$H^2$ regularity for divergence style right hand side.

Question

Let $u\in H^1(\Omega)$ satisfy $$\int_\Omega (A\nabla u)\cdot\nabla v\,dx=\int_\Omega g\cdot \nabla v\,dx\,\,\forall v\in H^1_0(\Omega)$$ where $g\in L^2(\Omega;\Bbb R^2)$, $A\in W^{1,\infty}(\Omega;\Bbb R^{2\times 2})$, $\Omega\subset\Bbb R^2$ bounded with sufficiently smooth boundary, can we show that $u\in H^2_{\operatorname{Loc}}(\Omega)$?

Answer 1

Your equation is $-\text{div} (A \nabla u)= \text{div} g$. The heuristic is that since $\nabla u$ and $g$ have the same 'strength' in the equation, $\nabla u$ will have the same regularity as $g$. In other words $H^ 1$ is as best as you can hope (at lest in these simple spaces).

As an example take $g\in L^ 2 \setminus H^ 1$, then the solution of $-\Delta u = \text{div}g \in H^{-1}\setminus L^2$ clearly has $u\notin H^ 2$.