I'm trying to solve the PDE $\triangle u=0$ in polar coordinates for $r<1$ using separation of variables.
My attempt:
I know that in polar coordinates the Laplace equation is given by $$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial ^2u}{\partial \theta^2}=0$$
Let $u(r,\theta)=R(r)\Theta(\theta)$. Then the Laplace equation becomes $$\frac{r^2R''(r)+rR'(r)}{R(r)}=-\frac{\Theta''(\theta)}{\Theta(\theta)}=k$$ for some constant $k$.
Solving the ODE for $\Theta$ then gives me $$\Theta(\theta)=A\cos(\sqrt{k}\theta)+B\sin(\sqrt{k}\theta)$$
I'm now told to use that $\Theta(0)=\Theta(2\pi)$ and $\Theta'(0)=\Theta'(2\pi)$.
Doing this leads me to the pair of equations $$A=A\cos(2\pi\sqrt{k})+B\sin(2\pi\sqrt{k})$$
$$B\sqrt{k}=B\sqrt{k}\cos(2\pi\sqrt{k})-A\sqrt{k}\sin(2\pi\sqrt{k})$$
I'm not sure how to move forward from here though.
Best Answer
Everything looks great so far!
The fact that $\Theta(\theta)$ is periodic with period $2\pi$ tells you that $\sqrt{k}$ must be a non-negative integer: $$\sqrt{k} = n , \ \ \ \ \ \ \ \ \ \ n\in \{ 0, 1, 2, 3, \dots \} $$ That's all!
Having observed this, you now need to solve the equation for $R(r)$: $$ r^2 R''(r) + r R'(r) = n^2 R(r).$$ I'll leave you to verify that the solutions are $$ R(r) = \begin{cases} C + D \ln r & n = 0, \\ Cr^n + Dr^{-n} & n = 1,2,3,\dots \end{cases} $$ [If you want to derive this from scratch, you could observe that the ODE is homogenous, which means that it is a good idea to try the substitution $r = e^t$.]
Now try to use the fact that you're working in the region $r < 1$ to rule out $\ln r$ and $r^{-n}$ as invalid solutions.
Finally take a linear superposition of all the solutions that you have found to obtain the most general solution. Good luck!