research article analytical solutions for the elastic circular rod...
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Research ArticleAnalytical Solutions for the Elastic Circular Rod NonlinearWave Boussinesq and Dispersive Long Wave Equations
Shi Jing1 and Yan Xin-li2
1 School of Science Changrsquoan University Xirsquoan 710064 China2 School of Science Xirsquoan University of Architecture and Technology Xirsquoan 710055 China
Correspondence should be addressed to Shi Jing shijing-sj163com
Received 31 December 2013 Revised 23 March 2014 Accepted 25 March 2014 Published 22 April 2014
Academic Editor Michael Meylan
Copyright copy 2014 S Jing and Y Xin-li This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
The solving processes of the homogeneous balance method Jacobi elliptic function expansion method fixed point method andmodified mapping method are introduced in this paper By using four different methods the exact solutions of nonlinear waveequation of a finite deformation elastic circular rod Boussinesq equations and dispersive long wave equations are studied In thediscussion the more physical specifications of these nonlinear equations have been identified and the results indicated that thesemethods (especially the fixed point method) can be used to solve other similar nonlinear wave equations
1 Introduction
Nonlinear partial differential equations are widely used todescribe complex phenomena in various fields of sciencessuch as biology chemistry communication and especiallymany branches of physics like condensedmatter physics fieldtheory fluid dynamics plasma physics and optics and soforth In this paper we consider nonlinear wave equation of afinite deformation elastic circular rod Boussinesq equationsand dispersive long wave equations with same physicalbehaviorThe elastic circular rod is one important componentin the structures in which the dynamics of these componentsare governed by double nonlinearity and double dispersionwave equation [1] The Boussinesq equation which was firstintroduced in 1871 arises in several physical applicationsThedynamics of shallow water waves which are seen in variousplaces like sea beaches lakes and rivers are governed by theBoussinesq equation In recent years there has been muchinterest in some variants of the Boussinesq systems [2ndash6]These coupledBoussinesq equations [7] arise in shallowwaterwaves for two-layered fluid flow This situation occurs whenthere is an accidental oil spill from a ship which results ina layer of oil floating above the layer of water The (2 + 1)-dimensional dispersive long wave equations [8 9] were firstderived by Boiti as a compatibility condition for a ldquoweakrdquo
Lax pair A good understanding of the solutions for theseequations is very helpful to coastal and civil engineers inapplying the nonlinear water model to coastal harbor design
In the field of nonlinear science to find exact solutionsfor a nonlinear system is one of the most fundamentaland significant studies The evaluation of exact solutionsof nonlinear wave has complicated nonlinear defects suchequations are often very difficult to be solved Althoughintensive investigations have made significant progress inrecent years manymethods have been proposed to constructexact solutions such as Weierstrass elliptic function methodthe homogeneous balance method sine-cosine method thenonlinear transformation method the hyperbolic tangentfunctions finite expansion improvedmapping approach andfurther extended tanh method
The homogeneous balance method is a powerful tool tofind solitary wave solutions of nonlinear partial differentialequations Fan introduced the homogeneous balancemethodinto the search for Backlund transformations and obtainedmore solutions [10]
The mapping method is a very effective direct method toconstruct exact solutions of nonlinear equations Zhang et almake use of the auxiliary equation and the expanded map-ping methods to find the new exact periodic solutions for(2 + 1)-dimensional dispersive long wave equations [11]
Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 487571 10 pageshttpdxdoiorg1011552014487571
2 Journal of Applied Mathematics
The basic idea of the fixed point methods consists infinding an iteration function which generates successiveapproximations to the solution [12]
Many periodic solutions have been recently expressed interms of various Jacobi-elliptic functions for a wide class ofnonlinear evolution equations which have been obtained bymeans of Jacobi elliptic expansion method [13]
The longitudinal wave equation of a finite deformationelastic circular rod Boussinesq equations and dispersive longwave equations are nonlinear partial differential equationsof different scientific field We find that they have the samecharacteristics In this paper the analytical solutions of thedifferential equations for the elastic circular rod Boussinesqequations and dispersive long wave equations are solved byusing homogeneous balance method Jacobi elliptic functionmethod fixed point method andmodifiedmappingmethodThemore physical specifications of these nonlinear equationshave been identified and the results indicated
2 Three Types of Nonlinear Wave Equations
The longitudinal wave equation of a finite deformation elasticcircular rod is [1]
119906119905119905minus 1198882
0119906119909119909= [
3119864
21205881199062+119864
21205881199063+]21198772
2(119906119905119905minus 1198882
1119906119909119909)]119909119909
(1)
where 1198880= radic119864120588 is the longitudinal wave velocity for a linear
elastic rod and 1198881= radic120583120588 is the shear wave velocity 120588 is the
density of thematerial ] is the Poisson ratio119877 is the diameterof bar 119864 is Yongrsquos modulus of material 120583 is the elastic shearmodulus of material 119905 is the time variable
Make the traveling wave transformation
119906 (119909 119905) = 119906 (120585) 120585 = 119896 (119909 minus 120582119905) (2)
Substituting (2) into (1) and integrating it with respect to120585 twice we can obtain
119896211990610158401015840+ 1205731119906 + 12057321199062+ 12057331199063= 119873 (3)
where 119873 is the second integral constant The first integralconstant is zero and
1205731=
2 (1205822 minus 11988820)
]21198772 (1205822 minus 11988821) 120573
2=
311988820
]21198772 (1205822 minus 11988821)
1205733=
11988820
]21198772 (1205822 minus 11988821)
(4)
The variant Boussinesq equations were discussed in [7]They are coupling wave equations which can be expressed asfollows
119867119905+ (119867119906)
119909+ 119906119909119909119909
= 0
119906119905+ 119867119909+ 119906119906119909= 0
(5)
where 119906(119909 119905) is the velocity and 119867(119909 119905) is the total depthMake the traveling wave transformation
119867(119909 119905) = 119867 (120585) 119906 (119909 119905) = 119906 (120585) 120585 = 119896 (119909 minus 120582119905)
(6)
Substituting (6) into (5) and integrating with respect to 120585we have
minus120582119867 +119867119906 + 119896211990610158401015840= 1198861 119867 = 120582119906 minus
1
21199062+ 1198871 (7)
where 1198861 1198871are integral constants
Substituting the second formula of (7) into the first oneand assuming120573
1= 1198871minus1205822120573
2= (32)120582120573
3= minus(12)119873 = 119886
1+
1205821198871 we can obtain (3) So the solutions of (7) are equivalent
to that of (3) and the second formula of (7)(2 + 1)-dimension dispersive long wave equations were
discussed in [8 9] They can be written as follows
119906119905119910+ ℎ119909119909+1
2(1199062)119909119910= 0
ℎ119905+ (ℎ119906 + 119906 + 119906
119909119910)119909= 0
(8)
Suppose that the traveling wave solutions for (8) can begiven in the following forms
119906 (119909 119910 119905) = 119906 (120585) ℎ (119909 119910 119905) = ℎ (120585)
120585 = 119909 + 119899119910 + 120596119905(9)
Substituting (9) into (8) we have
11989911990610158401015840+ (1 minus 119899120596
2) 119906 minus
3
21198991205961199062minus119899
21199063= 1198731
ℎ = minus119899120596119906 minus1
21198991199062
(10)
where1198731is the second integral constant
Obviously the first formula of (10) and (3) have the samecharacteristics On the other hand the solutions of (10) and(7) are samewith each other except for the coefficientsThere-fore in following parts the solutions of (5) are presented
3 Homogeneous Balance Method
In [14 15] the homogeneous balance method is used tosolve nonlinear wave equations For example the followingequation was solved in [15]
119906119905+ 6119906119906
119909+ 119906119909119909119909
= 0 (11)
The exact solution of the following corresponding equa-tion was given by using the same method
119906119905minus 119863119906119909119909+ 120582 (119906
3+ 1205721199062+ 120573119906) = 0 (12)
To solve (5) the homogeneous balance method wasimproved in this paper
Journal of Applied Mathematics 3
Now suppose that
119906 (119909 119905) = 1198911015840120596119909+ 119887 119867 (119909 119905) = 119892
10158401205962
119909+ 119888 (13)
where119891(120596)119892(120596) and120596(119909 119905) are the undetermined functionsand 119887 119888 are undetermined constants
Substituting (13) into (5) the equations can be rewrittenas follows
(119891101584010158401198921015840+ 119891101584011989210158401015840+ 119891(4)) 1205964
119909+ 61198911015840101584010158401205962
1199091205962119909
+ 11989110158401015840(1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909) + 11989210158401015840(1198871205963
119909+ 1205962
119909120596119905)
+ 3119891101584011989210158401205962
1199091205962119909+ 1198911015840(1198881205962119909+ 1205964119909)
+ 21198921015840(1198871205961199091205962119909+ 120596119909120596119909119905) = 0
(11989210158401015840+ 119891101584011989110158401015840) 1205963
119909+ 11989110158401015840(120596119909120596119905+ 1198871205962
119909) + 119891101584021205961199091205962119909
+ 1198911015840(120596119909119905+ 1198871205962119909) + 2119892
10158401205961199091205962119909= 0
(14)
where 1205962119909= 120596119909119909 1205963119909= 120596119909119909119909
1205964119909= 120596119909119909119909119909
In (14) let
119891101584010158401198921015840+ 119891101584011989210158401015840+ 119891(4)
= 0 11989210158401015840+ 119891101584011989110158401015840= 0 (15)
Integrating the second formula of (15) we have
1198921015840+1
211989110158402= 0 (16)
Furthermore we can obtain
1198921015840= minus
1
211989110158402 119892
10158401015840= minus119891101584011989110158401015840 (17)
Substituting (17) into the first formula of (15) we have
119891(4)minus3
21198911015840211989110158401015840= 0 (18)
Integrating (18) we have
119891101584010158401015840minus1
211989110158403= 0 (19)
We know the solution of (19) is
119891 = 2 ln120596 (20)
Obviously we have
11989110158402= minus2119891
10158401015840 119891
101584011989110158401015840= minus119891101584010158401015840
11989110158401198921015840= minus119891101584010158401015840 119891
10158403= 2119891101584010158401015840
(21)
Substituting (17) and (21) into (14) the equations can berewritten as follows
1205962
119909(31205962119909+ 119887120596119909+ 120596119905) 119891101584010158401015840
+ (1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909+ 2119887120596
1199091205962119909+ 2120596119909120596119909119905) 11989110158401015840
+ (1198881205962119909+ 1205964119909) 1198911015840= 0
120596119909(120596119905+ 119887120596119909) 11989110158401015840+ (120596119909119905+ 1198871205962119909) 1198911015840= 0
(22)
Letting all coefficients of 119891 in (22) be zero we have31205962119909+ 119887120596119909+ 120596119905= 0
1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909+ 2119887120596
1199091205962119909+ 2120596119909120596119909119905= 0
1198881205962119909+ 1205964119909= 0
120596119905+ 119887120596119909= 0 120596
119909119905+ 1198871205962119909= 0
(23)
From first and fifth second formula of (23) it can beobtained that
120596 = 120572 (119905) 119909 + 120573 (119905) (24)where 120572(119905) is undetermined function and 120573(119905) is arbitraryfunction
From second formula of (23) we have
120572 (119905) = 119890minus(1198882)119905+119896
(25)where 119896 is the integral constant
Therefore from (13) (20) (24) and (25) we know thesolutions of (5) are
1199061(119909 119905) =
2120572 (119905)
120572 (119905) 119909 + 120573 (119905)+ 119887
1198671(119909 119905) =
minus21205722(119905)
[120572 (119905) 119909 + 120573 (119905)]2+ 119888
(26)
4 Jacobi Elliptic Function Method
In [16] the NLS equation and Zakharov equation werestudied by using the Jacobi elliptic function method In thispaper (1) and (5) are discussed by using this method
41 Jacobi Elliptic Sine Function Method Let119906 = 1198860+ 1198861sn120585 (27)
We know119889119906
119889120585= 1198861cn120585dn120585
1198892119906
1198891205852= minus (1 + 119898
2) 1198861sn120585 + 21198982119886
1sn3120585
1199062= 1198862
0+ 211988601198861sn120585 + 1198862
1sn2120585
1199063= 1198863
0+ 31198862
01198861sn120585 + 3119886
01198862
1sn2120585 + 1198863
1sn3120585
(28)
where cn120585 and dn120585 are the Jacobi elliptic cosine functionand the third type Jacobi elliptic function respectively and119898 (0 lt 119898 lt 1) is module
Substituting (27) into (3) we have
12057311198860+ 12057321198862
0+ 12057331198863
0minus 119873
+ 1198861[minus1198962(1 + 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] sn120585
+ 1198862
1(1205732+ 312057331198860) sn2120585 + 119886
1(211989621198982+ 12057331198862
1) sn3120585 = 0
(29)
4 Journal of Applied Mathematics
Let the coefficients of all derivatives of sn120585 be zero and wehave
1198860=minus1205732
31205733
1198861= plusmn
radic2119896119898
radicminus1205733
1198962=
1
1 + 1198982(1205731+ 212057321198860+ 312057331198862
0)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(30)
Therefore the solution of (3) is
1199062(119909 119905) = 119886
0+ 1198861sn119896 (119909 minus 120582119905) (31)
where 1198860 1198861 and 119896 are denoted by (30) and 120582 is arbitrary
constantWhen119898 rarr 1 sn120585 rarr tanh 120585
1199062(119909 119905) = 119886
0+ 1198861tanh 119896 (119909 minus 120582119905) (32)
Thus the solutions of (5) are (32) and the followingformula
1198672(119909 119905) = 119887
1+ 1205821199062minus1
21199062
2 (33)
Assuming 120582 = minus1 1198871= 15 120573
1= 05 120573
2= minus15 and
1205733= minus05 it can be obtained that 119886
0= minus1 119886
1= 2 and 119896 =
1 The solutions of (5) are 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and119867(119909 119905) = 2 minus 2 tanh2(119909 + 119905) The solitary wave and behaviorof the solutions 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and 119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) are shown in Figures 1 and 2 respectivelyfor 0 le 119905 le 1 and minus10 le 119909 le 10 The waveform is similar tothe result in [17]
42 Jacobi Elliptic Cosine Function Method Let
119906 = 1198860+ 1198861cn120585 (34)
Thus
1199061015840= minus1198861sn120585dn120585 119906
10158401015840= (2119898
2minus 1) 119886
1cn120585 minus 21198982cn3120585
1199062= 1198862
0+ 211988601198861cn120585 + 1198862
1cn2120585
1199063= 1198863
0+ 31198862
01198861cn120585 + 3119886
01198862
1cn2120585 + 1198863
1cn3120585
(35)
Substituting (35) into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0minus 119873
+ 1198861[1198962(21198982minus 1) + 120573
1+ 212057321198860+ 312057331198862
0] cn120585
+ 1198862
1(1205732+ 312057331198860) cn2120585
+ 1198861(minus211989621198982+ 12057331198862
1) cn3120585 = 0
(36)
05
10
0
1
0
10
minus10
minus5
minus3
minus2
minus1
minus25
minus2
minus15
minus1
minus05
05
05
Figure 1 The solitary wave and behavior of the solutions 119906(119909 119905) =minus1 + 2 tanh(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Figure 2 The solitary wave and behavior of the solutions119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Let the coefficients of all derivatives of cn120585 be zero and wehave
1198860= minus
1205732
31205733
1198861= plusmn119896119898radic
2
1205733
1198962=
12057322minus 312057311205733
31205733(21198982 minus 1)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(37)
Therefore the solution of (3) is
1199063(119909 119905) = 119886
0+ 1198861cn119896 (119909 minus 120582119905) (38)
where 1198860 1198861and 119896 are denoted by (37) and 120582 is arbitrary
constantThe solutions of (7) are (38) and the following formula
1198673(119909 119905) = 119887
1+ 1205821199063(119909 119905) minus
1
21199062
3(119909 119905) (39)
Journal of Applied Mathematics 5
05
10
0
11
2
3
2
3
22
24
26
28
12
14
05
15
25
16
18
minus10
minus5
Figure 3 The solitary wave and behavior of the solutions 119906(119909 119905) =1 + 2 sech(119909 minus 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Taking 119898 rarr 1 cn120585 rarr sech120585 (38) can be rewritten asfollows
1199063(119909 119905) = 119886
0+ 1198861sech119896 (119909 minus 120582119905) (40)
Assuming 120582 = 1 1205731= 05 120573
2= minus15 and 120573
3= 05 it can
be obtained that 1198860= 1 1198861= 2 and 119896 = 1 The solution of (3)
is 119906(119909 119905) = 1+2 sech(119909minus119905)The solitary wave and behavior ofthe solutions are shown in Figure 3 respectively for 0 le 119905 le 1
and minus10 le 119909 le 10
43 Third Kind of Jacobi Elliptic Function Method Let
119906 = 1198860+ 1198861dn120585 (41)
Thus
1199061015840= minus119898
21198861sn120585 119906
10158401015840= (2 minus 119898
2) 1198861dn120585 minus 2119886
1dn3120585
1199063= 1198863
0+ 31198862
01198861dn120585 + 3119886
01198862
1dn2120585 + 1198863
1dn3120585
(42)
Substituting them into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] dn120585
+ 1198862
1(1205732+ 312057331198860) dn2120585 + 119886
1(minus21198962+ 1198862
11205733) dn3120585 = 0
(43)
Letting the coefficients of all derivatives of dn120585 be zerowe have
1198860=minus1205732
31205733
1198861= plusmn119896radic
2
1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(44)
Therefore the solution of (3) is
1199064(119909 119905) = 119886
0+ 1198861dn119896 (119909 minus 120582119905) (45)
where 1198860 1198861 and 119896 are denoted by (44) and 120582 is arbitrary
constantThus the solutions of (7) are (45) and the following
formula
1198674(119909 119905) = 119887
1+ 1205821199064(119909 119905) minus
1
21199062
4(119909 119905) (46)
Letting119898 rarr 1 then (45) equals (38)
44 Jacobi Elliptic Function cs120585Method Let
119906 = 1198860+ 1198861cs120585 cs120585 equiv cn120585
sn120585 (47)
Then
1199061015840= minus1198861(1 + cs2120585) dn120585
11990610158401015840= (2 minus 119898
2) 1198861cs120585 + 2119886
1cs3120585
(48)
Substituting (47) and (48) into (3) we can obtain
119863 = 12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] cs120585
+ 1198862
1(1205732+ 312057331198860) cs2120585 + 119886
1(21198962+ 1198862
11205733) cs3120585
(49)
Let the coefficients in above formula be zero and we have
1198860=minus1205732
31205733
1198861= plusmn1198962radic
2
minus1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(50)
According to (47) the solution of (3) is
1199065(119909 119905) = 119886
0+ 1198861cs119896 (119909 minus 120582119905) (51)
where 1198860 1198861 and 119896 are denoted by (50) and 120582 is arbitrary
constantTherefore the solutions of (7) are (51) and the following
formula
1198675= 1198871+ 1205821199065(119909 119905) minus
1
21199062
5(119909 119905) (52)
When119898 rarr 1
1199065(119909 119905) = 119886
0+ 1198861csch119896 (119909 minus 120582119905) (53)
Assuming 1205731= (1 minus 1198991205962)119899 120573
2= minus(32)120596 120573
3= minus(12)
119873 = minus(1198731119899) and 119896 = 1 we can obtain (3) Choosing
6 Journal of Applied Mathematics
02
40
24
0
5
10
15
0
5
10
minus4 minus4
minus10
minus15
minus2
minus2
minus5
minus10
minus5
Figure 4The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 csch(119909 minus 23119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
02
4
02
4
0
10
20
30
40
50
0
5
10
15
20
25
30
35
40
45
minus4 minus4
minus10
minus2 minus2
Figure 5The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =minus13+43 csch2(119909minus23119910+119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
119899 = minus(23) 120596 = 1 then 1205731
= minus(52) 1205732
= minus(32)Substituting 120573
1 1205732 1205733into (50) it can be obtained that
1198860= minus1 119886
1= 2 and 119896 = 1 The solutions of (8) are
119906(119909 119910 119905) = minus1 + 2 csch(119909 minus (23)119910 + 119905) and ℎ(119909 119910 119905) =
minus(13) + (43)csch2(119909 minus (23)119910 + 119905) The solitary wave andbehavior of the solutions 119906(119909 119905) = minus1+ 2 csch(119909 minus (23)119910 + 119905)and ℎ(119909 119910 119905) = minus(13)+(43)csch2(119909minus(23)119910+119905) are shownin Figures 4 and 5 respectively for 119905 = 0 minus4 le 119909 le 4minus4 le 119910 le 4 and 119909 minus (23)119910 = 0
5 Fixed Point Method
In general fixed point theory can be divided into two typesThe first type is only used to discuss the existence of solutionThe second type is used not only to discuss the existence anduniqueness of solution but also to search the fixed point Weare more interested in the second type
Lemma 1 (see [18ndash20]) Suppose 119864 is a real Banach spacewhich has normal cone Consider 119906
0 V0isin 119864 119906
0lt V0 The
operator 119860 [1199060 V0] times [119906
0 V0] rarr 119864 is mixed monotone and
satisfies
(1)
1199060le 119860 (119906
0 V0) 119860 (V
0 1199060) le V0
(54)
(2) for all 119906 V (1199060le 119906 le V le V
0) exist constant 120572 isin (0 1)
satisfies
119860 (V 119906) minus 119860 (119906 V) le 120572 V minus 119906 (55)
Then there exists a unique 119906 (119906 isin [1199060 V0]) which satisfies
119860 (119906 119906) = 119906 (56)
where 119906 is called the fixed point of 119860In [1199060 V0] for all 119908
0 letting 119908
119899= 119860(119908
119899minus1 119908119899minus1
) 119899 =
1 2 we have
119906 = lim119899rarrinfin
119908119899 (57)
In this paper 119864 is taken as a continuous functionspace 119862[0 1] which is defined in the closed interval [0 1]Introducing equivalent norm it is easy to know that 119862[0 1]is real Banach space which has normal cone Let
1199060= 0 V
0= 1 (58)
Integrating (3) with respect to 120585 we can obtain
119889119906
119889120585= minus119896minus2int120585
0
[1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889120585 + 119873120585 + 119862
1
(59)
where 1198621is integral constant
Integrating the above formula with respect to 120585 and usingthe formula in [12] we have
119906 = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(60)
where 1198622is integral constant
If 1205732gt 0 120573
3gt 0 (ie 1205822 gt 119888
1) and 120573
1lt 0 (ie 1205822 lt 1198882
0)
suppose
119860 (119906 V) = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
2V2 (119904) + 120573
3V3 (119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(61)
Theorem 2 In (61) supposing 1205732gt 0 120573
3gt 0 120573
1lt 0 119862
1ge 0
1198622ge 0 and 119896minus2 le 119873(120573
2+1205733) le 2(1+119862
1+1198622)(1205732+1205733minus1205731)
there exists unique fixed point 119906 isin [1199060 V0] for the operator 119860
which is defined by (61) Then (57) is true
Journal of Applied Mathematics 7
Proof Obviously 119860 [1199060 V0] times [119906
0 V0] rarr 119862[0 1] is mixed
monotone Consider
119860 (1199060 V0) = 119860 (0 1)
=1
2[119873 minus 119896
minus2(1205732+ 1205733)] 1205852+ 1198621120585 + 1198622ge 0 = 119906
0
119860 (V0 1199060) =
1
2(119873 minus 120573
1119896minus2) 1205852+ 1198621120585 + 1198622
le1
2119873(1 minus
1205731
1205732+ 1205733
) + 1198621+ 1198622le 1 = V
0
(62)
Thus 119860 satisfies the term (1) of Lemma 1For all 119906 V isin [119906
0 V0] 119906 le V
119860 (V 119906) minus 119860 (119906 V)
= minus119896minus2int120585
0
(120585 minus 119904)
times [1205731V (119904) + 120573
21199062(119904) + 120573
21199063(119904)
minus1205731119906 (119904) minus 120573
2V2 (119904) minus 120573
3V3 (119904)] 119889119904
= 119896minus2int120585
0
(120585 minus 119904) 119890119872119904119890minus119872119904
[V (119904) minus 119906 (119904)]
times minus 1205731+ 1205732[V (119904) + 119906 (119904)]
+1205733[V2 (119904) + 119906 (119904) V (119904) + 1199062 (119904)] 119889119904
le (minus1205731+ 21205732+ 31205733) 119896minus2V minus 119906int
120585
0
119890119872119878119889119904
leminus1205731+ 21205732+ 31205733
1198962119872V minus 119906 119890119872120585
(63)
Then we have
119860 (V 119906) minus 119860 (119906 V) le 120572 (64)
where 120572 = (minus1205731+ 21205732+ 31205733)(1198962119872) and when 119872 is large
enough 120572 isin (0 1)Therefore operator119860which is defined by (61) satisfies the
term (2) of Lemma 1 The proof is completed
Obviously the fixed point of the operator defined by (61)is equivalent to the solution of (3) We denote
1199066(119909 119905) = 119906 (120585) (65)
By using the classical iteration algorithms such as Manniteration method Ishikawa iteration method and Nooriteration method the solutions can be obtained Then thetraveling wave solutions of (5) are (65) and the followingformula
1198676(119909 119905) = 119887
1+ 1205821199066(119909 119905) minus
1
21199062
6(119909 119905) (66)
6 Modified Mapping Method
The modified mapping method was introduced and thenonlinear evolution equation was solved in [21 22] In[22] several hundreds of Jacobi elliptic function expansionsolutions were obtained We also use this method in thispaper
Let
119906 = 1198600+ 1198601119891 + 119861
1119891minus1 (67)
where 11989110158401015840 = 119901119891 + 1199021198913 hArr 11989110158402 = 119903 + 1199011198912 + (12)1199021198914 119903 119901 and119902 are constants
Thus
1199061015840= 11986011198911015840minus 1198611119891minus21198911015840
11990610158401015840= 1198601119901119891 + 119860
11199021198913+ 1198611119901119891minus1+ 21198611119903119891minus3
1199062= 1198602
0+ 211986011198611+ 211986001198601119891 + 2119860
01198611119891minus1+ 1198602
11198912+ 1198612
1119891minus2
1199063= 1198603
0+ 6119860011986011198611+ 31198601(1198602
0+ 11986011198611) 119891
+ 311986001198602
11198912+ 1198603
11198913+ 31198611(1198602
0+ 11986011198611) 119891minus1
+ 311986001198612
1119891minus2+ 1198613
1119891minus3
(68)
Substituting (68) into (3) we have
12057311198600+ 1205732(1198602
0+ 211986011198611) + 1205733(1198603
0+ 6119860011986011198611) minus 119873
+ 1198601[1198962119901 + 1205731+ 12057321198600+ 31205733(1198602
0+ 11986011198611)] 119891
+ 1198602
1(1205732+ 312057331198600) 1198912+ 1198601(1198962119902 + 12057331198602
1) 1198913
+ 1198611[1198962119901 + 1205731+ 212057321198600+ 31205733(1198602
0+ 11986011198611)] 119891minus1
+ 1198612
1(1205732+ 312057331198600) 119891minus2+ 1198611(21198962119903 + 12057331198612
1) 119891minus3= 0
(69)
Notice that we get the same algebraic equations when thecoefficients of 119891 119891minus1 and 1198912 119891minus2 in (69) are zero Thereforethere are only five algebra equations Solve them and we have
1198600=minus1205732
31205733
1198601= plusmn119896radic
minus119902
1205733
1198611= plusmn119896radic
minus2119903
1205733
119896 = plusmnradic12057322minus 312057311205733minus 31198962120573
3radic2119902119903
31205733119901
(70)
In order to make 1198601 1198611be real number 119902 and 119903 should
be different signs with 1205733
(i) Choosing 119903 = 1119901 = minus2 and 119902 = 2 we have119891 = tanh 120585Then the solitary wave solutions of (3) are
1199067(119909 119905) = 119860
0+ 1198601tanh 120585 + 119861
1tanhminus1120585 (71)
8 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Journal of Applied Mathematics
The basic idea of the fixed point methods consists infinding an iteration function which generates successiveapproximations to the solution [12]
Many periodic solutions have been recently expressed interms of various Jacobi-elliptic functions for a wide class ofnonlinear evolution equations which have been obtained bymeans of Jacobi elliptic expansion method [13]
The longitudinal wave equation of a finite deformationelastic circular rod Boussinesq equations and dispersive longwave equations are nonlinear partial differential equationsof different scientific field We find that they have the samecharacteristics In this paper the analytical solutions of thedifferential equations for the elastic circular rod Boussinesqequations and dispersive long wave equations are solved byusing homogeneous balance method Jacobi elliptic functionmethod fixed point method andmodifiedmappingmethodThemore physical specifications of these nonlinear equationshave been identified and the results indicated
2 Three Types of Nonlinear Wave Equations
The longitudinal wave equation of a finite deformation elasticcircular rod is [1]
119906119905119905minus 1198882
0119906119909119909= [
3119864
21205881199062+119864
21205881199063+]21198772
2(119906119905119905minus 1198882
1119906119909119909)]119909119909
(1)
where 1198880= radic119864120588 is the longitudinal wave velocity for a linear
elastic rod and 1198881= radic120583120588 is the shear wave velocity 120588 is the
density of thematerial ] is the Poisson ratio119877 is the diameterof bar 119864 is Yongrsquos modulus of material 120583 is the elastic shearmodulus of material 119905 is the time variable
Make the traveling wave transformation
119906 (119909 119905) = 119906 (120585) 120585 = 119896 (119909 minus 120582119905) (2)
Substituting (2) into (1) and integrating it with respect to120585 twice we can obtain
119896211990610158401015840+ 1205731119906 + 12057321199062+ 12057331199063= 119873 (3)
where 119873 is the second integral constant The first integralconstant is zero and
1205731=
2 (1205822 minus 11988820)
]21198772 (1205822 minus 11988821) 120573
2=
311988820
]21198772 (1205822 minus 11988821)
1205733=
11988820
]21198772 (1205822 minus 11988821)
(4)
The variant Boussinesq equations were discussed in [7]They are coupling wave equations which can be expressed asfollows
119867119905+ (119867119906)
119909+ 119906119909119909119909
= 0
119906119905+ 119867119909+ 119906119906119909= 0
(5)
where 119906(119909 119905) is the velocity and 119867(119909 119905) is the total depthMake the traveling wave transformation
119867(119909 119905) = 119867 (120585) 119906 (119909 119905) = 119906 (120585) 120585 = 119896 (119909 minus 120582119905)
(6)
Substituting (6) into (5) and integrating with respect to 120585we have
minus120582119867 +119867119906 + 119896211990610158401015840= 1198861 119867 = 120582119906 minus
1
21199062+ 1198871 (7)
where 1198861 1198871are integral constants
Substituting the second formula of (7) into the first oneand assuming120573
1= 1198871minus1205822120573
2= (32)120582120573
3= minus(12)119873 = 119886
1+
1205821198871 we can obtain (3) So the solutions of (7) are equivalent
to that of (3) and the second formula of (7)(2 + 1)-dimension dispersive long wave equations were
discussed in [8 9] They can be written as follows
119906119905119910+ ℎ119909119909+1
2(1199062)119909119910= 0
ℎ119905+ (ℎ119906 + 119906 + 119906
119909119910)119909= 0
(8)
Suppose that the traveling wave solutions for (8) can begiven in the following forms
119906 (119909 119910 119905) = 119906 (120585) ℎ (119909 119910 119905) = ℎ (120585)
120585 = 119909 + 119899119910 + 120596119905(9)
Substituting (9) into (8) we have
11989911990610158401015840+ (1 minus 119899120596
2) 119906 minus
3
21198991205961199062minus119899
21199063= 1198731
ℎ = minus119899120596119906 minus1
21198991199062
(10)
where1198731is the second integral constant
Obviously the first formula of (10) and (3) have the samecharacteristics On the other hand the solutions of (10) and(7) are samewith each other except for the coefficientsThere-fore in following parts the solutions of (5) are presented
3 Homogeneous Balance Method
In [14 15] the homogeneous balance method is used tosolve nonlinear wave equations For example the followingequation was solved in [15]
119906119905+ 6119906119906
119909+ 119906119909119909119909
= 0 (11)
The exact solution of the following corresponding equa-tion was given by using the same method
119906119905minus 119863119906119909119909+ 120582 (119906
3+ 1205721199062+ 120573119906) = 0 (12)
To solve (5) the homogeneous balance method wasimproved in this paper
Journal of Applied Mathematics 3
Now suppose that
119906 (119909 119905) = 1198911015840120596119909+ 119887 119867 (119909 119905) = 119892
10158401205962
119909+ 119888 (13)
where119891(120596)119892(120596) and120596(119909 119905) are the undetermined functionsand 119887 119888 are undetermined constants
Substituting (13) into (5) the equations can be rewrittenas follows
(119891101584010158401198921015840+ 119891101584011989210158401015840+ 119891(4)) 1205964
119909+ 61198911015840101584010158401205962
1199091205962119909
+ 11989110158401015840(1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909) + 11989210158401015840(1198871205963
119909+ 1205962
119909120596119905)
+ 3119891101584011989210158401205962
1199091205962119909+ 1198911015840(1198881205962119909+ 1205964119909)
+ 21198921015840(1198871205961199091205962119909+ 120596119909120596119909119905) = 0
(11989210158401015840+ 119891101584011989110158401015840) 1205963
119909+ 11989110158401015840(120596119909120596119905+ 1198871205962
119909) + 119891101584021205961199091205962119909
+ 1198911015840(120596119909119905+ 1198871205962119909) + 2119892
10158401205961199091205962119909= 0
(14)
where 1205962119909= 120596119909119909 1205963119909= 120596119909119909119909
1205964119909= 120596119909119909119909119909
In (14) let
119891101584010158401198921015840+ 119891101584011989210158401015840+ 119891(4)
= 0 11989210158401015840+ 119891101584011989110158401015840= 0 (15)
Integrating the second formula of (15) we have
1198921015840+1
211989110158402= 0 (16)
Furthermore we can obtain
1198921015840= minus
1
211989110158402 119892
10158401015840= minus119891101584011989110158401015840 (17)
Substituting (17) into the first formula of (15) we have
119891(4)minus3
21198911015840211989110158401015840= 0 (18)
Integrating (18) we have
119891101584010158401015840minus1
211989110158403= 0 (19)
We know the solution of (19) is
119891 = 2 ln120596 (20)
Obviously we have
11989110158402= minus2119891
10158401015840 119891
101584011989110158401015840= minus119891101584010158401015840
11989110158401198921015840= minus119891101584010158401015840 119891
10158403= 2119891101584010158401015840
(21)
Substituting (17) and (21) into (14) the equations can berewritten as follows
1205962
119909(31205962119909+ 119887120596119909+ 120596119905) 119891101584010158401015840
+ (1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909+ 2119887120596
1199091205962119909+ 2120596119909120596119909119905) 11989110158401015840
+ (1198881205962119909+ 1205964119909) 1198911015840= 0
120596119909(120596119905+ 119887120596119909) 11989110158401015840+ (120596119909119905+ 1198871205962119909) 1198911015840= 0
(22)
Letting all coefficients of 119891 in (22) be zero we have31205962119909+ 119887120596119909+ 120596119905= 0
1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909+ 2119887120596
1199091205962119909+ 2120596119909120596119909119905= 0
1198881205962119909+ 1205964119909= 0
120596119905+ 119887120596119909= 0 120596
119909119905+ 1198871205962119909= 0
(23)
From first and fifth second formula of (23) it can beobtained that
120596 = 120572 (119905) 119909 + 120573 (119905) (24)where 120572(119905) is undetermined function and 120573(119905) is arbitraryfunction
From second formula of (23) we have
120572 (119905) = 119890minus(1198882)119905+119896
(25)where 119896 is the integral constant
Therefore from (13) (20) (24) and (25) we know thesolutions of (5) are
1199061(119909 119905) =
2120572 (119905)
120572 (119905) 119909 + 120573 (119905)+ 119887
1198671(119909 119905) =
minus21205722(119905)
[120572 (119905) 119909 + 120573 (119905)]2+ 119888
(26)
4 Jacobi Elliptic Function Method
In [16] the NLS equation and Zakharov equation werestudied by using the Jacobi elliptic function method In thispaper (1) and (5) are discussed by using this method
41 Jacobi Elliptic Sine Function Method Let119906 = 1198860+ 1198861sn120585 (27)
We know119889119906
119889120585= 1198861cn120585dn120585
1198892119906
1198891205852= minus (1 + 119898
2) 1198861sn120585 + 21198982119886
1sn3120585
1199062= 1198862
0+ 211988601198861sn120585 + 1198862
1sn2120585
1199063= 1198863
0+ 31198862
01198861sn120585 + 3119886
01198862
1sn2120585 + 1198863
1sn3120585
(28)
where cn120585 and dn120585 are the Jacobi elliptic cosine functionand the third type Jacobi elliptic function respectively and119898 (0 lt 119898 lt 1) is module
Substituting (27) into (3) we have
12057311198860+ 12057321198862
0+ 12057331198863
0minus 119873
+ 1198861[minus1198962(1 + 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] sn120585
+ 1198862
1(1205732+ 312057331198860) sn2120585 + 119886
1(211989621198982+ 12057331198862
1) sn3120585 = 0
(29)
4 Journal of Applied Mathematics
Let the coefficients of all derivatives of sn120585 be zero and wehave
1198860=minus1205732
31205733
1198861= plusmn
radic2119896119898
radicminus1205733
1198962=
1
1 + 1198982(1205731+ 212057321198860+ 312057331198862
0)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(30)
Therefore the solution of (3) is
1199062(119909 119905) = 119886
0+ 1198861sn119896 (119909 minus 120582119905) (31)
where 1198860 1198861 and 119896 are denoted by (30) and 120582 is arbitrary
constantWhen119898 rarr 1 sn120585 rarr tanh 120585
1199062(119909 119905) = 119886
0+ 1198861tanh 119896 (119909 minus 120582119905) (32)
Thus the solutions of (5) are (32) and the followingformula
1198672(119909 119905) = 119887
1+ 1205821199062minus1
21199062
2 (33)
Assuming 120582 = minus1 1198871= 15 120573
1= 05 120573
2= minus15 and
1205733= minus05 it can be obtained that 119886
0= minus1 119886
1= 2 and 119896 =
1 The solutions of (5) are 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and119867(119909 119905) = 2 minus 2 tanh2(119909 + 119905) The solitary wave and behaviorof the solutions 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and 119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) are shown in Figures 1 and 2 respectivelyfor 0 le 119905 le 1 and minus10 le 119909 le 10 The waveform is similar tothe result in [17]
42 Jacobi Elliptic Cosine Function Method Let
119906 = 1198860+ 1198861cn120585 (34)
Thus
1199061015840= minus1198861sn120585dn120585 119906
10158401015840= (2119898
2minus 1) 119886
1cn120585 minus 21198982cn3120585
1199062= 1198862
0+ 211988601198861cn120585 + 1198862
1cn2120585
1199063= 1198863
0+ 31198862
01198861cn120585 + 3119886
01198862
1cn2120585 + 1198863
1cn3120585
(35)
Substituting (35) into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0minus 119873
+ 1198861[1198962(21198982minus 1) + 120573
1+ 212057321198860+ 312057331198862
0] cn120585
+ 1198862
1(1205732+ 312057331198860) cn2120585
+ 1198861(minus211989621198982+ 12057331198862
1) cn3120585 = 0
(36)
05
10
0
1
0
10
minus10
minus5
minus3
minus2
minus1
minus25
minus2
minus15
minus1
minus05
05
05
Figure 1 The solitary wave and behavior of the solutions 119906(119909 119905) =minus1 + 2 tanh(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Figure 2 The solitary wave and behavior of the solutions119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Let the coefficients of all derivatives of cn120585 be zero and wehave
1198860= minus
1205732
31205733
1198861= plusmn119896119898radic
2
1205733
1198962=
12057322minus 312057311205733
31205733(21198982 minus 1)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(37)
Therefore the solution of (3) is
1199063(119909 119905) = 119886
0+ 1198861cn119896 (119909 minus 120582119905) (38)
where 1198860 1198861and 119896 are denoted by (37) and 120582 is arbitrary
constantThe solutions of (7) are (38) and the following formula
1198673(119909 119905) = 119887
1+ 1205821199063(119909 119905) minus
1
21199062
3(119909 119905) (39)
Journal of Applied Mathematics 5
05
10
0
11
2
3
2
3
22
24
26
28
12
14
05
15
25
16
18
minus10
minus5
Figure 3 The solitary wave and behavior of the solutions 119906(119909 119905) =1 + 2 sech(119909 minus 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Taking 119898 rarr 1 cn120585 rarr sech120585 (38) can be rewritten asfollows
1199063(119909 119905) = 119886
0+ 1198861sech119896 (119909 minus 120582119905) (40)
Assuming 120582 = 1 1205731= 05 120573
2= minus15 and 120573
3= 05 it can
be obtained that 1198860= 1 1198861= 2 and 119896 = 1 The solution of (3)
is 119906(119909 119905) = 1+2 sech(119909minus119905)The solitary wave and behavior ofthe solutions are shown in Figure 3 respectively for 0 le 119905 le 1
and minus10 le 119909 le 10
43 Third Kind of Jacobi Elliptic Function Method Let
119906 = 1198860+ 1198861dn120585 (41)
Thus
1199061015840= minus119898
21198861sn120585 119906
10158401015840= (2 minus 119898
2) 1198861dn120585 minus 2119886
1dn3120585
1199063= 1198863
0+ 31198862
01198861dn120585 + 3119886
01198862
1dn2120585 + 1198863
1dn3120585
(42)
Substituting them into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] dn120585
+ 1198862
1(1205732+ 312057331198860) dn2120585 + 119886
1(minus21198962+ 1198862
11205733) dn3120585 = 0
(43)
Letting the coefficients of all derivatives of dn120585 be zerowe have
1198860=minus1205732
31205733
1198861= plusmn119896radic
2
1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(44)
Therefore the solution of (3) is
1199064(119909 119905) = 119886
0+ 1198861dn119896 (119909 minus 120582119905) (45)
where 1198860 1198861 and 119896 are denoted by (44) and 120582 is arbitrary
constantThus the solutions of (7) are (45) and the following
formula
1198674(119909 119905) = 119887
1+ 1205821199064(119909 119905) minus
1
21199062
4(119909 119905) (46)
Letting119898 rarr 1 then (45) equals (38)
44 Jacobi Elliptic Function cs120585Method Let
119906 = 1198860+ 1198861cs120585 cs120585 equiv cn120585
sn120585 (47)
Then
1199061015840= minus1198861(1 + cs2120585) dn120585
11990610158401015840= (2 minus 119898
2) 1198861cs120585 + 2119886
1cs3120585
(48)
Substituting (47) and (48) into (3) we can obtain
119863 = 12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] cs120585
+ 1198862
1(1205732+ 312057331198860) cs2120585 + 119886
1(21198962+ 1198862
11205733) cs3120585
(49)
Let the coefficients in above formula be zero and we have
1198860=minus1205732
31205733
1198861= plusmn1198962radic
2
minus1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(50)
According to (47) the solution of (3) is
1199065(119909 119905) = 119886
0+ 1198861cs119896 (119909 minus 120582119905) (51)
where 1198860 1198861 and 119896 are denoted by (50) and 120582 is arbitrary
constantTherefore the solutions of (7) are (51) and the following
formula
1198675= 1198871+ 1205821199065(119909 119905) minus
1
21199062
5(119909 119905) (52)
When119898 rarr 1
1199065(119909 119905) = 119886
0+ 1198861csch119896 (119909 minus 120582119905) (53)
Assuming 1205731= (1 minus 1198991205962)119899 120573
2= minus(32)120596 120573
3= minus(12)
119873 = minus(1198731119899) and 119896 = 1 we can obtain (3) Choosing
6 Journal of Applied Mathematics
02
40
24
0
5
10
15
0
5
10
minus4 minus4
minus10
minus15
minus2
minus2
minus5
minus10
minus5
Figure 4The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 csch(119909 minus 23119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
02
4
02
4
0
10
20
30
40
50
0
5
10
15
20
25
30
35
40
45
minus4 minus4
minus10
minus2 minus2
Figure 5The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =minus13+43 csch2(119909minus23119910+119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
119899 = minus(23) 120596 = 1 then 1205731
= minus(52) 1205732
= minus(32)Substituting 120573
1 1205732 1205733into (50) it can be obtained that
1198860= minus1 119886
1= 2 and 119896 = 1 The solutions of (8) are
119906(119909 119910 119905) = minus1 + 2 csch(119909 minus (23)119910 + 119905) and ℎ(119909 119910 119905) =
minus(13) + (43)csch2(119909 minus (23)119910 + 119905) The solitary wave andbehavior of the solutions 119906(119909 119905) = minus1+ 2 csch(119909 minus (23)119910 + 119905)and ℎ(119909 119910 119905) = minus(13)+(43)csch2(119909minus(23)119910+119905) are shownin Figures 4 and 5 respectively for 119905 = 0 minus4 le 119909 le 4minus4 le 119910 le 4 and 119909 minus (23)119910 = 0
5 Fixed Point Method
In general fixed point theory can be divided into two typesThe first type is only used to discuss the existence of solutionThe second type is used not only to discuss the existence anduniqueness of solution but also to search the fixed point Weare more interested in the second type
Lemma 1 (see [18ndash20]) Suppose 119864 is a real Banach spacewhich has normal cone Consider 119906
0 V0isin 119864 119906
0lt V0 The
operator 119860 [1199060 V0] times [119906
0 V0] rarr 119864 is mixed monotone and
satisfies
(1)
1199060le 119860 (119906
0 V0) 119860 (V
0 1199060) le V0
(54)
(2) for all 119906 V (1199060le 119906 le V le V
0) exist constant 120572 isin (0 1)
satisfies
119860 (V 119906) minus 119860 (119906 V) le 120572 V minus 119906 (55)
Then there exists a unique 119906 (119906 isin [1199060 V0]) which satisfies
119860 (119906 119906) = 119906 (56)
where 119906 is called the fixed point of 119860In [1199060 V0] for all 119908
0 letting 119908
119899= 119860(119908
119899minus1 119908119899minus1
) 119899 =
1 2 we have
119906 = lim119899rarrinfin
119908119899 (57)
In this paper 119864 is taken as a continuous functionspace 119862[0 1] which is defined in the closed interval [0 1]Introducing equivalent norm it is easy to know that 119862[0 1]is real Banach space which has normal cone Let
1199060= 0 V
0= 1 (58)
Integrating (3) with respect to 120585 we can obtain
119889119906
119889120585= minus119896minus2int120585
0
[1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889120585 + 119873120585 + 119862
1
(59)
where 1198621is integral constant
Integrating the above formula with respect to 120585 and usingthe formula in [12] we have
119906 = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(60)
where 1198622is integral constant
If 1205732gt 0 120573
3gt 0 (ie 1205822 gt 119888
1) and 120573
1lt 0 (ie 1205822 lt 1198882
0)
suppose
119860 (119906 V) = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
2V2 (119904) + 120573
3V3 (119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(61)
Theorem 2 In (61) supposing 1205732gt 0 120573
3gt 0 120573
1lt 0 119862
1ge 0
1198622ge 0 and 119896minus2 le 119873(120573
2+1205733) le 2(1+119862
1+1198622)(1205732+1205733minus1205731)
there exists unique fixed point 119906 isin [1199060 V0] for the operator 119860
which is defined by (61) Then (57) is true
Journal of Applied Mathematics 7
Proof Obviously 119860 [1199060 V0] times [119906
0 V0] rarr 119862[0 1] is mixed
monotone Consider
119860 (1199060 V0) = 119860 (0 1)
=1
2[119873 minus 119896
minus2(1205732+ 1205733)] 1205852+ 1198621120585 + 1198622ge 0 = 119906
0
119860 (V0 1199060) =
1
2(119873 minus 120573
1119896minus2) 1205852+ 1198621120585 + 1198622
le1
2119873(1 minus
1205731
1205732+ 1205733
) + 1198621+ 1198622le 1 = V
0
(62)
Thus 119860 satisfies the term (1) of Lemma 1For all 119906 V isin [119906
0 V0] 119906 le V
119860 (V 119906) minus 119860 (119906 V)
= minus119896minus2int120585
0
(120585 minus 119904)
times [1205731V (119904) + 120573
21199062(119904) + 120573
21199063(119904)
minus1205731119906 (119904) minus 120573
2V2 (119904) minus 120573
3V3 (119904)] 119889119904
= 119896minus2int120585
0
(120585 minus 119904) 119890119872119904119890minus119872119904
[V (119904) minus 119906 (119904)]
times minus 1205731+ 1205732[V (119904) + 119906 (119904)]
+1205733[V2 (119904) + 119906 (119904) V (119904) + 1199062 (119904)] 119889119904
le (minus1205731+ 21205732+ 31205733) 119896minus2V minus 119906int
120585
0
119890119872119878119889119904
leminus1205731+ 21205732+ 31205733
1198962119872V minus 119906 119890119872120585
(63)
Then we have
119860 (V 119906) minus 119860 (119906 V) le 120572 (64)
where 120572 = (minus1205731+ 21205732+ 31205733)(1198962119872) and when 119872 is large
enough 120572 isin (0 1)Therefore operator119860which is defined by (61) satisfies the
term (2) of Lemma 1 The proof is completed
Obviously the fixed point of the operator defined by (61)is equivalent to the solution of (3) We denote
1199066(119909 119905) = 119906 (120585) (65)
By using the classical iteration algorithms such as Manniteration method Ishikawa iteration method and Nooriteration method the solutions can be obtained Then thetraveling wave solutions of (5) are (65) and the followingformula
1198676(119909 119905) = 119887
1+ 1205821199066(119909 119905) minus
1
21199062
6(119909 119905) (66)
6 Modified Mapping Method
The modified mapping method was introduced and thenonlinear evolution equation was solved in [21 22] In[22] several hundreds of Jacobi elliptic function expansionsolutions were obtained We also use this method in thispaper
Let
119906 = 1198600+ 1198601119891 + 119861
1119891minus1 (67)
where 11989110158401015840 = 119901119891 + 1199021198913 hArr 11989110158402 = 119903 + 1199011198912 + (12)1199021198914 119903 119901 and119902 are constants
Thus
1199061015840= 11986011198911015840minus 1198611119891minus21198911015840
11990610158401015840= 1198601119901119891 + 119860
11199021198913+ 1198611119901119891minus1+ 21198611119903119891minus3
1199062= 1198602
0+ 211986011198611+ 211986001198601119891 + 2119860
01198611119891minus1+ 1198602
11198912+ 1198612
1119891minus2
1199063= 1198603
0+ 6119860011986011198611+ 31198601(1198602
0+ 11986011198611) 119891
+ 311986001198602
11198912+ 1198603
11198913+ 31198611(1198602
0+ 11986011198611) 119891minus1
+ 311986001198612
1119891minus2+ 1198613
1119891minus3
(68)
Substituting (68) into (3) we have
12057311198600+ 1205732(1198602
0+ 211986011198611) + 1205733(1198603
0+ 6119860011986011198611) minus 119873
+ 1198601[1198962119901 + 1205731+ 12057321198600+ 31205733(1198602
0+ 11986011198611)] 119891
+ 1198602
1(1205732+ 312057331198600) 1198912+ 1198601(1198962119902 + 12057331198602
1) 1198913
+ 1198611[1198962119901 + 1205731+ 212057321198600+ 31205733(1198602
0+ 11986011198611)] 119891minus1
+ 1198612
1(1205732+ 312057331198600) 119891minus2+ 1198611(21198962119903 + 12057331198612
1) 119891minus3= 0
(69)
Notice that we get the same algebraic equations when thecoefficients of 119891 119891minus1 and 1198912 119891minus2 in (69) are zero Thereforethere are only five algebra equations Solve them and we have
1198600=minus1205732
31205733
1198601= plusmn119896radic
minus119902
1205733
1198611= plusmn119896radic
minus2119903
1205733
119896 = plusmnradic12057322minus 312057311205733minus 31198962120573
3radic2119902119903
31205733119901
(70)
In order to make 1198601 1198611be real number 119902 and 119903 should
be different signs with 1205733
(i) Choosing 119903 = 1119901 = minus2 and 119902 = 2 we have119891 = tanh 120585Then the solitary wave solutions of (3) are
1199067(119909 119905) = 119860
0+ 1198601tanh 120585 + 119861
1tanhminus1120585 (71)
8 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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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
Journal of Applied Mathematics 3
Now suppose that
119906 (119909 119905) = 1198911015840120596119909+ 119887 119867 (119909 119905) = 119892
10158401205962
119909+ 119888 (13)
where119891(120596)119892(120596) and120596(119909 119905) are the undetermined functionsand 119887 119888 are undetermined constants
Substituting (13) into (5) the equations can be rewrittenas follows
(119891101584010158401198921015840+ 119891101584011989210158401015840+ 119891(4)) 1205964
119909+ 61198911015840101584010158401205962
1199091205962119909
+ 11989110158401015840(1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909) + 11989210158401015840(1198871205963
119909+ 1205962
119909120596119905)
+ 3119891101584011989210158401205962
1199091205962119909+ 1198911015840(1198881205962119909+ 1205964119909)
+ 21198921015840(1198871205961199091205962119909+ 120596119909120596119909119905) = 0
(11989210158401015840+ 119891101584011989110158401015840) 1205963
119909+ 11989110158401015840(120596119909120596119905+ 1198871205962
119909) + 119891101584021205961199091205962119909
+ 1198911015840(120596119909119905+ 1198871205962119909) + 2119892
10158401205961199091205962119909= 0
(14)
where 1205962119909= 120596119909119909 1205963119909= 120596119909119909119909
1205964119909= 120596119909119909119909119909
In (14) let
119891101584010158401198921015840+ 119891101584011989210158401015840+ 119891(4)
= 0 11989210158401015840+ 119891101584011989110158401015840= 0 (15)
Integrating the second formula of (15) we have
1198921015840+1
211989110158402= 0 (16)
Furthermore we can obtain
1198921015840= minus
1
211989110158402 119892
10158401015840= minus119891101584011989110158401015840 (17)
Substituting (17) into the first formula of (15) we have
119891(4)minus3
21198911015840211989110158401015840= 0 (18)
Integrating (18) we have
119891101584010158401015840minus1
211989110158403= 0 (19)
We know the solution of (19) is
119891 = 2 ln120596 (20)
Obviously we have
11989110158402= minus2119891
10158401015840 119891
101584011989110158401015840= minus119891101584010158401015840
11989110158401198921015840= minus119891101584010158401015840 119891
10158403= 2119891101584010158401015840
(21)
Substituting (17) and (21) into (14) the equations can berewritten as follows
1205962
119909(31205962119909+ 119887120596119909+ 120596119905) 119891101584010158401015840
+ (1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909+ 2119887120596
1199091205962119909+ 2120596119909120596119909119905) 11989110158401015840
+ (1198881205962119909+ 1205964119909) 1198911015840= 0
120596119909(120596119905+ 119887120596119909) 11989110158401015840+ (120596119909119905+ 1198871205962119909) 1198911015840= 0
(22)
Letting all coefficients of 119891 in (22) be zero we have31205962119909+ 119887120596119909+ 120596119905= 0
1198881205962
119909+ 31205962
2119909+ 41205961199091205963119909+ 2119887120596
1199091205962119909+ 2120596119909120596119909119905= 0
1198881205962119909+ 1205964119909= 0
120596119905+ 119887120596119909= 0 120596
119909119905+ 1198871205962119909= 0
(23)
From first and fifth second formula of (23) it can beobtained that
120596 = 120572 (119905) 119909 + 120573 (119905) (24)where 120572(119905) is undetermined function and 120573(119905) is arbitraryfunction
From second formula of (23) we have
120572 (119905) = 119890minus(1198882)119905+119896
(25)where 119896 is the integral constant
Therefore from (13) (20) (24) and (25) we know thesolutions of (5) are
1199061(119909 119905) =
2120572 (119905)
120572 (119905) 119909 + 120573 (119905)+ 119887
1198671(119909 119905) =
minus21205722(119905)
[120572 (119905) 119909 + 120573 (119905)]2+ 119888
(26)
4 Jacobi Elliptic Function Method
In [16] the NLS equation and Zakharov equation werestudied by using the Jacobi elliptic function method In thispaper (1) and (5) are discussed by using this method
41 Jacobi Elliptic Sine Function Method Let119906 = 1198860+ 1198861sn120585 (27)
We know119889119906
119889120585= 1198861cn120585dn120585
1198892119906
1198891205852= minus (1 + 119898
2) 1198861sn120585 + 21198982119886
1sn3120585
1199062= 1198862
0+ 211988601198861sn120585 + 1198862
1sn2120585
1199063= 1198863
0+ 31198862
01198861sn120585 + 3119886
01198862
1sn2120585 + 1198863
1sn3120585
(28)
where cn120585 and dn120585 are the Jacobi elliptic cosine functionand the third type Jacobi elliptic function respectively and119898 (0 lt 119898 lt 1) is module
Substituting (27) into (3) we have
12057311198860+ 12057321198862
0+ 12057331198863
0minus 119873
+ 1198861[minus1198962(1 + 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] sn120585
+ 1198862
1(1205732+ 312057331198860) sn2120585 + 119886
1(211989621198982+ 12057331198862
1) sn3120585 = 0
(29)
4 Journal of Applied Mathematics
Let the coefficients of all derivatives of sn120585 be zero and wehave
1198860=minus1205732
31205733
1198861= plusmn
radic2119896119898
radicminus1205733
1198962=
1
1 + 1198982(1205731+ 212057321198860+ 312057331198862
0)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(30)
Therefore the solution of (3) is
1199062(119909 119905) = 119886
0+ 1198861sn119896 (119909 minus 120582119905) (31)
where 1198860 1198861 and 119896 are denoted by (30) and 120582 is arbitrary
constantWhen119898 rarr 1 sn120585 rarr tanh 120585
1199062(119909 119905) = 119886
0+ 1198861tanh 119896 (119909 minus 120582119905) (32)
Thus the solutions of (5) are (32) and the followingformula
1198672(119909 119905) = 119887
1+ 1205821199062minus1
21199062
2 (33)
Assuming 120582 = minus1 1198871= 15 120573
1= 05 120573
2= minus15 and
1205733= minus05 it can be obtained that 119886
0= minus1 119886
1= 2 and 119896 =
1 The solutions of (5) are 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and119867(119909 119905) = 2 minus 2 tanh2(119909 + 119905) The solitary wave and behaviorof the solutions 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and 119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) are shown in Figures 1 and 2 respectivelyfor 0 le 119905 le 1 and minus10 le 119909 le 10 The waveform is similar tothe result in [17]
42 Jacobi Elliptic Cosine Function Method Let
119906 = 1198860+ 1198861cn120585 (34)
Thus
1199061015840= minus1198861sn120585dn120585 119906
10158401015840= (2119898
2minus 1) 119886
1cn120585 minus 21198982cn3120585
1199062= 1198862
0+ 211988601198861cn120585 + 1198862
1cn2120585
1199063= 1198863
0+ 31198862
01198861cn120585 + 3119886
01198862
1cn2120585 + 1198863
1cn3120585
(35)
Substituting (35) into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0minus 119873
+ 1198861[1198962(21198982minus 1) + 120573
1+ 212057321198860+ 312057331198862
0] cn120585
+ 1198862
1(1205732+ 312057331198860) cn2120585
+ 1198861(minus211989621198982+ 12057331198862
1) cn3120585 = 0
(36)
05
10
0
1
0
10
minus10
minus5
minus3
minus2
minus1
minus25
minus2
minus15
minus1
minus05
05
05
Figure 1 The solitary wave and behavior of the solutions 119906(119909 119905) =minus1 + 2 tanh(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Figure 2 The solitary wave and behavior of the solutions119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Let the coefficients of all derivatives of cn120585 be zero and wehave
1198860= minus
1205732
31205733
1198861= plusmn119896119898radic
2
1205733
1198962=
12057322minus 312057311205733
31205733(21198982 minus 1)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(37)
Therefore the solution of (3) is
1199063(119909 119905) = 119886
0+ 1198861cn119896 (119909 minus 120582119905) (38)
where 1198860 1198861and 119896 are denoted by (37) and 120582 is arbitrary
constantThe solutions of (7) are (38) and the following formula
1198673(119909 119905) = 119887
1+ 1205821199063(119909 119905) minus
1
21199062
3(119909 119905) (39)
Journal of Applied Mathematics 5
05
10
0
11
2
3
2
3
22
24
26
28
12
14
05
15
25
16
18
minus10
minus5
Figure 3 The solitary wave and behavior of the solutions 119906(119909 119905) =1 + 2 sech(119909 minus 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Taking 119898 rarr 1 cn120585 rarr sech120585 (38) can be rewritten asfollows
1199063(119909 119905) = 119886
0+ 1198861sech119896 (119909 minus 120582119905) (40)
Assuming 120582 = 1 1205731= 05 120573
2= minus15 and 120573
3= 05 it can
be obtained that 1198860= 1 1198861= 2 and 119896 = 1 The solution of (3)
is 119906(119909 119905) = 1+2 sech(119909minus119905)The solitary wave and behavior ofthe solutions are shown in Figure 3 respectively for 0 le 119905 le 1
and minus10 le 119909 le 10
43 Third Kind of Jacobi Elliptic Function Method Let
119906 = 1198860+ 1198861dn120585 (41)
Thus
1199061015840= minus119898
21198861sn120585 119906
10158401015840= (2 minus 119898
2) 1198861dn120585 minus 2119886
1dn3120585
1199063= 1198863
0+ 31198862
01198861dn120585 + 3119886
01198862
1dn2120585 + 1198863
1dn3120585
(42)
Substituting them into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] dn120585
+ 1198862
1(1205732+ 312057331198860) dn2120585 + 119886
1(minus21198962+ 1198862
11205733) dn3120585 = 0
(43)
Letting the coefficients of all derivatives of dn120585 be zerowe have
1198860=minus1205732
31205733
1198861= plusmn119896radic
2
1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(44)
Therefore the solution of (3) is
1199064(119909 119905) = 119886
0+ 1198861dn119896 (119909 minus 120582119905) (45)
where 1198860 1198861 and 119896 are denoted by (44) and 120582 is arbitrary
constantThus the solutions of (7) are (45) and the following
formula
1198674(119909 119905) = 119887
1+ 1205821199064(119909 119905) minus
1
21199062
4(119909 119905) (46)
Letting119898 rarr 1 then (45) equals (38)
44 Jacobi Elliptic Function cs120585Method Let
119906 = 1198860+ 1198861cs120585 cs120585 equiv cn120585
sn120585 (47)
Then
1199061015840= minus1198861(1 + cs2120585) dn120585
11990610158401015840= (2 minus 119898
2) 1198861cs120585 + 2119886
1cs3120585
(48)
Substituting (47) and (48) into (3) we can obtain
119863 = 12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] cs120585
+ 1198862
1(1205732+ 312057331198860) cs2120585 + 119886
1(21198962+ 1198862
11205733) cs3120585
(49)
Let the coefficients in above formula be zero and we have
1198860=minus1205732
31205733
1198861= plusmn1198962radic
2
minus1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(50)
According to (47) the solution of (3) is
1199065(119909 119905) = 119886
0+ 1198861cs119896 (119909 minus 120582119905) (51)
where 1198860 1198861 and 119896 are denoted by (50) and 120582 is arbitrary
constantTherefore the solutions of (7) are (51) and the following
formula
1198675= 1198871+ 1205821199065(119909 119905) minus
1
21199062
5(119909 119905) (52)
When119898 rarr 1
1199065(119909 119905) = 119886
0+ 1198861csch119896 (119909 minus 120582119905) (53)
Assuming 1205731= (1 minus 1198991205962)119899 120573
2= minus(32)120596 120573
3= minus(12)
119873 = minus(1198731119899) and 119896 = 1 we can obtain (3) Choosing
6 Journal of Applied Mathematics
02
40
24
0
5
10
15
0
5
10
minus4 minus4
minus10
minus15
minus2
minus2
minus5
minus10
minus5
Figure 4The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 csch(119909 minus 23119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
02
4
02
4
0
10
20
30
40
50
0
5
10
15
20
25
30
35
40
45
minus4 minus4
minus10
minus2 minus2
Figure 5The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =minus13+43 csch2(119909minus23119910+119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
119899 = minus(23) 120596 = 1 then 1205731
= minus(52) 1205732
= minus(32)Substituting 120573
1 1205732 1205733into (50) it can be obtained that
1198860= minus1 119886
1= 2 and 119896 = 1 The solutions of (8) are
119906(119909 119910 119905) = minus1 + 2 csch(119909 minus (23)119910 + 119905) and ℎ(119909 119910 119905) =
minus(13) + (43)csch2(119909 minus (23)119910 + 119905) The solitary wave andbehavior of the solutions 119906(119909 119905) = minus1+ 2 csch(119909 minus (23)119910 + 119905)and ℎ(119909 119910 119905) = minus(13)+(43)csch2(119909minus(23)119910+119905) are shownin Figures 4 and 5 respectively for 119905 = 0 minus4 le 119909 le 4minus4 le 119910 le 4 and 119909 minus (23)119910 = 0
5 Fixed Point Method
In general fixed point theory can be divided into two typesThe first type is only used to discuss the existence of solutionThe second type is used not only to discuss the existence anduniqueness of solution but also to search the fixed point Weare more interested in the second type
Lemma 1 (see [18ndash20]) Suppose 119864 is a real Banach spacewhich has normal cone Consider 119906
0 V0isin 119864 119906
0lt V0 The
operator 119860 [1199060 V0] times [119906
0 V0] rarr 119864 is mixed monotone and
satisfies
(1)
1199060le 119860 (119906
0 V0) 119860 (V
0 1199060) le V0
(54)
(2) for all 119906 V (1199060le 119906 le V le V
0) exist constant 120572 isin (0 1)
satisfies
119860 (V 119906) minus 119860 (119906 V) le 120572 V minus 119906 (55)
Then there exists a unique 119906 (119906 isin [1199060 V0]) which satisfies
119860 (119906 119906) = 119906 (56)
where 119906 is called the fixed point of 119860In [1199060 V0] for all 119908
0 letting 119908
119899= 119860(119908
119899minus1 119908119899minus1
) 119899 =
1 2 we have
119906 = lim119899rarrinfin
119908119899 (57)
In this paper 119864 is taken as a continuous functionspace 119862[0 1] which is defined in the closed interval [0 1]Introducing equivalent norm it is easy to know that 119862[0 1]is real Banach space which has normal cone Let
1199060= 0 V
0= 1 (58)
Integrating (3) with respect to 120585 we can obtain
119889119906
119889120585= minus119896minus2int120585
0
[1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889120585 + 119873120585 + 119862
1
(59)
where 1198621is integral constant
Integrating the above formula with respect to 120585 and usingthe formula in [12] we have
119906 = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(60)
where 1198622is integral constant
If 1205732gt 0 120573
3gt 0 (ie 1205822 gt 119888
1) and 120573
1lt 0 (ie 1205822 lt 1198882
0)
suppose
119860 (119906 V) = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
2V2 (119904) + 120573
3V3 (119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(61)
Theorem 2 In (61) supposing 1205732gt 0 120573
3gt 0 120573
1lt 0 119862
1ge 0
1198622ge 0 and 119896minus2 le 119873(120573
2+1205733) le 2(1+119862
1+1198622)(1205732+1205733minus1205731)
there exists unique fixed point 119906 isin [1199060 V0] for the operator 119860
which is defined by (61) Then (57) is true
Journal of Applied Mathematics 7
Proof Obviously 119860 [1199060 V0] times [119906
0 V0] rarr 119862[0 1] is mixed
monotone Consider
119860 (1199060 V0) = 119860 (0 1)
=1
2[119873 minus 119896
minus2(1205732+ 1205733)] 1205852+ 1198621120585 + 1198622ge 0 = 119906
0
119860 (V0 1199060) =
1
2(119873 minus 120573
1119896minus2) 1205852+ 1198621120585 + 1198622
le1
2119873(1 minus
1205731
1205732+ 1205733
) + 1198621+ 1198622le 1 = V
0
(62)
Thus 119860 satisfies the term (1) of Lemma 1For all 119906 V isin [119906
0 V0] 119906 le V
119860 (V 119906) minus 119860 (119906 V)
= minus119896minus2int120585
0
(120585 minus 119904)
times [1205731V (119904) + 120573
21199062(119904) + 120573
21199063(119904)
minus1205731119906 (119904) minus 120573
2V2 (119904) minus 120573
3V3 (119904)] 119889119904
= 119896minus2int120585
0
(120585 minus 119904) 119890119872119904119890minus119872119904
[V (119904) minus 119906 (119904)]
times minus 1205731+ 1205732[V (119904) + 119906 (119904)]
+1205733[V2 (119904) + 119906 (119904) V (119904) + 1199062 (119904)] 119889119904
le (minus1205731+ 21205732+ 31205733) 119896minus2V minus 119906int
120585
0
119890119872119878119889119904
leminus1205731+ 21205732+ 31205733
1198962119872V minus 119906 119890119872120585
(63)
Then we have
119860 (V 119906) minus 119860 (119906 V) le 120572 (64)
where 120572 = (minus1205731+ 21205732+ 31205733)(1198962119872) and when 119872 is large
enough 120572 isin (0 1)Therefore operator119860which is defined by (61) satisfies the
term (2) of Lemma 1 The proof is completed
Obviously the fixed point of the operator defined by (61)is equivalent to the solution of (3) We denote
1199066(119909 119905) = 119906 (120585) (65)
By using the classical iteration algorithms such as Manniteration method Ishikawa iteration method and Nooriteration method the solutions can be obtained Then thetraveling wave solutions of (5) are (65) and the followingformula
1198676(119909 119905) = 119887
1+ 1205821199066(119909 119905) minus
1
21199062
6(119909 119905) (66)
6 Modified Mapping Method
The modified mapping method was introduced and thenonlinear evolution equation was solved in [21 22] In[22] several hundreds of Jacobi elliptic function expansionsolutions were obtained We also use this method in thispaper
Let
119906 = 1198600+ 1198601119891 + 119861
1119891minus1 (67)
where 11989110158401015840 = 119901119891 + 1199021198913 hArr 11989110158402 = 119903 + 1199011198912 + (12)1199021198914 119903 119901 and119902 are constants
Thus
1199061015840= 11986011198911015840minus 1198611119891minus21198911015840
11990610158401015840= 1198601119901119891 + 119860
11199021198913+ 1198611119901119891minus1+ 21198611119903119891minus3
1199062= 1198602
0+ 211986011198611+ 211986001198601119891 + 2119860
01198611119891minus1+ 1198602
11198912+ 1198612
1119891minus2
1199063= 1198603
0+ 6119860011986011198611+ 31198601(1198602
0+ 11986011198611) 119891
+ 311986001198602
11198912+ 1198603
11198913+ 31198611(1198602
0+ 11986011198611) 119891minus1
+ 311986001198612
1119891minus2+ 1198613
1119891minus3
(68)
Substituting (68) into (3) we have
12057311198600+ 1205732(1198602
0+ 211986011198611) + 1205733(1198603
0+ 6119860011986011198611) minus 119873
+ 1198601[1198962119901 + 1205731+ 12057321198600+ 31205733(1198602
0+ 11986011198611)] 119891
+ 1198602
1(1205732+ 312057331198600) 1198912+ 1198601(1198962119902 + 12057331198602
1) 1198913
+ 1198611[1198962119901 + 1205731+ 212057321198600+ 31205733(1198602
0+ 11986011198611)] 119891minus1
+ 1198612
1(1205732+ 312057331198600) 119891minus2+ 1198611(21198962119903 + 12057331198612
1) 119891minus3= 0
(69)
Notice that we get the same algebraic equations when thecoefficients of 119891 119891minus1 and 1198912 119891minus2 in (69) are zero Thereforethere are only five algebra equations Solve them and we have
1198600=minus1205732
31205733
1198601= plusmn119896radic
minus119902
1205733
1198611= plusmn119896radic
minus2119903
1205733
119896 = plusmnradic12057322minus 312057311205733minus 31198962120573
3radic2119902119903
31205733119901
(70)
In order to make 1198601 1198611be real number 119902 and 119903 should
be different signs with 1205733
(i) Choosing 119903 = 1119901 = minus2 and 119902 = 2 we have119891 = tanh 120585Then the solitary wave solutions of (3) are
1199067(119909 119905) = 119860
0+ 1198601tanh 120585 + 119861
1tanhminus1120585 (71)
8 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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OptimizationJournal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Journal of Applied Mathematics
Let the coefficients of all derivatives of sn120585 be zero and wehave
1198860=minus1205732
31205733
1198861= plusmn
radic2119896119898
radicminus1205733
1198962=
1
1 + 1198982(1205731+ 212057321198860+ 312057331198862
0)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(30)
Therefore the solution of (3) is
1199062(119909 119905) = 119886
0+ 1198861sn119896 (119909 minus 120582119905) (31)
where 1198860 1198861 and 119896 are denoted by (30) and 120582 is arbitrary
constantWhen119898 rarr 1 sn120585 rarr tanh 120585
1199062(119909 119905) = 119886
0+ 1198861tanh 119896 (119909 minus 120582119905) (32)
Thus the solutions of (5) are (32) and the followingformula
1198672(119909 119905) = 119887
1+ 1205821199062minus1
21199062
2 (33)
Assuming 120582 = minus1 1198871= 15 120573
1= 05 120573
2= minus15 and
1205733= minus05 it can be obtained that 119886
0= minus1 119886
1= 2 and 119896 =
1 The solutions of (5) are 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and119867(119909 119905) = 2 minus 2 tanh2(119909 + 119905) The solitary wave and behaviorof the solutions 119906(119909 119905) = minus1 + 2 tanh(119909 + 119905) and 119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) are shown in Figures 1 and 2 respectivelyfor 0 le 119905 le 1 and minus10 le 119909 le 10 The waveform is similar tothe result in [17]
42 Jacobi Elliptic Cosine Function Method Let
119906 = 1198860+ 1198861cn120585 (34)
Thus
1199061015840= minus1198861sn120585dn120585 119906
10158401015840= (2119898
2minus 1) 119886
1cn120585 minus 21198982cn3120585
1199062= 1198862
0+ 211988601198861cn120585 + 1198862
1cn2120585
1199063= 1198863
0+ 31198862
01198861cn120585 + 3119886
01198862
1cn2120585 + 1198863
1cn3120585
(35)
Substituting (35) into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0minus 119873
+ 1198861[1198962(21198982minus 1) + 120573
1+ 212057321198860+ 312057331198862
0] cn120585
+ 1198862
1(1205732+ 312057331198860) cn2120585
+ 1198861(minus211989621198982+ 12057331198862
1) cn3120585 = 0
(36)
05
10
0
1
0
10
minus10
minus5
minus3
minus2
minus1
minus25
minus2
minus15
minus1
minus05
05
05
Figure 1 The solitary wave and behavior of the solutions 119906(119909 119905) =minus1 + 2 tanh(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Figure 2 The solitary wave and behavior of the solutions119867(119909 119905) =2 minus 2 tanh2(119909 + 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Let the coefficients of all derivatives of cn120585 be zero and wehave
1198860= minus
1205732
31205733
1198861= plusmn119896119898radic
2
1205733
1198962=
12057322minus 312057311205733
31205733(21198982 minus 1)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(37)
Therefore the solution of (3) is
1199063(119909 119905) = 119886
0+ 1198861cn119896 (119909 minus 120582119905) (38)
where 1198860 1198861and 119896 are denoted by (37) and 120582 is arbitrary
constantThe solutions of (7) are (38) and the following formula
1198673(119909 119905) = 119887
1+ 1205821199063(119909 119905) minus
1
21199062
3(119909 119905) (39)
Journal of Applied Mathematics 5
05
10
0
11
2
3
2
3
22
24
26
28
12
14
05
15
25
16
18
minus10
minus5
Figure 3 The solitary wave and behavior of the solutions 119906(119909 119905) =1 + 2 sech(119909 minus 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Taking 119898 rarr 1 cn120585 rarr sech120585 (38) can be rewritten asfollows
1199063(119909 119905) = 119886
0+ 1198861sech119896 (119909 minus 120582119905) (40)
Assuming 120582 = 1 1205731= 05 120573
2= minus15 and 120573
3= 05 it can
be obtained that 1198860= 1 1198861= 2 and 119896 = 1 The solution of (3)
is 119906(119909 119905) = 1+2 sech(119909minus119905)The solitary wave and behavior ofthe solutions are shown in Figure 3 respectively for 0 le 119905 le 1
and minus10 le 119909 le 10
43 Third Kind of Jacobi Elliptic Function Method Let
119906 = 1198860+ 1198861dn120585 (41)
Thus
1199061015840= minus119898
21198861sn120585 119906
10158401015840= (2 minus 119898
2) 1198861dn120585 minus 2119886
1dn3120585
1199063= 1198863
0+ 31198862
01198861dn120585 + 3119886
01198862
1dn2120585 + 1198863
1dn3120585
(42)
Substituting them into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] dn120585
+ 1198862
1(1205732+ 312057331198860) dn2120585 + 119886
1(minus21198962+ 1198862
11205733) dn3120585 = 0
(43)
Letting the coefficients of all derivatives of dn120585 be zerowe have
1198860=minus1205732
31205733
1198861= plusmn119896radic
2
1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(44)
Therefore the solution of (3) is
1199064(119909 119905) = 119886
0+ 1198861dn119896 (119909 minus 120582119905) (45)
where 1198860 1198861 and 119896 are denoted by (44) and 120582 is arbitrary
constantThus the solutions of (7) are (45) and the following
formula
1198674(119909 119905) = 119887
1+ 1205821199064(119909 119905) minus
1
21199062
4(119909 119905) (46)
Letting119898 rarr 1 then (45) equals (38)
44 Jacobi Elliptic Function cs120585Method Let
119906 = 1198860+ 1198861cs120585 cs120585 equiv cn120585
sn120585 (47)
Then
1199061015840= minus1198861(1 + cs2120585) dn120585
11990610158401015840= (2 minus 119898
2) 1198861cs120585 + 2119886
1cs3120585
(48)
Substituting (47) and (48) into (3) we can obtain
119863 = 12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] cs120585
+ 1198862
1(1205732+ 312057331198860) cs2120585 + 119886
1(21198962+ 1198862
11205733) cs3120585
(49)
Let the coefficients in above formula be zero and we have
1198860=minus1205732
31205733
1198861= plusmn1198962radic
2
minus1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(50)
According to (47) the solution of (3) is
1199065(119909 119905) = 119886
0+ 1198861cs119896 (119909 minus 120582119905) (51)
where 1198860 1198861 and 119896 are denoted by (50) and 120582 is arbitrary
constantTherefore the solutions of (7) are (51) and the following
formula
1198675= 1198871+ 1205821199065(119909 119905) minus
1
21199062
5(119909 119905) (52)
When119898 rarr 1
1199065(119909 119905) = 119886
0+ 1198861csch119896 (119909 minus 120582119905) (53)
Assuming 1205731= (1 minus 1198991205962)119899 120573
2= minus(32)120596 120573
3= minus(12)
119873 = minus(1198731119899) and 119896 = 1 we can obtain (3) Choosing
6 Journal of Applied Mathematics
02
40
24
0
5
10
15
0
5
10
minus4 minus4
minus10
minus15
minus2
minus2
minus5
minus10
minus5
Figure 4The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 csch(119909 minus 23119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
02
4
02
4
0
10
20
30
40
50
0
5
10
15
20
25
30
35
40
45
minus4 minus4
minus10
minus2 minus2
Figure 5The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =minus13+43 csch2(119909minus23119910+119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
119899 = minus(23) 120596 = 1 then 1205731
= minus(52) 1205732
= minus(32)Substituting 120573
1 1205732 1205733into (50) it can be obtained that
1198860= minus1 119886
1= 2 and 119896 = 1 The solutions of (8) are
119906(119909 119910 119905) = minus1 + 2 csch(119909 minus (23)119910 + 119905) and ℎ(119909 119910 119905) =
minus(13) + (43)csch2(119909 minus (23)119910 + 119905) The solitary wave andbehavior of the solutions 119906(119909 119905) = minus1+ 2 csch(119909 minus (23)119910 + 119905)and ℎ(119909 119910 119905) = minus(13)+(43)csch2(119909minus(23)119910+119905) are shownin Figures 4 and 5 respectively for 119905 = 0 minus4 le 119909 le 4minus4 le 119910 le 4 and 119909 minus (23)119910 = 0
5 Fixed Point Method
In general fixed point theory can be divided into two typesThe first type is only used to discuss the existence of solutionThe second type is used not only to discuss the existence anduniqueness of solution but also to search the fixed point Weare more interested in the second type
Lemma 1 (see [18ndash20]) Suppose 119864 is a real Banach spacewhich has normal cone Consider 119906
0 V0isin 119864 119906
0lt V0 The
operator 119860 [1199060 V0] times [119906
0 V0] rarr 119864 is mixed monotone and
satisfies
(1)
1199060le 119860 (119906
0 V0) 119860 (V
0 1199060) le V0
(54)
(2) for all 119906 V (1199060le 119906 le V le V
0) exist constant 120572 isin (0 1)
satisfies
119860 (V 119906) minus 119860 (119906 V) le 120572 V minus 119906 (55)
Then there exists a unique 119906 (119906 isin [1199060 V0]) which satisfies
119860 (119906 119906) = 119906 (56)
where 119906 is called the fixed point of 119860In [1199060 V0] for all 119908
0 letting 119908
119899= 119860(119908
119899minus1 119908119899minus1
) 119899 =
1 2 we have
119906 = lim119899rarrinfin
119908119899 (57)
In this paper 119864 is taken as a continuous functionspace 119862[0 1] which is defined in the closed interval [0 1]Introducing equivalent norm it is easy to know that 119862[0 1]is real Banach space which has normal cone Let
1199060= 0 V
0= 1 (58)
Integrating (3) with respect to 120585 we can obtain
119889119906
119889120585= minus119896minus2int120585
0
[1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889120585 + 119873120585 + 119862
1
(59)
where 1198621is integral constant
Integrating the above formula with respect to 120585 and usingthe formula in [12] we have
119906 = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(60)
where 1198622is integral constant
If 1205732gt 0 120573
3gt 0 (ie 1205822 gt 119888
1) and 120573
1lt 0 (ie 1205822 lt 1198882
0)
suppose
119860 (119906 V) = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
2V2 (119904) + 120573
3V3 (119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(61)
Theorem 2 In (61) supposing 1205732gt 0 120573
3gt 0 120573
1lt 0 119862
1ge 0
1198622ge 0 and 119896minus2 le 119873(120573
2+1205733) le 2(1+119862
1+1198622)(1205732+1205733minus1205731)
there exists unique fixed point 119906 isin [1199060 V0] for the operator 119860
which is defined by (61) Then (57) is true
Journal of Applied Mathematics 7
Proof Obviously 119860 [1199060 V0] times [119906
0 V0] rarr 119862[0 1] is mixed
monotone Consider
119860 (1199060 V0) = 119860 (0 1)
=1
2[119873 minus 119896
minus2(1205732+ 1205733)] 1205852+ 1198621120585 + 1198622ge 0 = 119906
0
119860 (V0 1199060) =
1
2(119873 minus 120573
1119896minus2) 1205852+ 1198621120585 + 1198622
le1
2119873(1 minus
1205731
1205732+ 1205733
) + 1198621+ 1198622le 1 = V
0
(62)
Thus 119860 satisfies the term (1) of Lemma 1For all 119906 V isin [119906
0 V0] 119906 le V
119860 (V 119906) minus 119860 (119906 V)
= minus119896minus2int120585
0
(120585 minus 119904)
times [1205731V (119904) + 120573
21199062(119904) + 120573
21199063(119904)
minus1205731119906 (119904) minus 120573
2V2 (119904) minus 120573
3V3 (119904)] 119889119904
= 119896minus2int120585
0
(120585 minus 119904) 119890119872119904119890minus119872119904
[V (119904) minus 119906 (119904)]
times minus 1205731+ 1205732[V (119904) + 119906 (119904)]
+1205733[V2 (119904) + 119906 (119904) V (119904) + 1199062 (119904)] 119889119904
le (minus1205731+ 21205732+ 31205733) 119896minus2V minus 119906int
120585
0
119890119872119878119889119904
leminus1205731+ 21205732+ 31205733
1198962119872V minus 119906 119890119872120585
(63)
Then we have
119860 (V 119906) minus 119860 (119906 V) le 120572 (64)
where 120572 = (minus1205731+ 21205732+ 31205733)(1198962119872) and when 119872 is large
enough 120572 isin (0 1)Therefore operator119860which is defined by (61) satisfies the
term (2) of Lemma 1 The proof is completed
Obviously the fixed point of the operator defined by (61)is equivalent to the solution of (3) We denote
1199066(119909 119905) = 119906 (120585) (65)
By using the classical iteration algorithms such as Manniteration method Ishikawa iteration method and Nooriteration method the solutions can be obtained Then thetraveling wave solutions of (5) are (65) and the followingformula
1198676(119909 119905) = 119887
1+ 1205821199066(119909 119905) minus
1
21199062
6(119909 119905) (66)
6 Modified Mapping Method
The modified mapping method was introduced and thenonlinear evolution equation was solved in [21 22] In[22] several hundreds of Jacobi elliptic function expansionsolutions were obtained We also use this method in thispaper
Let
119906 = 1198600+ 1198601119891 + 119861
1119891minus1 (67)
where 11989110158401015840 = 119901119891 + 1199021198913 hArr 11989110158402 = 119903 + 1199011198912 + (12)1199021198914 119903 119901 and119902 are constants
Thus
1199061015840= 11986011198911015840minus 1198611119891minus21198911015840
11990610158401015840= 1198601119901119891 + 119860
11199021198913+ 1198611119901119891minus1+ 21198611119903119891minus3
1199062= 1198602
0+ 211986011198611+ 211986001198601119891 + 2119860
01198611119891minus1+ 1198602
11198912+ 1198612
1119891minus2
1199063= 1198603
0+ 6119860011986011198611+ 31198601(1198602
0+ 11986011198611) 119891
+ 311986001198602
11198912+ 1198603
11198913+ 31198611(1198602
0+ 11986011198611) 119891minus1
+ 311986001198612
1119891minus2+ 1198613
1119891minus3
(68)
Substituting (68) into (3) we have
12057311198600+ 1205732(1198602
0+ 211986011198611) + 1205733(1198603
0+ 6119860011986011198611) minus 119873
+ 1198601[1198962119901 + 1205731+ 12057321198600+ 31205733(1198602
0+ 11986011198611)] 119891
+ 1198602
1(1205732+ 312057331198600) 1198912+ 1198601(1198962119902 + 12057331198602
1) 1198913
+ 1198611[1198962119901 + 1205731+ 212057321198600+ 31205733(1198602
0+ 11986011198611)] 119891minus1
+ 1198612
1(1205732+ 312057331198600) 119891minus2+ 1198611(21198962119903 + 12057331198612
1) 119891minus3= 0
(69)
Notice that we get the same algebraic equations when thecoefficients of 119891 119891minus1 and 1198912 119891minus2 in (69) are zero Thereforethere are only five algebra equations Solve them and we have
1198600=minus1205732
31205733
1198601= plusmn119896radic
minus119902
1205733
1198611= plusmn119896radic
minus2119903
1205733
119896 = plusmnradic12057322minus 312057311205733minus 31198962120573
3radic2119902119903
31205733119901
(70)
In order to make 1198601 1198611be real number 119902 and 119903 should
be different signs with 1205733
(i) Choosing 119903 = 1119901 = minus2 and 119902 = 2 we have119891 = tanh 120585Then the solitary wave solutions of (3) are
1199067(119909 119905) = 119860
0+ 1198601tanh 120585 + 119861
1tanhminus1120585 (71)
8 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Journal of Applied Mathematics 5
05
10
0
11
2
3
2
3
22
24
26
28
12
14
05
15
25
16
18
minus10
minus5
Figure 3 The solitary wave and behavior of the solutions 119906(119909 119905) =1 + 2 sech(119909 minus 119905) for 0 ⩽ 119905 ⩽ 1 and minus10 ⩽ 119909 ⩽ 10
Taking 119898 rarr 1 cn120585 rarr sech120585 (38) can be rewritten asfollows
1199063(119909 119905) = 119886
0+ 1198861sech119896 (119909 minus 120582119905) (40)
Assuming 120582 = 1 1205731= 05 120573
2= minus15 and 120573
3= 05 it can
be obtained that 1198860= 1 1198861= 2 and 119896 = 1 The solution of (3)
is 119906(119909 119905) = 1+2 sech(119909minus119905)The solitary wave and behavior ofthe solutions are shown in Figure 3 respectively for 0 le 119905 le 1
and minus10 le 119909 le 10
43 Third Kind of Jacobi Elliptic Function Method Let
119906 = 1198860+ 1198861dn120585 (41)
Thus
1199061015840= minus119898
21198861sn120585 119906
10158401015840= (2 minus 119898
2) 1198861dn120585 minus 2119886
1dn3120585
1199063= 1198863
0+ 31198862
01198861dn120585 + 3119886
01198862
1dn2120585 + 1198863
1dn3120585
(42)
Substituting them into (3) we can obtain
12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] dn120585
+ 1198862
1(1205732+ 312057331198860) dn2120585 + 119886
1(minus21198962+ 1198862
11205733) dn3120585 = 0
(43)
Letting the coefficients of all derivatives of dn120585 be zerowe have
1198860=minus1205732
31205733
1198861= plusmn119896radic
2
1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(44)
Therefore the solution of (3) is
1199064(119909 119905) = 119886
0+ 1198861dn119896 (119909 minus 120582119905) (45)
where 1198860 1198861 and 119896 are denoted by (44) and 120582 is arbitrary
constantThus the solutions of (7) are (45) and the following
formula
1198674(119909 119905) = 119887
1+ 1205821199064(119909 119905) minus
1
21199062
4(119909 119905) (46)
Letting119898 rarr 1 then (45) equals (38)
44 Jacobi Elliptic Function cs120585Method Let
119906 = 1198860+ 1198861cs120585 cs120585 equiv cn120585
sn120585 (47)
Then
1199061015840= minus1198861(1 + cs2120585) dn120585
11990610158401015840= (2 minus 119898
2) 1198861cs120585 + 2119886
1cs3120585
(48)
Substituting (47) and (48) into (3) we can obtain
119863 = 12057311198860+ 12057321198862
0+ 12057331198863
0
minus 119873 + 1198861[1198962(2 minus 119898
2) + 1205731+ 212057321198860+ 312057331198862
0] cs120585
+ 1198862
1(1205732+ 312057331198860) cs2120585 + 119886
1(21198962+ 1198862
11205733) cs3120585
(49)
Let the coefficients in above formula be zero and we have
1198860=minus1205732
31205733
1198861= plusmn1198962radic
2
minus1205733
1198962=
1
1198982 minus 2(312057311205733minus 12057322
31205733
)
119873 = 12057311198860+ 12057321198862
0+ 12057331198863
0
(50)
According to (47) the solution of (3) is
1199065(119909 119905) = 119886
0+ 1198861cs119896 (119909 minus 120582119905) (51)
where 1198860 1198861 and 119896 are denoted by (50) and 120582 is arbitrary
constantTherefore the solutions of (7) are (51) and the following
formula
1198675= 1198871+ 1205821199065(119909 119905) minus
1
21199062
5(119909 119905) (52)
When119898 rarr 1
1199065(119909 119905) = 119886
0+ 1198861csch119896 (119909 minus 120582119905) (53)
Assuming 1205731= (1 minus 1198991205962)119899 120573
2= minus(32)120596 120573
3= minus(12)
119873 = minus(1198731119899) and 119896 = 1 we can obtain (3) Choosing
6 Journal of Applied Mathematics
02
40
24
0
5
10
15
0
5
10
minus4 minus4
minus10
minus15
minus2
minus2
minus5
minus10
minus5
Figure 4The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 csch(119909 minus 23119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
02
4
02
4
0
10
20
30
40
50
0
5
10
15
20
25
30
35
40
45
minus4 minus4
minus10
minus2 minus2
Figure 5The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =minus13+43 csch2(119909minus23119910+119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
119899 = minus(23) 120596 = 1 then 1205731
= minus(52) 1205732
= minus(32)Substituting 120573
1 1205732 1205733into (50) it can be obtained that
1198860= minus1 119886
1= 2 and 119896 = 1 The solutions of (8) are
119906(119909 119910 119905) = minus1 + 2 csch(119909 minus (23)119910 + 119905) and ℎ(119909 119910 119905) =
minus(13) + (43)csch2(119909 minus (23)119910 + 119905) The solitary wave andbehavior of the solutions 119906(119909 119905) = minus1+ 2 csch(119909 minus (23)119910 + 119905)and ℎ(119909 119910 119905) = minus(13)+(43)csch2(119909minus(23)119910+119905) are shownin Figures 4 and 5 respectively for 119905 = 0 minus4 le 119909 le 4minus4 le 119910 le 4 and 119909 minus (23)119910 = 0
5 Fixed Point Method
In general fixed point theory can be divided into two typesThe first type is only used to discuss the existence of solutionThe second type is used not only to discuss the existence anduniqueness of solution but also to search the fixed point Weare more interested in the second type
Lemma 1 (see [18ndash20]) Suppose 119864 is a real Banach spacewhich has normal cone Consider 119906
0 V0isin 119864 119906
0lt V0 The
operator 119860 [1199060 V0] times [119906
0 V0] rarr 119864 is mixed monotone and
satisfies
(1)
1199060le 119860 (119906
0 V0) 119860 (V
0 1199060) le V0
(54)
(2) for all 119906 V (1199060le 119906 le V le V
0) exist constant 120572 isin (0 1)
satisfies
119860 (V 119906) minus 119860 (119906 V) le 120572 V minus 119906 (55)
Then there exists a unique 119906 (119906 isin [1199060 V0]) which satisfies
119860 (119906 119906) = 119906 (56)
where 119906 is called the fixed point of 119860In [1199060 V0] for all 119908
0 letting 119908
119899= 119860(119908
119899minus1 119908119899minus1
) 119899 =
1 2 we have
119906 = lim119899rarrinfin
119908119899 (57)
In this paper 119864 is taken as a continuous functionspace 119862[0 1] which is defined in the closed interval [0 1]Introducing equivalent norm it is easy to know that 119862[0 1]is real Banach space which has normal cone Let
1199060= 0 V
0= 1 (58)
Integrating (3) with respect to 120585 we can obtain
119889119906
119889120585= minus119896minus2int120585
0
[1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889120585 + 119873120585 + 119862
1
(59)
where 1198621is integral constant
Integrating the above formula with respect to 120585 and usingthe formula in [12] we have
119906 = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(60)
where 1198622is integral constant
If 1205732gt 0 120573
3gt 0 (ie 1205822 gt 119888
1) and 120573
1lt 0 (ie 1205822 lt 1198882
0)
suppose
119860 (119906 V) = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
2V2 (119904) + 120573
3V3 (119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(61)
Theorem 2 In (61) supposing 1205732gt 0 120573
3gt 0 120573
1lt 0 119862
1ge 0
1198622ge 0 and 119896minus2 le 119873(120573
2+1205733) le 2(1+119862
1+1198622)(1205732+1205733minus1205731)
there exists unique fixed point 119906 isin [1199060 V0] for the operator 119860
which is defined by (61) Then (57) is true
Journal of Applied Mathematics 7
Proof Obviously 119860 [1199060 V0] times [119906
0 V0] rarr 119862[0 1] is mixed
monotone Consider
119860 (1199060 V0) = 119860 (0 1)
=1
2[119873 minus 119896
minus2(1205732+ 1205733)] 1205852+ 1198621120585 + 1198622ge 0 = 119906
0
119860 (V0 1199060) =
1
2(119873 minus 120573
1119896minus2) 1205852+ 1198621120585 + 1198622
le1
2119873(1 minus
1205731
1205732+ 1205733
) + 1198621+ 1198622le 1 = V
0
(62)
Thus 119860 satisfies the term (1) of Lemma 1For all 119906 V isin [119906
0 V0] 119906 le V
119860 (V 119906) minus 119860 (119906 V)
= minus119896minus2int120585
0
(120585 minus 119904)
times [1205731V (119904) + 120573
21199062(119904) + 120573
21199063(119904)
minus1205731119906 (119904) minus 120573
2V2 (119904) minus 120573
3V3 (119904)] 119889119904
= 119896minus2int120585
0
(120585 minus 119904) 119890119872119904119890minus119872119904
[V (119904) minus 119906 (119904)]
times minus 1205731+ 1205732[V (119904) + 119906 (119904)]
+1205733[V2 (119904) + 119906 (119904) V (119904) + 1199062 (119904)] 119889119904
le (minus1205731+ 21205732+ 31205733) 119896minus2V minus 119906int
120585
0
119890119872119878119889119904
leminus1205731+ 21205732+ 31205733
1198962119872V minus 119906 119890119872120585
(63)
Then we have
119860 (V 119906) minus 119860 (119906 V) le 120572 (64)
where 120572 = (minus1205731+ 21205732+ 31205733)(1198962119872) and when 119872 is large
enough 120572 isin (0 1)Therefore operator119860which is defined by (61) satisfies the
term (2) of Lemma 1 The proof is completed
Obviously the fixed point of the operator defined by (61)is equivalent to the solution of (3) We denote
1199066(119909 119905) = 119906 (120585) (65)
By using the classical iteration algorithms such as Manniteration method Ishikawa iteration method and Nooriteration method the solutions can be obtained Then thetraveling wave solutions of (5) are (65) and the followingformula
1198676(119909 119905) = 119887
1+ 1205821199066(119909 119905) minus
1
21199062
6(119909 119905) (66)
6 Modified Mapping Method
The modified mapping method was introduced and thenonlinear evolution equation was solved in [21 22] In[22] several hundreds of Jacobi elliptic function expansionsolutions were obtained We also use this method in thispaper
Let
119906 = 1198600+ 1198601119891 + 119861
1119891minus1 (67)
where 11989110158401015840 = 119901119891 + 1199021198913 hArr 11989110158402 = 119903 + 1199011198912 + (12)1199021198914 119903 119901 and119902 are constants
Thus
1199061015840= 11986011198911015840minus 1198611119891minus21198911015840
11990610158401015840= 1198601119901119891 + 119860
11199021198913+ 1198611119901119891minus1+ 21198611119903119891minus3
1199062= 1198602
0+ 211986011198611+ 211986001198601119891 + 2119860
01198611119891minus1+ 1198602
11198912+ 1198612
1119891minus2
1199063= 1198603
0+ 6119860011986011198611+ 31198601(1198602
0+ 11986011198611) 119891
+ 311986001198602
11198912+ 1198603
11198913+ 31198611(1198602
0+ 11986011198611) 119891minus1
+ 311986001198612
1119891minus2+ 1198613
1119891minus3
(68)
Substituting (68) into (3) we have
12057311198600+ 1205732(1198602
0+ 211986011198611) + 1205733(1198603
0+ 6119860011986011198611) minus 119873
+ 1198601[1198962119901 + 1205731+ 12057321198600+ 31205733(1198602
0+ 11986011198611)] 119891
+ 1198602
1(1205732+ 312057331198600) 1198912+ 1198601(1198962119902 + 12057331198602
1) 1198913
+ 1198611[1198962119901 + 1205731+ 212057321198600+ 31205733(1198602
0+ 11986011198611)] 119891minus1
+ 1198612
1(1205732+ 312057331198600) 119891minus2+ 1198611(21198962119903 + 12057331198612
1) 119891minus3= 0
(69)
Notice that we get the same algebraic equations when thecoefficients of 119891 119891minus1 and 1198912 119891minus2 in (69) are zero Thereforethere are only five algebra equations Solve them and we have
1198600=minus1205732
31205733
1198601= plusmn119896radic
minus119902
1205733
1198611= plusmn119896radic
minus2119903
1205733
119896 = plusmnradic12057322minus 312057311205733minus 31198962120573
3radic2119902119903
31205733119901
(70)
In order to make 1198601 1198611be real number 119902 and 119903 should
be different signs with 1205733
(i) Choosing 119903 = 1119901 = minus2 and 119902 = 2 we have119891 = tanh 120585Then the solitary wave solutions of (3) are
1199067(119909 119905) = 119860
0+ 1198601tanh 120585 + 119861
1tanhminus1120585 (71)
8 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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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 Journal of Applied Mathematics
02
40
24
0
5
10
15
0
5
10
minus4 minus4
minus10
minus15
minus2
minus2
minus5
minus10
minus5
Figure 4The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 csch(119909 minus 23119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
02
4
02
4
0
10
20
30
40
50
0
5
10
15
20
25
30
35
40
45
minus4 minus4
minus10
minus2 minus2
Figure 5The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =minus13+43 csch2(119909minus23119910+119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
119899 = minus(23) 120596 = 1 then 1205731
= minus(52) 1205732
= minus(32)Substituting 120573
1 1205732 1205733into (50) it can be obtained that
1198860= minus1 119886
1= 2 and 119896 = 1 The solutions of (8) are
119906(119909 119910 119905) = minus1 + 2 csch(119909 minus (23)119910 + 119905) and ℎ(119909 119910 119905) =
minus(13) + (43)csch2(119909 minus (23)119910 + 119905) The solitary wave andbehavior of the solutions 119906(119909 119905) = minus1+ 2 csch(119909 minus (23)119910 + 119905)and ℎ(119909 119910 119905) = minus(13)+(43)csch2(119909minus(23)119910+119905) are shownin Figures 4 and 5 respectively for 119905 = 0 minus4 le 119909 le 4minus4 le 119910 le 4 and 119909 minus (23)119910 = 0
5 Fixed Point Method
In general fixed point theory can be divided into two typesThe first type is only used to discuss the existence of solutionThe second type is used not only to discuss the existence anduniqueness of solution but also to search the fixed point Weare more interested in the second type
Lemma 1 (see [18ndash20]) Suppose 119864 is a real Banach spacewhich has normal cone Consider 119906
0 V0isin 119864 119906
0lt V0 The
operator 119860 [1199060 V0] times [119906
0 V0] rarr 119864 is mixed monotone and
satisfies
(1)
1199060le 119860 (119906
0 V0) 119860 (V
0 1199060) le V0
(54)
(2) for all 119906 V (1199060le 119906 le V le V
0) exist constant 120572 isin (0 1)
satisfies
119860 (V 119906) minus 119860 (119906 V) le 120572 V minus 119906 (55)
Then there exists a unique 119906 (119906 isin [1199060 V0]) which satisfies
119860 (119906 119906) = 119906 (56)
where 119906 is called the fixed point of 119860In [1199060 V0] for all 119908
0 letting 119908
119899= 119860(119908
119899minus1 119908119899minus1
) 119899 =
1 2 we have
119906 = lim119899rarrinfin
119908119899 (57)
In this paper 119864 is taken as a continuous functionspace 119862[0 1] which is defined in the closed interval [0 1]Introducing equivalent norm it is easy to know that 119862[0 1]is real Banach space which has normal cone Let
1199060= 0 V
0= 1 (58)
Integrating (3) with respect to 120585 we can obtain
119889119906
119889120585= minus119896minus2int120585
0
[1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889120585 + 119873120585 + 119862
1
(59)
where 1198621is integral constant
Integrating the above formula with respect to 120585 and usingthe formula in [12] we have
119906 = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
21199062(119904) + 120573
31199063(119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(60)
where 1198622is integral constant
If 1205732gt 0 120573
3gt 0 (ie 1205822 gt 119888
1) and 120573
1lt 0 (ie 1205822 lt 1198882
0)
suppose
119860 (119906 V) = minus119896minus2int120585
0
(120585 minus 119904) [1205731119906 (119904) + 120573
2V2 (119904) + 120573
3V3 (119904)] 119889119904
+1
21198731205852+ 1198621120585 + 1198622
(61)
Theorem 2 In (61) supposing 1205732gt 0 120573
3gt 0 120573
1lt 0 119862
1ge 0
1198622ge 0 and 119896minus2 le 119873(120573
2+1205733) le 2(1+119862
1+1198622)(1205732+1205733minus1205731)
there exists unique fixed point 119906 isin [1199060 V0] for the operator 119860
which is defined by (61) Then (57) is true
Journal of Applied Mathematics 7
Proof Obviously 119860 [1199060 V0] times [119906
0 V0] rarr 119862[0 1] is mixed
monotone Consider
119860 (1199060 V0) = 119860 (0 1)
=1
2[119873 minus 119896
minus2(1205732+ 1205733)] 1205852+ 1198621120585 + 1198622ge 0 = 119906
0
119860 (V0 1199060) =
1
2(119873 minus 120573
1119896minus2) 1205852+ 1198621120585 + 1198622
le1
2119873(1 minus
1205731
1205732+ 1205733
) + 1198621+ 1198622le 1 = V
0
(62)
Thus 119860 satisfies the term (1) of Lemma 1For all 119906 V isin [119906
0 V0] 119906 le V
119860 (V 119906) minus 119860 (119906 V)
= minus119896minus2int120585
0
(120585 minus 119904)
times [1205731V (119904) + 120573
21199062(119904) + 120573
21199063(119904)
minus1205731119906 (119904) minus 120573
2V2 (119904) minus 120573
3V3 (119904)] 119889119904
= 119896minus2int120585
0
(120585 minus 119904) 119890119872119904119890minus119872119904
[V (119904) minus 119906 (119904)]
times minus 1205731+ 1205732[V (119904) + 119906 (119904)]
+1205733[V2 (119904) + 119906 (119904) V (119904) + 1199062 (119904)] 119889119904
le (minus1205731+ 21205732+ 31205733) 119896minus2V minus 119906int
120585
0
119890119872119878119889119904
leminus1205731+ 21205732+ 31205733
1198962119872V minus 119906 119890119872120585
(63)
Then we have
119860 (V 119906) minus 119860 (119906 V) le 120572 (64)
where 120572 = (minus1205731+ 21205732+ 31205733)(1198962119872) and when 119872 is large
enough 120572 isin (0 1)Therefore operator119860which is defined by (61) satisfies the
term (2) of Lemma 1 The proof is completed
Obviously the fixed point of the operator defined by (61)is equivalent to the solution of (3) We denote
1199066(119909 119905) = 119906 (120585) (65)
By using the classical iteration algorithms such as Manniteration method Ishikawa iteration method and Nooriteration method the solutions can be obtained Then thetraveling wave solutions of (5) are (65) and the followingformula
1198676(119909 119905) = 119887
1+ 1205821199066(119909 119905) minus
1
21199062
6(119909 119905) (66)
6 Modified Mapping Method
The modified mapping method was introduced and thenonlinear evolution equation was solved in [21 22] In[22] several hundreds of Jacobi elliptic function expansionsolutions were obtained We also use this method in thispaper
Let
119906 = 1198600+ 1198601119891 + 119861
1119891minus1 (67)
where 11989110158401015840 = 119901119891 + 1199021198913 hArr 11989110158402 = 119903 + 1199011198912 + (12)1199021198914 119903 119901 and119902 are constants
Thus
1199061015840= 11986011198911015840minus 1198611119891minus21198911015840
11990610158401015840= 1198601119901119891 + 119860
11199021198913+ 1198611119901119891minus1+ 21198611119903119891minus3
1199062= 1198602
0+ 211986011198611+ 211986001198601119891 + 2119860
01198611119891minus1+ 1198602
11198912+ 1198612
1119891minus2
1199063= 1198603
0+ 6119860011986011198611+ 31198601(1198602
0+ 11986011198611) 119891
+ 311986001198602
11198912+ 1198603
11198913+ 31198611(1198602
0+ 11986011198611) 119891minus1
+ 311986001198612
1119891minus2+ 1198613
1119891minus3
(68)
Substituting (68) into (3) we have
12057311198600+ 1205732(1198602
0+ 211986011198611) + 1205733(1198603
0+ 6119860011986011198611) minus 119873
+ 1198601[1198962119901 + 1205731+ 12057321198600+ 31205733(1198602
0+ 11986011198611)] 119891
+ 1198602
1(1205732+ 312057331198600) 1198912+ 1198601(1198962119902 + 12057331198602
1) 1198913
+ 1198611[1198962119901 + 1205731+ 212057321198600+ 31205733(1198602
0+ 11986011198611)] 119891minus1
+ 1198612
1(1205732+ 312057331198600) 119891minus2+ 1198611(21198962119903 + 12057331198612
1) 119891minus3= 0
(69)
Notice that we get the same algebraic equations when thecoefficients of 119891 119891minus1 and 1198912 119891minus2 in (69) are zero Thereforethere are only five algebra equations Solve them and we have
1198600=minus1205732
31205733
1198601= plusmn119896radic
minus119902
1205733
1198611= plusmn119896radic
minus2119903
1205733
119896 = plusmnradic12057322minus 312057311205733minus 31198962120573
3radic2119902119903
31205733119901
(70)
In order to make 1198601 1198611be real number 119902 and 119903 should
be different signs with 1205733
(i) Choosing 119903 = 1119901 = minus2 and 119902 = 2 we have119891 = tanh 120585Then the solitary wave solutions of (3) are
1199067(119909 119905) = 119860
0+ 1198601tanh 120585 + 119861
1tanhminus1120585 (71)
8 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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
Journal of Applied Mathematics 7
Proof Obviously 119860 [1199060 V0] times [119906
0 V0] rarr 119862[0 1] is mixed
monotone Consider
119860 (1199060 V0) = 119860 (0 1)
=1
2[119873 minus 119896
minus2(1205732+ 1205733)] 1205852+ 1198621120585 + 1198622ge 0 = 119906
0
119860 (V0 1199060) =
1
2(119873 minus 120573
1119896minus2) 1205852+ 1198621120585 + 1198622
le1
2119873(1 minus
1205731
1205732+ 1205733
) + 1198621+ 1198622le 1 = V
0
(62)
Thus 119860 satisfies the term (1) of Lemma 1For all 119906 V isin [119906
0 V0] 119906 le V
119860 (V 119906) minus 119860 (119906 V)
= minus119896minus2int120585
0
(120585 minus 119904)
times [1205731V (119904) + 120573
21199062(119904) + 120573
21199063(119904)
minus1205731119906 (119904) minus 120573
2V2 (119904) minus 120573
3V3 (119904)] 119889119904
= 119896minus2int120585
0
(120585 minus 119904) 119890119872119904119890minus119872119904
[V (119904) minus 119906 (119904)]
times minus 1205731+ 1205732[V (119904) + 119906 (119904)]
+1205733[V2 (119904) + 119906 (119904) V (119904) + 1199062 (119904)] 119889119904
le (minus1205731+ 21205732+ 31205733) 119896minus2V minus 119906int
120585
0
119890119872119878119889119904
leminus1205731+ 21205732+ 31205733
1198962119872V minus 119906 119890119872120585
(63)
Then we have
119860 (V 119906) minus 119860 (119906 V) le 120572 (64)
where 120572 = (minus1205731+ 21205732+ 31205733)(1198962119872) and when 119872 is large
enough 120572 isin (0 1)Therefore operator119860which is defined by (61) satisfies the
term (2) of Lemma 1 The proof is completed
Obviously the fixed point of the operator defined by (61)is equivalent to the solution of (3) We denote
1199066(119909 119905) = 119906 (120585) (65)
By using the classical iteration algorithms such as Manniteration method Ishikawa iteration method and Nooriteration method the solutions can be obtained Then thetraveling wave solutions of (5) are (65) and the followingformula
1198676(119909 119905) = 119887
1+ 1205821199066(119909 119905) minus
1
21199062
6(119909 119905) (66)
6 Modified Mapping Method
The modified mapping method was introduced and thenonlinear evolution equation was solved in [21 22] In[22] several hundreds of Jacobi elliptic function expansionsolutions were obtained We also use this method in thispaper
Let
119906 = 1198600+ 1198601119891 + 119861
1119891minus1 (67)
where 11989110158401015840 = 119901119891 + 1199021198913 hArr 11989110158402 = 119903 + 1199011198912 + (12)1199021198914 119903 119901 and119902 are constants
Thus
1199061015840= 11986011198911015840minus 1198611119891minus21198911015840
11990610158401015840= 1198601119901119891 + 119860
11199021198913+ 1198611119901119891minus1+ 21198611119903119891minus3
1199062= 1198602
0+ 211986011198611+ 211986001198601119891 + 2119860
01198611119891minus1+ 1198602
11198912+ 1198612
1119891minus2
1199063= 1198603
0+ 6119860011986011198611+ 31198601(1198602
0+ 11986011198611) 119891
+ 311986001198602
11198912+ 1198603
11198913+ 31198611(1198602
0+ 11986011198611) 119891minus1
+ 311986001198612
1119891minus2+ 1198613
1119891minus3
(68)
Substituting (68) into (3) we have
12057311198600+ 1205732(1198602
0+ 211986011198611) + 1205733(1198603
0+ 6119860011986011198611) minus 119873
+ 1198601[1198962119901 + 1205731+ 12057321198600+ 31205733(1198602
0+ 11986011198611)] 119891
+ 1198602
1(1205732+ 312057331198600) 1198912+ 1198601(1198962119902 + 12057331198602
1) 1198913
+ 1198611[1198962119901 + 1205731+ 212057321198600+ 31205733(1198602
0+ 11986011198611)] 119891minus1
+ 1198612
1(1205732+ 312057331198600) 119891minus2+ 1198611(21198962119903 + 12057331198612
1) 119891minus3= 0
(69)
Notice that we get the same algebraic equations when thecoefficients of 119891 119891minus1 and 1198912 119891minus2 in (69) are zero Thereforethere are only five algebra equations Solve them and we have
1198600=minus1205732
31205733
1198601= plusmn119896radic
minus119902
1205733
1198611= plusmn119896radic
minus2119903
1205733
119896 = plusmnradic12057322minus 312057311205733minus 31198962120573
3radic2119902119903
31205733119901
(70)
In order to make 1198601 1198611be real number 119902 and 119903 should
be different signs with 1205733
(i) Choosing 119903 = 1119901 = minus2 and 119902 = 2 we have119891 = tanh 120585Then the solitary wave solutions of (3) are
1199067(119909 119905) = 119860
0+ 1198601tanh 120585 + 119861
1tanhminus1120585 (71)
8 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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 Journal of Applied Mathematics
0
5
02
4
0
2
4
6
minus6
minus5
minus4
minus3
minus2
minus1
0
1
2
3
4
minus4
minus2minus5
minus8
minus6
minus4
minus2
Figure 6The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + 2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
0
5
02
412
13
14
15
16
17
18
13
135
14
145
15
155
16
165
17
minus4
minus2
minus5
Figure 7The solitarywave and behavior of the solutions ℎ(119909 119910 119905) =7 + 4 tanh2(119909 minus 2119910 + 119905) + 4 tanhminus2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0= minus1
1198601= 2 and 119861
1= 2 The solutions of (8) are 119906(119909 119910 119905) = minus1 +
2 tanh(119909 minus 2119910 + 119905) + 2 tanhminus1(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 7 +
4 tanh2(119909minus2119910+119905)+4 tanhminus2(119909minus2119910+119905)The solitary wave andbehavior of them are shown in Figures 6 and 7 respectivelyfor 119905 = 0 minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
(ii) Choosing 119903 = 1 119901 = minus(1 + 1198982) and 119902 = 21198982 thus
119891 = sn120585 (72)
where119898 (0 lt 119898 lt 1) is themodule of Jacobi elliptic functionTherefore the expansion solutions of (3) are
1199068(119909 119905) = 119860
0+ 1198601sn120585 + 119861
1snminus1120585 (73)
Letting119898 rarr 1 then (73) equals (71)
0
5
02
4
0
2
4
0
1
2
3
minus6
minus4
minus2
minus4
minus2minus5
minus5
minus4
minus3
minus2
minus1
Figure 8The solitarywave and behavior of the solutions119906(119909 119910 119905) =minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 + sech(119909 minus 2119910 +119905)) tanh(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4 and minus4 ⩽ 119910 ⩽ 4
(iii) Choosing 119903 = 14 119901 = (1198982 minus 2)2 and 119902 = 11989822 wehave
119891 =sn120585
(1 + dn120585) (74)
Therefore the solutions of (3) are
1199069(119909 119905) = 119860
0+
1198601sn120585
1 + dn120585+1198611(1 + dn120585)sn120585
(75)
where 1198600 1198601 1198611of 119906119894(119894 = 7 8 9) are denoted by (70)
Taking119898 rarr 1 (75) can be rewritten as follows
1199069(119909 119905) = 119860
0+1198601tanh 120585
1 + sech120585+1198611(1 + sech120585)tanh 120585
(76)
Assuming 119899 = minus2 120596 = 1 119896 = 1 and 1205733= minus(12) it
can be obtained that 1205731= minus(32) 120573
2= minus(32) 119860
0=
minus1 1198601= 1 and 119861
1= 1 The solutions of (8) are 119906(119909
119910 119905) = minus1 + tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905)) + (1 +
sech(119909 minus 2119910 + 119905)) tanh(119909 minus 2119910 + 119905) and ℎ(119909 119910 119905) = 1 +
[tanh(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))]2 + [(1 + sech(119909 minus
2119910 + 119905)) tanh(119909 minus 2119910 + 119905)]2 The solitary wave and behaviorof them are shown in Figures 8 and 9 respectively for 119905 = 0minus4 le 119909 le 4 minus4 le 119910 le 4 and 119909 minus 2119910 = 0
7 Summary and Conclusions
In summary by using different methods nonlinear waveequation of a finite deformation elastic circular rod Boussi-nesq equations and dispersive long wave equations aresolved in this paper and several analytical solutions areobtained Kink soliton solutions are obtained when we usethe homogenous balancemethodThe Jacobi elliptic functionmethod is utilized and periodic solutions are obtainedWhen119898 rarr 1 the solutions reduce to solitary solution Theirwave forms are the bell type kink type exotic type and
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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
Journal of Applied Mathematics 9
minus20
24
02
40
5
10
15
20
4
6
8
10
12
14
16
minus4 minus4
minus2
Figure 9 The solitary wave and behavior of the solutionsℎ(119909 119910 119905) = 1 + tanh2(119909 minus 2119910 + 119905)(1 + sech(119909 minus 2119910 + 119905))2 +(1 + sech(119909 minus 2119910 + 119905))2tanh2(119909 minus 2119910 + 119905) for 119905 = 0 minus4 ⩽ 119909 ⩽ 4and minus4 ⩽ 119910 ⩽ 4
peakon type By using themodifiedmappingmethod peakonsolutions of (3) (5) (8) are obtained On the other handafter proving the existence and uniqueness of fixed point ofoperator 119860 we can use the classical iteration algorithms toget the solutions The method has wide application and canbe used to solve other nonlinear wave equations Fixed pointmethodmay be significant and important for the explanationof some special physical problems whose analytical solutioncannot be obtained The results indicate the following Firstthe natural phenomena and the physical properties of theequations are abundant which should be further discoveredSecond there are many kinds of methods for solving theseequations Third the arbitrary constant can be determinedby the initial and boundary conditions and we can obtain theexact solutions for certain engineering or scientific problemsIt is shown that the methods proposed in this paper forthree types of nonlinear wave equations are feasible fordetermining exact solutions of other nonlinear equations
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
This work was supported by the Special Fund for BasicScientific Research of Central Colleges (Grant noCHD2009JC140)
References
[1] Z F Liu and S Y Zhang ldquoSolitary waves in finite deforma-tion elastic circular rodrdquo Journal of Applied Mathematics andMechanics vol 27 no 10 pp 1255ndash1260 2006
[2] E V Krishnan S Kumar and A Biswas ldquoSolitons andother nonlinear waves of the Boussinesq equationrdquo NonlinearDynamics An International Journal of Nonlinear Dynamics andChaos in Engineering Systems vol 70 no 2 pp 1213ndash1221 2012
[3] A Biswas M Song H Triki et al ldquoSolitons Shockwaves con-servation laws and bifurcation analysis of boussinesq equationwith power law nonlinearity and dual-dispersionrdquo Journal ofApplied Mathematics and Information Science vol 8 no 3 pp949ndash957 2014
[4] A Jarsquoafar M Jawad M D Petkovic and A Biswas ldquoSolitonsolutions to a few coupled nonlinear wave equations by tanhmethodrdquo The Iranian Journal of Science and Technology A vol37 no 2 pp 109ndash115 2013
[5] J F Zhang ldquoMulti-solitary wave solutions for variant Boussi-nesq equations and Kupershmidt equationsrdquo Journal of AppliedMathematics and Mechanics vol 21 no 2 pp 171ndash175 2000
[6] M Zhang Q Liu J Wang and K Wu ldquoA new supersymmetricclassical Boussinesq equationrdquo Chinese Physics B vol 17 no 1pp 10ndash16 2008
[7] Y B Yuan D M Pu and S M Li ldquoBifurcations of travellingwave solutions in variant Boussinesq equationsrdquo Journal ofApplied Mathematics and Mechanics vol 27 no 6 pp 716ndash7262006
[8] Sirendaoreji and S Jiong ldquoAuxiliary equation method forsolving nonlinear partial differential equationsrdquo Physics LettersA vol 309 no 5-6 pp 387ndash396 2003
[9] Taogetusang and Sirendaoreji ldquoNew types of exact solitarywavesolutions for (2 + 1)-dimensional dispersive long-wave equa-tions and combined KdV-Burgers equationrdquo Chinese Journal ofEngineering Mathematics vol 23 no 5 pp 943ndash946 2006
[10] E G Fan and H Q Zhang ldquoSolitary wave solutions of a classof nonlinear wave equationsrdquo Acta Physica Sinica vol 46 no 7pp 1254ndash1258 1997
[11] W L Zhang G J Wu M Zhang J M Wang and J H HanldquoNew exact periodic solutions to (2+1)-dimensional dispersivelong wave equationsrdquo Chinese Physics B vol 17 no 4 pp 1156ndash1164 2008
[12] Z Bartoszewski ldquoSolving boundary value problems for delaydifferential equations by a fixed-point methodrdquo Journal ofComputational and Applied Mathematics vol 236 no 6 pp1576ndash1590 2011
[13] Z T Fu S K Liu S D Liu and Q Zhao ldquoNew Jacobi ellipticfunction expansion and new periodic solutions of nonlinearwave equationsrdquo Physics Letters A vol 290 no 1-2 pp 72ndash762001
[14] M L Wang Y B Zhou and Z B Li ldquoApplication of a homoge-neous balancemethod to exact solutions of nonlinear equationsinmathematical physicsrdquo Physics Letters A General Atomic andSolid State Physics vol 216 no 1ndash5 pp 67ndash75 1996
[15] E G Fan andH Q Zhang ldquoThe homogeneous balancemethodfor solving nonlinear soliton equationsrdquoActa Physica Sinica vol47 no 3 pp 353ndash362 1998
[16] S D Liu Z T Fu S K Liu and Q Zhao ldquoEnvelopingperiodic solutions to nonlinear wave equations with Jacobielliptic functionsrdquo Acta Physica Sinica vol 51 no 4 pp 718ndash722 2002
[17] A J M Jawad M D Petkovic P Laketa and A BiswasldquoDynamics of shallow water waves with Boussinesq equationrdquoScientia Iranica B vol 20 no 1 pp 179ndash184 2013
[18] X L Yan ldquoExistence and uniqueness theorems of solution forsymmetric contracted operator equations and their applica-tionsrdquoChinese Science Bulletin vol 36 no 10 pp 800ndash805 1991
10 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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 Journal of Applied Mathematics
[19] J P Hao and X L Yan ldquoExact solution of large deformationbasic equations of circular membrane under central forcerdquoJournal of Applied Mathematics and Mechanics vol 27 no 10pp 1169ndash1172 2006
[20] X L Yan Exact Solution of Nonlinear Equations EconomicScientic Press Beijing China 2004
[21] Y-Z Peng ldquoExact solutions for some nonlinear partial differen-tial equationsrdquo Physics Letters A vol 314 no 5-6 pp 401ndash4082003
[22] L X Gong ldquoSome new exact solutions via the Jacobi ellipticfunctions to the nonlinear Schrodinger equationrdquo Acta PhysicaSinica vol 55 no 9 pp 4414ndash4419 2006
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