﻿ 利用exp(-G(ξ))方法和拟设函数法求Sharma-Tasso-Olver方程精确解 The Exact Solutions of the Sharma-Tasso-Olver Equation Using exp(-G(ξ)) Method and Ansatz Method

Vol. 08  No. 02 ( 2019 ), Article ID: 28843 , 8 pages
10.12677/AAM.2019.82025

The Exact Solutions of the Sharma-Tasso-Olver Equation Using exp(-G(ξ)) Method and Ansatz Method

Fengyan Zhang1,2, Dongxue Li3, Junmei Wang1

1School of Mathematics and Statistics, Shandong Normal University, Jinan Shandong

2School of Mathematics and Statistics, Shandong Linyi University, Linyi Shandong

3Dezhou No. 10 Middle School, Dezhou Shandong

Received: Jan. 22nd, 2019; accepted: Feb. 6th, 2019; published: Feb. 13th, 2019

ABSTRACT

Using the traveling wave transformation and homogeneous balance simplified the Sharma-Tasso-Olver equation to obtain the reduced ordinary differential equations, using the $exp\left(-G\left(\xi \right)\right)$ -method to get the trigonometric function solution, hyperbolic function solution and the rational function solution. In addition, an exact soliton solution is provided by using the Ansatz method.

Keywords:Sharma-Tasso-Olver Equation, $exp\left(-G\left(\xi \right)\right)$ -Method, Soliton Solution, Ansatz Method

1山东师范大学，数学与统计学院，山东 济南

2临沂大学，数学与统计学院，山东 临沂

3德州市第十中学，山东 德州

1. 引言

${u}_{t}+3\epsilon {u}^{2}{u}_{x}+3\epsilon {u}_{x}^{2}+3\epsilon u{u}_{xx}+\epsilon {u}_{xxx}=0$ (1)

2. $exp\left(-G\left(\xi \right)\right)$ 方法

$u\left(x,t\right)=u\left(\xi \right)$ , $\xi =kx-wt$ ,

$-w{u}^{\prime }+3\epsilon k{u}^{2}{u}^{\prime }+3\epsilon {k}^{2}{\left({u}^{\prime }\right)}^{2}+3\epsilon {k}^{2}u{u}^{″}+\epsilon {k}^{3}{u}^{‴}=0,$

$-wu+\epsilon k{u}^{3}+3\epsilon {k}^{2}u{u}^{\prime }+\epsilon {k}^{3}{u}^{″}=0.$ (2)

$u\left(\xi \right)=\underset{n=0}{\overset{m}{\sum }}{a}_{n}{\left(\mathrm{exp}\left(-G\left(\xi \right)\right)\right)}^{n},$ (3)

${G}^{\prime }\left(\xi \right)=\mathrm{exp}\left(-G\left(\xi \right)\right)+\mu \mathrm{exp}\left(G\left(\xi \right)\right)+\lambda ,$ (4)

$u\left(\xi \right)={a}_{0}+{a}_{1}\mathrm{exp}\left(-G\left(\xi \right)\right).$ (5)

$\mathrm{exp}{\left(-G\left(\xi \right)\right)}^{3}:2{k}^{3}\epsilon {a}_{1}-3\epsilon {k}^{2}{a}_{1}^{2}+k\epsilon {a}_{1}^{3}=0,$ (6)

$\mathrm{exp}{\left(-G\left(\xi \right)\right)}^{2}:{k}^{3}\epsilon {a}_{1}-{k}^{2}\epsilon \lambda {a}_{1}^{2}+{k}^{2}\epsilon {a}_{1}^{2}{a}_{0}=0,$ (7)

$\mathrm{exp}\left(-G\left(\xi \right)\right):{k}^{3}\epsilon {\lambda }^{2}{a}_{1}+2{k}^{3}\epsilon \mu {a}_{1}-3{k}^{2}\epsilon \mu {a}_{1}^{2}-3{k}^{2}\epsilon \lambda {a}_{1}{a}_{0}+3k\epsilon {a}_{1}{a}_{0}^{2}-w{a}_{1}=0,$ (8)

$\mathrm{exp}{\left(-G\left(\xi \right)\right)}^{0}:{k}^{3}\epsilon \lambda \mu {a}_{1}+3{k}^{2}\epsilon \mu {a}_{0}+k\epsilon {a}_{0}^{3}-w{a}_{0}=0.$ (9)

① 当 ${\lambda }^{2}-4\mu >0$ ，且 $\mu \ne 0$ 时，

${u}_{1.1}=\frac{2k\mu }{-\sqrt{{\lambda }^{2}-4\mu }\mathrm{tanh}\left(\sqrt{{\lambda }^{2}-4\mu }/2\left(\xi +C\right)\right)-\lambda }+\frac{\lambda k}{2}$(如图1)

② 当 ${\lambda }^{2}-4\mu <0$ ，且 $\mu \ne 0$ 时，

${u}_{1.2}=\frac{2k\mu }{\sqrt{4\mu -{\lambda }^{2}}\mathrm{tan}\left(\sqrt{4\mu -{\lambda }^{2}}/2\left(\xi +C\right)\right)-\lambda }+\frac{\lambda k}{2}$(如图2)

③ 当 ${\lambda }^{2}-4\mu >0$ ，且 $\mu =0,\lambda \ne 0$ 时，

${u}_{1.3}=\frac{\lambda k}{\mathrm{cosh}\left(\lambda \left(\xi +C\right)\right)+\mathrm{sinh}\left(\lambda \left(\xi +C\right)\right)-1}+\frac{\lambda k}{2}$(如图3)

④ 当 ${\lambda }^{2}-4\mu =0$ ，且 $\mu \ne 0,\lambda \ne 0$ 时，

${u}_{1.4}=-\frac{{\lambda }^{2}k\left(\xi +C\right)}{2\lambda \left(\xi +C\right)+4}+\frac{\lambda k}{2}$(如图4)

⑤ 当 ${\lambda }^{2}-4\mu =0$ ，且 $\mu =0,\lambda =0$ 时，

${u}_{1.5}=\frac{k}{\xi +C}+\frac{\lambda k}{2}$(如图5)

Figure 1. Hyperbolic function solution ${u}_{1.1}$

Figure 2. Trigonometric function solution ${u}_{1.2}$

Figure 3. Hyperbolic function solution ${u}_{1.3}$

Figure 4. Rational function solution ${u}_{1.4}$

Figure 5. Rational function solution ${u}_{1.5}$

① 当 ${\lambda }^{2}-4\mu >0$ ，且 $\mu \ne 0$ 时，

${u}_{2.1}=\frac{4k\mu }{-\sqrt{{\lambda }^{2}-4\mu }\mathrm{tanh}\left(\sqrt{{\lambda }^{2}-4\mu }/2\left(\xi +C\right)\right)-\lambda }+\lambda k$

② 当 ${\lambda }^{2}-4\mu <0$ ，且 $\mu \ne 0$ 时，

${u}_{2.2}=\frac{4k\mu }{\sqrt{4\mu -{\lambda }^{2}}\mathrm{tan}\left(\sqrt{4\mu -{\lambda }^{2}}/2\left(\xi +C\right)\right)-\lambda }+\lambda k$

③ 当 ${\lambda }^{2}-4\mu >0$ ，且 $\mu =0,\lambda \ne 0$ 时，

${u}_{2.3}=\frac{2\lambda k}{\mathrm{cosh}\left(\lambda \left(\xi +C\right)\right)+\mathrm{sinh}\left(\lambda \left(\xi +C\right)\right)-1}+\lambda k$

④ 当 ${\lambda }^{2}-4\mu =0$ ，且 $\mu \ne 0,\lambda \ne 0$ 时，

${u}_{2.4}=-\frac{{\lambda }^{2}k\left(\xi +C\right)}{\lambda \left(\xi +C\right)+2}+\lambda k$

⑤ 当 ${\lambda }^{2}-4\mu =0$ ，且 $\mu =0,\lambda =0$ 时，

${u}_{2.5}=\frac{2k}{\xi +C}+\lambda k$

3. 拟设双曲函数法

$u\left(x,t\right)=A\mathrm{tan}{h}^{″}\tau$$\tau =B\left(x-vt\right)$(10)

${u}_{t}=-nvAB\left({\mathrm{tanh}}^{n-1}\tau -{\mathrm{tanh}}^{n+1}\tau \right),$ (11)

${u}_{x}=nvAB\left({\mathrm{tanh}}^{n-1}\tau -{\mathrm{tanh}}^{n+1}\tau \right),,$ (12)

${u}_{xx}=n\left(n-1\right)A{B}^{2}{\mathrm{tanh}}^{n-2}\tau -2{n}^{2}A{B}^{2}{\mathrm{tanh}}^{n}\tau +n\left(n+1\right)A{B}^{2}{\mathrm{tanh}}^{n+2}\tau ,$ (13)

$\begin{array}{c}{u}_{xxx}=n\left(n-1\right)\left(n-2\right)A{B}^{3}{\mathrm{tanh}}^{n-3}\tau -\left[n\left(n-1\right)\left(n-2\right)+2{n}^{3}\right]A{B}^{3}{\mathrm{tanh}}^{n-1}\tau \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left[n\left(n+1\right)\left(n+2\right)+2{n}^{3}\right]A{B}^{3}{\mathrm{tanh}}^{n+1}\tau -n\left(n+1\right)\left(n+2\right)A{B}^{3}{\mathrm{tanh}}^{n+3}\tau ,\end{array}$ (14)

$\begin{array}{l}\text{ }-nvAB\left({\mathrm{tanh}}^{n-1}\tau -{\mathrm{tanh}}^{n+1}\tau \right)+3\epsilon nv{A}^{2}B\left({\mathrm{tanh}}^{2n-1}\tau -{\mathrm{tanh}}^{2n+1}\tau \right)\\ \text{ }+3\epsilon {\left[nvAB\left({\mathrm{tanh}}^{n-1}\tau -{\mathrm{tanh}}^{n+1}\tau \right)\right]}^{2}+3\epsilon {A}^{2}{B}^{2}\left[n\left(n-1\right){\mathrm{tanh}}^{2n-2}\tau \\ \text{ }-2{n}^{2}{\mathrm{tanh}}^{2n}\tau n\left(n+1\right){\mathrm{tanh}}^{2n+2}\tau \right]+\epsilon \left[n\left(n-1\right)\left(n-2\right)A{B}^{3}{\mathrm{tanh}}^{n-3}\tau \\ \text{ }-\left(n\left(n-1\right)\left(n-2\right)+2{n}^{3}\right)A{B}^{3}{\mathrm{tanh}}^{n-1}\tau +\left(n\left(n+1\right)\left(n+2\right)+2{n}^{3}\right)A{B}^{3}{\mathrm{tanh}}^{n+1}\tau \\ \text{ }-n\left(n+1\right)\left(n+2\right)A{B}^{3}{\mathrm{tanh}}^{n+3}\tau .\end{array}$ (15)

$B=A$ 或者 $B=\frac{A}{2}$$v=\epsilon {A}^{2}$

$u\left(x,t\right)=A\mathrm{tanh}\left(B\left(x-vt\right)\right)$

4. 结论

The Exact Solutions of the Sharma-Tasso-Olver Equation Using exp(-G(ξ)) Method and Ansatz Method[J]. 应用数学进展, 2019, 08(02): 219-226. https://doi.org/10.12677/AAM.2019.82025

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