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Abstract and Applied Analysis
Volume 2010, Article ID 241898, 9 pages
http://dx.doi.org/10.1155/2010/241898
Research Article

A Bäcklund Transformation for the Burgers Hierarchy

School of Mathematical Sciences, Yangzhou University, Yangzhou 225002, China

Received 19 October 2009; Revised 2 February 2010; Accepted 14 March 2010

Academic Editor: John Mallet-Paret

Copyright © 2010 Xifang Cao and Chuanyou Xu. This 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.

Abstract

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

𝑃1=1,𝑃0(𝑢)=𝑢,(1.1) and for 𝑗1, define the differential expressions 𝑃𝑗(𝑢,,𝜕𝑗𝑥𝑢) recursively as follows:

𝑃𝑗𝑢,,𝜕𝑗𝑥𝑢=𝑢+𝜕𝑥𝑃𝑗1𝑢,,𝜕𝑥𝑗1𝑢.(1.2) Then the Burgers hierarchy is defined by

𝑢𝑡=𝜕𝑥𝑃𝑗𝑢,,𝜕𝑗𝑥𝑢,𝑗1.(1.3) The first few members of the hierarchy (1.3) are

𝑢𝑡=2𝑢𝑢𝑥+𝑢𝑥𝑥,𝑢(1.4)𝑡=3𝑢2𝑢𝑥+3𝑢2𝑥+3𝑢𝑢𝑥𝑥+𝑢𝑥𝑥𝑥𝑢,(1.5)𝑡=4𝑢3𝑢𝑥+12𝑢𝑢2𝑥+6𝑢2𝑢𝑥𝑥+10𝑢𝑥𝑢𝑥𝑥+4𝑢𝑢𝑥𝑥𝑥+𝑢𝑥𝑥𝑥𝑥,(1.6) with (1.4) being just the Burgers equation.

There is much literature on the Burgers hierarchy. Olver [1] derived the hierarchy (1.3) from the point of view of infinitely many symmetries. The work in [2] showed that the Cole-Hopf transformation

𝑤𝑤𝑢=𝑥𝑤(1.7) transforms solutions of the linear equation

𝑤𝑡=𝜕𝑥𝑗+1𝑤(1.8) to that of (1.3). With the help of the Cole-Hopf transformation (1.9), Taflin [3] and Tasso [4] 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. [5] 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 [6] 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 [7], when studying the surfaces of constant negative curvature, that the sine-Gordon equation

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

𝑣𝑥=𝑢𝑥2𝜆sin𝑢+𝑣2,𝑣𝑡=𝑢𝑡+2𝜆sin𝑢𝑣2(1.10) 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:

1𝑣(𝑥,𝑡)=4arctanexp𝛼𝜆𝑥𝜆𝑡,(1.11) 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 [8]

𝑢𝑡=6𝑢𝑢𝑥+𝑢𝑥𝑥𝑥,(1.12) modified KdV equation [9]

𝑢𝑡=𝑢2𝑢𝑥+𝑢𝑥𝑥𝑥,(1.13) Burgers equation (1.4) [10], and a generalized Burgers equation [11]

𝑢𝑡+𝑏(𝑡)𝑢𝑢𝑥+𝑎(𝑡)𝑢𝑥𝑥=0,(1.14) 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., [1215]). 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 𝑣𝑢=𝑣+𝑥.𝜆+𝑣(2.1) Then 𝑃𝑗𝑢,,𝜕𝑗𝑥𝑢=𝜆𝑃𝑗𝑣,,𝜕𝑗𝑥𝑣+𝑃𝑗+1𝑣,,𝜕𝑥𝑗+1𝑣𝜆+𝑣,𝑗1.(2.2)

Proof. We use induction to prove (2.2).
First, for 𝑗=1, 𝑃1𝑢,𝑢𝑥=𝑢2+𝑢𝑥=𝑣𝑣+𝑥𝜆+𝑣2+𝑣𝑥𝑣𝑥2(𝜆+𝑣)2+𝑣𝑥𝑥=𝜆𝑣𝜆+𝑣2+𝑣𝑥+𝑣3+3𝑣𝑣𝑥+𝑣𝑥𝑥=𝜆+𝑣𝜆𝑃1𝑣,𝑣𝑥+𝑃2𝑣,𝑣𝑥,𝑣𝑥𝑥.𝜆+𝑣(2.3) So (2.2) is true for 𝑗=1.
Next, fix a 𝑘>1, and assume that (2.2) is true for 𝑗=𝑘1. Then 𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢=𝑣𝑣+𝑥𝜆+𝑣+𝜕𝑥𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢=𝑣𝑣+𝑥𝜆+𝑣+𝜕𝑥𝜆𝑃𝑘1𝑣,,𝜕𝑥𝑘1𝑣+𝑃𝑘𝑣,,𝜕𝑘𝑥𝑣=𝜆+𝑣𝑣+𝜕𝑥𝜆𝑃𝑘1𝑣,,𝜕𝑥𝑘1𝑣+𝑃𝑘𝑣,,𝜕𝑘𝑥𝑣=𝜆+𝑣𝜆𝑃𝑘𝑣,,𝜕𝑘𝑥𝑣+𝑃𝑘+1𝑣,,𝜕𝑥𝑘+1𝑣;𝜆+𝑣(2.4) 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 𝑣𝑣𝑥𝑣=(𝜆+𝑣)(𝑢𝑣),𝑡=(𝜆+𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢(2.5) 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)(𝑢𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢+(𝜆+𝑣)𝜕𝑥𝑃𝑗𝑣,,𝜕𝑗𝑥𝑣(𝜆+𝑣)2𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢,𝑣(2.6)𝑡𝑥=(𝜆+𝑣)(𝑢𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢+(𝜆+𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝜕𝑥𝑃𝑘𝑣,,𝜕𝑗𝑥𝑣𝑣(𝜆+𝑣)𝑗1𝑘=0(𝜆)𝑗1𝑘𝜕𝑥𝑃𝑘𝑣,,𝜕𝑗𝑥𝑣(𝜆+𝑣)2(𝑢𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘1𝑢,,𝜕𝑘𝑥𝑢=(𝜆+𝑣)(𝑢𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢+(𝜆+𝑣)𝜕𝑥𝑃𝑗𝑣,,𝜕𝑗𝑥𝑣(𝜆+𝑣)2𝑗1𝑘=0(𝜆)𝑗1𝑘𝜕𝑥𝑃𝑘𝑣,,𝜕𝑗𝑥𝑣(𝜆+𝑣)2(𝑢𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘1𝑢,,𝜕𝑘𝑥𝑢=(𝜆+𝑣)(𝑢𝑣)𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢+(𝜆+𝑣)𝜕𝑥𝑃𝑗𝑣,,𝜕𝑗𝑥𝑣(𝜆+𝑣)2𝑗𝑘=0(𝜆)𝑗𝑘𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢;(2.7) therefore 𝑣𝑥𝑡=𝑣𝑡𝑥; that is, (2.5) is an integrable system associated with (1.3).
From the first equation of (2.5) we have 𝑣𝑢=𝑣+𝑥.𝜆+𝑣(2.8) So 𝑃0(𝑣𝑢)𝑣=𝑥=𝜕𝜆+𝑣𝑥𝑃0(𝑣).𝜆+𝑣(2.9) On the other hand, by (2.2) 𝑃𝑘𝑢,,𝜕𝑘𝑥𝑢𝑣𝑃𝑘1𝑢,,𝜕𝑥𝑘1𝑢=𝜆𝜕𝑥𝑃𝑘1𝑣,,𝜕𝑥𝑘1𝑣+𝜕𝑥𝑃𝑘𝑣,,𝜕𝑘𝑥𝑣𝜆+𝑣,𝑘1.(2.10) Substituting (2.9) and (2.10) into the second equation of (2.5) yields 𝑣𝑡=𝜕𝑥𝑃𝑗𝑣,,𝜕𝑗𝑥𝑣;(2.11) 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

𝑣𝑒(𝑥,𝑡)=(𝜆+𝑢)d𝑥𝑒𝜆(𝜆+𝑢)d𝑥d𝑥𝜆𝑐(𝑡)𝑒(𝜆+𝑢)d𝑥,d𝑥+𝑐(𝑡)(3.1) 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 𝑃𝑗𝑢,,𝜕𝑗𝑥𝑢1for𝑗1.(3.2) So (2.5) becomes 𝑣𝑥𝑣=(𝜆+𝑣)(1𝑣),𝑡=(𝜆+𝑣)(1𝑣)1(𝜆)𝑗+1.1+𝜆(3.3) Solving (3.3) gives the following solution of (1.3): 𝑒𝑣(𝑥,𝑡)=(1+𝜆)𝑥+(1+(1)𝑗𝜆𝑗+1)𝑡+𝜆𝑒𝑐𝑒(1+𝜆)𝑥+(1+(1)𝑗𝜆𝑗+1)𝑡𝑒𝑐,(3.4) 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), 1𝑢(𝑥,𝑡)=𝑥(3.5) is a solution of (1.3). Then from (1.2) we have 𝑃𝑗𝑢,,𝜕𝑗𝑥𝑢0for𝑗1.(3.6) So (2.5) becomes 𝑣𝑥1=(𝜆+𝑣)𝑥,𝑣𝑣𝑡=(𝜆+𝑣)(𝜆)𝑗1𝑥𝑣(𝜆)𝑗1𝑣𝑥.(3.7) 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.(3.8)

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

Example 3.3. By the Cole-Hopf transformation (1.7), 2𝑢(𝑥,𝑡)=𝑥(3.9) is a solution of (1.3) for 𝑗2. Then from (1.2) we have 𝑃1𝑢,𝑢𝑥=2𝑥2,𝑃𝑗𝑢,,𝜕𝑗𝑥𝑢0for𝑗2.(3.10) So (2.5) becomes 𝑣𝑥2=(𝜆+𝑣)𝑥,𝑣𝑣𝑡=(𝜆+𝑣)(𝜆)𝑗2𝑥𝑣+2(𝜆)𝑗11𝑥2𝑣𝑥2(𝜆)𝑗2𝑣𝑥2.(3.11) Solving (3.11) gives the following solution of (1.3) for 𝑗2: 𝑣(𝑥,𝑡)=2𝜆(1+𝜆𝑥)𝑒𝜆(𝑥+(𝜆)𝑗𝑡)+𝜆𝑒𝑐22𝜆𝑥+𝜆2𝑥2𝑒𝜆(𝑥+(𝜆)𝑗𝑡)𝑒𝑐.(3.12)

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

Example 3.4. By the Cole-Hopf transformation (1.7), 3𝑢(𝑥,𝑡)=𝑥(3.13) is a solution of (1.3) for 𝑗3. Then from (1.2) we have 𝑃1𝑢,𝑢𝑥=6𝑥2,𝑃2𝑢,𝑢𝑥,𝑢𝑥𝑥=6𝑥3,𝑃𝑗𝑢,,𝜕𝑗𝑥𝑢0for𝑗3.(3.14) So (2.5) becomes 𝑣𝑥3=(𝜆+𝑣)𝑥,𝑣𝑣𝑡=(𝜆+𝑣)(𝜆)𝑗3𝑥𝑣+3(𝜆)𝑗12𝑥2𝑣𝑥+6(𝜆)𝑗21𝑥3𝑣𝑥26(𝜆)𝑗3𝑣𝑥3.(3.15) Solving (3.15) gives the following solution of (1.3) for 𝑗3: 𝑣(𝑥,𝑡)=3𝜆22𝜆𝑥+𝜆2𝑥2𝑒𝜆(𝑥+(𝜆)𝑗𝑡)+𝜆𝑒𝑐6+6𝜆𝑥3𝜆2𝑥2+𝜆3𝑥3𝑒𝜆(𝑥+(𝜆)𝑗𝑡)𝑒𝑐.(3.16)

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

Remark 3.5. In general, for an arbitrary positive integer 𝑘, 𝑘𝑢(𝑥,𝑡)=𝑥(3.17) 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)=𝜕𝑓𝑥,𝑥2,,𝑥𝑘𝑒/𝜕𝑥𝜆(𝑥+(𝜆)𝑗𝑡)+𝜆𝑒𝑐𝑓𝑥,𝑥2,,𝑥𝑘𝑒𝜆(𝑥+(𝜆)𝑗𝑡)𝑒𝑐,(3.18) where 𝑓𝑥,𝑥2,,𝑥𝑘=(1)𝑘𝑘!+(1)𝑘1𝑘!𝜆𝑥+(1)𝑘2𝑘!𝜆2!2𝑥2+𝑘𝜆𝑘1𝑥𝑘1+𝜆𝑘𝑥𝑘.(3.19)

Acknowledgment

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

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