how can I prove that $u_{tt}=c^{2}u_{xx}$ can be reduced to $u_{\xi \eta }=0$ can you tell me how can I prove that the wave equation given by :
$\; \; \; \; \; \; \; \; \; \; \; u_{tt}=c^{2}u_{xx}$
can be reduced to : 
$\; \; \; \; \; \; \; \; \; \; \; u_{\xi \eta }=0$
And show that the general solution of $\;u(x,t)$ could be written as :
$\; \; \; \; \; \; \; \; \; \; \;u(x,t)=f(x-ct)+g(x+ct)$
For the first part I thought maybe substituing for :
$\; \;\; \;\; \;\; \;\; \;\; \;\xi = x-ct\; \;$ and $\; \;\eta =x+ct$
But i got stuck maybe I'm doing something wrong. Thanks in advance for your help.
 A: The chain rule will be your best friend on this one.  Using the new variables you defined:
$$\begin{align}
u_{t} &= u_{\xi}\xi_{t} + u_{\eta}\eta_{t} = -cu_{\xi}+cu_{\eta}\\
u_{tt} &= u_{t\xi}\xi_{t} + u_{t\eta}\eta_{t} = -c(-cu_{\xi \xi} + cu_{\eta\xi}) + c(-cu_{\xi \eta} + cu_{\eta \eta})\\
u_{x}&= u_{\xi}\xi_{x} + u_{\eta}\eta_{x} = u_{\xi} + u_{\eta}\\
u_{xx} &= u_{\xi\xi}\xi_{x} + u_{\eta\eta}\eta_{x} = u_{x\xi} + u_{x\eta}
\end{align}
$$
Putting it together then:
$$\begin{align}
-c(-cu_{\xi \xi} + cu_{\eta\xi}) + c(-cu_{\xi \eta} + cu_{\eta \eta}) &= c^{2}(u_{\xi\xi} + u_{\eta\eta})\\
c^{2}u_{\xi\xi}-c^{2}u_{\eta\xi}-c^{2}u_{\xi\eta}+c^{2}u_{\eta\eta} &= c^{2}u_{\xi\xi} + c^{2}u_{\eta\eta}
\end{align}
$$
This then reduces to $$u_{\xi\eta} = 0.$$  This implies that $$u_{\xi} = F(\xi)$$ where $F$ is an arbitrary $C^{1}$ function and therefore $$u(\xi,\eta) = f(\xi) + g(\eta)$$ where $f$ and $g$ are arbitrary $C^{1}$ functions.  Substituting back to $x$ and $t$ gives us $$u(x,y) = f(x-ct) + g(x+ct).$$
A: Hint: your equation is equivalent to
$$
u_{tt} - c^2 u_{xx} = \left(\frac{\partial^2}{\partial t^2}- c^2\frac{\partial^2}{\partial x^2}\right) u = 0.
$$
This expression can be rewritten as
$$
\left(\frac{\partial}{\partial t}- c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t}+ c\frac{\partial}{\partial x}\right) u
$$
