问题

一维波动方程是指

\[
\frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial^2 u}{\partial x^2}.
\]

这里 $u(x,t)$ 是关于坐标 $x$ 和时间 $t$ 的二元函数, 且一般假设是 $C^2$ 的, 此时 $\frac{\partial^2 u}{\partial x\partial t}=\frac{\partial^2 u}{\partial t\partial x}$. 从而

\[
\begin{split}
\biggl(\frac{\partial^2 u}{\partial t^2}-a^2\frac{\partial^2 u}{\partial x^2}\biggr)&=\biggl(\frac{\partial}{\partial t}+a\frac{\partial}{\partial x}\biggr)\biggl(\frac{\partial}{\partial t}-a\frac{\partial}{\partial x}\biggr)u\\
&=\biggl(\frac{\partial}{\partial t}-a\frac{\partial}{\partial x}\biggr)\biggl(\frac{\partial}{\partial t}+a\frac{\partial}{\partial x}\biggr)u
\end{split}
\]

显然, 满足 $(\frac{\partial}{\partial t}-a\frac{\partial}{\partial x})u\equiv C$ 或 $(\frac{\partial}{\partial t}-a\frac{\partial}{\partial x})u\equiv C$ 的函数 $u(x,t)$ 是波动方程的解.

加上初值条件 $u(x,0)=\varphi(x)$ 和 $u'_t(x,0)=\psi(x)$, 就成为一维波动方程的柯西问题.

\[
\begin{cases}
\dfrac{\partial^2 u}{\partial t^2}-a^2\dfrac{\partial^2 u}{\partial x^2}=0,\\
u(x,0)=\varphi(x),\quad u'_t(x,0)=\psi(x).
\end{cases}
\]