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

Authors: Lin Jianming; Ding Jie; Yuan Wenjun
Journal: Abstract and Applied Analysis
 (2011)

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## 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.


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## Body

## 1. Introduction

Letα be a constant. We consider the Sharma-Tasso-Olver (STO) equation(1.1)ut+α(u3)x+32α(u2)xx+αuxxx=0.This paper is concerned with the STO equation [1] and [2–17].Attention has been focused on STO equation (1.1) in [1] and [2–4] 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,(1.2)F(x,t,u(x,t))=0,
in two independent variables, x and t. Following [18, 20], the Laurent expansion of the solution u(x,t),(1.3)u(x,t)=ϕα(x,t)∑k=0∞uk(x,t)ϕk(x,t),
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 stepsStep 1 (leading order analysis).
Determine the (negative) integerα 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 selectedα 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).
Substitutingj=k at which k is not equal r into the cycle formula, we have nonlinear equation fk(uk⋯u0,ϕ⋯)=0(1<k<r). At resonance levels, ur should be arbitrary, and then we are deducing a nonlinear equation g(ur-1⋯u0,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(2.1)u=ϕα∑j=0∞ujϕj,
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 (2.2)u=∑j=0∞ujϕj-1.
Expressing (1.1), we have that (2.3)ut+3αu2ux+3αux2+3αuuxx+αuxxx=0.
Differentiating (2.2), we obtain that
(2.4)ut=(j-1)∑j=0∞ujϕj-2ϕt+∑j=0∞uj,tϕj-1,ux=(j-1)∑j=0∞ujϕj-2ϕx+∑j=0∞uj,xϕj-1,uxx=(j-1)∑j=0∞ujϕj-2ϕxx+2(j-1)∑j=0∞uj,xϕj-2ϕx+(j-1)(j-2)∑j=0∞ujϕj-3ϕx2+∑j=0∞uj,xxϕj-1,uxxx=3(j-1)∑j=0∞uj,xϕj-2ϕxx+(j-1)∑j=0∞ujϕj-2ϕxxx+3(j-1)(j-2)∑j=0∞ujϕj-3ϕxϕxx+3(j-1)∑j=0∞uj,xxϕj-2ϕx+∑j=0∞uj,xxxϕj-1+3(j-1)(j-2)∑j=0∞uj,xϕj-3ϕx2+(j-1)(j-2)(j-3)∑j=0∞ujϕj-4ϕx3.
Substituting (2.4) into (2.3), we get the cycle formula on uj(2.5)(j-3)uj-2ϕt+uj-3,t+3α∑m=0j-n∑n=0juj-m-num[(n-1)unϕx+un-1,x]+3α∑m=0j[(j-m-1)uj-mϕx+uj-m-1,x][(m-1)umϕx+um-1,x]+3α∑m=0juj-m[(m-2)um-1ϕxx+2(m-2)um-1,xϕx+(m-1)(m-2)umϕx2+um-2,xx]+α(j-3)ujϕx3[3(j-3)uj-2,xϕxx+(j-3)uj-2ϕxxx+3(j-2)(j-3)uj-1ϕxϕxx+3(j-2)(j-3)uj-1,xϕx2+uj-3,xxx+3(j-3)uj-2,xxϕx+(j-1)(j-2)(j-3)ujϕx3]=0.
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).
Substitutingu0=ϕx into (2.5), we have the following equation on uj:
(2.6)(j+1)(j-1)(j-3)αujϕx3=Fj(uj-1⋯u0,ϕt,ϕx,ϕxx,ϕxxx⋯)(j=1,2,…).
By (2.6) we see that u-1, u1 and u3 are arbitrary. Hence j=3 are resonances. Moreover the compatibility conditions will be deduced by (2.5) or (2.6).
Settingj=2 and j=3, we infer that
(2.7)ϕt+3α(u12ϕx+u2ϕx2+u1,xϕx+u1ϕxx)+αϕxxx=0,(2.8)∂∂x[ϕt+3α(u12ϕx+u2ϕx2+u1,xϕx+u1ϕxx)+αϕxxx]=0.
It is easy to see that (2.8) does so if (2.7) holds. So u3 is arbitrary.
Takingj=4,u3=0 and letting u2=u4=0 in (2.6), we have that
(2.9)u1,t+3αu12u1,x+3αu1,x2+3αu1u1,xx+αu1,xxx=0.
By (2.6), noting that u2=0, u3=0 and u4=0, we can infer that
(2.10)uj=0(j≥2).
Thus, we obtain from the Bäckland transformation of the STO equation that(2.11)u=ϕxϕ+u1,
where u1 is the seed solution of (1.1), the u1 and ϕ satisfy the following condition:
(2.12)ϕt+3α(u12ϕx+u1,xϕx+u1ϕxx)+αϕxxx=0.Case 2 (u0=2ϕx).
Rewrite (2.5) as the following form on uj(2.13)(j+1)(j+2)(j-3)αujϕx3=Gj(uj-1⋯u0,ϕt,ϕx,ϕxx,ϕxxx⋯),(j=1,2,…).
From (2.13) we know that u3 is arbitrary. That is, j=3 are resonances.
Takingj=1, j=2, and j=3 in (2.13), we get that
(2.14)u1=-ϕxxϕx,(2.15)ϕt+6αu2ϕx2-3αu1ϕxx-2αϕxxx=0,(2.16)∂∂x[ϕt+6αu2ϕx2-3αu1ϕxx-2αϕxxx]=0.
It is easy to see that (2.15) implies (2.16). So u3 is arbitrary. Equation (1.1) satisfies Painlevé property.
Takingj=4, u3=0 and letting u2=u4=0 in (2.5) or (2.13), we have that
(2.17)u1,t+3αu12u1,x+3αu1,x2+3αu1u1,xx+αu1,xxx=0.
Substitutingu1=-ϕxx/ϕx into the above equation, we deduce the second condition below
(2.18)-ϕx3ϕxxt+ϕx2ϕxxϕxt+18αϕxx4-30αϕxϕxx2ϕxxx+7αϕx2ϕxxϕxxxx+6αϕx2ϕxxx2-αϕx3ϕxxxxx=0
By (2.5) or (2.13), noting that u2=0, u3=0, and u4=0, we can easy to have that
(2.19)uj=0(j≥2).
Thus, we obtain from the other Bäckland transformation of the STO equation that(2.20)u=2ϕxϕ-ϕxxϕx,
where ϕ(x,t) satisfies the following conditions:
(2.21)ϕxϕt+3αϕxx2-2αϕxϕxxx=0,-ϕx3ϕxxt+ϕx2ϕxxϕxt+18αϕxx4-30αϕxϕxx2ϕxxx+7αϕx2ϕxxϕxxxx+6αϕx2ϕxxx2-αϕx3ϕxxxxx=0.


## 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 solutionu1=0 in the Bäcklund transformation (2.11), (2.12) becomes
(3.1)ϕt+αϕxxx=0.
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)
(3.2)u1(x,t)=2k1x+k2k1x2+k2x+r.
Substituting the ansatzϕ2=acosh(kx+ωt+r)+bsinh(kx+ωt+r)+c into (3.1) will produce the dispersion relation between ω and k,
(3.3)ω+αk3=0,
and the new exact solutions are that
(3.4)u2(x,t)=aksinh⁡⁡(kx-αk3t+r)+bkcosh⁡⁡(kx-αk3t+r)acosh⁡⁡(kx-αk3t+r)+bsinh⁡⁡(kx-αk3t+r)+c.
Taking the seed solutionu1=c for (1.1) will get
(3.5)ϕt+3α(c2ϕx+cϕxx)+αϕxxx=0.
Observing (3.5), we have ϕ3=kx-3αc2kt+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.
Substitutingϕ3, ϕ4, ϕ5 and u1=c into the Bäcklund transformation (2.11), we obtain the exact solutions for STO equation (1.1)
(3.6)u3(x,t)=kkx-3αc2kt+r+c,u4(x,t)=kekx-(3αc2k+3αck2+αk3)t+ra+ekx-(3αc2k+3αck2+αk3)t+r+c,u5(x,t)=ksinh⁡⁡[kx-(3αc2k+3αck2+αk3)t+r]+kcosh⁡⁡[kx-(3αc2k+3αck2+αk3)t+r]sinh⁡[kx-(3αc2k+3αck2+αk3)t+r]+cosh⁡⁡[kx-(3αc2k+3αck2+αk3)t+r]+b+c.
Taking the seed solutionu1=1/x for STO equation (1.1), (2.12) becomes
(3.7)ϕt+3αϕxxx+αϕxxx=0.
Substitute the ansatzϕ6=k1x3+k3x-24αk1t+b into (2.11), yielding the solutions for the STO equation (1.1)
(3.8)u6(x,t)=3k1x2+k3k1x3+k3x-24αk1t+b+1x.Remark 3.1.
Whena=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,
(3.9)ω=-αk3.
Thus, (2.20) yields the new exact solutions for the STO equation (1.1)
(3.10)u7=2ksinh⁡⁡[kx-αk3t+r]+kcosh⁡⁡[kx-αk3t+r]sinh⁡⁡[kx-αk3t+r]+cosh⁡⁡[kx-αk3t+r]+b-k.
Substituting ansatzϕ2=ekx+ωt+r+b into (2.21) and balancing the coefficients, we have the new exact solutions for the STO equation (1.1)
(3.11)u7(x,t)=2kekx-αk3t+rekx-αk3t+r+b-k.



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