partial differential equation-exercice let in $\mathbb{R}^2$ the equation 
$$
\dfrac{\partial^2 u(x,t)}{\partial t^2} - \dfrac{\partial^2 u(x,t)}{\partial x^2} = 0
$$
We put: 
$
\begin{cases}
x=\xi + \eta\\
t=\xi- \eta
\end{cases}
$
and $u(x,t)=\tilde{u}(\xi,\eta)$.
1. How we prove that $\dfrac{\partial^2 \tilde{u}}{\partial \xi \partial \eta}(\xi,\eta)=0$?
2. How we deduce that the solution of the differential equation is $u(x,t)= f(x+t)+g(x-t)$?
I'm lost, thank you for the help.
 A: You can think of it like this: you can treat differential operators with constant coefficients like polynomials. So it boils down to the following computation: $$\frac{\partial \widetilde{u}}{\partial \xi} = \frac{\partial}{\partial \xi} u\left(\xi+\eta,\xi - \eta\right) = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t},$$and in the same way $$\frac{\partial \widetilde{u}}{\partial \eta} =  \frac{\partial u}{\partial x} - \frac{\partial u}{\partial t}. $$Meaning we have: $$\frac{\partial}{\partial \xi} = \frac{\partial}{\partial x}+\frac{\partial}{\partial t}, \quad \frac{\partial}{\partial \eta} = \frac{\partial}{\partial x}-\frac{\partial}{\partial t},$$and so: \begin{align}  \frac{\partial^2u}{\partial x^2}-\frac{\partial^2 u}{\partial t^2} &= \left(\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial t^2}\right)u = \left(\frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)\left(\frac{\partial}{\partial x}-\frac{\partial}{\partial t}\right)u \\ &= \frac{\partial}{\partial \xi}\frac{\partial}{\partial \eta}u = \frac{\partial^2u}{\partial \xi \partial \eta}.\end{align}
Now $$\frac{\partial^2u}{\partial \xi\partial \eta}=0 \implies u(\xi, \eta) = f(\xi) + g(\eta) \implies u(x,t) =f(x+t) + g(x-t).$$We're identifying $u$ with $\widetilde{u}$ all the way.
A: Let $\tilde{u}=v$.
$$
\dfrac{\partial^2 v(\xi,\eta)}{\partial t^2} - \dfrac{\partial^2 v(\xi,\eta)}{\partial x^2} = 0
$$
Note that:
$$\xi=\frac{x+t}{2}$$
$$\eta=\frac{x-t}{2}$$
Now use chain rule
$$
\dfrac{\partial v(\xi,\eta)}{\partial t} =\dfrac{\partial v(\xi,\eta)}{\partial \xi}\dfrac{\partial \xi}{\partial t}+\dfrac{\partial v(\xi,\eta)}{\partial \eta}\dfrac{\partial \eta}{\partial t}=\frac{1}{2}\dfrac{\partial v(\xi,\eta)}{\partial \xi}-\frac{1}{2}\dfrac{\partial v(\xi,\eta)}{\partial \eta}
$$
Use chain rule again for
$$
\dfrac{\partial^2 v(\xi,\eta)}{\partial t^2} 
$$
Do the same with
$$
\dfrac{\partial^2 v(\xi,\eta)}{\partial x^2} 
$$
Plug your results into the PDE and check if it is zero.
A: One must convert the derivative operators first,
\begin{align}
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial t} = -u_\xi + u_\eta
\end{align}
and likewise
\begin{align}
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x } = u_\xi + u_\eta
\end{align}
where I've used $\xi = x-t$ and $\eta = x+t$ because these coordinates make the work is a bit more clear.  By doing so, you are now looking at this PDE along the characteristics ($\xi$ and $\eta$), which is usually a simplification compared to the normal problem.  
Computing the second derivative is straightforward,
\begin{align}
\partial_{tt} = \partial_{\xi\xi} - 2\partial_{\xi\eta} + \partial_{\eta\eta},\\
\partial_{xx} = \partial_{\xi\xi} + 2\partial_{\xi\eta} + \partial_{\eta\eta}.
\end{align}
Upon substitution into the PDE, you get your result,
\begin{equation}
4u_{\xi\eta} = 0.
\end{equation}
What type of function is zero after differentiating with respect to $\xi$ once and w.r.t $\eta$ once? Only a function of the form $u = f(\xi) + g(\eta).$
