Let $n=2$ or $3$ and $\Omega \subset \mathbb{R}^n$ be an open bounded domain. Let suppose that $\Omega$ is divided in two subdomains $\Omega_1$ and $\Omega_2$ and we define $\Gamma = \partial \Omega_1 \cap \partial \Omega_2$, $\Gamma_1 = \partial \Omega_1 \setminus \Gamma$ and $\Gamma_2 = \partial \Omega_2 \setminus \Gamma$. Consider the following Poisson transmission problem

\begin{equation} \left\{ \begin{array}{rccl} -\mu_1 \Delta u_1 & = & f_1 & \text{in $\Omega_1$}, \\ -\mu_2 \Delta u_2 & = & f_2 & \text{in $\Omega_2$}, \\ u_1 & = & 0 & \text{on $\Gamma_1$}, \\ u_2 & = & 0 & \text{on $\Gamma_2$}, \\ u_1 & = & u_2 & \text{on $\Gamma$}, \\ \mu_1 u_1 \cdot n_1 & = & - \mu_2 u_2 \cdot n_2 & \text{on $\Gamma$}, \end{array} \right. \end{equation}

where $n_i$ is the exterior normal of $\Omega_i$, for $i\in\{1,2\}$, and $f_i \in L^2(\Omega_i)$, for $i\in\{1,2\}$.

By classical arguments it is easy to show that the weak formulation of this problem is well-posed in the sens that there is a unique solution $(u_1,u_2)$ in $H^1_0(\Omega)$ (Lax-Milgram theorem).

Moreover, if all boundaries are smooth enough, we know that this solution is partially of regularity $H^2$ in $\Omega_1$ and $\Omega_2$, i.e. $u_i \in H^2(\Omega_i)\cap H^1_{\Gamma_i}(\Omega_i)$ for $i\in\{1,2\}$. And this result is also true for more general elliptic problems (M. Costabel, M. Dauge, and S. Nicaise. Corner Singularities and Analytic Regularity for Linear Elliptic Systems. Part I: Smooth domains. 2010.). However, obviously, this solution is not globally of regularity $H^2$ in the entire domain $\Omega$, because of the jump of the gradient through $\Gamma$.

**Questions:**
Do we have similar results when boundaries are not smooth ? Typically, what happen when $\Omega$ is a square in 2D (or a cube in 3D) and $\Gamma$ is the vertical segment (plane in 3D) cutting the square (or cube) in half ?