The G′G -expansion method for some nonlinear evolution equations

The G′G -expansion method for some nonlinear evolution equations

Applied Mathematics and Computation 217 (2010) 384–391 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepag...

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Applied Mathematics and Computation 217 (2010) 384–391

Contents lists available at ScienceDirect

Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

The

G0  G

-expansion method for some nonlinear evolution equations

S. Kutluay, A. Esen *, O. Tasbozan Department of Mathematics, Inonu University, Faculty of Arts and Science, 44280 Malatya, Turkey

a r t i c l e

i n f o

a b s t r a c t  0 In this paper, the GG -expansion method is applied to the Liouville, sine–Gordon and new coupled MKdV equations to obtain their some generalized exact travelling wave solutions. Ó 2010 Elsevier Inc. All rights reserved.

Keywords:  0 G G -expansion method Liouville equation Sine–Gordon equation Coupled MKdV equation

1. Introduction Mathematical modeling of many physical phenomena in various fields of physics and engineering generally leads to nonlinear ordinary or partial differential equations. It is known that investigating and constructing exact solutions of these equations are of great importance in applied mathematics. Therefore, in recent years, many effective methods such as sine–cosine method [1], tanh function method [2], variational iteration method [3], homotopy perturbation method [4–6], Exp-function method [7,8], F-expansion method [9,10], and others have been proposed for obtaining exact solutions to nonlinear partial differential equations. Some of these methods use transformations to reduce nonlinear equations into more simple equations, some others of methods use a trial function in an iterative scheme which converges rapidly to the exact solution and also there are still some other methods which consist of looking for the solution of the nonlinear evolution equations (NLEEs) considered as a polynomial in variable satisfying a subsidiary nonlinear ordinary differential equation.  0 Recently, Wang et al. [11] have proposed a method called the GG -expansion method to obtain exact travelling wave solutions of NLEEs arising in fluid dynamics, plasma, elastic media, optical fibers, etc. They successfully applied the method to find the travelling wave solutions involving parameters of the Korteweg–de Vries, the modified Korteweg–de Vries, the var 0 iant Boussinesq equations and the Hirota–Satsuma equations. The GG -expansion method has also successfully been applied G0  to various NLEEs  0 [12–16]. The key idea of the G -expansion method is to express the solution of NLLEs a finite series of the polynomial in GG satisfying the second order linear subsidiary ordinary differential equation G00 + kG0 + lG = 0. Before applying the method to the NLEEs considered in this paper, let us give the solution procedure of the method. 2. The basic idea behind the

G0  -expansion method G

To illustrate the basic idea behind this method, we consider the following nonlinear partial differential equation with only two independent variables x and t, and dependent variable u

Nðu; ut ; ux ; utt ; uxt ; uxx ; . . .Þ ¼ 0:

ð1Þ

Using the travelling wave transformation

u ¼ uðnÞ;

n ¼ x  wt;

* Corresponding author. E-mail address: [email protected] (A. Esen). 0096-3003/$ - see front matter Ó 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2010.05.073

ð2Þ

S. Kutluay et al. / Applied Mathematics and Computation 217 (2010) 384–391

385

Eq. (1) reduces to an ordinary differential equation (ODE) in the form

  N uðnÞ; wu0 ðnÞ; u0 ðnÞ; w2 u00 ðnÞ; wu00 ðnÞ; u00 ðnÞ; . . . ¼ 0:

ð3Þ

G0 

The G -expansion method is based on the assumption that travelling wave solutions of Eq. (3) can be expressed by a poly 0 nomial in GG as

uðnÞ ¼

 0 j n X G aj ; G j¼0

an – 0;

ð4Þ

with G = G(n) satisfying the second order linear ODE

G00 þ kG0 þ lG ¼ 0;

ð5Þ

where aj (j = 0, 1, 2,. . ., n), k and l are constants to be determined later. u(n) can be determined explicitly by using the following steps: Step 1. By considering the homogeneous balance between the highest nonlinear terms and the highest order derivatives of u(n) in Eq. (3), the positive integer n in Eq. (4) is determined.  0 Step 2. By substituting (4) with Eq. (5) into Eq. (3) and collecting all terms with the same power of GG together, the left hand side of Eq. (3) is converted into a polynomial. After setting each coefficient of this polynomial to zero, we obtain a set of algebraic equations in terms of aj (j = 0, 1, 2,. . ., n), w, k and l. Step 3. Solving the system of algebraic equations with the aid of symbolic computation and then substituting the results with the general solutions of Eq. (5) into Eq. (4) gives travelling wave solutions of Eq. (3). To illustrate the effectiveness and convenience of the method, we apply it to the Liouville, sine–Gordon and new coupled MKdV equations described in the next section. 3. Applications of the

G 0  -expansion method G

3.1. The Liouville equation We first consider the Liouville equation [17]

uxt þ eu ¼ 0:

ð6Þ

To look for travelling wave solutions of Eq. (6), we use the wave transformation n = x  wt and change Eq. (6) into the form of an ODE

wu00 þ eu ¼ 0:

ð7Þ

Using the transformation

u ¼ lnv ;

ð8Þ

Eq. (7) reduces to nonlinear ODE in the form

wðvv 00  v 02 Þ þ v 3 ¼ 0;

ð9Þ

where the prime denotes differentiation with respect to n. We now suppose that the solution of Eq. (9) is given by Eq. (4). In 00 order to determine the value of integer n, we balance vv with v3, then we get 2n + 2 = 3n which yields n = 2. Therefore, the 0 solution of (9) can be expressed by a polynomial in GG as follows:

v ðnÞ ¼ a0 þ a1

 0  0 2 G G þ a2 ; G G

a2 – 0:

ð10Þ

 0 i Substituting Eq. (10) along with (5) into Eq. (9), and setting the coefficients of all powers of GG ði ¼ 0; 1; . . . ; 6Þ to zero, we obtain a system of nonlinear algebraic equations for a0, a1, a2 and w. Solving the resulting system with the help of Maple, we have the following set of solution:

a0 ¼ 2wl;

a1 ¼ 2wk;

a2 ¼ 2w;

w ¼ w;

ð11Þ

where k and l are arbitrary constants. Inserting Eq. (11) into (10), we obtain

 0  0 2 G G : þ 2w G G

v ðnÞ ¼ 2wl þ 2wk

ð12Þ

Substituting the general solutions of Eq. (5) into Eq. (12) we have three types of travelling wave solutions of Eq. (9) as follows:

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S. Kutluay et al. / Applied Mathematics and Computation 217 (2010) 384–391

When k2  4l > 0

80 9 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 > > 2 2 1 1 < = c cosh  4 l sinh  4 l k n þ c k n 1 2 1 2 2 C qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A  1 : v 1 ðnÞ ¼ wðk2  4lÞ>B @ > 2 : c1 sinh 1 k2  4ln þ c2 cosh 1 k2  4ln ; 2 2

ð13Þ

When k2  4l < 0

80 9 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 > > 2 2 1 1 < = c sin l  k cos l  k 4 n þ c 4 n 1 2 1 2 2 C qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ 1 v 2 ðnÞ ¼  wðk2  4lÞ>B @ > 2 : c1 cos 1 4l  k2 n þ c2 sin 1 4l  k2 n ; 2 2

ð14Þ

and when k2  4l = 0

v 3 ðnÞ ¼

2wc22 ðc1 þ c2 nÞ2

ð15Þ

;

where n = x  wt, c1 and c2 arbitrary constants. In particular, if we choose c2 – 0; c21 < c22 , then the solution (13) give the solitary wave solution

v 1 ðx; tÞ ¼ 

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 1 wðk2  4lÞsech ðx  wtÞ k2  4l þ n0 ; 2 2

ð16Þ

1

where k2  4l > 0; n0 ¼ tanh ðcc12 Þ and the solution (14) give the travelling wave solution

v 2 ðx; tÞ ¼ 

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 wðk2  4lÞcsc2 ðx  wtÞ 4l  k2 þ n1 ; 2 2

ð17Þ

where k2  4l < 0; n1 ¼ tan1 ðcc12 Þ and recall that u(x, t) = lnv(x, t), hence we obtain the travelling wave solutions of the Liouville equation from Eqs. (16) and (17) as follows, respectively

 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 1 ðx  wtÞ k2  4l þ n0 ; u1 ðx; tÞ ¼ ln  wðk2  4lÞsech 2 2

w < 0;

ð18Þ

1

where k2  4l > 0 and n0 ¼ tanh ðcc12 Þ,



 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 ðx  wtÞ 4l  k2 þ n1 ; u2 ðx; tÞ ¼ ln  wðk2  4lÞcsc2 2 2

w > 0;

ð19Þ

where k2  4l < 0; n1 ¼ tan1 ðcc12 Þ and the third travelling wave solution from Eq. (15) is as follows

"

u3 ðx; tÞ ¼ ln

#

2wc22

ðc1 þ c2 ðx  wtÞÞ2

w > 0;

;

ð20Þ

where c1 and c2 are free parameters. Among these solutions, it is possible to obtain the solution given by Wazwaz [17] by taking n0 = 0 and some manipulation in Eq. (18). It should be noted that the other solutions derived here do not appear in Ref. [17]. 3.2. The sine–Gordon equation We secondly consider sine–Gordon equation [18–20]

utt  uxx þ sin u ¼ 0: In order to apply the 2

G0  G

ð21Þ

method, we use the wave transformation n = x  wt and change Eq. (21) into the form

00

ðw  1Þu þ sin u ¼ 0:

ð22Þ

We next use the transformation

v ¼ eiu ;

ð23Þ

so that

sin u ¼ which gives

v  v 1 2i

;

cos u ¼

v þ v 1 2

;

ð24Þ

S. Kutluay et al. / Applied Mathematics and Computation 217 (2010) 384–391

u ¼ arccos



v þ v 1 2

:

387

ð25Þ

This transformation will change Eq. (22) into the ODE in the form

2ðw2  1Þðv 00 v  ðv 0 Þ2 Þ þ v 3  v ¼ 0;

ð26Þ

where the prime denotes differentiation with respect to n. We now suppose that the solution of Eq. (26) is given by Eq. (4). In order to determine the value of integer n, we balance vv00 with v3, we get 2n + 2 = 3n which yields n = 2. Thus the solution of Eq. (26) is exactly the same with Eq. (10).  0 i Substituting Eq. (10) along with (5) into (26) and setting the coefficients of all powers of GG ði ¼ 0; 1; . . . ; 6Þ to zero, we obtain a system of nonlinear algebraic equations for a0, a1, a2 and w. Solving the resulting system with the help of Maple, we have the following sets of solutions:

a0 ¼

k2 k 2  4l

a1 ¼

;

4k k 2  4l

;

a2 ¼

4 k2  4l

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 w¼ 1 2 k  4l

;

ð27Þ

and

a0 ¼ 

k2 k 2  4l

a1 ¼ 

;

4k k2  4l

;

a2 ¼ 

4 k 2  4l

;

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 : w¼ 1þ 2 k  4l

ð28Þ

Inserting Eqs. (27) and (28) into Eq. (10), we get

v 1 ðnÞ ¼

k2 2

k  4l

þ

 0  0 2 G 4 G þ 2 k  4l G k  4l G 4k

2

ð29Þ

and

v 2 ðnÞ ¼ 

k2 k2  4l



 0  0 2 G 4 G ;  k 2  4l G k2  4l G 4k

ð30Þ

respectively. Using the general solutions of (5), we obtain travelling wave solutions of Eq. (26) as follows When k2  4l > 0

0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 2 1 1 c cosh  4 l sinh k n þ c k2  4lnC 1 2 2 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ; v 1 ðx; tÞ ¼ B @ c1 sinh 12 k2  4ln þ c2 cosh 12 k2  4ln where n ¼ x 

ð31Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1  k2 4 t and l

0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 2 1 1 c cosh  4 l sinh k n þ c k2  4lnC 1 2 2 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ; v 2 ðx; tÞ ¼ B @ c1 sinh 12 k2  4ln þ c2 cosh 12 k2  4ln where n ¼ x 

ð32Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 þ k2 4 t; c1 and c2 arbitrary constants. l

When k2  4l < 0

0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 2 1 1 c sin l  k cos 4 n þ c 4l  k2 nC 1 2 2 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ; v 3 ðx; tÞ ¼ B @ c1 cos 12 4l  k2 n þ c2 sin 12 4l  k2 n where n ¼ x 

ð33Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1  k2 4 t and l

0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 2 2 1 1 Bc1 sin 2 4l  k n þ c2 cos 2 4l  k nC qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ; v 4 ðx; tÞ ¼ @ c1 cos 12 4l  k2 n þ c2 sin 12 4l  k2 n

ð34Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 where n ¼ x  ð 1 þ k2 4 Þt; c1 and c2 arbitrary constants. In particular, if we choose c2 –0; c21 < c22 , then the solutions (31)– l (34) give the travelling wave solutions

 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1 k2  4ln þ n0 ; 2  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 1 k2  4ln þ n0 ; v 2 ðx; tÞ ¼ tanh 2

v 1 ðx; tÞ ¼ tanh2

ð35Þ ð36Þ

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S. Kutluay et al. / Applied Mathematics and Computation 217 (2010) 384–391 1

where n0 ¼ tanh ðcc12 Þ; k2  4l > 0 and

 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1 4l  k2 n þ n1 ; 2  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1 4l  k 2 n þ n 1 ; v 4 ðx; tÞ ¼ cot2 2

v 3 ðx; tÞ ¼ cot2

where n1 ¼ tan1

  c1 c2

ð37Þ ð38Þ

. Using Eq. (25), we get the following travelling wave solutions of Eq. (21)

 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 

1 2 1 2 1 ; tanh k2  4ln þ n0 þ coth k2  4ln þ n0 2 2 2  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 

1 2 1 2 1 ; tanh k2  4ln þ n0 þ coth k2  4ln þ n0 u2 ðx; tÞ ¼ arccos  2 2 2  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 

1 1 1 u3 ðx; tÞ ¼ arccos  tan2 4l  k2 n þ n1 þ cot2 4l  k2 n þ n1 2 2 2

u1 ðx; tÞ ¼ arccos

ð39Þ ð40Þ ð41Þ

and

u4 ðx; tÞ ¼ arccos

 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 

1 1 1 ; tan2 4l  k2 n þ n1 þ cot2 4l  k2 n þ n1 2 2 2

ð42Þ

1

where n0 ¼ tanh ðcc12 Þ and n1 ¼ tan1 ðcc12 Þ. Comparing our results with Wazwaz’s results in Ref. [18], we can see that the results are the same, if we chose n0 and n1 zero, for appropriate parameters. 3.3. The new coupled MKdV equation We finally consider the new coupled MKdV equation [21]

  1 uxxx  3u2 ux þ 3ðvv x Þx þ 3 uv 2 x ; 2   ¼ v xxx  3ðv ux Þx þ 6uv ux þ 3 u2  v 2 v x ;

ut ¼

vt

ð43Þ

where Eq. (43) becomes the MKdV equation for v = 0. In Ref. [20], Wu et al. take into account a 4  4 matrix spectral problem with three potentials and obtain a new hierarchy of nonlinear evolution equations. In Ref. [22], Cao et al. proposed some kinds of soliton solutions for Eq. (43). Substituting the wave transformation u(x, t) = u(n), v(x, t) = v(n), n = x  wt into Eq. (43), integrating the resulting equation with respect to n yields the ODEs

1 00 u  u3 þ 3vv 0 þ 3uv 2 ; 2 c2  wv ¼ v 00  3v u0  v 3 þ 3u2 v ;

c1  wu ¼

ð44Þ

where c1 and c2 are integration constants that are to be determined later. Considering the homogenous balance between u00 and uv2 and between v00 and u2v in Eq. (44), we get n + 2 = n + 2m and m + 2 = 2n + m which yield n = m = 1, then we suppose that

 0 G ; G  0 G v ðnÞ ¼ b0 þ b1 : G uðnÞ ¼ a0 þ a1

ð45Þ

 0 i Substituting Eq. (45) along with (5) into Eq. (44), collecting the coefficients of GG (i = 0, 1, 2, 3) for each equation and setting it to zero, we obtain a system of nonlinear algebraic equations. Solving the resulting system, we have the following sets of solutions: 2

3

c1 ¼ k2 b0 þ 6kb0  8b0 þ

1 lk  2b0 l; 2

1 lk þ 2b0 l; 2   1 2 1 2 k þ 2l  12kb0 þ 24b0 ; a0 ¼ k  b0 ; w¼ 4 2 2

3

c2 ¼ k2 b0  6kb0 þ 8b0 

and

ð46Þ a1 ¼

1 ; 2

b1 ¼

1 2

S. Kutluay et al. / Applied Mathematics and Computation 217 (2010) 384–391

1 lk þ 2b0 l; 2 1 2 3 c2 ¼ k2 b0 þ 6kb0 þ 8b0 þ lk þ 2b0 l; 2  1 2 1 2 k þ 2l þ 12kb0 þ 24b0 ; a0 ¼ k þ b0 ; w¼ 4 2 2

389

3

c1 ¼ k2 b0 þ 6kb0 þ 8b0 þ

ð47Þ a1 ¼

1 ; 2

1 b1 ¼  : 2

Inserting Eqs. (46) and (47) into (45), we get

  1 G0 k þ  b0 ; 2 G 2   1 G0 þ b0 v 1 ðnÞ ¼ 2 G

ð48Þ

  1 G0 k þ þ b0 ; 2 G 2   1 G0 þ b0 ; v 2 ðnÞ ¼  2 G

ð49Þ

u1 ðnÞ ¼

and

u2 ðnÞ ¼

respectively. Using the general solutions of (5), we obtain travelling wave solutions of Eq. (44) as follows: when k2  4l > 0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2  4l BA1 cosh 12 k2  4ln þ A2 sinh 12 k2  4lnC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ  b0 ; u1 ðnÞ ¼ @ 4 4 A1 sinh 12 k2  4ln þ A2 cosh 12 k2  4ln 0 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2  4l BA1 cosh 12 k2  4ln þ A2 sinh 12 k2  4lnC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A  þ b0 ; v 1 ðnÞ ¼ @ 4 4 A1 sinh 12 k2  4ln þ A2 cosh 12 k2  4ln

ð50Þ

  2 where n ¼ x  14 k2 þ 2l  12kb0 þ 24b0 t and

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2  4l BA1 cosh 12 k2  4ln þ A2 sinh 12 k2  4lnC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ þ b0 ; u2 ðnÞ ¼ @ 4 4 A1 sinh 12 k2  4ln þ A2 cosh 12 k2  4ln qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2  4l BA1 cosh 12 k2  4ln þ A2 sinh 12 k2  4lnC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ þ b0 ; v 2 ðnÞ ¼  @ 4 4 A1 sinh 12 k2  4ln þ A2 cosh 12 k2  4ln

ð51Þ

  2 where n ¼ x  14 k2 þ 2l þ 12kb0 þ 24b0 t, A1 and A2 arbitrary constants. When k2  4l < 0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4l  k2 BA1 sin 12 4l  k2 n þ A2 cos 12 4l  k2 nC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ  b0 ; u3 ðnÞ ¼ @ 4 4 A1 cos 12 4l  k2 n þ A2 sin 12 4l  k2 n 0 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4l  k2 BA1 sin 12 4l  k2 n þ A2 cos 12 4l  k2 nC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A  þ b0 ; v 3 ðnÞ ¼ @ 4 4 A1 cos 12 4l  k2 n þ A2 sin 12 4l  k2 n

ð52Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4l  k2 BA1 sin 12 4l  k2 n þ A2 cos 12 4l  k2 nC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ þ b0 ; u4 ðnÞ ¼ @ 4 4 A1 cos 12 4l  k2 n þ A2 sin 12 4l  k2 n 0 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4l  k2 BA1 sin 12 4l  k2 n þ A2 cos 12 4l  k2 nC k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ þ b0 ; v 4 ðnÞ ¼  @ 4 4 A1 cos 12 4l  k2 n þ A2 sin 12 4l  k2 n

ð53Þ

  2 where n ¼ x  14 k2 þ 2l  12kb0 þ 24b0 t and

  2 where n ¼ x  14 k2 þ 2l þ 12kb0 þ 24b0 t, A1 and A2 arbitrary constants.

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S. Kutluay et al. / Applied Mathematics and Computation 217 (2010) 384–391

When k2  4l = 0

  1 2A2 k þ  b0 ; k 4 2 A1 þ A2 n   1 2A2 v 5 ðnÞ ¼  k  þ b0 ; 4 A1 þ A2 n

ð54Þ

  1 2A2 k k þ þ b0 ; 4 2 A1 þ A2 n   1 2A2 v 6 ðnÞ ¼ k  þ b0 ; 4 A1 þ A2 n

ð55Þ

u5 ðnÞ ¼ 

  2 where n ¼ x  14 k2 þ 2l  12kb0 þ 24b0 t and

u6 ðnÞ ¼ 

  2 where n ¼ x  14 k2 þ 2l þ 12kb0 þ 24b0 t; A1 and A2 arbitrary constants. In particular, if we choose A2 – 0; A21 < A22 , then the solutions (50)–(53) lead to the travelling wave solutions as follows: when k2  4l > 0

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 k þ  b0 ; k  k2  4l tanh k2  4lðx  wtÞ þ n0 4 2 2   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 þ b0 ; k2  4lðx  wtÞ þ n0 v 1 ðx; tÞ ¼  k  k2  4l tanh 4 2

ð56Þ

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 k þ þ b0 ; k  k2  4l tanh k2  4lðx  wtÞ þ n0 4 2 2   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 þ b0 ; k2  4lðx  wtÞ þ n0 v 2 ðx; tÞ ¼ k  k2  4l tanh 4 2

ð57Þ

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 k þ  b0 ; k  4l  k2 cot 4l  k2 ðx  wtÞ þ n1 4 2 2   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 þ b0 ; 4l  k2 ðx  wtÞ þ n1 v 3 ðx; tÞ ¼  k  4l  k2 cot 4 2

ð58Þ

  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 k þ þ b0 ; k  4l  k2 cot 4l  k2 ðx  wtÞ þ n1 4 2 2   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 þ b0 ; 4l  k2 ðx  wtÞ þ n1 v 4 ðx; tÞ ¼ k  4l  k2 cot 4 2

ð59Þ

u1 ðx; tÞ ¼ 

  2 where w ¼ 14 k2 þ 2l  12kb0 þ 24b0 and

u2 ðx; tÞ ¼ 

    2 1 A1 where w ¼ 14 k2 þ 2l þ 12kb0 þ 24b0 ; n0 ¼ tanh . A2 When k2  4l < 0

u3 ðx; tÞ ¼ 

  2 where w ¼ 14 k2 þ 2l  12kb0 þ 24b0 and

u4 ðx; tÞ ¼ 

    2 where w ¼ 14 k2 þ 2l þ 12kb0 þ 24b0 ; n1 ¼ tan1 AA12 . 4. Conclusion

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