how can I prove that $u_{tt}=c^{2}u_{xx}$ can be reduced to $u_{\xi \eta }=0$

Question

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.

Answer 1

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 $$ \left(\frac{\partial}{\partial t}- c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t}+ c\frac{\partial}{\partial x}\right) u $$

Answer 2

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 $$u(x,y) = f(x-ct) + g(x+ct).$$