I found this task, and I find it quite hard so solve: For the 3D polarcoordinates it is: $$ \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} r \sqrt{1- \tau^2} \cos(\phi)\\ r \sqrt{1- \tau^2} \sin( \phi ) \\ r \tau \end{pmatrix} $$
with $ r \in \mathbb{R}_0^+, \phi \in [ 0, 2 \pi[, \tau \in[-1,1] $ ($ r = \cos( \theta) , \theta \in [0, \pi], \tau = \sin( \theta ), \theta \in [ - \frac{ \pi}2,\frac{ \pi}{2}] $
For the 2D Laplace Operator it is :
$$ \triangle = \frac{ \partial^2}{ \partial r^2}+ \frac2r \frac{ \partial }{ \partial r } + \frac1{r^2} ( -2 \tau \frac{ \partial }{ \partial \tau}+ (1- \tau^2) \frac{ \partial^2}{ \partial \tau^2}+ \frac{1}{1- \tau^2} \frac{ \partial^2 }{\partial \phi^2})$$
I want to solve the 2D Wave Equation $$ \frac{ \partial^2 u}{ \partial t^2}= \alpha^2 \triangle u $$ with the separation method $$ u(t,r, \phi , \tau)= v(t) \omega(r) f( \phi) g( \tau) $$
How can I determine, that it comes to those ODE's:
$ 0= v''+ \omega^2 v , \omega \in \mathbb{ R} $, constant
$0=r^2 \omega''+2rw'+ ( \frac{ \omega^2 }{ \alpha^2} r^2- \lambda) \omega, \lambda \in \mathbb{R} $, constant
$0= f''- \mu f, \mu \in \mathbb{R} $ constant
$ 0= (1- \tau )^2 g''- 2 \tau g'+ ( \lambda + \frac{ \mu}{1- \tau^2}) g $
thanks for any help !