A Bäcklund Transformation for the Burgers Hierarchy

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

𝑃₋₁=1,𝑃₀(𝑢)=𝑢,(1.1) and for 𝑗≥1, define the differential expressions 𝑃_𝑗(𝑢,…,𝜕^𝑗𝑥𝑢) recursively as follows:

𝑃_𝑗𝑢,…,𝜕^𝑗𝑥𝑢=𝑢+𝜕_𝑥𝑃_𝑗−1𝑢,…,𝜕^𝑥𝑗−1𝑢.(1.2) Then the Burgers hierarchy is defined by

𝑢_𝑡=𝜕_𝑥𝑃_𝑗𝑢,…,𝜕^𝑗𝑥𝑢,𝑗≥1.(1.3) The first few members of the hierarchy (1.3) are

𝑢𝑢_𝑡=2𝑢𝑢_𝑥+𝑢_𝑥𝑥,(1.4)_𝑡𝑢=3𝑢²𝑢_𝑥+3𝑢^2𝑥+3𝑢𝑢_𝑥𝑥+𝑢_𝑥𝑥𝑥,(1.5)_𝑡=4𝑢³𝑢_𝑥+12𝑢𝑢^2𝑥+6𝑢²𝑢_𝑥𝑥+10𝑢_𝑥𝑢_𝑥𝑥+4𝑢𝑢_𝑥𝑥𝑥+𝑢_{𝑥𝑥𝑥𝑥},(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

𝑤⟼𝑢=𝑤^𝑥𝑤(1.7) transforms solutions of the linear equation

𝑤_𝑡=𝜕^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

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

𝑣_𝑥=𝑢_𝑥−2𝜆sin𝑢+𝑣𝑣_𝑡2,=−𝑢_𝑡+2𝜆sin𝑢−2𝑣(1.10) is integrable; moreover, 𝑣 also solves (1.9). So (1.10) gives a transformation 𝑢↦𝑣, 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,(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]

𝑢_𝑡=6𝑢𝑢_𝑥+𝑢_𝑥𝑥𝑥,(1.12) modified KdV equation [9]

𝑢_𝑡=𝑢²𝑢_𝑥+𝑢_𝑥𝑥𝑥,(1.13) Burgers equation (1.4) [10], and a generalized Burgers equation [11]

𝑢_𝑡+𝑏(𝑡)𝑢𝑢_𝑥+𝑎(𝑡)𝑢_𝑥𝑥=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., [12–15]). 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 𝑃_𝑗 have the following property.

Theorem 2.1. For an arbitrary constant 𝜆, let 𝑢=𝑣+𝑣^𝑥𝜆+𝑣.(2.1) Then 𝑃_𝑗𝑢,…,𝜕^𝑗𝑥𝑢=𝜆𝑃^𝑗𝑣,…,𝜕^𝑗𝑥𝑣+𝑃_𝑗+1𝑣,…,𝜕^𝑥𝑗+1𝑣𝜆+𝑣,𝑗≥1.(2.2)

Proof. We use induction to prove (2.2).
First, for 𝑗=1, 𝑃₁𝑢,𝑢_𝑥=𝑢²+𝑢_𝑥=𝑣+𝑣^𝑥𝜆+𝑣²+𝑣_𝑥−𝑣^𝑥2(𝜆+𝑣)²+𝑣^𝑥𝑥=𝜆𝑣𝜆+𝑣₂+𝑣_𝑥+𝑣³+3𝑣𝑣_𝑥+𝑣_𝑥𝑥=𝜆𝑃₁𝜆+𝑣𝑣,𝑣_𝑥+𝑃₂𝑣,𝑣_𝑥,𝑣_𝑥𝑥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=𝑘−1. Then 𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢=𝑣+𝑣^𝑥𝜆+𝑣+𝜕^𝑥𝑃^𝑘−1𝑢,…,𝜕^𝑥𝑘−1=𝑣+𝑣_𝑥𝑢𝜆+𝑣+𝜕^𝑥𝜆𝑃^𝑘−1𝑣,…,𝜕^𝑥𝑘−1𝑣+𝑃_𝑘𝑣,…,𝜕^𝑘𝑥𝑣=𝑣+𝜕_𝑥𝜆+𝑣𝜆𝑃_𝑘−1𝑣,…,𝜕^𝑥𝑘−1𝑣+𝑃_𝑘𝑣,…,𝜕^𝑘𝑥𝑣=𝜆𝑃_𝑘𝜆+𝑣𝑣,…,𝜕^𝑘𝑥𝑣+𝑃_𝑘+1𝑣,…,𝜕^𝑥𝑘+1𝑣𝜆+𝑣;(2.4) that is, (2.2) is valid for 𝑗=𝑘.
Therefore, (2.2) is always true for 𝑗≥1.

Now we state our main result.

Theorem 2.2. If 𝑢 is a solution of (1.3), then the system on 𝑣𝑣𝑣_𝑥=(𝜆+𝑣)(𝑢−𝑣),_𝑡=(𝜆+𝑣)^𝑗_𝑘=0(−𝜆)^𝑗−𝑘𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢(2.5) is integrable; moreover, 𝑣 also satisfies (1.3). Therefore, (2.5) defines a Bäcklund transformation 𝑢↦𝑣, in a unified form, for each member of the Burgers hierarchy (1.3).

Proof. By (1.3) and (2.5) we have 𝑣_𝑥𝑡=(𝜆+𝑣)(𝑢−𝑣)^𝑗_𝑘=0+(𝜆+𝑣)𝜕_𝑥(−𝜆)^𝑗−𝑘𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢−(𝜆+𝑣)_2𝑗𝑃_𝑗𝑣,…,𝜕^𝑗𝑥𝑣_𝑘=0𝑣(−𝜆)^𝑗−𝑘𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢,(2.6)_𝑡𝑥=(𝜆+𝑣)(𝑢−𝑣)^𝑗_𝑘=0+(𝜆+𝑣)_𝑗(−𝜆)^𝑗−𝑘𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢_𝑘=0−𝑣(𝜆+𝑣)(_𝑗−1−𝜆)^𝑗−𝑘𝜕_𝑥𝑃_𝑘𝑣,…,𝜕^𝑗𝑥𝑣_𝑘=0−(𝜆+𝑣)₂(−𝜆)^{𝑗−1−𝑘}𝜕_𝑥𝑃_𝑘𝑣,…,𝜕^𝑗𝑥𝑣(𝑢−𝑣)^𝑗_𝑘=0=(𝜆+𝑣)(𝑢−𝑣)_𝑗(−𝜆)^𝑗−𝑘𝑃_𝑘−1𝑢,…,𝜕^𝑘𝑥𝑢_𝑘=0+(𝜆+𝑣)𝜕_𝑥(−𝜆)^𝑗−𝑘𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢𝑃_𝑗𝑣,…,𝜕^𝑗𝑥𝑣−(𝜆+𝑣)^2𝑗−1_𝑘=0−(𝜆+𝑣)₂(−𝜆)^{𝑗−1−𝑘}𝜕_𝑥𝑃_𝑘𝑣,…,𝜕^𝑗𝑥𝑣(𝑢−𝑣)^𝑗_𝑘=0=(𝜆+𝑣)(𝑢−𝑣)_𝑗(−𝜆)^𝑗−𝑘𝑃_𝑘−1𝑢,…,𝜕^𝑘𝑥𝑢_𝑘=0+(𝜆+𝑣)𝜕_𝑥(−𝜆)^𝑗−𝑘𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢−(𝜆+𝑣)_2𝑗𝑃_𝑗𝑣,…,𝜕^𝑗𝑥𝑣_𝑘=0(−𝜆)^𝑗−𝑘𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢;(2.7) therefore 𝑣_𝑥𝑡=𝑣_𝑡𝑥; that is, (2.5) is an integrable system associated with (1.3).
From the first equation of (2.5) we have 𝑢=𝑣+𝑣^𝑥𝜆+𝑣.(2.8) So 𝑃₀(𝑢)−𝑣=𝑣^𝑥𝜆+𝑣=𝜕^𝑥𝑃₀(𝑣)𝜆+𝑣.(2.9) On the other hand, by (2.2) 𝑃_𝑘𝑢,…,𝜕^𝑘𝑥𝑢−𝑣𝑃_𝑘−1𝑢,…,𝜕^𝑥𝑘−1𝑢=𝜆𝜕^𝑥𝑃_𝑘−1𝑣,…,𝜕^𝑥𝑘−1𝑣+𝜕_𝑥𝑃_𝑘𝑣,…,𝜕^𝑘𝑥𝑣𝜆+𝑣,𝑘≥1.(2.10) Substituting (2.9) and (2.10) into the second equation of (2.5) yields 𝑣_𝑡=𝜕_𝑥𝑃_𝑗𝑣,…,𝜕^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

𝑣(𝑥,𝑡)=𝑒^{∫(𝜆+𝑢)d𝑥}−𝜆∫𝑒^{∫(𝜆+𝑢)d𝑥}d𝑥−𝜆𝑐(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 𝑃_𝑗𝑢,…,𝜕^𝑗𝑥𝑢≡1for𝑗≥1.(3.2) So (2.5) becomes 𝑣𝑣_𝑥=(𝜆+𝑣)(1−𝑣),_𝑡=(𝜆+𝑣)(1−𝑣)1−(−𝜆)^𝑗+11+𝜆.(3.3) Solving (3.3) gives the following solution of (1.3): 𝑣(𝑥,𝑡)=𝑒^{(1+𝜆)𝑥+(1+(−1)𝑗𝜆𝑗+1)𝑡}+𝜆𝑒^𝑐𝑒^{(1+𝜆)𝑥+(1+(−1)𝑗𝜆𝑗+1)𝑡}−𝑒^𝑐,(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), 𝑢(𝑥,𝑡)=1𝑥(3.5) is a solution of (1.3). Then from (1.2) we have 𝑃_𝑗𝑢,…,𝜕^𝑗𝑥𝑢≡0for𝑗≥1.(3.6) So (2.5) becomes 𝑣_𝑥=(𝜆+𝑣)1𝑣𝑥−𝑣_𝑡,=(𝜆+𝑣)(−𝜆)^𝑗1𝑥−𝑣−(−𝜆)^𝑗−1𝑣𝑥.(3.7) Solving (3.7) gives the following solution of (1.3): 𝑣(𝑥,𝑡)=𝜆𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}+𝜆𝑒^𝑐(−1+𝜆𝑥)𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}−𝑒^𝑐.(3.8)

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

Example 3.3. By the Cole-Hopf transformation (1.7), 𝑢(𝑥,𝑡)=2𝑥(3.9) is a solution of (1.3) for 𝑗≥2. Then from (1.2) we have 𝑃₁𝑢,𝑢_𝑥=2𝑥²,𝑃_𝑗𝑢,…,𝜕^𝑗𝑥𝑢≡0for𝑗≥2.(3.10) So (2.5) becomes 𝑣_𝑥=(𝜆+𝑣)2𝑣𝑥−𝑣_𝑡,=(𝜆+𝑣)(−𝜆)^𝑗2𝑥−𝑣+2(−𝜆)^𝑗−1𝑥1₂−𝑣𝑥−2(−𝜆)^𝑗−2𝑣𝑥².(3.11) Solving (3.11) gives the following solution of (1.3) for 𝑗≥2: 𝑣(𝑥,𝑡)=2𝜆(−1+𝜆𝑥)𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}+𝜆𝑒^𝑐2−2𝜆𝑥+𝜆²𝑥²𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}−𝑒^𝑐.(3.12)

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

Example 3.4. By the Cole-Hopf transformation (1.7), 𝑢(𝑥,𝑡)=3𝑥(3.13) is a solution of (1.3) for 𝑗≥3. Then from (1.2) we have 𝑃₁𝑢,𝑢_𝑥=6𝑥²,𝑃₂𝑢,𝑢_𝑥,𝑢_𝑥𝑥=6𝑥³,𝑃_𝑗𝑢,…,𝜕^𝑗𝑥𝑢≡0for𝑗≥3.(3.14) So (2.5) becomes 𝑣_𝑥=(𝜆+𝑣)3𝑣𝑥−𝑣_𝑡,=(𝜆+𝑣)(−𝜆)^𝑗3𝑥−𝑣+3(−𝜆)^𝑗−1𝑥2₂−𝑣𝑥+6(−𝜆)^𝑗−2𝑥1₃−𝑣𝑥²−6(−𝜆)^𝑗−3𝑣𝑥³.(3.15) Solving (3.15) gives the following solution of (1.3) for 𝑗≥3: 𝑣(𝑥,𝑡)=3𝜆2−2𝜆𝑥+𝜆²𝑥²𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}+𝜆𝑒^𝑐−6+6𝜆𝑥−3𝜆²𝑥²+𝜆³𝑥³𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}−𝑒^𝑐.(3.16)

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(3.17) is a solution of (1.3) for 𝑗≥𝑘. Substituting (3.17) into (2.5) gives the following solution of (1.3) for 𝑗≥𝑘: 𝑣(𝑥,𝑡)=𝜕𝑓𝑥,𝑥²,…,𝑥^𝑘/𝜕𝑥𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}+𝜆𝑒^𝑐𝑓𝑥,𝑥²,…,𝑥^𝑘𝑒^{𝜆(𝑥+(−𝜆)𝑗𝑡)}−𝑒^𝑐,(3.18) where 𝑓𝑥,𝑥²,…,𝑥^𝑘=(−1)^𝑘𝑘!+(−1)^𝑘−1𝑘!𝜆𝑥+(−1)^𝑘−2𝑘!2!𝜆²𝑥²+⋯−𝑘𝜆^𝑘−1𝑥^𝑘−1+𝜆^𝑘𝑥^𝑘.(3.19)

Acknowledgment

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