I am attempting to find an analytical solution, in terms of general Green functions, of the 3D poisson equation over a finite cylinder, with mixed boundary conditions. I have Neumann boundary conditions on the two faces of the cylinder and a Dirichlet boundary condition on the outer radius. The problem under question is: $$ \nabla^2 T_{(\mathbf{r})} + q_{(\mathbf{r})} = 0 $$ with boundary conditions: $$ \frac{\partial T}{\partial z}=f_1(\mathbf{r}) \;\;\;\;\;\;\;\;\;\; \text{on} \;\;\;\;\;\; z=0$$ $$ \frac{\partial T}{\partial z}=f_2(\mathbf{r}) \;\;\;\;\;\;\;\;\;\; \text{on} \;\;\;\;\;\; z=t$$ $$ T=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{on} \;\;\;\;\; \rho=r_0 $$ I found the papers
- https://iopscience.iop.org/article/10.1088/0508-3443/18/1/314
- https://www.sciencedirect.com/science/article/abs/pii/0017931065900438
The [1] solves a more general problem with convective boundary conditions. My problem shall in theory be found as a special case of the general solution, but in the paper it is reported that in case $h_1=h_2=h_3=0$ the solution must be treated differently. My case is instead $h_1=h_2=0$ and $h_3\rightarrow \infty$. Is my understanding that all equation actually still apply but the number of roots is different. In the second paper[2] the same problem with only Neumann conditions is solved instead, which is in fact the special case $h_1=h_2=h_3=0$.
A total of three Green functions are present, one for each driving term. For $f_1$ and $f_2$, two possible solutions are given and it is convenient for my case to use the Hankel transform one (Eq. 70[2], Eq. 78[2]). The third boundary condition needs to be changed in my case. I would claim that to do that I must simply change Eq. 35.a)[2] and instead take $\mu_{km}$ as the root of the equation: $$ J_{k(\mu_{km}r_0)} = 0 $$ which does not involve a differentiation as it would in the Neumann case. For the rest, the same exact integral equation shall apply.
For the volumetric term, $q_{(r)}$, the procedure based on Hankel transform is not reported in [2] but it is mentioned that it is as well applicable as for the two other driving terms. Comparing with the paper [1], I would claim the Dirichlet boundary condition can be achieved by taking $h_3 \rightarrow \infty$ in Eq.17[1] and $h_1=h_2=0$ for the determination of the root $\nu_k$, which then simplifies considerably. Using this and performing the limit in 39b[1] shall give the Green function for my boundary conditions. The reasoning is further confirmed by trying to derive the Neumann Green function which was instead reported in [2] using the same method. In that case, $h_3=0$ and a similar expression of the kernel to the one reported in 52a-[2] is actually found, the difference being only due to different symbolism used.
Would the described procedure be correct also more formally? What caveats shall be considered other than the possible different number of roots? Finally, how could I verify if all the assembly of equation is correct, rather than numerically? I attempted computing the Laplacian and verifying all the boundary condition of the resulting integrals but couldn't continue.
Edit
I have verified the approach of simply changing root for $f_2$ terms by using a known analytical solution. For this to work, Eq.(65.b) [2] must be re-derived, since Eq.35a[2] was used to simplify the integral in Eq.65b[2]. The term for $m=0$ in the solution drops, so the solution becomes: \begin{equation} T_{2(\mathbf{r})} = \int\int f_{2(\mathbf{r}_i)} \mathcal{G}_{(\mathbf{r},\mathbf{r}_i)} d^2\mathbf{r}_i \end{equation} where,using radial coordinates in domain and co-domain: \begin{equation} \mathcal{G}_{(\mathbf{r},\mathbf{r}_i)} = \frac{2}{\pi}\sum_{k=0}^{\infty}\sum_{m=1}^{\infty}\frac{\cos{\left(k \left(\varphi - \varphi_i\right) \right)} \cosh{\left({\mu}_{km}z \right)} J_{k}\left(\rho {\mu}_{km}\right) J_{k}\left(\rho_i {\mu}_{km}\right)}{(1+{\delta}_{k0})\left( r_{0}^{2} {\mu}_{km} \right) \sinh{\left({\mu}_{km}t \right)} J^{2}_{|k-1|}\left(r_{0} {\mu}_{km}\right)} \end{equation} Similar equation can be derived for the $f_{1(\mathbf{r})}$ term.
Formula is partially validated by using a known result for gaussian on-axis pumping, which leads once integrated over the thickness to the identity: \begin{equation} \frac{\ln{\left(\frac{r_{0}^{2}}{\rho^{2}} \right)} - \operatorname{E}_{1}\left(\frac{2 \rho^{2}}{w^{2}}\right) + \operatorname{E}_{1}\left(\frac{2 r_{0}^{2}}{w^{2}}\right)}{4 \pi} = \sum_{m=0}^{\infty} \frac{e^{- \frac{\mu_{km}^{2} w^{2}}{8}} J_{0}\left(\mu_{km} \rho\right)}{\pi \mu_{km}^{2} r_{0}^{2} J^{2}_{1}\left(\mu_{km} r_{0}\right)} \end{equation} which I numerically verified to hold.