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I am trying to find the eigenfrequencies of waves in a cylinder, or put into equations: $$\frac{1}{c^2}\frac{\partial^2}{\partial t^2}u = \Delta u$$ with $u = u(t,r,\theta,z)$. Giving, in cylindrical coordinates:$$r^2\omega^2c^{-2}u+r^2\partial_r^2u+r\partial_ru+\partial_{\theta}^2u+r^2\partial_z^2u=0$$ My boundary conditions are:$$u(R,z,t,\theta)=u(r,0,t,\theta)=0$$ with $R$ the radius of my cylinder. My thoughts were to look for a separation of variables and writing it (after a scaling of the r variable $r\rightarrow r\sqrt{\omega^2c^{-2}-k_z^2}$, with $\partial_z^2u=-k_z^2u$ and $k_z$ may be take continuous or discrete values) as:$$(r^2-m^2)u+r^2\partial_r^2u+r\partial_ru=0$$ with $m\in \mathbb{N}$ (because of the periodicity in the $\theta$ coordinate. And this is not more than a Bessel equation with known solutions, and the next step would be to look for the zeros of the specific Bessel functions to get my eigenvalues. But I am still uncertain about my $k_z$ because, thus far, it is not constrained even if I have in mind that my cylinder is finite, and that I will have to add another boundary condition: $u(r,e,t,\theta)=0$, with $e$ the height of the new cylinder. My question is: can I still have a discrete spectrum if I relax this boundary condition?

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When you cylinder is finite in $z$ and you apply boundary conditions on each end of the cylinder in $z$, you will in fact get a discrete set of eigenvalues and eigenfunctions in $z$. Because of the nature of the Laplacian in cylindrical coordinates, you may perform a separation of variables as you mention above so that these eigenfunctions in $z$ satisfy the simple equation above. The $k_z$ are found from the boundary conditions, viz:

$$u(r,\theta,z,t) = R(r) \Theta(\theta) Z(z) T(t)$$

$$Z''(z) + k_z^2 Z=0 \implies Z(z) = A \cos{\left ( k_z z \right )} + B \sin{\left (k_z z \right )} $$

If you impose boundary conditions $Z(0) = Z(L) = 0$, where $L$ is the length of the cylinder, then you get $k_z = n \pi/L$, with eigenfunctions $Z_n(z) = \sin{(n \pi z/L)}$; other BC's will impose other values on $k_z$ and other eigenfunctions. Of course, $k_z$ is coupled with the radial, angular, and temporal eigenvalues by way of the separation constant.

For an infinite cylinder, i.e., as $L \rightarrow \infty$, the spectrum becomes continuous and $Z$ is best represented as a Fourier transform of some spectrum function $\hat{Z}(k_z)$.

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Thank you rlgordonma, I doubted of the legitimacy of separating the variables. – Learning is a mess Jan 30 '13 at 11:15

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