$\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\text{ implies } \frac{\partial^2 u}{\partial z \partial y}=0$ I have the following question:

Show that if $z=x+ct$ and $y=x-ct$ then:
$$\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\text{ implies } \frac{\partial^2 u}{\partial z \partial y}=0$$

If I had $u$ it would follow trivially I imagine, but I don't have $u$. It would seem that $u$ is linear in $t,x$ and since $t,x$ are linear in $y,z$ it seems logically consistent, but I am not sure what to do. Perhaps this is related to harmonic functions.
What should I first do to get this started?

Perhaps I notice that $c=\frac{z-x}{t} \text{ and } c=\frac{x-y}{t}$
and thus $\frac{z-x}{t} =\frac{x-y}{t}\implies x-y=z-x\implies 2x=y+z$
 A: Compute the $(x,t)$ derivatives in terms of the new variables $(z,y)$ by means of your given transformation. Then, the chain rule reads:
\begin{align}
\frac{\partial u}{\partial t} = &  \frac{\partial u}{\partial  z } \frac{\partial z }{\partial t } +  \frac{\partial u}{\partial  y } \frac{\partial y }{\partial t } = c \, (u_z - u_y) \\
\frac{\partial u}{\partial x} = &  \frac{\partial u}{\partial  z } \frac{\partial z }{\partial x } +  \frac{\partial u}{\partial  y } \frac{\partial y }{\partial x } = u_z + u_y  \\
\frac{\partial^2 u}{\partial t^2} = & \frac{\partial}{\partial t}\left( c \, (u_z - u_y) \right) = c^2 (u_{zz} -2 u_{zy} + u_{yy})
\end{align}
Can you derive the second $x$-derivative and substitute back in the original PDE?
Hope this helps. Cheers!
A: Write the variables $x$ and $t$ in terms of $z$ and $y$. Next use the chain rule by writing
$$u(x,t) = u(\phi(z,y), \psi(z,y)).$$
Next compute
$$\frac{\partial}{\partial y} u(\phi(z,y), \psi(z,y))$$
and
$$\frac{\partial}{\partial z} \left( \frac{\partial}{\partial y} u(\phi(z,y), \psi(z,y)) \right).$$
You will use the equivalent form of the wave equation given by
$$\frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} =0.$$
