阿多米安分解法解MKdV方程组

参考文献：MCM 2008,47,1035 The decomposition method for solving the coupled modified KdV equations

George Adomian (1922--1996)

方法

广义修正Kdv方程组：

\begin{equation} \begin{split} & L_tu + L_xu + M_1(u, v) = 0 \\ & L_tv + L_xv + M_2(u, v) = 0 \end{split} \label{MKdV} \end{equation}

其中$L_t=\frac{\partial }{\partial t}$、$L_x=\frac{\partial^3 }{\partial x^3}$为线性算符。方程两边作用逆算符$L_t^{-1}=\int_0^t(\bullet)dt$，得：

\begin{equation} u(x,t)=u(x,0)-L_t^{-1}[L_xu + M_1(u, v)] \label{usolform} \end{equation}

\begin{equation} v(x,t)=v(x,0)-L_t^{-1}[L_xv + M_2(u, v)] \label{vsolform} \end{equation}

假设未知函数$u(x,t)$和$v(x,t)$可写为无穷级数形式：

\begin{equation} u(x,t)=\sum_{n=0}^{\infty}u_n(x,t),v(x,t)=\sum_{n=0}^{\infty}v_n(x,t) \label{uvser} \end{equation}

非线性算符$M_1$和$M_2$可写为阿多米安多项式的无穷级数：

\begin{equation} M_1(u, v)=\sum_{n=0}^{\infty}A_n ,M_2(u, v)=\sum_{n=0}^{\infty}B_n(x,t) \label{MAdo} \end{equation}

阿多米安多项式由下式给出：

\begin{equation} A_n(ut_0,u_1,\cdots,u_n;v_0,v_1,\cdots,v_n)=\frac{1}{n!}\frac{d^n}{d\lambda^n}\left . M\left ( \sum_{k=0}^n\lambda^ku_k,\sum_{k=0}^n\lambda^kv_k\right )\right |_{\lambda=0} \label{Ado} \end{equation}

方程的解满足如下递归关系：

\begin{equation} \begin{split} & u_0(x,t)=u(x,0),u_{n+1}=-L_t^{-1}[L_xu_{n} + A_n],n\ge 0 \\ & v_0(x,t)=v(x,0),v_{n+1}=-L_t^{-1}[L_xv_{n} + B_n],n\ge 0 \end{split} \label{uvrec} \end{equation}

算例

方程为

\begin{equation} \begin{split} & u_t+3u^2u_x-3(uv)_x=F(x,t),u(x,0)=f(x) \\ & v_t-3vv_x+3(uv)_x=G(x,t),v(x,0)-3u^2v_x=g(x) \end{split} \label{exeqs} \end{equation}

令

\begin{equation} F(x,t)=\frac{1}{2}u_{xxx}+\frac{3}{2}v_{xx}-3 u_x \label{F} \end{equation}

\begin{equation} G(x,t)=-v_{xxx}+3v_x \label{G} \end{equation}

\begin{equation} f(x)=\frac{1}{2}+\tanh(x) \label{f} \end{equation}

\begin{equation} g(x)=1+\tanh(x) \label{g} \end{equation}

首先给出各非线性项的阿多米安多项式。

\begin{equation} \begin{split} & u^2u_x = (u_0+u_1+u_2+u_3+\cdots)(u_0+u_1+u_2+u_3+\cdots)\\ &(u_{0x}+u_{1x}+u_{2x}+u_{3x}+\cdots) = \sum_{n=0}^{\infty}A_n \\ & A_0 = u_0^2u_{0x} \\ & A_1 = 2u_0u_1u_{0x}+u_{1x}u_0^2 \\ & A_2 = u_0^2u_{2x}+2u_0 u_1u_{1x}+2u_0 u_2u_{0x}+u_1^2u_{0x} \\ & A_3 = u_0^2u_{3x}+2u_0u_1u_{2x}+2u_0u_2u_{1x}+2u_0u_3u_{0x}+u_1^2u_{1x}+2u_1u_2u_{0x} \end{split} \label{An} \end{equation}

（注意到，左边$A_n$下标正好是右边各变量下标之和。）

\begin{equation} \begin{split} & vv_x = (v_0+v_1+v_2+v_3+\cdots)(v_{0x}+v_{1x}+v_{2x}+v_{3x}+\cdots) \\ & = \sum_{n=0}^{\infty}B_n \\ & B_0 = v_0 v_{0x} \\ & B_1 = v_0 v_{1x}+v_1 v_{0x} \\ & B_2 = v_0 v_{2x}+v_1 v_{1x}+v_2 v_{0x} \\ & B_3 = v_0 v_{3x}+v_1 v_{2x}+v_2 v_{1x}+v_3 v_{0x} \end{split} \label{Bn} \end{equation}

\begin{equation} \begin{split} & (uv)_x = (u_0+u_1+u_2+u_3+\cdots)(v_{0x}+v_{1x}+v_{2x}+v_{3x}+\cdots)+\\ & (u_{0x}+u_{1x}+u_{2x}+u_{3x}+\cdots)(v_0+v_1+v_2+v_3+\cdots) \\ & = \sum_{n=0}^{\infty}C_n \\ & C_0 = u_0 v_{0x}+u_{0x}v_0 \\ & C_1 = u_0 v_{1x}+u_1 v_{0x}+u_{0x}v_1+u_{1x}v_0 \\ & C_2 = u_0 v_{2x}+u_1 v_{1x}+u_2 v_{0x}+u_{0x}v_2+u_{1x}v_1+u_{2x}v_0 \\ & C_3 = u_0 v_{3x}+u_1 v_{2x}+u_2 v_{1x}+u_3 v_{0x}+u_{0x}v_3+u_{1x}v_2+u_{2x}v_1+u_{3x}v_0 \end{split} \label{Cn} \end{equation}

\begin{equation} \begin{split} & u_xv_x = (u_{0x}+u_{1x}+u_{2x}+u_{3x}+\cdots)(v_{0x}+v_{1x}+v_{2x}+v_{3x}+\cdots) \\ & = \sum_{n=0}^{\infty}H_n \\ & H_0 = u_{0x} v_{0x} \\ & H_1 = u_{0x} v_{1x}+u_{1x} v_{0x} \\ & H_2 = u_{0x} v_{2x}+u_{1x} v_{1x}+u_{2x} v_{0x} \\ & H_3 = u_{0x} v_{3x}+u_{1x} v_{2x}+u_{2x} v_{1x}+u_{3x} v_{0x} \end{split} \label{Hn} \end{equation}

\begin{equation} \begin{split} & u^2v_x = (u_0+u_1+u_2+u_3+\cdots)(u_0+u_1+u_2+u_3+\cdots)\\ &(v_{0x}+v_{1x}+v_{2x}+v_{3x}+\cdots) = \sum_{n=0}^{\infty}D_n \\ & D_0 = u_0^2v_{0x} \\ & D_1 = 2u_0u_1v_{0x}+v_{1x}u_0^2 \\ & D_2 = u_0^2v_{2x}+2u_0 u_1v_{1x}+2u_0 u_2v_{0x}+u_1^2v_{0x} \\ & D_3 = u_0^2v_{3x}+2u_0u_1v_{2x}+2u_0u_2v_{1x}+2u_0u_3v_{0x}+u_1^2v_{1x}+2u_1u_2v_{0x} \end{split} \label{Dn} \end{equation}

由递归关系\eqref{uvrec}式，得级数解各项：

\begin{equation} u_0= \frac{1}{2}+\tanh(x) \label{u0} \end{equation}

\begin{equation} v_0= 1+\tanh(x) \label{v0} \end{equation}

\begin{equation} \begin{split} u_1=& \int_0^t \left (\frac{1}{2}u_{0xxx}+\frac{3}{2}v_{0xx}-3 u_{0x}-3A_0 +3C_0 \right ) dt \\ =&(-1+\tanh^2x)\frac{t}{4} \end{split} \label{u1} \end{equation}

\begin{equation} \begin{split} v_1=& \int_0^t \left ( -v_{0xxx}+3v_{0x}-3 B_0 -3 H_0 +3 D_0 \right ) dt \\ =&(-1+\tanh^2x)\frac{t}{4} \end{split} \label{v1} \end{equation}

\begin{equation} \begin{split} u_2=& \int_0^t \left (\frac{1}{2}u_{1xxx}+\frac{3}{2}v_{1xx}-3 u_{1x}-3A_1 +3C_1 \right ) dt \\ =&(-\tanh x+\tanh^3x)\left ( \frac{t}{4} \right)^2 \end{split} \label{u2} \end{equation}

\begin{equation} \begin{split} v_2=& \int_0^t \left ( -v_{1xxx}+3v_{1x}-3 B_1 -3 H_1 +3 D_1 \right ) dt \\ =&(-\tanh x+\tanh^3x)\left ( \frac{t}{4} \right)^2 \end{split} \label{v2} \end{equation}

仿此过程，可以计算更多项，各项加在一起，可得精确解

\begin{equation} u(x,t)=\sum_{n=0}^{\infty}u_n(x,t)=\frac{1}{2}+\tanh(x+t/4) \label{uexact} \end{equation}

\begin{equation} v(x,t)=\sum_{n=0}^{\infty}v_n(x,t)=1+\tanh(x+t/4) \label{vexact} \end{equation}