\renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\erf}{\operatorname{erf}} \newcommand{\dag}{\dagger} \newcommand{\const}{\mathrm{const}} \newcommand{\arcsinh}{\operatorname{arcsinh}}
Consider Helmholtz equation in the disc (recall that such equation is obtained from wave equation after separation of t from spatial variables): \begin{equation} v_{rr} + r^{-1}v_r - r^{-2}v_{\theta\theta}=-\lambda v\qquad r\le a. \label{eq-8.2.1} \end{equation} Separating variables v=R(r)\Theta(\theta) we arrive to \begin{equation*} \frac{r^2 R'' + rR'+\lambda r^2R}{R}+\frac{\Theta''}{\Theta}=0 \end{equation*} and therefore \begin{align} & \Theta''=-\mu \Theta,\label{eq-8.2.2}\\ & r^2 R'' +rR' + (\lambda r^2 -\mu )R=0\label{eq-8.2.3} \end{align} and \mu =-l^2, \Theta =e^{\pm in\theta} and \begin{equation} r^2 R'' +rR' + (\lambda r^2 -l^2 )R=0. \label{eq-8.2.4} \end{equation} As \lambda=1 it is Bessel equation and solutions are Bessel functions J_l and Y_l which which are called Bessel functions of the 1st kind and of Bessel functions of the 2nd kind, respectively, and the former are regular at 0. Therefore R=J_l (r\sqrt{\lambda}) and plugging into Dirichlet or Neumann boundary conditions we get respectively \begin{align} &J_l (a\sqrt{\lambda})=0, \label{eq-8.2.5}\\ &J'_l (a\sqrt{\lambda})=0, \label{eq-8.2.6} \end{align} and then \lambda = z_{l,n}^2a^{-2} and \lambda = w_{l,n}^2a^{-2} respectively where z_{l,n} and w_{l,n} are n-th zero of J_l or J_l' respectively.
Remark 1. Bessel functions are elementary functions only for half-integer l=\frac{1}{2},\frac{3}{2},\frac{5}{2},\ldots when they are related to spherical Bessel functions.
Consider Laplace equation in the cylinder \{r\le a, 0\le z\le b\} with homogeneous Dirichlet (or Neumann, etc) boundary conditions: \begin{align} &u_{rr}+r^{-1}u_r + r^{-2}u_{\theta\theta}+u_{zz}=-\omega^2 u, \label{eq-8.mathjax/2.7.5}\\ &u|_{z=0}=u|_{z=b}=0, \label{eq-8.2.8}\\ &u|_{r=a}= 0.\label{eq-8.2.9} \end{align} Separating Z from r,\theta u=Z(z)v(r,\theta) we get \begin{equation*} \frac{\Lambda v}{v} + \frac {Z''}{Z}=-\omega^2 v \end{equation*} and then Z''=- \beta Z, and \Lambda v:= v_{rr}+r^{-1}v_r+ r^{-2}v_{\theta\theta}=-\lambda v with \lambda =\omega^2 -\beta and from boundary conditions to Z we have \beta=\pi^2 m^2 b^{-2} and separating r,\theta: v=R(r)\Theta(\theta) we arrive like in the previous Subsection to (\ref{eq-8.2.4}). One can prove that there are no nontrivial solutions as \lambda\le 0 and therefore \lambda>0 and everything is basically reduced to the previous Subsection.
Exercise 1. Do it in detail.
Consider Laplace equation in the cylinder \{r\le a, 0\le z\le b\} with homogeneous Dirichlet (or Neumann, etc) boundary conditions on the top and bottom leads and non-homogeneous condition on the lateral boundary: \begin{align} &u_{rr}+r^{-1}u_r + r^{-2}u_{\theta\theta}+u_{zz}=-\omega^2 u, \label{eq-8.2.10}\\ &u|_{z=0}=u|_{z=b}=0, \label{eq-8.2.11}\\ &u|_{r=a}= g(z,\theta).\label{eq-8.2.12} \end{align} Separating Z from r,\theta Separating Z from r,\theta u=Z(z)v(r,\theta) we get \begin{equation*} \frac{\Lambda v}{v} + \frac {Z''}{Z}=0 \end{equation*} and then Z''=- \beta Z, and \Lambda v:= v_{rr}+r^{-1}v_r+ (-\beta+ r^{-2}v_{\theta\theta})=0 and from boundary conditions to Z we have \beta=\pi^2 m^2 b^{-2} and separating r,\theta: v=R(r)\Theta(\theta) we arrive like in the previous Subsection to \begin{equation} r^2 R'' +rR' + (-\beta r^2 -l^2 )R=0. \label{eq-8.2.13} \end{equation} However now \beta>0 and we do not need to satisfy homogeneous condition as r=a (on the contrary, we do not want it to have non-trivial solutions.
Then we use modified Bessel functions I_l and K_l and R= CI_l (r\sqrt{\beta}).