A Sobolev space on a perforated domain with extension by a solid vector field Presentation :
Let $\Omega \subset \mathbb{R}^3$ an open bounded regular set and $B=B(a,r)$ a ball such as $\bar{B} \subset \Omega$.
I'm studying the following space :
$$V=\{v|_{\Omega \setminus B} \in (H^1(\Omega \setminus B))^3 \ | \ \operatorname{div}(v) = 0 \text{ in } \Omega \setminus B \ | \ \gamma_0(v)=0 \text{ a.e on }  \partial \Omega \ | \ \exists (b,c) \in \mathbb{R}^3 \times \mathbb{R}^3 \text{ such as } v|_{B}(x)=b + c \wedge (x-a) \}$$
where $\gamma_0 : (H^1(\Omega \setminus B))^3 \rightarrow (L^2(\partial \Omega \cup \partial B))^3$ is the sobolev trace application on $\Omega$ and the divergence condition should be understand in the distribution sense :
$$\int\limits_{\Omega \setminus B} v \cdot \nabla \varphi=0, \ \forall \varphi \in C^\infty_c(\Omega \setminus B).$$
Concretely, the function $v \in V$ is a divergence free element in $(H^1(\Omega \setminus B))^3$, which is worth $0$ on $\partial \Omega$, that has been extended in the ball $B$ by a solid vector field $x \mapsto b + c \wedge (x-a)$ so as to get a function $v$ which is at least in $L^2(\Omega)$. I want to study the property of this extension.
My question
I was wondering if it were possible to show that an element $v \in V$ is  in $(H^1_0(\Omega))^3$ and is divergence free in $\Omega$.
My attempt so far
I want to show that the weak derivatives of $v$ are elements of $L^2(\Omega)$. However, since I have a jump at the interface $\partial B$, i'm not sure I will be able to have this kind of regularity.
For the divergence free condition, I know that the divergence of a solid vector field $x \mapsto b + c \wedge (x-a)$ is $0$ pointwise, so the weak divergence is equal to $0$ both in $\Omega \setminus B$ and $B$, but again I have an issue with the jump at $\partial B$.
Any help or references wich deal with such extension are welcomed, as always :). Feel free to ask me questions.
 A: I think i found the answer to my question. Let's modify the definition of my space to function defined only in $\Omega \setminus B$ :
$$V=\{v \in (H^1(\Omega \setminus B))^3 \ | \ \operatorname{div}(v) = 0 \text{ in } \Omega \setminus B \ | \ \gamma_0(v)=0 \text{ a.e on }  \partial \Omega \ | \ \exists (b,c) \in \mathbb{R}^3 \times \mathbb{R}^3 \text{ such as } \gamma_0(v)(x)=b + c \wedge (x-a) \text{ a.e on }  \partial B  \}$$
Let $v \in V$, we extend $v$ in the inclusion $B$ by it's value on $\partial B$, noted $ s: x \mapsto b + c \wedge (x-a)$. The extended vector $v$ is indeed a distribution in $D'(\Omega)$ since it is locally integrable on $\Omega$.
Now, let $\varphi \in D(\Omega)$, and compute for $j \in \{1,2,3 \}$ the following vector :
\begin{align}
<\partial_j v | \varphi > & = - < v | \partial_j \varphi> \ \text{ by definition}\\
& = - \int_{\Omega} v \ \partial_j \varphi =  - \int_{\Omega \setminus B} v \ \partial_j \varphi - \int_{B} s \ \partial_j \varphi\\
& = \int_{\Omega \setminus B} \underbrace{\partial_j v}_{\in L^2(\Omega \setminus B)^
3} \ \varphi -\int_{\partial B} v \ \varphi \ n_j -\underbrace{\int_{\partial \Omega} v \ \varphi \ n_j}_{=0} + \int_B \underbrace{\partial_j s}_{\in L^2(B)^3} \ \varphi - \int_{\partial B} s \ \varphi \ \tilde{n}_j
\end{align}
where $n_j$ is the $j$ composant of the intern normal vector for $\partial B$, and $\tilde{n}_j$ is the $j$ composant of the extern normal vector for $\partial B$, therefore $n_j = - \tilde{n}_j$ and we finally have :
$$<\partial_j v | \varphi >= \int_{\Omega} w \ \varphi, \quad \forall \varphi \in D(\Omega)$$
with $w = \partial_j v \ \mathbb{I}_{\Omega \setminus B} + \partial_j s \ \mathbb{I}_{B} \in (L^2(\Omega))^3$, and therefore $v \in (H^1(\Omega))^3$.
Similar computation yields the result for the divergence !
