In the book Mixed Finite Element Methods and Applications by Boffi, Brezzi, and Fortin there is a pretty long discussion about why the Raviart-Thomas (RT) projection is only defined for functions in $H^1$ (Remark 2.5.1). In addition, For DG methods there are similar projections, only defined for functions in $H^1.$

To set the stage for my question, I introduce a linear elliptic PDE. Let $\Omega\in \mathbb{R}^d$ with $d\in \{2,3\}$ be a bounded polyhedral domain. Suppose a source function $\sigma\colon \Omega\to \mathbb{R}$, is in $L^2(\Omega)$, the Dirichlet boundary condition $u_d\colon \partial\Omega_D\to \mathbb{R}$, and the Neumann boundary condition $\vec{g}_N\cdot\vec{\eta}\colon \partial\Omega_N \to \mathbb{R}$ are all functions with enough regularity such that the following PDE is well defined.

Our PDE is \begin{align*} 0 &= \vec{q} + \vec{\nabla} u&& x\in \Omega,\\ \sigma &= \vec{\nabla} \cdot \vec{q}&& x\in \Omega,\\ u_d &= u&& x\in \partial\Omega_D,\\ \vec{g}_N\cdot \vec{n}&= \vec{q}\cdot \vec{n} && x\in \partial\Omega_N. \end{align*}

What are weak restrictions that we need to place on the data (the domain, the source term, and the boundary conditions) so that $\vec{q}\in H^1(\Omega)$? Am I correct in assuming that we need some sort of elliptic regularity?

References to the literature are appreciated. I have many papers dealing with elliptic regularity but they all deal with even simpler PDEs (homogenous BC, smooth boundary, no Neumann data, etc).


I don't think you get this from your setup. If $\sigma$ is only in $L^2$, then the equation $\nabla \cdot q = \sigma$ would suggest that you only get $q \in H_\text{div}$, regardless of what boundary conditions you have or how smooth the domain's boundary is.

  • 1
    $\begingroup$ Yes, that is the crux of my question. We only are sure that $q\in H_{div}$ in the current setup. However, as a special case, if $\sigma$ and $u_d$ and $g_n$ are all zero then $u=0$ and $\vec{q}=0$ so $\vec{q}\in H^1$, I am looking for less drastic conditions that ensure $\vec{q}\in H^1$. $\endgroup$ – fred Apr 13 '16 at 18:58
  • $\begingroup$ To get this, variational formulations are typically of not much use. But you can of course get this from the classical (strong) formulation in some cases. In particular, you know that the solution is the solution of the Laplace equation, so assuming sufficiently regularity of domain and forcing terms, you know that at least in the interior, the solution $u$ is in fact $C^\infty$, and so is its gradient $q$. $\endgroup$ – Wolfgang Bangerth Apr 13 '16 at 21:32

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