I am attempting to find a unique solution of the following Dirichlet Problem: $$\begin{cases} u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta}=0, \quad r \in [0, 1) \\ u(1, \theta) = 4 - 9 \sin(\theta) \\ u(r, \theta + 2 \pi) = u(r, \theta) \end{cases} $$ By using Separation of Variables, we can assume a solution of the form $ u(r, \theta) = R(r) \cdot \Theta(\theta)$, from which we obtain: $$ \frac{r^2 R'' + r R'}{R} = - \frac{\Theta''}{\Theta} = n^2 $$ Solving each seperately, we obtain: $$ - \Theta'' = n^2 \Theta $$ $$ \Theta(\theta) = a_n \sin(n \theta) + b_n \cos(n \theta) $$ and $$ r^2 R'' + r R' = n^2 R $$ If we assume $ R(r) = r^k$, $$ r^2 (k^2 - k) r^{k-2} + r(k) r^{k-1} - n^2 r^k = 0 $$ $$ \left(k^2 - n^2 \right) r^k = 0 $$ $$ R(r) = c_n r^{+n} + d_n r^{-n} $$ And if $n^2 = 0$, $$r^2 R'' + r R' = 0 \to r P' + P = 0$$ $$ P = \frac{e}{r} $$ $$ R(r) = e \ln(r) + f $$
So in total, $R(r) = e \ln(r) + f + c_n r^{+n} + d_n r^{-n} $? and therefore $$ u(r, \theta) = e \ln(r) + f + \sum_n \left[ c_n r^{+n} + d_n r^{-n} \right] \left[ a_n \sin(n \theta) + b_n \cos(n \theta) \right] $$
Using the $u(r, \theta + 2 \pi) = u(r, \theta)$ condition, $n$ has to be a whole number, to maintain periodicity. Additionally, since $u(r, \theta)$ is defined on $ r \in [0, 1) $, $e = d_n = 0$. It can also be seen by the $u(1, \theta) = 4 - 9 \sin(\theta) $ boundary condition that $n = 1$, $ f = 4 $, $c_1 = -9 $, $a_1$ is redundant, and $b_1 = 0$.
So in conclusion, $$ u(r, \theta) = 4 - 9 r \sin(\theta) $$
I just wanted to verify that this was correct.