Solving PDE on finite domain Given the PDE, $\Delta u=x$ in the region $x^2+y^2<1$. And $\frac{\partial u}{\partial r}=y$ on $x^2+y^2=1$. I am supposed to find all solutions.
The only machinery I know for finding solutions on bounded domains is separation of variables. But I only know how to do separation of variables for homogeneous problems. So I found the solution to $\Delta u=0$ is $u(x,y)=y$. But it is unclear how to extend this to inhomogeneous case. (It amounts to solving $\Delta u=x$ with $\frac{\partial u}{\partial r}=0$) 
Maybe there some transform that turns $\Delta u=x$ into Laplace's equation? Thank-you.
 A: One approach goes as follows. First, you make the boundary condition homogeneous. This can be done by introducing $v=u-g$, where $g$ is some function that satisfies $\partial_r g = y$ at the boundary of the disk. The function $v$ must then satisfy $\Delta v = x-\Delta g$. Now you use separation of variables. The idea is to expand the right hand side $f(x,y)=x-\Delta g$ in terms of the eigenfunctions of the Neumann Laplacian on the disk. So let $\phi_n$ be those eigenfunctions, and let
$$
f = \sum_n b_n\phi_n.
$$
We look for the solution in the form
$$
v = \sum_n a_n\phi_n.
$$
Then the equation is
$$
\Delta v = \sum_n a_n\Delta \phi_n = \sum_n a_n\lambda_n\phi_n=\sum_n b_n\phi_n,
$$
and so
$$
a_n=b_n/\lambda_n,
$$
that is
$$
v = \sum_n \frac{b_n}{\lambda_n} \phi_n.
$$
Note that we need $b_0=0$ because $\lambda_0=0$ for the Neumann Laplacian, and by a related reason the solution $v$ is unique up to a constant.
A: I'll start by decomposing problem as follows,
\begin{cases}
\Delta u=x\\
\partial u/\partial r=0
\end{cases}
\begin{cases}
\Delta v=0\\
\partial v/\partial r=y
\end{cases}
The second problem is easy. You can do it with seperation
of variables or just observe that the solution is $v=y$. For the
first problem we can start by converting to polar coordinates,
$$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta\theta}=r\cos\theta$$
Now it is natural to guess the solution is like $r^{3}\cos\theta$.
Plugging this in yields that $\frac{1}{8}r^{3}\cos\theta$ works,
and then you need to fix the BC so we have $u=\frac{1}{8}r^{3}\cos\theta-\frac{3}{8}r\cos\theta$.
Adding $u$ and $v$ yields the solution,
$$\frac{x}{8}\left(x^{2}+y^{2}-3\right)+y + C$$ 
Now since I added the constant $C$, we have all solutions. This is because solution to Poisson with Neumann data is unique up to a constant. (Easy energy method to show this)
