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Abstract and Applied Analysis
Volume 2011, Article ID 935710, 8 pages
http://dx.doi.org/10.1155/2011/935710
Research Article

Bäcklund Transformation and New Exact Solutions of the Sharma-Tasso-Olver Equation

1School of Economy and Management, Guangzhou University of Chinese Medicine, Guangzhou 510006, China
2Department of Mathematics, Shandong University, Jinan 250100, China
3School of Mathematics and Information Science, Guangzhou University, Guangzhou 510006, China

Received 20 February 2011; Accepted 23 March 2011

Academic Editor: Svatoslav Staněk

Copyright © 2011 Lin Jianming et al. 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

The Sharma-Tasso-Olver (STO) equation is investigated. The Painlevé analysis is efficiently used for analytic study of this equation. The Bäcklund transformations and some new exact solutions are formally derived.

1. Introduction

Let 𝛼 be a constant. We consider the Sharma-Tasso-Olver (STO) equation𝑢𝑡𝑢+𝛼3𝑥+32𝛼𝑢2𝑥𝑥+𝛼𝑢𝑥𝑥𝑥=0.(1.1)

This paper is concerned with the STO equation [1] and [217].

Attention has been focused on STO equation (1.1) in [1] and [24] and the references therein due to its appearance in scientific applications. In [1], the tanh method, the extended tanh method, and other ansatz involving hyperbolic and exponential functions are efficiently used for the analytic study of this equation. And some new solitons and kinks solutions are formally derived. The proposed schemes are reliable and manageable. In [2], this equation was handled by using the Cole-Hopf transformations method. The simple symmetry reduction procedure is repeatedly used in [3] to obtain exact solutions where soliton fission and fusion were examined. However, in [4], the soliton fission and fusion were examined thoroughly by using Hirota's direct method and the Bäcklund transformations method. In [6], the authors show that symmetry constraints do not always yield exact solutions through analyzing the STO equation. The generalized Kaup-Newell-type hierarchy of nonlinear evolution equations is explicitly related to STO equation from [9]. Using the improved tanh function method in [10], the STO equation with its fission and fusion has some exact solutions. In [11] some exact solution of the STO equation are given by implying a generalized tanh function method for approximating some solutions which have been known. In [12], the method of third-order mode coupling is applied to the STO equation. In [13], the same exact explicit solutions of the STO equation, as this work are derived by using an extension of the homogeneous balance method and the Bäcklund transformation. Recently in [7, 8], the authors have introduced some other discussions about new solution techniques, a multiple exp-function method and linear superposition principle applying to nonlinear equations, which rely on a close observation on relations between linear objects and nonlinear objects.

Many types of travelling waves are of particular interest in solitary wave theory. The solitons, which are localized travelling waves, asymptotically zero at large distances, the periodic solutions, the kink waves which rise or descend from one asymptotic state to another, are well-known types of travelling waves.

The objectives of this work are twofold. First, we seek to establish new solitons and kink solutions of distinct physical structures for the nonlinear equation (1.1). Second, we aim to implement many strategies to achieve our goal, namely, the Painlevé test [18, 19], hyperbolic functions ansatze, and exponential functions ansatze to obtain new exact solutions. In what follows, the Painlevé test will be reviewed briefly.

We give the three steps of the test for a single PDE, (𝑥,𝑡,𝑢(𝑥,𝑡))=0,(1.2) in two independent variables, 𝑥 and 𝑡. Following [18, 20], the Laurent expansion of the solution 𝑢(𝑥,𝑡),𝑢(𝑥,𝑡)=𝜙𝛼(𝑥,𝑡)𝑘=0𝑢𝑘(𝑥,𝑡)𝜙𝑘(𝑥,𝑡),(1.3) should be single valued in the neighborhood of a noncharacteristic, movable and singular manifold 𝜙(𝑥,𝑡), which can be viewed as the surface of the movable poles in the complex plane. In (1.3), 𝑢0(𝑥,𝑡)0,𝛼 is a negative integer and 𝑢𝑘(𝑥,𝑡) are analytic functions in a neighborhood of 𝜙(𝑥,𝑡).

The Painlevé test have the following steps

Step 1 (leading order analysis). Determine the (negative) integer 𝛼 and 𝑢0 by balancing the minimal power terms after the substitution of 𝑢=𝑢0𝜙𝛼 into the given PDE. There may be several branches for 𝑢0, and for each the next two steps must be performed.

Step 2 (determination of the resonances). For selected 𝛼 and 𝑢0, calculate the nonnegative integers 𝑟, called the resonances, at which arbitrary functions 𝑢𝑟 enter the series (1.3). To do so, substitute (1.3) into (1.1), and normalize all orders of 𝜙 with the minimal power term. Then we have the cycle formula on 𝑢𝑗, through reducing the formula, and the resonance 𝑟 can be obtained.

Step 3 (verification of the compatibility conditions). Substituting 𝑗=𝑘 at which 𝑘 is not equal 𝑟 into the cycle formula, we have nonlinear equation 𝑓𝑘(𝑢𝑘𝑢0,𝜙)=0(1<𝑘<𝑟). At resonance levels, 𝑢𝑟 should be arbitrary, and then we are deducing a nonlinear equation 𝑔(𝑢𝑟1𝑢0,𝜙)=0. If the equation 𝑓 implies 𝑔, then the compatibility condition is unconditionally satisfied.
An equation for which these three steps can be carried out consistently and unambiguously passes the Painlevé test. Equation (1.3) is called the truncated Painlevé expansion.

2. The Compatibility Conditions and Bäcklund Transform

Let 𝑢=𝜙𝛼𝑗=0𝑢𝑗𝜙𝑗,(2.1) where 𝜙=𝜙(𝑥,𝑡),𝑢𝑗=𝑢𝑗(𝑥,𝑡) are analytic functions on (𝑥,𝑡) in the neighborhood of the manifold 𝑀={(𝑥,𝑡)𝜙(𝑥,𝑡)=0}.

Substituting (2.1) into (1.1), we can get the 𝛼=1. Thus, (2.1) becomes 𝑢=𝑗=0𝑢𝑗𝜙𝑗1.(2.2) Expressing (1.1), we have that 𝑢𝑡+3𝛼𝑢2𝑢𝑥+3𝛼𝑢2𝑥+3𝛼𝑢𝑢𝑥𝑥+𝛼𝑢𝑥𝑥𝑥=0.(2.3) Differentiating (2.2), we obtain that 𝑢𝑡=(𝑗1)𝑗=0𝑢𝑗𝜙𝑗2𝜙𝑡+𝑗=0𝑢𝑗,𝑡𝜙𝑗1,𝑢𝑥=(𝑗1)𝑗=0𝑢𝑗𝜙𝑗2𝜙𝑥+𝑗=0𝑢𝑗,𝑥𝜙𝑗1,𝑢𝑥𝑥=(𝑗1)𝑗=0𝑢𝑗𝜙𝑗2𝜙𝑥𝑥+2(𝑗1)𝑗=0𝑢𝑗,𝑥𝜙𝑗2𝜙𝑥+(𝑗1)(𝑗2)𝑗=0𝑢𝑗𝜙𝑗3𝜙2𝑥+𝑗=0𝑢𝑗,𝑥𝑥𝜙𝑗1,𝑢𝑥𝑥𝑥=3(𝑗1)𝑗=0𝑢𝑗,𝑥𝜙𝑗2𝜙𝑥𝑥+(𝑗1)𝑗=0𝑢𝑗𝜙𝑗2𝜙𝑥𝑥𝑥+3(𝑗1)(𝑗2)𝑗=0𝑢𝑗𝜙𝑗3𝜙𝑥𝜙𝑥𝑥+3(𝑗1)𝑗=0𝑢𝑗,𝑥𝑥𝜙𝑗2𝜙𝑥+𝑗=0𝑢𝑗,𝑥𝑥𝑥𝜙𝑗1+3(𝑗1)(𝑗2)𝑗=0𝑢𝑗,𝑥𝜙𝑗3𝜙2𝑥+(𝑗1)(𝑗2)(𝑗3)𝑗=0𝑢𝑗𝜙𝑗4𝜙3𝑥.(2.4) Substituting (2.4) into (2.3), we get the cycle formula on 𝑢𝑗(𝑗3)𝑢𝑗2𝜙𝑡+𝑢𝑗3,𝑡+3𝛼𝑗𝑛𝑗𝑚=0𝑛=0𝑢𝑗𝑚𝑛𝑢𝑚(𝑛1)𝑢𝑛𝜙𝑥+𝑢𝑛1,𝑥+3𝛼𝑗𝑚=0(𝑗𝑚1)𝑢𝑗𝑚𝜙𝑥+𝑢𝑗𝑚1,𝑥(𝑚1)𝑢𝑚𝜙𝑥+𝑢𝑚1,𝑥+3𝛼𝑗𝑚=0𝑢𝑗𝑚(𝑚2)𝑢𝑚1𝜙𝑥𝑥+2(𝑚2)𝑢𝑚1,𝑥𝜙𝑥+(𝑚1)(𝑚2)𝑢𝑚𝜙2𝑥+𝑢𝑚2,𝑥𝑥+𝛼3(𝑗3)𝑢𝑗2,𝑥𝜙𝑥𝑥+(𝑗3)𝑢𝑗2𝜙𝑥𝑥𝑥+3(𝑗2)(𝑗3)𝑢𝑗1𝜙𝑥𝜙𝑥𝑥+3(𝑗2)(𝑗3)𝑢𝑗1,𝑥𝜙2𝑥+𝑢𝑗3,𝑥𝑥𝑥+3(𝑗3)𝑢𝑗2,𝑥𝑥𝜙𝑥+(𝑗1)(𝑗2)(𝑗3)𝑢𝑗𝜙3𝑥=0.(2.5) Taking 𝑗=0 in (2.5), we deduce that 𝑢0=𝜙𝑥 or 𝑢0=2𝜙𝑥.

We will get the Bäcklund transformations, according to the above two cases, respectively.

Case 1 (𝑢0=𝜙𝑥). Substituting 𝑢0=𝜙𝑥 into (2.5), we have the following equation on 𝑢𝑗: (𝑗+1)(𝑗1)(𝑗3)𝛼𝑢𝑗𝜙3𝑥=𝐹𝑗𝑢𝑗1𝑢0,𝜙𝑡,𝜙𝑥,𝜙𝑥𝑥,𝜙𝑥𝑥𝑥(𝑗=1,2,).(2.6) By (2.6) we see that 𝑢1, 𝑢1 and 𝑢3 are arbitrary. Hence 𝑗=3 are resonances. Moreover the compatibility conditions will be deduced by (2.5) or (2.6).
Setting 𝑗=2 and 𝑗=3, we infer that 𝜙𝑡𝑢+3𝛼21𝜙𝑥+𝑢2𝜙2𝑥+𝑢1,𝑥𝜙𝑥+𝑢1𝜙𝑥𝑥+𝛼𝜙𝑥𝑥𝑥=0,(2.7)𝜕𝜙𝜕𝑥𝑡𝑢+3𝛼21𝜙𝑥+𝑢2𝜙2𝑥+𝑢1,𝑥𝜙𝑥+𝑢1𝜙𝑥𝑥+𝛼𝜙𝑥𝑥𝑥=0.(2.8)
It is easy to see that (2.8) does so if (2.7) holds. So 𝑢3 is arbitrary.
Taking 𝑗=4,𝑢3=0 and letting 𝑢2=𝑢4=0 in (2.6), we have that 𝑢1,𝑡+3𝛼𝑢21𝑢1,𝑥+3𝛼𝑢21,𝑥+3𝛼𝑢1𝑢1,𝑥𝑥+𝛼𝑢1,𝑥𝑥𝑥=0.(2.9)
By (2.6), noting that 𝑢2=0, 𝑢3=0 and 𝑢4=0, we can infer that 𝑢𝑗=0(𝑗2).(2.10)
Thus, we obtain from the Bäckland transformation of the STO equation that 𝜙𝑢=𝑥𝜙+𝑢1,(2.11) where 𝑢1 is the seed solution of (1.1), the 𝑢1 and 𝜙 satisfy the following condition: 𝜙𝑡𝑢+3𝛼21𝜙𝑥+𝑢1,𝑥𝜙𝑥+𝑢1𝜙𝑥𝑥+𝛼𝜙𝑥𝑥𝑥=0.(2.12)

Case 2 (𝑢0=2𝜙𝑥). Rewrite (2.5) as the following form on 𝑢𝑗(𝑗+1)(𝑗+2)(𝑗3)𝛼𝑢𝑗𝜙3𝑥=𝐺𝑗𝑢𝑗1𝑢0,𝜙𝑡,𝜙𝑥,𝜙𝑥𝑥,𝜙𝑥𝑥𝑥,(𝑗=1,2,).(2.13) From (2.13) we know that 𝑢3 is arbitrary. That is, 𝑗=3 are resonances.
Taking 𝑗=1, 𝑗=2, and 𝑗=3 in (2.13), we get that 𝑢1𝜙=𝑥𝑥𝜙𝑥,𝜙(2.14)𝑡+6𝛼𝑢2𝜙2𝑥3𝛼𝑢1𝜙𝑥𝑥2𝛼𝜙𝑥𝑥𝑥𝜕=0,(2.15)𝜙𝜕𝑥𝑡+6𝛼𝑢2𝜙2𝑥3𝛼𝑢1𝜙𝑥𝑥2𝛼𝜙𝑥𝑥𝑥=0.(2.16)
It is easy to see that (2.15) implies (2.16). So 𝑢3 is arbitrary. Equation (1.1) satisfies Painlevé property.
Taking 𝑗=4, 𝑢3=0 and letting 𝑢2=𝑢4=0 in (2.5) or (2.13), we have that 𝑢1,𝑡+3𝛼𝑢21𝑢1,𝑥+3𝛼𝑢21,𝑥+3𝛼𝑢1𝑢1,𝑥𝑥+𝛼𝑢1,𝑥𝑥𝑥=0.(2.17)
Substituting 𝑢1=𝜙𝑥𝑥/𝜙𝑥 into the above equation, we deduce the second condition below 𝜙3𝑥𝜙𝑥𝑥𝑡+𝜙2𝑥𝜙𝑥𝑥𝜙𝑥𝑡+18𝛼𝜙4𝑥𝑥30𝛼𝜙𝑥𝜙2𝑥𝑥𝜙𝑥𝑥𝑥+7𝛼𝜙2𝑥𝜙𝑥𝑥𝜙𝑥𝑥𝑥𝑥+6𝛼𝜙2𝑥𝜙2𝑥𝑥𝑥𝛼𝜙3𝑥𝜙𝑥𝑥𝑥𝑥𝑥=0(2.18)
By (2.5) or (2.13), noting that 𝑢2=0, 𝑢3=0, and 𝑢4=0, we can easy to have that 𝑢𝑗=0(𝑗2).(2.19)
Thus, we obtain from the other Bäckland transformation of the STO equation that 𝑢=2𝜙𝑥𝜙𝜙𝑥𝑥𝜙𝑥,(2.20) where 𝜙(𝑥,𝑡) satisfies the following conditions: 𝜙𝑥𝜙𝑡+3𝛼𝜙2𝑥𝑥2𝛼𝜙𝑥𝜙𝑥𝑥𝑥=0,𝜙3𝑥𝜙𝑥𝑥𝑡+𝜙2𝑥𝜙𝑥𝑥𝜙𝑥𝑡+18𝛼𝜙4𝑥𝑥30𝛼𝜙𝑥𝜙2𝑥𝑥𝜙𝑥𝑥𝑥+7𝛼𝜙2𝑥𝜙𝑥𝑥𝜙𝑥𝑥𝑥𝑥+6𝛼𝜙2𝑥𝜙2𝑥𝑥𝑥𝛼𝜙3𝑥𝜙𝑥𝑥𝑥𝑥𝑥=0.(2.21)

3. New Exact Solutions for STO Equation

As it is well known that the Bäcklund transformation is one of the most effective methods for finding exact solutions of nonlinear partial differential equations. By (2.11) and (2.20), choose some seed solutions, and then we can get some new single travelling solitary wave solutions below, respectively.

Case 1 (for Bäcklund transformation (2.11)). Taking the seed solution 𝑢1=0 in the Bäcklund transformation (2.11), (2.12) becomes 𝜙𝑡+𝛼𝜙𝑥𝑥𝑥=0.(3.1)
The arbitrary multiple solitary wave solution of (1.1) can be easily written down. For the sake of simplicity, we only give some new single travelling solitary wave solutions here. Substituting the ansatz 𝜙1=𝑘1𝑥2+𝑘2𝑥+𝑟 which satisfies (3.1) into (2.11), and leads the new exact solutions of (1.1) 𝑢1(𝑥,𝑡)=2𝑘1𝑥+𝑘2𝑘1𝑥2+𝑘2𝑥+𝑟.(3.2)
Substituting the ansatz 𝜙2=𝑎cosh(𝑘𝑥+𝜔𝑡+𝑟)+𝑏sinh(𝑘𝑥+𝜔𝑡+𝑟)+𝑐 into (3.1) will produce the dispersion relation between 𝜔 and 𝑘, 𝜔+𝛼𝑘3=0,(3.3) and the new exact solutions are that 𝑢2(𝑥,𝑡)=𝑎𝑘sinh𝑘𝑥𝛼𝑘3𝑡+𝑟+𝑏𝑘cosh𝑘𝑥𝛼𝑘3𝑡+𝑟𝑎cosh𝑘𝑥𝛼𝑘3𝑡+𝑟+𝑏sinh𝑘𝑥𝛼𝑘3𝑡+𝑟+𝑐.(3.4)
Taking the seed solution 𝑢1=𝑐 for (1.1) will get 𝜙𝑡𝑐+3𝛼2𝜙𝑥+𝑐𝜙𝑥𝑥+𝛼𝜙𝑥𝑥𝑥=0.(3.5)
Observing (3.5), we have 𝜙3=𝑘𝑥3𝛼𝑐2𝑘𝑡+𝑟, 𝜙4=𝑎+𝑒𝑘𝑥(3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3)𝑡+𝑟 and 𝜙5=sinh[𝑘𝑥(3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3)𝑡+𝑟]+cosh[𝑘𝑥(3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3)𝑡+𝑟]+𝑏.
Substituting 𝜙3, 𝜙4, 𝜙5 and 𝑢1=𝑐 into the Bäcklund transformation (2.11), we obtain the exact solutions for STO equation (1.1) 𝑢3𝑘(𝑥,𝑡)=𝑘𝑥3𝛼𝑐2𝑢𝑘𝑡+𝑟+𝑐,4(𝑥,𝑡)=𝑘𝑒𝑘𝑥(3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3)𝑡+𝑟𝑎+𝑒𝑘𝑥(3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3)𝑡+𝑟𝑢+𝑐,5(𝑥,𝑡)=𝑘sinh𝑘𝑥3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3𝑡+𝑟+𝑘cosh𝑘𝑥3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3𝑡+𝑟sinh𝑘𝑥3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3𝑡+𝑟+cosh𝑘𝑥3𝛼𝑐2𝑘+3𝛼𝑐𝑘2+𝛼𝑘3𝑡+𝑟+𝑏+𝑐.(3.6)
Taking the seed solution 𝑢1=1/𝑥 for STO equation (1.1), (2.12) becomes 𝜙𝑡𝜙+3𝛼𝑥𝑥𝑥+𝛼𝜙𝑥𝑥𝑥=0.(3.7)
Substitute the ansatz 𝜙6=𝑘1𝑥3+𝑘3𝑥24𝛼𝑘1𝑡+𝑏 into (2.11), yielding the solutions for the STO equation (1.1) 𝑢6(𝑥,𝑡)=3𝑘1𝑥2+𝑘3𝑘1𝑥3+𝑘3𝑥24𝛼𝑘1+1𝑡+𝑏𝑥.(3.8)

Remark 3.1. When 𝑎=0 or 𝑏=0, the solution 𝑢2(𝑥,𝑡) is obtained in [1] by tanh method. In [4], the authors have obtained the 𝑢4(𝑥,𝑡). As for the case 𝑢0=2𝜙𝑥, the authors of [4, page 238] said that no new meaningful results can be obtained; we could here obtain some meaningful results.

Case 2 (for Bäcklund transformation (2.20)). Observe, (2.21), we have ansatz 𝜙1=sinh(𝑘𝑥+𝜔𝑡+𝑟)+cosh(𝑘𝑥+𝜔𝑡+𝑟). Substituting 𝜙1 into (2.21) and seting the coefficients to be equal to zero, we have the dispersion relation between 𝜔 and 𝑘, 𝜔=𝛼𝑘3.(3.9)
Thus, (2.20) yields the new exact solutions for the STO equation (1.1) 𝑢7=2𝑘sinh𝑘𝑥𝛼𝑘3𝑡+𝑟+𝑘cosh𝑘𝑥𝛼𝑘3𝑡+𝑟sinh𝑘𝑥𝛼𝑘3𝑡+𝑟+cosh𝑘𝑥𝛼𝑘3𝑡+𝑟+𝑏𝑘.(3.10)
Substituting ansatz 𝜙2=𝑒𝑘𝑥+𝜔𝑡+𝑟+𝑏 into (2.21) and balancing the coefficients, we have the new exact solutions for the STO equation (1.1) 𝑢7(𝑥,𝑡)=2𝑘𝑒𝑘𝑥𝛼𝑘3𝑡+𝑟𝑒𝑘𝑥𝛼𝑘3𝑡+𝑟+𝑏𝑘.(3.11)

Acknowledgments

The authors would like to express their hearty thanks to the Chern Institute of Mathematics of Nankai University for supplying the financial supports and wonderful work conditions. The authors wish to thank the referee for his or her very helpful comments and useful suggestions. This work was supported by the NSF of China (10771220), Doctorial Point Fund of National Education, Ministry of China (200810780002).

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