When solving the partial differential equations for a vibrating circular membrane:
PDE:
$$\frac{\partial^2 u}{\partial t^2} = c^2\nabla^2u$$
Subject to the following boundary coundtion:
$$u(a, \theta, t) = 0$$
Why is there a singularity at $$r = 0$$?
The singularity occurs because the Jacobian of the coordinate transformation from polar to rectangular is $r$, which vanishes at $r=0$. So it is an artifact of the non-faithful coordinate transformation.
For an orthogonal coordinate transformation in $3d$ from $(a,b,c)$ to $(x,y,z)$ with metric scale factors $m_a,m_b,m_c$, the Laplacian becomes $$ \Delta f= \frac{1}{m_a m_b m_c}\left[\frac{\partial}{\partial a}\left(\frac{m_b m_c}{m_a}\frac{\partial f}{\partial a}\right)+\frac{\partial }{\partial b}\left(\frac{m_a m_c}{m_b}\frac{\partial f}{\partial{b}}\right)+\frac{\partial f}{\partial c}\left(\frac{m_a m_b}{m_c}\frac{\partial f }{\partial c}\right)\right] $$ The Jacobian of such an orthogonal coordinate transformation is $J=m_am_bm_c$. There is a singularity at any point where one or more of the metric scale factors $m_a,m_b,m_c$ vanish.
For example, cylindrical coordinates is an orthogonal coordinate system in variables $r,\theta,z$, the scale factors are $m_r=1,m_{\theta}=r,m_z=1$. The Laplacian in cylindrical coordinates is \begin{align} \Delta f & = \frac{1}{r}\left[\frac{\partial}{\partial r}\left(r\frac{\partial f}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\frac{1}{r}\frac{\partial f}{\partial \theta}\right)+\frac{\partial}{\partial z}\left(r\frac{\partial f}{\partial z}\right)\right] \\ & =\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial f}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2}+\frac{\partial^2 f}{\partial z^2}. \end{align} If you leave out the $z$ dependence, there is a reduction to polar coordinates. There is a singularity at $r=0$ because the Jacobian $J=r$ vanishes at $r=0$. This allows a non-trivial singular solution $f(r,\theta,\phi)=\ln(r)$, which is a multiple of the so-called "fundamental" solution. This solution does not naturally arise in Cartesian coordinates, but does in cylindrical coordinates. In order to eliminate such a solution, a condition must be imposed at $r=0$, such as requiring boundedness of $f$ near $r=0$, which turns out to be the correct condition for eliminating the fundamental solution.