Laplace equation on a disk I have the Laplace equation $$\Delta u=\frac{1}{r} \frac{\partial}{\partial r } \left(r \frac{\partial u}{\partial r} \right)+\frac{1}{r^2} \frac{\partial u }{\partial \theta^2}=0$$ on a unit disk $$0<r \leq 1$$
We note that we have the boundary condition for R $$|u(0,\theta)|<\infty \rightarrow |R(0)|<\infty$$
and the boundary conditions for $\theta$ are $$\Theta(- \pi)=\Theta(+ \pi)$$ and $$\Theta'(- \pi)=\Theta'(+ \pi)$$
I assume the solution is of the form $$u(r,\theta)=R(r)\Theta(\theta)$$ and 
subbing this into $$\Delta u$$
I get 2 ODE's, $$-\frac{r}{R(r)}(rR'(r))'=k$$ and $$\frac{\Theta''(\theta)}{\Theta(\theta)}=k$$
I first look at the case for when $$k=p^2>0$$
This gives $$\Theta=ae^{p \theta}+be^{-p \theta}$$
So by subbing in the boundary conditions for $\theta$, I get $$ae^{p \pi}+be^{-p \pi}=ae^{-p \pi}+be^{p \pi}$$
Why does this give me no solution?
When I look at the case $$k=0$$ the corresponding ODE for R(r) gives $$R=c_1 \ln r+c_2$$
Why does this being subject to the boundary condition $|R(0)|$ give me $R=c_2$? and why do I get the solution for this case being $u_0 (r,\theta)=1$?
 A: 
Why does this give me no solution?

You have $ae^{p\pi}+be^{-p\pi}=ae^{-p\pi}+be^{p\pi}$. That is equivalent to $a(e^{p\pi}-e^{-p\pi})=b(e^{p\pi}-e^{-p\pi})$, which implies either $a=b$ or $e^{p\pi}=e^{-p\pi}\iff p=ki$, but in our case $p^2=k>0$, so that $k$ must be 0. If $p\neq0$ you have no solution, with $p=0$ you get a constant function which is found also in the $k=0$ case, so nothing from this case that you don't get from other cases. But if $p\neq0$, the solution becomes $\Theta(\theta)=\frac{a}{2}\cosh(p\theta)$, which satisfies one boundary condition (the one on $\Theta$, being the $\cosh$ an even function), but not the other, as $\Theta'(\theta)=\frac{ap}{2}\sinh(p\theta)$, which is odd.

Why does this being subject to the boundary condition $|R(0)|$ give me $R=c_2$?

Well, suppose otherwise. If $c_1\neq0$, then for $r\to0$ we have $|R(r)|\to+\infty$, but the boundary condition states otherwise. So $c_1=0$ and $R(r)=c_2$.

And why do I get the solution for this case being $u_0(r,\theta)=1$?

I'd rather say you get a constant function. The $R$ part we have already shown to be constant, and the $\theta$ part has a zero second derivative and must thus be $\Theta(\theta)=a\theta+b$, which satisfies the boundary conditions iff $a=0$, and ends up being forced by the boundary conditions to be $\Theta(\theta)\equiv b$, a constant. $u_0(r,\theta)=R(r)\Theta(\theta)=b\cdot c_2$, so it is constant. Why it is one, I wouldn't know. Maybe there is some kind of normalization imposed by whatever you are following.
Hope this answers you.
