# The Painlevé Tests, Bäcklund Transformation and Bilinear Form for the KdV Equation with a Self-Consistent Source

Authors: Yali Shen; Fengqin Zhang; Xiaomei Feng
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

The Painlevé property and Bäcklund transformation for the KdV equation with a self-consistent source are presented. By testing the equation, it is shown that the equation has the Painlevé property. In order to further prove its integrality, we give its bilinear form and construct its bilinear Bäcklund transformation by the Hirota's bilinear operator. And then the soliton solution of the equation is obtained, based on the proposed bilinear form.


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## Body

## 1. Introduction

It is well known that some nonlinear partial differential equations such as the soliton equations with self-consistent sources have important physical applications. In recent years, there are many ways for solving the soliton equations that can be used to the soliton equations with self-consistent sources as well. For example, the soliton solutions of some equations such as the KdV, AKNS, and nonlinear schrödinger equation with self-consistent sources are obtained through the inverse scattering method [1, 2]. In [3] a Darboux transformation, positon and negaton solutions to a Schrödinger self-consistent source equation are further constructed. Also, the binary Darboux transformations for the KdV hierarchies with self-consistent sources were presented in [4]. In addition to that, the Hirota bilinear method has been successfully used in the search for exact solutions of continuous and discrete systems, and also in the search for new integrable equations by testing for multisoliton solutions or Bäcklund transformation [5, 6]. Recently a bilinear Bäcklund transformation has been presented for a (3+1)-dimensional generalized KP equation. Meanwhile, two classes of exponential and rational traveling wave solutions with arbitrary wave numbers are computed by applying the proposed bilinear Bäcklund transformation (see [7] for details). It is a good reference for solving many high-dimensional soliton equations.Besides, the Painlevé analysis is a powerful tool for identifying the integrability of a nonlinear system. A partial different equation has the Painlevé property when the solutions of the partial different equation are single-valued about the movable, singularity manifold [8]. The basic thought as follows: if the singularity manifold is determined by ϕ(z1,z2,…,zn)=0 and u=u(z1,z2,…,zn) is a solution of the partial different equation, then we assume that(1.1)u=ϕα∑j=0∞ujϕj,
where(1.2)ϕ=ϕ(z1,z2,…,zn),uj=uj(z1,z2,…,zn),u0≠0,are analytic functions of(zj) in a neighborhood of the manifold ϕ=0 and α is an integer. Substitution of (1.1) into the partial different equation determines the values of α and defines the recursion relations for uj,j≠0,1,2,⋯. When the anatz (1.1) is correct, the pde is said to possess the Painlevé property and is conjectured to be integrable [9].Motivated by the previous works, we focus our attention on the following nonlinear partial differential equations (PDEs) which is expressed by(1.3)ut+uxxx+12uux=(φ2)x,φxx+2uφ=λφ,
where λ is an arbitrary constant. In fact, (1.3) is a reduced form of the KdV equation with a source by symmetry constraints [10, 11]. The main purpose of this paper is to demonstrate the connection between the Painlevé property and the Bäcklund transformation for (1.3). Moreover, we get the bilinear Bäcklund transformation and the exact solution for (1.3) by the Hirota bilinear method. Thus we further convince the integrability of the equation.The paper is organized as follows. In Section2, we investigate the Painlevé property for (1.3). By testing the equation it is shown that the equation has the Painlevé property. Furthermore, we obtain a Bäcklund transformation of (1.3). In Section 3, using the Hirota’s bilinear operator, we obtain its bilinear form and construct its bilinear Bäcklund transformation. And then its one-soliton solution is obtained. Finally, conclusion is given in Section 4.


## 2. PainlEvÉ Test

As we know, the basic Painlevé test for ODEs consists of the following steps [12].Step 1.
Identify all possible dominant balances, that is, all singularities of formu~u0(z-z0)μ.Step 2.
If all exponentsμ are integers, find the resonances where arbitrary constants can appear.Step 3.
If all resonances are integers, check the resonance conditions in each Laurent expansion.Conclusion.
If no obstruction is found in Steps1–3 for every dominant balances, then the Painlevé test is satisfied.
The above series may be substituted into the PDEs. Now we apply the above steps to (1.3). We will further give all possible solutions with integer resonances but without further analysis of the last cases. The expansions about the singular manifold have the forms:(2.1)u=∑j=0∞ajϕj+α,φ=∑j=0∞bjϕj+β.
To find the dominant balances, we are looking for leading order singular behaviour of the form
(2.2)u≈a0ϕα,φ≈b0ϕβ.
And the derivatives of (2.2) are given by
(2.3)ut≈a0αϕα-1ϕt,ux≈a0αϕα-1ϕx,uxxx≈a0α(α-1)(α-2)ϕα-3ϕx3,φx≈b0βϕβ-1ϕx,φxx≈b0β(β-1)ϕβ-2ϕx2.
Substituting (2.2)-(2.3) into (1.3), we get the following forms
(2.4)a0α(α-1)(α-2)ϕα-3ϕx3+12a02αϕxϕ2α-1=2b02βϕxϕ2β-1,b0β(β-1)ϕβ-2ϕx2+2a0b0ϕα+β=0.
Calculating and simplifying equation (2.4), we get the dominant balances
(2.5)u≈-3ϕx2ϕ-2,φ≈±6ϕx2ϕ-2,
where α=β=-2,a0=-3ϕx2,b0=±6ϕx2. So we complete the first step.
The second step in applying the Painlevé test is to find the resonances. To find the resonances numbersj, we substitute (2.1) into (1.3), and collecting terms of each order of ϕ, we obtain
ϕ-5:(2.6)-24a0ϕx3-24a02ϕx=-4b02ϕx.
By calculating equation (2.6), we get b02=6a0(a0+ϕx2).
ϕ-4:(2.7)18a0xϕx2-6a1ϕx3+18a0ϕxϕxx+12a0a0x-36a0a1ϕx=2b0b0x-6b0b1ϕx,6b0ϕx2+2a0b0=0.
From (2.6) and (2.7), we have a0=-3ϕx2, b0=±6ϕx2, 6b1=±(15ϕxx-17a1), where a0 and b0 are consistent with step one.
ϕ-3:(2.8)-a0ϕt-3a0xxϕx+3a1xϕx2-3a0xϕxx+3a1ϕxϕxx-a0ϕxxx+6a0a1x+6a1a0x-6a12ϕx-12a0a2ϕx=b0b1x+b1b0x-b12ϕx-2b0b2ϕx,-2b0xϕx+b1ϕx2-b0ϕxx+a0b1+a1b0=0.
From (2.7) and (2.8), we get a1=3ϕxx, b1=∓6ϕxx,
ϕ-2:(2.9)a0t-a1ϕt+a0xxx-3a1xxϕx-3a1xϕxx-a1ϕxxx+12a0a2x+12a1a1x+12a2a0x-12a0a3ϕx-12a1a2ϕx=2b0b2x+2b1b1x+2b2b0x-2b0b3ϕx-2b1b2ϕx,b0xx-2b1xϕx-b1ϕxx+2a0b2+2a1b1+2a2b0=λb0.
From (2.8) and (2.9), we get
(2.10)a2=15ϕxx2-20ϕxϕxxx-ϕxϕt+4λϕx220ϕx2,b2=±-15ϕxx2+20ϕxϕxxx-ϕxϕt-6λϕx210ϕx2,
ϕj-5:(2.11)aj-3,t+(j-4)aj-2ϕt+aj-3,xxx+3(j-4)aj-2,xxϕx+3(j-3)(j-4)aj-1,xϕx2+(j-2)(j-3)(j-4)ajϕx3+3(j-4)aj-2,xϕxx+3(j-3)(j-4)aj-1ϕxϕxx+(j-4)aj-2ϕxxx+12{a0aj-1,x+a1aj-2,x+⋅⋅⋅+aj-1a0,x+ϕx[(j-2)a0aj+(j-3)a1aj-1+⋅⋅⋅+2aj-4a4+aj-3a3-aj-1a1-2aja0]}=2{b0bj-1,x+b1bj-2,x+⋅⋅⋅+bj-1b0,x+ϕx[(j-2)b0bj+(j-3)b1bj-1+⋯+2bj-4b4+bj-3b3-bj-1b1-2bjb0]},
ϕj-4:(2.12)bj-2,xx+2(j-3)bj-1,xϕx+(j-2)(j-3)bjϕx2+(j-3)bj-1ϕxx+2a0bj+2a1bj-1+⋯+ajb0=λbj-2.
Substituting a0=-3ϕx2,b0=±6ϕx2 into (2.11) and (2.12), the coefficients of aj and bj may be rearranged to give
(2.13)[(j-2)(j-3)-36](j-4)ajϕx3+F(a0,a1,…,aj-1,ϕt,ϕx,ϕxx,…)=±12(j-4)bjϕx3+G(b0,b1,…,bj-1,ϕt,ϕx,ϕxx,…),(2.14)[(j-2)(j-3)-6]bjϕx2±12ajϕx2+H(a0,a1,…,aj-1,b0,b1,⋯,bj-1,ϕt,ϕx,ϕxx,…)=0,
simplifying (2.13)-(2.14), we have
(2.15)(j-4)[j(j-5)(j2-5j-30)+144](j2-5j-30)ajϕx3=-j(j-5)(j2-5j-30)F+j(j-5)(j2-5j-30)G∓12(j-4)(j2-5j-30)ϕxH,(2.16)(j-4)[j(j-5)(j2-5j-30)+144]bjϕx3=-(j2-5j-30)(j-4)ϕxH∓12G±12F.
There, it is found that the resonance occurs at j=4, so the second step is completed.
For the last step, we will check the resonance conditions. So we need to find the orders in the expansion (2.1) where arbitrary constants may appear:
ϕ-1:(2.17)a1t+a1xxx+12[a0a3x+a1a2x+a2a1x+a3a0x+ϕx(2a0a4+a1a3-a3a1-2a4a0)]=2[b0b3x+b1b2x+b2b1x+b3b0x+ϕx(2b0b4+b1b3-b3b1-2b4b0)],b1xx+2a0b3+2a1b2+2a2b1+2a3b0=λb1.
From (2.17), we know a4 and b4 are both arbitrary. Thus (1.3) possess the Painlevé property.
We now specialize (2.1) by setting the resonance functions a4=b4=0. Furthermore, we require a3=b3=0, it is easily demonstrated that aj=0,bj=0,j≥3 from the recursion relations.
Ifa2 and b2 satisfy(2.18)a2t+a2xxx+12a2a2x=(b22)x,b2xx+2a2b2=λb2,
we obtain a Bäcklund transformation of (1.3):
(2.19)u=3(ln⁡ϕ)xx+ũ,ũ≡a2,φ=-6(ln⁡ϕ)xx+φ̃,φ̃≡b2,
where we consider the case of b0=6ϕx2, b1=-6ϕxx moreover, u,φ and ũ,φ̃ satisfy (1.3) and
(2.20)3ũ(ln⁡ϕ)xx+φ̃(ln⁡ϕ)xx=θ(t),ϕtϕx=5λ+(15ϕxx2ϕx2-20ϕxxxϕx).
Many studies [9, 13] show that a new solution can usually be obtained from a given solution of an equation if the so-called Bäcklund transformation for the equation is found. Therefore, it is worth to find the Bäcklund transformation of an equation. In the next section, we will give the bilinear Bäcklund transformation of (1.3).


## 3. Bilinear Form

As we know, when you want to use Hirota method, the first thing you need to do is to rewrite the equation under consideration as the bilinear form [14]. This can be achieved for (1.3) by the following dependent variable transformation:(3.1)u=(ln⁡f)xx,φ=gf.
Equation (1.3) can be written into bilinear forms(3.2)(DxDt+Dx4)f⋅f=2g2,Dx2f⋅g=λfg,
where D is the well-known Hirota bilinear operator(3.3)DxmDtna⋅b=(∂∂x1-∂∂x2)m(∂∂t1-∂∂t2)na(x1,t1)b(x2,t2)|x1=x2,t1=t2..Now we will give the bilinear Bäcklund transformation of (1.3).Theorem 3.1.
Suppose that  (f, g) is a solution of (3.2), then  (f′, g′), satisfying the following relations:
(3.4)Dxg⋅f′-μfg′=0,(3.5)Dxf⋅g′-μgf′=0,(3.6)(Dx2-ν)f⋅f′=0,(3.7)(Dt+Dx3+3νDx)f⋅f′+1μgg′=0,
is another solution of (3.2), where μ and ν are arbitrary constants and μ≠0.Proof.
We consider the following:(3.8)P1=(DxDt+Dx4)f′⋅f′-2g′2,P2=Dx2f′⋅g′-λf′g′.
We will show that (3.4)–(3.7) imply P1=0 and P2=0. We first work on the case of P1. We will use various bilinear identities which, for convenience, are presented in the appendix:
(3.9)-P1ff=[(DxDt+Dx4)f⋅f-2g2]f′f′-[(DxDt+Dx4)f′⋅f′-2g′2]ff=(A.1)(A.2)2Dx(Dtf⋅f′)⋅ff′+2Dx[(Dx3f⋅f′)⋅ff′+3(Dx2f⋅f′)⋅(Dxf′⋅f)]+2g′g′ff-2ggf′f′=(3.6)2Dx(Dtf⋅f′)⋅ff′+2Dx[(Dx3f⋅f′)⋅ff′+3νff′⋅(Dxf′⋅f)]+2fg′fg′-2gf′gf′=(3.4)(3.5)2Dx[(Dt+Dx3+3νDx)f⋅f′]⋅ff′+21μ[(Dxg⋅f′)fg′-(Dxf⋅g′)gf′]=(A.3)2Dx[(Dt+Dx3+3νDx)f⋅f′]⋅ff′+21μDxgg′⋅ff′=2Dx[(Dt+Dx3+3νDx)f⋅f′+1μgg′]⋅ff′≡(3.7)0.Next we come to the second part of the proof:
(3.10)-P2fg=(Dx2f⋅g-λfg)f′g′-(Dx2f′⋅g′-λf′g′)fg=(A.4)Dx[(Dxf⋅g′)⋅f′g+fg′⋅(Dxf′⋅g)]=(3.4)(3.5)Dx(μgf′⋅f′g-fg′⋅μfg′)≡(A.5)0.
Thus we have completed the proof of Theorem 3.1.We will show that our Bäcklund transformation (3.4)–(3.7) supplies us with a Lax representation for (1.3). Suppose(3.11)v=(ln⁡f)xx,g=φf,f′=Xf,g′=Yg,
then the variables f, g, f′ and g′ can be eliminated from (3.4)–(3.7). The elimination results in(3.12)Xx=φxφX-μY,(3.13)Yx=-φxφY-μX,(3.14)Xxx=(ν-2ω)X,(3.15)Xt=-Xxxx-3(2ω+ν)Xx+1μφ2Y.
Therefore, we have the following.Theorem 3.2.
The compatibility condition of (3.12)–(3.15) is (1.3).  In fact, using the compatibility conditions Xxxt=Xtxx, one can obtain (1.3) where ν and ω satisfy ν-2ω=λ.Finally we will give the soliton solution of the equation (1.3) by the standard perturbation method:(3.16)u=[ln⁡(1+e2ξ)]xx,φ=2kβ(t)eξ1+e2ξ,whereξ=kx-4k3t+∫0tβ(s)ds+c, c, k are arbirtary constants and λ=k2.


## 4. Conclusion

In this paper, we investigate the Painlevé property for the KdV equation with a self-consistent source. By tests to the equation, it is shown that only the principal balance of the equation has the Painlevé property. While noninteger resonances are allowed with the weak extension of the Painlevé test [12]. We obtain the two different Bäcklund transformations. And then the soliton solution for (1.3) is given.



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