I am trying to solve the shape of an elastic sheet clamped at $r=1$ and $r=b<1$. $$\left\{ \begin{array}{c l} \Delta u = \rho(r,\phi) \quad (a<r<1)\\ u(a)=0\\ u(1)=1 \end{array}\right.$$
I have solved the solution for a case that has the rotational symmetry ($\partial u /\partial \phi = 0$): $$\left\{ \begin{array}{c l} \Delta u = g \quad (a<r<1)\\ u(a)=0\\ u(1)=1 \end{array}\right.$$ where $g$ is a constant.
With the polar form of the Laplacian $\Delta = \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 u}{\partial \phi ^2}=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)$, the solution is $$u(r,\phi)=\frac{g}{4} r^2 + C_1 \log{r} + C_2$$ where $C_1, C_2$ are determined by the two boundary conditions.
My question is how can I find the Green function $G(\mathbf{r,r'})$ of this problem to reproduce this result ($\rho (r,\phi)= g$) so that I can apply it to the original problem with general $\rho(r,\phi)$?
I am particularly interested in $\rho(r,\phi)=\delta(r-b)\delta(\phi)$ where $a<b<1$, which represents a point charge at $r=b$.
I really appreciate your attention!
Following Dylan's solution and choosing $\phi_0=0$, $A_n(r)$ can be calculated as \begin{equation} \begin{aligned} A_n(r)&=\frac{1}{n} \left[\Theta(r-r_0)\cdot \sinh{(n\log{(\frac{r}{r_0})})}+\frac{\sinh{(n\log{(\frac{a}{r})})}\sinh{(n\log{r_0})}}{\sinh{(n\log{a})}}\right]\\ &=\frac{1}{n} \left[\Theta (r-r_0) \cdot \left((\frac{r}{r_0})^n-(\frac{r}{r_0})^{-n}\right)+\frac{\left((\frac{a}{r})^n-(\frac{a}{r})^{-n}\right)\left(r_0^n-r_0^{-n}\right)}{a^n-a^{-n}}\right] \end{aligned} \end{equation} Since the source is even in the $\phi$ component, $B_n(r)=0$.
Here is a plot of the numerical result (cross section) for a problem with actual numbers. $u_c(r)$ uses the first 30 terms of the series.
$u_h(r)$ is the homogeneous solution. $u_m(r)$ is the part corresponding to the uniform loading in the in-homogeneous solution and $u_c(r)$ is the part corresponding to the point load.
You can split up the forcing function and the boundary into two different solutions
$$ \begin{cases} \Delta w = 0 \\ w(a,\phi) = 0 \\ w(1,\phi) = 1 \end{cases} \qquad \begin{cases} \Delta v = \rho(r,\phi) \\ v(a,\phi) = 0 \\ v(1,\phi) = 0 \end{cases} \qquad (r,\phi)\in [a,1] \times[0,2\pi] $$
Then $u = w + v$ solves the original problem.
The homogeneous problem is simple enough to solve, so I won't go into details here. You should get
$$ w(r) = -\frac{\ln(r/a)}{\ln a} $$
For the non-homogeneous problem, consider a solution of the form
$$ u(r,\phi) = A_0(r) + \sum_{n=1}^\infty \big[A_n(r)\cos(n\phi) + B_n(r)\sin(n\phi)\big] $$
where $\cos(n\phi)$ and $\sin(n\phi)$ are eigenfunctions of the Laplacian in the $\phi$ direction. Then
$$ \Delta u = \Delta_rA_0 + \sum_{n=1}^\infty \left[\left(\Delta_rA_n - \frac{n^2}{r^2}A_n \right)\cos(n\phi) + \left(\Delta_rB_n- \frac{n^2}{r^2}B_n \right)\sin(n\phi)\right] $$
where $\Delta_r = \frac{d^2}{dr^2} + \frac{1}{r}\frac{d}{dr}$
You can find the corresponding Fourier series of the forcing function in $[0,2\pi]$
$$ \rho(r,\phi) = \alpha_0(r) + \sum_{n=1}^\infty \big[\alpha_n(r)\cos(n\phi) + \beta_n(r)\sin(n\phi)\big] $$
where, from definitions
$$ \alpha_0(r) = \frac{1}{2\pi}\int_0^{2\pi} \rho(r,\phi) d\phi, \quad \alpha_n(r) = \frac{1}{\pi}\int_0^{2\pi} \rho(r,\phi) \cos(n\phi) d\phi $$ $$ \beta_n(r) = \frac{1}{\pi}\int_0^{2\pi} \rho(r,\phi) \sin(n\phi) d\phi $$
Then it's left to solve the family of ODEs
$$ \begin{cases} A_0'' + \dfrac{1}{r}A_0' = \alpha_0(r) \\ A_0(a) = A_0(1) = 0 \end{cases} $$ $$ \begin{cases} A_n'' + \dfrac{1}{r}A_n' - \dfrac{n^2}{r^2}A_n = \alpha_n(r) \\ A_n(a) = A_n(1) = 0 \end{cases} \qquad \begin{cases} B_n'' + \dfrac{1}{r}B_n' - \dfrac{n^2}{r^2}B_n = \beta_n(r) \\ B_n(a) = B_n(1) = 0 \end{cases} $$
A point source in polar coordinates is given by
\begin{align} \rho(r,\phi) &= \frac{1}{r}\delta(r-r_0)\delta(\phi-\phi_0) \\ &= \frac{1}{r}\delta(r-r_0)\left[\frac{1}{2\pi} + \frac{1}{\pi}\sum_{n=1}^\infty \cos(n\phi_0)\cos(n\phi) + \sin(n\phi_0)\sin(n\phi) \right] \end{align}
so it's more convenient (but not necessary) to scale the radial component in the solution to match those constants
$$ u(r,\phi) = \frac{1}{2\pi}A_0(r) + \frac{1}{\pi}\sum_{n=1}^\infty \big[A_n(r)\cos(n\phi_0)\cos(n\phi) + B_n(r)\sin(n\phi_0)\sin(n\phi)\big] $$
Following the same steps
$$ \begin{cases} A_0'' + \dfrac{1}{r}A_0' = \dfrac{1}{r}\delta(r-r_0) \\ \\ A_0(a) = A_0(1) = 0 \end{cases} $$
$$ \begin{cases} A_n'' + \dfrac{1}{r}A_n' - \dfrac{n^2}{r^2}A_n = \dfrac{1}{r}\delta(r-r_0) \\ \\ A_n(a) = A_n(1) = 0 \end{cases} \qquad \begin{cases} B_n'' + \dfrac{1}{r}B_n' - \dfrac{n^2}{r^2}B_n = \dfrac{1}{r}\delta(r-r_0) \\ \\ B_n(a) = B_n(1) = 0 \end{cases} $$
You can follow this example to solve for the radial components. I'll do the first one. Start with a piece-wise homogeneous solution that's zero on the boundary
$$ A_0(r) = \begin{cases} c_1\ln(r/a), & r < r_0 \\ c_2\ln r, & r > r_0 \end{cases} $$
Continuity gives
$$ A_0(r_0) = c_1 \ln(r_0/a) = c_2\ln r_0 $$
The jump condition gives
$$ A_0'(r_0^+) - A_0'(r_0^-) = \frac{c_2-c_1}{r_0} = \frac{1}{r} $$
Then
$$ c_1 = -\frac{\ln r_0}{\ln a}, \qquad c_2 = -\frac{\ln(r_0/a)}{\ln a} $$