Given $\frac{\partial^2u}{\partial t^2}=\frac{\partial^2u}{\partial x^2}$, $\xi=x+t,\eta=x-t$, show $\frac{\partial^2 u}{\partial\xi\partial\eta}=0$

Question

Given $u(x,t)$ that satisfies the following wave equation for all $x,t$:

\begin{align*} \frac{\partial^2 u}{\partial t^2} &= \frac{\partial^2 u}{\partial x^2} \end{align*}

Let $\xi = x + t$, $\eta = x - t$, $v(\xi, \eta) = u(x,t)$ so that:

\begin{align*} v(\xi, \eta) &= u(x,t) = u\left(\frac{\xi+\eta}{2}, \frac{\xi-\eta}{2}\right) = v(x+t, x-t) \\ \end{align*}

Show that:

\begin{align*} \frac{\partial^2 v}{\partial \xi \partial \eta} &= 0 \end{align*}

Hint: Use the change of variable formula.

I presume I'm overlooking something simple but how do I show that final equation? How would I use change of variables here?

Answer 1

You can use the chain rule:

$$\frac{\partial v}{\partial \eta} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial u}{\partial t}\frac{\partial t}{\partial \eta} = \frac12\frac{\partial u}{\partial x} - \frac12 \frac{\partial u}{\partial t}$$

Do this again and you will have your result.