# H(curl) conforming Nédélec-Elements to satisfy div(B)=0

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?

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.