Solutions of Laplace's Equation in the polar coordinates Question:
Consider the Laplace’s equation in the circular zone:
$$\Delta u = u_{rr}+\frac 1r u_{r}+\frac 1{r^2}u_{\theta \theta}, $$
where $(r,\theta)\in \mathbb{R}^2$ and $u(a,\theta)=b\cos^2\theta$. The center of the circular zone is the origin, radius $a$, and $b$ is a constant.
i) When does a solution exists ?
ii) Solve the boundary-value problem?
My Attempt
Let $u(r,\theta)=\phi(r)\xi(\theta).$
$$u(r,\theta)=\phi(r)\xi(\theta)$$
$$u_r=\phi'(r)\xi(\theta), u_{\theta\theta}=\phi(r)\xi''(\theta)$$
$$\implies \frac{1}{r}(r\phi'(r)\xi(\theta))+\frac{1}{r^2}\phi(r)\xi''(\theta)=0$$
$$\implies \frac{1}{r}(r\phi''(r)\xi(\theta) +\phi'(r))\xi(\theta) = -\frac{1}{r^2}\phi(r)\xi''(\theta)$$
$$\implies r(\frac{r\phi''(r) +\phi'(r)}{\phi(r)})=-\frac{\xi''(\theta)}{\xi(\theta)}=\lambda$$
We now have the equations:
$$(1):r^2\phi''(r) +r\phi'(r) - \mu^2\phi(r)=0,$$ 
$$(2):\xi''(\theta)+\mu^2\xi(\theta)=0.$$
if $\lambda=\mu^2 >0$:
$$\phi(r)=c_1r^{\mu}+c_2r^{-\mu}$$
$$\xi(\theta)=c_3\cos(\mu\theta)+c_4\sin(\mu\theta).$$
Then the corresponding solutions are of the form $$u(r,\theta)=(c_1r^{\mu}+c_2r^{-\mu})(c_3\cos(\mu\theta)+c_4\sin(\mu\theta))$$
If $\lambda=\mu^2=0$, we have 
$$\xi(\theta)=c_5+c_6\theta.$$ 
$$\phi(r)=c_7\log(r)+c_8$$
Then the corresponding solutions are of the form $$u(r,\theta)=(c_5+c_6\theta)(c_7\log(r)+c_8)$$.
If $\lambda=-\mu^2<0$, we have 
$$\xi(\theta)=c_9r^{\mu\theta}+c_{10}r^{-\mu\theta}.$$ 
$$\phi(r)=c_{11}\cos(\log(\mu r))+c_{12}\sin(\log(\mu r))$$
Then the corresponding solutions are of the form $$u(r,\theta)=(c_{11}\cos(\log(\mu r))+c_{12}\sin(\log(\mu r)))(c_9r^{\mu\theta}+c_{10}r^{-\mu\theta})$$
My Stacked Part
i) When does a solution exists ? Why?
ii) How can I find the solution by applying the boundary condition? What is the solution? 
I am new in Laplacian equations. Could you flesh something out pls.?
Thanks.
 A: You've got a good start here.  Here are some observations that should help you complete the solution:


*

*$u(r,0) = u(r, 2\pi)$ and $\partial u/\partial \theta|_{(r,0)} = \partial u/\partial \theta|_{(r,2\pi)}$ for all $r$, since $\theta = 0$ and $\theta = 2 \pi$ correspond to the same point in your domain.  This places constraints on the form of your function $\xi(\theta)$, and allows you to eliminate the vast majority of the possible values for $\lambda$ that you've listed above.

*$u(0,\theta)$ must be a finite value.  This allows you to eliminate quite a few of the possible forms for $\phi(r)$ that remain after you've done the previous step.

*Recall the trig identity that $\cos^2 \theta = \frac{1}{2} + \frac{1}{2} \cos 2 \theta$.
A: The problem you have stated is a Dirichlet problem for a circle which always has a unique solution. The general solution of $\nabla^2u=0$ given $u(a,\theta)=f(\theta)$ is
$$u(r,\theta)=\dfrac{a_0}{2}+\sum_{n=1}^\infty\dfrac{r^n}{a^n}(a_n\cos n\theta+b_n\sin n\theta)$$ where $a_n's$ and $b_n's$ are Fourier Coefficients of $f(\theta)$.   
Reference: Partial Differential Equations of Mathematical Physics by Tyn-Myint U.
