# Coupled PDE: a confusion in boundary condition setup

I have a coupled PDE problem(Poisson-Schrondinger system), i.e.

first I need to solve an eigenvalue problem (Schrodinger problem discretized by Galerkin method)

$$Ax=\lambda x, ~~~A=A(u)$$

the output $x$ is then used to compute some source term(charge) of a poisson equation(discretized on the same grid)

$$Ku=b, ~~~b=b(x)$$

The problems is: I would like to solve this problem in a neumann sense, which means I don't want to enforce x to be 0 at the boundary for the eigenvalue problem although it might be quite small. So I prefer Neumann B.C. for both PDE, then it seems to me that even if the matrix is singular $Ax=\lambda x$ works fine while $Ku=b$ cannot.

My solution is to set Neumann B.C. for $Ax=\lambda x$ and Dirichlet B.C. for the poisson problem. the result looks fine, and my questions are:

1. Ku=b won't work when K is singular because the solution is undetermined to a constant, am I right?

2. Why does the eigenvalue problem still works when the A matrix is singular? does it automatically throw away the null space?( I solve the eigenvalue problem by using matlab eigs)

3. the combination of Neumann and Dirichlet B.C. in this coupled problem still leads to the physical problem posed in a Neumann sense, right? (since x solved from $Ax=\lambda x$ satisfy Neumann and b=b(x) implicitly built this Neumann condition into a poisson problem)

3. The problem with $A$ and $x$ is different from the problem with $K$ and $u$. $x$ will transport whatever Neumann-ness it has to $b$, but $u$ will satisfy Dirichlet conditions you imposed on it since that what you imposed. I could say more if you posted the continuous problem from which this is all derived.
• I guess my point about the BCs is that you should pick the ones that make the most sense physically. You can always solve the problem with the singularity by imposing a constraint that the average (integral) of $u$ over the domain is zero. Jun 27 '14 at 13:20