Exact solutions for the nonlinear Schrödinger equation with variable coefficients using the generalized extended tanh-function, the sine–cosine and the exp-function methods

Exact solutions for the nonlinear Schrödinger equation with variable coefficients using the generalized extended tanh-function, the sine–cosine and the exp-function methods

Applied Mathematics and Computation 218 (2011) 2259–2268 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homep...

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Applied Mathematics and Computation 218 (2011) 2259–2268

Contents lists available at ScienceDirect

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

Exact solutions for the nonlinear Schrödinger equation with variable coefficients using the generalized extended tanh-function, the sine–cosine and the exp-function methods E.M.E. Zayed ⇑, M.A.M. Abdelaziz Mathematics Department, Faculty of Science, Zagazig University, Zagazig, Egypt

a r t i c l e

i n f o

Keywords: The generalized nonlinear Schrödinger equation with variable coefficients Generalized extended tanh-function method Sine–cosine method Exp-function method Exact solutions

a b s t r a c t In this article we find the exact traveling wave solutions of the generalized nonlinear Schrödinger (GNLS) equation with variable coefficients using three methods via the generalized extended tanh-function method, the sine–cosine method and the exp-function method. The main objective of this article is to compare the efficiency of these methods by delivering the exact traveling wave solutions of the proposed nonlinear equation. Ó 2011 Elsevier Inc. All rights reserved.

1. Introduction In the nonlinear science, many important phenomena in various fields can be described by the nonlinear evolution equations (NLEES). Searching for exact soliton solutions of NLEEs plays an important and a significant role in the study on the dynamics of those phenomena. With the development of soliton theory, many powerful methods for obtaining the exact solutions of NLEES have been presented, such as the extended tanh-function method [1–5], the tanh-sech method [6–8], the sine–cosine method [9–11], the homogeneous balance method [12,13], the exp-function method [14–17], the Jacobi elliptic function method [18–21], the F-expansion method [22], the homotopy perturbation method [23,24], the variational iteration method [25], the inverse scattering transformation method [26], the Bäcklund transformation method [27], the Hirota bilinear method [28,29] and so on. To our knowledge, most of the aforementioned methods are related to constant coefficients models. Recently, much attention has been paid to the variable-coefficient nonlinear equations which can describe many nonlinear phenomena more realistically than their constant-coefficient ones. In the present article, we discuss two important objectives. Firstly, we apply the generalized extended tanh-function method, the sine–cosine method and the exp-function method to find the exact solutions of the following generalized nonlinear Schrödinger (GNLS) equation with variable coefficients [30]:

1 iux þ bðxÞutt þ aðxÞujuj2  icðxÞu ¼ 0; 2

ð1:1Þ

pffiffiffiffiffiffiffi where u = u(x, t) is a real or complex valued function of x, t and i ¼ 1. Secondly, we are comparing the efficiency of these methods. The coefficients a(x), b(x) and c(x) in Eq. (1.1) are functions of the indicated variable x, satisfying the condition a(x)b(x) / c(x)2 in order that Eq. (1.1) has the given solutions presented in this paper.

⇑ Corresponding author. E-mail addresses: [email protected] (E.M.E. Zayed), [email protected] (M.A.M. Abdelaziz). 0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.07.043

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2. Description of the generalized extended tanh-function method Suppose that a nonlinear evolution equation is given by

Fðu; ut ; ux ; utt ; uxx ; . . .Þ ¼ 0;

ð2:1Þ

where u = u(x, t) is an unknown function, F is a polynomial in u and its partial derivatives, in which the highest order derivatives and non-linear terms are involved. In the following, we give the main steps of this method: Step 1. Using the generalized wave transformation

uðx; tÞ ¼ uðnÞ;

n ¼ pðxÞt þ qðxÞ;

ð2:2Þ

where p(x) and q(x) are differentiable functions of x to be determined. Then, Eq. (2.1) is reduced to the following ODE:

Qðu; pðxÞu0 ðnÞ; ½p0 ðxÞt þ q0 ðxÞu0 ðnÞ; pðxÞ2 u00 ðnÞ; . . .Þ ¼ 0;

ð2:3Þ

where Q is a polynomial in u and its total derivatives, while 0 denotes the derivative with respect to the indicated variable. Step 2. We suppose that Eq. (2.3) has the following formal solution:

uðnÞ ¼ a0 þ

N h i X ak Y k ðnÞ þ ak Y k ðnÞ ;

ð2:4Þ

i¼0

where N is a positive integer, and a0, ak, ak are constants, while Y(n) is given by

YðnÞ ¼ tanhðnÞ:

ð2:5Þ

The independent variable (2.5) leads to the following derivatives:

d d ¼ ð1  Y 2 Þ ; dn dY " # 2 2 d d 2Þ 2 d þ ð1  Y Þ 2 ; ¼ ð1  Y 2Y dY dn2 dY " # 3 2 3 d d 2Þ 2 2 d 2 2 d þ 6Yð1  Y Þ 2 þ ð1  Y Þ ¼ ð1  Y ð6Y  2Þ ; 3 dY dn3 dY dY

ð2:6Þ

and so on. Step 3. Determine the positive integer N in (2.4) by balancing the highest order derivatives and nonlinear terms in Eq. (2.3). Step 4. Substituting (2.4) along with (2.6) into (2.3) and equating the coefficients of tjYs(n) to zero, we get a system of algebraic equations. Step 5. Solving these algebraic equations by Maple or Mathematica, we get the values of a0, ak, ak, p(x) and q(x). Step 6. Substituting these values into (2.4) and (2.2), we can obtain the exact traveling wave solutions of Eq. (2.1). 3. On using the generalized extended tanh-function method to solve Eq. (1.1) Assume that the solution of Eq. (1.1) can be written in the form

uðx; tÞ ¼ tðx; tÞ exp½ihðx; tÞ;

ð3:1Þ

where t(x, t) and h(x, t) are amplitude and phase functions respectively. Substituting (3.1) into (1.1) and separating the real and imaginary parts, we obtain

1 thx þ bðxÞ½ttt  tðht Þ2  þ aðxÞt3 ¼ 0 2

ð3:2Þ

and

1 2

tx þ bðxÞ½2tt ht þ thtt   cðxÞt ¼ 0:

ð3:3Þ

Balancing ttt and t3 in Eq. (3.2), we have N = 1. We assume that Eqs. (3.2) and (3.3) have the following formal solutions

tðnÞ ¼ a0 þ a1 YðnÞ þ a1 Y 1 ðnÞ; 2

hðx; tÞ ¼ f ðxÞt þ gðxÞt þ hðxÞ:

ð3:4Þ ð3:5Þ

E.M.E. Zayed, M.A.M. Abdelaziz / Applied Mathematics and Computation 218 (2011) 2259–2268

2261

where f(x), g(x) and h(x) are differentiable functions to be determined. From (2.6) and (3.4) we have the derivatives

t0 ¼ ½a1 Y 2 ðnÞ þ a1 Y 2 ðnÞ þ a1 þ a1 ; t00 ¼ 2½a1 Y 3 ðnÞ þ a1 Y 3 ðnÞ  2½a1 YðnÞ þ a1 Y 1 ðnÞ; t000 ¼ 6½a1 Y 4 ðnÞ þ a1 Y 4 ðnÞ þ 8½a1 Y 2 ðnÞ þ a1 Y 2 ðnÞ  2½a1 þ a1 ;

ð3:6Þ ð3:7Þ ð3:8Þ

and so on. Substituting (3.4)–(3.8) along with (2.2) into (3.2) and (3.3), collecting the coefficients of tjYs(n), (i = 0, 1, 2, s = 3, 2, 1, 0, 1, 2, 3) and setting each coefficient to zero, we obtain the following system of algebraic equations:

t 0 Y 3 : a1 ½bðxÞpðxÞ2 þ a31 aðxÞ ¼ 0; t 0 Y 3 : a1 ½bðxÞpðxÞ2 þ a31 aðxÞ ¼ 0; t 0 Y 2 : 3a0 a21 aðxÞ ¼ 0; t0 Y

2

a1 ½q0 ðxÞ þ bðxÞpðxÞgðxÞ ¼ 0;

: 3a0 a21 aðxÞ ¼ 0;

a1 ½q0 ðxÞ þ bðxÞpðxÞgðxÞ ¼ 0; 1 0 t 0 Y 1 : a1 ½h ðxÞ þ bðxÞpðxÞ2 þ bðxÞgðxÞ2  3aðxÞða1 a1 þ a20 Þ ¼ 0; : a1 ½bðxÞf ðxÞ  cðxÞ ¼ 0; 2 1 0 2 0 1 t Y : a1 ½h ðxÞ þ bðxÞpðxÞ þ bðxÞgðxÞ2  3aðxÞða1 a1 þ a20 Þ ¼ 0; : a1 ½bðxÞf ðxÞ  cðxÞ ¼ 0; 2 1 0 t 0 Y 0 : a0 ½h ðxÞ þ bðxÞgðxÞ2  aðxÞð6a1 a1 þ a20 Þ ¼ 0; : ða1 þ a1 Þ½q0 ðxÞ þ bðxÞpðxÞgðxÞ þ a0 ½bðxÞf ðxÞ  cðxÞ ¼ 0; 2 t 1 Y 2 : a1 ½p0 ðxÞ þ 2bðxÞpðxÞf ðxÞ ¼ 0; t 1 Y 2 : a1 ½p0 ðxÞ þ 2bðxÞpðxÞf ðxÞ ¼ 0; t 1 Y 1 : a1 ½g 0 ðxÞ þ 2bðxÞgðxÞf ðxÞ ¼ 0; t 1 Y 1 : a1 ½g 0 ðxÞ þ 2bðxÞgðxÞf ðxÞ ¼ 0; t 1 Y 0 : a0 ½g 0 ðxÞ þ 2bðxÞgðxÞf ðxÞ ¼ 0;

: ða1 þ a1 Þ½p0 ðxÞ þ 2bðxÞpðxÞf ðxÞ ¼ 0;

t 2 Y 1 : a1 ½f 0 ðxÞ þ 2bðxÞf ðxÞ2  ¼ 0; t 2 Y 1 : a1 ½f 0 ðxÞ þ 2bðxÞf ðxÞ2  ¼ 0; t 2 Y 0 : a0 ½f 0 ðxÞ þ 2bðxÞf ðxÞ2  ¼ 0: ð3:9Þ Solving the algebraic Eq. (3.9) by Maple, we have the following cases of solutions: Case1.

a0 ¼ 0; hðxÞ ¼

a1 ¼ 0;

a1 ¼ a1 ;

1 2 ðc þ 2c22 Þf ðxÞ þ c4 ; 4 1

pðxÞ ¼ c2 f ðxÞ;

cðxÞ ¼ bðxÞf ðxÞ;

1 c1 c2 f ðxÞ þ c3 ; 2 c2 bðxÞf ðxÞ2 aðxÞ ¼ 2 2 ; a1

qðxÞ ¼

gðxÞ ¼ c1 f ðxÞ; ð3:10Þ

where ci(i = 15) are arbitrary constants, such that c2 – 0 and f(x) is given by

 Z  f ðxÞ 2 bðxÞdx þ c5 ¼ 1:

ð3:11Þ

In this case, the exact solution of Eq. (1.1) has the from:

   1 uðnÞ ¼ a1 cothðnÞ exp if ðxÞ t 2 þ c1 t þ ðc21 þ 2c22 Þ : 4

ð3:12Þ

Case2.

a0 ¼ 0; hðxÞ ¼

a1 ¼ a1 ;

a1 ¼ 0;

1 2 ðc þ 2c22 Þf ðxÞ þ c4 ; 4 1

pðxÞ ¼ c2 f ðxÞ;

cðxÞ ¼ bðxÞf ðxÞ;

1 c1 c2 f ðxÞ þ c3 ; 2 c2 bðxÞf ðxÞ2 aðxÞ ¼ 2 2 : a1

qðxÞ ¼

gðxÞ ¼ c1 f ðxÞ; ð3:13Þ

In this case, the exact solution of Eq. (1.1) has the from:

   1 uðnÞ ¼ a1 tanhðnÞ exp ifðxÞ t2 þ c1 t þ ðc21 þ 2c22 Þ : 4

ð3:14Þ

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Case3.

a0 ¼ 0; hðxÞ ¼

a1 ¼ a1 ;

a1 ¼ a1 ;

1 2 ðc  4c22 Þf ðxÞ þ c4 ; 4 1

pðxÞ ¼ c2 f ðxÞ;

cðxÞ ¼ bðxÞf ðxÞ;

1 c1 c2 f ðxÞ þ c3 ; 2 2 c bðxÞf ðxÞ2 aðxÞ ¼ 2 2 : a1 qðxÞ ¼

gðxÞ ¼ c1 f ðxÞ; ð3:15Þ

In this case, the exact solution of Eq. (1.1) has the from:

   1 uðnÞ ¼ a1 ½tanhðnÞ  cothðnÞ exp if ðxÞ t 2 þ c1 t þ ðc21  4c22 Þ : 4

ð3:16Þ

which can be written in the form

     ip 1 uðnÞ ¼ 2ia1 sech 2n þ exp if ðxÞ t2 þ c1 t þ ðc21  4c22 Þ ; 4 2

ð3:17Þ

where



1 c2 f ðxÞðc1 þ 2tÞ þ c3 : 2

ð3:18Þ

Remark 1. The following observations are a consequence of the remarks in the [31]: 1. We note that n given by (3.18) contains an arbitrary constant c3. So, it is always possible to phase shift n by an arbitrary constant amount. 2. We note the following identities: (i) tanhh + cothh = 2coth (2h), (ii) tanh h  coth h ¼ 2isechð2h þ i2pÞ, (iii) coth (ih) = icoth, (iv) tanh (ih) = itanh, (v) sech (ih) = sech. 3. The multiplier of the exp-function in the expressions (3.12), (3.14) and (3.17) can be written in the form AY where A is an arbitrary constant and Y = cothn in case 1, Y = tanhn in case 2 and Y = sechn in case 3 having the identity (ii) and the transformation c2 ! c22 . 4. If we use the identities (iii), (iv) and (v) and the transformation c2 ? ic2, then Y can also be cotn, tann and secn respectively. Remark 2. From cases 1, 2 and 3 we deduce that a(x) / b (x)f(x)2. But c(x) = b(x)f(x), then we conclude that

aðxÞbðxÞ / cðxÞ2 :

ð3:19Þ

4. Description of the sine–cosine method In this method, we admit the use of the relation (3.1), where t (n) has the form

tðnÞ ¼ kðxÞ sinm ðnÞ; jnj <

p 2

ð4:1Þ

or

p tðnÞ ¼ kðxÞ cosm ðnÞ; jnj < ; 2

ð4:2Þ

where m is a constant while k(x) is a differentiable function of x to be determined later. From (4.1) and (4.2) we have

tx ¼

  dkðxÞ dpðxÞ dqðxÞ m m1 sin ðnÞ þ mkðxÞ tþ sin ðnÞ cosðnÞ; dx dx dx

tt ¼ mkðxÞpðxÞ sinm1 ðnÞ cosðnÞ; ttt ¼ mkðxÞpðxÞ2 ½m sinm ðnÞ þ ðm  1Þ sinm2 ðnÞ

ð4:3Þ

E.M.E. Zayed, M.A.M. Abdelaziz / Applied Mathematics and Computation 218 (2011) 2259–2268

2263

or

tx ¼

  dkðxÞ dpðxÞ dqðxÞ cosm1 ðnÞ sinðnÞ; cosm ðnÞ  mkðxÞ tþ dx dx dx

tt ¼ mkðxÞpðxÞ cosm1 ðnÞ sinðnÞ;

ttt ¼ mkðxÞpðxÞ2 m cosm ðnÞ þ ðm  1Þ cosm2 ðnÞ :

ð4:4Þ

From (3.5) we have

df ðxÞ 2 dgðxÞ dhðxÞ t þ tþ ; dx dx dx ht ¼ 2f ðxÞt þ gðxÞ;

hx ¼

ð4:5Þ

htt ¼ 2f ðxÞ:

ð4:7Þ

ð4:6Þ

Substituting (4.1), (4.3) and (4.5)–(4.7) along with (3.5) or (4.2), (4.4) and (4.5)–(4.7) along with (3.5) into (3.2) and (3.3) a trigonometric equations are obtained with either tnsinR(n) or tncosR(n) terms so that the parameter R can be determined by comparing exponents. Equating the coefficients of tnsinR(n) and tncosR(n) to zero, we get a system of algebraic equations. Solving these algebraic equations by Maple or Mathematica, we get the values of m, k(x), p(x), q(x), f(x), g(x), and h(x). Substituting these values into (3.1), we can obtain the exact traveling wave solutions of Eq. (1.1). 5. On using the sine–cosine method to solve Eq. (1.1) In this section, we shall derive the exact solutions of Eq. (1.1) using the sine–cosine method. To this end, we first apply the cosine method to Eq. (1.1). Substituting (4.2), (4.4) and (4.5)–(4.7) along with (3.5) into (3.2) and (3.3), we have

  dkðxÞ dpðxÞ dqðxÞ cosm ðnÞ  mkðxÞ tþ cosm1 ðnÞ sinðnÞ  mkðxÞbðxÞpðxÞ½2f ðxÞt þ gðxÞ cosm1 ðnÞ sinðnÞ dx dx dx þ kðxÞbðxÞf ðxÞ cosm ðnÞ  kðxÞcðxÞ cosm ðnÞ ¼ 0

ð5:1Þ

df ðxÞ 2 dgðxÞ 1 1 t cosm ðnÞ  kðxÞ t cosm ðnÞ  m2 kðxÞbðxÞpðxÞ2 cosm ðnÞ þ mðm  1ÞkðxÞbðxÞpðxÞ2 cosm2 ðnÞ dx dx 2 2 dhðxÞ 1 2 2 2 m m m  kðxÞ cos ðnÞ  2kðxÞbðxÞf ðxÞ t cos ðnÞ  kðxÞbðxÞgðxÞ cos ðnÞ  2kðxÞbðxÞf ðxÞgðxÞt cosm ðnÞ dx 2 þ kðxÞ3 aðxÞ cos3m ðnÞ ¼ 0;

ð5:2Þ

and

 kðxÞ

which are satisfied only if the following conditions hold:

8 dkðxÞ > > > dx þ kðxÞbðxÞf ðxÞ  kðxÞcðxÞ ¼ 0; < > > > :

mkðxÞ dqðxÞ  mkðxÞbðxÞpðxÞgðxÞ ¼ 0; dx mkðxÞ dpðxÞ dx

ð5:3Þ

 2mkðxÞbðxÞpðxÞf ðxÞ ¼ 0

and

8 kðxÞ dfdxðxÞ  2kðxÞbðxÞf ðxÞ2 ¼ 0; > > > > > > > > kðxÞ dgðxÞ  2kðxÞbðxÞf ðxÞgðxÞ ¼ 0; > dx > > > > dhðxÞ < kðxÞ dx  12 m2 kðxÞbðxÞpðxÞ2 > > >  12 kðxÞbðxÞgðxÞ2 ¼ 0; > > > > > 1 > > mðm  1ÞkðxÞbðxÞpðxÞ2 þ kðxÞ3 aðxÞ ¼ 0; > 2 > > : m  2 ¼ 3m:

ð5:4Þ

Similar results are also obtained by using the sine method which are omitted here. Solving the algebraic Eqs. (5.3) and (5.4) by Maple, we have the following cases of solutions:

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Case1.

Z qðxÞ ¼ c1 c2 bðxÞdx þ c5 ; f ðxÞ ¼ 0; Z 1 gðxÞ ¼ c1 ; hðxÞ ¼  ðc21 þ c22 Þ bðxÞdx þ c4 ; cðxÞ ¼ cðxÞ; 2  2 R c2 2 cðxÞdx aðxÞ ¼  bðxÞe ; bðxÞ ¼ bðxÞ; m ¼ 1; c3 kðxÞ ¼ c3 e

R

cðxÞdx

;

pðxÞ ¼ c2 ;

ð5:5Þ

where ci(i = 15) are arbitrary constants, such that c1, c2 and c3 – 0. In this case, the exact solutions of Eq. (1.1) have the forms:

 Z   Z 1 uðnÞ ¼ c3 secðnÞ exp i c1 t  ðc21 þ c22 Þ bðxÞdx þ c4 þ cðxÞdx ; 2

ð5:6Þ

where

  Z n ¼ c2 t  c1 bðxÞdx þ c5 :

ð5:7Þ

Case2.

kðxÞ ¼

qffiffiffiffiffiffiffiffiffi R f ðxÞe cðxÞdx ;

pðxÞ ¼ c2 f ðxÞ; qðxÞ ¼

1 hðxÞ ¼ ðc21 þ c22 Þf ðxÞ þ c4 ; R4 aðxÞ ¼ c22 bðxÞf ðxÞe2 cðxÞdx ; bðxÞ ¼ bðxÞ; gðxÞ ¼ c1 f ðxÞ;

1 c1 c2 f ðxÞ þ c3 ; 2

cðxÞ ¼ cðxÞ

ð5:8Þ

m ¼ 1;

where ci(i = 14) are arbitrary constants, such that c1, c2 – 0 and f(x) has the same form (3.11). In this case, the exact solutions of Eq. (1.1) have the forms:

uðnÞ ¼

   Z  qffiffiffiffiffiffiffiffiffi 1 f ðxÞ secðnÞ exp if ðxÞ t 2 þ c1 t þ ðc21 þ c22 Þ þ cðxÞdx ; 4

ð5:9Þ

where n is given by (3.18). Remark 3 (i) When k(x) is a constant we deduce that the solution (5.9) of case 2 reduces to the solution (3.16) of case 3 in Section 3. R To show this, we see that if k(x) is a constant then, we deduce from the conditions (5.8) that f ðxÞ / e2 cðxÞdx ; aðxÞ / bðxÞf 2 ðxÞ and c(x) = b(x)f(x). Thus the conditions (5.8) of case 2 agree with the conditions (3.15) of case 3 in Section 3. (ii) By the use of the transformation c2 ? ic2 we get sec (n) ? sech (n). 6. Description of the exp-function method and its use to solve Eq. (1.1) The exp-function method was first proposed by He and Wu [14]. Using the generalized wave transformation (2.2), Eqs. (3.2) and (3.3) become

1 1 0 tðnÞft2 ½f 0 ðxÞ þ 2bðxÞf ðxÞ2  þ t½g 0 ðxÞ þ 2bðxÞf ðxÞgðxÞ þ bðxÞgðxÞ2 þ h ðxÞg þ bðxÞpðxÞ2 t00 ðnÞ þ aðxÞt3 ðnÞ ¼ 0 2 2

ð6:1Þ

t0 ðnÞftp0 ðxÞ þ q0 ðxÞ þ ½2tf ðxÞ þ gðxÞbðxÞpðxÞg þ tðnÞ½bðxÞf ðxÞ  cðxÞ ¼ 0:

ð6:2Þ

and

In this method, we admit the use of the relation (3.1), where t (n) has the form

Pd

tðnÞ ¼ Pqn¼c

an expðnnÞ ; expðmnÞ

m¼p bm

ð6:3Þ

where c, d, p and q are positive integers which are unknown to be determine later, an and bm are unknown constants. Eq. (6.3) can be written in the form

tðnÞ ¼

ac expðcnÞ þ . . . þ ad expðdnÞ : ap expðpnÞ þ . . . þ aq expðqnÞ

ð6:4Þ

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In order to determine values of c and p, we balance the linear term of the highest order derivatives t00 with the highest order nonlinear term t3 in Eq. (6.1). By simple calculations, we have

c1 exp½ðc þ 3pÞn þ . . . ; c2 exp½4pn þ    c3 exp½ð3c þ pÞn þ . . . ¼ ; c4 exp½4pn þ   

t00 ¼

ð6:5Þ

t3

ð6:6Þ

where ci are coefficients for simplicity. By balancing the highest order of exp-function in Eqs. (6.5) and (6.6), we have c + 3p = 3c + p which leads to

p ¼ c:

ð6:7Þ

Similarly, from the ansatz (6.4), we have

. . . þ d1 exp½ðd þ 3qÞn ;    þ d2 exp½4qn    þ d3 exp½ð3d þ qÞn ¼ ;    þ d4 exp½4qn

t00 ¼

ð6:8Þ

t3

ð6:9Þ

where di are coefficients for simplicity. By balancing the lowest order of exp-function in Eqs. (6.8) and (6.9), we have (d + 3q) = (3d + q) which leads to

q ¼ d:

ð6:10Þ

Choosing p = c = 1 and q = d = 1, Eq. (6.4) becomes

tðnÞ ¼

a1 expðnÞ þ a0 þ a1 expðnÞ : b1 expðnÞ þ b0 þ b1 expðnÞ

ð6:11Þ

Substituting Eq. (6.11) into Eqs. (6.1) and (6.2), we get 3 1 X ðri þ t 2 qi þ t ei Þ ¼ 0; A i¼3

2 1X ðli þ t xi Þ ¼ 0; B i¼2

ð6:12Þ

where

A ¼ ðb1 expðnÞ þ b0 þ b1 expðnÞÞ3 ;

B ¼ ðb1 expðnÞ þ b0 þ b1 expðnÞÞ2 ;

dhðxÞ 1 2  a1 b1 bðxÞgðxÞ2 þ a31 aðxÞ ¼ 0; dx 2 dhðxÞ 1 1 1 2 dhðxÞ 2 2 ¼ a0 b1  2a1 b1 b0  a1 b1 b0 bðxÞgðxÞ2  a0 b1 bðxÞgðxÞ2 þ 3a21 a0 aðxÞ  a1 b1 b0 bðxÞpðxÞ2 þ a0 b1 bðxÞpðxÞ2 ¼ 0; dx dx 2 2 2 dhðxÞ dhðxÞ 1 2 dhðxÞ 2 dhðxÞ 2 ¼ a1 b0  a1 b1  2a0 b1 b0  2a1 b1 b1  a1 b0 bðxÞgðxÞ2 dx dx dx dx 2 1 1 1 2 2  a1 b1 bðxÞgðxÞ2  a1 b1 b1 bðxÞgðxÞ2  a0 b1 b0 bðxÞgðxÞ2  a0 b1 b0 bðxÞpðxÞ2 þ a1 b0 bðxÞpðxÞ2  2a1 b1 b1 bðxÞpðxÞ2 2 2 2

r3 ¼ a1 b21 r2 r1

2

þ 2a1 b1 bðxÞpðxÞ2 þ 3a21 a1 aðxÞ þ 3a1 a20 aðxÞ ¼ 0;

r0 ¼ 2a0 b1 b1

dhðxÞ dhðxÞ dhðxÞ 1 2 dhðxÞ 2 2  2a1 b1 b0  2a1 b1 b0  a0 b0  a0 b0 bðxÞgðxÞ dx dx dx dx 2 2

2

2

 a1 b1 b0 bðxÞgðxÞ  a0 b1 b1 bðxÞgðxÞ  a1 b1 b0 bðxÞgðxÞ þ a30 aðxÞ þ 6a0 a1 a1 aðxÞ 3 3 2 2 2  3a0 b1 b1 bðxÞpðxÞ þ a1 b1 b0 bðxÞpðxÞ þ a1 b1 b0 bðxÞpðxÞ ¼ 0; 2 2 dhðxÞ dhðxÞ 1 2 dhðxÞ 2 dhðxÞ 2 2 r1 ¼ 2a1 b1 b1  2a0 b1 b0  a1 b1  a1 b0  a1 b1 bðxÞgðxÞ dx dx dx dx 2 1 1 2 2  a1 b0 bðxÞgðxÞ2  a1 b1 b1 bðxÞgðxÞ2  a0 b1 b0 bðxÞgðxÞ2 þ 3a1 a21 aðxÞ þ 3a1 a20 aðxÞ þ a1 b0 bðxÞpðxÞ2 2 2 1 2  a0 b0 b1 bðxÞpðxÞ2 þ 2a1 b1 bðxÞpðxÞ2  2a1 b1 b1 bðxÞpðxÞ2 ¼ 0; 2 dhðxÞ dhðxÞ 1 2 r2 ¼ a0 b21  2a1 b1 b0  a1 b1 b0 bðxÞgðxÞ2  a0 b1 bðxÞgðxÞ2 þ 3a0 a21 aðxÞ dx dx 2 1 1 1 2 2 dhðxÞ 2  a1 b1 b0 bðxÞpðxÞ2 þ a0 b1 bðxÞpðxÞ2 ¼ 0; r3 ¼ a1 b1  a1 b1 bðxÞgðxÞ2 þ a31 aðxÞ ¼ 0; 2 2 dx 2 df ðxÞ 2 df ðxÞ 2 2 df ðxÞ 2 2 2 2  2a1 b1 bðxÞf ðxÞ ¼ 0; t 2 q2 ¼ a0 b1  2a1 b1 b0  2a0 b1 bðxÞf ðxÞ  4a1 b1 b0 bðxÞf ðxÞ ¼ 0; t 2 q3 ¼ a1 b1 dx dx dx

2266

E.M.E. Zayed, M.A.M. Abdelaziz / Applied Mathematics and Computation 218 (2011) 2259–2268

df ðxÞ df ðxÞ df ðxÞ 2 df ðxÞ 2 2  a1 b0  2a1 b1 b1  2a0 b1 b0  4a1 b1 b1 bðxÞf ðxÞ  4a0 b1 b0 bðxÞf ðxÞ dx dx dx dx df ðxÞ 2 2 2 df ðxÞ  2a1 b1 bðxÞf ðxÞ2  2a1 b0 bðxÞf ðxÞ2 ¼ 0; t 2 q0 ¼ a0 b0  2a1 b1 b0 dx dx df ðxÞ df ðxÞ 2 2 2  2a1 b1 b0  2a0 b0 bðxÞf ðxÞ  4a1 b1 b0 bðxÞf ðxÞ  4a1 b1 b0 bðxÞf ðxÞ2  4a0 b1 b1 bðxÞf ðxÞ2 ¼ 0;  2a0 b1 b1 dx dx df ðxÞ df ðxÞ 2 df ðxÞ 2 df ðxÞ 2  a1 b0  2a1 b1 b1  2a0 b1 b0  4a1 b1 b1 bðxÞf ðxÞ ¼ a1 b1 dx dx dx dx 2 2  4a0 b1 b0 bðxÞf ðxÞ2  2a1 b1 bðxÞf ðxÞ2  2a1 b0 bðxÞf ðxÞ2 ¼ 0; 2

t2 q1 ¼ a1 b1

t2 q1

df ðxÞ df ðxÞ 2 2 2  2a1 b1 b0  4a1 b1 b0 bðxÞf ðxÞ  2a0 b1 bðxÞf ðxÞ ¼ 0; dx dx 2 df ðxÞ 2 2 dgðxÞ 2 t2 q3 ¼ a1 b1  2a1 b1 bðxÞf ðxÞ2 ¼ 0; te3 ¼ a1 b1  2a1 b1 bðxÞf ðxÞgðxÞ ¼ 0; dx dx dgðxÞ 2 dgðxÞ 2 te2 ¼ a0 b1  2a1 b1 b0  4a1 b1 b0 bðxÞf ðxÞgðxÞ  2a0 b1 bðxÞf ðxÞgðxÞ ¼ 0; dx dx dgðxÞ dgðxÞ 2 dgðxÞ 2 dgðxÞ te1 ¼ a1 b1  a1 b0  2a1 b1 b1  2a0 b1 b0  4a1 b1 b1 bðxÞf ðxÞgðxÞ  4a0 b1 b0 bðxÞf ðxÞgðxÞ dx dx dx dx dgðxÞ dgðxÞ 2 2 2 dgðxÞ  2a1 b1 bðxÞf ððxÞgðxÞ  2a1 b0 bðxÞf ððxÞgðxÞ ¼ 0; t e0 ¼ a0 b0  2a1 b1 b0  2a0 b1 b1 dx dx dx dgðxÞ 2  2a1 b1 b0  4a1 b1 b0 bðxÞf ðxÞgðxÞ  4a1 b1 b0 bðxÞf ðxÞgðxÞ  4a0 b1 b1 bðxÞf ðxÞgðxÞ  2a0 b0 bðxÞf ðxÞgðxÞ ¼ 0; dx dgðxÞ dgðxÞ 2 dgðxÞ 2 dgðxÞ te1 ¼ a1 b1  a1 b0  2a1 b1 b1  2a0 b1 b0  4a1 b1 b1 bðxÞf ðxÞgðxÞ  4a0 b1 b0 bðxÞf ðxÞgðxÞ dx dx dx dx dgðxÞ 2 2 2 dgðxÞ  2a1 b1 bðxÞf ðxÞgðxÞ  2a1 b0 bðxÞf ðxÞgðxÞ ¼ 0; t e2 ¼ a0 b1  2a1 b1 b0  4a1 b1 b0 bðxÞf ðxÞgðxÞ dx dx 2 2 dgðxÞ 2  2a0 b1 bðxÞf ðxÞgðxÞ ¼ 0; t e3 ¼ a1 b1  2a1 b1 bðxÞf ðxÞgðxÞ ¼ 0; l2 ¼ a1 b1 bðxÞf ðxÞ  a1 b1 cðxÞ ¼ 0; dx dqðxÞ dqðxÞ l1 ¼ a1 b0  a0 b1 þ a1 b0 bðxÞgðxÞpðxÞ  a0 b1 bðxÞgðxÞpðxÞ þ a0 b1 bðxÞf ðxÞ þ a1 b0 bðxÞf ðxÞ  a1 b0 cðxÞ  a0 b1 cðxÞ ¼ 0; dx dx dqðxÞ dqðxÞ l0 ¼ 2a1 b1  2a1 b1 þ a1 b1 bðxÞf ðxÞ þ a1 b1 bðxÞf ðxÞ þ a0 b0 bðxÞf ðxÞ þ 2a1 b1 bðxÞgðxÞpðxÞ dx dx dqðxÞ dqðxÞ  a1 b0 þ a0 b1 bðxÞgðxÞpðxÞ  2a1 b1 bðxÞgðxÞpðxÞ  a1 b1 cðxÞ  a1 b1 cðxÞ  a0 b0 cðxÞ ¼ 0; l1 ¼ a0 b1 dx dx  a1 b0 bðxÞgðxÞpðxÞ þ a0 b1 bðxÞf ðxÞ þ a1 b0 bðxÞf ðxÞ  a0 b1 bðxÞ  a1 b0 cðxÞ ¼ 0; 2

t2 q2 ¼ a0 b1

l2 ¼ a1 b1 bðxÞf ðxÞ  a1 b1 cðxÞ ¼ 0;

dpðxÞ dpðxÞ  a0 b1 þ 2a1 b0 bðxÞf ðxÞpðxÞ  2a0 b1 bðxÞf ðxÞpðxÞ ¼ 0; dx dx dpðxÞ dpðxÞ  2a1 b1 þ 4a1 b1 bðxÞf ðxÞpðxÞ  4a1 b1 bðxÞf ðxÞpðxÞ ¼ 0; t x0 ¼ 2a1 b1 dx dx dpðxÞ dpðxÞ t x1 ¼ a0 b1  a1 b0 þ 2a0 b1 bðxÞf ðxÞpðxÞ  2a1 b0 bðxÞf ðxÞpðxÞ ¼ 0: dx dx

t x1 ¼ a1 b0

ð6:13Þ

Solving the system (6.13) by Maple, we have the following cases of solutions: Case1.

a0 ¼ a0 ;

a1 ¼ a1 ;

a1 ¼ 0;

b0 ¼

a0 b1 ; a1

1 c1 c2 f ðxÞ þ c3 ; gðxÞ ¼ c1 f ðxÞ; 2 2 2 0 c b f ðxÞ aðxÞ ¼ 2 1 2 ; cðxÞ ¼ bðxÞf ðxÞ; 8a1 qðxÞ ¼

b1 ¼ b1 ;

hðxÞ ¼

b1 ¼ 0;

pðxÞ ¼ c2 f ðxÞ;

1 2 c þ 2c21 f ðxÞ þ c4 ; 8 2

ð6:14Þ

where ci(i = 14) are arbitrary constants, such that c1, c2 – 0 and f(x) has the same form (3.11). In this case, the exact solution of Eq. (1.1) has the from:

uðnÞ ¼

    a1 ða1 en þ a0 Þ 1 2 2 2 ; c þ c t þ þ 2c f ðxÞ þ c exp i t 1 4 1 b1 ða1 en  a0 Þ 8 2

which can be written in the form

ð6:15Þ

E.M.E. Zayed, M.A.M. Abdelaziz / Applied Mathematics and Computation 218 (2011) 2259–2268

uðnÞ ¼

      a1 nþ/ 1 exp i t 2 þ c1 t þ c22 þ 2c21 f ðxÞ þ c4 ; coth 2 8 b1

2267

ð6:16Þ

where e/ ¼ aa10 . Case2.

a0 ¼ 0;

a1 ¼ a1 ;

a1 ¼ a1 ;

b0 ¼ 0;

1 c1 c2 f ðxÞ þ c3 ; gðxÞ ¼ c1 f ðxÞ; 2 2 c2 b f 0 ðxÞ aðxÞ ¼ 2 1 2 ; cðxÞ ¼ bðxÞf ðxÞ: 2a1 qðxÞ ¼

b1 ¼ b1 ; hðxÞ ¼

b1 ¼

a1 b1 ; a1

pðxÞ ¼ c2 f ðxÞ;

1 2 c þ 2c22 f ðxÞ þ c4 ; 4 1

ð6:17Þ

In this case, the exact solution of Eq. (1.1) has the from:

uðnÞ ¼

    a1 ða1 en þ a1 en Þ 1 2 2 2 þ c t þ þ 2c ; c f ðxÞ þ c exp i t 1 4 2 b1 ða1 en  a1 en Þ 4 1

ð6:18Þ

which can be written in the form

uðnÞ ¼

    a1 1 cothðn þ /Þ exp i t2 þ c1 t þ c21 þ 2c22 f ðxÞ þ c4 ; 4 b1

ð6:19Þ

where e2/ ¼ aa11 . Case3.

a0 ¼ a0 ; a1 ¼ 0; a1 ¼ 0; b0 ¼ 0; b1 ¼ b1 ; b1 ¼ b1 ; pðxÞ ¼ c2 f ðxÞ; 1 1 qðxÞ ¼ c1 c2 f ðxÞ þ c3 ; gðxÞ ¼ c1 f ðxÞ; hðxÞ ¼ c21  c22 f ðxÞ þ c4 ; 2 4 2b b c2 f 0 ðxÞ aðxÞ ¼  1 12 2 ; cðxÞ ¼ bðxÞf ðxÞ: a0

ð6:20Þ

In this case, the exact solution of Eq. (1.1) has the from:

uðnÞ ¼

    a0 1 2 2 2 ; exp i t þ c t þ  2c c f ðxÞ þ c 1 4 1 2 4 b1 en þ b1 en

ð6:21Þ

which can be written in the form

    a0 1 uðnÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi sechðn þ /Þ exp i t 2 þ c1 t þ c21  2c22 f ðxÞ þ c4 ; 4 2 b1 b1

ð6:22Þ

where e2/ ¼ bb11 . Case4.

a0 ¼ a0 ;

a1 ¼ a1 ;

pðxÞ ¼ c2 f ðxÞ; hðxÞ ¼

a1 ¼ a1 ;

qðxÞ ¼

b0 ¼

1 c1 c2 f ðxÞ þ c3 ; 2

1 2 c2 þ 2c21 f ðxÞ þ c4 ; 8

b1

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a20  4a1 a1 a1

;

b1 ¼ b1 ;

b1 ¼

a1 b1 ; a1

2

gðxÞ ¼ c1 f ðxÞ;

aðxÞ ¼

c22 b1 f 0 ðxÞ ; 8a21

ð6:23Þ

cðxÞ ¼ bðxÞf ðxÞ:

In this case, the exact solution of Eq. (1.1) has the from:

uðnÞ ¼

    a1 ða1 en þ a0 þ a1 en Þ 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 exp i t2 þ c1 t þ c22 þ 2c21 f ðxÞ þ c4 ; 8 b1 a1 en  a20  4a1 a1  a1 en

ð6:24Þ

which can be written in the form

uðnÞ ¼ where e/ ¼

      a1 nþ/ 1 exp i t 2 þ c1 t þ c22 þ 2c21 f ðxÞ þ c4 ; coth 2 8 b1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2

a0 

a0 4a1 a1 2a1

and n is given by (3.18).

ð6:25Þ

2268

E.M.E. Zayed, M.A.M. Abdelaziz / Applied Mathematics and Computation 218 (2011) 2259–2268

Remark 4. From cases 1–4, we deduce that a(x) / f0 (x). But c(x) = b(x)f(x) where f(x) is given by (3.11). Differentiating (3.11), we get f0 (x) / b (x)f(x)2. Consequently we have

aðxÞbðxÞ / cðxÞ2 ;

ð6:26Þ

which is equivalent to (3.19). Remark 5. Each of these solutions is just a disguised version of one or other of the solutions of Section 3. Remark 6. All solutions of this article have been checked with Maple by putting them back into the original Eq. (1.1).

7. Conclusions In this article, the generalized extended tanh-function method, the sine–cosine method and the exp-function method have been applied to find the exact traveling wave solutions of the generalized nonlinear Schrödinger (GNLS) equation with variable coefficients. We have shown in remarks 2 and 4 that if the coefficients a(x), b (x) and c(x) in Eq. (1.1) satisfying the condition a (x)b(x) / c(x)2 then Eq. (1.1) has the solutions given in this article. On comparing the solutions constructed in Sections 3, 5 and 6 using the above three methods respectively, we have noted that, when k(x) is not a constant, the solution of Eq. (1.1) in Section 5 are different from those obtained in Sections 3 and 6. But, when k(x) is a constant we have shown in remark 3 that, the solution of case 2 in Section 5 reduces to the solution of case 3 in Section 3. 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