$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$
Consider Helmholtz equation in the disk (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 $v=R(r)\Phi(\phi)$ we arrive to \begin{equation*} \frac{r^2 R'' + rR'+\lambda r^2R}{R}+\frac{Phi''}{|Phi}=0 \end{equation*} and therefore \begin{align} & \Phi''=-\mu Phi,\label{eq-8.2.2}\\ & r^2 R'' +rR' + (\lambda r^2 -\mu )R=0\label{eq-8.2.3} \end{align} and $\mu =-l^2$, $\Phi =e^{\pm in\theta}$ 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 the 2nd kind, respectively, and the former are regular at $0$. Therefore $R=J_l (r\sqrt{\lambda})$ and plugging into Dirichlet or Neiumann 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 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-8.2.7}\\ &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 $r,\phi$: $v=R(r)\Phi(\phi)$ 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 $r,\phi$: $v=R(r)\Phi(\phi)$ 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 $R= CI_l (r\sqrt{\beta})$.