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