- I am trying to compute the Green's function $\mathrm{G}\left(x,x'\right)$, $\ \nabla_{x}^{2}\mathrm{G}\left(x,x'\right) = 4\pi\,\delta\left(x - x'\right)$ and $\mathrm{G}\left(x,x'\right) = 0$ for $x$ on the cylinder.
- Textbook solutions use cylindrical coordinates, $\left(\rho,\phi,z\right)$, but they involve series which diverge for $\rho = \rho'$ and $z = z'$.
- In addition to divergence on the circle, the series converges very slowly near the circle.
- It is not surprising that the series diverge as they expand the Green's functions as a Fourier series in $\phi$, and the circle contains the $1/\left\vert x - x'\right\vert$ singularity.
For the record, my favorite "textbook solution" for a cylinder or radius $1$ is $$ \mathrm{G}\left(x,x'\right) = \sum_{\nu = -\infty}^{\infty}\sum_{{\large i} = 1}^{\infty} A_{\large\nu i}\,\mathrm{e}^{\mathrm{i}\nu \left(\phi -\phi'\right)}\,\mathrm{e}^{-X_{\LARGE\nu i}\left\vert z -z'\right\vert}\, \mathrm{J}_{\nu}\left(X_{\large \nu i}\rho\right) \mathrm{J}_{\nu}\left(X_{\nu i}\rho'\right) $$ where $X_{\large\nu i}$ is the $i'$th zero of Bessel function $\mathrm{J}_{\nu}$ and
$$ A_{\large\nu i} = \frac{2}{X_{\large\nu i}}\, \frac{1}{\mathrm{J}_{\nu + 1}^{2}\left(X_{\large\nu i}\right)} $$
This series converges well when $z-z'$ is not too small but it fails to converge if $z = z'$.
Another "textbook solution" is
$G(x, x\prime)= \frac{2}{\pi}\sum_{\nu=-\infty}^{\nu=\infty}\int_0^\infty dk\:e^{i\nu (\phi-\phi\prime)}\cos(k(z-z\prime))K_\nu(k\:\rho)I_\nu(k\:\rho\prime)-K_\nu(k)I_\nu(k\rho)I_\nu(\rho\prime)/I_\nu(k)$ when $\rho > \rho\prime$ and
$G(x, x\prime)= \frac{2}{\pi}\sum_{\nu=-\infty}^{\nu=\infty}\int_0^\infty dk\:e^{i\nu (\phi-\phi\prime)}\cos(k(z-z\prime))K_\nu(k\:\rho\prime)I_\nu(k\:\rho)-K_\nu(k)I_\nu(k\rho)I_\nu(\rho\prime)/I_\nu(k)$ when $\rho \leq \rho\prime$
Where $K$ and $I$ are modified Bessel functions and the last term is to satisfy the boundary condition at $\rho=1$. This solution converges for $\rho - \rho\prime$ not too small, but diverges when $\rho=\rho\prime$.
My Attempt at a Solution near the Circle:
I reasoned that the divergence of $1/|x-x\prime|$ was responsible for the divergences, and so I tried
$G(x, x\prime)= 1/|x-x\prime| - H(x,x\prime)$, where $H$ is a homogeneous solution: $ \nabla_x^2 G(x, x\prime) = 0$, which fixes the boundary condition, i.e. $H(x,x\prime) = 1/|x-x\prime|$ on the cylinder.
I put $H(x,x\prime)=\sum_{\nu=-\infty}^{\nu=\infty}\int_0^{\infty}dk\:B(\nu,k, \rho\prime )e^{i\nu(\phi-\phi\prime)}\cos(k\:(z-z\prime))I_\nu(k\:\rho)$
$1/|x-x\prime|=((z-z\prime)^2+\rho^2+\rho\prime^2-2\rho\rho\prime \cos(\phi-\phi\prime))^{-1/2}$ and on the cylinder $\rho=1$.
Where $B(\nu,k, \rho\prime )= \frac{1}{2\pi^2}\int_{-\pi}^\pi d\theta \int_{-\infty}^\infty dz \frac{\cos(kz)\cos(\nu \theta)}{z^2+1+rho\prime^2-2\rho\prime \cos(\theta)}I_\nu^{-1}(k)$
We can use the identity $\int_0^\infty dz \frac{\cos(a\:z)}{1+z^2}=K_0(a)$ to do the z integration, then
$B(\nu,k, \rho\prime )= \frac{1}{\pi^2}\int_{-\pi}^\pi d\theta \cos(\nu \theta))K_0(k*(1+\rho\prime^2-2\rho\prime \cos(\theta))^{1/2})I_\nu^{-1}(k)$
I found that this solution works well when $\rho\prime$ is not too close to 1, but I am unhappy with the integrals needed to compute $B$. I was hoping for a method that would work numerically, but standard numerical integration routines (i.e. Mathematica ) give a lot of warnings and are very slow. If this approach is to work a better way to do the $B$ integrals is needed.