8.2. Helmholtz equation in the disk

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

Separation of variable in polar and cylindrical coordinates


  1. Helmholtz equation in the disk
  2. Helmholtz equation in the cylinder
  3. Laplace equation in the cylinder

Helmholtz equation in the disk

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.

Helmholtz equation in the cylinder

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.

Laplace equation in the cylinder

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 $R= CI_l (r\sqrt{\beta})$.


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