We want to solve $r^2u''_{rr}+ru'_r+u''_{\theta \theta}=0$ where $0\leq \theta <2\pi$ and $0\leq r \leq 2$, given that $u(2,\theta)=\cos(2\theta)$.
I managed to work out a simple solution, but when I checked it, I saw that it does not agree with $r^2u''_{rr}+ru'_r+u''_{\theta \theta}=0$ and I'd like to know where is the mistake.
My solution
Let $u(r,\theta)=R(r)T(\theta)$. Then from $r^2u''_{rr}+ru'_r+u''_{\theta \theta}=0$ we have $\frac{r^2R''(r)+rR'(r)}{R(r)}=-\frac{T''(\theta)}{T(\theta)}$
On the left hand side we have a function of just $r$, on the right hand just $\theta$, so they must both be scalar $\frac{r^2R''(r)+rR'(r)}{R(r)}=-\frac{T''(\theta)}{T(\theta)}=\lambda$
From this we have $T''(\theta)+\lambda T(\theta)=0$, but since $u$ is periodic with respect to $\theta$ with period of $2\pi$, we must have $T(0)=T(2\pi)$ and $T'(0)=T'(2\pi)$. This is a sturm-liouville problem whos solution is given by:
$\lambda_k=k^2$ and $T_k(\theta)=A_k\cos(k\theta)+B_k\sin(k\theta)$. where $k=1,2,...$ and $T_0=1$.
But we also had $\frac{r^2R''(r)+rR'(r)}{R(r)}=\lambda=k^2$, so $r^2R''_k(r)+rR'_k(r)-k^2R_k(r)=0$
If we guess a solution $R(r)=r^\alpha$ then we have $\alpha(\alpha-1)\alpha-k^2=\alpha^2-k^2=0$ so $\alpha=\pm k$ and so $R_k(r)=C_kr^k+D_kr^{-k}$ when $k\neq 0$, and $R_0=C_0+D_0\ln(r)$.
So overall our solution will be of the form $u(r,\theta)=T_0(\theta)R_0(r)+\sum_{k=1}^{\infty}T_k(\theta)R_k(r)$ which equals to
$u(r,\theta)=C_0+D_0\ln(r)+\sum_{k=1}^{\infty}[A_k\cos(k\theta)+B_k\sin(k\theta)](C_kr^k+D_kr^{-k})$
We impose another restriction that $\lim_{r\to 0}u(r,\theta)$ will be bounded. it will diverge to infinity or negative infinity. This can only happen when $D_0=0$ and $D_k=0$, so we have:
$u(r,\theta)=C_0+\sum_{k=1}^{\infty}[A_k\cos(k\theta)+B_k\sin(k\theta)]C_kr^k$
To make things simple, since we don't know either $A_k,B_k,C_k$ let's just say $C_k=1$ and find $A_k,B_k$ that all the conditions are met. So overall:
$u(r,\theta)=C_0+\sum_{k=1}^{\infty}[A_k\cos(k\theta)+B_k\sin(k\theta)]r^k$. but we also had a condition that $u(2,\theta)=\cos(2\theta)$. so
$u(2,\theta)=C_0+\sum_{k=1}^{\infty}[A_k\cos(k\theta)+B_k\sin(k\theta)]2^k=\cos(2\theta)$. This can only hold if $C_0=0$, $B_k=0$ for all $k$. In addition we must have $A_k=0$ when $k\neq 2$, and we must also have $2^2A_2=1$, so $A_2=\frac{1}{4}$ so my answer is:
$u(r,\theta)=\frac{1}{4}r^2\cos(2\theta)$. That's the solution I got.
Indeed it is true that $\lim_{r \to 0}u(r,\theta)=0$ is bounded.
And it is also true that $u(2,\theta)=\frac{1}{4}4\cos(2\theta)=\cos(2\theta)$. So these conditions are met.
However, we have $u'_r=\frac{1}{2}r\cos(2\theta)$, $u''_{rr}=\cos(2\theta)$ and $u''_{\theta \theta}=-r^2\cos(2\theta)$. so:
$r^2u''_{rr}+ru'_{r}+u''_{\theta \theta}=\frac{1}{2}r^2\cos(2\theta)$.
And it should be zero. So I'm really not sure what's going on.
$$ u_{rr} = \frac{1}{2}\cos(2\theta) \neq \cos (2\theta) $$