it is well known, that for a non-convex domain $\Omega$ the space $H^1(\Omega, \mathbb{R}²)$ is a proper subset of $H(curl) \cap H(div)$. Here, $H(curl) = \{v \in L²(\Omega)², \nabla\times v = \partial_1v_2-\partial_2v_1 \in L²(\Omega)\} $ and $H(div)=\{v \in L²(\Omega)², \text{div } v = \partial_1v_1+\partial_2v_2 \in L²(\Omega)\}$.
I am looking for an example, where $ v \in H(curl) \cap H(div)$, but not in $H^1(\Omega, \mathbb{R}²)$. Furthermore, I want to have the constraint div v=0. Lets consider $\Omega=(-0.5,0.5)²\backslash [0,0.5]²$.
In Paper, there might be an example given by equation (5.2) $v=\nabla \times (r^\frac{2}{3} \text{ cos}(\frac{2}{3} \theta-\frac{\pi}{3}))$. Is this one an example, I am looking for?
Thank you very much for your help.
The example you give from the paper is a little bit obscure to me. I propose another solution/perspective, which is quite general (but does not provide an explicit expression, at least in principle) for what concerns counter-examples in the context of $H(\text{curl}),H(\text{div})$. The main idea is to go back to the Laplace equation, which is very easy to deal with, and everything is known about it.
Consider the (weak) solution $u \in H^1(\Omega)$ of the boundary value problem
$$ \begin{cases} \Delta u = 0 \qquad \text{in } \Omega \\ u = g \qquad \text{on } \ \partial \Omega, \end{cases} $$ where $g \in H^{1/2}(\partial \Omega)$. It is well-known that $u \notin H^2(\Omega)$ if the domain is not enough regular (non-convex, with non-smooth boundary...). Define now
$$ \mathbf{w} := \nabla u; $$ then $\mathbf{w} \in \mathbf{L}^2(\Omega) $ since $u \in H^1(\Omega)$, $\operatorname{curl} \mathbf{w} = \mathbf{0}$ since $\mathbf{w}$ is a gradient and $\operatorname{div} \mathbf{w} = \operatorname{div}(\nabla u)= \Delta u = 0$ by construction.
On the other hand, it can happen that $\mathbf{w} \notin (H^1(\Omega))^3$, because $u$ is not necessarily in $H^2(\Omega)$: see remark above.