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A note on the improved (G/G)(G^{\prime}/G)- expansion method

M. S. Abdel Latif Mathematics and Engineering Physics Dept., Faculty of Engineering,
Mansoura University, Mansoura, Egypt.
Abstract

In this paper, we show that the improved (G/G)(G^{\prime}/G)- expansion method is equivalent to the tanh method and gives the same exact solutions of nonlinear partial differential equations.

keywords:
Exact solution , Improved (G/G)(G^{\prime}/G)- expansion method , tanh method

1 Introduction

Recently, Many methods have been proposed for obtaining exact traveling wave solutions of partial differential equations. An example of these methods is the (G/G)(G^{\prime}/G)- expansion method [1] which is used to obtain traveling wave solutions of many models (see for example [2, 3]). Also, many improved and extended versions of this method have been proposed to get more exact solutions of partial differential equations (see for example [4, 5, 6]).

Many papers are published to comment on the classical version of the (G/G)(G^{\prime}/G)- expansion method. For example, the equivalence between the (G/G)(G^{\prime}/G)- expansion method and the tanh method is proved in [7, 8, 9]. Moreover, in [10], it is shown that the (G/G)(G^{\prime}/G)- expansion method is a specific form of the simplest equation method [11].

The improved (G/G)(G^{\prime}/G)- expansion method [5] is used to obtain new exact solutions of some models [12, 13]. In this paper, we show that this improved (G/G)(G^{\prime}/G)- expansion method is equivalent to the tanh method and doesn’t give any new exact solutions of nonlinear partial differential equations.

2 The tanh method [8]

In this section, we give the detailed description of the tanh method. Suppose that a nonlinear evolution equation (NLEE) with independent variable uu and two independent variables xx and tt is given by

H(u,ut,ux,utt,uxx,uxt,)=0,H(u,u_{t},u_{x},u_{tt},u_{xx},u_{xt},...)=0, (1)

where, HH is a polynomial in u(x,t)u(x,t) and its various partial derivatives, in which the highest order derivatives and nonlinear terms are involved. To determine uu explicitly, one can follow the following steps:

Step 1: Use the traveling wave transformation:

u=u(ξ),ξ=xνt,u=u(\xi),\quad\xi=x-\nu t, (2)

where, ν\nu is a constant to be determined latter. Then, the NLEE (1) is reduced to a nonlinear ordinary differential equation (NLODE) for u=u(ξ)u=u(\xi):

H(u,u,u′′,u′′′,)=0.H(u,u^{\prime},u^{\prime\prime},u^{\prime\prime\prime},...)=0. (3)

step 2: Suppose that the NLODE (3) has the following solution:

u=i=0nbi(tanh(k(ξξ0))i,u=\sum_{i=0}^{n}b_{i}(\tanh(k(\xi-\xi_{0}))^{i}, (4)

where, k,bi(i=0,,n)k,b_{i}(i=0,...,n) are constants to be determined later, ξ0\xi_{0} is an arbitrary constant and nn is a positive integer to be determined in step 3.

Step 3: Determine the positive integer nn by balancing the highest order derivatives and nonlinear terms in Eq. (3).

Step 4: Substituting Eq. (4) into Eq. (3) and equating expressions of different power of (tanh(k(ξξ0))i(\tanh(k(\xi-\xi_{0}))^{i} to zero, we obtain coefficients bib_{i} and the parameter kk.

Step 5: Substituting bib_{i} and kk into Eq. (4), we can obtain the explicit solutions of Eq. (1) immediately.

3 The improved (G/G)(G^{\prime}/G)- expansion method [5]

In this section, we give the detailed description of the improved (G’/G) -expansion method. To determine uu in Eq. (1) explicitly using the improved (G/G)(G^{\prime}/G)- expansion method, one can follow the following five steps:

Step 1: Use the traveling wave transformation (2) to reduce the NLEE (1) to the NLODE (3)

Step 2: Suppose that the NLODE (3) has the following solution:

u=i=nnai(G/G)i(1+σ(G/G))i=i=nnai((G/G)1+σ(G/G))i,u=\sum_{i=-n}^{n}\frac{a_{i}\left(G^{\prime}/G\right)^{i}}{\left(1+\sigma\left(G^{\prime}/G\right)\right)^{i}}=\sum_{i=-n}^{n}a_{i}\left(\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}\right)^{i}, (5)

where, σ\sigma and ai(i=n,,n)a_{i}(i=-n,...,n) are constants to be determined later, nn is a positive integer, and G=G(ξ)G=G(\xi) satisfies the following second order linear ordinary differential equation(LODE):

G′′+μG=0,G^{\prime\prime}+\mu G=0, (6)

where, μ\mu is a real constant. The general solutions of Eq. (6) can be listed as follows. When μ<0\mu<0, we obtain the hyperbolic function solution of Eq. (6)

G=A1cosh(μξ)+A2sinh(μξ),G=A_{1}\cosh(\sqrt{-\mu}\xi)+A_{2}\sinh(\sqrt{-\mu}\xi), (7)

where, A1A_{1} and A2A_{2} are arbitrary constants. When μ>0\mu>0, we obtain the trigonometric function solution of Eq. (6)

G=A1cos(μξ)+A2sin(μξ),G=A_{1}\cos(\sqrt{\mu}\xi)+A_{2}\sin(\sqrt{\mu}\xi), (8)

where, A1A_{1} and A2A_{2} are arbitrary constants. When μ=0\mu=0, we obtain the linear solution of Eq. (6)

G=A1+A2ξ,G=A_{1}+A_{2}\xi, (9)

where, A1A_{1} and A2A_{2} are arbitrary constants

Step 3: Determine the positive integer nn by balancing the highest order derivatives and nonlinear terms in Eq. (3).

Step 4: Substituting (5) along with (6) into Eq. (6) and then setting all the coefficients of (G/G)k,(k=1,2,3,)\left(G^{\prime}/G\right)^{k},(k=1,2,3,...)of the resulting systems numerator to zero, yields a set of over-determined nonlinear algebraic equations for ν,σ\nu,\sigma and ai(i=n,,n)a_{i}(i=-n,...,n).

Step 5: Assuming that the constants ν,σ\nu,\sigma and ai(i=n,,n)a_{i}(i=-n,...,n) can be obtained by solving the algebraic equations in Step 4, then substituting these constants and the known general solutions of Eq. (6) into (5), we can obtain the explicit solutions of Eq. (1) immediately.

4 Equivalence of the two methods

In the second step of the improved (G/G)(G^{\prime}/G)- expansion method let y=(G/G)1+σ(G/G)y=\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}, Eqs. (5), (6) are transformed into

u=i=nnaiyi,u=\sum_{i=-n}^{n}a_{i}y^{i}, (10)
y+(1+μσ2)y22σμy+μ=0.y^{\prime}+(1+\mu\sigma^{2})y^{2}-2\sigma\mu y+\mu=0. (11)

The general solution of Eq. (11) is given by [8]

y=α+βtanh(μ(ξξ0)),y=\alpha+\beta\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right), (12)

where, α=σμ1+μσ2\alpha=\frac{\sigma\mu}{1+\mu\sigma^{2}} and β=μ1+μσ2\beta=\frac{\sqrt{-\mu}}{1+\mu\sigma^{2}}.

Substituting solution (12) into expansion (10) we have

u=i=nnai(α+βtanh(μ(ξξ0)))i=u1+u2,u=\sum_{i=-n}^{n}a_{i}\left(\alpha+\beta\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right)\right)^{i}=u_{1}+u_{2}, (13)

where,

u1=i=0nai(α+βtanh(μ(ξξ0)))i=i=0nbi(tanh(μ(ξξ0)))i,u_{1}=\sum_{i=0}^{n}a_{i}\left(\alpha+\beta\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right)\right)^{i}=\sum_{i=0}^{n}b_{i}\left(\tanh\left(-\mu(\xi-\xi_{0})\right)\right)^{i}, (14)
u2=i=n1ai(α+βtanh(μ(ξξ0)))i=i=1nai(1α+βtanh(μ(ξξ0)))i=i=1nai(αα2β2αα2β2+1α+βtanh(μ(ξξ0)))i=i=1nai(αα2β2+ββ2α2βα+tanh(μ(ξξ0))1+βαtanh(μ(ξξ0)))i=i=1nai(αα2β2+ββ2α2tanh(μ(ξξ0)+z))i,z=tanh1βα,u_{2}=\sum_{i=-n}^{-1}a_{i}\left(\alpha+\beta\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right)\right)^{i}=\\ \sum_{i=1}^{n}a_{-i}\left(\frac{1}{\alpha+\beta\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right)}\right)^{i}=\\ \sum_{i=1}^{n}a_{-i}\left(\frac{\alpha}{\alpha^{2}-\beta^{2}}-\frac{\alpha}{\alpha^{2}-\beta^{2}}+\frac{1}{\alpha+\beta\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right)}\right)^{i}=\\ \sum_{i=1}^{n}a_{-i}\left(\frac{\alpha}{\alpha^{2}-\beta^{2}}+\frac{\beta}{\beta^{2}-\alpha^{2}}\frac{\frac{\beta}{\alpha}+\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right)}{1+\frac{\beta}{\alpha}\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})\right)}\right)^{i}=\\ \sum_{i=1}^{n}a_{-i}\left(\frac{\alpha}{\alpha^{2}-\beta^{2}}+\frac{\beta}{\beta^{2}-\alpha^{2}}\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})+z\right)\right)^{i},\quad z=\tanh^{-1}\frac{\beta}{\alpha}, (15)

therefore, u2u_{2} may be rewritten as

u2=i=0nci(tanh(μ(ξξ0)+z))i,u_{2}=\sum_{i=0}^{n}c_{i}\left(\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})+z\right)\right)^{i}, (16)

hence,

u=u1+u2=i=0nbi(tanh(μ(ξξ0)))i+i=0nci(tanh(μ(ξξ0)+z))i,u=u_{1}+u_{2}=\sum_{i=0}^{n}b_{i}\left(\tanh\left(-\mu(\xi-\xi_{0})\right)\right)^{i}+\sum_{i=0}^{n}c_{i}\left(\tanh\left(\sqrt{-\mu}(\xi-\xi_{0})+z\right)\right)^{i}, (17)

since ξ0\xi_{0} is an arbitrary constant, then we can write

u=i=0ndi(tanh(μ(ξξ1)))i,u=\sum_{i=0}^{n}d_{i}\left(\tanh\left(-\mu(\xi-\xi_{1})\right)\right)^{i}, (18)

where, ξ1\xi_{1} is an arbitrary constant. It is now clear that the two methods will give the same solutions expressed in terms of the tanh function.

5 Another proof of equivalence

In this section we will proof that The solution formula (5) will give solutions in the form of the tanh function and the rational function only.

Case 1: When μ<0\mu<0, we have

GG=μA2cosh(μξ)+A1sinh(μξ)A1cosh(μξ)+A2sinh(μξ)=μ1+A1A2tanh(μξ)A1A2+tanh(μξ)=μcoth(μξ+d),d=tanh1A1A2,\frac{G^{\prime}}{G}=\sqrt{-\mu}\frac{A_{2}\cosh(\sqrt{-\mu}\xi)+A_{1}\sinh(\sqrt{-\mu}\xi)}{A_{1}\cosh(\sqrt{-\mu}\xi)+A_{2}\sinh(\sqrt{-\mu}\xi)}=\sqrt{-\mu}\frac{1+\frac{A_{1}}{A_{2}}\tanh(\sqrt{-\mu}\xi)}{\frac{A_{1}}{A_{2}}+\tanh(\sqrt{-\mu}\xi)}\\ =\sqrt{-\mu}\coth(\sqrt{-\mu}\xi+d),\quad d=\tanh^{-1}\frac{A_{1}}{A_{2}}, (19)
1+σ(G/G)(G/G)=G+σGG=GG+σ=1μtanh(μξ+d)+σ,\frac{1+\sigma\left(G^{\prime}/G\right)}{\left(G^{\prime}/G\right)}=\frac{G+\sigma G^{\prime}}{G^{\prime}}=\frac{G}{G^{\prime}}+\sigma=\frac{1}{\sqrt{-\mu}}\tanh(\sqrt{-\mu}\xi+d)+\sigma, (20)
j=n1aj((G/G)1+σ(G/G))j=j=1naj(1+σ(G/G)(G/G))j=j=1naj(1μtanh(μξ+d)+σ)j=j=0nbj(tanh(μξ+d))j,\sum_{j=-n}^{-1}a_{j}\left(\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}\right)^{j}=\sum_{j=1}^{n}a_{-j}\left(\frac{1+\sigma\left(G^{\prime}/G\right)}{\left(G^{\prime}/G\right)}\right)^{j}\\ =\sum_{j=1}^{n}a_{-j}\left(\frac{1}{\sqrt{-\mu}}\tanh(\sqrt{-\mu}\xi+d)+\sigma\right)^{j}=\sum_{j=0}^{n}b_{j}(\tanh(\sqrt{-\mu}\xi+d))^{j}, (21)
(G/G)1+σ(G/G)=GG+σG=μtanh(μξ+d)+σμ=(μσ1+μσ2μσ1+μσ2+μtanh(μξ+d)+σμ)=(μσ1+μσ2+μ1+μσ21σμ+tanh(μξ+d)1+1σμtanh(μξ+d))=μσ1+μσ2+μ1+μσ2tanh(μξ+d+k),k=tanh11σμ,\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}=\frac{G^{\prime}}{G+\sigma G^{\prime}}=\frac{\sqrt{-\mu}}{\tanh(\sqrt{-\mu}\xi+d)+\sigma\sqrt{-\mu}}\\ =\left(\frac{\mu\sigma}{1+\mu\sigma^{2}}-\frac{\mu\sigma}{1+\mu\sigma^{2}}+\frac{\sqrt{-\mu}}{\tanh(\sqrt{-\mu}\xi+d)+\sigma\sqrt{-\mu}}\right)\\ =\left(\frac{\mu\sigma}{1+\mu\sigma^{2}}+\frac{\sqrt{-\mu}}{1+\mu\sigma^{2}}\frac{\frac{1}{\sigma\sqrt{-\mu}}+\tanh(\sqrt{-\mu}\xi+d)}{1+\frac{1}{\sigma\sqrt{-\mu}}\tanh(\sqrt{-\mu}\xi+d)}\right)\\ =\frac{\mu\sigma}{1+\mu\sigma^{2}}+\frac{\sqrt{-\mu}}{1+\mu\sigma^{2}}\tanh(\sqrt{-\mu}\xi+d+k),\quad k=\tanh^{-1}\frac{1}{\sigma\sqrt{\mu}}, (22)
j=1naj((G/G)1+σ(G/G))j=j=1naj(μσ1+μσ2+μ1+μσ2tanh(μξ+d+k))j=j=0ncj(tanh(μξ+d+k))j,\sum_{j=1}^{n}a_{j}\left(\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}\right)^{j}=\sum_{j=1}^{n}a_{j}\left(\frac{\mu\sigma}{1+\mu\sigma^{2}}+\frac{\sqrt{-\mu}}{1+\mu\sigma^{2}}\tanh(\sqrt{-\mu}\xi+d+k)\right)^{j}\\ =\sum_{j=0}^{n}c_{j}(\tanh(\sqrt{-\mu}\xi+d+k))^{j}, (23)

So, in this case (when μ<0\mu<0) Eq. (5) can be rewritten as

u=j=nnaj((G/G)1+σ(G/G))j=j=0nbj(tanh(μξ+d))j+j=0ncj(tanh(μξ+d+k))j=j=0nej(tanh(μ(ξξ0))j,u=\sum_{j=-n}^{n}a_{j}\left(\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}\right)^{j}=\sum_{j=0}^{n}b_{j}(\tanh(\sqrt{-\mu}\xi+d))^{j}+\\ \sum_{j=0}^{n}c_{j}(\tanh(\sqrt{-\mu}\xi+d+k))^{j}=\sum_{j=0}^{n}e_{j}(\tanh(\sqrt{-\mu}(\xi-\xi_{0}))^{j}, (24)

where, ξ0\xi_{0} is an arbitrary constant.

Case 2: When μ>0\mu>0, we have

GG=μA2cos(μξ)A1sin(μξ)A1cos(μξ)+A2sin(μξ)=μ1A1A2tan(μξ)A1A2+tan(μξ)=μcot(μξ+d),d=tan1A1A2,\frac{G^{\prime}}{G}=\sqrt{\mu}\frac{A_{2}\cos(\sqrt{\mu}\xi)-A_{1}\sin(\sqrt{\mu}\xi)}{A_{1}\cos(\sqrt{\mu}\xi)+A_{2}\sin(\sqrt{\mu}\xi)}=\sqrt{\mu}\frac{1-\frac{A_{1}}{A_{2}}\tan(\sqrt{\mu}\xi)}{\frac{A_{1}}{A_{2}}+\tan(\sqrt{\mu}\xi)}\\ =\sqrt{\mu}\cot(\sqrt{\mu}\xi+d),\quad d=\tan^{-1}\frac{A_{1}}{A_{2}}, (25)
1+σ(G/G)(G/G)=G+σGG=GG+σ=1μtan(μξ+d)+σ,\frac{1+\sigma\left(G^{\prime}/G\right)}{\left(G^{\prime}/G\right)}=\frac{G+\sigma G^{\prime}}{G^{\prime}}=\frac{G}{G^{\prime}}+\sigma=\frac{1}{\sqrt{\mu}}\tan(\sqrt{\mu}\xi+d)+\sigma, (26)
j=n1aj((G/G)1+σ(G/G))j=j=1naj(1+σ(G/G)(G/G))j=j=1naj(1μtan(μξ+d)+σ)j=j=0nbj(tan(μξ+d))j,\sum_{j=-n}^{-1}a_{j}\left(\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}\right)^{j}=\sum_{j=1}^{n}a_{-j}\left(\frac{1+\sigma\left(G^{\prime}/G\right)}{\left(G^{\prime}/G\right)}\right)^{j}\\ =\sum_{j=1}^{n}a_{-j}\left(\frac{1}{\sqrt{\mu}}\tan(\sqrt{\mu}\xi+d)+\sigma\right)^{j}=\sum_{j=0}^{n}b_{j}(\tan(\sqrt{\mu}\xi+d))^{j}, (27)
(G/G)1+σ(G/G)=GG+σG=μtan(μξ+d)+σμ=(μσ1μσ2+μσ1μσ2+μtan(μξ+d)+σμ)=(μσ1μσ2+μ1μσ21σμ+tan(μξ+d)1+1σμtan(μξ+d))=μσ1+μσ2+μ1μσ2tan(μξ+dk),k=tan11σμ,\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}=\frac{G^{\prime}}{G+\sigma G^{\prime}}=\frac{\sqrt{\mu}}{\tan(\sqrt{\mu}\xi+d)+\sigma\sqrt{\mu}}\\ =\left(-\frac{\mu\sigma}{1-\mu\sigma^{2}}+\frac{\mu\sigma}{1-\mu\sigma^{2}}+\frac{\sqrt{\mu}}{\tan(\sqrt{\mu}\xi+d)+\sigma\sqrt{\mu}}\right)\\ =\left(-\frac{\mu\sigma}{1-\mu\sigma^{2}}+\frac{\sqrt{\mu}}{1-\mu\sigma^{2}}\frac{\frac{1}{\sigma\sqrt{\mu}}+\tan(\sqrt{\mu}\xi+d)}{1+\frac{1}{\sigma\sqrt{\mu}}\tan(\sqrt{\mu}\xi+d)}\right)\\ =-\frac{\mu\sigma}{1+\mu\sigma^{2}}+\frac{\sqrt{\mu}}{1-\mu\sigma^{2}}\tan(\sqrt{\mu}\xi+d-k),\quad k=\tan^{-1}\frac{1}{\sigma\sqrt{\mu}}, (28)
j=1naj((G/G)1+σ(G/G))j=j=1naj(μσ1μσ2+μ1μσ2tan(μξ+dk))j=j=0ncj(tan(μξ+dk))j,\sum_{j=1}^{n}a_{j}\left(\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}\right)^{j}=\sum_{j=1}^{n}a_{j}\left(-\frac{\mu\sigma}{1-\mu\sigma^{2}}+\frac{\sqrt{\mu}}{1-\mu\sigma^{2}}\tan(\sqrt{\mu}\xi+d-k)\right)^{j}\\ =\sum_{j=0}^{n}c_{j}(\tan(\sqrt{\mu}\xi+d-k))^{j}, (29)

So, in this case (when μ>0\mu>0) Eq. (5) can be rewritten as

u=j=nnaj((G/G)1+σ(G/G))j=j=0nbj(tan(μξ+d))j+j=0ncj(tan(μξ+dk))j=j=0nej(tan(μ(ξξ0)))j,u=\sum_{j=-n}^{n}a_{j}\left(\frac{\left(G^{\prime}/G\right)}{1+\sigma\left(G^{\prime}/G\right)}\right)^{j}=\sum_{j=0}^{n}b_{j}(\tan(\sqrt{\mu}\xi+d))^{j}\\ +\sum_{j=0}^{n}c_{j}(\tan(\sqrt{\mu}\xi+d-k))^{j}=\sum_{j=0}^{n}e_{j}(\tan(\sqrt{\mu}(\xi-\xi_{0})))^{j}, (30)

where, ξ0\xi_{0} is an arbitrary constant. By considering the formula [14]

tan(iα)=itanh(α),i=1,\tan(i\alpha)=i\tanh(\alpha),\quad i=\sqrt{-1}, (31)

Eq. (30) can be reformulated as

u=j=0nej(tan(μ(ξξ0)))j=j=0nej(tan(iμ(ξξ0)))j=j=0nej(itanh(μ(ξξ0)))j=j=0nfj(tanh(μ(ξξ0)))j,μ<0,u=\sum_{j=0}^{n}e_{j}(\tan(\sqrt{\mu}(\xi-\xi_{0})))^{j}=\sum_{j=0}^{n}e_{j}(-\tan(i\sqrt{-\mu}(\xi-\xi_{0})))^{j}=\\ \sum_{j=0}^{n}e_{j}(-i\tanh(\sqrt{-\mu}(\xi-\xi_{0})))^{j}=\sum_{j=0}^{n}f_{j}(\tanh(\sqrt{-\mu}(\xi-\xi_{0})))^{j},\quad\mu<0, (32)

which is equivalent to the solution (24) in case 1.

Case 3. When μ=0\mu=0, in this case we will simply obtain the rational solution.

6 Conclusion

It is shown that the improved (G/G)(G^{\prime}/G)- expansion method is equivalent to the tanh method and doesn’t give any new exact solutions of nonlinear partial differential equations.

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