$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}\implies \frac{\partial^2 v}{\partial \xi\partial \eta}=0$

Question

I'm reading "Fourier analysis, an introduction" by Elias M. And Rami Shakarchi and in chapter one it said:

The function $u(x,t)$ satisfied the equation $$\dfrac{\partial^2 u}{\partial x^2}=\dfrac{\partial^2 u}{\partial t^2}$$

Now let's set $2$ new variables: $\xi=x+t$ and $\eta=x-t$.

We now define the function $v$ such that $v(\xi,\eta)=u(x,t)$.

I understand this but in the book from this they conclude $2$ things that I'm failed to understand how:

$$\dfrac{\partial^2 v}{\partial \xi\partial \eta}=0$$ And $$v(\xi,\eta)=F(\xi)+G(\eta)$$($F,G$ are twice differentiable functions)

How from the change of variables they got to this? I would be happy with either a hint or an answer, thank you very much.

Answer 1

Hint: Use the chain rule

$$\frac{\partial}{\partial x} = \frac{\partial \xi}{\partial x} \frac{\partial}{\partial \xi} + \frac{\partial \eta}{\partial x} \frac{\partial}{\partial \eta},$$ and $$\frac{\partial}{\partial t} = \frac{\partial \xi}{\partial t} \frac{\partial}{\partial \xi} + \frac{\partial \eta}{\partial t} \frac{\partial}{\partial \eta}.$$