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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 $k \nabla p \in H_{\text{div}}(\Omega)$?

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

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