Neumann BV problem on disk (weak vs classical solution)

Question

I am tring to solve

$\bigtriangleup u =-1$ such that the normal derivative vanishes at the boundary where the domain is the unit disc.

In polar coordinates I got I got $u(r)=-1/4 r^{2} +1/2 \ln(r)$

as a solution. Does this qualify as a weak solution (since it has a singularity).

Are there any smooth solutions?

Answer 1

There are no solutions, weak or smooth. The homogeneous Neumann condition is incompatible with the Poisson equation $\Delta u = f$ unless $\int_\Omega f =0$.

For your $u$, the weak Laplacian has a delta function component at the origin, which violates the equation.