<s>[INST] Show that the wave equation takes the form $\frac{\partial^2 u}{\partial r \partial s}$ Show that the wave equation $\frac{\partial^2 u}{\partial t^2} - a^2 \frac{\partial^2 u}{\partial x^2} = 0$ takes the form $\frac{\partial^2 u}{\partial r \partial s} = 0$ under the change of variable $r = x + at$, $s = x - at$. 
I don't know how to proceed it. 
 [/INST] First for $r =x + at$ and $s=x-at$, we have $x=\frac{1}{2}(r+s)$ and $t=\frac{1}{2a}(r-s)$. Using chain rule for partial derivatives. 
For $u:=u(r,s)$, we have 
\begin{align*}
u_s &= u_x \frac{dx}{ds} + u_t \frac{dt}{ds} = \frac{1}{2}u_x -\frac{1}{2a} u_t \\
&\Rightarrow u_{rs} = \frac{1}{2}\bigg(u_{xx} \frac{dx}{dr} +u_{xt} \frac{dt}{dr}\bigg) -\frac{1}{2a}\bigg( u_{tt}\frac{dt}{dr} + u_{tx} \frac{dx}{dr}\bigg) \\
&= \frac{1}{2}\bigg(\frac{1}{2}u_{xx} +\frac{1}{2a}u_{xt}\bigg) - \frac{1}{2a}\bigg(\frac{1}{2a}u_{tt}  + \frac{1}{2}u_{tx}\bigg) \\
&= \frac{1}{4}u_{xx} -\frac{1}{4a^2}u_{tt} 
\end{align*}
So $\displaystyle \frac{\partial^2 u}{\partial r\partial s}=0 \iff a^2 \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial t^2}=0 $.
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