# Regularity of the Divergence of Weak Solutions to Elliptic PDEs

Let $\Omega$ be a bounded domain in $\mathbb{R}^2$ and define $$H_{\text{div}} (\Omega)= \{\boldsymbol{u}: \Omega \to \mathbb{R}^2: u_i \in L^2(\Omega), ~\text{and}~ \nabla \cdot \boldsymbol{u} \in L^2(Ω)\}.$$

Considering the elliptic problem $$\begin{cases} -\nabla \cdot (k \nabla p) = f & \text{in } \Omega\\ p = 0 & \text{on } \partial \Omega \end{cases}$$ and its associated variational formulation: Find $p \in H_0^1(\Omega)$ such that $$\int_{\Omega} k \nabla p \cdot \nabla u ~\mathrm{d} x = \int_{\Omega} f u ~\mathrm{d} x.$$

Let $f \in L^2(\Omega)$ and $k$ be strictly positive and regular enough so that the Lax-Milgram theorem can be applied to establish solutions to the variational formulation.

If I consider divergence in the sense of weak derivatives then can I write with integration by parts: $$\int_{\Omega} f u ~\mathrm{d} x = \int_{\Omega} k \nabla p \cdot \nabla u ~\mathrm{d} x = \int_{\Omega} \left(-\nabla \cdot(k \nabla p) \right) u ~\mathrm{d}x.$$ and conclude not only that $p \in H_0^1(\Omega)$ but also $k \nabla p \in H_{\text{div}}(\Omega)$?

• Yes, this is correct. – daw Feb 1 '18 at 12:44
• Thanks daw! I think I got myself turned around by thinking of $H_{\text{div}}$ as $H^2$. – Steve Feb 1 '18 at 19:37