I am trying to solve the general equation for cylindrical symmetric waves:
$$\frac1{c^2}\frac{\partial^2u}{\partial t^2}= \frac1r\frac{\partial}{\partial r}(r\frac{\partial}{\partial r}u)$$ with $u = u(r,t)$.

I was expecting that by plugging a function of the form $u(r,t) = \frac{1}{\sqrt{r}}f(r,t)$ I would arrive with something a nice plane wave form for this $f(r,t)$ but I don't. My idea was that for spherically symmetric waves we would put $u = \frac1R f$ as the energy goes as $\frac{1}{r^2}$, but with cylindrical propagation it goes as $\frac1r$. What am I thinking or doing wrong? Does someone know the common solution for this equation?

Edit:

After having thought a bit more about it, I am now trying to get the eigenmodes, so to resolve: $$\frac{\omega^2}{c^2}u + \frac{\partial}{\partial r^2}u + \frac1r \frac{\partial}{\partial r}u=0$$
which, after multiplying by $r^2$ looks like a Bessel diff. equation:
$$x^2y''+xy'+(x^2-p^2)y=0$$
for $p=0$.
But which operation can I do to get rid of the $\frac{\omega^2}{c^2}$ factor. Even after that, I am not sure how to ling the eigenvalues of the modes to the zeros of the Bessel functions, given that I have the boundary condition $u(R,t) = 0$ for $R$ the diameter of my cylinder.

This post imported from StackExchange Mathematics at 2014-06-02 20:24 (UCT), posted by SE-user Learning is a mess