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!


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.

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  • $\begingroup$ Thanks for your reply. I have a following question. If F(x,y) = F(x), which means the source term only related to x, n=1,2,3,... will disappear because of Fourier cosine series. And the solution might be the n=0 term. But, this term could not be able to deal with because of the singular point in the cosine integral. Could we derived the answer in this situation? $\endgroup$ – H.Webb Apr 7 '16 at 15:21
  • $\begingroup$ @H.Webb : The underlying reason for the ambiguity is that the constant function is a solution of your equation where $F=0$, if you ignore the condition at $\infty$. So, without some additional constraint, the problem is not well posed. That's the reason for the condition at $\infty$. These kinds of things can happen when the only boundary conditions involve derivatives. Even on a finite domain, the constant solution is a problem if the only conditions are on the normal derivatives, for example. $\endgroup$ – Disintegrating By Parts Apr 7 '16 at 16:47

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