１次元波動方程式の一般解の検証

波動方程式（電場 E の場合）

$(\frac {\partial^{2} }{\partial x^2} - \frac {1}{c^{2}} \frac {\partial^{2} }{\partial t^2}) E = 0$

一般解

$E=F(x-ct)+G(x+ct)$

前進波： $F(x-ct)$ 後退波： $G(x+ct)$

検証

$(\frac {\partial^{2} }{\partial x^2} - \frac {1}{c^{2}} \frac {\partial^{2} }{\partial t^2}) E$

$=(\frac {\partial^{2} }{\partial x^2} - \frac {1}{c^{2}} \frac {\partial^{2} }{\partial t^2}) \bigl\{F(x-ct)+G(x+ct)\bigr\}$

$=\frac {\partial^{2} }{\partial x^2}F(x-ct) - \frac {1}{c^{2}} \frac {\partial^{2} }{\partial t^2}F(x-ct) +\frac {\partial^{2} }{\partial x^2}G(x+ct) - \frac {1}{c^{2}} \frac {\partial^{2} }{\partial t^2}G(x+ct)$

$=\frac {\partial }{\partial x}F'(x-ct) - \frac {1}{c^{2}} \frac {\partial}{\partial t}(-c)F'(x-ct) +\frac {\partial }{\partial x}G'(x+ct) - \frac {1}{c^{2}} \frac {\partial }{\partial t}(+c)G'(x+ct)$