# The Two-Variable(G′/G,1/G)-Expansion Method for Solving the Nonlinear KdV-mKdV Equation

Authors: E. M. E. Zayed; M. A. M. Abdelaziz
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

We apply the two-variable (G′/G, 1/G)-expansion method to construct new exact traveling wave solutions with parameters of the nonlinear (1+1)-dimensional KdV-mKdV equation. This method can be thought of as the generalization of the well-known (G′/G)-expansion method given recently by M. Wang et al. When the parameters are replaced by special values, the well-known solitary wave solutions of this equation are rediscovered from the traveling waves. It is shown that the proposed method provides a more general powerful mathematical tool for solving nonlinear evolution equations in mathematical physics.


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## Body

## 1. Introduction

In the recent years, investigations of exact solutions to nonlinear PDEs play an important role in the study of nonlinear physical phenomena. Many powerful methods have been presented, such as the inverse scattering transform method [1], the Hirota method [2], the truncated Painlevē expansion method [3–6], the Backlund transform method [7, 8], the exp-function method [9–13], the tanh function method [14–17], the Jacobi elliptic function expansion method [18–20], the original (G′/G)-expansion method [21–29], the two-variable (G′/G,1/G)-expansion method [30], and the first integral method [31]. The key idea of the original (G′/G)-expansion method is that the exact solutions of nonlinear PDEs can be expressed by a polynomial in one variable (G′/G) in which G=G(ξ) satisfies the second ordinary differential equation G′′(ξ)+λG′(ξ)+μG(ξ)=0, where λ and μ are constants. In this paper, we will use the two-variable (G′/G,1/G)-expansion method, which can be considered as an extension of the original (G′/G)-expansion method. The key idea of the two-variable (G′/G,1/G)-expansion method is that the exact traveling wave solutions of nonlinear PDEs can be expressed by a polynomial in the two variables (G′/G) and (1/G), in which G=G(ξ) satisfies a second-order linear ODE, namely G′′(ξ)+λG(ξ)=μ, where λ and μ are constants. The degree of this polynomial can be determined by considering the homogeneous balance between the highest-order derivatives and nonlinear terms in the given nonlinear PDEs, while the coefficients of this polynomial can be obtained by solving a set of algebraic equations resulted from the process of using the method. Recently, Li et al. [30] have applied the two-variable (G′/G,1/G)-expansion method and determined the exact solutions of Zakharov equations.The objective of this paper is to apply the two-variable(G′/G,1/G)-expansion method to find the exact traveling wave solutions of the following nonlinear (1+1)-dimensional KdV-mKdV equation:(1.1)ut+αuux+βu2ux+uxxx=0,
where α and β are nonzero constants. This equation may describe the wave propagation of the bound particle, sound wave, and thermal pulse. It is the most popular soliton equation and often exists in practical problems, such as fluid physics and quantum field theory. Recently, Zayed and Gepreel [23] have found the exact solutions of (1.1) using the original (G′/G)-expansion method.


## 2. Description of the Two-Variable(G′/G,1/G)-Expansion Method

Before we describe the main steps of this method, we need the following remarks (see [30]):Remark 2.1.
If we consider the second-order linear ODE(2.1)G′′(ξ)+λG(ξ)=μ
and set ϕ=G′/G and ψ=1/G, then we get
(2.2)ϕ′=-ϕ2+μψ-λ,ψ′=-ϕψ.Remark 2.2.
Ifλ<0, then the general solution of (2.1) is
(2.3)G(ξ)=A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)+μλ,
where A1 and A2 are arbitrary constants. Consequently, we have
(2.4)ψ2=-λλ2σ+μ2(ϕ2-2μψ+λ),
where σ=A12-A22.Remark 2.3.
Ifλ>0, then the general solution of (2.1) is
(2.5)G(ξ)=A1sin⁡(ξλ)+A2cos⁡(ξλ)+μλ,
and hence
(2.6)ψ2=-λλ2σ-μ2(ϕ2-2μψ+λ),
where σ=A12+A22.Remark 2.4.
Ifλ=0, then the general solution of (2.1) is
(2.7)G(ξ)=μ2ξ2+A1ξ+A2,
and hence
(2.8)ψ2=1A12-2μA2(ϕ2-2μψ).
Suppose we have the following NLPDEs in the form:
(2.9)F(u,ut,ux,uxx,utt,…)=0,
where F is a polynomial in u and its partial derivatives. In the following, we give the main steps of the two-variable (G′/G,1/G)-expansion method [30].Step 1.
The traveling wave variable(2.10)u(x,t)=u(ξ),ξ=x-Vt
reduces (2.9) to an ODE in the form
(2.11)P(u,u′,u′′,…)=0,
where V is a constant and P is a polynomial in u and its total derivatives, while ′=d/dξ.Step 2.
Suppose that the solutions of (2.11) can be expressed by a polynomial in the two variables ϕ and ψ as follows:
(2.12)u(ξ)=∑i=0Naiϕi+∑i=1Nbiϕi-1ψ,
where ai(i=0,1,…,N) and bi(i=1,…,N) are constants to be determined later.Step 3.
Determine the positive integerN in  (2.12) by using the homogeneous balance between the highest-order derivatives and the nonlinear terms in (2.11).Step 4.
Substituting (2.12) into (2.11) along with (2.2) and (2.4), the left-hand side of (2.11) can be covered into a polynomial in ϕ and ψ,in which the degree of ψ is not longer than 1. Equating each coefficient of this polynomial to zero yields a system of algebraic equations that can be solved by using the Maple or Mathematica to get the values of ai,bi,V,μ,A1,A2, and λ where λ<0. Thus, we get the exact solutions in terms of the hyperbolic functions.Step 5.
Similar to Step4, substituting (2.12) into (2.11) along with (2.2) and (2.6) for λ>0 (or (2.2) and (2.8) for λ=0), we obtain the exact solutions of (2.11) expressed by trigonometric functions (or by rational functions), respectively.


## 3. An Application

In this section, we apply the method described in Section2 to find the exact traveling wave solutions of the nonlinear (1+1)-dimensional KdV-mKdV Equation (1.1). To this end, we see that the traveling wave variable (2.10) permits us to convert (1.1) into the following ODE:(3.1)-Vu′+αuu′+βu2u′+u′′′=0.
By balancing u′′′ with u2u′ in (3.1), we get N=1. Consequently, we get(3.2)u(ξ)=a0+a1ϕ(ξ)+b1ψ(ξ),
where a0,a1,and b1 are constants to be determined later. There are three cases to be discussed as follows.Case 1.
Hyperbolic function solutions(λ<0).
Ifλ<0, substituting (3.2) into (3.1) and using (2.2) and (2.4), the left-hand side of (3.1) becomes a polynomial in ϕ and ψ. Setting the coefficients of this polynomial to zero yields a system of algebraic equations in a0,a1,b1,μ,σ, and λ as follows:(3.3)ϕ4:12a1λ2σμ2+6a1μ4+βa13λ4σ2+2βa13λ2σμ2+βa13μ4+6a1λ4σ2-3βa1b12λ3σ-3βa1b12λμ2=0,ϕ3:4βa0a12λ2σμ2+2βa0a12λ4σ2+αa12λ4σ2-αb12λμ2-2βb13λ2μ-αb12λ3σ-2βa0b12λ3σ+6b1μ3λ+2βa12b1λ3μσ+2βa12b1λμ3+αa12μ4+2αa12λ2σμ2+6b1μλ3σ-2βa0b12λμ2+2βa0a12μ4=0,ϕ3ψ:6b1λ4σ2-βb13λ3σ+3βa12b1μ4+6βa12b1λ2σμ2+6b1μ4+12b1λ2σμ2-βb13λμ2+3βa12b1λ4σ2=0,ϕ2:-Va1λ4σ2+αb1λ3a1μσ+αa0a1μ4+βa02a1μ4+βa13λ5σ2+βa13λμ4+13a1λ3σμ2-Va1μ4-2Va1λ2σμ2+5a1λμ4+8a1λ5σ2+2βa13λ3σμ2+βa02a1λ4σ2+αa0a1λ4σ2+αb1λa1μ3-2βb12λ2a1μ2+2βa0b1λa1μ3-4βb12λ4a1σ+2αa0a1λ2σμ2+2βa02a1λ2σμ2+2βa0b1λ3a1μσ=0,ϕ2ψ:-βa13μ5+7βb12λ3a1μσ-12a1μλ4σ2-2βa13μ3λ2σ-βa13μλ4σ2+2αa1b1μ4+4βa0a1b1λ4σ2+8βa0a1b1λ2σμ2-24a1μ3λ2σ+4αa1b1λ2σμ2+4βa0a1b1μ4+7βb12λa1μ3+2αa1b1λ4σ2-12a1μ5=0,ϕ1:6b1μ3λ2-αb12λ4σ+2βa0a12λ5σ2+αa12λ5σ2+αa12λμ4+2αa12λ3σμ2-αb12λ2μ2-2βb13λ3μ+2βa12b1λ2μ3+2βa0a12λμ4+6b1μλ4σ-2βa0b12λ2μ2+4βa0a12λ3σμ2-2βa0b12λ4σ+2βa12b1λ4μσ=0,ϕ1ψ:-βb13λ4σ+αa0b1μ4+βa02b1μ4-Vb1λ4σ2+3βb13λ2μ2-2b1λ3σμ2-2βa0a12μ5+2αb12λμ3-αa12μ5-Vb1μ4+5b1λ5σ2-7b1λμ4+αa0b1λ4σ2+βa02b1λ4σ2+2αa0b1λ2σμ2-2αa12μ3λ2σ-αa12μλ4σ2-2Vb1λ2σμ2-2βa0a12μλ4σ2-4βa0a12μ3λ2σ+2βa02b1λ2σμ2+2βa12b1λ5σ2-2βa12b1λμ4+2αb12λ3μσ+4βa0b12λ3μσ+4βa0b12λμ3=0,ϕ0:a1λ4σμ2-Va1λ5σ2-Va1λμ4+αa0a1λ5σ2+αa0a1λμ4-a1λ2μ4+2a1λ6σ2+αb1λ4a1μσ+βb12λ3a1μ2-βb12λ5a1σ+βa02a1λ5σ2-2Va1λ3σμ2+2βa02a1λ3σμ2+αb1λ2a1μ3+2βa0b1λ2a1μ3+2αa0a1λ3σμ2+2βa0b1λ4a1μσ+βa02a1λμ4=0,ϕ0ψ:-αb1a1λμ4-βa02a1μλ4σ2-βb12λ2a1μ3+a1μ5λ-αa0a1μ5+3βb12λ4a1μσ+αb1a1λ5σ2-2αa0a1μ3λ2σ-βa02a1μ5+2Va1μ3λ2σ+Va1μλ4σ2+Va1μ5-2βa02a1μ3λ2σ-4a1μ3λ3σ+2βa0b1a1λ5σ2-2βa0b1a1λμ4-5a1μλ5σ2-αa0a1μλ4σ2=0.
Solving the algebraic equations (3.3) by the Maple or Mathematica, we get the following results.Result 1.
We have(3.4)a0=-12β(α±μ6βλλ2σ+μ2),a1=0,b1=±3(λ2σ+μ2)2βλ,V=-4σβλ3+σα2λ2-2βλμ2+α2μ24β(λ2σ+μ2),σ=A12-A22.
From (2.3) and (3.2) and (3.4),we deduce the traveling wave solution of (1.1) as follows:
(3.5)u(ξ)=-12β(α±μ6βλλ2σ+μ2)±3(λ2σ+μ2)2βλ×(1A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)+μ/λ),
where
(3.6)ξ=x+(4σβλ3+σα2λ2-2βλμ2+α2μ24β(λ2σ+μ2))t.
In particular, by settingA1=0,A2>0, and μ=0 in (3.5), we have the solitary solution
(3.7)u(ξ)=-α2β±-3λ2βsech(ξ-λ),
while, if A2=0,A1>0, and μ=0, then we have the solitary solution
(3.8)u(ξ)=-α2β±3λ2βcsch(ξ-λ).Result 2.
We have(3.9)a0=-α2β,a1=±-32β,b1=±6(λ2σ+μ2)βλV=2βλ-α24β,σ=A12-A22.
In this result, we deduce the traveling wave solution of (1.1) as follows:
(3.10)u(ξ)=-α2β±1β[A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)+μ/λ]×{3λ2[A1cosh⁡(ξ-λ)+A2sinh⁡(ξ-λ)]+6(λ2σ+μ2)λ},
where
(3.11)ξ=x-(2βλ-α24β)t.
In particular, by settingA1=0,A2>0, and μ=0 in (3.10), we have the solitary solution
(3.12)u(ξ)=-α2β±3λ2βtanh(ξ-λ)±-6λβsech(ξ-λ),
while, if A2=0,A1>0, and μ=0, then we have the solitary solution
(3.13)u(ξ)=-α2β±3λ2βcoth⁡(ξ-λ)±6λβcsch(ξ-λ).Case 2.
Trigonometric function solutions(λ>0).
Ifλ>0, substituting (3.2) into (3.1) and using (2.2) and (2.6), we get a polynomial in ϕ and ψ. We vanish each coefficient of this polynomial to get the following algebraic equations.(3.14)ϕ4:-βa13μ4-βa13λ4σ2+2βa13λ2σμ2-6a1μ4-6a1λ4σ2+12a1λ2σμ2-3βa1b12λ3σ+3βa1b12λμ2=0,ϕ3:6b1μλ3σ+αb12λμ2-2βa12b1λμ3-6b1μ3λ+4βa0a12λ2σμ2+2βb13λ2μ-2βa0b12λ3σ+2βa0b12λμ2-αa12λ4σ2-αa12μ4-αb12λ3σ-2βa0a12λ4σ2-2βa0a12μ4+2αa12λ2σμ2+2βa12b1λ3μσ=0,ϕ3ψ:-3βa12b1μ4+6βa12b1λ2σμ2-6b1λ4σ2-6b1μ4-3βa12b1λ4σ2+12b1λ2σμ2-βb13λ3σ+βb13λμ2=0,ϕ2:Va1λ4σ2-αa0a1μ4-βa02a1μ4-βa13λ5σ2-βa13λμ4-5a1λμ4+13a1λ3σμ2+Va1μ4-8a1λ5σ2-βa02a1λ4σ2+2βa13λ3σμ2-αa0a1λ4σ2+2βa02a1λ2σμ2+2βb12λ2a1μ2+αb1λ3a1μσ-αb1λa1μ3+2βa0b1λ3a1μσ+2αa0a1λ2σμ2-2Va1λ2σμ2-4βb12λ4a1σ-2βa0b1λa1μ3=0,ϕ2ψ:βa13μ5-24a1μ3λ2σ+12a1μ5+12a1μλ4σ2+βa13μλ4σ2+7βb12λ3a1μσ-4βa0a1b1λ4σ2-2βa13μ3λ2σ-4βa0a1b1μ4-2αa1b1μ4+4αa1b1λ2σμ2+8βa0a1b1λ2σμ2-7βb12λa1μ3-2αa1b1λ4σ2=0,ϕ1:-αb12λ4σ-αa12λ5σ2-αa12λμ4-2βa0a12λμ4-2βa12b1λ2μ3+αb12λ2μ2+2βb13λ3μ+2βa12b1λ4μσ+6b1μλ4σ-6b1μ3λ2+4βa0a12λ3σμ2-2βa0a12λ5σ2+2βa0b12λ2μ2-2βa0b12λ4σ+2αa12λ3σμ2=0,ϕ1ψ:-βb13λ4σ-βa02b1μ4+Vb1λ4σ2-αa0b1μ4-3βb13λ2μ2-2b1λ3σμ2+2βa0a12μ5-2αb12λμ3+Vb1μ4+αa12μ5-5b1λ5σ2+7b1λμ4-βa02b1λ4σ2-αa0b1λ4σ2+αa12μλ4σ2+2βa02b1λ2σμ2+2βa0a12μλ4σ2-4βa0a12μ3λ2σ-2αa12μ3λ2σ-2Vb1λ2σμ2+2αa0b1λ2σμ2-2βa12b1λ5σ2+2βa12b1λμ4+2αb12λ3μσ+4βa0b12λ3μσ-4βa0b12λμ3=0,ϕ0:a1λ4σμ2+Va1λ5σ2+Va1λμ4+αb1λ4a1μσ-βb12λ3a1μ2+a1λ2μ4-2a1λ6σ2-βa02a1λ5σ2-βa02a1λμ4-αa0a1λ5σ2-αa0a1λμ4+2βa02a1λ3σμ2-2Va1λ3σμ2-αb1λ2a1μ3+2αa0a1λ3σμ2-2βa0b1λ2a1μ3-βb12λ5a1σ+2βa0b1λ4a1μσ=0,ϕ0ψ:-Va1μ5-Va1μλ4σ2-a1μ5λ-αb1a1λ5σ2+αb1a1λμ4+βa02a1μ5+βb12λ2a1μ3+5a1μλ5σ2+βa02a1μλ4σ2+3βb12λ4a1μσ+αa0a1μ5+2Va1μ3λ2σ-4a1μ3λ3σ+αa0a1μλ4σ2-2βa0b1a1λ5σ2+2βa0b1a1λμ4-2αa0a1μ3λ2σ-2βa02a1μ3λ2σ=0.
Solving the algebraic equations (3.14) by the Maple or Mathematica, we obtain the following results.Result 1.
We have(3.15)a0=-1β(α2±3μβλ6(μ2-λ2σ)),a1=0,b1=±6(μ2-λ2σ)βλ,V=-4σβλ3+σα2λ2+2βλμ2-α2μ24β(λ2σ-μ2),σ=A12+A22.
From (2.5), (3.2), and (3.15), we deduce the traveling wave solution of (1.1) as follows:
(3.16)u(ξ)=-1β(α2±3μβλ6(μ2-λ2σ))±6(μ2-λ2σ)βλ×(1A1sin⁡(ξλ)+A2cos⁡(ξλ)+μ/λ),
where
(3.17)ξ=x+(4σβλ3+σα2λ2+2βλμ2-α2μ24β(λ2σ-μ2))t.
In particular, by settingA1=0,A2>0, and μ=0 in (3.16), we have the periodic solution
(3.18)u(ξ)=-α2β±-6λβsec(ξλ),
while, if A2=0,A1>0, and μ=0, then we have the periodic solution
(3.19)u(ξ)=-α2β±-6λβcsc(ξλ).Result 2.
We have(3.20)a0=-α2β,a1=±-32β,b1=±3(μ2-λ2σ)2βλ,V=2βλ-α24β,σ=A12+A22.
In this result, we deduce the traveling wave solution of (1.1) as follows:
(3.21)u(ξ)=-α2β±-32β(1A1sin⁡(ξλ)+A2cos⁡(ξλ)+μ/λ)×{λ[A1cos⁡(ξλ)-A2sin⁡(ξλ)]+λ2σ-μ2λ},
where
(3.22)ξ=x-2βλ-α24βt.
In particular, by settingA1=0,A2>0, and μ=0 in (3.21), we have the periodic solution
(3.23)u(ξ)=-α2β±-3λ2β[-tan(ξλ)+sec(ξλ)],
while, if A2=0,A1>0, and μ=0, then we have the periodic solution
(3.24)u(ξ)=-α2β±-3λ2β[cot(ξλ)+csc(ξλ)].Case 3.
Rational function solutions(λ=0).
Ifλ=0, substituting (3.2) into (3.1) and using (2.2) and (2.8), we get a polynomial in ϕ and ψ. Setting each coefficient of this polynomial to zero yields the following algebraic equations:(3.25)ϕ4:βa13A14-4βa13A12μA2-24a1A12μA2+24a1μ2A22+4βa13μ2A22+6a1A14+3βa1b12A12-6βa1b12μA2=0,ϕ3:-4βa0b12μA2-6b1μA12+4αa12μ2A22+2βa0a12A14+8βa0a12μ2A22-2αb12μA2+12b1μ2A2+2βa0b12A12-2βb13μ+αa12A14-2βa12b1μA12-8βa0a12A12μA2+αb12A12+4βa12b1μ2A2-4αa12A12μA2=0,ϕ3ψ:12βa12b1μ2A22+6b1A14+βb13A12-2βb13μA2+3βa12b1A14-12βa12b1A12μA2-24b1A12μA2+24b1μ2A22=0,ϕ2:αa0a1A14+βa02a1A14+2βb12a1μ2-4Va1μ2A22-Va1A14+3a1μ2A12-6a1μ3A2-αb1a1μA12+4Va1A12μA2-4βa02a1A12μA2+4βa02a1μ2A22-4αa0a1A12μA2+4αa0a1μ2A22+2αb1a1μ2A2-2βa0b1a1μA12+4βa0b1a1μ2A2=0,ϕ2ψ:-48a1μ3A22+48a1μ2A12A2-12a1μA14-7βb12a1μA12+4βa13μ2A12A2-βa13μA14+8αa1b1μ2A22+14βb12a1μ2A2+2αa1b1A14-8αa1b1A12μA2-4βa13μ3A22+4βa0a1b1A14-16βa0a1b1A12μA2+16βa0a1b1μ2A22=0,ϕ1ψ:αa0b1A14-αa12μA14+βa02b1A14-4Vb1μ2A22-4αa12μ3A22-2αb12μA12+4αb12μ2A2-Vb1A14+4βb13μ2+12b1μ2A12-24b1μ3A2+4Vb1A12μA2+4αa0b1μ2A22-4αa0b1a12μA2-2βa0a12μA14-4βa02b1A12μA2+4βa02b1μ2A22+4αa12μ2A12A2-4βa0b12μA12+8βa0b12μ2A2+4βa12b1μ2A12+8βa0a12μ2A12A2-8βa0a12μ3A22-8βa12b1μ3A2=0,ϕ0ψ:Va1μA14-4βb12a1μ3+4Va1μ3A22-6a1μ3A12+12a1μ4A2+4αa0a1μ2A12A2-αa0a1μA14-βa02a1μA14-4αa0a1μ3A22+4βa02a1μ2A12A2-4βa02a1μ3A22-4Va1μ2A12A2-4αb1a1μ3A2+4βa0b1a1μ2A12+2αb1a1μ2A12-8βa0b1a1μ3A2=0.
Solving the algebraic equations (3.25) by the Maple or Mathematica, we obtain the following results.Result 1.
We have(3.26)a0=-1β(α2±3μβ6(2μA2-A12)),a1=0,b1=±6(2μA2-A12)β,V=-α2A12+6βμ2-2α2μA24β(A12-2μA2).
From (2.7), (3.2) and (3.26), we deduce the traveling wave solution of (1.1) as follows:
(3.27)u(ξ)=-1β(α2±3μβ6(2μA2-A12))±6(2μA2-A12)β(1(μ/2)ξ2+A1ξ+A2),
where
(3.28)ξ=x+(α2A12+6βμ2-2α2μA24β(A12-2μA2))t.Result 2.
We have(3.29)a0=-α2β,a1=±-32β,b1=±3(2μA2-A12)2βV=-α24β.
In this result, we deduce the traveling wave solution of (1.1) as follows:
(3.30)u(ξ)=-α2β±-32β(μξ+A1+A12-2μA2(μ/2)ξ2+A1ξ+A2),
where
(3.31)ξ=x+α24βt.Remark 3.1.
All solutions of this paper have been checked with Maple by putting them back into the original equation (1.1).


## 4. Conclusions

In this paper, the(G′/G,1/G)-expansion method was employed to obtain some new as well as some known solutions of a selected nonlinear equation, namely, the (1+1)-dimensional KdV-mKdV equation. As the two parameters A1 and A2 take special values, we obtain the solitary wave solutions. When μ=0 and bi=0 in (2.1) and (2.12), the two-variable (G′/G,1/G)-expansion method reduces to the original (G′/G)-expansion method. So, the two-variable (G′/G,1/G)-expansion method is an extension of the original (G′/G)-expansion method. The proposed method in this paper is more effective and more general than the original (G′/G)-expansion method because it gives exact solutions in more general forms. In summary, the advantage of the two-variable (G′/G,1/G)-expansion method over the original (G′/G)-expansion method is that the solutions using the first method recover the solutions using the second one.



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