AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation24189810.1155/2010/241898241898Research ArticleA Bäcklund Transformation for the Burgers HierarchyCaoXifang1XuChuanyou1Mallet-ParetJohn1School of Mathematical SciencesYangzhou UniversityYangzhou 225002Chinayzu.edu.cn20101603201020101910200902022010140320102010Copyright © 2010This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We give a Bäcklund transformation in a unified form for each member in the Burgers hierarchy. By applying the Bäcklund transformation to the trivial solutions, we generate some solutions of the Burgers hierarchy.

1. Introduction

Let

P-1=1,P0(u)=u, and for j1, define the differential expressions Pj(u,,xju) recursively as follows:

Pj(u,,xju)=(u+x)Pj-1(u,,xj-1u). Then the Burgers hierarchy is defined by

ut=xPj(u,,xju),j1. The first few members of the hierarchy (1.3) are

ut=2uux+uxx,ut=3u2  ux+3ux2+3uuxx+uxxx,ut=4u3ux+12uux2+6u2uxx+10uxuxx+4uuxxx+uxxxx, with (1.4) being just the Burgers equation.

There is much literature on the Burgers hierarchy. Olver  derived the hierarchy (1.3) from the point of view of infinitely many symmetries. The work in  showed that the Cole-Hopf transformation

wu=wxw transforms solutions of the linear equation

wt=xj+1w to that of (1.3). With the help of the Cole-Hopf transformation (1.9), Taflin  and Tasso  showed, respectively, that the Burgers equation (1.4) and the second member (1.5) of the hierarchy (1.3) can be written in the Hamiltonian form. More recently, Talukdar et al.  constructed an appropriate Lagrangian by solving the inverse problem of variational calculus and then Hamiltonized (1.5) to get the relevant Poisson structure. Furthermore, they pointed out that their method is applicable to each member of (1.3). Pickering  proved explicitly that each member of (1.3) passes the Weiss-Tabor-Carnevale Painlevé test.

This paper is devoted to the study of Bäcklund transformation for the Burgers hierarchy. Bäcklund transformation was named after the Swedish mathematical physicist and geometer Albert Victor Bäcklund(1845-1922), who found in 1883 , when studying the surfaces of constant negative curvature, that the sine-Gordon equation

uxt=sinu has the following property: if u solves (1.9), then for an arbitrary non-zero constant λ, the system on v

vx=ux-2λsinu+v2,vt=-ut+2λsinu-v2 is integrable; moreover, v also solves (1.9). So (1.10) gives a transformation uv, now called Bäcklund transformation, which takes one solution of (1.9) into another. For example, substituting the trivial solution u(x,t)0 into (1.10) yields one-soliton solution:

v(x,t)=4arctanexp(α-λx-1λt), where α is an arbitrary constant. By repeating this procedure one can get multiple-soliton solutions. Some other nonlinear partial differential equations (PDEs), such as KdV equation 

ut=6uux+uxxx, modified KdV equation 

ut=u2ux+uxxx, Burgers equation (1.4) , and a generalized Burgers equation 

ut+b(t)uux+a(t)uxx=0, also possess Bäcklund transformations. Now Bäcklund transformation has become a useful tool for generating solutions to certain nonlinear PDEs. Much literature is devoted to searching Bäcklund transformations for some nonlinear PDEs (see, e.g., ). In this paper, we give a Bäcklund transformation for each member in the Burgers hierarchy. As an application, by applying our Bäcklund transformation to the trivial solutions, we generate some new solutions of (1.3).

2. Bäcklund Transformation

First, the differential expressions Pj have the following property.

Theorem 2.1.

For an arbitrary constant λ, let u=v+vxλ+v. Then Pj(u,,xju)=λPj(v,,xjv)+Pj+1(v,,xj+1v)λ+v,j1.

Proof.

We use induction to prove (2.2).

First, for j=1, P1(u,ux)=u2+ux=(v+vxλ+v)2+vx-vx2(λ+v)2+vxxλ+v=λ(v2+vx)+v3+3vvx+vxxλ+v=λP1(v,vx)+P2(v,vx,vxx)λ+v. So (2.2) is true for j=1.

Next, fix a k>1, and assume that (2.2) is true for j=k-1. Then Pk(u,,xku)=(v+vxλ+v+x)Pk-1(u,,xk-1u)=(v+vxλ+v+x)λPk-1(v,,xk-1v)+Pk(v,,xkv)λ+v=(v+x)(λPk-1(v,,xk-1v)+Pk(v,,xkv))λ+v=λPk(v,,xkv)+Pk+1(v,,xk+1v)λ+v; that is, (2.2) is valid for j=k.

Therefore, (2.2) is always true for j1.

Now we state our main result.

Theorem 2.2.

If u is a solution of (1.3), then the system on vvx=(λ+v)(u-v),vt=(λ+v)k=0j(-λ)j-k(Pk(u,,xku)-vPk-1(u,,xk-1u)) is integrable; moreover, v also satisfies (1.3). Therefore, (2.5) defines a Bäcklund transformation uv, in a unified form, for each member of the Burgers hierarchy (1.3).

Proof.

By (1.3) and (2.5) we have vxt=(λ+v)(u-v)k=0j(-λ)j-k(Pk(u,,xku)-vPk-1(u,,xk-1u))+(λ+v)xPj(v,,xjv)-(λ+v)2k=0j(-λ)j-k(Pk(u,,xku)-vPk-1(u,,xk-1u)),vtx=(λ+v)(u-v)k=0j(-λ)j-k(Pk(u,,xku)-vPk-1(u,,xk-1u))+(λ+v)k=0j(-λ)j-kxPk(v,,xjv)-v(λ+v)k=0j-1(-λ)j-1-kxPk(v,,xjv)-(λ+v)2(u-v)k=0j(-λ)j-kPk-1(u,,xku)=(λ+v)(u-v)k=0j(-λ)j-k(Pk(u,,xku)-vPk-1(u,,xk-1u))+(λ+v)xPj(v,,xjv)-(λ+v)2k=0j-1(-λ)j-1-kxPk(v,,xjv)-(λ+v)2(u-v)k=0j(-λ)j-kPk-1(u,,xku)=(λ+v)(u-v)k=0j(-λ)j-k(Pk(u,,xku)-vPk-1(u,,xk-1u))+(λ+v)xPj(v,,xjv)-(λ+v)2k=0j(-λ)j-k(Pk(u,,xku)-vPk-1(u,,xk-1u)); therefore vxt=vtx; that is, (2.5) is an integrable system associated with (1.3).

From the first equation of (2.5) we have u=v+vxλ+v. So P0(u)-v=vxλ+v=xP0(v)λ+v. On the other hand, by (2.2) Pk(u,,xku)-vPk-1(u,,xk-1u)=λxPk-1(v,,xk-1v)+xPk(v,,xkv)λ+v,k1. Substituting (2.9) and (2.10) into the second equation of (2.5) yields vt=xPj(v,,xjv); that is, v also satisfies the Burgers hierarchy (1.3).

3. Exact Solutions

In this section we always assume that λ is an arbitrary nonzero constant.

From a known solution u of (1.3), the first equation of (2.5) gives

v(x,t)=e(λ+u)dx-λe(λ+u)dxdx-λc(t)e(λ+u)dxdx+c(t), with the “integration constant" c(t) satisfying a first-order ordinary differential equation determined by the second equation of (2.5).

Example 3.1.

Take the trivial solution u(x,t)1 of (1.3). Then from (1.2) we have Pj(u,,xju)1for  j1. So (2.5) becomes vx=(λ+v)(1-v),vt=(λ+v)(1-v)(1-(-λ)j+1)1+λ. Solving (3.3) gives the following solution of (1.3): v(x,t)=e(1+λ)x+(1+(-1)jλj+1)t+λece(1+λ)x+(1+(-1)jλj+1)t-ec, where c is an arbitrary constant.

Note that (3.4) is a traveling wave solution.

Example 3.2.

By the Cole-Hopf transformation (1.7), u(x,t)=1x is a solution of (1.3). Then from (1.2) we have Pj(u,,xju)0for  j1. So (2.5) becomes vx=(λ+v)(1x-v),vt=(λ+v)((-λ)j(1x-v)-(-λ)j-1vx). Solving (3.7) gives the following solution of (1.3): v(x,t)=λeλ(x+(-λ)jt)+λec(-1+λx)eλ(x+(-λ)jt)-ec.

Note that (3.8) is not a traveling wave solution.

Example 3.3.

By the Cole-Hopf transformation (1.7), u(x,t)=2x is a solution of (1.3) for j2. Then from (1.2) we have P1(u,ux)=2x2,Pj(u,,xju)0for  j2. So (2.5) becomes vx=(λ+v)(2x-v),vt=(λ+v)((-λ)j(2x-v)+2(-λ)j-1(1x2-vx)-2(-λ)j-2vx2). Solving (3.11) gives the following solution of (1.3) for j2: v(x,t)=2λ(-1+λx)eλ(x+(-λ)jt)+λec(2-2λx+λ2x2)eλ(x+(-λ)jt)-ec.

Note that (3.12) is not a traveling wave solution.

Example 3.4.

By the Cole-Hopf transformation (1.7), u(x,t)=3x is a solution of (1.3) for j3. Then from (1.2) we have P1(u,ux)=6x2,P2(u,ux,uxx)=6x3,Pj(u,,xju)0for  j3. So (2.5) becomes vx=(λ+v)(3x-v),vt=(λ+v)((-λ)j(3x-v)+3(-λ)j-1(2x2-vx)+6(-λ)j-2(1x3-vx2)-6(-λ)j-3vx3). Solving (3.15) gives the following solution of (1.3) for j3: v(x,t)=3λ(2-2λx+λ2x2)eλ(x+(-λ)jt)+λec(-6+6λx-3λ2x2+λ3x3)eλ(x+(-λ)jt)-ec.

Note that (3.16) is not a traveling wave solution.

Remark 3.5.

In general, for an arbitrary positive integer k, u(x,t)=kx is a solution of (1.3) for jk. Substituting (3.17) into (2.5) gives the following solution of (1.3) for jk: v(x,t)=(f(x,x2,,xk)/x)eλ(x+(-λ)jt)+λecf(x,x2,,xk)eλ(x+(-λ)jt)-ec, where f(x,x2,,xk)=(-1)kk!+(-1)k-1k!λx+(-1)k-2k!2!λ2x2+-kλk-1xk-1+λkxk.

Acknowledgment

This work is supported by the National Natural Science Foundation of China through the Grant no. 10571149.

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