# Solutions and Painlevé Property for the KdV Equation with Self-Consistent Source

Authors: Yali Shen; Ruoxia Yao
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

In this paper, the polynomial solutions in terms of Jacobi’s elliptic functions of the KdV equation with a self-consistent source (KdV-SCS) are presented. The extended(G′/G)-expansion method is utilized to obtain exact traveling wave solutions of the KdV-SCS, which finally are expressed in terms of the hyperbolic function, the trigonometric function, and the rational function. Meanwhile we find the Lie point symmetry and Lie symmetry group and give several group-invariant solutions for the KdV-SCS. Finally, we supplement the results of the Painlevé property in our previous work and get the Bäcklund transformations of the KdV-SCS.


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## Body

## 1. Introduction

It is well known that the soliton equations with self-consistent sources (SESCSs) can exhibit abundant nonlinear dynamics compared to soliton equations themselves and have important physical applications [1]. These SESCSs are usually used to describe interactions between different solitary waves and are relevant in some problems related with hydrodynamics, solid state physics, or plasma physics [2–4]. The sources appear in solitary waves with nonconstant velocity and lead to a variety of dynamics of physical models [2]. For example, the KdV equation with a self-consistent source (KdV-SCS) describes the interaction of long and short capillary-gravity waves [5, 6]. During the past four decades or so searching for explicit solutions of nonlinear evolution equations by using various different methods is the main goal for many researchers. Many powerful methods to construct exact solutions of nonlinear evolution equations have been established and developed, which can be used to deal with the SESCSs as well. For instance, some equations with self-consistent sources have been studied by the inverse scattering method [5, 7], Darboux transformation method [8, 9], and Hirota method [10–12].In recent years, many effectively straightforward methods have been proposed such as the Jacobi elliptic function expansion method [13, 14], the tanh-function expansion method [15, 16], the F-expansion method [17, 18], and the (G′/G)-expansion method [19–21]. Motivated by the previous works, we focus our attention on the following nonlinear partial differential equations (PDEs):(1a)ut+uxxx+12uux=v2x,(1b)vxx+2uv=λv,where λ is an arbitrary constant. In fact, (1a)-(1b) is a reduced form of the KdV equation with source [22, 23].The general KdV-SCS has been discussed from various aspects, such that, with Wronskian technique, the mixed rational-soliton solutions for the KdV-SCS are obtained [22]. The complexion solutions of the KdV-SCS are presented by the Darboux transformation [23]. The soliton solutions for the KdV hierarchy with self-consistent sources are obtained by the inverse scattering method [5]. The integration of KdV-SCS and higher KdV-SCS in the class of periodic functions are, respectively, studied in [24, 25]. However, the Jacobi’s elliptic function solutions, the group-invariant solutions by the Lie group approach [26], and the extended (G′/G)-expansion method for the KdV-SCS have not been presented.This paper is organized as follows. In Section2, we construct the polynomial solutions in terms of Jacobi’s elliptic functions for (1a)-(1b). In Section 3, using the extended (G′/G)-expansion method, we obtain the exact traveling wave solutions of (1a)-(1b). In Section 4, we give the group-invariant solutions of (1a)-(1b) by the Lie group approach. In Section 5, the results of the Painlevé property for (1a)-(1b) are supplemented. Section 6 is a brief conclusion.


## 2. Jacobi’s Elliptic Function Solutions

In this section we mainly construct the polynomial solutions of (1a)-(1b) in terms of Jacobi’s elliptic functions [27].Consider the following tripled Riccati equations:(2)f′=gh,g′=-fh,h′=-M2fg,where ′=d/dξ and modulus 0<M<1, which have three solutions as listed(3)f=snξ;M,g=cnξ;M,h=dnξ;M.Solutions (3) satisfy(4)sn2ξ;M=1-cn2ξ;M,dn2ξ;M=1-M2sn2ξ;M.Step  J1. Transform (1a)-(1b) to ODEs.To begin with, by using the travelling wave transformation,(5)ξ=kx+ωt,where ξ is referred to the traveling wave variable and k and ω represent the amplitude and velocity of the traveling wave, respectively, and setting U(ξ)=u(x,t), V(ξ)=v(x,t), system (1a)-(1b) is transformed into the following nonlinear ordinary differential equations (ODEs):(6a)ωU+6kU2-kV2+k3U′′=0,(6b)k2V′′+2UV-λV=0,while ′=d/dξ.Step  J2. Determine the expressions of the polynomial solutions.Suppose that the solution of (6a)-(6b) can be expressed as the following finite series:(7)Uξ=∑i=0maifi+∑i=1mbigfi-1+∑i=1mcihfi-1,Vξ=∑i=0nAifi+∑i=1nBigfi-1+∑i=1nCihfi-1,where ai, bi, ci, Ai, Bi, and Ci are constants to be determined later and the positive integers m,n are determined by balancing the highest nonlinear terms and the highest-order partial derivative terms in (6a)-(6b) [28]. Then we get m=n=2, and(8)Uξ=a0+a1f+a2f2+b1g+b2gf+c1h+c2hf,Vξ=A0+A1f+A2f2+B1g+B2gf+C1h+C2hf.Step  J3. Derive the algebraic system for the coefficients ai, bi, ci, Ai, Bi, and Ci.Substituting (8) into (6a)-(6b) and repeatedly applying (4) and (2), and collecting all terms of same power of f,g,h together and setting each coefficient of the polynomials to zero, we get a system of algebraic equations for the unknowns ai, bi, ci, Ai, Bi, Ci, M, k, and ω, which on solving gives five sets of solutions for an algebraic system omitted here.Step  J4. Build and test the Jacobi’s elliptic function solutions.Substituting the above five sets of solutions separately into (8) and replacing f, g, h with (3), we get the Jacobi’s elliptic function solutions for (1a)-(1b) as follows.Solution 1: (9) u x , t = 1 2 3 k 2 λ - λ A 0 + 12 k 4 - 3 k 2 A 0 3 k 2 - A 0 - 3 4 A 0 4 k 2 - A 0 3 k 2 - A 0 s n 2 ξ ; M , v x , t = A 0 - 3 2 A 0 4 k 2 - A 0 3 k 2 - A 0 s n 2 ξ ; M , M = ± 1 k 4 k 2 A 0 - A 0 2 12 k 2 - 4 A 0 , A 0 = 30 k 3 + 6 k λ + ω ± - 300 k 6 + 36 k 2 λ 2 + 12 k λ ω + ω 2 10 k .Solution 2: (10) u x , t = 1 2 3 k 2 λ + λ A 0 + 12 k 4 + 3 k 2 A 0 3 k 2 + A 0 + 3 4 A 0 4 k 2 + A 0 3 k 2 + A 0 s n 2 ξ ; M , v x , t = A 0 - 3 2 A 0 4 k 2 + A 0 3 k 2 + A 0 s n 2 ξ ; M , M = ± 1 k - 4 k 2 A 0 + A 0 2 12 k 2 + 4 A 0 , A 0 = - 30 k 3 + 6 k λ + ω ± - 300 k 6 + 36 k 2 λ 2 + 12 k λ ω + ω 2 10 k .Solution 3: (11) u x , t = k 2 M 2 2 + λ 2 - k 2 M 2 s n 2 ξ ; M , v x , t = ± ω k + 2 k 4 M 2 + 6 k 2 λ - 4 k 4 d n ξ ; M .Solution 4: (12) u x , t = k 2 2 + λ 2 - k 2 M 2 s n 2 ξ ; M , v x , t = ± 2 k 4 - 4 k 4 M 2 + ω k + 6 k 2 λ M c n ξ ; M .Solution 5: (13) u x , t = 1 2 k 2 + 1 2 k 2 M 2 + λ 2 - k 2 M 2 s n 2 ξ ; M , v x , t = ± - k 2 k 3 + 2 k 3 M 2 + ω + 6 k λ M s n ξ ; M ,where ξ=kx+ωt, k, ω are arbitrary constants and 0<M<1.According tosn(ξ;0)=sin⁡(ξ), sn(ξ;1)=tanh⁡(ξ), cn(ξ;0)=cos⁡(ξ), cn(ξ;1)=sech(ξ), and cn(mξ;1/m)=dn(ξ;m), the above solutions can be expressed in terms of hyperbolic functions and the trigonometric functions. When setting M=1 in (9), we get A0=2k2 and A0=6k2. Substituting A0=2k2 into (9) and applying tanh2ξ+sech2ξ=1, we get the following typical travelling wave solution from (9): (14)ux,t=λ2+3k2sech2kx+ωt,vx,t=-4k2+6k2sech2kx+ωt.Likewise, when making M=1 in (10)-(13), we can also get solutions in terms of hyperbolic functions. It shows that the Jacobi’s elliptic function expansion method is more general than that of the hyperbolic function expansion method [29].


## 3. Extended(G′/G)-Expansion Method to KdV-SCS

### 3.1. Description of the Extended(G′/G)-Expansion Method

Let us have a look at the extended(G′/G)-expansion method briefly. For a given nonlinear PDE,(15)Fu,ut,ux,uxt,uxx,utt,…=0,where u=u(x,t) and F is a polynomial about u(x,t) and its various partial derivatives.To begin with, using the following traveling wave transformation,(16)ux,t=Uξ,ξ=kx+ωt,Equation (15) reduces to the following ODE:(17)PU,U′,U′′,…,Un=0,where ′=d/dξ.Suppose that the solution of (17) can be expressed as a finite series in (G′/G)(18)Uξ=∑i=0maiG′Gi,am≠0,m∈N,where we let G=G(ξ) satisfy the following ODE instead of that form in [30](19)GG′′=αG′2+βGG′+γG2,where ai(i=0,1,2,…,m), α, β, γ are constants to be determined later, and the positive integer m can be determined the same as before.Substituting (18) along with (19) into (17) and equating the coefficients of each power of G′/G to zeros, we obtain a system of algebraic equations for the unknowns ai(i=1,2,…,m), k, and ω. Then we can determine the unknowns.Equation (19) possesses solutions listed below.Whenβ2-4(α-1)γ>0, α≠1,(20)G′G=β2-2α+β2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ.Whenβ2-4(α-1)γ<0, α≠1,(21)G′G=β2-2α+-β2+4αγ-4γ-C1sin⁡1/2-β2+4αγ-4γξ+C2cos⁡1/2-β2+4αγ-4γξ2-2αC1cos⁡1/2-β2+4αγ-4γξ+C2sin⁡1/2-β2+4αγ-4γξ.Whenβ2-4(α-1)γ=0, α≠1,(22)G′G=-ξβ+2+C1β2ξα-ξ+C1α-C1.In (20)-(22), C1, C2 are arbitrary constants.Therefore, by the sign of the discriminantβ2-4(α-1)γ, we can obtain the exact solutions of (15).


### 3.2. Application of the KdV-SCS

In this section, we demonstrate the extendedG′/G-expansion method on the KdV-SCS (1a)-(1b).Introducing the travelling wave transformations(23)ux,t=Uξ,vx,t=Vξ,where ξ=kx+ωt, k, ω will be determined later. Similar to the strategy in Step J1, substituting U(ξ)=U, V(ξ)=V into ((1a)-(1b)), one can transform (1a)-(1b) into nonlinear ODEs as (6a)-(6b).Now, we make the ansatzU(ξ)=∑i=0maiG′/Gi, V(ξ)=∑i=0nbiG′/Gi for the solutions of (6a)-(6b). Likewise, using the homogeneous balance method we obtain m=n=2. Therefore, the solutions of (6a)-(6b) have the following extended forms:(24)Uξ=a0+a1G′G+a2G′G2,Vξ=b0+b1G′G+b2G′G2.Substituting (24) and (19) into (6a)-(6b) and collecting all terms with the same power of G′/G together, the left-hand sides of (6a)-(6b) are converted into other polynomials in G′/G. Equating each coefficient of the polynomials to zero yields a set of simultaneous algebraic equations for k, ω, a0, a1, a2, b0, b1, and b2 omitted here. Solving the algebraic system leads to several types of traveling wave solutions under various parameter constraints for the KdV-SCS ((1a)-(1b)).Case 1.
β 2 - 4 α γ + 4 γ > 0, α≠1.
In this case, we get five sets traveling wave solutions of the KdV-SCS (1a)-(1b).Solution 1. The first set of the unknowns under this condition reads(25)k=k,ω=k4k2γα-k2β2-4k2γ-6λ+8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ,a0=-k2γα+k2γ+λ2,a1=-βα-1k2,a2=-1-α2k2,b0=±12βk-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ,b1=±-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λα-1k,b2=0.Substituting (25) into (24), we get the following solution expression equations:(26a)Uξ=-k2γα+k2γ+λ2-βα-1k2G′G-1-α2k2G′G2,(26b)Vξ=±12βk-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ±-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λα-1kG′G.Whence, from the ansatz (26a)-(26b) together with (20), the first set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(27a)ux,t=-k2γα+k2γ+λ2-βα-1k2β2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α-1-α2k2β2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,(27b)vx,t=±12βk-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ±-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λα-1kβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α,where C1 and C2 are arbitrary constants and the traveling wave variable is(28)ξ=kx+k4k2γα-k2β2-4k2γ-6λ+8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ+8k2λγα-2k2β2λ-8k2λγ-6λ22t.Solution 2. The second set of the unknowns under this condition reads(29)k=±λ4αγ-β2-4γ,ω=∓λ4αγ-β2-4γλ,a0=-λ2γα+β2-2γ24αγ-β2-4γ,a1=-3βλα-14αγ-β2-4γ,a2=-3λα-124αγ-β2-4γ,b0=±λ2γα+β2-2γ4αγ-β2-4γ,b1=±6βλα-14αγ-β2-4γ,b2=±6λα-124αγ-β2-4γ.Substituting (29) into (24), we get the following solution expression equations:(30a)Uξ=-λ2γα+β2-2γ24αγ-β2-4γ-3βλα-14αγ-β2-4γG′G-3λα-124αγ-β2-4γG′G2,(30b)Vξ=±λ2γα+β2-2γ4αγ-β2-4γ±6βλα-14αγ-β2-4γG′G±6λα-124αγ-β2-4γG′G2.Whence, from the ansatz (30a)-(30b) together with (20), the second set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(31a)
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,where C1 and C2 are arbitrary constants and the traveling wave variable is(32)ξ=±λ4αγ-β2-4γx∓λ4αγ-β2-4γλtwithλ<0.Solution 3. The third set of the unknowns under this condition reads(33)k=±-λ4αγ-β2-4γ,ω=∓-λ4αγ-β2-4γλ,a0=3γλα-14αγ-β2-4γ,a1=3βλα-14αγ-β2-4γ,a2=3λα-124αγ-β2-4γ,b0=±6γλα-14αγ-β2-4γ,b1=±6βλα-14αγ-β2-4γ,b2=±6λα-124αγ-β2-4γ.Substituting (33) into (24), we get the following solution expression equations:(34a)Uξ=3γλα-14αγ-β2-4γ+3βλα-14αγ-β2-4γG′G+3λα-124αγ-β2-4γG′G2,(34b)Vξ=±6γλα-14αγ-β2-4γ±6βλα-14αγ-β2-4γG′G±6λα-124αγ-β2-4γG′G2.Whence, from the ansatz (34a)-(34b) together with (20), the third set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(35a)ux,t=3γλα-14αγ-β2-4γ+3βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α+3λα-124αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,(35b)vx,t=±6γλα-14αγ-β2-4γ±6βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α±6λα-124αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,where C1 and C2 are arbitrary constants and the traveling wave variable is(36)ξ=±-λ4αγ-β2-4γx∓-λ4αγ-β2-4γλtwithλ>0.Solution 4. The fourth set of the unknowns under this condition reads(37)k=±3λ8αγ-2β2-8γ,ω=±323λ8αγ-2β2-8γλ,a0=-λ5γα+β2-5γ24αγ-β2-4γ,a1=-9βλα-124αγ-β2-4γ,a2=-9λα-1224αγ-β2-4γ,b0=±3λ2γα+β2-2γ24αγ-β2-4γ,b1=±9βλα-14αγ-β2-4γ,b2=±9λα-124αγ-β2-4γ.Substituting (37) into (24), we get the following solution expression equations:(38a)Uξ=-λ5γα+β2-5γ24αγ-β2-4γ-9βλα-124αγ-β2-4γG′G-9λα-1224αγ-β2-4γG′G2,(38b)Vξ=±3λ2γα+β2-2γ24αγ-β2-4γ±9βλα-14αγ-β2-4γG′G±9λα-124αγ-β2-4γG′G2.Whence, from the ansatz (38a)-(38b) together with (20), the fourth set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads (39a)ux,t=-λ5γα+β2-5γ24αγ-β2-4γ-9βλα-124αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α-9λα-1224αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,(39b)vx,t=±3λ2γα+β2-2γ24αγ-β2-4γ±9βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α±9λα-124αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α2,where C1 and C2 are arbitrary constants and the traveling wave variable is(40)ξ=±3λ8αγ-2β2-8γx±323λ8αγ-2β2-8γλtwithλ<0.Solution 5. The fifth set of the unknowns under this condition reads(41)k=±-3λ8αγ-2β2-8γ,ω=±32-3λ8αγ-2β2-8γλ,a0=λ14γα+β2-14γ44αγ-β2-4γ,a1=9βλα-124αγ-β2-4γ,a2=9λα-1224αγ-β2-4γ,b0=±9γλα-14αγ-β2-4γ,b1=±9βλα-14αγ-β2-4γ,b2=±9λα-124αγ-β2-4γ.Substituting (41) into (24), we get the following solution expression equations:(42a)Uξ=λ14γα+β2-14γ44αγ-β2-4γ+9βλα-124αγ-β2-4γG′G+9λα-1224αγ-β2-4γG′G2,(42b)Vξ=±9γλα-14αγ-β2-4γ±9βλα-14αγ-β2-4γG′G±9λα-124αγ-β2-4γG′G2.Whence, from the ansatz (42a)-(42b) together with (20), the fifth set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(43a)ux,t=λ14γα+β2-14γ44αγ-β2-4γ+9βλα-124αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α+9λα-1224αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α2,(43b)vx,t=±9γλα-14αγ-β2-4γ±9βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α±9λα-124αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α2,where C1 and C2 are arbitrary constants and the traveling wave variable is(44)ξ=±-3λ8αγ-2β2-8γx±32-3λ8αγ-2β2-8γλtwithλ>0.Case 2.
β 2 - 4 α γ + 4 γ < 0, α≠1.In this case, each set of the unknowns and the correspondingU(ξ) and V(ξ) on this occasion consists with that in the case of β2-4αγ+4γ>0. We only need to replace G′/G of each group of solutions in Case 1 to (21). So the other five sets of traveling wave solutions are obtained and denoted by solutions 6-10.Solution 11. The set of the unknowns under this condition reads(45)k=±-a2α-1,ω=∓3-a2λα-1,a0=142α2λ-4αλ+2λ+β2a2α-12,a1=βa2α-1,b0=±123λa2βα-1,b1=±3λa2,b2=0.Substituting (45) into (24) together with (22), we get the eleventh set of travelling wave solutions in terms of rational functions to (1a)-(1b) as follows:(46a)ux,t=142α2λ-4αλ+2λ+β2a2α-12-12βa2ξβ+2+C1βα-1ξα-ξ+C1α-C1+14a2ξβ+2+C1β2ξα-ξ+C1α-C12,(46b)vx,t=±123λa2βα-1∓123λa2ξβ+2+C1βξα-ξ+C1α-C1,where C1 is arbitrary constant and a2<0 and the traveling wave variable is(47)ξ=±-a2α-1x∓3-a2λα-1twithλ<0.


## 3.1. Description of the Extended(G′/G)-Expansion Method

Let us have a look at the extended(G′/G)-expansion method briefly. For a given nonlinear PDE,(15)Fu,ut,ux,uxt,uxx,utt,…=0,where u=u(x,t) and F is a polynomial about u(x,t) and its various partial derivatives.To begin with, using the following traveling wave transformation,(16)ux,t=Uξ,ξ=kx+ωt,Equation (15) reduces to the following ODE:(17)PU,U′,U′′,…,Un=0,where ′=d/dξ.Suppose that the solution of (17) can be expressed as a finite series in (G′/G)(18)Uξ=∑i=0maiG′Gi,am≠0,m∈N,where we let G=G(ξ) satisfy the following ODE instead of that form in [30](19)GG′′=αG′2+βGG′+γG2,where ai(i=0,1,2,…,m), α, β, γ are constants to be determined later, and the positive integer m can be determined the same as before.Substituting (18) along with (19) into (17) and equating the coefficients of each power of G′/G to zeros, we obtain a system of algebraic equations for the unknowns ai(i=1,2,…,m), k, and ω. Then we can determine the unknowns.Equation (19) possesses solutions listed below.Whenβ2-4(α-1)γ>0, α≠1,(20)G′G=β2-2α+β2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ.Whenβ2-4(α-1)γ<0, α≠1,(21)G′G=β2-2α+-β2+4αγ-4γ-C1sin⁡1/2-β2+4αγ-4γξ+C2cos⁡1/2-β2+4αγ-4γξ2-2αC1cos⁡1/2-β2+4αγ-4γξ+C2sin⁡1/2-β2+4αγ-4γξ.Whenβ2-4(α-1)γ=0, α≠1,(22)G′G=-ξβ+2+C1β2ξα-ξ+C1α-C1.In (20)-(22), C1, C2 are arbitrary constants.Therefore, by the sign of the discriminantβ2-4(α-1)γ, we can obtain the exact solutions of (15).


## 3.2. Application of the KdV-SCS

In this section, we demonstrate the extendedG′/G-expansion method on the KdV-SCS (1a)-(1b).Introducing the travelling wave transformations(23)ux,t=Uξ,vx,t=Vξ,where ξ=kx+ωt, k, ω will be determined later. Similar to the strategy in Step J1, substituting U(ξ)=U, V(ξ)=V into ((1a)-(1b)), one can transform (1a)-(1b) into nonlinear ODEs as (6a)-(6b).Now, we make the ansatzU(ξ)=∑i=0maiG′/Gi, V(ξ)=∑i=0nbiG′/Gi for the solutions of (6a)-(6b). Likewise, using the homogeneous balance method we obtain m=n=2. Therefore, the solutions of (6a)-(6b) have the following extended forms:(24)Uξ=a0+a1G′G+a2G′G2,Vξ=b0+b1G′G+b2G′G2.Substituting (24) and (19) into (6a)-(6b) and collecting all terms with the same power of G′/G together, the left-hand sides of (6a)-(6b) are converted into other polynomials in G′/G. Equating each coefficient of the polynomials to zero yields a set of simultaneous algebraic equations for k, ω, a0, a1, a2, b0, b1, and b2 omitted here. Solving the algebraic system leads to several types of traveling wave solutions under various parameter constraints for the KdV-SCS ((1a)-(1b)).Case 1.
β 2 - 4 α γ + 4 γ > 0, α≠1.
In this case, we get five sets traveling wave solutions of the KdV-SCS (1a)-(1b).Solution 1. The first set of the unknowns under this condition reads(25)k=k,ω=k4k2γα-k2β2-4k2γ-6λ+8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ,a0=-k2γα+k2γ+λ2,a1=-βα-1k2,a2=-1-α2k2,b0=±12βk-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ,b1=±-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λα-1k,b2=0.Substituting (25) into (24), we get the following solution expression equations:(26a)Uξ=-k2γα+k2γ+λ2-βα-1k2G′G-1-α2k2G′G2,(26b)Vξ=±12βk-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ±-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λα-1kG′G.Whence, from the ansatz (26a)-(26b) together with (20), the first set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(27a)ux,t=-k2γα+k2γ+λ2-βα-1k2β2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α-1-α2k2β2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,(27b)vx,t=±12βk-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ±-8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λα-1kβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α,where C1 and C2 are arbitrary constants and the traveling wave variable is(28)ξ=kx+k4k2γα-k2β2-4k2γ-6λ+8k2λγα-2k2β2λ-8k2λγ-6λ24k2γα-k2β2-4k2γ-2λ+8k2λγα-2k2β2λ-8k2λγ-6λ22t.Solution 2. The second set of the unknowns under this condition reads(29)k=±λ4αγ-β2-4γ,ω=∓λ4αγ-β2-4γλ,a0=-λ2γα+β2-2γ24αγ-β2-4γ,a1=-3βλα-14αγ-β2-4γ,a2=-3λα-124αγ-β2-4γ,b0=±λ2γα+β2-2γ4αγ-β2-4γ,b1=±6βλα-14αγ-β2-4γ,b2=±6λα-124αγ-β2-4γ.Substituting (29) into (24), we get the following solution expression equations:(30a)Uξ=-λ2γα+β2-2γ24αγ-β2-4γ-3βλα-14αγ-β2-4γG′G-3λα-124αγ-β2-4γG′G2,(30b)Vξ=±λ2γα+β2-2γ4αγ-β2-4γ±6βλα-14αγ-β2-4γG′G±6λα-124αγ-β2-4γG′G2.Whence, from the ansatz (30a)-(30b) together with (20), the second set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(31a)
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α
2
,where C1 and C2 are arbitrary constants and the traveling wave variable is(32)ξ=±λ4αγ-β2-4γx∓λ4αγ-β2-4γλtwithλ<0.Solution 3. The third set of the unknowns under this condition reads(33)k=±-λ4αγ-β2-4γ,ω=∓-λ4αγ-β2-4γλ,a0=3γλα-14αγ-β2-4γ,a1=3βλα-14αγ-β2-4γ,a2=3λα-124αγ-β2-4γ,b0=±6γλα-14αγ-β2-4γ,b1=±6βλα-14αγ-β2-4γ,b2=±6λα-124αγ-β2-4γ.Substituting (33) into (24), we get the following solution expression equations:(34a)Uξ=3γλα-14αγ-β2-4γ+3βλα-14αγ-β2-4γG′G+3λα-124αγ-β2-4γG′G2,(34b)Vξ=±6γλα-14αγ-β2-4γ±6βλα-14αγ-β2-4γG′G±6λα-124αγ-β2-4γG′G2.Whence, from the ansatz (34a)-(34b) together with (20), the third set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(35a)ux,t=3γλα-14αγ-β2-4γ+3βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α+3λα-124αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,(35b)vx,t=±6γλα-14αγ-β2-4γ±6βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α±6λα-124αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,where C1 and C2 are arbitrary constants and the traveling wave variable is(36)ξ=±-λ4αγ-β2-4γx∓-λ4αγ-β2-4γλtwithλ>0.Solution 4. The fourth set of the unknowns under this condition reads(37)k=±3λ8αγ-2β2-8γ,ω=±323λ8αγ-2β2-8γλ,a0=-λ5γα+β2-5γ24αγ-β2-4γ,a1=-9βλα-124αγ-β2-4γ,a2=-9λα-1224αγ-β2-4γ,b0=±3λ2γα+β2-2γ24αγ-β2-4γ,b1=±9βλα-14αγ-β2-4γ,b2=±9λα-124αγ-β2-4γ.Substituting (37) into (24), we get the following solution expression equations:(38a)Uξ=-λ5γα+β2-5γ24αγ-β2-4γ-9βλα-124αγ-β2-4γG′G-9λα-1224αγ-β2-4γG′G2,(38b)Vξ=±3λ2γα+β2-2γ24αγ-β2-4γ±9βλα-14αγ-β2-4γG′G±9λα-124αγ-β2-4γG′G2.Whence, from the ansatz (38a)-(38b) together with (20), the fourth set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads (39a)ux,t=-λ5γα+β2-5γ24αγ-β2-4γ-9βλα-124αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α-9λα-1224αγ-β2-4γβ2-4αγ+4γC1sinh⁡1/2β2-4αγ+4γξ+C2cosh⁡1/2β2-4αγ+4γξ2-2αC1cosh⁡1/2β2-4αγ+4γξ+C2sinh⁡1/2β2-4αγ+4γξ+β2-2α2,(39b)vx,t=±3λ2γα+β2-2γ24αγ-β2-4γ±9βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α±9λα-124αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α2,where C1 and C2 are arbitrary constants and the traveling wave variable is(40)ξ=±3λ8αγ-2β2-8γx±323λ8αγ-2β2-8γλtwithλ<0.Solution 5. The fifth set of the unknowns under this condition reads(41)k=±-3λ8αγ-2β2-8γ,ω=±32-3λ8αγ-2β2-8γλ,a0=λ14γα+β2-14γ44αγ-β2-4γ,a1=9βλα-124αγ-β2-4γ,a2=9λα-1224αγ-β2-4γ,b0=±9γλα-14αγ-β2-4γ,b1=±9βλα-14αγ-β2-4γ,b2=±9λα-124αγ-β2-4γ.Substituting (41) into (24), we get the following solution expression equations:(42a)Uξ=λ14γα+β2-14γ44αγ-β2-4γ+9βλα-124αγ-β2-4γG′G+9λα-1224αγ-β2-4γG′G2,(42b)Vξ=±9γλα-14αγ-β2-4γ±9βλα-14αγ-β2-4γG′G±9λα-124αγ-β2-4γG′G2.Whence, from the ansatz (42a)-(42b) together with (20), the fifth set of traveling wave solutions in terms of hyperbolic functions to (1a)-(1b) reads(43a)ux,t=λ14γα+β2-14γ44αγ-β2-4γ+9βλα-124αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α+9λα-1224αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α2,(43b)vx,t=±9γλα-14αγ-β2-4γ±9βλα-14αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α±9λα-124αγ-β2-4γβ2-4αγ+4γC1sinh1/2β2-4αγ+4γξ+C2cosh1/2β2-4αγ+4γξ2-2αC1cosh1/2β2-4αγ+4γξ+C2sinh1/2β2-4αγ+4γξ+β2-2α2,where C1 and C2 are arbitrary constants and the traveling wave variable is(44)ξ=±-3λ8αγ-2β2-8γx±32-3λ8αγ-2β2-8γλtwithλ>0.Case 2.
β 2 - 4 α γ + 4 γ < 0, α≠1.In this case, each set of the unknowns and the correspondingU(ξ) and V(ξ) on this occasion consists with that in the case of β2-4αγ+4γ>0. We only need to replace G′/G of each group of solutions in Case 1 to (21). So the other five sets of traveling wave solutions are obtained and denoted by solutions 6-10.Solution 11. The set of the unknowns under this condition reads(45)k=±-a2α-1,ω=∓3-a2λα-1,a0=142α2λ-4αλ+2λ+β2a2α-12,a1=βa2α-1,b0=±123λa2βα-1,b1=±3λa2,b2=0.Substituting (45) into (24) together with (22), we get the eleventh set of travelling wave solutions in terms of rational functions to (1a)-(1b) as follows:(46a)ux,t=142α2λ-4αλ+2λ+β2a2α-12-12βa2ξβ+2+C1βα-1ξα-ξ+C1α-C1+14a2ξβ+2+C1β2ξα-ξ+C1α-C12,(46b)vx,t=±123λa2βα-1∓123λa2ξβ+2+C1βξα-ξ+C1α-C1,where C1 is arbitrary constant and a2<0 and the traveling wave variable is(47)ξ=±-a2α-1x∓3-a2λα-1twithλ<0.


## 4. Group-Invariant Solution

As that in [31], one can suppose that the system of (1a)-(1b) admits a set of one-parameter (ε) Lie group of point transformations:(48)x∗=x,t∗=t,u∗=u+εσ1x,t,u,v+Oϵ2,v∗=v+εσ2x,t,u,v+Oϵ2.As known, σ1 and σ2 are called the classical Lie point symmetry of (1a)-(1b) which satisfy(49)V1=σ1∂∂u,σ1=Ux,t,u,v-Xx,t,u,v∂∂xu-Tx,t,u,v∂∂tu,V2=σ2∂∂v,σ2=Vx,t,u,v-Xx,t,u,v∂∂xv-Tx,t,u,v∂∂tv,where X(x,t,u,v), T(x,t,u,v), U(x,t,u,v), and V(x,t,u,v) are infinitesimals, which can be obtained by solving a determining equations, and V1, V2 are the corresponding Lie point symmetry generators of group (48). Obviously, σ1 and σ2 satisfy(50)∂∂εFu+εσ1ε=0=0,∂∂εFv+εσ2ε=0=0.Theorem 1.
The determining equations possess the following properties:(i) U ( x , t , u , v ) and V(x,t,u,v) are linear functions about u and v, respectively, T(x,t,u,v) is a function only about t, and X(x,t,u,v) is independent of u and v.(ii) X ( x , t , u , v ), T(x,t,u,v), U(x,t,u,v), and V(x,t,u,v) satisfy the following equations:(51)-3∂∂xX+∂∂tT,∂2∂x2X-2∂2∂x∂vV,-3∂2∂x∂uU+3∂2∂x2X,4v∂∂xX+2v∂∂vV-2∂∂uUv+2V,-∂3∂x3U-∂∂tU+2v∂∂xV-12u∂∂xU,-24∂∂xXu+∂∂tX+∂3∂x3X-3∂3∂x2∂uU-12U,-4∂∂xXuv+2∂∂xXλv+2∂∂vVuv-∂∂vVλv-2vU-2uV+λV-∂2∂x2V,∂∂uT,∂∂vT,∂∂xT,∂∂vU,∂∂uV,∂∂uX,∂∂vX,∂2∂v2V,∂3∂x∂u2U,which on solving yields (52)Xx,t,u,v=x+12λtθ13+θ3,Tx,t,u,v=θ1t+θ2,Ux,t,u,v=-2u-λθ13,Vx,t,u,v=-23vθ1.Once the infinitesimalsX(x,t,u,v), T(x,t,u,v), U(x,t,u,v), and V(x,t,u,v) are determined, we can obtain the following Lie symmetry groups.In Lie symmetry group ifθ1≠0, (53)x1ϵ=x-6λt+3θ3-18λθ2θ1eθ1/3ϵ+6λt+6λθ2θ1eθ1ϵ+12λθ2-3θ3θ1,t1ϵ=t+θ2θ1eθ1ϵ-θ2θ1,u1ϵ=u-λ2e-2θ1/3ϵ+λ2,v1ϵ=ve-2θ1/3ϵ.Hence, solving the well-known characteristic equations gives birth to the first group-invariant solution (the so-called similarity solution)(54)ux,t=12λ+Uξθ1t+θ22/3,vx,t=Vξθ1t+θ22/3, where the similarity variable is (55)ξ=xθ1t+θ23-6λtθ1t+θ23+3θ3-18λθ2θ1t+θ23θ1 and U(ξ), V(ξ) satisfy(56a)U′′′=-12UU′+2VV′+23θ1U-13-θ1t+θ22/3xθ1+6θ1t+θ22/3θ1λt-3θ1t+θ22/3θ3+18λθ1t+θ22/3θ2θ1t+θ2U′,(56b)V′′=-2UV,Whenθ1=0, we obtain another Lie symmetry group: (57)x1ϵ=x+θ3ϵ,t1ϵ=t+θ2ϵ,u1ϵ=u,v1ϵ=v. Meanwhile the second group-invariant solution is characterized by u(x,t)=U(ξ),v(x,t)=V(ξ), and U(ξ),V(ξ) satisfy(58a)U′′′=θ3θ2U′-12UU′+2VV′,(58b)V′′=-2UV+λV,where the similarity variable (59)ξ=x-θ3θ2t.


## 5. The Painlevé Property of the KdV-SCS

We have explored the Painlevé property of (1a)-(1b) in [32], where we only gave two principal branches. Now we obtain other principal branches, and on the basis of that we give more exact expansions of (1a)-(1b).The expansions of (1a)-(1b) about the singular manifold have the forms,(60)ux,t=∑j=0∞ujx,tϕx,tj+μ,vx,t=∑j=0∞vjx,tϕx,tj+ν.By using the WTC Painlevé test [33], we first obtain the following three principal branches:(61)i:u~-ϕx2ϕ-2,v~v0ϕ-1,v0arbitrary;ii:u~-3ϕx2ϕ-2,v~6ϕx2ϕ-2;iii:u~-3ϕx2ϕ-2,v~-6ϕx2ϕ-2.Principal branch (i): it turns out that the branch (i) is the only principal branch, with resonances(62)r=-1,0,3,4,6. The resonance -1 is always present, since it corresponds to the arbitrariness of ϕ, while r=0 comes from the arbitrary constant v0 in the leading order term of the expansion for v; the other three values arise from arbitrary coefficients higher up in the series for u, v, so that altogether there should be five arbitrary constants appearing in these Laurent series.Next referring to the procedure in [34], we give the truncated expansion forms for (i) as follows:(63a)u=ln⁡ϕxx+u~,u~≡u2,(63b)v=v0ϕ+v~,v~≡v1,where u, v, ϕ, u2, v0, and v1 all are functions of x,t.We can take the coefficients in the expansions (60) to be functions of t only, as this is referred to the ‘reduced ansatz’ of Kruskal [35]. So we get the coefficients of expansions (60) for (i) as follows:(64)u0t=-1,v0t=v0t;u1t=0,v1t=0;u2t=112ψ′t-112v0t2,v2t=-λ2v0t+112v0tψ′t-112v0t3;u3t=0,v3t=v3t;u4t=u4t,v4t=-λ28v0t+λ24v0tψ′t-λ24v0t3-1288v0tψ′t2+1144ψ′tv0t3-1288v0t5-12v0tu4t;u5t=-13v0tv3t+172ψ′′t-136v0tv0′t,v5t=-160v3tψ′t+112v3tv0t2-1360v0tψ′′t+1180v0t2v0′t+λ10v3t;u6t=u6t,v6t=λ2288v0tψ′t-λ2288v0t3-λ1728v0tψ′t2+λ864v0t3ψ′t-λ1728v0t5+131104v0tψ′t3-110368v0t3ψ′t2+110368v0t5ψ′t-131104v0t7-1216v0tu4tψ′t+1216u4tv0t3+λ36u4tv0t-λ3144v0t-19u6tv0t,where ψ(t) is an arbitrary function. As can be seen from the above expressions, the resonance conditions at r=3,4,6 corresponding to v3(t), u4(t), and u6(t) are satisfied.Nonprincipal branch (ii): the second branch (ii) has resonances (65)r=-1,-3,4,6,8.The presence of r=-3 means that this is a nonprincipal branch. Then there should be four arbitrary constants appearing in these Laurent series.We give the truncated expansion forms for(ii) as follows:(66a)u=3ln⁡ϕxx+u~,u~≡u2,(66b)v=-6ln⁡ϕxx+v~,v~≡v2,where u, v, ϕ, u2, and v2 all are functions of x,t.Referring to the ‘reduced ansatz’ of Kruskal, we get the coefficients of expansions (60) for (ii) as follows:(67)u0t=-3,v0t=6;u1t=0,v1t=0;u2t=120ψ′t+λ5,v2t=-3λ5+110ψ′t;u3t=0,v3t=0;u4t=u4t,v4t=-3λ100ψ′t+1400ψ′t2+9λ2100+3u4t;u5t=0,v5t=1240ψ′′t;u6t=-124u4tψ′t+λ4u4t-12v6t-9λ24000ψ′t+3λ8000ψ′t2-148000ψ′t3+9λ32000,v6t=v6t;u7t=λ2000ψ′′t+7120u4′t-112000ψ′′tψ′t,v7t=-120u4′t-λ4000ψ′′t+124000ψ′′tψ′t;u8t=-12u4t2-2v8t-λ300ψ′tu4t+13600u4tψ′t2+3λ240000ψ′t2-λ120000ψ′t3-3λ310000ψ′t+λ2100u4t+12880000ψ′t4+9λ420000,v8t=v8t.As can be seen from the above expressions, the resonance conditions atr=4,6,8 corresponding to u4(t), v6(t), and v8(t) are satisfied.Nonprincipal branch (iii): the third branch (iii) has resonances (68)r=-1,-3,4,6,8.We give the truncated expansion forms for(iii) as follows:(69a)u=3ln⁡ϕxx+u~,u~≡u2,(69b)v=6ln⁡ϕxx+v~,v~≡v2,where u, v, ϕ, u2, and v2 all are functions of x,t.Referring to the ‘reduced ansatz’ of Kruskal, we get the coefficients of expansions (60) for (iii) as follows:(70)u0t=-3,v0t=-6;u1t=0,v1t=0;u2t=120ψ′t+λ5,v2t=3λ5-110ψ′t;u3t=0,v3t=0;u4t=u4t,v4t=3λ100ψ′t-1400ψ′t2-9λ2100-3u4t;u5t=0,v5t=-1240ψ′′t;u6t=u6t,v6t=112u4tψ′t-λ2u4t+2u6t+9λ22000ψ′t-3λ4000ψ′t2+124000ψ′t3-9λ31000;u7t=λ2000ψ′′t+7120u4′t-112000ψ′′tψ′t,v7t=120u4′t+λ4000ψ′′t-124000ψ′′tψ′t;u8t=-12u4t2+2v8t-λ300ψ′tu4t+13600u4tψ′t2+3λ240000ψ′t2-λ120000ψ′t3-3λ310000ψ′t+λ2100u4t+12880000ψ′t4+9λ420000,v8t=v8t.As seen from the above expression, the resonance conditions at r=4,6,8 corresponding to u4(t), u6(t), and v8(t) are satisfied.


## 6. Conclusion

As demonstrated above, we get the polynomial solutions of the KdV equation with a self-consistent source (KdV-SCS) which are expressed in terms of Jacobi’s elliptic functions. The extended(G′/G)-expansion method has been successfully applied in this paper to deal with the new exact traveling wave solutions of the KdV-SCS. As a result, the hyperbolic function solutions, the trigonometric function solutions, and the rational function solutions with arbitrary parameters are obtained. The arbitrary parameters imply that those corresponding solutions have abundant local structures. Meanwhile, we give the reduction forms and the group-invariant solution of the KdV-SCS. By the WTC Painlevé test method, we show that the KdV-SCS passes the Painlevé test. And the truncated expansion form gives the Bäcklund transformations of (1a)-(1b). In fact, these methods are also readily applicable to a large variety of nonlinear partial differential equations; we indeed can obtain some new analytical solutions for many nonlinear differential equations. And it is very satisfying to see that more analytical solutions for the physically interesting equation (1a)-(1b) can be obtained by these most fundamental but widely applicable approaches. As far as we know, our solutions have not been reported in previous literature.



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*Source: 4014382-2018-11-25.xml*