Schrodinger equation solution regularity in two dimensions Consider the following Schrodinger equation on $\Omega = (0,2)^2 \subset \mathbb R^2$:
$$(*)\qquad (-\Delta + V(x) ) u(x) = \lambda u(x), \quad \left.u\right\vert_{x\in\partial \Omega} = 0 $$
with non-continuous potential,
$$V(x) = \begin{cases} 0,& x \in S \\ -V_0 \gg 1,& x\in \Omega \setminus S \end{cases}, \qquad S= (0,1)^2\subset \Omega.$$
We know for the 1D analog (finite square well with Dirichlet boundary conditions on an interval $I$), the solution is $C^1(I)$. What can be said about the regularity for $(*)$? I tried integrating,
$$-\int_\Omega \Delta u~dx = \int_{\partial \Omega} \nabla u \cdot n~ ds  = \int_\Omega (\lambda - V(x)) u~dx,$$
but couldn't see how to obtain a statement on the regularity of $u$. Any help? I understand there might be a pole in the gradient at the point $x_0 = (1,1)^T$, but what about on $\Omega\setminus \{x_0\}$? I couldn't find any literature on this problem.
 A: Let us assume that $u \in H^1_0(\Omega)$ is a weak solution to $(\ast)$ ($H^1_0(\Omega)$ is an example of a Sobolev space which are usually the appropriate spaces for constructing solutions to PDEs).
I assume that $V_0$ and $\lambda$ are constants. Since $-\Delta u = \lambda u$ in $S$ and $-\Delta u +V_0u=\lambda u$ in $\Omega \setminus \overline S$, it follows from standard elliptic regularity theory that $u$ is analytic in $S \cup (\Omega \setminus \overline S)$. A reference for this is Ch. 6.3 in Partial Differential Equations by Lawrence Evans, but pretty much any textbook on regularity theory of elliptic PDE will prove this.
Note that even though $u$ is analytic in $S \cup (\Omega \setminus \overline S)$ this doesn't mean $u$ is analytic in $\Omega$ - since $V$ is discontinuous across $\partial S$ this will translate to lower regularity of $u$ across $\partial S$. However, we can still say something about the regularity of $u$ in $\Omega$.
Indeed, it follows from the Harnack inequality of Evans and Safonov (see see ch. 9 in Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger particularly Cor 9.29), we have that $u \in C^\alpha (\overline \Omega)$ for some $\alpha \in (0,1)$. Next, re-write $(\ast)$ as: $$ \tag{$\ast\ast$}-\Delta u = f:=\lambda u -Vu, $$ and we know that $f \in L^\infty(\Omega)$. From the Calderon-Zygmund theory of elliptic PDE (see ch. 9 in Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger), this implies that $ u \in W^{2,p}_{\mathrm{loc}}(\Omega)$ for all $p> 1$ which, by Sobolev embedding, implies that $ u \in C^{1,\gamma}_{\mathrm{loc}}(\Omega)$ for all $\gamma \in (0,1)$. This is just about as good as you can get since $u$ is definitely not in $C^2(\Omega)$ - if you had that $u\in C^2(\Omega)$ then the left hand side of $(\ast\ast)$ is continuous but the right hand side is discontinuous. The only way you could improve this regularity is by showing $u\in C^{1,1}_{\mathrm{loc}}(\Omega)$ - you could possibly obtain this by applying some of the techniques used to prove regularity of solutions to Free Boundary Problems, for example see Regularity of Free Boundaries in Obstacle-Type Problems by Petrosyan, Shahgholian, and Uraltseva.

To summarise: You can say $u$ is analytic on $S \cup (\Omega \setminus \overline S)$ and is $ u \in C^{1,\gamma}_{\mathrm{loc}}(\Omega)$ for all $\gamma \in (0,1)$. It is also possible that $ u \in C^{1,1}_{\mathrm{loc}}(\Omega)$, but no better.
