阿多米安分解法解初值问题



George Adomian (1922--1996)

方法

常微分方程可写为如下形式:

\begin{equation} Lu(t)+Ru(t)+N(u(t))=f(t) \label{ODE} \end{equation}

其中$L$为最高阶可逆求导算符,$R$为线性微分算符,$N$为非线性算符,$f(t)$为一个函数,$u(t)$为待求未知函数。

将\eqref{ODE}移项,得

\begin{equation} Lu(t)=-Ru(t)-N(u(t))+f(t) \label{ODET} \end{equation}

则方程的解为如下形式:

\begin{equation} u(t)=\Psi_0(t)+L^{-1}f(t)-L^{-1}Ru(t)-L^{-1}N(u(t)) \label{ODEsolform} \end{equation}

其中$\Psi_0(t)$来自初始条件,对于常见的求导算符,有

\begin{equation} \Psi_0(t)= \begin{cases} u(0), & L=\frac{d}{dt} \\ u(0)+xu'(0), & L=\frac{d^2}{dt^2} \\ u(0)+xu'(0)+\frac{1}{2!}x^2u''(0), & L=\frac{d^3}{dt^3} \\ u(0)+xu'(0)+\frac{1}{2!}x^2u''(0)+\frac{1}{3!}x^3u'''(0), & L=\frac{d^4}{dt^4} \end{cases} \label{Psi0} \end{equation}

方程的解可表示为如下级数:

\begin{equation} u(t) = \sum_{n = 0}^{\infty} u_n(t) = u_0(t) + u_1(t) + u_2(t) + \cdots \label{user} \end{equation}

令$N(u(t))=F(u(t))$,$F(u)$在$u(t)=u_0(t)$附近做泰勒展开:

\begin{equation} \begin{split} F(u)=&F(u_0)+F'(u_0)(u-u_0)+F''(u_0)\frac{(u-u_0)^2}{2!} \\ &+F'''(u_0)\frac{(u-u_0)^3}{3!}+\cdots \\ =&F(u_0)+F'(u_0)(u_1+u_2+\cdots)+F''(u_0)\frac{(u_1+u_2+\cdots)^2}{2!}\\ &+F'''(u_0)\frac{(u_1+u_2+\cdots)^3}{3!}+\cdots \\ =& A_0+A_1+A_2+A_3+\cdots \\ =& \sum_{n=0}^{\infty} A_n \end{split} \label{FTaylor} \end{equation}

其中$A_n$为阿多米安多项式,前8项阿多米安多项式表达式为:

\begin{eqnarray} A_0 &=& F( u_0 ) , \\ A_1 &=& u_1 F' (u_0 ) , \\ A_2 &=& u_2 F' (u_0 ) + \frac{1}{2!}\, u_1^2 F'' (u_0 ) , \\ A_3 &=& u_3 F' (u_0 ) + \frac{1}{3!}\, u_1^3 F''' (u_0 ) + \frac{1}{2!} \,u_1 u_2 F'' (u_0 ) , \\ A_4 &=& u_4 F' (u_0 ) + \left[ \frac{1}{2!}\, u_2^2 + u_1 u_3 \right] F'' (u_0 ) + \frac{1}{2!} \,u_1^2 u_2 F''' (u_0 ) + \frac{1}{4!} \, u_1^2 F'''' (u_0 ) , \\ A_5 &=& u_5 F' (u_0 ) + F'' (u_0 ) \left( u_1 u_4 + u_2 u_3 \right) + \frac{1}{2!}\,F''' (u_0 ) \left( u_1^2 u_3 + u_1 u_2^2 \right) + \frac{1}{3!}\, F^{(4)} (u_0 ) \, u_1^3 u_2 + \frac{1}{5!} \, f^{(5)} (u_0 ) \, u_1^5 , \\ A_6 &=& u_6 F' (u_0 ) + F'' (u_0 ) \left( u_1 u_5 + u_2 u_4 + u_3^2 \right) + \frac{1}{3!}\,F''' (u_0 ) \left( 3\,u_1^2 u_4 + u_2^3 + 6\,u_1 u_2 u_3 \right) + \frac{1}{4!}\,F^{(4)} (u_0 ) \left( 4\,u_1^3 + 6\,u_1^2 u_2^2 \right) + \frac{1}{4!}\,F^{(5)} (u_0 )\, u_1^4 u_2 + \frac{1}{6!} \,F^{(6)} (u_0 ) \, u_1^6 , \\ A_7 &=& u_7 F' (u_0 ) + F'' (u_0 ) \left( u_1 u_6 + u_2 u_5 + u_3 u_4 \right) + \frac{1}{2}\,F''' (u_0 ) \left( u_1^2 u_5 + u_1 u_3^2 + u_3 u_2^2 + 2\,u_1 u_2 u_4 \right) + \frac{1}{3!} \,F^{(4)} (u_0 ) \left( u_1^3 u_4 +3\, u_1^2 u_2 u_3 + u_1 u_2^3 \right) + \frac{1}{4!}\,F^{(5)} (u_0 ) \left( u_1^4 u_3 + 2\,u_1^3 u_2^2 \right) + \frac{1}{6!} \,F^{(6)} (u_0 ) \,u_1^5 u_2 + \frac{1}{7!} \,F^{(7)} (u_0 ) \,u_1^7 , \\ A_8 &=& u_8 F' (u_0 ) + F'' (u_0 ) \left( u_1 u_7 + u_2 u_6 + u_3 u_5 + u_4^2 /2 \right) + \frac{1}{2}\,F''' (u_0 ) \left( u_1^2 u_6 + u_2^2 u_4 + u_2 u_3^2 + 2\,u_1 u_2 u_5 + 2\, u_1 u_3 u_4 \right) + \frac{1}{3!} \,f^{(4)} (u_0 ) \left( u_1^3 u_5 +3\, u_1^2 u_2 u_4 + 3\, u_1 u_2^2 u_3 + 3\,u_1^2 u_3^2 /2 + u_2^4 /4 \right) + \frac{1}{4!}\,F^{(5)} (u_0 ) \left( u_1^4 u_4 + 4\,u_1^3 u_2 u_3 + 2\, u_1^2 u_2^3 \right) + \frac{1}{6!} \,F^{(6)} (u_0 ) \left( u_1^5 u_3 + 15\, u_1^4 u_2^2 \right) + \frac{1}{6!} \,F^{(7)} (u_0 ) \,u_1^6 u_2 + \frac{1}{8!} \,f^{(8)} (u_0 ) \, u_1^8 , \\ \vdots &=& \vdots \label{Aex} \end{eqnarray}

可浓缩为一个统一的表达式:

\begin{equation} A_n \left( u_0 , u_1 , \ldots , u_n \right) = \frac{1}{n!} \, \frac{{\text d}^n}{{\text d} \lambda^n} \left[ F \left( x, \sum_{k = 0}^{\infty} u_k \lambda^k \right) \right]_{\lambda =0} , \qquad n=0,1,2,\ldots \label{Anexp} \end{equation}

将\eqref{user}和\eqref{FTaylor}式代入\eqref{ODEsolform},得

\begin{equation} \sum_{n = 0}^{\infty} u_n(t)=\Psi_0(t)+L^{-1}f(t)-L^{-1}R\sum_{n = 0}^{\infty} u_n(t)-L^{-1}\sum_{n = 0}^{\infty} A_n(t) \label{ODEsoluser} \end{equation}

方程的级数解\eqref{user}中各项满足如下递归关系:

\begin{equation} \begin{split} u_0(t)=&\Psi_0(t)+L^{-1}f(t) \\ u_k(t)=&-L^{-1}Ru_{k-1}(t)-L^{-1} A_{k-1}(t) \end{split} \label{unexp} \end{equation}

举例



例题1 $u'=u^2,u(0)=1$

对照标准形式\eqref{ODE}式,$L=\frac{d}{dt}$,$Ru=0$,$N(u)=-u^2$。

$N(u)=-u^2$的阿多米安多项式前四项为:

\begin{equation*} \begin{split} A_0=&-u_0^2 \\ A_1=&-2u_0u_1 \\ A_2=&-2u_0u_2-u_1^2 \\ A_3=&-2u_0u_3-2u_1u_2\\ \vdots \end{split} \end{equation*}

级数解前5项依次为

\begin{equation*} \begin{split} u_0=& u(0) =1 \\ u_1=& -\int_0^t A_0(s) ds = \int_0^t u_0^2 ds = \int_0^t ds = t \\ u_2=& -\int_0^t A_1(s) ds = \int_0^t 2u_0u_1 ds = \int_0^t 2s ds = t^2 \\ u_3=& -\int_0^t A_2(s) ds = \int_0^t (2u_0u_2+u_1^2) ds = \int_0^t 3s^2 ds = t^3 \\ u_4=& -\int_0^t A_3(s) ds = \int_0^t (2u_0u_3+2u_1u_2) ds = \int_0^t 4s^3 ds = t^4 \\ \vdots \end{split} \end{equation*}

方程的解为:

\begin{equation*} u(t) = \sum_{n = 0}^{\infty} u_n(t) = 1 + t + t^2+ t^3+t^4+ \cdots = \frac{1}{1-t^2} \end{equation*}

与精确解一致。

例题2 $uu'-t=0,u(0)=1$

方程整理为:

\begin{equation*} u'-\frac{t}{u}=0 \end{equation*}

两边积分,方程的解:

\begin{equation*} u(t)=\sum_{n=0}^{\infty}u_n(t) =1+\int_0^t s\sum_{n=0}^{\infty}A_nds \end{equation*}

$A_n$是$N(u)=\frac{1}{u}$的阿多米安多项式:

\begin{equation*} \begin{split} A_0=&\frac{1}{u_0} \\ A_1=&-\frac{u_1}{u_0^2} \\ A_2=&\frac{u_1^2}{u_0^3}-\frac{u_2}{u_0^2}\\ \vdots \end{split} \end{equation*}

级数解中各项

\begin{equation*} \begin{split} u_0=&1 \\ u_1=&\int_0^t sA_0ds=\frac{t^2}{2} \\ u_2=&\int_0^t sA_1ds=-\int_0^t \frac{s^3}{2}ds=-\frac{t^4}{8} \\ u_3=&\int_0^t sA_2ds=\frac{t^6}{16} \\ \vdots \end{split} \end{equation*}

方程的解:

\begin{equation*} u(t)=\sum_{n=0}^{\infty}u_n(t) =1+\frac{t^2}{2}-\frac{t^4}{8}+\frac{t^6}{16}+\cdots =\sqrt{t^2+1} \end{equation*}

得到精确解。

例题3 $u'-\frac{u^2}{1-tu}=0,u(0)=1$

方程可整理为

\begin{equation*} u'=tuu'+u^2 \end{equation*}

两边积分,方程的解:

\begin{equation*} u(t)=\sum_{n=0}^{\infty}u_n(t) =1+\int_0^t s\sum_{n=0}^{\infty}A_nds+\int_0^t \sum_{n=0}^{\infty}B_nds \end{equation*}

$A_n$是$uu'$的阿多米安多项式,即

\begin{equation*} \begin{split} \sum_{n=0}^{\infty}A_n=&uu'=(u_0+u_1+u_2+u_3+\cdots)(u'_0+u'_1+u'_2+u'_3+\cdots) \\ =& (u_0u'_0)+(u_0u'_1+u_1u'_0)+(u_0u'_2+u_1u'_1+u_2u'_0) \end{split} \end{equation*}

于是,

\begin{equation*} \begin{split} A_0=&u_0u'_0\\ A_1=&u_0u'_1+u_1u'_0 \\ A_2=&u_0u'_2+u_1u'_1+u_2u'_0\\ \vdots \end{split} \end{equation*}

$B_n$是$u^2$的阿多米安多项式:

\begin{equation*} \begin{split} B_0=&u_0^2\\ B_1=&2u_1u_0 \\ B_2=&2u_0u_2+u_1^2\\ \vdots \end{split} \end{equation*}

方程级数解各项

\begin{equation*} \begin{split} u_0=&1 \\ u_1=&\int_0^t sA_0ds+\int_0^t B_0ds=t \\ u_2=&\int_0^t sA_1ds+\int_0^t B_1ds=\frac{3}{2}t^2 \\ u_3=&\int_0^t sA_2ds+\int_0^t B_2ds=\frac{8}{3}t^3\\ \vdots \end{split} \end{equation*}

方程的解:

\begin{equation*} u(t)=\sum_{n=0}^{\infty}u_n(t) =1+t+\frac{3}{2}t^2+\frac{8}{3}t^3+\cdots \end{equation*}

例题4 $ u''(t)-u'(t)^2+u(t)^2=e^t, u(0)=u'(0)=1$



方程可整理为

\begin{equation*} u''(t)=e^t+u'(t)^2-u(t)^2 \end{equation*}

两边积分一次,得:

\begin{equation*} u'(t)-u'(0)=e^t-1+\int_0^t u'(s)^2 ds-\int_0^t u(s)^2ds \end{equation*}

两边再积分一次:

\begin{equation*} u(t)-u(0)-tu'(0)=e^t-t-1+\int_0^t\int_0^{s} u'(\tau)^2 d\tau ds-\int_0^t\int_0^{s} u(\tau)^2d\tau ds \end{equation*}

即:

\begin{equation*} \sum_{n=0}^{\infty}u_n(t) =e^t+\int_0^t\int_0^{s} \sum_{n=0}^{\infty}A_n d\tau ds-\int_0^t\int_0^{s} \sum_{n=0}^{\infty}B_nd\tau ds \end{equation*}

$A_n$ 是$u'^2$的阿多米安多项式,$B_n$是$u^2$的阿多米安多项式。

级数解第一项$u_0=e^t$,第二项为:

\begin{equation*} u_1=\int_0^t\int_0^{s} (e^{\tau})^2 d\tau ds-\int_0^t\int_0^{s} (e^{\tau})^2d\tau ds =0 \end{equation*}

这意味着$u_n=0,n=1,2,\cdots$,因此方程的解为:

\begin{equation*} u(t)=e^t \end{equation*}

标签: 微分方程, 阿多米安分解法

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