7
$\begingroup$

Most authors are very clear that it's very dangerous to just use $\mathrm{H}(curl)$ conforming edge elements, which are divergence free, to satisfy $\mathrm{div}(\mathbf{B})=0$ and implement this condition in no other way. I just don't understand why. Could anyone explain in which case this could fail or point me to some good source explaining in which case $\mathrm{H}(curl)$ conforming Nedelec Elements would fail to satisfy $\mathrm{div}(\mathbf{B})=0$ and why?

$\endgroup$
3

2 Answers 2

4
$\begingroup$

One issue (and this is mentioned by Mur in the first paper you linked in the comments above) is the fact that, while these edge functions provide tangential field continuity across interfaces and zero divergence within each element, they also allow for discontinuities in normal fields across interfaces. This behavior is non-physical, giving rise to artificial surface charges which produce errors in finite element solutions.

$\endgroup$
3
$\begingroup$

The answer you are looking for is found in the notion of weak divergence. Recalling some basic facts about distributions, we say that a function $u$ has a weak divergence if for any smooth function with compact support $\varphi \in C^{\infty}_0(\Omega)$, $$ -\int_\Omega u\cdot\nabla\varphi\in\mathbb{R}. $$ Then by density arguments the weak divergence can be seen as a functional inside the dual space of $H^1_0(\Omega)$¸namely $H^{-1}(\Omega)$, where for all $\varphi\in H^1_0(\Omega)$ $$ \DeclareMathOperator{\Div}{div} \varphi\mapsto\ <\Div u,\varphi> := -\int_\Omega u\cdot\nabla\varphi.$$ The norm in this space is given by $$ \|\Div u\|_{-1} := \sup_{\varphi\in H^1_0(\Omega)} \frac{<\Div u,\varphi>}{\|\nabla\varphi\|_{L^2(\Omega)}} $$ and it is by this norm that the weak divergence of a function $u$ in the Nédélec space can be large even though $\Div u|_K\equiv 0$ for an element $K$ in the mesh. We can approximate this norm by choosing a conforming finite element subspace, e.g. Lagrange elements, $P_h\subset H^1_0(\Omega)$. See this nice paper for a more in-depth overview.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.