# Cylindrical waves

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.

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• I pretty much found my answer in there: vixra.org/pdf/0908.0045v1.pdf An amplitude decreasing as $\frac{1}{\sqrt{r}}$ is true asymptotically, once you can neglect the energy spread through the basis of your cylinder. – Learning is a mess Jan 29 '13 at 14:54
• Look up the fundamental solution to the wave equation in various dimensions. The "cylindrical symmetry" case is essentially the same as the situation in 2 spatial dimensions. When the spatial dimension is even strong Huygens' principle fails and there is no simple reduction to the plane-wave form. When the spatial dimension is odd, there is a reduction, but the formula is much more complicated in general compared to the 3 dimensional case. – Willie Wong Jan 29 '13 at 16:14
• Thanks for the input Willie. I just updated my question and sharpened my question. – Learning is a mess Jan 29 '13 at 18:32
• To see that this is a Bessel DE, try rescaling your $r$. (i.e. substitute $r' = kr$ where $k = \omega/c$) – Wouter Jan 29 '13 at 18:53
• That's right. On a sidenote: in this case there's no problem applying the boundary conditions because they are zero. However in general, if the BC's are e.g. $u(R,t) = f(t)$, you'd have to fourier transform those as well before applying them. (or inverse-fourier transform your solution and apply the original BC's) – Wouter Jan 29 '13 at 19:04