Find $u$ such that $\frac{\partial^2u}{\partial x²}+\frac{\partial^2u}{\partial y^2}=1$ in $r<a$ and $u=0$ in $r=a$.
My attempt
The equation in polar coordinates is given by:
$$ \frac{\partial^2u}{\partial r²}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta²} = 1$$
I'm looking for a solution that depends only on $r$. Therefore:
$$ \frac{\partial^2u}{\partial r²}+\frac{1}{r}\frac{\partial u}{\partial r} = 1 \Rightarrow r\frac{\partial^2u}{\partial r²}+\frac{\partial u}{\partial r} = r\Rightarrow \frac{\partial (ru)}{\partial r} = r $$
Integrating twice in $r$, I found that:
$$ u(r,\theta) = \frac{r^2}{4} + A\ln(r) + B $$
Applying the boundary condition we get:
$$ \frac{a^2}{4} + A\ln(a) + B = 0 $$
I'm stuck here. I can't find the constants $A$ and $B$. Furthermore, the solution of this Poisson equation is unique (by maximum principle). It looks like we have a family of solutions. What am I doing wrong?
Since $r \mapsto \ln r$ is not bounded near $r=0$, one has $A=0$.
Now we get $$ \frac{a^2}{4} + B = 0$$ so $$ B = - \frac{a^2}{4}$$.