0

(5)

where 119901 gt 0 120573 ge 119901 and 120574 ge 119901 On an interesting examplein [5] the author established the oscillations provided at leastone of constants appearing in the coefficients is sufficientlylarge (see also [6] for 119901 = 1) which is presented here in thenext section as a particular case of our main results

On the properties of some classes of the one-dimensionalquasilinear differential equations with 120601-Laplacian operatorswe refer reader to [14ndash20] and the references therein Aboutthe applications of second-order functional differential equa-tions in the mathematical description of certain phenomenain physics technics and biology (oscillation in a vacuumtube interaction of an oscillator with an energy sourcecoupled oscillators in electronics chemistry and ecologyrelativistic motion of a mass in a central field ship course

stabilization moving of the tip of a growing plant etc) wesuggest reading Kolmanovskii and Myshkis book [21]

2 Main Results and Examples

Let 120601 = 120601(V) be a function which appears in the first term of(1) such that

120601 isin 1198621

(RR) 120601 is odd and increasing function on R

(6)

120601 (V) V ge 1003816100381610038161003816120601 (V)1003816100381610038161003816(119901+1)119901

forallV isin R and some 119901 gt 0 (7)

These two assumptions are fulfilled respectively inthe linear case (119903(119905)1199091015840(119905))

1015840 where 120601(V) equiv V and 119901 =1 in the one-dimensional 119901-Laplacian quasilinear case(119903(119905)|1199091015840(119905)|

119901minus1

1199091015840(119905))1015840

where 120601(V) equiv |V|119901minus1V and 119901 gt 0 andin the mean prescribed curvature quasilinear case

[119903 (119905) 1199091015840

(119905) (1 + 11990910158402

(119905))minus12

]1015840

(8)

where 120601(V) = V(1 + V2)minus12 and 119901 = 1 The importance of theprescribed mean curvature quasilinear differential operatorslies in the capillarity type problems in fluid mechanics flux-limited diffusion phenomena and prescribedmean curvatureproblems see for instance [14 22 23]

The function 119891 = 119891(119906) which appears in the second termof (1) satisfies

119891 (119906) is odd function on R

119891 (119906)

119906119901ge 119870 gt 0 forall119906 gt 0 and some 119870 isin R

(9)

where 119901 gt 0 is from assumption (7)In the first two theorems below the exponents 120572

119894 satisfy

1205721ge sdot sdot sdot ge 120572

119898gt 119901 gt 120572

119898+1ge sdot sdot sdot ge 120572

119899gt 0 119898 isin N

120572119894gt 120572119894+1 119894 isin 1 119899 minus 1

there exists (119899 + 1)-tuple (1205780 1205781 120578

119899)

such that 0 lt 120578119894lt 1

119899

sum119894=1

120578119894lt 1 120578

0= 1 minus

119899

sum119894=1

120578119894

119899

sum119894=1

120572119894120578119894= 119901

(10)

where 119901 is from assumption (7) For instance if 119899 = 2 1205721=

52 119901 = 1 and 1205722= 12 then 120578

0= 1205781= 1205782= 13 On

assumption (10) see for instance [9 Lemma 1]Unlike recently published oscillation criteria for the linear

and half-linear second-order forced functional differential

Abstract and Applied Analysis 3

equations our oscillation criterion is only based on thefollowing elementary integral inequality

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905 ge 1

119895 isin 1 2

(11)

where 1198861lt 1198871le 1198862lt 1198872 119901 is from (7) 120587

119901= (119901(119901 +

1))120587 sin(119901120587(119901 + 1)) 120582119895

gt 0 and functions 119876119895(119905) and

119877119895(119905) are explicitly expressed by the coefficients of (1) just like

it is done in the next three main results

Theorem 1 (delay equation) One assumes (6) (7) (9) and(10) Let 119903(119905) be a nondecreasing positive function on [119905

0infin)

Let ℎ119894(119905) = 120591

119894(119905) le 119905 on [119905

0infin) and lim

119905rarrinfin120591119894(119905) = infin 119894 isin

1 2 119899 Let for every 119879 ge 1199050there exist 119886

1 1198871 1198862 1198872 119879 le

1198861lt 1198871le 120591min(1198862) le 119886

2lt 1198872such that

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [120591min (1198861) 1198871] cup [120591min (1198862) 1198872] (12)

119890 (119905) le 0 119900119899 [120591min (1198861) 1198871]

119890 (119905) ge 0 119900119899 [120591min (1198862) 1198872] (13)

where 120591min(119905) = min1205911(119905) 1205912(119905) 120591

119899(119905) Equation (1) is

oscillatory provided there are two real parameters 1205821 1205822gt 0

such that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

(14)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

120572119894120578119894

(15)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

120578119894appearing respectively in (7) (9) and (10)

As we can see in (13) the forcing term 119890(119905) can bean oscillatory function The main property of any intervaloscillation criterion (see [8]) is that the coefficients of theconsidered equation do not satisfy some conditions on thewhole [119905

0infin) than only on intervals [119886

1 1198871] cup [119886

2 1198872] The

main consequence of Theorem 1 is the following oscillationcriterion for (1) with 119903(119905) equiv 1 and 119903

119894(119905) equiv 0 119894 isin 1 2 119899

Corollary 2 One assumes (6) (7) (10) (12) and (13) Thenequation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905)) = 119890 (119905) (16)

is oscillatory provided

119901119901(119901+1)

2120587119901

int119887119895

119886119895

119876119895(119905) 119889119905 ge ( max

119905isin[119886119895 119887119895]119876119895(119905))

119901(119901+1)

gt 0

119895 isin 1 2

(17)

where 119876119895(119905) is defined in (15)

In particular for 119901 = 1 previous corollary takes thefollowing form

Corollary 3 Let (10) hold with 119901 = 1 120591119894(119905) le 119905 on [119905

0infin)

and lim119905rarrinfin

120591119894(119905) = infin 119894 isin 1 2 119899 Let for every 119879 ge

1199050there exist 119886

1 1198871 1198862 1198872 119879 le 119886

1lt 1198871le 120591min(1198862) le 119886

2lt

1198872such that assumptions (12) and (13) hold with 119903

119894(119905) equiv 0

Equation

11990910158401015840

(119905) +119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905)) = 119890 (119905) (18)

is oscillatory provided

1

120587int119887119895

119886119895

119876119895(119905) 119889119905 ge radic max

119905isin[119886119895 119887119895]119876119895(119905) gt 0 119895 isin 1 2 (19)

where 119876119895(119905) is defined in (15)

In the next examples and remarks we discuss the applica-tion of oscillation criteria from Corollary 3 and [1 Theorem22] on the following equation

11990910158401015840+ 1198981sin (119905) 1003816100381610038161003816119909 (1205911 (119905))

10038161003816100381610038161205721 sgn119909 (120591

1(119905))

+ 1198982cos (119905) 1003816100381610038161003816119909 (1205912 (119905))

10038161003816100381610038161205722 sgn119909 (120591

2(119905)) = minus119890

0cos (2119905)

(20)

where 1205721gt 1 gt 120572

2gt 0 120591

1(119905) = 120591

2(119905) = 119905 minus 1205878 and 119898

1 1198982

and 1198900are positive constants

Example 4 As a consequence of Corollary 3 we show inthis example that (20) is oscillatory provided the constants1198981 1198982 and 119890

0satisfy the following simple inequality

119868119895

120587radic1205781198901205780

01198981205781

11198981205782

2ge 1 119895 isin 1 2 (21)

where 120578119894satisfy (10) 120578 = prod

2

119894=0120578minus120578119894

119894 and 119868

1 1198682gt 0 are two real

numbers defined by

1198681= int1205874

1205878

119882(119905)119905 minus 1205878

119905119889119905

1198682= int1205872

31205878

119882(119905)119905 minus 31205878

119905 minus 1205874119889119905

119882 (119905) = |cos 2119905|1205780 |sin 119905|1205781 |cos 119905|1205782

(22)

Indeed let 120587119901= 1205872 120591min(119905) = 120591

1(119905) = 120591

2(119905) = 119905 minus

1205878 [1198861 1198871] = [1205878 + 2119896120587 1205874 + 2119896120587][119886

2 1198872] = [31205878 +

4 Abstract and Applied Analysis

2119896120587 1205872 + 2119896120587] 119896 isin N 1199021(119905) = sin(119905) 119902

2(119905) = cos(119905)

and 119890(119905) = minus1198900cos(2119905) Firstly it is elementary to check that

the required inequalities (12) and (13) are satisfied Next weprove that required inequality (19) immediately follows fromassumption (21) On the first hand we estimate from abovethe term on the right hand side in (19) using that120572

11205781+12057221205782=

1

119876119895(119905) = 120578119890

1205780

01198981205781

11198981205782

2119882(119905)

119905 minus 119886119895

119905 minus 119886119895+ 1205878

le 1205781198901205780

01198981205781

11198981205782

2

119905 isin [119886119895 119887119895]

(23)

that is

0 lt radic max119905isin[119886119895 119887119895]

119876119895(119905) le radic120578119890

1205780

01198981205781

11198981205782

2 119895 isin 1 2 (24)

On the other hand we calculate the integral on the lefthand side in (19)

int1198871

1198861

1198761(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205874+2119896120587

1205878+2119896120587

119882(119905)119905 minus (1205878 + 2119896120587)

119905 minus (1205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198681

int1198872

1198862

1198762(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205872+2119896120587

31205878+2119896120587

119882(119905)119905 minus (31205878 + 2119896120587)

119905 minus (31205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198682

(25)

Now from previous two equalities and inequalities (21)and (24) we conclude

1

120587int119887119895

119886119895

119876119895(119905) 119889119905 =

119868119895

1205871205781198901205780

01198981205781

11198981205782

2

ge radic1205781198901205780

01198981205781

11198981205782

2ge radic max119905isin[119886119895 119887119895]

119876119895(119905) gt 0

(26)

Thus assumption (21) proves desired inequality (19) andthus Corollary 2 verifies that (20) is oscillatory provided (21)holds

Remark 5 Oneway to obtain the inequality (21) is to supposefor instance that at least one of the constants 119898

1 1198982

and 1198900is large enough

Remark 6 We can consider the slightly more general equa-tion than (20)

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) 1003816100381610038161003816119909 (119905 minus 120591

1)10038161003816100381610038161205721 sgn119909 (119905 minus 120591

1)

+ 1198982cos (119905) 1003816100381610038161003816119909 (119905 minus 120591

2)10038161003816100381610038161205722 sgn119909 (119905 minus 120591

2) = minus119890

0cos (2119905)

(27)

where 1205911= 1205912= 1205878 and 120601 = 120601(V) satisfy (6) and (7)

and 1205721gt 119901 gt 120572

2gt 0 In a similar way as in Example 4

one can show that (27) is oscillatory provided at least one ofthe constants 119898

1 1198982 and 119890

0is large enough

Example 7 In this example we present an application of[1 Theorem 22] for getting another condition on theconstants 119898

1 1198982 and 119890

0for oscillation of (20) Let 120591

1=

1205912= 1205878

[1198861 1198881] = [

120587

8+ 2119896120587

3120587

16+ 2119896120587]

[1198881 1198871] = [

3120587

16+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

3120587

8+ 2119896120587

7120587

16+ 2119896120587]

[1198882 1198872] = [

7120587

16+ 2119896120587

120587

2+ 2119896120587]

(28)

1199021(119905) = 119898

1sin(119905) 119902

2(119905) = 119898

2cos(119905) and 119890(119905) = minus119890

0cos(2119905)

It is clear that 119902119894(119905) ge 0 on [119886

1minus 120591119894 1198871] cup [119886

2minus 120591119894 1198872] 119894 =

1 2 119890(119905) le 0 on [1198861minus 120591119894 1198871] and 119890(119905) ge 0 on [119886

2minus

120591119894 1198872] 119894 = 1 2 Therefore we may apply [1 Theorem 22] on

(20) provided the following inequality holds

1

119867119895(119888119895 119886119895)int119888119895

119886119895

(119876119895(119905)119867119895(119905 119886119895) minus 1) 119889119905

+1

119867119895(119887119895 119888119895)int119887119895

119888119895

(119876119895(119905)119867119895(119887119895 119905) minus 1) 119889119905 gt 0

119895 isin 1 2

(29)

Now if we put 1198671(119905 119904) = 119867

2(119905 119904) = (119905 minus 119904)2 and ℎ

1198951

(119905 119904) = ℎ1198952(119905 119904) = 1 for 119905 gt 119904 then previous inequality takes

the following concrete form

int312058716+2119896120587

1205878+2119896120587

1198761(119905) (119905 minus

120587

8minus 2119896120587)

2

119889119905

+ int1205874+2119896120587

312058716+2119896120587

1198761(119905) (119905 minus

120587

4minus 2119896120587)

2

119889119905 gt120587

8

int712058716+2119896120587

31205878+2119896120587

1198762(119905) (119905 minus

3120587

8minus 2119896120587)

2

119889119905

+ int1205872+2119896120587

712058716+2119896120587

1198762(119905) (119905 minus

120587

2minus 2119896120587)

2

119889119905 gt120587

8

(30)

Similarly as in Example 4 we can calculate previousintegrals and conclude by [1 Theorem 22] that (20) isoscillatory provided the following inequalities hold

1205781198901205780

01198981205781

11198981205782

2(11986811+ 11986812) gt

120587

8

1205781198901205780

01198981205781

11198981205782

2(11986821+ 11986822) gt

120587

8

(31)

Abstract and Applied Analysis 5

where

11986811= int312058716

1205878

119882(119905)(119905 minus 1205878)3

119905119889119905

11986812= int1205874

312058716

119882(119905)(119905 minus 1205878) (119905 minus 1205874)2

119905119889119905

11986821= int712058716

31205878

119882(119905)(119905 minus 31205878)3

119905 minus 1205874119889119905

11986822= int1205872

712058716

119882(119905)(119905 minus 31205878) (119905 minus 1205872)2

119905 minus 1205874119889119905

(32)

and119882(119905) = | cos 2119905|1205780 | sin 119905|1205781 | cos 119905|1205782

We leave to the reader to compare and verify whichone of the two inequalities (21) and (31) is simpler to beverified Also similar to the previous example it is possibleto get corresponding inequalities analogously to (31) for othercriteria published in the papers cited in the references

Remark 8 In [1 Section 3] the authors give an application of[1 Theorem 22] to (20) where 120591

1(119905) = 119905 minus 120591

1 1205912(119905) = 119905 minus

1205912 1205911= 1205878 and 120591

2= 1205874 However the following choice

(see [1 Section 3])

[1198861 1198881] = [2119896120587

120587

8+ 2119896120587]

[1198881 1198871] = [

120587

8+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

120587

4+ 2119896120587

3120587

8+ 2119896120587]

[1198882 1198872] = [

3120587

8+ 2119896120587

120587

2+ 2119896120587]

(33)

is not correct since the desired conditions 119902119894(119905) ge 0 on [119886

1minus

120591119894 1198871]cup[1198862minus120591119894 1198872] 119894 = 1 2 are not fulfilled Hence we suggest

reader to use the intervals proposed in the previous example

Open Question 9 The well-known variational techniquebased on the generalized Philosrsquo 119867-function has been usedin [1ndash11] to obtain some oscillation criteria for (1) where thesecond-order quasilinear differential operator (119903(119905)120601(1199091015840(119905)))1015840

is linear or half-linear Is it possible to use this technique inthe case of any function 120601(V) satisfying general conditions (6)and (7)

Theorem 10 (advanced equation) Under assumptions (6)(7) (9) and (10) let 119903(119905) be a nonincreasing positive functionon [1199050infin) and ℎ

119894(119905) = 120590

119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 2 119899

Let for every 119879 ge 1199050there exist numbers 119886

1 1198871 1198862 1198872 such that

119879 le 1198861lt 1198871le 120590max(1198871) le 119886

2lt 1198872and

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [1198861 120590max (1198871)] cup [119886

2 120590max (1198872)]

119890 (119905) le 0 119900119899 [1198861 120590max (1198871)]

119890 (119905) ge 0 119900119899 [1198862 120590max (1198872)]

(34)

where 120590max(119905) = max1205901(119905) 120590

119899(119905) Equation (1) is oscilla-

tory provided there are two real parameters 1205821 1205822gt 0 such

that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(35)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(119887119895)minus120590119894(119905)

120590119894(119887119895) minus 119905

)

120572119894120578119894

(36)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

and 120578119894appearing respectively in (7) (9) and (10)

Now analogously with (16) we consider the oscillation ofthe advanced equation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(37)

As a consequence of Theorem 10 we have the followingcriterion

Corollary 11 One assumes (6) (7) (10) and (34) Then (37)is oscillatory provided condition (17) is satisfied where 119876

119895(119905) is

defined in (36)

As the third case we consider second-order functionaldifferential equations with both delay and advanced argu-ments

Theorem 12 (delay-advanced equation) One assumes (6)(7) and (9) 120572

119894gt 119901 for 119894 isin 1 2 119899 and 119898 isin N 1 lt

119898 lt 119899 Let 119903(119905) equiv 119888119900119899119904119905 gt 0 on [1199050infin) ℎ

119894(119905) = 120591

119894(119905) le

119905 for 119894 isin 1 119898 and ℎ119894(119905) = 120590

119894(119905) ge 119905 for 119894 isin 119898 +

1 119899 on [1199050infin) Let 119903

119894(119905) 119902119894(119905) and 119890(119905) satisfy (12)-(13)

for 119894 isin 1 119898 and (34) for 119894 isin 119898 + 1 119899 Equation (1)is oscillatory provided there are two real parameters 120582

1 1205822gt

0 such that (11) is fulfilled where

119877119895(119905) = 119870

119898

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+ 119870119899

sum119894=119898+1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(38)

6 Abstract and Applied Analysis

119876119895(119905) =

119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(39)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870 appear-

ing respectively in (7) and (9)

Analogously with (16) and (37) we consider the oscilla-tion of delay-advanced equation

(120601 (1199091015840

(119905)))1015840

+119898

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905))

+119899

sum119894=119898+1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(40)

As a consequence of Theorem 12 we derive the followingoscillation criterion for (40)

Corollary 13 Let all assumptions of Theorem 12 hold withrespect to 119903

119894(119905) equiv 0 and 119891(119906) equiv 0 Equation (40) is oscillatory

provided condition (17) is satisfied where 119876119895(119905) is defined in

(39)

In Zafer [5] (see also [6]) the author has illustratedits main oscillation result on the following example of thesecond-order functional differential equations with delay andadvanced arguments Since |119909|120572minus1119909 = |119909|120572 sgn119909 it can berewritten in the form

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(41)

where 1198981 1198982 and 119890

0are positive constants 120591(119905) = 119905 minus 1205875

120590(119905) = 119905 + 12058720 and 1205721 1205722gt 0 It has been proved that

previous equation is oscillatory provided at least one of theconstants 119898

1 1198982 and 119890

0is large enough Here according

to Corollary 13 we can repeat this interesting conclusion forslightly complicated equation but with 120572

1 1205722gt 1

Example 14 We consider the following 120601-Laplacian func-tional differential equation with delay and advanced argu-ments

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(42)

where 120601 = 120601(V) satisfies (6) (7) 120591(119905) = 119905 minus 1205910 120590(119905) = 119905 + 120590

0

and 1205910 1205900isin [0 1205874) We claim that the previous equation

is oscillatory provided at least one of the constants 1198981 1198982

and 1198900is large enough Indeed it is enough to show that

condition (17) is fulfilled by using Corollary 13 It can be donein a similarway as in Example 4with (20) andCorollary 3Weleave it to the reader

3 Auxiliary Results QualitativeProperties of Concave-Like Functions

Let 119886 lt 119887 be two arbitrary real numbers It is known that if119909(119905) is a concave and smooth enough function on (119886 119887) thatis if 11990910158401015840(119905) le 0 for all 119905 isin (119886 119887) then 119909(119905) minus 119909(119904) ge 1199091015840(119905)(119905 minus 119904)for all 119904 119905 isin (119886 119887) 119904 lt 119905 Moreover if 119909(119905) gt 0 for all 119905 isin (119886 119887)then from previous inequality we obtain

1199091015840 (119905)

119909 (119905)le

1

119905 minus 119904forall119904 119905 isin (119886 119887) 119904 lt 119905 (43)

However often we do not have any information aboutthe sign of the second-order linear differential operator in(1) except only in a particular case when 119903(119905) equiv 1 and120601(V) equiv V Hence the main goal of this section is to findsome sufficient conditions on 119903(119905) and 120601(V) such that theassumption ldquo(119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)rdquo implies thedesired inequality (43) It is done in the following results

Lemma 15 Let 120601 = 120601(V) satisfy (6) Let 0 lt 119903(119904) le 119903(119905) forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862([119886 119887)R) such that 119909(119905) gt 0 for all 119905 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (119886 119887)

119905ℎ1198901198991199091015840 (119905)

119909 (119905)le

1

119905 minus 119886forall119905 isin (119886 119887)

(44)

Proof From assumption (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)it follows that

119903 (119905) 120601 (1199091015840

(119905)) le 119903 (120585) 120601 (1199091015840

(120585)) forall120585 119905 isin (119886 119887) 120585 le 119905

(45)

To the end of this proof let 119905 isin (119886 119887) be fixed Thanks to(6) if 120601(1199091015840(119905)) le 0 then 1199091015840(119905) le 0 too because 120601 is odd andincreasing In this case since 119903(119904) gt 0 and 119909(119904) gt 0 on (119886 119887)it implies that

1199091015840 (119905)

119909 (119905)le 0 lt

1

119905 minus 119886forall119905 isin (119886 119887) (46)

Thus it remains to show (44) for the case of 120601(1199091015840(119905)) ge 0From 0 lt 119903(120585) le 119903(119905) for all 120585 isin (119886 119887) 120585 le 119905 we obtain119903(119905)119903(120585) ge 1 which together with 120601(1199091015840(119905)) ge 0 and (45) gives

120601 (1199091015840

(119905)) le119903 (119905)

119903 (120585)120601 (1199091015840

(119905)) le 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 le 119905

(47)

Abstract and Applied Analysis 7

Acting on (47) with 120601minus1 and using that 120601minus1 is an increas-ing function because of (6) we obtain

1199091015840

(120585) ge 1199091015840

(119905) forall120585 isin (119886 119887) 120585 le 119905 (48)

Next since 119909(119905) gt 0 on (119886 119887) and 119909 isin 119862([119886 119887)R) wehave 119909(119886) ge 0 From the mean-value theorem for 119909(119905) on(119886 119905) we get a 120585 isin (119886 119905) depending on 119886 and 119905 such that119909(119905) minus 119909(119886) = 1199091015840(120585)(119905 minus 119886) and since 119909(119886) ge 0 from (48)we obtain

119909 (119905) ge 1199091015840

(120585) (119905 minus 119886) ge 1199091015840

(119905) (119905 minus 119886) (49)

which proves the desired inequality in (44)

In the advanced case of (1) that is when ℎ119894(119905) = 120590

119894(119905) ge 119905

we have the analogous result to Lemma 15

Lemma 16 Let 120601 = 120601(V) satisfy (6) Let 119903(119904) ge 119903(119905) gt 0 forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862((119886 119887]R) such that 119909(119904) gt 0 for all 119904 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (119886 119887)

119905ℎ1198901198991199091015840 (119904)

119909 (119904)ge minus

1

119887 minus 119904forall119904 isin (119886 119887)

(50)

Proof From assumption (119903(119904)120601(1199091015840(119904)))1015840 le 0 for all 119904 isin (119886 119887)we have

119903 (119904) 120601 (1199091015840

(119904)) ge 119903 (120585) 120601 (1199091015840

(120585)) forall119904 120585 isin (119886 119887) 119904 le 120585

(51)

To the end of this proof let 119904 isin (119886 119887) be fixed If120601(1199091015840(119904)) ge0 then 1199091015840(119904) ge 0 because 120601 is odd and increasing and sinceby assumption 119909(119905) gt 0 for all 119905 isin (119886 119887) we have that

1199091015840 (119904)

119909 (119904)ge 0 ge minus

1

119887 minus 119904forall119904 isin (119886 119887) (52)

which proves (50) in this case Let now 120601(1199091015840(119904)) le 0 Itimplies that 1199091015840(119904) le 0 and 120601(1199091015840(120585)) le 0 because of (51) and119903(119904) gt 0 for all 119904 isin (119886 119887) Hence from 119903(120585) le 119903(119904) for all120585 isin (119886 119887) 120585 ge 119904 from (51) and 120601(1199091015840(120585)) le 0 we especiallyconclude that

120601 (1199091015840

(119904)) ge119903 (120585)

119903 (119904)120601 (1199091015840

(120585)) ge 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 ge 119904

(53)

Acting on (53) with 120601minus1 we obtain

1199091015840

(119904) = 120601minus1(120601 (1199091015840

(119904))) ge 120601minus1(120601 (1199091015840

(120585))) = 1199091015840

(120585)

forall120585 isin (119886 119887) 120585 ge 119904(54)

By the Lagrangersquos mean-value theorem there exists a 120585 isin(119904 119887) depending on 119904 119887 such that 119909(119887) minus 119909(119904) = 1199091015840(120585)(119887 minus 119904)Since 119909(119904) gt 0 for all 119904 isin (119886 119887) and 119909 isin 119862((119886 119887]R) from(54) we obtain

minus119909 (119904) le 119909 (119887) minus 119909 (119904) = 1199091015840

(120585) (119887 minus 119904) le 1199091015840

(119904) (119887 minus 119904) (55)

which proves the desired inequality

Statement (44) will be frequently used in the followingform

Corollary 17 Let 120601 = 120601(V) satisfy (6) 120591(119905) le 119905 and 0 lt119903(119904) le 119903(119905) for all 119904 119905 119905

0le 119904 le 119905 Let 119886

1 1198871be two arbitrary

real numbers such that 1199050le 120591(119886

1) le 119886

1lt 1198871 Then for any

function 119909 isin 1198622((120591(1198861) 1198871)R) cap 119862([120591(119886

1) 1198871)R) such that

119909(119905) gt 0 for all 119905 isin (120591(1198861) 1198871) the following statement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (120591 (1198861) 1198871)

119905ℎ119890119899119909 (120591 (119905))

119909 (119905)ge120591 (119905) minus 120591 (119886

1)

119905 minus 120591 (1198861)

forall119905 isin (1198861 1198871)

(56)

Proof By assumptions of this corollary we have 120591(119905) le 119905119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))

1015840

le 0 for all 119905 isin (120591(1198861) 1198871) In

particular from statement (44) applied on (119886 119887) = (120591(1198861) 1198871)

we get

1199091015840 (119905)

119909 (119905)le

1

119905 minus 120591 (1198861)

forall119905 isin (120591 (1198861) 1198871) (57)

Integrating this inequality over the interval [120591(119905) 119905] for all119905 isin (1198861 1198871) we obtain

119909 (119905)

119909 (120591 (119905))le

119905 minus 120591 (1198861)

120591 (119905) minus 120591 (1198861)

forall119905 isin (1198861 1198871) (58)

which proves the desired inequality in (56)

Statement (50) will appear in the following form

Corollary 18 Let 1198861lt 1198871be two arbitrary real numbers Let

120601 = 120601(V) satisfy (6) 120590(119905) ge 119905 119903(119905) gt 0 and 119903(119904) ge 119903(119905) forall 119904 119905 119904 le 119905 Then for any function 119909 isin 1198622((119886

1 120590(1198871))R) cap

119862((1198861 120590(1198871)]R) such that 119909(119905) gt 0 for all 119905 isin (119886

1 120590(1198871)) the

following statement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (1198861 120590 (1198871))

119905ℎ119890119899119909 (120590 (119904))

119909 (119904)ge120590 (1198871) minus 120590 (119904)

120590 (1198871) minus 119904

forall119904 isin (1198861 1198871)

(59)

Proof Analogously with the proof of Corollary 17 we justneed to use the second inequality in (50) on (119886 119887) =(1198861 120590(1198871)) and integrate it over the interval [119904 120590(119904)] for all

119904 isin (1198861 1198871)

Next by Corollary 17 and the arithmetic-geometric meaninequality

119899

sum119894=0

120578119894119906119894ge119899

prod119894=0

119906120578119894

119894 119906119894ge 0 120578

119894gt 0 (60)

we can prove the following proposition

Proposition 19 (with delay arguments) Let 0 lt 119903(119904) le 119903(119905)for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

equations our oscillation criterion is only based on thefollowing elementary integral inequality

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905 ge 1

119895 isin 1 2

(11)

where 1198861lt 1198871le 1198862lt 1198872 119901 is from (7) 120587

119901= (119901(119901 +

1))120587 sin(119901120587(119901 + 1)) 120582119895

gt 0 and functions 119876119895(119905) and

119877119895(119905) are explicitly expressed by the coefficients of (1) just like

it is done in the next three main results

Theorem 1 (delay equation) One assumes (6) (7) (9) and(10) Let 119903(119905) be a nondecreasing positive function on [119905

0infin)

Let ℎ119894(119905) = 120591

119894(119905) le 119905 on [119905

0infin) and lim

119905rarrinfin120591119894(119905) = infin 119894 isin

1 2 119899 Let for every 119879 ge 1199050there exist 119886

1 1198871 1198862 1198872 119879 le

1198861lt 1198871le 120591min(1198862) le 119886

2lt 1198872such that

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [120591min (1198861) 1198871] cup [120591min (1198862) 1198872] (12)

119890 (119905) le 0 119900119899 [120591min (1198861) 1198871]

119890 (119905) ge 0 119900119899 [120591min (1198862) 1198872] (13)

where 120591min(119905) = min1205911(119905) 1205912(119905) 120591

119899(119905) Equation (1) is

oscillatory provided there are two real parameters 1205821 1205822gt 0

such that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

(14)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

120572119894120578119894

(15)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

120578119894appearing respectively in (7) (9) and (10)

As we can see in (13) the forcing term 119890(119905) can bean oscillatory function The main property of any intervaloscillation criterion (see [8]) is that the coefficients of theconsidered equation do not satisfy some conditions on thewhole [119905

0infin) than only on intervals [119886

1 1198871] cup [119886

2 1198872] The

main consequence of Theorem 1 is the following oscillationcriterion for (1) with 119903(119905) equiv 1 and 119903

119894(119905) equiv 0 119894 isin 1 2 119899

Corollary 2 One assumes (6) (7) (10) (12) and (13) Thenequation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905)) = 119890 (119905) (16)

is oscillatory provided

119901119901(119901+1)

2120587119901

int119887119895

119886119895

119876119895(119905) 119889119905 ge ( max

119905isin[119886119895 119887119895]119876119895(119905))

119901(119901+1)

gt 0

119895 isin 1 2

(17)

where 119876119895(119905) is defined in (15)

In particular for 119901 = 1 previous corollary takes thefollowing form

Corollary 3 Let (10) hold with 119901 = 1 120591119894(119905) le 119905 on [119905

0infin)

and lim119905rarrinfin

120591119894(119905) = infin 119894 isin 1 2 119899 Let for every 119879 ge

1199050there exist 119886

1 1198871 1198862 1198872 119879 le 119886

1lt 1198871le 120591min(1198862) le 119886

2lt

1198872such that assumptions (12) and (13) hold with 119903

119894(119905) equiv 0

Equation

11990910158401015840

(119905) +119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905)) = 119890 (119905) (18)

is oscillatory provided

1

120587int119887119895

119886119895

119876119895(119905) 119889119905 ge radic max

119905isin[119886119895 119887119895]119876119895(119905) gt 0 119895 isin 1 2 (19)

where 119876119895(119905) is defined in (15)

In the next examples and remarks we discuss the applica-tion of oscillation criteria from Corollary 3 and [1 Theorem22] on the following equation

11990910158401015840+ 1198981sin (119905) 1003816100381610038161003816119909 (1205911 (119905))

10038161003816100381610038161205721 sgn119909 (120591

1(119905))

+ 1198982cos (119905) 1003816100381610038161003816119909 (1205912 (119905))

10038161003816100381610038161205722 sgn119909 (120591

2(119905)) = minus119890

0cos (2119905)

(20)

where 1205721gt 1 gt 120572

2gt 0 120591

1(119905) = 120591

2(119905) = 119905 minus 1205878 and 119898

1 1198982

and 1198900are positive constants

Example 4 As a consequence of Corollary 3 we show inthis example that (20) is oscillatory provided the constants1198981 1198982 and 119890

0satisfy the following simple inequality

119868119895

120587radic1205781198901205780

01198981205781

11198981205782

2ge 1 119895 isin 1 2 (21)

where 120578119894satisfy (10) 120578 = prod

2

119894=0120578minus120578119894

119894 and 119868

1 1198682gt 0 are two real

numbers defined by

1198681= int1205874

1205878

119882(119905)119905 minus 1205878

119905119889119905

1198682= int1205872

31205878

119882(119905)119905 minus 31205878

119905 minus 1205874119889119905

119882 (119905) = |cos 2119905|1205780 |sin 119905|1205781 |cos 119905|1205782

(22)

Indeed let 120587119901= 1205872 120591min(119905) = 120591

1(119905) = 120591

2(119905) = 119905 minus

1205878 [1198861 1198871] = [1205878 + 2119896120587 1205874 + 2119896120587][119886

2 1198872] = [31205878 +

4 Abstract and Applied Analysis

2119896120587 1205872 + 2119896120587] 119896 isin N 1199021(119905) = sin(119905) 119902

2(119905) = cos(119905)

and 119890(119905) = minus1198900cos(2119905) Firstly it is elementary to check that

the required inequalities (12) and (13) are satisfied Next weprove that required inequality (19) immediately follows fromassumption (21) On the first hand we estimate from abovethe term on the right hand side in (19) using that120572

11205781+12057221205782=

1

119876119895(119905) = 120578119890

1205780

01198981205781

11198981205782

2119882(119905)

119905 minus 119886119895

119905 minus 119886119895+ 1205878

le 1205781198901205780

01198981205781

11198981205782

2

119905 isin [119886119895 119887119895]

(23)

that is

0 lt radic max119905isin[119886119895 119887119895]

119876119895(119905) le radic120578119890

1205780

01198981205781

11198981205782

2 119895 isin 1 2 (24)

On the other hand we calculate the integral on the lefthand side in (19)

int1198871

1198861

1198761(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205874+2119896120587

1205878+2119896120587

119882(119905)119905 minus (1205878 + 2119896120587)

119905 minus (1205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198681

int1198872

1198862

1198762(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205872+2119896120587

31205878+2119896120587

119882(119905)119905 minus (31205878 + 2119896120587)

119905 minus (31205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198682

(25)

Now from previous two equalities and inequalities (21)and (24) we conclude

1

120587int119887119895

119886119895

119876119895(119905) 119889119905 =

119868119895

1205871205781198901205780

01198981205781

11198981205782

2

ge radic1205781198901205780

01198981205781

11198981205782

2ge radic max119905isin[119886119895 119887119895]

119876119895(119905) gt 0

(26)

Thus assumption (21) proves desired inequality (19) andthus Corollary 2 verifies that (20) is oscillatory provided (21)holds

Remark 5 Oneway to obtain the inequality (21) is to supposefor instance that at least one of the constants 119898

1 1198982

and 1198900is large enough

Remark 6 We can consider the slightly more general equa-tion than (20)

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) 1003816100381610038161003816119909 (119905 minus 120591

1)10038161003816100381610038161205721 sgn119909 (119905 minus 120591

1)

+ 1198982cos (119905) 1003816100381610038161003816119909 (119905 minus 120591

2)10038161003816100381610038161205722 sgn119909 (119905 minus 120591

2) = minus119890

0cos (2119905)

(27)

where 1205911= 1205912= 1205878 and 120601 = 120601(V) satisfy (6) and (7)

and 1205721gt 119901 gt 120572

2gt 0 In a similar way as in Example 4

one can show that (27) is oscillatory provided at least one ofthe constants 119898

1 1198982 and 119890

0is large enough

Example 7 In this example we present an application of[1 Theorem 22] for getting another condition on theconstants 119898

1 1198982 and 119890

0for oscillation of (20) Let 120591

1=

1205912= 1205878

[1198861 1198881] = [

120587

8+ 2119896120587

3120587

16+ 2119896120587]

[1198881 1198871] = [

3120587

16+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

3120587

8+ 2119896120587

7120587

16+ 2119896120587]

[1198882 1198872] = [

7120587

16+ 2119896120587

120587

2+ 2119896120587]

(28)

1199021(119905) = 119898

1sin(119905) 119902

2(119905) = 119898

2cos(119905) and 119890(119905) = minus119890

0cos(2119905)

It is clear that 119902119894(119905) ge 0 on [119886

1minus 120591119894 1198871] cup [119886

2minus 120591119894 1198872] 119894 =

1 2 119890(119905) le 0 on [1198861minus 120591119894 1198871] and 119890(119905) ge 0 on [119886

2minus

120591119894 1198872] 119894 = 1 2 Therefore we may apply [1 Theorem 22] on

(20) provided the following inequality holds

1

119867119895(119888119895 119886119895)int119888119895

119886119895

(119876119895(119905)119867119895(119905 119886119895) minus 1) 119889119905

+1

119867119895(119887119895 119888119895)int119887119895

119888119895

(119876119895(119905)119867119895(119887119895 119905) minus 1) 119889119905 gt 0

119895 isin 1 2

(29)

Now if we put 1198671(119905 119904) = 119867

2(119905 119904) = (119905 minus 119904)2 and ℎ

1198951

(119905 119904) = ℎ1198952(119905 119904) = 1 for 119905 gt 119904 then previous inequality takes

the following concrete form

int312058716+2119896120587

1205878+2119896120587

1198761(119905) (119905 minus

120587

8minus 2119896120587)

2

119889119905

+ int1205874+2119896120587

312058716+2119896120587

1198761(119905) (119905 minus

120587

4minus 2119896120587)

2

119889119905 gt120587

8

int712058716+2119896120587

31205878+2119896120587

1198762(119905) (119905 minus

3120587

8minus 2119896120587)

2

119889119905

+ int1205872+2119896120587

712058716+2119896120587

1198762(119905) (119905 minus

120587

2minus 2119896120587)

2

119889119905 gt120587

8

(30)

Similarly as in Example 4 we can calculate previousintegrals and conclude by [1 Theorem 22] that (20) isoscillatory provided the following inequalities hold

1205781198901205780

01198981205781

11198981205782

2(11986811+ 11986812) gt

120587

8

1205781198901205780

01198981205781

11198981205782

2(11986821+ 11986822) gt

120587

8

(31)

Abstract and Applied Analysis 5

where

11986811= int312058716

1205878

119882(119905)(119905 minus 1205878)3

119905119889119905

11986812= int1205874

312058716

119882(119905)(119905 minus 1205878) (119905 minus 1205874)2

119905119889119905

11986821= int712058716

31205878

119882(119905)(119905 minus 31205878)3

119905 minus 1205874119889119905

11986822= int1205872

712058716

119882(119905)(119905 minus 31205878) (119905 minus 1205872)2

119905 minus 1205874119889119905

(32)

and119882(119905) = | cos 2119905|1205780 | sin 119905|1205781 | cos 119905|1205782

We leave to the reader to compare and verify whichone of the two inequalities (21) and (31) is simpler to beverified Also similar to the previous example it is possibleto get corresponding inequalities analogously to (31) for othercriteria published in the papers cited in the references

Remark 8 In [1 Section 3] the authors give an application of[1 Theorem 22] to (20) where 120591

1(119905) = 119905 minus 120591

1 1205912(119905) = 119905 minus

1205912 1205911= 1205878 and 120591

2= 1205874 However the following choice

(see [1 Section 3])

[1198861 1198881] = [2119896120587

120587

8+ 2119896120587]

[1198881 1198871] = [

120587

8+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

120587

4+ 2119896120587

3120587

8+ 2119896120587]

[1198882 1198872] = [

3120587

8+ 2119896120587

120587

2+ 2119896120587]

(33)

is not correct since the desired conditions 119902119894(119905) ge 0 on [119886

1minus

120591119894 1198871]cup[1198862minus120591119894 1198872] 119894 = 1 2 are not fulfilled Hence we suggest

reader to use the intervals proposed in the previous example

Open Question 9 The well-known variational techniquebased on the generalized Philosrsquo 119867-function has been usedin [1ndash11] to obtain some oscillation criteria for (1) where thesecond-order quasilinear differential operator (119903(119905)120601(1199091015840(119905)))1015840

is linear or half-linear Is it possible to use this technique inthe case of any function 120601(V) satisfying general conditions (6)and (7)

Theorem 10 (advanced equation) Under assumptions (6)(7) (9) and (10) let 119903(119905) be a nonincreasing positive functionon [1199050infin) and ℎ

119894(119905) = 120590

119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 2 119899

Let for every 119879 ge 1199050there exist numbers 119886

1 1198871 1198862 1198872 such that

119879 le 1198861lt 1198871le 120590max(1198871) le 119886

2lt 1198872and

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [1198861 120590max (1198871)] cup [119886

2 120590max (1198872)]

119890 (119905) le 0 119900119899 [1198861 120590max (1198871)]

119890 (119905) ge 0 119900119899 [1198862 120590max (1198872)]

(34)

where 120590max(119905) = max1205901(119905) 120590

119899(119905) Equation (1) is oscilla-

tory provided there are two real parameters 1205821 1205822gt 0 such

that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(35)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(119887119895)minus120590119894(119905)

120590119894(119887119895) minus 119905

)

120572119894120578119894

(36)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

and 120578119894appearing respectively in (7) (9) and (10)

Now analogously with (16) we consider the oscillation ofthe advanced equation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(37)

As a consequence of Theorem 10 we have the followingcriterion

Corollary 11 One assumes (6) (7) (10) and (34) Then (37)is oscillatory provided condition (17) is satisfied where 119876

119895(119905) is

defined in (36)

As the third case we consider second-order functionaldifferential equations with both delay and advanced argu-ments

Theorem 12 (delay-advanced equation) One assumes (6)(7) and (9) 120572

119894gt 119901 for 119894 isin 1 2 119899 and 119898 isin N 1 lt

119898 lt 119899 Let 119903(119905) equiv 119888119900119899119904119905 gt 0 on [1199050infin) ℎ

119894(119905) = 120591

119894(119905) le

119905 for 119894 isin 1 119898 and ℎ119894(119905) = 120590

119894(119905) ge 119905 for 119894 isin 119898 +

1 119899 on [1199050infin) Let 119903

119894(119905) 119902119894(119905) and 119890(119905) satisfy (12)-(13)

for 119894 isin 1 119898 and (34) for 119894 isin 119898 + 1 119899 Equation (1)is oscillatory provided there are two real parameters 120582

1 1205822gt

0 such that (11) is fulfilled where

119877119895(119905) = 119870

119898

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+ 119870119899

sum119894=119898+1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(38)

6 Abstract and Applied Analysis

119876119895(119905) =

119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(39)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870 appear-

ing respectively in (7) and (9)

Analogously with (16) and (37) we consider the oscilla-tion of delay-advanced equation

(120601 (1199091015840

(119905)))1015840

+119898

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905))

+119899

sum119894=119898+1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(40)

As a consequence of Theorem 12 we derive the followingoscillation criterion for (40)

Corollary 13 Let all assumptions of Theorem 12 hold withrespect to 119903

119894(119905) equiv 0 and 119891(119906) equiv 0 Equation (40) is oscillatory

provided condition (17) is satisfied where 119876119895(119905) is defined in

(39)

In Zafer [5] (see also [6]) the author has illustratedits main oscillation result on the following example of thesecond-order functional differential equations with delay andadvanced arguments Since |119909|120572minus1119909 = |119909|120572 sgn119909 it can berewritten in the form

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(41)

where 1198981 1198982 and 119890

0are positive constants 120591(119905) = 119905 minus 1205875

120590(119905) = 119905 + 12058720 and 1205721 1205722gt 0 It has been proved that

previous equation is oscillatory provided at least one of theconstants 119898

1 1198982 and 119890

0is large enough Here according

to Corollary 13 we can repeat this interesting conclusion forslightly complicated equation but with 120572

1 1205722gt 1

Example 14 We consider the following 120601-Laplacian func-tional differential equation with delay and advanced argu-ments

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(42)

where 120601 = 120601(V) satisfies (6) (7) 120591(119905) = 119905 minus 1205910 120590(119905) = 119905 + 120590

0

and 1205910 1205900isin [0 1205874) We claim that the previous equation

is oscillatory provided at least one of the constants 1198981 1198982

and 1198900is large enough Indeed it is enough to show that

condition (17) is fulfilled by using Corollary 13 It can be donein a similarway as in Example 4with (20) andCorollary 3Weleave it to the reader

3 Auxiliary Results QualitativeProperties of Concave-Like Functions

Let 119886 lt 119887 be two arbitrary real numbers It is known that if119909(119905) is a concave and smooth enough function on (119886 119887) thatis if 11990910158401015840(119905) le 0 for all 119905 isin (119886 119887) then 119909(119905) minus 119909(119904) ge 1199091015840(119905)(119905 minus 119904)for all 119904 119905 isin (119886 119887) 119904 lt 119905 Moreover if 119909(119905) gt 0 for all 119905 isin (119886 119887)then from previous inequality we obtain

1199091015840 (119905)

119909 (119905)le

1

119905 minus 119904forall119904 119905 isin (119886 119887) 119904 lt 119905 (43)

However often we do not have any information aboutthe sign of the second-order linear differential operator in(1) except only in a particular case when 119903(119905) equiv 1 and120601(V) equiv V Hence the main goal of this section is to findsome sufficient conditions on 119903(119905) and 120601(V) such that theassumption ldquo(119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)rdquo implies thedesired inequality (43) It is done in the following results

Lemma 15 Let 120601 = 120601(V) satisfy (6) Let 0 lt 119903(119904) le 119903(119905) forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862([119886 119887)R) such that 119909(119905) gt 0 for all 119905 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (119886 119887)

119905ℎ1198901198991199091015840 (119905)

119909 (119905)le

1

119905 minus 119886forall119905 isin (119886 119887)

(44)

Proof From assumption (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)it follows that

119903 (119905) 120601 (1199091015840

(119905)) le 119903 (120585) 120601 (1199091015840

(120585)) forall120585 119905 isin (119886 119887) 120585 le 119905

(45)

To the end of this proof let 119905 isin (119886 119887) be fixed Thanks to(6) if 120601(1199091015840(119905)) le 0 then 1199091015840(119905) le 0 too because 120601 is odd andincreasing In this case since 119903(119904) gt 0 and 119909(119904) gt 0 on (119886 119887)it implies that

1199091015840 (119905)

119909 (119905)le 0 lt

1

119905 minus 119886forall119905 isin (119886 119887) (46)

Thus it remains to show (44) for the case of 120601(1199091015840(119905)) ge 0From 0 lt 119903(120585) le 119903(119905) for all 120585 isin (119886 119887) 120585 le 119905 we obtain119903(119905)119903(120585) ge 1 which together with 120601(1199091015840(119905)) ge 0 and (45) gives

120601 (1199091015840

(119905)) le119903 (119905)

119903 (120585)120601 (1199091015840

(119905)) le 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 le 119905

(47)

Abstract and Applied Analysis 7

Acting on (47) with 120601minus1 and using that 120601minus1 is an increas-ing function because of (6) we obtain

1199091015840

(120585) ge 1199091015840

(119905) forall120585 isin (119886 119887) 120585 le 119905 (48)

Next since 119909(119905) gt 0 on (119886 119887) and 119909 isin 119862([119886 119887)R) wehave 119909(119886) ge 0 From the mean-value theorem for 119909(119905) on(119886 119905) we get a 120585 isin (119886 119905) depending on 119886 and 119905 such that119909(119905) minus 119909(119886) = 1199091015840(120585)(119905 minus 119886) and since 119909(119886) ge 0 from (48)we obtain

119909 (119905) ge 1199091015840

(120585) (119905 minus 119886) ge 1199091015840

(119905) (119905 minus 119886) (49)

which proves the desired inequality in (44)

In the advanced case of (1) that is when ℎ119894(119905) = 120590

119894(119905) ge 119905

we have the analogous result to Lemma 15

Lemma 16 Let 120601 = 120601(V) satisfy (6) Let 119903(119904) ge 119903(119905) gt 0 forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862((119886 119887]R) such that 119909(119904) gt 0 for all 119904 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (119886 119887)

119905ℎ1198901198991199091015840 (119904)

119909 (119904)ge minus

1

119887 minus 119904forall119904 isin (119886 119887)

(50)

Proof From assumption (119903(119904)120601(1199091015840(119904)))1015840 le 0 for all 119904 isin (119886 119887)we have

119903 (119904) 120601 (1199091015840

(119904)) ge 119903 (120585) 120601 (1199091015840

(120585)) forall119904 120585 isin (119886 119887) 119904 le 120585

(51)

To the end of this proof let 119904 isin (119886 119887) be fixed If120601(1199091015840(119904)) ge0 then 1199091015840(119904) ge 0 because 120601 is odd and increasing and sinceby assumption 119909(119905) gt 0 for all 119905 isin (119886 119887) we have that

1199091015840 (119904)

119909 (119904)ge 0 ge minus

1

119887 minus 119904forall119904 isin (119886 119887) (52)

which proves (50) in this case Let now 120601(1199091015840(119904)) le 0 Itimplies that 1199091015840(119904) le 0 and 120601(1199091015840(120585)) le 0 because of (51) and119903(119904) gt 0 for all 119904 isin (119886 119887) Hence from 119903(120585) le 119903(119904) for all120585 isin (119886 119887) 120585 ge 119904 from (51) and 120601(1199091015840(120585)) le 0 we especiallyconclude that

120601 (1199091015840

(119904)) ge119903 (120585)

119903 (119904)120601 (1199091015840

(120585)) ge 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 ge 119904

(53)

Acting on (53) with 120601minus1 we obtain

1199091015840

(119904) = 120601minus1(120601 (1199091015840

(119904))) ge 120601minus1(120601 (1199091015840

(120585))) = 1199091015840

(120585)

forall120585 isin (119886 119887) 120585 ge 119904(54)

By the Lagrangersquos mean-value theorem there exists a 120585 isin(119904 119887) depending on 119904 119887 such that 119909(119887) minus 119909(119904) = 1199091015840(120585)(119887 minus 119904)Since 119909(119904) gt 0 for all 119904 isin (119886 119887) and 119909 isin 119862((119886 119887]R) from(54) we obtain

minus119909 (119904) le 119909 (119887) minus 119909 (119904) = 1199091015840

(120585) (119887 minus 119904) le 1199091015840

(119904) (119887 minus 119904) (55)

which proves the desired inequality

Statement (44) will be frequently used in the followingform

Corollary 17 Let 120601 = 120601(V) satisfy (6) 120591(119905) le 119905 and 0 lt119903(119904) le 119903(119905) for all 119904 119905 119905

0le 119904 le 119905 Let 119886

1 1198871be two arbitrary

real numbers such that 1199050le 120591(119886

1) le 119886

1lt 1198871 Then for any

function 119909 isin 1198622((120591(1198861) 1198871)R) cap 119862([120591(119886

1) 1198871)R) such that

119909(119905) gt 0 for all 119905 isin (120591(1198861) 1198871) the following statement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (120591 (1198861) 1198871)

119905ℎ119890119899119909 (120591 (119905))

119909 (119905)ge120591 (119905) minus 120591 (119886

1)

119905 minus 120591 (1198861)

forall119905 isin (1198861 1198871)

(56)

Proof By assumptions of this corollary we have 120591(119905) le 119905119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))

1015840

le 0 for all 119905 isin (120591(1198861) 1198871) In

particular from statement (44) applied on (119886 119887) = (120591(1198861) 1198871)

we get

1199091015840 (119905)

119909 (119905)le

1

119905 minus 120591 (1198861)

forall119905 isin (120591 (1198861) 1198871) (57)

Integrating this inequality over the interval [120591(119905) 119905] for all119905 isin (1198861 1198871) we obtain

119909 (119905)

119909 (120591 (119905))le

119905 minus 120591 (1198861)

120591 (119905) minus 120591 (1198861)

forall119905 isin (1198861 1198871) (58)

which proves the desired inequality in (56)

Statement (50) will appear in the following form

Corollary 18 Let 1198861lt 1198871be two arbitrary real numbers Let

120601 = 120601(V) satisfy (6) 120590(119905) ge 119905 119903(119905) gt 0 and 119903(119904) ge 119903(119905) forall 119904 119905 119904 le 119905 Then for any function 119909 isin 1198622((119886

1 120590(1198871))R) cap

119862((1198861 120590(1198871)]R) such that 119909(119905) gt 0 for all 119905 isin (119886

1 120590(1198871)) the

following statement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (1198861 120590 (1198871))

119905ℎ119890119899119909 (120590 (119904))

119909 (119904)ge120590 (1198871) minus 120590 (119904)

120590 (1198871) minus 119904

forall119904 isin (1198861 1198871)

(59)

Proof Analogously with the proof of Corollary 17 we justneed to use the second inequality in (50) on (119886 119887) =(1198861 120590(1198871)) and integrate it over the interval [119904 120590(119904)] for all

119904 isin (1198861 1198871)

Next by Corollary 17 and the arithmetic-geometric meaninequality

119899

sum119894=0

120578119894119906119894ge119899

prod119894=0

119906120578119894

119894 119906119894ge 0 120578

119894gt 0 (60)

we can prove the following proposition

Proposition 19 (with delay arguments) Let 0 lt 119903(119904) le 119903(119905)for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

2119896120587 1205872 + 2119896120587] 119896 isin N 1199021(119905) = sin(119905) 119902

2(119905) = cos(119905)

and 119890(119905) = minus1198900cos(2119905) Firstly it is elementary to check that

the required inequalities (12) and (13) are satisfied Next weprove that required inequality (19) immediately follows fromassumption (21) On the first hand we estimate from abovethe term on the right hand side in (19) using that120572

11205781+12057221205782=

1

119876119895(119905) = 120578119890

1205780

01198981205781

11198981205782

2119882(119905)

119905 minus 119886119895

119905 minus 119886119895+ 1205878

le 1205781198901205780

01198981205781

11198981205782

2

119905 isin [119886119895 119887119895]

(23)

that is

0 lt radic max119905isin[119886119895 119887119895]

119876119895(119905) le radic120578119890

1205780

01198981205781

11198981205782

2 119895 isin 1 2 (24)

On the other hand we calculate the integral on the lefthand side in (19)

int1198871

1198861

1198761(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205874+2119896120587

1205878+2119896120587

119882(119905)119905 minus (1205878 + 2119896120587)

119905 minus (1205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198681

int1198872

1198862

1198762(119905) 119889119905

= 1205781198901205780

01198981205781

11198981205782

2int1205872+2119896120587

31205878+2119896120587

119882(119905)119905 minus (31205878 + 2119896120587)

119905 minus (31205878 + 2119896120587) + 1205878119889119905

= 1205781198901205780

01198981205781

11198981205782

21198682

(25)

Now from previous two equalities and inequalities (21)and (24) we conclude

1

120587int119887119895

119886119895

119876119895(119905) 119889119905 =

119868119895

1205871205781198901205780

01198981205781

11198981205782

2

ge radic1205781198901205780

01198981205781

11198981205782

2ge radic max119905isin[119886119895 119887119895]

119876119895(119905) gt 0

(26)

Thus assumption (21) proves desired inequality (19) andthus Corollary 2 verifies that (20) is oscillatory provided (21)holds

Remark 5 Oneway to obtain the inequality (21) is to supposefor instance that at least one of the constants 119898

1 1198982

and 1198900is large enough

Remark 6 We can consider the slightly more general equa-tion than (20)

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) 1003816100381610038161003816119909 (119905 minus 120591

1)10038161003816100381610038161205721 sgn119909 (119905 minus 120591

1)

+ 1198982cos (119905) 1003816100381610038161003816119909 (119905 minus 120591

2)10038161003816100381610038161205722 sgn119909 (119905 minus 120591

2) = minus119890

0cos (2119905)

(27)

where 1205911= 1205912= 1205878 and 120601 = 120601(V) satisfy (6) and (7)

and 1205721gt 119901 gt 120572

2gt 0 In a similar way as in Example 4

one can show that (27) is oscillatory provided at least one ofthe constants 119898

1 1198982 and 119890

0is large enough

Example 7 In this example we present an application of[1 Theorem 22] for getting another condition on theconstants 119898

1 1198982 and 119890

0for oscillation of (20) Let 120591

1=

1205912= 1205878

[1198861 1198881] = [

120587

8+ 2119896120587

3120587

16+ 2119896120587]

[1198881 1198871] = [

3120587

16+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

3120587

8+ 2119896120587

7120587

16+ 2119896120587]

[1198882 1198872] = [

7120587

16+ 2119896120587

120587

2+ 2119896120587]

(28)

1199021(119905) = 119898

1sin(119905) 119902

2(119905) = 119898

2cos(119905) and 119890(119905) = minus119890

0cos(2119905)

It is clear that 119902119894(119905) ge 0 on [119886

1minus 120591119894 1198871] cup [119886

2minus 120591119894 1198872] 119894 =

1 2 119890(119905) le 0 on [1198861minus 120591119894 1198871] and 119890(119905) ge 0 on [119886

2minus

120591119894 1198872] 119894 = 1 2 Therefore we may apply [1 Theorem 22] on

(20) provided the following inequality holds

1

119867119895(119888119895 119886119895)int119888119895

119886119895

(119876119895(119905)119867119895(119905 119886119895) minus 1) 119889119905

+1

119867119895(119887119895 119888119895)int119887119895

119888119895

(119876119895(119905)119867119895(119887119895 119905) minus 1) 119889119905 gt 0

119895 isin 1 2

(29)

Now if we put 1198671(119905 119904) = 119867

2(119905 119904) = (119905 minus 119904)2 and ℎ

1198951

(119905 119904) = ℎ1198952(119905 119904) = 1 for 119905 gt 119904 then previous inequality takes

the following concrete form

int312058716+2119896120587

1205878+2119896120587

1198761(119905) (119905 minus

120587

8minus 2119896120587)

2

119889119905

+ int1205874+2119896120587

312058716+2119896120587

1198761(119905) (119905 minus

120587

4minus 2119896120587)

2

119889119905 gt120587

8

int712058716+2119896120587

31205878+2119896120587

1198762(119905) (119905 minus

3120587

8minus 2119896120587)

2

119889119905

+ int1205872+2119896120587

712058716+2119896120587

1198762(119905) (119905 minus

120587

2minus 2119896120587)

2

119889119905 gt120587

8

(30)

Similarly as in Example 4 we can calculate previousintegrals and conclude by [1 Theorem 22] that (20) isoscillatory provided the following inequalities hold

1205781198901205780

01198981205781

11198981205782

2(11986811+ 11986812) gt

120587

8

1205781198901205780

01198981205781

11198981205782

2(11986821+ 11986822) gt

120587

8

(31)

Abstract and Applied Analysis 5

where

11986811= int312058716

1205878

119882(119905)(119905 minus 1205878)3

119905119889119905

11986812= int1205874

312058716

119882(119905)(119905 minus 1205878) (119905 minus 1205874)2

119905119889119905

11986821= int712058716

31205878

119882(119905)(119905 minus 31205878)3

119905 minus 1205874119889119905

11986822= int1205872

712058716

119882(119905)(119905 minus 31205878) (119905 minus 1205872)2

119905 minus 1205874119889119905

(32)

and119882(119905) = | cos 2119905|1205780 | sin 119905|1205781 | cos 119905|1205782

We leave to the reader to compare and verify whichone of the two inequalities (21) and (31) is simpler to beverified Also similar to the previous example it is possibleto get corresponding inequalities analogously to (31) for othercriteria published in the papers cited in the references

Remark 8 In [1 Section 3] the authors give an application of[1 Theorem 22] to (20) where 120591

1(119905) = 119905 minus 120591

1 1205912(119905) = 119905 minus

1205912 1205911= 1205878 and 120591

2= 1205874 However the following choice

(see [1 Section 3])

[1198861 1198881] = [2119896120587

120587

8+ 2119896120587]

[1198881 1198871] = [

120587

8+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

120587

4+ 2119896120587

3120587

8+ 2119896120587]

[1198882 1198872] = [

3120587

8+ 2119896120587

120587

2+ 2119896120587]

(33)

is not correct since the desired conditions 119902119894(119905) ge 0 on [119886

1minus

120591119894 1198871]cup[1198862minus120591119894 1198872] 119894 = 1 2 are not fulfilled Hence we suggest

reader to use the intervals proposed in the previous example

Open Question 9 The well-known variational techniquebased on the generalized Philosrsquo 119867-function has been usedin [1ndash11] to obtain some oscillation criteria for (1) where thesecond-order quasilinear differential operator (119903(119905)120601(1199091015840(119905)))1015840

is linear or half-linear Is it possible to use this technique inthe case of any function 120601(V) satisfying general conditions (6)and (7)

Theorem 10 (advanced equation) Under assumptions (6)(7) (9) and (10) let 119903(119905) be a nonincreasing positive functionon [1199050infin) and ℎ

119894(119905) = 120590

119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 2 119899

Let for every 119879 ge 1199050there exist numbers 119886

1 1198871 1198862 1198872 such that

119879 le 1198861lt 1198871le 120590max(1198871) le 119886

2lt 1198872and

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [1198861 120590max (1198871)] cup [119886

2 120590max (1198872)]

119890 (119905) le 0 119900119899 [1198861 120590max (1198871)]

119890 (119905) ge 0 119900119899 [1198862 120590max (1198872)]

(34)

where 120590max(119905) = max1205901(119905) 120590

119899(119905) Equation (1) is oscilla-

tory provided there are two real parameters 1205821 1205822gt 0 such

that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(35)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(119887119895)minus120590119894(119905)

120590119894(119887119895) minus 119905

)

120572119894120578119894

(36)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

and 120578119894appearing respectively in (7) (9) and (10)

Now analogously with (16) we consider the oscillation ofthe advanced equation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(37)

As a consequence of Theorem 10 we have the followingcriterion

Corollary 11 One assumes (6) (7) (10) and (34) Then (37)is oscillatory provided condition (17) is satisfied where 119876

119895(119905) is

defined in (36)

As the third case we consider second-order functionaldifferential equations with both delay and advanced argu-ments

Theorem 12 (delay-advanced equation) One assumes (6)(7) and (9) 120572

119894gt 119901 for 119894 isin 1 2 119899 and 119898 isin N 1 lt

119898 lt 119899 Let 119903(119905) equiv 119888119900119899119904119905 gt 0 on [1199050infin) ℎ

119894(119905) = 120591

119894(119905) le

119905 for 119894 isin 1 119898 and ℎ119894(119905) = 120590

119894(119905) ge 119905 for 119894 isin 119898 +

1 119899 on [1199050infin) Let 119903

119894(119905) 119902119894(119905) and 119890(119905) satisfy (12)-(13)

for 119894 isin 1 119898 and (34) for 119894 isin 119898 + 1 119899 Equation (1)is oscillatory provided there are two real parameters 120582

1 1205822gt

0 such that (11) is fulfilled where

119877119895(119905) = 119870

119898

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+ 119870119899

sum119894=119898+1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(38)

6 Abstract and Applied Analysis

119876119895(119905) =

119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(39)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870 appear-

ing respectively in (7) and (9)

Analogously with (16) and (37) we consider the oscilla-tion of delay-advanced equation

(120601 (1199091015840

(119905)))1015840

+119898

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905))

+119899

sum119894=119898+1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(40)

As a consequence of Theorem 12 we derive the followingoscillation criterion for (40)

Corollary 13 Let all assumptions of Theorem 12 hold withrespect to 119903

119894(119905) equiv 0 and 119891(119906) equiv 0 Equation (40) is oscillatory

provided condition (17) is satisfied where 119876119895(119905) is defined in

(39)

In Zafer [5] (see also [6]) the author has illustratedits main oscillation result on the following example of thesecond-order functional differential equations with delay andadvanced arguments Since |119909|120572minus1119909 = |119909|120572 sgn119909 it can berewritten in the form

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(41)

where 1198981 1198982 and 119890

0are positive constants 120591(119905) = 119905 minus 1205875

120590(119905) = 119905 + 12058720 and 1205721 1205722gt 0 It has been proved that

previous equation is oscillatory provided at least one of theconstants 119898

1 1198982 and 119890

0is large enough Here according

to Corollary 13 we can repeat this interesting conclusion forslightly complicated equation but with 120572

1 1205722gt 1

Example 14 We consider the following 120601-Laplacian func-tional differential equation with delay and advanced argu-ments

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(42)

where 120601 = 120601(V) satisfies (6) (7) 120591(119905) = 119905 minus 1205910 120590(119905) = 119905 + 120590

0

and 1205910 1205900isin [0 1205874) We claim that the previous equation

is oscillatory provided at least one of the constants 1198981 1198982

and 1198900is large enough Indeed it is enough to show that

condition (17) is fulfilled by using Corollary 13 It can be donein a similarway as in Example 4with (20) andCorollary 3Weleave it to the reader

3 Auxiliary Results QualitativeProperties of Concave-Like Functions

Let 119886 lt 119887 be two arbitrary real numbers It is known that if119909(119905) is a concave and smooth enough function on (119886 119887) thatis if 11990910158401015840(119905) le 0 for all 119905 isin (119886 119887) then 119909(119905) minus 119909(119904) ge 1199091015840(119905)(119905 minus 119904)for all 119904 119905 isin (119886 119887) 119904 lt 119905 Moreover if 119909(119905) gt 0 for all 119905 isin (119886 119887)then from previous inequality we obtain

1199091015840 (119905)

119909 (119905)le

1

119905 minus 119904forall119904 119905 isin (119886 119887) 119904 lt 119905 (43)

However often we do not have any information aboutthe sign of the second-order linear differential operator in(1) except only in a particular case when 119903(119905) equiv 1 and120601(V) equiv V Hence the main goal of this section is to findsome sufficient conditions on 119903(119905) and 120601(V) such that theassumption ldquo(119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)rdquo implies thedesired inequality (43) It is done in the following results

Lemma 15 Let 120601 = 120601(V) satisfy (6) Let 0 lt 119903(119904) le 119903(119905) forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862([119886 119887)R) such that 119909(119905) gt 0 for all 119905 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (119886 119887)

119905ℎ1198901198991199091015840 (119905)

119909 (119905)le

1

119905 minus 119886forall119905 isin (119886 119887)

(44)

Proof From assumption (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)it follows that

119903 (119905) 120601 (1199091015840

(119905)) le 119903 (120585) 120601 (1199091015840

(120585)) forall120585 119905 isin (119886 119887) 120585 le 119905

(45)

To the end of this proof let 119905 isin (119886 119887) be fixed Thanks to(6) if 120601(1199091015840(119905)) le 0 then 1199091015840(119905) le 0 too because 120601 is odd andincreasing In this case since 119903(119904) gt 0 and 119909(119904) gt 0 on (119886 119887)it implies that

1199091015840 (119905)

119909 (119905)le 0 lt

1

119905 minus 119886forall119905 isin (119886 119887) (46)

Thus it remains to show (44) for the case of 120601(1199091015840(119905)) ge 0From 0 lt 119903(120585) le 119903(119905) for all 120585 isin (119886 119887) 120585 le 119905 we obtain119903(119905)119903(120585) ge 1 which together with 120601(1199091015840(119905)) ge 0 and (45) gives

120601 (1199091015840

(119905)) le119903 (119905)

119903 (120585)120601 (1199091015840

(119905)) le 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 le 119905

(47)

Abstract and Applied Analysis 7

Acting on (47) with 120601minus1 and using that 120601minus1 is an increas-ing function because of (6) we obtain

1199091015840

(120585) ge 1199091015840

(119905) forall120585 isin (119886 119887) 120585 le 119905 (48)

Next since 119909(119905) gt 0 on (119886 119887) and 119909 isin 119862([119886 119887)R) wehave 119909(119886) ge 0 From the mean-value theorem for 119909(119905) on(119886 119905) we get a 120585 isin (119886 119905) depending on 119886 and 119905 such that119909(119905) minus 119909(119886) = 1199091015840(120585)(119905 minus 119886) and since 119909(119886) ge 0 from (48)we obtain

119909 (119905) ge 1199091015840

(120585) (119905 minus 119886) ge 1199091015840

(119905) (119905 minus 119886) (49)

which proves the desired inequality in (44)

In the advanced case of (1) that is when ℎ119894(119905) = 120590

119894(119905) ge 119905

we have the analogous result to Lemma 15

Lemma 16 Let 120601 = 120601(V) satisfy (6) Let 119903(119904) ge 119903(119905) gt 0 forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862((119886 119887]R) such that 119909(119904) gt 0 for all 119904 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (119886 119887)

119905ℎ1198901198991199091015840 (119904)

119909 (119904)ge minus

1

119887 minus 119904forall119904 isin (119886 119887)

(50)

Proof From assumption (119903(119904)120601(1199091015840(119904)))1015840 le 0 for all 119904 isin (119886 119887)we have

119903 (119904) 120601 (1199091015840

(119904)) ge 119903 (120585) 120601 (1199091015840

(120585)) forall119904 120585 isin (119886 119887) 119904 le 120585

(51)

To the end of this proof let 119904 isin (119886 119887) be fixed If120601(1199091015840(119904)) ge0 then 1199091015840(119904) ge 0 because 120601 is odd and increasing and sinceby assumption 119909(119905) gt 0 for all 119905 isin (119886 119887) we have that

1199091015840 (119904)

119909 (119904)ge 0 ge minus

1

119887 minus 119904forall119904 isin (119886 119887) (52)

which proves (50) in this case Let now 120601(1199091015840(119904)) le 0 Itimplies that 1199091015840(119904) le 0 and 120601(1199091015840(120585)) le 0 because of (51) and119903(119904) gt 0 for all 119904 isin (119886 119887) Hence from 119903(120585) le 119903(119904) for all120585 isin (119886 119887) 120585 ge 119904 from (51) and 120601(1199091015840(120585)) le 0 we especiallyconclude that

120601 (1199091015840

(119904)) ge119903 (120585)

119903 (119904)120601 (1199091015840

(120585)) ge 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 ge 119904

(53)

Acting on (53) with 120601minus1 we obtain

1199091015840

(119904) = 120601minus1(120601 (1199091015840

(119904))) ge 120601minus1(120601 (1199091015840

(120585))) = 1199091015840

(120585)

forall120585 isin (119886 119887) 120585 ge 119904(54)

By the Lagrangersquos mean-value theorem there exists a 120585 isin(119904 119887) depending on 119904 119887 such that 119909(119887) minus 119909(119904) = 1199091015840(120585)(119887 minus 119904)Since 119909(119904) gt 0 for all 119904 isin (119886 119887) and 119909 isin 119862((119886 119887]R) from(54) we obtain

minus119909 (119904) le 119909 (119887) minus 119909 (119904) = 1199091015840

(120585) (119887 minus 119904) le 1199091015840

(119904) (119887 minus 119904) (55)

which proves the desired inequality

Statement (44) will be frequently used in the followingform

Corollary 17 Let 120601 = 120601(V) satisfy (6) 120591(119905) le 119905 and 0 lt119903(119904) le 119903(119905) for all 119904 119905 119905

0le 119904 le 119905 Let 119886

1 1198871be two arbitrary

real numbers such that 1199050le 120591(119886

1) le 119886

1lt 1198871 Then for any

function 119909 isin 1198622((120591(1198861) 1198871)R) cap 119862([120591(119886

1) 1198871)R) such that

119909(119905) gt 0 for all 119905 isin (120591(1198861) 1198871) the following statement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (120591 (1198861) 1198871)

119905ℎ119890119899119909 (120591 (119905))

119909 (119905)ge120591 (119905) minus 120591 (119886

1)

119905 minus 120591 (1198861)

forall119905 isin (1198861 1198871)

(56)

Proof By assumptions of this corollary we have 120591(119905) le 119905119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))

1015840

le 0 for all 119905 isin (120591(1198861) 1198871) In

particular from statement (44) applied on (119886 119887) = (120591(1198861) 1198871)

we get

1199091015840 (119905)

119909 (119905)le

1

119905 minus 120591 (1198861)

forall119905 isin (120591 (1198861) 1198871) (57)

Integrating this inequality over the interval [120591(119905) 119905] for all119905 isin (1198861 1198871) we obtain

119909 (119905)

119909 (120591 (119905))le

119905 minus 120591 (1198861)

120591 (119905) minus 120591 (1198861)

forall119905 isin (1198861 1198871) (58)

which proves the desired inequality in (56)

Statement (50) will appear in the following form

Corollary 18 Let 1198861lt 1198871be two arbitrary real numbers Let

120601 = 120601(V) satisfy (6) 120590(119905) ge 119905 119903(119905) gt 0 and 119903(119904) ge 119903(119905) forall 119904 119905 119904 le 119905 Then for any function 119909 isin 1198622((119886

1 120590(1198871))R) cap

119862((1198861 120590(1198871)]R) such that 119909(119905) gt 0 for all 119905 isin (119886

1 120590(1198871)) the

following statement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (1198861 120590 (1198871))

119905ℎ119890119899119909 (120590 (119904))

119909 (119904)ge120590 (1198871) minus 120590 (119904)

120590 (1198871) minus 119904

forall119904 isin (1198861 1198871)

(59)

Proof Analogously with the proof of Corollary 17 we justneed to use the second inequality in (50) on (119886 119887) =(1198861 120590(1198871)) and integrate it over the interval [119904 120590(119904)] for all

119904 isin (1198861 1198871)

Next by Corollary 17 and the arithmetic-geometric meaninequality

119899

sum119894=0

120578119894119906119894ge119899

prod119894=0

119906120578119894

119894 119906119894ge 0 120578

119894gt 0 (60)

we can prove the following proposition

Proposition 19 (with delay arguments) Let 0 lt 119903(119904) le 119903(119905)for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

where

11986811= int312058716

1205878

119882(119905)(119905 minus 1205878)3

119905119889119905

11986812= int1205874

312058716

119882(119905)(119905 minus 1205878) (119905 minus 1205874)2

119905119889119905

11986821= int712058716

31205878

119882(119905)(119905 minus 31205878)3

119905 minus 1205874119889119905

11986822= int1205872

712058716

119882(119905)(119905 minus 31205878) (119905 minus 1205872)2

119905 minus 1205874119889119905

(32)

and119882(119905) = | cos 2119905|1205780 | sin 119905|1205781 | cos 119905|1205782

We leave to the reader to compare and verify whichone of the two inequalities (21) and (31) is simpler to beverified Also similar to the previous example it is possibleto get corresponding inequalities analogously to (31) for othercriteria published in the papers cited in the references

Remark 8 In [1 Section 3] the authors give an application of[1 Theorem 22] to (20) where 120591

1(119905) = 119905 minus 120591

1 1205912(119905) = 119905 minus

1205912 1205911= 1205878 and 120591

2= 1205874 However the following choice

(see [1 Section 3])

[1198861 1198881] = [2119896120587

120587

8+ 2119896120587]

[1198881 1198871] = [

120587

8+ 2119896120587

120587

4+ 2119896120587]

[1198862 1198882] = [

120587

4+ 2119896120587

3120587

8+ 2119896120587]

[1198882 1198872] = [

3120587

8+ 2119896120587

120587

2+ 2119896120587]

(33)

is not correct since the desired conditions 119902119894(119905) ge 0 on [119886

1minus

120591119894 1198871]cup[1198862minus120591119894 1198872] 119894 = 1 2 are not fulfilled Hence we suggest

reader to use the intervals proposed in the previous example

Open Question 9 The well-known variational techniquebased on the generalized Philosrsquo 119867-function has been usedin [1ndash11] to obtain some oscillation criteria for (1) where thesecond-order quasilinear differential operator (119903(119905)120601(1199091015840(119905)))1015840

is linear or half-linear Is it possible to use this technique inthe case of any function 120601(V) satisfying general conditions (6)and (7)

Theorem 10 (advanced equation) Under assumptions (6)(7) (9) and (10) let 119903(119905) be a nonincreasing positive functionon [1199050infin) and ℎ

119894(119905) = 120590

119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 2 119899

Let for every 119879 ge 1199050there exist numbers 119886

1 1198871 1198862 1198872 such that

119879 le 1198861lt 1198871le 120590max(1198871) le 119886

2lt 1198872and

119903119894(119905) ge 0 119902

119894(119905) ge 0

119900119899 [1198861 120590max (1198871)] cup [119886

2 120590max (1198872)]

119890 (119905) le 0 119900119899 [1198861 120590max (1198871)]

119890 (119905) ge 0 119900119899 [1198862 120590max (1198872)]

(34)

where 120590max(119905) = max1205901(119905) 120590

119899(119905) Equation (1) is oscilla-

tory provided there are two real parameters 1205821 1205822gt 0 such

that (11) is fulfilled where

119877119895(119905) = 119870

119899

sum119894=1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(35)

119876119895(119905)

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(119887119895)minus120590119894(119905)

120590119894(119887119895) minus 119905

)

120572119894120578119894

(36)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870

and 120578119894appearing respectively in (7) (9) and (10)

Now analogously with (16) we consider the oscillation ofthe advanced equation

(120601 (1199091015840

(119905)))1015840

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(37)

As a consequence of Theorem 10 we have the followingcriterion

Corollary 11 One assumes (6) (7) (10) and (34) Then (37)is oscillatory provided condition (17) is satisfied where 119876

119895(119905) is

defined in (36)

As the third case we consider second-order functionaldifferential equations with both delay and advanced argu-ments

Theorem 12 (delay-advanced equation) One assumes (6)(7) and (9) 120572

119894gt 119901 for 119894 isin 1 2 119899 and 119898 isin N 1 lt

119898 lt 119899 Let 119903(119905) equiv 119888119900119899119904119905 gt 0 on [1199050infin) ℎ

119894(119905) = 120591

119894(119905) le

119905 for 119894 isin 1 119898 and ℎ119894(119905) = 120590

119894(119905) ge 119905 for 119894 isin 119898 +

1 119899 on [1199050infin) Let 119903

119894(119905) 119902119894(119905) and 119890(119905) satisfy (12)-(13)

for 119894 isin 1 119898 and (34) for 119894 isin 119898 + 1 119899 Equation (1)is oscillatory provided there are two real parameters 120582

1 1205822gt

0 such that (11) is fulfilled where

119877119895(119905) = 119870

119898

sum119894=1

119903119894(119905) (

120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+ 119870119899

sum119894=119898+1

119903119894(119905) (

120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(38)

6 Abstract and Applied Analysis

119876119895(119905) =

119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(39)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870 appear-

ing respectively in (7) and (9)

Analogously with (16) and (37) we consider the oscilla-tion of delay-advanced equation

(120601 (1199091015840

(119905)))1015840

+119898

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905))

+119899

sum119894=119898+1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(40)

As a consequence of Theorem 12 we derive the followingoscillation criterion for (40)

Corollary 13 Let all assumptions of Theorem 12 hold withrespect to 119903

119894(119905) equiv 0 and 119891(119906) equiv 0 Equation (40) is oscillatory

provided condition (17) is satisfied where 119876119895(119905) is defined in

(39)

In Zafer [5] (see also [6]) the author has illustratedits main oscillation result on the following example of thesecond-order functional differential equations with delay andadvanced arguments Since |119909|120572minus1119909 = |119909|120572 sgn119909 it can berewritten in the form

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(41)

where 1198981 1198982 and 119890

0are positive constants 120591(119905) = 119905 minus 1205875

120590(119905) = 119905 + 12058720 and 1205721 1205722gt 0 It has been proved that

previous equation is oscillatory provided at least one of theconstants 119898

1 1198982 and 119890

0is large enough Here according

to Corollary 13 we can repeat this interesting conclusion forslightly complicated equation but with 120572

1 1205722gt 1

Example 14 We consider the following 120601-Laplacian func-tional differential equation with delay and advanced argu-ments

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(42)

where 120601 = 120601(V) satisfies (6) (7) 120591(119905) = 119905 minus 1205910 120590(119905) = 119905 + 120590

0

and 1205910 1205900isin [0 1205874) We claim that the previous equation

is oscillatory provided at least one of the constants 1198981 1198982

and 1198900is large enough Indeed it is enough to show that

condition (17) is fulfilled by using Corollary 13 It can be donein a similarway as in Example 4with (20) andCorollary 3Weleave it to the reader

3 Auxiliary Results QualitativeProperties of Concave-Like Functions

Let 119886 lt 119887 be two arbitrary real numbers It is known that if119909(119905) is a concave and smooth enough function on (119886 119887) thatis if 11990910158401015840(119905) le 0 for all 119905 isin (119886 119887) then 119909(119905) minus 119909(119904) ge 1199091015840(119905)(119905 minus 119904)for all 119904 119905 isin (119886 119887) 119904 lt 119905 Moreover if 119909(119905) gt 0 for all 119905 isin (119886 119887)then from previous inequality we obtain

1199091015840 (119905)

119909 (119905)le

1

119905 minus 119904forall119904 119905 isin (119886 119887) 119904 lt 119905 (43)

However often we do not have any information aboutthe sign of the second-order linear differential operator in(1) except only in a particular case when 119903(119905) equiv 1 and120601(V) equiv V Hence the main goal of this section is to findsome sufficient conditions on 119903(119905) and 120601(V) such that theassumption ldquo(119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)rdquo implies thedesired inequality (43) It is done in the following results

Lemma 15 Let 120601 = 120601(V) satisfy (6) Let 0 lt 119903(119904) le 119903(119905) forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862([119886 119887)R) such that 119909(119905) gt 0 for all 119905 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (119886 119887)

119905ℎ1198901198991199091015840 (119905)

119909 (119905)le

1

119905 minus 119886forall119905 isin (119886 119887)

(44)

Proof From assumption (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)it follows that

119903 (119905) 120601 (1199091015840

(119905)) le 119903 (120585) 120601 (1199091015840

(120585)) forall120585 119905 isin (119886 119887) 120585 le 119905

(45)

To the end of this proof let 119905 isin (119886 119887) be fixed Thanks to(6) if 120601(1199091015840(119905)) le 0 then 1199091015840(119905) le 0 too because 120601 is odd andincreasing In this case since 119903(119904) gt 0 and 119909(119904) gt 0 on (119886 119887)it implies that

1199091015840 (119905)

119909 (119905)le 0 lt

1

119905 minus 119886forall119905 isin (119886 119887) (46)

Thus it remains to show (44) for the case of 120601(1199091015840(119905)) ge 0From 0 lt 119903(120585) le 119903(119905) for all 120585 isin (119886 119887) 120585 le 119905 we obtain119903(119905)119903(120585) ge 1 which together with 120601(1199091015840(119905)) ge 0 and (45) gives

120601 (1199091015840

(119905)) le119903 (119905)

119903 (120585)120601 (1199091015840

(119905)) le 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 le 119905

(47)

Abstract and Applied Analysis 7

Acting on (47) with 120601minus1 and using that 120601minus1 is an increas-ing function because of (6) we obtain

1199091015840

(120585) ge 1199091015840

(119905) forall120585 isin (119886 119887) 120585 le 119905 (48)

Next since 119909(119905) gt 0 on (119886 119887) and 119909 isin 119862([119886 119887)R) wehave 119909(119886) ge 0 From the mean-value theorem for 119909(119905) on(119886 119905) we get a 120585 isin (119886 119905) depending on 119886 and 119905 such that119909(119905) minus 119909(119886) = 1199091015840(120585)(119905 minus 119886) and since 119909(119886) ge 0 from (48)we obtain

119909 (119905) ge 1199091015840

(120585) (119905 minus 119886) ge 1199091015840

(119905) (119905 minus 119886) (49)

which proves the desired inequality in (44)

In the advanced case of (1) that is when ℎ119894(119905) = 120590

119894(119905) ge 119905

we have the analogous result to Lemma 15

Lemma 16 Let 120601 = 120601(V) satisfy (6) Let 119903(119904) ge 119903(119905) gt 0 forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862((119886 119887]R) such that 119909(119904) gt 0 for all 119904 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (119886 119887)

119905ℎ1198901198991199091015840 (119904)

119909 (119904)ge minus

1

119887 minus 119904forall119904 isin (119886 119887)

(50)

Proof From assumption (119903(119904)120601(1199091015840(119904)))1015840 le 0 for all 119904 isin (119886 119887)we have

119903 (119904) 120601 (1199091015840

(119904)) ge 119903 (120585) 120601 (1199091015840

(120585)) forall119904 120585 isin (119886 119887) 119904 le 120585

(51)

To the end of this proof let 119904 isin (119886 119887) be fixed If120601(1199091015840(119904)) ge0 then 1199091015840(119904) ge 0 because 120601 is odd and increasing and sinceby assumption 119909(119905) gt 0 for all 119905 isin (119886 119887) we have that

1199091015840 (119904)

119909 (119904)ge 0 ge minus

1

119887 minus 119904forall119904 isin (119886 119887) (52)

which proves (50) in this case Let now 120601(1199091015840(119904)) le 0 Itimplies that 1199091015840(119904) le 0 and 120601(1199091015840(120585)) le 0 because of (51) and119903(119904) gt 0 for all 119904 isin (119886 119887) Hence from 119903(120585) le 119903(119904) for all120585 isin (119886 119887) 120585 ge 119904 from (51) and 120601(1199091015840(120585)) le 0 we especiallyconclude that

120601 (1199091015840

(119904)) ge119903 (120585)

119903 (119904)120601 (1199091015840

(120585)) ge 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 ge 119904

(53)

Acting on (53) with 120601minus1 we obtain

1199091015840

(119904) = 120601minus1(120601 (1199091015840

(119904))) ge 120601minus1(120601 (1199091015840

(120585))) = 1199091015840

(120585)

forall120585 isin (119886 119887) 120585 ge 119904(54)

By the Lagrangersquos mean-value theorem there exists a 120585 isin(119904 119887) depending on 119904 119887 such that 119909(119887) minus 119909(119904) = 1199091015840(120585)(119887 minus 119904)Since 119909(119904) gt 0 for all 119904 isin (119886 119887) and 119909 isin 119862((119886 119887]R) from(54) we obtain

minus119909 (119904) le 119909 (119887) minus 119909 (119904) = 1199091015840

(120585) (119887 minus 119904) le 1199091015840

(119904) (119887 minus 119904) (55)

which proves the desired inequality

Statement (44) will be frequently used in the followingform

Corollary 17 Let 120601 = 120601(V) satisfy (6) 120591(119905) le 119905 and 0 lt119903(119904) le 119903(119905) for all 119904 119905 119905

0le 119904 le 119905 Let 119886

1 1198871be two arbitrary

real numbers such that 1199050le 120591(119886

1) le 119886

1lt 1198871 Then for any

function 119909 isin 1198622((120591(1198861) 1198871)R) cap 119862([120591(119886

1) 1198871)R) such that

119909(119905) gt 0 for all 119905 isin (120591(1198861) 1198871) the following statement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (120591 (1198861) 1198871)

119905ℎ119890119899119909 (120591 (119905))

119909 (119905)ge120591 (119905) minus 120591 (119886

1)

119905 minus 120591 (1198861)

forall119905 isin (1198861 1198871)

(56)

Proof By assumptions of this corollary we have 120591(119905) le 119905119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))

1015840

le 0 for all 119905 isin (120591(1198861) 1198871) In

particular from statement (44) applied on (119886 119887) = (120591(1198861) 1198871)

we get

1199091015840 (119905)

119909 (119905)le

1

119905 minus 120591 (1198861)

forall119905 isin (120591 (1198861) 1198871) (57)

Integrating this inequality over the interval [120591(119905) 119905] for all119905 isin (1198861 1198871) we obtain

119909 (119905)

119909 (120591 (119905))le

119905 minus 120591 (1198861)

120591 (119905) minus 120591 (1198861)

forall119905 isin (1198861 1198871) (58)

which proves the desired inequality in (56)

Statement (50) will appear in the following form

Corollary 18 Let 1198861lt 1198871be two arbitrary real numbers Let

120601 = 120601(V) satisfy (6) 120590(119905) ge 119905 119903(119905) gt 0 and 119903(119904) ge 119903(119905) forall 119904 119905 119904 le 119905 Then for any function 119909 isin 1198622((119886

1 120590(1198871))R) cap

119862((1198861 120590(1198871)]R) such that 119909(119905) gt 0 for all 119905 isin (119886

1 120590(1198871)) the

following statement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (1198861 120590 (1198871))

119905ℎ119890119899119909 (120590 (119904))

119909 (119904)ge120590 (1198871) minus 120590 (119904)

120590 (1198871) minus 119904

forall119904 isin (1198861 1198871)

(59)

Proof Analogously with the proof of Corollary 17 we justneed to use the second inequality in (50) on (119886 119887) =(1198861 120590(1198871)) and integrate it over the interval [119904 120590(119904)] for all

119904 isin (1198861 1198871)

Next by Corollary 17 and the arithmetic-geometric meaninequality

119899

sum119894=0

120578119894119906119894ge119899

prod119894=0

119906120578119894

119894 119906119894ge 0 120578

119894gt 0 (60)

we can prove the following proposition

Proposition 19 (with delay arguments) Let 0 lt 119903(119904) le 119903(119905)for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

119876119895(119905) =

119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(119886119895)

119905 minus 120591119894(119886119895)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(119887119895) minus 120590119894(119905)

120590119894(119887119895) minus 119905

)

119901

(39)

for 119905 isin [119886119895 119887119895] 119895 isin 1 2 and positive constants 119901 119870 appear-

ing respectively in (7) and (9)

Analogously with (16) and (37) we consider the oscilla-tion of delay-advanced equation

(120601 (1199091015840

(119905)))1015840

+119898

sum119894=1

119902119894(119905)

1003816100381610038161003816119909 (120591119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120591

119894(119905))

+119899

sum119894=119898+1

119902119894(119905)

1003816100381610038161003816119909 (120590119894 (119905))1003816100381610038161003816120572119894 sgn119909 (120590

119894(119905)) = 119890 (119905)

(40)

As a consequence of Theorem 12 we derive the followingoscillation criterion for (40)

Corollary 13 Let all assumptions of Theorem 12 hold withrespect to 119903

119894(119905) equiv 0 and 119891(119906) equiv 0 Equation (40) is oscillatory

provided condition (17) is satisfied where 119876119895(119905) is defined in

(39)

In Zafer [5] (see also [6]) the author has illustratedits main oscillation result on the following example of thesecond-order functional differential equations with delay andadvanced arguments Since |119909|120572minus1119909 = |119909|120572 sgn119909 it can berewritten in the form

(100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119901minus1

1199091015840

(119905))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(41)

where 1198981 1198982 and 119890

0are positive constants 120591(119905) = 119905 minus 1205875

120590(119905) = 119905 + 12058720 and 1205721 1205722gt 0 It has been proved that

previous equation is oscillatory provided at least one of theconstants 119898

1 1198982 and 119890

0is large enough Here according

to Corollary 13 we can repeat this interesting conclusion forslightly complicated equation but with 120572

1 1205722gt 1

Example 14 We consider the following 120601-Laplacian func-tional differential equation with delay and advanced argu-ments

(120601 (1199091015840

(119905)))1015840

+ 1198981sin (119905) |119909 (120591 (119905))|1205721 sgn119909 (120591 (119905))

+ 1198982cos (119905) |119909 (120590 (119905))|1205722 sgn119909 (120590 (119905)) = minus119890

0cos (2119905)

(42)

where 120601 = 120601(V) satisfies (6) (7) 120591(119905) = 119905 minus 1205910 120590(119905) = 119905 + 120590

0

and 1205910 1205900isin [0 1205874) We claim that the previous equation

is oscillatory provided at least one of the constants 1198981 1198982

and 1198900is large enough Indeed it is enough to show that

condition (17) is fulfilled by using Corollary 13 It can be donein a similarway as in Example 4with (20) andCorollary 3Weleave it to the reader

3 Auxiliary Results QualitativeProperties of Concave-Like Functions

Let 119886 lt 119887 be two arbitrary real numbers It is known that if119909(119905) is a concave and smooth enough function on (119886 119887) thatis if 11990910158401015840(119905) le 0 for all 119905 isin (119886 119887) then 119909(119905) minus 119909(119904) ge 1199091015840(119905)(119905 minus 119904)for all 119904 119905 isin (119886 119887) 119904 lt 119905 Moreover if 119909(119905) gt 0 for all 119905 isin (119886 119887)then from previous inequality we obtain

1199091015840 (119905)

119909 (119905)le

1

119905 minus 119904forall119904 119905 isin (119886 119887) 119904 lt 119905 (43)

However often we do not have any information aboutthe sign of the second-order linear differential operator in(1) except only in a particular case when 119903(119905) equiv 1 and120601(V) equiv V Hence the main goal of this section is to findsome sufficient conditions on 119903(119905) and 120601(V) such that theassumption ldquo(119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)rdquo implies thedesired inequality (43) It is done in the following results

Lemma 15 Let 120601 = 120601(V) satisfy (6) Let 0 lt 119903(119904) le 119903(119905) forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862([119886 119887)R) such that 119909(119905) gt 0 for all 119905 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (119886 119887)

119905ℎ1198901198991199091015840 (119905)

119909 (119905)le

1

119905 minus 119886forall119905 isin (119886 119887)

(44)

Proof From assumption (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (119886 119887)it follows that

119903 (119905) 120601 (1199091015840

(119905)) le 119903 (120585) 120601 (1199091015840

(120585)) forall120585 119905 isin (119886 119887) 120585 le 119905

(45)

To the end of this proof let 119905 isin (119886 119887) be fixed Thanks to(6) if 120601(1199091015840(119905)) le 0 then 1199091015840(119905) le 0 too because 120601 is odd andincreasing In this case since 119903(119904) gt 0 and 119909(119904) gt 0 on (119886 119887)it implies that

1199091015840 (119905)

119909 (119905)le 0 lt

1

119905 minus 119886forall119905 isin (119886 119887) (46)

Thus it remains to show (44) for the case of 120601(1199091015840(119905)) ge 0From 0 lt 119903(120585) le 119903(119905) for all 120585 isin (119886 119887) 120585 le 119905 we obtain119903(119905)119903(120585) ge 1 which together with 120601(1199091015840(119905)) ge 0 and (45) gives

120601 (1199091015840

(119905)) le119903 (119905)

119903 (120585)120601 (1199091015840

(119905)) le 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 le 119905

(47)

Abstract and Applied Analysis 7

Acting on (47) with 120601minus1 and using that 120601minus1 is an increas-ing function because of (6) we obtain

1199091015840

(120585) ge 1199091015840

(119905) forall120585 isin (119886 119887) 120585 le 119905 (48)

Next since 119909(119905) gt 0 on (119886 119887) and 119909 isin 119862([119886 119887)R) wehave 119909(119886) ge 0 From the mean-value theorem for 119909(119905) on(119886 119905) we get a 120585 isin (119886 119905) depending on 119886 and 119905 such that119909(119905) minus 119909(119886) = 1199091015840(120585)(119905 minus 119886) and since 119909(119886) ge 0 from (48)we obtain

119909 (119905) ge 1199091015840

(120585) (119905 minus 119886) ge 1199091015840

(119905) (119905 minus 119886) (49)

which proves the desired inequality in (44)

In the advanced case of (1) that is when ℎ119894(119905) = 120590

119894(119905) ge 119905

we have the analogous result to Lemma 15

Lemma 16 Let 120601 = 120601(V) satisfy (6) Let 119903(119904) ge 119903(119905) gt 0 forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862((119886 119887]R) such that 119909(119904) gt 0 for all 119904 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (119886 119887)

119905ℎ1198901198991199091015840 (119904)

119909 (119904)ge minus

1

119887 minus 119904forall119904 isin (119886 119887)

(50)

Proof From assumption (119903(119904)120601(1199091015840(119904)))1015840 le 0 for all 119904 isin (119886 119887)we have

119903 (119904) 120601 (1199091015840

(119904)) ge 119903 (120585) 120601 (1199091015840

(120585)) forall119904 120585 isin (119886 119887) 119904 le 120585

(51)

To the end of this proof let 119904 isin (119886 119887) be fixed If120601(1199091015840(119904)) ge0 then 1199091015840(119904) ge 0 because 120601 is odd and increasing and sinceby assumption 119909(119905) gt 0 for all 119905 isin (119886 119887) we have that

1199091015840 (119904)

119909 (119904)ge 0 ge minus

1

119887 minus 119904forall119904 isin (119886 119887) (52)

which proves (50) in this case Let now 120601(1199091015840(119904)) le 0 Itimplies that 1199091015840(119904) le 0 and 120601(1199091015840(120585)) le 0 because of (51) and119903(119904) gt 0 for all 119904 isin (119886 119887) Hence from 119903(120585) le 119903(119904) for all120585 isin (119886 119887) 120585 ge 119904 from (51) and 120601(1199091015840(120585)) le 0 we especiallyconclude that

120601 (1199091015840

(119904)) ge119903 (120585)

119903 (119904)120601 (1199091015840

(120585)) ge 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 ge 119904

(53)

Acting on (53) with 120601minus1 we obtain

1199091015840

(119904) = 120601minus1(120601 (1199091015840

(119904))) ge 120601minus1(120601 (1199091015840

(120585))) = 1199091015840

(120585)

forall120585 isin (119886 119887) 120585 ge 119904(54)

By the Lagrangersquos mean-value theorem there exists a 120585 isin(119904 119887) depending on 119904 119887 such that 119909(119887) minus 119909(119904) = 1199091015840(120585)(119887 minus 119904)Since 119909(119904) gt 0 for all 119904 isin (119886 119887) and 119909 isin 119862((119886 119887]R) from(54) we obtain

minus119909 (119904) le 119909 (119887) minus 119909 (119904) = 1199091015840

(120585) (119887 minus 119904) le 1199091015840

(119904) (119887 minus 119904) (55)

which proves the desired inequality

Statement (44) will be frequently used in the followingform

Corollary 17 Let 120601 = 120601(V) satisfy (6) 120591(119905) le 119905 and 0 lt119903(119904) le 119903(119905) for all 119904 119905 119905

0le 119904 le 119905 Let 119886

1 1198871be two arbitrary

real numbers such that 1199050le 120591(119886

1) le 119886

1lt 1198871 Then for any

function 119909 isin 1198622((120591(1198861) 1198871)R) cap 119862([120591(119886

1) 1198871)R) such that

119909(119905) gt 0 for all 119905 isin (120591(1198861) 1198871) the following statement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (120591 (1198861) 1198871)

119905ℎ119890119899119909 (120591 (119905))

119909 (119905)ge120591 (119905) minus 120591 (119886

1)

119905 minus 120591 (1198861)

forall119905 isin (1198861 1198871)

(56)

Proof By assumptions of this corollary we have 120591(119905) le 119905119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))

1015840

le 0 for all 119905 isin (120591(1198861) 1198871) In

particular from statement (44) applied on (119886 119887) = (120591(1198861) 1198871)

we get

1199091015840 (119905)

119909 (119905)le

1

119905 minus 120591 (1198861)

forall119905 isin (120591 (1198861) 1198871) (57)

Integrating this inequality over the interval [120591(119905) 119905] for all119905 isin (1198861 1198871) we obtain

119909 (119905)

119909 (120591 (119905))le

119905 minus 120591 (1198861)

120591 (119905) minus 120591 (1198861)

forall119905 isin (1198861 1198871) (58)

which proves the desired inequality in (56)

Statement (50) will appear in the following form

Corollary 18 Let 1198861lt 1198871be two arbitrary real numbers Let

120601 = 120601(V) satisfy (6) 120590(119905) ge 119905 119903(119905) gt 0 and 119903(119904) ge 119903(119905) forall 119904 119905 119904 le 119905 Then for any function 119909 isin 1198622((119886

1 120590(1198871))R) cap

119862((1198861 120590(1198871)]R) such that 119909(119905) gt 0 for all 119905 isin (119886

1 120590(1198871)) the

following statement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (1198861 120590 (1198871))

119905ℎ119890119899119909 (120590 (119904))

119909 (119904)ge120590 (1198871) minus 120590 (119904)

120590 (1198871) minus 119904

forall119904 isin (1198861 1198871)

(59)

Proof Analogously with the proof of Corollary 17 we justneed to use the second inequality in (50) on (119886 119887) =(1198861 120590(1198871)) and integrate it over the interval [119904 120590(119904)] for all

119904 isin (1198861 1198871)

Next by Corollary 17 and the arithmetic-geometric meaninequality

119899

sum119894=0

120578119894119906119894ge119899

prod119894=0

119906120578119894

119894 119906119894ge 0 120578

119894gt 0 (60)

we can prove the following proposition

Proposition 19 (with delay arguments) Let 0 lt 119903(119904) le 119903(119905)for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

Acting on (47) with 120601minus1 and using that 120601minus1 is an increas-ing function because of (6) we obtain

1199091015840

(120585) ge 1199091015840

(119905) forall120585 isin (119886 119887) 120585 le 119905 (48)

Next since 119909(119905) gt 0 on (119886 119887) and 119909 isin 119862([119886 119887)R) wehave 119909(119886) ge 0 From the mean-value theorem for 119909(119905) on(119886 119905) we get a 120585 isin (119886 119905) depending on 119886 and 119905 such that119909(119905) minus 119909(119886) = 1199091015840(120585)(119905 minus 119886) and since 119909(119886) ge 0 from (48)we obtain

119909 (119905) ge 1199091015840

(120585) (119905 minus 119886) ge 1199091015840

(119905) (119905 minus 119886) (49)

which proves the desired inequality in (44)

In the advanced case of (1) that is when ℎ119894(119905) = 120590

119894(119905) ge 119905

we have the analogous result to Lemma 15

Lemma 16 Let 120601 = 120601(V) satisfy (6) Let 119903(119904) ge 119903(119905) gt 0 forall 119904 119905 isin (119886 119887) 119904 le 119905 For any function 119909 isin 1198622((119886 119887)R) cap119862((119886 119887]R) such that 119909(119904) gt 0 for all 119904 isin (119886 119887) the followingstatement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (119886 119887)

119905ℎ1198901198991199091015840 (119904)

119909 (119904)ge minus

1

119887 minus 119904forall119904 isin (119886 119887)

(50)

Proof From assumption (119903(119904)120601(1199091015840(119904)))1015840 le 0 for all 119904 isin (119886 119887)we have

119903 (119904) 120601 (1199091015840

(119904)) ge 119903 (120585) 120601 (1199091015840

(120585)) forall119904 120585 isin (119886 119887) 119904 le 120585

(51)

To the end of this proof let 119904 isin (119886 119887) be fixed If120601(1199091015840(119904)) ge0 then 1199091015840(119904) ge 0 because 120601 is odd and increasing and sinceby assumption 119909(119905) gt 0 for all 119905 isin (119886 119887) we have that

1199091015840 (119904)

119909 (119904)ge 0 ge minus

1

119887 minus 119904forall119904 isin (119886 119887) (52)

which proves (50) in this case Let now 120601(1199091015840(119904)) le 0 Itimplies that 1199091015840(119904) le 0 and 120601(1199091015840(120585)) le 0 because of (51) and119903(119904) gt 0 for all 119904 isin (119886 119887) Hence from 119903(120585) le 119903(119904) for all120585 isin (119886 119887) 120585 ge 119904 from (51) and 120601(1199091015840(120585)) le 0 we especiallyconclude that

120601 (1199091015840

(119904)) ge119903 (120585)

119903 (119904)120601 (1199091015840

(120585)) ge 120601 (1199091015840

(120585)) forall120585 isin (119886 119887) 120585 ge 119904

(53)

Acting on (53) with 120601minus1 we obtain

1199091015840

(119904) = 120601minus1(120601 (1199091015840

(119904))) ge 120601minus1(120601 (1199091015840

(120585))) = 1199091015840

(120585)

forall120585 isin (119886 119887) 120585 ge 119904(54)

By the Lagrangersquos mean-value theorem there exists a 120585 isin(119904 119887) depending on 119904 119887 such that 119909(119887) minus 119909(119904) = 1199091015840(120585)(119887 minus 119904)Since 119909(119904) gt 0 for all 119904 isin (119886 119887) and 119909 isin 119862((119886 119887]R) from(54) we obtain

minus119909 (119904) le 119909 (119887) minus 119909 (119904) = 1199091015840

(120585) (119887 minus 119904) le 1199091015840

(119904) (119887 minus 119904) (55)

which proves the desired inequality

Statement (44) will be frequently used in the followingform

Corollary 17 Let 120601 = 120601(V) satisfy (6) 120591(119905) le 119905 and 0 lt119903(119904) le 119903(119905) for all 119904 119905 119905

0le 119904 le 119905 Let 119886

1 1198871be two arbitrary

real numbers such that 1199050le 120591(119886

1) le 119886

1lt 1198871 Then for any

function 119909 isin 1198622((120591(1198861) 1198871)R) cap 119862([120591(119886

1) 1198871)R) such that

119909(119905) gt 0 for all 119905 isin (120591(1198861) 1198871) the following statement holds

119894119891 (119903 (119905) 120601 (1199091015840

(119905)))1015840

le 0 forall119905 isin (120591 (1198861) 1198871)

119905ℎ119890119899119909 (120591 (119905))

119909 (119905)ge120591 (119905) minus 120591 (119886

1)

119905 minus 120591 (1198861)

forall119905 isin (1198861 1198871)

(56)

Proof By assumptions of this corollary we have 120591(119905) le 119905119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))

1015840

le 0 for all 119905 isin (120591(1198861) 1198871) In

particular from statement (44) applied on (119886 119887) = (120591(1198861) 1198871)

we get

1199091015840 (119905)

119909 (119905)le

1

119905 minus 120591 (1198861)

forall119905 isin (120591 (1198861) 1198871) (57)

Integrating this inequality over the interval [120591(119905) 119905] for all119905 isin (1198861 1198871) we obtain

119909 (119905)

119909 (120591 (119905))le

119905 minus 120591 (1198861)

120591 (119905) minus 120591 (1198861)

forall119905 isin (1198861 1198871) (58)

which proves the desired inequality in (56)

Statement (50) will appear in the following form

Corollary 18 Let 1198861lt 1198871be two arbitrary real numbers Let

120601 = 120601(V) satisfy (6) 120590(119905) ge 119905 119903(119905) gt 0 and 119903(119904) ge 119903(119905) forall 119904 119905 119904 le 119905 Then for any function 119909 isin 1198622((119886

1 120590(1198871))R) cap

119862((1198861 120590(1198871)]R) such that 119909(119905) gt 0 for all 119905 isin (119886

1 120590(1198871)) the

following statement holds

119894119891 (119903 (119904) 120601 (1199091015840

(119904)))1015840

le 0 forall119904 isin (1198861 120590 (1198871))

119905ℎ119890119899119909 (120590 (119904))

119909 (119904)ge120590 (1198871) minus 120590 (119904)

120590 (1198871) minus 119904

forall119904 isin (1198861 1198871)

(59)

Proof Analogously with the proof of Corollary 17 we justneed to use the second inequality in (50) on (119886 119887) =(1198861 120590(1198871)) and integrate it over the interval [119904 120590(119904)] for all

119904 isin (1198861 1198871)

Next by Corollary 17 and the arithmetic-geometric meaninequality

119899

sum119894=0

120578119894119906119894ge119899

prod119894=0

119906120578119894

119894 119906119894ge 0 120578

119894gt 0 (60)

we can prove the following proposition

Proposition 19 (with delay arguments) Let 0 lt 119903(119904) le 119903(119905)for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

satisfy (10) Let 120591119894(119905) le 119905 on [119905

0infin) 119894 isin 1 119899 and

120591min(119905) = min1205911(119905) 120591

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge

0 on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 1198871)R) cap 119862([120591min(1198861) 1198871)R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (120591min(1198861) 1198871) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

119905 isin (1198861 1198871)

(61)

Proof By inequality (60) in particular for 119906119894= 120578minus1119894119902119894(119905)

(119909(120591119894(119905)))120572119894 and 119906

0= 120578minus10119890(119905) together with (56) we obtain

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

=1

119909119901 (119905)[119899

sum119894=1

120578119894(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894) + 120578

0(120578minus1

0|119890 (119905)|)]

ge1

119909119901 (119905)

119899

prod119894=1

(120578minus1

119894119902119894(119905) (119909 (120591

119894(119905)))120572119894)120578119894

(120578minus1

0|119890 (119905)|)

1205780

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

(119909 (119905))sum119899

119894=1120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894prod119899

119894=1(119909 (120591119894(119905)))120572119894120578119894

prod119899

119894=1(119909 (119905))120572119894120578119894

= (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(119909 (120591119894(119905))

119909 (119905))

120572119894120578119894

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

120572119894120578119894

(62)

which proves this proposition

Proposition20 (with advanced arguments) Let 119903(119904) ge 119903(119905) gt0 for all 119904 119905 such that 119905

0le 119904 lt 119905 and let 120601 = 120601(V) satisfy

(6) Let exponents 120572119894 and the (119899 + 1)-tuple (120578

0 1205781 120578

119899)

satisfy (10) Let 120590119894(119905) ge 119905 on [119905

0infin) 119894 isin 1 119899 and

120590max(119905) = max1205901(119905) 120590

119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0

on [1198861 1198871] where 119905

0le 1198861lt 1198871 Then for any function 119909 isin

1198622((1198861 120590max(1198871))R) cap 119862((119886

1 120590max(1198871)]R) such that 119909(119905) gt 0

and (119903(119905)120601(1199091015840(119905)))1015840 le 0 for all 119905 isin (1198861 120590max(1198871)) we have

1

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus 119890 (119905)]

ge (120578minus1

0|119890 (119905)|)

1205780

119899

prod119894=1

(120578minus1

119894119902119894(119905))120578119894

119899

prod119894=1

(120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

120572119894120578119894

119905 isin (1198861 1198871)

(63)

Proof It is very similar to the proof of Proposition 19 butinstead of Corollary 17 we need to use Corollary 18

According to Corollary 17 we are able to prove the nextuseful proposition in which we also use the well-knownYoung inequality119883119884 ge 0

119883120574+ (120574 minus 1) 119884

120574ge 120574119883119884

120574minus1 120574 gt 1 (64)

Proposition 21 (with delay-advanced arguments) Let 119903(119905) equiv119888119900119899119904119905 gt 0 on [119905

0infin) real numbers 120572

119894gt 119901 gt 0 for all

119894 = 1 2 119899 and119898 isin N 1 lt 119898 lt 119899 Let 120601 = 120601(V) satisfy (6)120591119894(119905) le 119905 119894 isin 1 119898 and 120590

119894(119905) ge 119905 119894 isin 119898 + 1 119899

on [1199050infin) and 120591min(119905) = min120591

1(119905) 120591

119898(119905) 120590max(119905) =

max120590119898+1

(119905) 120590119899(119905) Let 119890(119905) le 0 and 119902

119894(119905) ge 0 on

[1198861 1198871] where 119905

0le 1198861

lt 1198871 Then for any function 119909 isin

1198622((120591min(1198861) 120590max(1198871))R) cap 119862([120591min(1198861) 120590max(1198871)]R) suchthat 119909(119905) gt 0 and (119903(119905)120601(1199091015840(119905)))1015840 le 0 on (120591min(1198861) 120590max(1198871))we have

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

+1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119898

sum119894=1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

+119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(65)

for all 119905 isin (1198861 1198871)

Proof First of all by inequality (64) in particular for 120574 =120572119894119901 and

119883 = (119902119894(119905))119901120572119894119909119901(120591119894(119905)) 119884 = (

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

(66)

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

together with (56) we obtain

1

119909119901 (119905)[119898

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus

1

2119890 (119905)]

=1

119909119901 (119905)

119898

sum119894=1

[(119902119894(119905))119901120572119894119909119901(120591119894(119905))]120572119894119901

+(120572119894

119901minus 1)[(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

119901120572119894

]

120572119894119901

ge1

119909119901 (119905)

119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

119909119901(120591119894(119905))

=119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

[119909 (120591119894(119905))

119909 (119905)]

119901

ge119898

sum119894=1

120572119894

119901(119902119894(119905))119901120572119894(

119901 |119890 (119905)|

2 (120572119894minus 119901)119898

)

(120572119894minus119901)120572119894

times (120591119894(119905) minus 120591

119894(1198861)

119905 minus 120591119894(1198861)

)

119901

(67)

In the same way one can show that

1

119909119901 (119905)[119899

sum119894=119898+1

119902119894(119905) (119909 (120590

119894(119905)))120572119894 minus

1

2119890 (119905)]

ge119899

sum119894=119898+1

120572119894

119901119902119901120572119894

119894(119905) (

119901 |119890 (119905)|

2 (120572119894minus 119901) (119899 minus 119898 minus 1)

)

(120572119894minus119901)120572119894

times (120590119894(1198871) minus 120590119894(119905)

120590119894(1198871) minus 119905

)

119901

(68)

The previous two inequalities prove this proposition

4 Proof of Main Results

Proof of Theorem 1 If the assertion of this theorem does nothold then there is a nonoscillatory solution 119909(119905) of (1) suchthat 119909(119905) = 0 for all 119905 ge 119879 and some 119879 ge 119905

0 Moreover it is

enough to work only with the case 119909(119905) gt 0 and 119890(119905) le 0 on[120591min(1198861) 1198871] since the second case 119909(119905) lt 0 and minus119890(119905) le 0 on[120591min(1198862) 1198872] can be transformed into the first one Indeedmultiplying equation (1) by minus1 and using assumptions that120601(V) and119891(119906) are odd functions we have that (1) is equivalentto

(119903 (119905) 120601 (minus1199091015840

(119905)))1015840

+119899

sum119894=1

119903119894(119905) 119891 (minus119909 (120591

119894(119905)))

+119899

sum119894=1

119902119894(119905)

1003816100381610038161003816(minus119909 (120591119894 (119905)))1003816100381610038161003816120572119894 sgn (minus119909 (120591

119894(119905))) = minus119890 (119905)

119905 ge 1199050

(69)

The proof is outlined in the following three steps

Step 1 Next for any 1205821gt 0 the following function is well

defined

1205961(119905) = minus

1205821119903 (119905) 120601 (1199091015840 (119905))

119909119901 (119905) 119905 isin [119886

1 1198871] (70)

and 1205961isin 1198621([119886

1 1198871)R) Since 120582

1gt 0 and 119903(119905) gt 0 on

[1198861 1198871] from the previous equality we get

10038161003816100381610038161003816120601 (1199091015840

(119905))10038161003816100381610038161003816 =

10038161003816100381610038161205961 (119905)1003816100381610038161003816

1205821119903 (119905)

119909119901

(119905)

1205961015840

1(119905) = minus

1205821(119903 (119905) 120601 (1199091015840 (119905)))

1015840

119909119901 (119905)+1199011205821119903 (119905) 120601 (1199091015840 (119905))

119909119901+1 (119905)1199091015840

(119905)

(71)

Using (1) and assumptions (7) and (9) from the secondequality of (71) we get

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821119870119899

sum119894=1

119903119894(119905) (

119909 (120591119894(119905))

119909 (119905))

119901

+1205821

119909119901 (119905)[119899

sum119894=1

119902119894(119905) (119909 (120591

119894(119905)))120572119894 minus 119890 (119905)]

(72)

where 119905 isin (1198861 1198871) Next using assumptions (12) and (13) in (1)

for all 119905 isin (120591min(1198861) 1198871) we easily conclude that 119909(119905) satisfiesall assumptions of Corollary 17 and Proposition 19 Hencefrom (56) (61) and (72) we observe

119889

1198891199051205961(119905) ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205961 (119905)1003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin (1198861 1198871)

(73)

Step 2 In this step we need the next elementary proposition

Proposition 22 Let 119901 gt 0 and let 120587119901be a positive number

defined by

120587119901=

120587

((119901 + 1) 119901) sin (119901120587 (119901 + 1)) (74)

Then there is an increasing odd function 119910 = 119910(119904) 119910 isin

1198621((minus120587119901 120587119901)R) such that

1199101015840

(119904) = 1 +1003816100381610038161003816119910 (119904)

1003816100381610038161003816(119901+1)119901

119904 isin (minus120587119901 120587119901)

119910 (0) = 0 119910 (120587119901) = infin

(75)

Moreover for 119901 = 1 we have 120587119901= 1205872 and we can take

119910(119904) = tan(119904) 119904 isin (minus1205872 1205872)

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

Proof Let 119911 = 119911(119905) be a function defined by

119911 (119905) = int119905

0

1

1 + |120591|(119901+1)119901119889120591 119905 isin R (76)

It is not difficult to see that 119911(infin) = 120587119901(see [24])

and 119911 is a bijection from R on interval (minus119911(infin) 119911(infin)) andthat the function 119910(119904) determined by the formula 119910(119904) =

119911minus1(119904) where 119911minus1(119904) is the inverse function of 119911(119905) satisfiesall properties given in (75)

Let now 119904119895

isin (minus1205872 1205872) be such that 119910(119904119895) =

1205961(119886119895) (such an 119904

119895exists since 119910(119904) is a bijection from

(minus1205872 1205872) to (minusinfininfin)) 119869 = [1198861 1198871] cup [119886

2 1198872] and let

1198620(119905) be a function defined by

1198620(119905) =

1

2120587119901

min119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(77)

Because of (11) we have 1198880119895= int119887119895

119886119895

1198620(120591)119889120591 ge 1 Hence

2120587119901

1198880119895

1198620(119905) le min

119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

119905 isin 119869

(78)

Next let 1198811(119905) and 119881

2(119905) be two functions defined by

119881119895(119905) = 119904

119895+2120587119901

1198880119895

int119905

119886119895

1198620(120591) 119889120591 119905 isin [119886

119895 119887119895] 119895 isin 1 2

(79)

Since 1198620(119905) ge 0 for all 119905 isin 119869 we have 119881

119895(119886119895) = 119904119895lt 120587119901

and 119881119895(119887119895) gt 2120587

119901+ 119904119895 Since 119881

119895(119905) is continuous it gives the

existence of 119879lowast119895isin (119886119895 119887119895) such that 119881

119895(119879lowast119895) = 120587

119901 Hence the

function 120596119895(119905) = tan(119881

119895(119905)) 119905 isin (119886

119895 119879lowast119895) satisfies

120596119895(119886119895) = 119910 (119881

119895(119886119895)) = 119910 (119904

119895) = 1205961(119886119895)

120596119895(119879lowast

119895) = 119910 (119881

119895(119879lowast

119895)) = 119910 (120587

119901) = infin

(80)

and because of (78) we obtain

1205961015840

119895(119905) = 119910

1015840(119881119895(119905)) 119881

1015840

119895(119905)

= (1 +10038161003816100381610038161003816119910 (119881119895 (119905))

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

= (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)1198811015840

119895(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)2120587119901

1198880119895

1198620(119905)

le (1 +10038161003816100381610038161003816120596119895 (119905)

10038161003816100381610038161003816(119901+1)119901

)

timesmin119901

(1205821119903 (119905))1119901

1205821(1198771(119905) + 119876

1(119905))

(81)

that is

1205961015840

119895(119905) le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161003816120596119895 (119905)10038161003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

119905 isin [119886119895 119879lowast

119895)

(82)

Step 3We claim that

1205961(119905) le 120596

1(119905) on [119886

1 119879lowast

1) (83)

In order to prove inequality (83) we need the followingproposition

Proposition 23 Let 119886 lt 119879lowast and 120593 120595 isin 1198621([119886 119879lowast)R) be twofunctions satisfying on [119886 119879lowast) respectively

1205931015840le

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205931003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

1205951015840ge

119901

(1205821119903 (119905))1119901

10038161003816100381610038161205951003816100381610038161003816(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905))

(84)

Then the following statement holds

120593 (119886) le 120595 (119886) 119894119898119901119897119894119890119904 120593 (119905) le 120595 (119905) forall119905 isin [119886 119879lowast) (85)

Proof If we denote by

119865 (119905 119906) =119901

(1205821119903 (119905))1119901

119906(119901+1)119901

+ 1205821(1198771(119905) + 119876

1(119905)) (86)

then (84) can be rewritten in the form 1205931015840 le 119865(119905 120593) and 1205951015840 ge119865(119905 120595) on [119886 119879lowast) Also it is not difficult to checkthat 119865(119905 119906) is locally Lipschitz function in the secondvariable and bounded in the first variable on bounded inter-val [119886 119879lowast) Hence we may apply the pointwise comparisonprinciple from [12 Lemma 19] and consequently weconclude that statement (85) holds which proves thisproposition

Next we proceed with the proof of inequality (83) By(80) we know that 120596

119895(119886119895) = 1205961(119886119895) which together with (73)

and (82) ensures that we may apply Proposition 23 Now thedesired inequality (83) immediately follows from statement(85)

Next from the second equality of (80) and inequality (83)we conclude that 120596

1(119879lowast1) = infin It contradicts the fact that

1205961isin 1198621([119886

1 1198871)) Hence 119909(119905) can not be oscillatory as it is

supposed at the beginning of this proof and we conclude that(1) is oscillatory

Proof of Corollary 2 From (16) we see that 119903(119905) equiv1 and 119903

119894(119905) equiv 0 Next to the end of this proof let 120582

119895be

defined by

120582119895= (

119901

max[119886119895 119887119895]

119876119895(119905)

)

119901(119901+1)

119895 isin 1 2 (87)

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 11

Now we can rewrite (17) in the form

1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1 (88)

Since max[119886119895 119887119895]

119876119895(119905) le 119876

119895(119905) it is clear that

119901

1205821119901

119895

ge 120582119895119876119895(119905) 119905 isin [119886

119895 119887119895] (89)

and so

min

119901

1205821119901

119895

120582119895119876119895(119905)

= 120582119895119876119895(119905) (90)

Together with assumption (88) we show that

1

2120587119901

int119887119895

119886119895

min

119901

(120582119895119903 (119905))1119901

120582119895(119877119895(119905) + 119876

119895(119905))

119889119905

=1

2120587119901

int119887119895

119886119895

min

119901

1205821119901

119895

120582119895119876119895(119905)

119889119905

=1

2120587119901

int119887119895

119886119895

120582119895119876119895(119905) 119889119905 ge 1

(91)

It proves that desired condition (11) is fulfilled Thus thiscorollary follows fromTheorem 1

Proofs of Theorems 10 and 12 It follows by the same lineof arguments as in the proof of Theorem 1 but instead ofProposition 19 we use Propositions 20 and 21 respectively

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291 Theauthor has dedicated this work to his professors AsimMusaefendicmdashGymnasium Maglaj Svetozar Kurepa NikolaSarapa Branko Najman and Darko Zubrinicmdashgraduateand postgraduate studies on Faculty of Sciences Univer-sity of Zagreb Jean-Pierre Puel and Jacques-Louis Lionsmdashpostdoctoral study in Paris 199697

References

[1] Y Bai and L Liu ldquoNew oscillation criteria for second-orderdelay differential equations with mixed nonlinearitiesrdquoDiscreteDynamics in Nature and Society vol 2010 Article ID 796256 9pages 2010

[2] S Murugadass E Thandapani and S Pinelas ldquoOscillationcriteria for forced second-order mixed type quasilinear delaydifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2010 article 73 9 pages 2010

[3] T S Hassan L Erbe and A Peterson ldquoForced oscillation ofsecond order differential equations with mixed nonlinearitiesrdquoActa Mathematica Scientia B vol 31 no 2 pp 613ndash626 2011

[4] C Li and S Chen ldquoOscillation of second-order functionaldifferential equations with mixed nonlinearities and oscillatorypotentialsrdquo Applied Mathematics and Computation vol 210 no2 pp 504ndash507 2009

[5] A Zafer ldquoInterval oscillation criteria for second order super-half linear functional differential equations with delay andadvanced argumentsrdquoMathematische Nachrichten vol 282 no9 pp 1334ndash1341 2009

[6] A F Guvenilir andA Zafer ldquoSecond-order oscillation of forcedfunctional differential equations with oscillatory potentialsrdquoComputers amp Mathematics with Applications vol 51 no 9-10pp 1395ndash1404 2006

[7] A F Guvenilir ldquoInterval oscillation of second-order functionaldifferential equations with oscillatory potentialsrdquo NonlinearAnalysis Theory Methods amp Applications vol 71 no 12 ppe2849ndashe2854 2009

[8] Q Kong ldquoInterval criteria for oscillation of second-orderlinear ordinary differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 229 no 1 pp 258ndash270 1999

[9] Y G Sun and J S W Wong ldquoOscillation criteria for secondorder forced ordinary differential equations with mixed non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 334 no 1 pp 549ndash560 2007

[10] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[11] L H Erbe Q Kong and B G Zhang Theory for FunctionalDifferential Equations vol 190 ofPure andAppliedMathematicsMarcel Dekker New York NY USA 1994

[12] M Pasic ldquoFite-Wintner-Leighton-type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[13] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[14] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001Reprint of the 1998 edition

[15] P L de Napoli and J P Pinasco ldquoA Lyapunov inequalityfor monotone quasilinear operatorsrdquo Differential and IntegralEquations vol 18 no 10 pp 1193ndash1200 2005

[16] M Feng ldquoPeriodic solutions for prescribed mean curvatureLienard equation with a deviating argumentrdquo Nonlinear Analy-sis Real World Applications vol 13 no 3 pp 1216ndash1223 2012

[17] I Rachunkova and M Tvrdy ldquoSecond-order periodic problemwith 120601-Laplacian and impulsesrdquo Nonlinear Analysis TheoryMethods and Applications vol 63 no 5ndash7 pp e257ndashe266 2005

[18] L Ferracuti and F Papalini ldquoBoundary-value problems forstrongly non-linear multivalued equations involving different120601-Laplaciansrdquo Advances in Differential Equations vol 14 no 5-6 pp 541ndash566 2009

[19] I Yermachenko and F Sadyrbaev ldquoQuasilinearization tech-nique for 120601-Laplacian type equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2012 Article ID975760 11 pages 2012

[20] J Henderson A Ouahab and S Youcefi ldquoExistence andtopological structure of solution sets for 120601-Laplacian impulsivedifferential equationsrdquo Electronic Journal of Differential Equa-tions vol 2012 article 56 16 pages 2012

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Abstract and Applied Analysis

[21] V Kolmanovskii and A Myshkis Introduction to the Theoryand Applications of Functional Differential Equations vol 463 ofMathematics and Its Applications KluwerAcademicDordrechtThe Netherlands 1999

[22] R Finn Equilibrium Capillary Surface Springer New York NYUSA 1986

[23] E Giusti Minimal Surfaces and Functions of Bounded Varia-tions Birkhauser Basel Switzerland 1984

[24] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of