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

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.

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:𝑢_𝑡+𝛼𝑢𝑢_𝑥+𝛽𝑢²𝑢_𝑥+𝑢_𝑥𝑥𝑥=0, (1.1) 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 𝐺^′′(𝜉)+𝜆𝐺(𝜉)=𝜇 (2.1) and set 𝜙=𝐺^′/𝐺 and 𝜓=1/𝐺, then we get 𝜙^′=−𝜙²+𝜇𝜓−𝜆,𝜓^′=−𝜙𝜓. (2.2)

Remark 2.2. If 𝜆<0, then the general solution of (2.1) is 𝐺(𝜉)=𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√+𝜇−𝜆𝜆, (2.3) where 𝐴₁ and 𝐴₂ are arbitrary constants. Consequently, we have 𝜓²=−𝜆𝜆²𝜎+𝜇²𝜙²−2𝜇𝜓+𝜆, (2.4) where 𝜎=𝐴²₁−𝐴²₂.

Remark 2.3. If 𝜆>0, then the general solution of (2.1) is 𝐺(𝜉)=𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆+𝜇𝜆, (2.5) and hence 𝜓²=−𝜆𝜆²𝜎−𝜇²𝜙²−2𝜇𝜓+𝜆, (2.6) where 𝜎=𝐴²₁+𝐴²₂.

Remark 2.4. If 𝜆=0, then the general solution of (2.1) is 𝐺(𝜉)=𝜇2𝜉²+𝐴₁𝜉+𝐴₂, (2.7) and hence 𝜓²=1𝐴²₁−2𝜇𝐴₂𝜙²−2𝜇𝜓. (2.8) Suppose we have the following NLPDEs in the form: 𝐹𝑢,𝑢_𝑡,𝑢_𝑥,𝑢_𝑥𝑥,𝑢_𝑡𝑡,…=0, (2.9) 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 𝑢(𝑥,𝑡)=𝑢(𝜉),𝜉=𝑥−𝑉𝑡 (2.10) reduces (2.9) to an ODE in the form 𝑃𝑢,𝑢^′,𝑢^′′,…=0, (2.11) 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: 𝑢(𝜉)=^𝑁_𝑖=0𝑎_𝑖𝜙^𝑖+^𝑁_𝑖=1𝑏_𝑖𝜙^𝑖−1𝜓, (2.12) where 𝑎_𝑖(𝑖=0,1,…,𝑁) and 𝑏_𝑖(𝑖=1,…,𝑁) are constants to be determined later.

Step 3. Determine the positive integer 𝑁 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 𝑎_𝑖,𝑏_𝑖,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 Step 4, 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 Section 2 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:−𝑉𝑢^′+𝛼𝑢𝑢^′+𝛽𝑢²𝑢^′+𝑢^′′′=0. (3.1) By balancing 𝑢^′′′ with 𝑢²𝑢^′ in (3.1), we get 𝑁=1. Consequently, we get𝑢(𝜉)=𝑎₀+𝑎₁𝜙(𝜉)+𝑏₁𝜓(𝜉), (3.2) 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).
If 𝜆<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:𝜙⁴∶12𝑎₁𝜆²𝜎𝜇²+6𝑎₁𝜇⁴+𝛽𝑎³₁𝜆⁴𝜎²+2𝛽𝑎³₁𝜆²𝜎𝜇²+𝛽𝑎³₁𝜇⁴+6𝑎₁𝜆⁴𝜎²−3𝛽𝑎₁𝑏²₁𝜆³𝜎−3𝛽𝑎₁𝑏²₁𝜆𝜇²𝜙=0,₃∶4𝛽𝑎₀𝑎²₁𝜆²𝜎𝜇²+2𝛽𝑎₀𝑎²₁𝜆⁴𝜎²+𝛼𝑎²₁𝜆⁴𝜎²−𝛼𝑏²₁𝜆𝜇²−2𝛽𝑏³₁𝜆²𝜇−𝛼𝑏²₁𝜆³𝜎−2𝛽𝑎₀𝑏²₁𝜆³𝜎+6𝑏₁𝜇³𝜆+2𝛽𝑎²₁𝑏₁𝜆³𝜇𝜎+2𝛽𝑎²₁𝑏₁𝜆𝜇³+𝛼𝑎²₁𝜇⁴+2𝛼𝑎²₁𝜆²𝜎𝜇²+6𝑏₁𝜇𝜆³𝜎−2𝛽𝑎₀𝑏²₁𝜆𝜇²+2𝛽𝑎₀𝑎²₁𝜇⁴𝜙=0,₃𝜓∶6𝑏₁𝜆⁴𝜎²−𝛽𝑏³₁𝜆³𝜎+3𝛽𝑎²₁𝑏₁𝜇⁴+6𝛽𝑎²₁𝑏₁𝜆²𝜎𝜇²+6𝑏₁𝜇⁴+12𝑏₁𝜆²𝜎𝜇²−𝛽𝑏³₁𝜆𝜇²+3𝛽𝑎²₁𝑏₁𝜆⁴𝜎²𝜙=0,₂∶−𝑉𝑎₁𝜆⁴𝜎²+𝛼𝑏₁𝜆³𝑎₁𝜇𝜎+𝛼𝑎₀𝑎₁𝜇⁴+𝛽𝑎²₀𝑎₁𝜇⁴+𝛽𝑎³₁𝜆⁵𝜎²+𝛽𝑎³₁𝜆𝜇⁴+13𝑎₁𝜆³𝜎𝜇²−𝑉𝑎₁𝜇⁴−2𝑉𝑎₁𝜆²𝜎𝜇²+5𝑎₁𝜆𝜇⁴+8𝑎₁𝜆⁵𝜎²+2𝛽𝑎³₁𝜆³𝜎𝜇²+𝛽𝑎²₀𝑎₁𝜆⁴𝜎²+𝛼𝑎₀𝑎₁𝜆⁴𝜎²+𝛼𝑏₁𝜆𝑎₁𝜇³−2𝛽𝑏²₁𝜆²𝑎₁𝜇²+2𝛽𝑎₀𝑏₁𝜆𝑎₁𝜇³−4𝛽𝑏²₁𝜆⁴𝑎₁𝜎+2𝛼𝑎₀𝑎₁𝜆²𝜎𝜇²+2𝛽𝑎²₀𝑎₁𝜆²𝜎𝜇²+2𝛽𝑎₀𝑏₁𝜆³𝑎₁𝜙𝜇𝜎=0,₂𝜓∶−𝛽𝑎³₁𝜇⁵+7𝛽𝑏²₁𝜆³𝑎₁𝜇𝜎−12𝑎₁𝜇𝜆⁴𝜎²−2𝛽𝑎³₁𝜇³𝜆²𝜎−𝛽𝑎³₁𝜇𝜆⁴𝜎²+2𝛼𝑎₁𝑏₁𝜇⁴+4𝛽𝑎₀𝑎₁𝑏₁𝜆⁴𝜎²+8𝛽𝑎₀𝑎₁𝑏₁𝜆²𝜎𝜇²−24𝑎₁𝜇³𝜆²𝜎+4𝛼𝑎₁𝑏₁𝜆²𝜎𝜇²+4𝛽𝑎₀𝑎₁𝑏₁𝜇⁴+7𝛽𝑏²₁𝜆𝑎₁𝜇³+2𝛼𝑎₁𝑏₁𝜆⁴𝜎²−12𝑎₁𝜇⁵𝜙=0,₁∶6𝑏₁𝜇³𝜆²−𝛼𝑏²₁𝜆⁴𝜎+2𝛽𝑎₀𝑎²₁𝜆⁵𝜎²+𝛼𝑎²₁𝜆⁵𝜎²+𝛼𝑎²₁𝜆𝜇⁴+2𝛼𝑎²₁𝜆³𝜎𝜇²−𝛼𝑏²₁𝜆²𝜇²−2𝛽𝑏³₁𝜆³𝜇+2𝛽𝑎²₁𝑏₁𝜆²𝜇³+2𝛽𝑎₀𝑎²₁𝜆𝜇⁴+6𝑏₁𝜇𝜆⁴𝜎−2𝛽𝑎₀𝑏²₁𝜆²𝜇²+4𝛽𝑎₀𝑎²₁𝜆³𝜎𝜇²−2𝛽𝑎₀𝑏²₁𝜆⁴𝜎+2𝛽𝑎²₁𝑏₁𝜆⁴𝜙𝜇𝜎=0,₁𝜓∶−𝛽𝑏³₁𝜆⁴𝜎+𝛼𝑎₀𝑏₁𝜇⁴+𝛽𝑎²₀𝑏₁𝜇⁴−𝑉𝑏₁𝜆⁴𝜎²+3𝛽𝑏³₁𝜆²𝜇²−2𝑏₁𝜆³𝜎𝜇²−2𝛽𝑎₀𝑎²₁𝜇⁵+2𝛼𝑏²₁𝜆𝜇³−𝛼𝑎²₁𝜇⁵−𝑉𝑏₁𝜇⁴+5𝑏₁𝜆⁵𝜎²−7𝑏₁𝜆𝜇⁴+𝛼𝑎₀𝑏₁𝜆⁴𝜎²+𝛽𝑎²₀𝑏₁𝜆⁴𝜎²+2𝛼𝑎₀𝑏₁𝜆²𝜎𝜇²−2𝛼𝑎²₁𝜇³𝜆²𝜎−𝛼𝑎²₁𝜇𝜆⁴𝜎²−2𝑉𝑏₁𝜆²𝜎𝜇²−2𝛽𝑎₀𝑎²₁𝜇𝜆⁴𝜎²−4𝛽𝑎₀𝑎²₁𝜇³𝜆²𝜎+2𝛽𝑎²₀𝑏₁𝜆²𝜎𝜇²+2𝛽𝑎²₁𝑏₁𝜆⁵𝜎²−2𝛽𝑎²₁𝑏₁𝜆𝜇⁴+2𝛼𝑏²₁𝜆³𝜇𝜎+4𝛽𝑎₀𝑏²₁𝜆³𝜇𝜎+4𝛽𝑎₀𝑏²₁𝜆𝜇³𝜙=0,₀∶𝑎₁𝜆⁴𝜎𝜇²−𝑉𝑎₁𝜆⁵𝜎²−𝑉𝑎₁𝜆𝜇⁴+𝛼𝑎₀𝑎₁𝜆⁵𝜎²+𝛼𝑎₀𝑎₁𝜆𝜇⁴−𝑎₁𝜆²𝜇⁴+2𝑎₁𝜆⁶𝜎²+𝛼𝑏₁𝜆⁴𝑎₁𝜇𝜎+𝛽𝑏²₁𝜆³𝑎₁𝜇²−𝛽𝑏²₁𝜆⁵𝑎₁𝜎+𝛽𝑎²₀𝑎₁𝜆⁵𝜎²−2𝑉𝑎₁𝜆³𝜎𝜇²+2𝛽𝑎²₀𝑎₁𝜆³𝜎𝜇²+𝛼𝑏₁𝜆²𝑎₁𝜇³+2𝛽𝑎₀𝑏₁𝜆²𝑎₁𝜇³+2𝛼𝑎₀𝑎₁𝜆³𝜎𝜇²+2𝛽𝑎₀𝑏₁𝜆⁴𝑎₁𝜇𝜎+𝛽𝑎²₀𝑎₁𝜆𝜇⁴𝜙=0,₀𝜓∶−𝛼𝑏₁𝑎₁𝜆𝜇⁴−𝛽𝑎²₀𝑎₁𝜇𝜆⁴𝜎²−𝛽𝑏²₁𝜆²𝑎₁𝜇³+𝑎₁𝜇⁵𝜆−𝛼𝑎₀𝑎₁𝜇⁵+3𝛽𝑏²₁𝜆⁴𝑎₁𝜇𝜎+𝛼𝑏₁𝑎₁𝜆⁵𝜎²−2𝛼𝑎₀𝑎₁𝜇³𝜆²𝜎−𝛽𝑎²₀𝑎₁𝜇⁵+2𝑉𝑎₁𝜇³𝜆²𝜎+𝑉𝑎₁𝜇𝜆⁴𝜎²+𝑉𝑎₁𝜇⁵−2𝛽𝑎²₀𝑎₁𝜇³𝜆²𝜎−4𝑎₁𝜇³𝜆³𝜎+2𝛽𝑎₀𝑏₁𝑎₁𝜆⁵𝜎²−2𝛽𝑎₀𝑏₁𝑎₁𝜆𝜇⁴−5𝑎₁𝜇𝜆⁵𝜎²−𝛼𝑎₀𝑎₁𝜇𝜆⁴𝜎²=0. (3.3)
Solving the algebraic equations (3.3) by the Maple or Mathematica, we get the following results.

Result 1. We have𝑎₀1=−2𝛽𝛼±𝜇6𝛽𝜆𝜆²𝜎+𝜇²,𝑎₁=0,𝑏₁=±3𝜆²𝜎+𝜇²,2𝛽𝜆𝑉=−4𝜎𝛽𝜆³+𝜎𝛼²𝜆²−2𝛽𝜆𝜇²+𝛼²𝜇²4𝛽𝜆²𝜎+𝜇²,𝜎=𝐴²₁−𝐴²₂. (3.4)
From (2.3) and (3.2) and (3.4),we deduce the traveling wave solution of (1.1) as follows: 𝑢(𝜉)=−12𝛽𝛼±𝜇6𝛽𝜆𝜆²𝜎+𝜇²±3𝜆²𝜎+𝜇²×12𝛽𝜆𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√,−𝜆+𝜇/𝜆 (3.5) where 𝜉=𝑥+4𝜎𝛽𝜆³+𝜎𝛼²𝜆²−2𝛽𝜆𝜇²+𝛼²𝜇²4𝛽𝜆²𝜎+𝜇²𝑡. (3.6)
In particular, by setting 𝐴₁=0,𝐴₂>0, and 𝜇=0 in (3.5), we have the solitary solution 𝑢(𝜉)=−𝛼±2𝛽−3𝜆2𝛽sech𝜉√−𝜆, (3.7) while, if 𝐴₂=0,𝐴₁>0, and 𝜇=0, then we have the solitary solution 𝑢(𝜉)=−𝛼±2𝛽3𝜆2𝛽csch𝜉√−𝜆. (3.8)

Result 2. We have 𝑎₀=−𝛼2𝛽,𝑎₁=±−32𝛽,𝑏₁=±6𝜆²𝜎+𝜇²𝛽𝜆𝑉=2𝛽𝜆−𝛼²4𝛽,𝜎=𝐴²₁−𝐴²₂. (3.9) In this result, we deduce the traveling wave solution of (1.1) as follows: 𝑢(𝜉)=−𝛼±12𝛽𝛽𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√×−𝜆+𝜇/𝜆3𝜆2𝐴₁cosh𝜉√−𝜆+𝐴₂sinh𝜉√−𝜆+6𝜆²𝜎+𝜇²𝜆, (3.10) where 𝜉=𝑥−2𝛽𝜆−𝛼²4𝛽𝑡. (3.11)
In particular, by setting 𝐴₁=0,𝐴₂>0, and 𝜇=0 in (3.10), we have the solitary solution 𝑢(𝜉)=−𝛼±2𝛽3𝜆𝜉√2𝛽tanh−𝜆±−6𝜆𝛽sech𝜉√−𝜆, (3.12) while, if 𝐴₂=0,𝐴₁>0, and 𝜇=0, then we have the solitary solution 𝑢(𝜉)=−𝛼±2𝛽3𝜆𝜉√2𝛽coth−𝜆±6𝜆𝛽csch𝜉√−𝜆. (3.13)

Case 2. Trigonometric function solutions (𝜆>0).
If 𝜆>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.𝜙⁴∶−𝛽𝑎³₁𝜇⁴−𝛽𝑎³₁𝜆⁴𝜎²+2𝛽𝑎³₁𝜆²𝜎𝜇²−6𝑎₁𝜇⁴−6𝑎₁𝜆⁴𝜎²+12𝑎₁𝜆²𝜎𝜇²−3𝛽𝑎₁𝑏²₁𝜆³𝜎+3𝛽𝑎₁𝑏²₁𝜆𝜇²𝜙=0,₃∶6𝑏₁𝜇𝜆³𝜎+𝛼𝑏²₁𝜆𝜇²−2𝛽𝑎²₁𝑏₁𝜆𝜇³−6𝑏₁𝜇³𝜆+4𝛽𝑎₀𝑎²₁𝜆²𝜎𝜇²+2𝛽𝑏³₁𝜆²𝜇−2𝛽𝑎₀𝑏²₁𝜆³𝜎+2𝛽𝑎₀𝑏²₁𝜆𝜇²−𝛼𝑎²₁𝜆⁴𝜎²−𝛼𝑎²₁𝜇⁴−𝛼𝑏²₁𝜆³𝜎−2𝛽𝑎₀𝑎²₁𝜆⁴𝜎²−2𝛽𝑎₀𝑎²₁𝜇⁴+2𝛼𝑎²₁𝜆²𝜎𝜇²+2𝛽𝑎²₁𝑏₁𝜆³𝜙𝜇𝜎=0,₃𝜓∶−3𝛽𝑎²₁𝑏₁𝜇⁴+6𝛽𝑎²₁𝑏₁𝜆²𝜎𝜇²−6𝑏₁𝜆⁴𝜎²−6𝑏₁𝜇⁴−3𝛽𝑎²₁𝑏₁𝜆⁴𝜎²+12𝑏₁𝜆²𝜎𝜇²−𝛽𝑏³₁𝜆³𝜎+𝛽𝑏³₁𝜆𝜇²𝜙=0,₂∶𝑉𝑎₁𝜆⁴𝜎²−𝛼𝑎₀𝑎₁𝜇⁴−𝛽𝑎²₀𝑎₁𝜇⁴−𝛽𝑎³₁𝜆⁵𝜎²−𝛽𝑎³₁𝜆𝜇⁴−5𝑎₁𝜆𝜇⁴+13𝑎₁𝜆³𝜎𝜇²+𝑉𝑎₁𝜇⁴−8𝑎₁𝜆⁵𝜎²−𝛽𝑎²₀𝑎₁𝜆⁴𝜎²+2𝛽𝑎³₁𝜆³𝜎𝜇²−𝛼𝑎₀𝑎₁𝜆⁴𝜎²+2𝛽𝑎²₀𝑎₁𝜆²𝜎𝜇²+2𝛽𝑏²₁𝜆²𝑎₁𝜇²+𝛼𝑏₁𝜆³𝑎₁𝜇𝜎−𝛼𝑏₁𝜆𝑎₁𝜇³+2𝛽𝑎₀𝑏₁𝜆³𝑎₁𝜇𝜎+2𝛼𝑎₀𝑎₁𝜆²𝜎𝜇²−2𝑉𝑎₁𝜆²𝜎𝜇²−4𝛽𝑏²₁𝜆⁴𝑎₁𝜎−2𝛽𝑎₀𝑏₁𝜆𝑎₁𝜇³𝜙=0,₂𝜓∶𝛽𝑎³₁𝜇⁵−24𝑎₁𝜇³𝜆²𝜎+12𝑎₁𝜇⁵+12𝑎₁𝜇𝜆⁴𝜎²+𝛽𝑎³₁𝜇𝜆⁴𝜎²+7𝛽𝑏²₁𝜆³𝑎₁𝜇𝜎−4𝛽𝑎₀𝑎₁𝑏₁𝜆⁴𝜎²−2𝛽𝑎³₁𝜇³𝜆²𝜎−4𝛽𝑎₀𝑎₁𝑏₁𝜇⁴−2𝛼𝑎₁𝑏₁𝜇⁴+4𝛼𝑎₁𝑏₁𝜆²𝜎𝜇²+8𝛽𝑎₀𝑎₁𝑏₁𝜆²𝜎𝜇²−7𝛽𝑏²₁𝜆𝑎₁𝜇³−2𝛼𝑎₁𝑏₁𝜆⁴𝜎²𝜙=0,₁∶−𝛼𝑏²₁𝜆⁴𝜎−𝛼𝑎²₁𝜆⁵𝜎²−𝛼𝑎²₁𝜆𝜇⁴−2𝛽𝑎₀𝑎²₁𝜆𝜇⁴−2𝛽𝑎²₁𝑏₁𝜆²𝜇³+𝛼𝑏²₁𝜆²𝜇²+2𝛽𝑏³₁𝜆³𝜇+2𝛽𝑎²₁𝑏₁𝜆⁴𝜇𝜎+6𝑏₁𝜇𝜆⁴𝜎−6𝑏₁𝜇³𝜆²+4𝛽𝑎₀𝑎²₁𝜆³𝜎𝜇²−2𝛽𝑎₀𝑎²₁𝜆⁵𝜎²+2𝛽𝑎₀𝑏²₁𝜆²𝜇²−2𝛽𝑎₀𝑏²₁𝜆⁴𝜎+2𝛼𝑎²₁𝜆³𝜎𝜇²𝜙=0,₁𝜓∶−𝛽𝑏³₁𝜆⁴𝜎−𝛽𝑎²₀𝑏₁𝜇⁴+𝑉𝑏₁𝜆⁴𝜎²−𝛼𝑎₀𝑏₁𝜇⁴−3𝛽𝑏³₁𝜆²𝜇²−2𝑏₁𝜆³𝜎𝜇²+2𝛽𝑎₀𝑎²₁𝜇⁵−2𝛼𝑏²₁𝜆𝜇³+𝑉𝑏₁𝜇⁴+𝛼𝑎²₁𝜇⁵−5𝑏₁𝜆⁵𝜎²+7𝑏₁𝜆𝜇⁴−𝛽𝑎²₀𝑏₁𝜆⁴𝜎²−𝛼𝑎₀𝑏₁𝜆⁴𝜎²+𝛼𝑎²₁𝜇𝜆⁴𝜎²+2𝛽𝑎²₀𝑏₁𝜆²𝜎𝜇²+2𝛽𝑎₀𝑎²₁𝜇𝜆⁴𝜎²−4𝛽𝑎₀𝑎²₁𝜇³𝜆²𝜎−2𝛼𝑎²₁𝜇³𝜆²𝜎−2𝑉𝑏₁𝜆²𝜎𝜇²+2𝛼𝑎₀𝑏₁𝜆²𝜎𝜇²−2𝛽𝑎²₁𝑏₁𝜆⁵𝜎²+2𝛽𝑎²₁𝑏₁𝜆𝜇⁴+2𝛼𝑏²₁𝜆³𝜇𝜎+4𝛽𝑎₀𝑏²₁𝜆³𝜇𝜎−4𝛽𝑎₀𝑏²₁𝜆𝜇³𝜙=0,₀∶𝑎₁𝜆⁴𝜎𝜇²+𝑉𝑎₁𝜆⁵𝜎²+𝑉𝑎₁𝜆𝜇⁴+𝛼𝑏₁𝜆⁴𝑎₁𝜇𝜎−𝛽𝑏²₁𝜆³𝑎₁𝜇²+𝑎₁𝜆²𝜇⁴−2𝑎₁𝜆⁶𝜎²−𝛽𝑎²₀𝑎₁𝜆⁵𝜎²−𝛽𝑎²₀𝑎₁𝜆𝜇⁴−𝛼𝑎₀𝑎₁𝜆⁵𝜎²−𝛼𝑎₀𝑎₁𝜆𝜇⁴+2𝛽𝑎²₀𝑎₁𝜆³𝜎𝜇²−2𝑉𝑎₁𝜆³𝜎𝜇²−𝛼𝑏₁𝜆²𝑎₁𝜇³+2𝛼𝑎₀𝑎₁𝜆³𝜎𝜇²−2𝛽𝑎₀𝑏₁𝜆²𝑎₁𝜇³−𝛽𝑏²₁𝜆⁵𝑎₁𝜎+2𝛽𝑎₀𝑏₁𝜆⁴𝑎₁𝜙𝜇𝜎=0,₀𝜓∶−𝑉𝑎₁𝜇⁵−𝑉𝑎₁𝜇𝜆⁴𝜎²−𝑎₁𝜇⁵𝜆−𝛼𝑏₁𝑎₁𝜆⁵𝜎²+𝛼𝑏₁𝑎₁𝜆𝜇⁴+𝛽𝑎²₀𝑎₁𝜇⁵+𝛽𝑏²₁𝜆²𝑎₁𝜇³+5𝑎₁𝜇𝜆⁵𝜎²+𝛽𝑎²₀𝑎₁𝜇𝜆⁴𝜎²+3𝛽𝑏²₁𝜆⁴𝑎₁𝜇𝜎+𝛼𝑎₀𝑎₁𝜇⁵+2𝑉𝑎₁𝜇³𝜆²𝜎−4𝑎₁𝜇³𝜆³𝜎+𝛼𝑎₀𝑎₁𝜇𝜆⁴𝜎²−2𝛽𝑎₀𝑏₁𝑎₁𝜆⁵𝜎²+2𝛽𝑎₀𝑏₁𝑎₁𝜆𝜇⁴−2𝛼𝑎₀𝑎₁𝜇³𝜆²𝜎−2𝛽𝑎²₀𝑎₁𝜇³𝜆²𝜎=0. (3.14)
Solving the algebraic equations (3.14) by the Maple or Mathematica, we obtain the following results.

Result 1. We have 𝑎₀=−1𝛽𝛼2±3𝜇𝛽𝜆6𝜇²−𝜆²𝜎,𝑎₁=0,𝑏₁=±6𝜇²−𝜆²𝜎,𝛽𝜆𝑉=−4𝜎𝛽𝜆³+𝜎𝛼²𝜆²+2𝛽𝜆𝜇²−𝛼²𝜇²4𝛽𝜆²𝜎−𝜇²,𝜎=𝐴²₁+𝐴²₂. (3.15)
From (2.5), (3.2), and (3.15), we deduce the traveling wave solution of (1.1) as follows: 𝑢(𝜉)=−1𝛽𝛼2±3𝜇𝛽𝜆6𝜇²−𝜆²𝜎±6𝜇²−𝜆²𝜎×1𝛽𝜆𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆,+𝜇/𝜆 (3.16) where 𝜉=𝑥+4𝜎𝛽𝜆³+𝜎𝛼²𝜆²+2𝛽𝜆𝜇²−𝛼²𝜇²4𝛽𝜆²𝜎−𝜇²𝑡. (3.17)
In particular, by setting 𝐴₁=0,𝐴₂>0, and 𝜇=0 in (3.16), we have the periodic solution 𝑢(𝜉)=−𝛼±2𝛽−6𝜆𝛽sec𝜉√𝜆, (3.18) while, if 𝐴₂=0,𝐴₁>0, and 𝜇=0, then we have the periodic solution 𝑢(𝜉)=−𝛼±2𝛽−6𝜆𝛽csc𝜉√𝜆. (3.19)

Result 2. We have 𝑎₀=−𝛼2𝛽,𝑎₁=±−32𝛽,𝑏₁=±3𝜇²−𝜆²𝜎,2𝛽𝜆𝑉=2𝛽𝜆−𝛼²4𝛽,𝜎=𝐴²₁+𝐴²₂. (3.20)
In this result, we deduce the traveling wave solution of (1.1) as follows: 𝑢(𝜉)=−𝛼±2𝛽−312𝛽𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆×√+𝜇/𝜆𝜆𝐴₁cos𝜉√𝜆−𝐴₂sin𝜉√𝜆+𝜆²𝜎−𝜇²𝜆, (3.21) where 𝜉=𝑥−2𝛽𝜆−𝛼²4𝛽𝑡. (3.22)
In particular, by setting 𝐴₁=0,𝐴₂>0, and 𝜇=0 in (3.21), we have the periodic solution 𝑢(𝜉)=−𝛼±2𝛽−3𝜆𝜉√2𝛽−tan𝜆+sec𝜉√𝜆, (3.23) while, if 𝐴₂=0,𝐴₁>0, and 𝜇=0, then we have the periodic solution 𝑢(𝜉)=−𝛼±2𝛽−3𝜆2𝛽cot𝜉√𝜆+csc𝜉√𝜆. (3.24)

Case 3. Rational function solutions (𝜆=0).
If 𝜆=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:𝜙⁴∶𝛽𝑎³₁𝐴⁴₁−4𝛽𝑎³₁𝐴²₁𝜇𝐴₂−24𝑎₁𝐴²₁𝜇𝐴₂+24𝑎₁𝜇²𝐴²₂+4𝛽𝑎³₁𝜇²𝐴²₂+6𝑎₁𝐴⁴₁+3𝛽𝑎₁𝑏²₁𝐴²₁−6𝛽𝑎₁𝑏²₁𝜇𝐴₂𝜙=0,₃∶−4𝛽𝑎₀𝑏²₁𝜇𝐴₂−6𝑏₁𝜇𝐴²₁+4𝛼𝑎²₁𝜇²𝐴²₂+2𝛽𝑎₀𝑎²₁𝐴⁴₁+8𝛽𝑎₀𝑎²₁𝜇²𝐴²₂−2𝛼𝑏²₁𝜇𝐴₂+12𝑏₁𝜇²𝐴₂+2𝛽𝑎₀𝑏²₁𝐴²₁−2𝛽𝑏³₁𝜇+𝛼𝑎²₁𝐴⁴₁−2𝛽𝑎²₁𝑏₁𝜇𝐴²₁−8𝛽𝑎₀𝑎²₁𝐴²₁𝜇𝐴₂+𝛼𝑏²₁𝐴²₁+4𝛽𝑎²₁𝑏₁𝜇²𝐴₂−4𝛼𝑎²₁𝐴²₁𝜇𝐴₂𝜙=0,₃𝜓∶12𝛽𝑎²₁𝑏₁𝜇²𝐴²₂+6𝑏₁𝐴⁴₁+𝛽𝑏³₁𝐴²₁−2𝛽𝑏³₁𝜇𝐴₂+3𝛽𝑎²₁𝑏₁𝐴⁴₁−12𝛽𝑎²₁𝑏₁𝐴²₁𝜇𝐴₂−24𝑏₁𝐴²₁𝜇𝐴₂+24𝑏₁𝜇²𝐴²₂𝜙=0,₂∶𝛼𝑎₀𝑎₁𝐴⁴₁+𝛽𝑎²₀𝑎₁𝐴⁴₁+2𝛽𝑏²₁𝑎₁𝜇²−4𝑉𝑎₁𝜇²𝐴²₂−𝑉𝑎₁𝐴⁴₁+3𝑎₁𝜇²𝐴²₁−6𝑎₁𝜇³𝐴₂−𝛼𝑏₁𝑎₁𝜇𝐴²₁+4𝑉𝑎₁𝐴²₁𝜇𝐴₂−4𝛽𝑎²₀𝑎₁𝐴²₁𝜇𝐴₂+4𝛽𝑎²₀𝑎₁𝜇²𝐴²₂−4𝛼𝑎₀𝑎₁𝐴²₁𝜇𝐴₂+4𝛼𝑎₀𝑎₁𝜇²𝐴²₂+2𝛼𝑏₁𝑎₁𝜇²𝐴₂−2𝛽𝑎₀𝑏₁𝑎₁𝜇𝐴²₁+4𝛽𝑎₀𝑏₁𝑎₁𝜇²𝐴₂𝜙=0,₂𝜓∶−48𝑎₁𝜇³𝐴²₂+48𝑎₁𝜇²𝐴²₁𝐴₂−12𝑎₁𝜇𝐴⁴₁−7𝛽𝑏²₁𝑎₁𝜇𝐴²₁+4𝛽𝑎³₁𝜇²𝐴²₁𝐴₂−𝛽𝑎³₁𝜇𝐴⁴₁+8𝛼𝑎₁𝑏₁𝜇²𝐴²₂+14𝛽𝑏²₁𝑎₁𝜇²𝐴₂+2𝛼𝑎₁𝑏₁𝐴⁴₁−8𝛼𝑎₁𝑏₁𝐴²₁𝜇𝐴₂−4𝛽𝑎³₁𝜇³𝐴²₂+4𝛽𝑎₀𝑎₁𝑏₁𝐴⁴₁−16𝛽𝑎₀𝑎₁𝑏₁𝐴²₁𝜇𝐴₂+16𝛽𝑎₀𝑎₁𝑏₁𝜇²𝐴²₂𝜙=0,₁𝜓∶𝛼𝑎₀𝑏₁𝐴⁴₁−𝛼𝑎²₁𝜇𝐴⁴₁+𝛽𝑎²₀𝑏₁𝐴⁴₁−4𝑉𝑏₁𝜇²𝐴²₂−4𝛼𝑎²₁𝜇³𝐴²₂−2𝛼𝑏²₁𝜇𝐴²₁+4𝛼𝑏²₁𝜇²𝐴₂−𝑉𝑏₁𝐴⁴₁+4𝛽𝑏³₁𝜇²+12𝑏₁𝜇²𝐴²₁−24𝑏₁𝜇³𝐴₂+4𝑉𝑏₁𝐴²₁𝜇𝐴₂+4𝛼𝑎₀𝑏₁𝜇²𝐴²₂−4𝛼𝑎₀𝑏₁𝑎²₁𝜇𝐴₂−2𝛽𝑎₀𝑎²₁𝜇𝐴⁴₁−4𝛽𝑎²₀𝑏₁𝐴²₁𝜇𝐴₂+4𝛽𝑎²₀𝑏₁𝜇²𝐴²₂+4𝛼𝑎²₁𝜇²𝐴²₁𝐴₂−4𝛽𝑎₀𝑏²₁𝜇𝐴²₁+8𝛽𝑎₀𝑏²₁𝜇²𝐴₂+4𝛽𝑎²₁𝑏₁𝜇²𝐴²₁+8𝛽𝑎₀𝑎²₁𝜇²𝐴²₁𝐴₂−8𝛽𝑎₀𝑎²₁𝜇³𝐴²₂−8𝛽𝑎²₁𝑏₁𝜇³𝐴₂𝜙=0,₀𝜓∶𝑉𝑎₁𝜇𝐴⁴₁−4𝛽𝑏²₁𝑎₁𝜇³+4𝑉𝑎₁𝜇³𝐴²₂−6𝑎₁𝜇³𝐴²₁+12𝑎₁𝜇⁴𝐴₂+4𝛼𝑎₀𝑎₁𝜇²𝐴²₁𝐴₂−𝛼𝑎₀𝑎₁𝜇𝐴⁴₁−𝛽𝑎²₀𝑎₁𝜇𝐴⁴₁−4𝛼𝑎₀𝑎₁𝜇³𝐴²₂+4𝛽𝑎²₀𝑎₁𝜇²𝐴²₁𝐴₂−4𝛽𝑎²₀𝑎₁𝜇³𝐴²₂−4𝑉𝑎₁𝜇²𝐴²₁𝐴₂−4𝛼𝑏₁𝑎₁𝜇³𝐴₂+4𝛽𝑎₀𝑏₁𝑎₁𝜇²𝐴²₁+2𝛼𝑏₁𝑎₁𝜇²𝐴²₁−8𝛽𝑎₀𝑏₁𝑎₁𝜇³𝐴₂=0. (3.25)
Solving the algebraic equations (3.25) by the Maple or Mathematica, we obtain the following results.

Result 1. We have 𝑎₀=−1𝛽𝛼2±3𝜇𝛽62𝜇𝐴₂−𝐴²₁,𝑎₁=0,𝑏₁=±62𝜇𝐴₂−𝐴²₁𝛽,𝛼𝑉=−²𝐴²₁+6𝛽𝜇²−2𝛼²𝜇𝐴₂4𝛽𝐴²₁−2𝜇𝐴₂. (3.26)
From (2.7), (3.2) and (3.26), we deduce the traveling wave solution of (1.1) as follows: 𝑢(𝜉)=−1𝛽𝛼2±3𝜇𝛽62𝜇𝐴₂−𝐴²₁±62𝜇𝐴₂−𝐴²₁𝛽1𝜇/2𝜉²+𝐴₁𝜉+𝐴₂, (3.27) where 𝛼𝜉=𝑥+²𝐴²₁+6𝛽𝜇²−2𝛼²𝜇𝐴₂4𝛽𝐴²₁−2𝜇𝐴₂𝑡. (3.28)

Result 2. We have 𝑎₀=−𝛼2𝛽,𝑎₁=±−32𝛽,𝑏₁=±32𝜇𝐴₂−𝐴²₁2𝛽𝑉=−𝛼².4𝛽 (3.29)
In this result, we deduce the traveling wave solution of (1.1) as follows: 𝑢(𝜉)=−𝛼±2𝛽−32𝛽𝜇𝜉+𝐴₁+𝐴²₁−2𝜇𝐴₂𝜇/2𝜉²+𝐴₁𝜉+𝐴₂, (3.30) where 𝛼𝜉=𝑥+²4𝛽𝑡. (3.31)

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 𝑏_𝑖=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.

Acknowledgment

The authors wish to thank the referees for their suggestions and comments on this paper.