Stuck on solving a non homogenous (linear) PDE

[fluidistic]

I fear this discussion has become rather narrowly focused and probably lacks broad interest. I suggest you ask a moderator to close this thread and then we can continue our discussion via PF messaging.

In this case can't we just continue in private without closing the thread, just in case someone else pops up with insights we could have missed? It will be closed after 2 years anyway.

 

[renormalize]

I've resolved the question posed by the OP so I'm posting my solution to benefit future readers of this thread.
@fluidistic seeks an analytic solution to the 2D Poisson equation:$$-\nabla^{2}\psi\equiv-\frac{\partial^{2}\psi}{\partial r^{2}}-\frac{1}{r}\frac{\partial\psi}{\partial r}-\frac{1}{r^{2}}\frac{\partial^{2}\psi}{\partial\theta^{2}}=\frac{A\cos2\theta}{r^{2}}\tag{1}$$where $A$ is a constant and the stream-function $\psi\left(r,\theta\right)$ must vanish on the boundary of the quarter-annulus; i.e.:$$\psi\left(r_{i},\theta\right)=\psi\left(r_{o},\theta\right)=\psi\left(r,0\right)=\psi\left(r,\frac{\pi}{2}\right)=0\tag{2a,b,c,d,}$$Evidently the particular solution $\psi=\psi_{p}\equiv\frac{1}{4}A\cos 2\theta$ satisfies (1) but, alas, it fails all of the above boundary conditions (2). But since we're free to superpose $\psi_p$ with any homogeneous solution $H$, where $\nabla^{2}H=0$, we choose to add $H=h\left(\theta\right)\equiv\frac{1}{4}A\left(\frac{4\theta}{\pi}-1\right)$ to define a new particular solution:$$\psi_{p^{\prime}}\equiv\psi_{p}+h\left(\theta\right)=\frac{1}{4}A\left(\cos2\theta+\frac{4\theta}{\pi}-1\right)\tag{3}$$that does satisfy the two angular conditions (2c,2d). Next, we Fourier expand (3) into a sine series to get:$$\psi_{p^{\prime}}=\frac{A}{2\pi}\sum_{n=1}^{\infty}\frac{\sin4n\theta}{n\left(4n^{2}-1\right)}\tag{4}$$Finally, noting that $H=h_{4n}\left(r,\theta\right)\equiv\frac{\sin4n\theta\left(r^{8n}+r_{i}^{4n}r_{o}^{4n}\right)}{r^{4n}\left(r_{i}^{4n}+r_{o}^{4n}\right)}$ is a homogeneous solution for any integer $n$, we make the substitution:$$\sin4n\theta\rightarrow\sin4n\theta-h_{4n}\left(r,\theta\right)=\sin4n\theta-\frac{\sin4n\theta\left(r^{8n}+r_{i}^{4n}r_{o}^{4n}\right)}{r^{4n}\left(r_{i}^{4n}+r_{o}^{4n}\right)}=-\frac{\sin4n\theta\left(r^{4n}-r_{i}^{4n}\right)\left(r^{4n}-r_{o}^{4n}\right)}{r^{4n}\left(r_{i}^{4n}+r_{o}^{4n}\right)}\tag{5}$$into (4) to arrive at the full analytic solution for the stream-function:$$\psi\left(r,\theta\right)=-\frac{A}{2\pi}\sum_{n=1}^{\infty}\frac{\sin4n\theta\left(r^{4n}-r_{i}^{4n}\right)\left(r^{4n}-r_{o}^{4n}\right)}{n\left(4n^{2}-1\right)r^{4n}\left(r_{i}^{4n}+r_{o}^{4n}\right)}\tag{6}$$This manifestly satisfies all four radial and angular boundary conditions (2).
Because performing practical computations necessitates keeping only a finite number of terms, we denote by $\psi_N$ the truncation of the infinite series in (6) to order $N$ and examine the form of the analytic stream-function for modest values of $N$. For example, setting $N=4$ yields the approximate analytic solution:
\begin{align}
\psi\left(r,\theta\right)\approx\psi_{4}\left(r,\theta\right) & = -\frac{A}{2\pi}\left[\frac{\sin4\theta\left(r^{4}-r_{i}^{4}\right)\left(r^{4}-r_{o}^{4}\right)}{3r^{4}\left(r_{i}^{4}+r_{o}^{4}\right)}+\frac{\sin8\theta\left(r^{8}-r_{i}^{8}\right)\left(r^{8}-r_{o}^{8}\right)}{30r^{8}\left(r_{i}^{8}+r_{o}^{8}\right)}\right. \nonumber \\
& \left.+\frac{\sin12\theta\left(r^{12}-r_{i}^{12}\right)\left(r^{12}-r_{o}^{12}\right)}{105r^{12}\left(r_{i}^{12}+r_{o}^{12}\right)}+\frac{\sin16\theta\left(r^{16}-r_{i}^{16}\right)\left(r^{16}-r_{o}^{16}\right)}{252r^{16}\left(r_{i}^{16}+r_{o}^{16}\right)}\right] \nonumber\tag{7}
\end{align}A comparison of the finite-element numerical solution $\psi_{FE}$ of eqs.(1,2) to the analytic-approximation $\psi_4$ looks like this:

Thus, $\psi_4$ offers a reasonably good analytic approximation to the full numerical stream-function $\psi_{FE}$.