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更新：28 MAR 2016
以波动方程为例
\(\dfrac{\partial^2u}{\partial t^2}=a^2\dfrac{\partial^2 u}{\partial x^2}+f(x,t),\qquad 0<x<l,\quad t>0\)
边界条件：齐次
\(u|_{x=0}=u|_{x=l}=0,\qquad t>0\)
初始条件：任意（最后用到Fourier变换）
\(u|_{t=0}=\varphi(x),\ \left.\dfrac{\partial u}{\partial t}\right|_{t=0}=\psi(x),\qquad 0 \leqslant x \leqslant l\)
解法：分解待求函数\(u(x,t)\)。设
\(u(x,t)=v(x,t)+w(x,t)\)
将方程非齐次项归结到\(v(x,t)\)，将初始条件归结到\(w(x,t)\)，即
对于\(v(x,t)\)
\(\dfrac{\partial^2v}{\partial t^2}=a^2\dfrac{\partial^2 v}{\partial x^2}+f(x,t),\qquad 0<x<l,\quad t>0\)
\(v|_{x=0}=v|_{x=l}=0,\qquad t>0\)
\(v|_{t=0}=0,\ \left.\dfrac{\partial v}{\partial t}\right|_{t=0}=0,\qquad 0 \leqslant x \leqslant l\)
对于\(w(x,t)\)
\(\dfrac{\partial^2w}{\partial t^2}=a^2\dfrac{\partial^2 w}{\partial x^2}+f(x,t),\qquad 0<x<l,\quad t>0\)
\(w|_{x=0}=w|_{x=l}=0,\qquad t>0\)
\(w|_{t=0}=\varphi(x),\ \left.\dfrac{\partial w}{\partial t}\right|_{t=0}=\psi(x),\qquad 0 \leqslant x \leqslant l\)
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原文地址：http://www.cnblogs.com/fnight/p/5327973.html
