What is the meaning of my partially solved PDE? 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?
 A: 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
$$
      \theta(2\pi-\theta)=f(1,\theta)=\sum_{n=1}^{\infty}A_n \sin(n\theta), \\
        A_n=\frac{\int_0^{2\pi}\theta(2\pi-\theta)\sin(n\theta)d\theta}{\int_0^{2\pi}\sin^2(n\theta)d\theta},\;\; n=1,2,3,\cdots.
$$
