I am working on the PDE $$\bigtriangleup u = 0,\ u(1,\ \theta) = 2\pi\theta - \theta^2$$
I plugged $u = R(r)T(\theta)$ into the polar Laplacian formula, and ended up with $\frac{\frac{\partial^2 T}{\partial \theta^2}}{T} = -\frac{r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r}}{R} = \lambda$. My solutions to the Sturm-Liouville problems are $R = Ar^{-\sqrt{-\lambda}} + Br^{\sqrt{-\lambda}}$ and $T = Ce^{-\sqrt\lambda \theta} + De^{\sqrt\lambda \theta}$ (I have ruled out the $\lambda = 0$ case). Thus I have $$u = (Ar^{-\sqrt{-\lambda}} + Br^{\sqrt{-\lambda}})(Ce^{-\sqrt\lambda \theta} + De^{\sqrt\lambda \theta})$$
I have heard several explanations for the next step, but the way it makes sense in my head (correct me if I'm wrong) is that we know the domain is a disk, because the given boundary condition describes a shape with constant radius which implies a circle, and a disk domain in turn implies that $u(r,\ \theta) = u(r,\ \theta + 2m\pi)$ for all $r$, $\theta$, and $m$, where $m$ is an arbitrary integer. This is like having an extra boundary condition for every $m$, and plugging into general solution yields $(Ar^{-\sqrt{-\lambda}} + Br^{\sqrt{-\lambda}})(Ce^{-\sqrt\lambda \theta} + De^{\sqrt\lambda \theta}) = (Ar^{-\sqrt{-\lambda}} + Br^{\sqrt{-\lambda}})(Ce^{-\sqrt\lambda (\theta + 2m\pi)} + De^{\sqrt\lambda (\theta + 2m\pi)})$. Since this must hold for all $r$, not just the special case where $Ar^{-\sqrt{-\lambda}} + Br^{\sqrt{-\lambda}} = 0$, I can divide by this factor to get $$Ce^{-\sqrt\lambda \theta} + De^{\sqrt\lambda \theta} = Ce^{-\sqrt\lambda (\theta + 2m\pi)} + De^{\sqrt\lambda (\theta + 2m\pi)}$$
The most promising solution to this equation seems to come from the case in which $e^{-2m\pi \sqrt\lambda} = 1$ and $e^{2m\pi \sqrt\lambda} = 1$. Taking the complex natural logarithms and rearranging yields $\lambda = -\frac{n^2}{m^2}$, where $n$ is an arbitrary integer, for both equations.
When I plug this value of $\lambda$ into the general solution and apply the explicitly given boundary condition, I will have a strange Fourier problem. It will not quite be a Fourier Series, because the frequencies will not be integers, but will not quite be a Fourier Transform, because the frequencies will not be the spectrum of real numbers. The frequencies will instead be the set of rationals $\frac{n}{m}$. This seems to occur because there are two different quantizations in the problem. The complex natural logarithm introduces a quantization as it normally does in separation of variables problems, and the domain introduces a quantization in virtue of the natural periodicity with which one revisits locations while walking around a circle. What do I make of this result from both conceptual and computational perspectives?
There is another interesting solution to this equation, $\lambda = \frac{(ln(\frac{C}{D}) + 2n \pi i)^2}{4(\theta + m\pi)^2}$. This makes the frequency spectrum even weirder. What is the meaning of this? Is there some obscure answer to the PDE found by utilizing this solution?
The Laplacian in polar coordinates is $$ \nabla^2f(r,\theta)= f_{rr}+\frac{1}{r}f_r +\frac{1}{r^2}u_{\theta\theta}. $$ The equation you want to solve is $\nabla^2f(r,\theta)=0$. To use separation of variables, assume $f(r,\theta)=R(r)\Theta(\theta)$, plug this in, divide both sides by $f(r,\theta)$, and multiply by $r$ to separate variables: $$ \frac{R''}{R}+\frac{1}{r}\frac{R'}{R}+\frac{1}{r^2}\frac{\Theta''}{\Theta}=0 \\ r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Theta''}{\Theta} $$ Therefore, there is a separation parameter $\lambda$ such that $$ r^2\frac{R''}{R}+r\frac{R'}{R}=\lambda,\;\; \lambda=-\frac{\Theta''}{\Theta} $$ The solutions in $\Theta$ must be periodic in $[0,2\pi]$, which imposes the restriction that $\lambda=n^2$ for $n=0,\pm 1,\pm 2,\pm 3,\cdots$, and $\Theta_n(\theta)=e^{in\theta}$. The corresponding solutions in $R$ must satisfy $$ r^2R_n''+rR_n'-n^2R_n=0. $$ The solutions in $r$ can have the form $r^{\lambda}$ where $$ r^2 \lambda(\lambda-1)r^{\lambda-2}+r \lambda r^{\lambda-1}-n^2 r^{\lambda}=0 \\ \lambda(\lambda-1)+\lambda-n^2=0 \\ \lambda^2 -n^2 =0 \\ \lambda =n,\;\;\; n=0,\pm 1,\pm 2,\pm 3,\cdots. $$ The negative powers are not useful because of the corresponding singularities at $r=0$. The general solution is $$ f(r,\theta) = \sum_{n=0}^{\infty}r^n(A_n\sin(n\theta)+B_n\cos(n\theta)). $$ (The term $\sin(n\theta)$ vanishes identically for $n=0$.) The requirement that $$ f(1,\theta)=2\pi\theta-\theta^2=\theta(2\pi-\theta) $$ forces $B_n=0$ for all $n=0,1,2,3,\cdots$, and gives $$ f(r,\theta)=\sum_{n=1}^{\infty}A_nr^n\sin(n\theta). $$ The $A_n$ are then determined by $$ \theta(2\pi-\theta)=f(1,\theta)=\sum_{n=1}^{\infty}A_n \sin(n\theta), \\ A_n=\frac{\int_0^{2\pi}\theta(2\pi-\theta)\sin(n\theta)d\theta}{\int_0^{2\pi}\sin^2(n\theta)d\theta},\;\; n=1,2,3,\cdots. $$