I am working on a problem which shows interior regularity may fail if you have non-Lipschitz coefficients.
Let $\Omega = [−1,1]$ and $K = [−1/2,1/2] \in \Omega$. Define $a(x)=1+1_K(x)$ and consider the weak solution $u \in H_0^1(\Omega)$ to the equation $$-\text{div}(a\nabla u)=1 \text{ on } \Omega, \ \ \ u=0 \text{ on }\partial{\Omega}.$$ Show that $u\notin H^2(\hat{\Omega})$ for an open neighbourhood $\hat{\Omega}$ of $K$.
The weak formulation is $\int a\Delta u \Delta v=\int v$ for all $v\in H_0^1(\Omega)$. I think I need to show that $u''$ has infinite $L^2$-norm on $\hat{\Omega}$. The problem should be that $a$ is a step function, it is $2$ on $K$ and $1$ on $\Omega - K$ and so it has a jump at $-1/2$ and $1/2$. Why is this such a big problem?
In the weak formulation, split the integral into parts on $K$ and $\Omega\setminus K$, derive strong formulations separately on these sets together with suitable boundary conditions on the boundary between $K$ and $\Omega\setminus K$. Then you can solve for $u$, it will be piecewise quadratic.