# 新的双辅助微分方程展开法求色散水波方程的精确解A New Double-Auxiliary Differential Equation Expansion Method for Exact Solutions of Dispersive Water Wave Equations

DOI: 10.12677/AAM.2020.93044, PDF, HTML, XML, 下载: 93  浏览: 150

Abstract: Based on the existing auxiliary differential equations and G'/G-expansion method, this paper con-structs a new double auxiliary differential equation expansion method, and uses the double auxil-iary differential equation expansion method to solve new exact solutions of dispersive water wave equations: hyperbolic function formal solution, rational function formal solution, trigonometric function formal solution, hyperbolic function and trigonometric function mixed form solution, trigonometric function and rational function mixed form solution. The method can also be applied to solve other nonlinear equations.

1. 引言

2. 双辅助方程展开法

$\left\{\begin{array}{l}P\left(u,v,{u}_{t},{v}_{t},{u}_{x},{v}_{x},\cdots \right)=0\\ H\left(u,v,{u}_{t},{v}_{t},{u}_{x},{v}_{x},\cdots \right)=0\end{array}$ (1)

$\left\{\begin{array}{l}u\left(x,t\right)=\underset{k=0}{\overset{m}{\sum }}\underset{i+j=k}{\sum }{a}_{ij}\left(x,t\right){\varphi }^{i}\left(\xi \right){\left(\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}\right)}^{j},\\ v\left(x,t\right)=\underset{k=0}{\overset{n}{\sum }}\underset{i+j=k}{\sum }{b}_{ij}\left(x,t\right){\varphi }^{i}\left(\xi \right){\left(\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}\right)}^{j},\end{array}$ (2)

${\varphi }^{\prime }\left(\xi \right)=A+B\varphi \left(\xi \right)+C{\varphi }^{2}\left(\xi \right)$ (3)

${G}^{″}\left(\theta \right)+\lambda {G}^{\prime }\left(\theta \right)+\mu G\left(\theta \right)=0$ (4)

${\varphi }_{1}\left(\xi \right)=-\frac{B+\sqrt{\Delta }}{2C}\mathrm{tanh}\left(\frac{\sqrt{\Delta }}{2}\xi \right)$ (5)

${\varphi }_{2}\left(\xi \right)=-\frac{B+\sqrt{\Delta }}{2C}\mathrm{coth}\left(\frac{\sqrt{\Delta }}{2}\xi \right)$ (6)

$\Delta ={B}^{2}-4AC<0$ 时，方程(3)的解为：

${\varphi }_{3}\left(\xi \right)=-\frac{B-\sqrt{-\Delta }}{2C}\mathrm{tan}\left(\frac{\sqrt{-\Delta }}{2}\xi \right)$ (7)

${\varphi }_{4}\left(\xi \right)=-\frac{B-\sqrt{-\Delta }}{2C}\mathrm{cot}\left(\frac{\sqrt{-\Delta }}{2}\xi \right)$ (8)

$A=B=0,C\ne 0$ 时，方程(3)的解为：

${\varphi }_{5}\left(\xi \right)=\frac{-1}{C\xi +{H}_{1}}$ (9)

${\Delta }_{1}={\lambda }^{2}-4\mu >0$ 时，方程(4)的解为：

$\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}=-\frac{\lambda }{2}+\frac{\sqrt{{\Delta }_{1}}}{2}\frac{{c}_{1}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}$ (10)

${\Delta }_{1}={\lambda }^{2}-4\mu <0$ 时，方程(4)的解为：

$\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}=-\frac{\lambda }{2}+\frac{\sqrt{-{\Delta }_{1}}}{2}\frac{-{c}_{1}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}$ (11)

${\Delta }_{1}={\lambda }^{2}-4\mu =0$ 时，方程(4)的解为：

$\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}=-\frac{\lambda }{2}+\frac{{c}_{2}}{{c}_{1}+{c}_{2}\theta }$ (12)

3. 色散水波方程的解

$\left\{\begin{array}{l}{u}_{t}={v}_{xxx}+2{\left(uv\right)}_{x}=0\\ {v}_{t}={u}_{x}+2v{v}_{x}=0\end{array}$ (13)

$\left\{\begin{array}{l}u\left(x,t\right)={a}_{0}\left(t\right)+{a}_{1}\left(t\right)\varphi \left(\xi \right)+{a}_{2}\left(t\right)\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}+{a}_{3}\left(t\right)\varphi \left(\xi \right)\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}\\ v\left(x,t\right)={b}_{0}\left(t\right)+{b}_{1}\left(t\right)\varphi \left(\xi \right)+{b}_{2}\left(t\right)\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}\end{array}$ (14)

${\varphi }^{i}\left(\xi \right){\left(\frac{{G}^{\prime }\left(\theta \right)}{G\left(\theta \right)}\right)}^{j}\left(i,j=0,1,2,3\right)$ 的多项式，令每一项幂系数为零，得到一个关于 ${a}_{0}\left(t\right),{a}_{1}\left(t\right),{a}_{2}\left(t\right),{b}_{0}\left(t\right),{b}_{1}\left(t\right),$

${b}_{2}\left(t\right),{k}_{1},{k}_{2},{\varpi }_{1}$${\varpi }_{2}$ 的线性方程组，借助Maple求解此方程组，可得到：

${k}_{1}={k}_{1},{k}_{2}=\frac{{k}_{1}B}{\lambda }$

${\varpi }_{1}=1,{\varpi }_{2}=\frac{B}{\lambda }$

${a}_{0}\left(t\right)=\frac{1}{4}\frac{{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu },{a}_{1}\left(t\right)=-\frac{{b}_{2}{k}_{2}^{3}\lambda \mu \left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}$ (15)

${a}_{2}\left(t\right)=-\frac{1}{4}\frac{{b}_{2}{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu {b}_{0}},{a}_{3}\left(t\right)=\frac{{b}_{2}{k}_{2}^{3}{\lambda }^{2}\left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}$

${b}_{0}\left(t\right)={b}_{0}\left(t\right),{b}_{1}\left(t\right)={b}_{1}\left(t\right),{b}_{2}\left(t\right)={b}_{2}\left(t\right)$

$\begin{array}{c}{u}_{1}\left(x,t\right)=\frac{1}{4}\frac{{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu }-\frac{{b}_{2}{k}_{2}^{3}\lambda \mu \left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{B+\sqrt{\Delta }}{2C}\mathrm{coth}\left(\frac{\sqrt{\Delta }}{2}\xi \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{1}{4}\frac{{b}_{2}{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu {b}_{0}}\left(\frac{\lambda }{2}-\frac{\sqrt{{\Delta }_{1}}}{2}\frac{{c}_{1}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{b}_{2}{k}_{2}^{3}{\lambda }^{2}\left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{B+\sqrt{\Delta }}{2C}\mathrm{coth}\left(\frac{\sqrt{\Delta }}{2}\xi \right)\left(\frac{\lambda }{2}-\frac{\sqrt{{\Delta }_{1}}}{2}\frac{{c}_{1}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}\right)\end{array}$ (16)

$\begin{array}{c}{v}_{1}\left(x,t\right)={b}_{0}\left(t\right)-{b}_{1}\left(t\right)\frac{B+\sqrt{\Delta }}{2C}\mathrm{coth}\left(\frac{\sqrt{\Delta }}{2}\xi \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{b}_{2}\left(t\right)\left(\frac{\lambda }{2}-\frac{\sqrt{{\Delta }_{1}}}{2}\frac{{c}_{1}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}\right)\end{array}$ (17)

(16)与(17)为色散水波方程的双曲函数形式解。

$\begin{array}{c}{u}_{2}\left(x,t\right)=\frac{1}{4}\frac{{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu }+\frac{{b}_{2}{k}_{2}^{3}\lambda \mu \left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{B-\sqrt{-\Delta }}{2C}\mathrm{tan}\left(\frac{\sqrt{-\Delta }}{2}\xi \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{1}{4}\frac{{b}_{2}{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu {b}_{0}}\left(\frac{\lambda }{2}-\frac{\sqrt{{\Delta }_{1}}}{2}\frac{{c}_{1}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{b}_{2}{k}_{2}^{3}{\lambda }^{2}\left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{B-\sqrt{-\Delta }}{2C}\mathrm{tan}\left(\frac{\sqrt{-\Delta }}{2}\xi \right)\left(\frac{\lambda }{2}-\frac{\sqrt{{\Delta }_{1}}}{2}\frac{{c}_{1}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}\right)\end{array}$ (18)

$\begin{array}{c}{v}_{2}\left(x,t\right)={b}_{0}\left(t\right)-{b}_{1}\left(t\right)\frac{B-\sqrt{-\Delta }}{2C}\mathrm{tan}\left(\frac{\sqrt{-\Delta }}{2}\xi \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{b}_{2}\left(t\right)\left(\frac{\lambda }{2}-\frac{\sqrt{{\Delta }_{1}}}{2}\frac{{c}_{1}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cosh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sinh}\left(\frac{\sqrt{{\Delta }_{1}}}{2}\theta \right)}\right)\end{array}$ (19)

(18)与(19)为色散水波方程的双曲函数与三角函数混合作用解。

$\begin{array}{c}{u}_{3}\left(x,t\right)=\frac{1}{4}\frac{{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu }+\frac{{b}_{2}{k}_{2}^{3}\lambda \mu \left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{1}{C\xi +{H}_{1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{1}{4}\frac{{b}_{2}{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu {b}_{0}}\left(\frac{\lambda }{2}-\frac{\sqrt{-{\Delta }_{1}}}{2}\frac{-{c}_{1}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{b}_{2}{k}_{2}^{3}{\lambda }^{2}\left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{1}{C\xi +{H}_{1}}\left(\frac{\lambda }{2}-\frac{\sqrt{-{\Delta }_{1}}}{2}\frac{-{c}_{1}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}\right)\end{array}$ (20)

${v}_{3}\left(x,t\right)={b}_{0}\left(t\right)-{b}_{1}\left(t\right)\frac{1}{C\xi +{H}_{1}}+{b}_{2}\left(t\right)\left(-\frac{\lambda }{2}+\frac{\sqrt{-{\Delta }_{1}}}{2}\frac{-{c}_{1}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}{{c}_{1}\mathrm{cos}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)+{c}_{2}\mathrm{sin}\left(\frac{\sqrt{-{\Delta }_{1}}}{2}\theta \right)}\right)$ (21)

(20)与(21)为色散水波方程的三角函数与有理函数混合作用解。

$\begin{array}{c}{u}_{4}\left(x,t\right)=\frac{1}{4}\frac{{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu }+\frac{{b}_{2}{k}_{2}^{3}\lambda \mu \left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{1}{C\xi +{H}_{1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{1}{4}\frac{{b}_{2}{k}_{2}^{2}\lambda \left(1+5\mu \right)}{\mu {b}_{0}}\left(-\frac{\lambda }{2}+\frac{{c}_{2}}{{c}_{1}+{c}_{2}\theta }\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{{b}_{2}{k}_{2}^{3}{\lambda }^{2}\left(1+5\mu \right)}{4\mu {b}_{0}{k}_{1}A}\frac{1}{C\xi +{H}_{1}}\left(-\frac{\lambda }{2}+\frac{{c}_{2}}{{c}_{1}+{c}_{2}\theta }\right)\end{array}$ (22)

${v}_{4}\left(x,t\right)={b}_{0}\left(t\right)-{b}_{1}\left(t\right)\frac{1}{C\xi +{H}_{1}}+{b}_{2}\left(t\right)\left(-\frac{\lambda }{2}+\frac{{c}_{2}}{{c}_{1}+{c}_{2}\theta }\right)$ (23)

(22)与(23)为色散水波方程的有理函数解。

4. 结论

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