Poisson equation in semi-infinite domain Im trying to solve the following Poisson equation:
$$\nabla^2\phi = F(x,y)$$
$$for\ x\in(0,\infty)\ and\ y\in(0,L)$$
$$\frac{\partial\phi(x,0)}{\partial y}=0\ , \ \frac{\partial\phi(x,L)}{\partial y}=0\ , \ \frac{\partial\phi(0,y)}{\partial x}=0$$
$$\phi(x,y)\rightarrow0\  uniformly\ as \ x \rightarrow\infty$$
$$$$
I want to solve this using the Fourier Transform or the separation of variable or the Green function. But I'm having trouble understanding since the equation is not homogeneous. And the boundary conditions make lots of trouble too.
Any help and suggestions are appreciated!
 A: The functions $\{ \cos(n\pi x/L) \}_{n=0}^{\infty}$ are the eigenfunction solutions of
$$
          u'' = \lambda u,\;\;\; u'(0)=0,\; u'(L)=0.
$$
So these functions form an orthogonal basis of $L^2[0,2\pi]$. And the functions $\{ \cos(sx)\}_{s \ge 0}$ are eigenfunctions on $[0,\infty)$. You should be able to represent any $g\in L^2([0,\infty)\times[0,L])$ function as
$$
        u(x,y) = \sum_{n=0}^{\infty}\left(\int_{0}^{\infty}C_{u}(n,s)\cos(sy)ds\right)\cos(n\pi x/L),
$$
where $C(n,s)$ is determined in the usual way:
$$
     C_{u}(n,s)= \frac{2}{L}\int_{0}^{L}\left(\frac{2}{\pi}\int_{0}^{\infty}u(x,y)\cos(sy)dy\right)\cos(n\pi x/L)dx.
$$
(For $n=0$ the outer constant must be adjusted to be $\frac{1}{L}$.) Then
$$
    (-n^2\pi^2/L^2-s^2)C_{u}(n,s)=C_{F}(n,s)
$$
The $n=0$ term does not come into play for you because of the condition at $\infty$, which is good because you wouldn't be able to deal with the $n=0$ equation otherwise. So, you only have to consider the above for $n=1,2,3,\cdots$. Once you know $C_{F}(n,s)$--which may be computed from the above--the solution $u(x,y)$ is determined.
