The problem:
Consider Laplace's equation $$\nabla^2u=\frac{1}{r}(ru_r)_r + \frac{1}{r^2}u_{\theta\theta}=0$$ on the annulus ${(r,\theta)}: r \in (\frac{1}{2},2),\theta \in[0,2\pi]$. Find all separable solutions $u(r,\theta)=\phi(r)\xi(\theta)$
Here is my work for this problem:
$$u(r,\theta)=\phi(r)\xi(\theta)$$ $$u_r=\phi'(r)\xi(\theta), u_{\theta\theta}=\phi(r)\xi''(\theta)$$ $$\implies \frac{1}{r}(r\phi'(r)\xi(\theta))+\frac{1}{r^2}\phi(r)\xi''(\theta)=0$$ $$\implies \frac{1}{r}(r\phi''(r)\xi(\theta) +\phi'(r))\xi(\theta) = -\frac{1}{r^2}\phi(r)\xi''(\theta)$$ $$\implies r(\frac{r\phi''(r) +\phi'(r)}{\phi(r)})=-\frac{\xi''(\theta)}{\xi(\theta)}=\mu^2$$ where $\mu^2 \geq 0$ so $\xi$ will be a periodic function. We now have the equations: $$(1):r^2\phi''(r) +r\phi'(r) - \mu^2\phi(r)=0,$$ $$(2):\xi''(\theta)+\mu^2\xi(\theta)=0.$$ Solving for (2), we obtain for $\mu >0$: $$\xi(\theta)=a_1cos(\mu\theta)+a_2sin(\mu\theta).$$ If $\mu=0$, we have $$\xi(\theta)=a_1+a_2\theta.$$ Solving for (1), we obtain for $\mu >0$ $$\phi(r)=b_1r^{\mu}+b_2r^{-\mu}$$ If $\mu=0$, we obtain $$\phi(r)=b_1log(r)+b_2$$
The only boundary condition is that r is bounded by zero (as we cannot have negative radius). Otherwise, we are given nothing but intervals. So to the overall question: where do I go from here to find the separable solutions? I know the answer is right around the corner I just cannot get around the corner :(
You have a few small typos, but they seem insignificant.
As the function $\xi(\theta)$ has to be $2\pi$-periodic we obtain the following:
If $\mu=0$ we must have $a_2=0$ in your solution, so that $\xi(\theta)=a_1$. Multiplying this with the corresponding $\phi$ gives solutions of the form $A_0 \log r+B_0$.
If $\mu>0$, in order to make $\xi$ $2\pi$-periodic we must have that $\mu=n$ is a (positive) integer. The corresponding solutions are of the form $(A_n r^n+B_nr^{-n})(C_n \cos n \theta+D_n \sin n \theta)$.