Let $\Omega \subset \mathbb{R}^3$ be an open but not simply connected domain and let $v \: \colon \Omega \to \mathbb{R}^3$ be a continuously differentiable vector field. Assume that $\textrm{curl} \, v = 0$ in $\Omega$. As $\Omega$ is not simply connected, we cannot conclude that there is a function $\varphi \in C^2(\Omega, \mathbb{R})$ such that $v = \nabla \varphi$.

However, if we additionally assume that for all smooth loops $\gamma \: \colon [0, 1] \to \Omega$ the line integral \begin{equation} \int_{\gamma} v(r) \cdot \textrm{d}r = 0, \qquad (1) \end{equation} then we find $\varphi \in C^2(\Omega, \mathbb{R})$ such that $ v = \nabla \varphi$.

My question is now:

Let $\Omega \subset \mathbb{R}^3$ be an open but not simply connected domain and given a vector field $ v \in L^2(\Omega, \mathbb{R}^3)$ with $\textrm{curl} \, v = 0$ in $\Omega$. Is there a similar condition like $(1)$ that ensures the existence of a function $\varphi \in H^1(\Omega)$ such that $v = \nabla \varphi$?

Thanks in advance!


The general condition is that if for all divergence free compactly supported smooth vector fields, that is $f \in \mathcal{D}(\Omega,\mathbb{R}^3)$ with $div(f)=0$, $v \in L^2(\Omega,\mathbb{R}^3)$ satisfies, \begin{equation} \int_\Omega v.f dx = 0 \end{equation}
then there will exist $\phi \in H^1(\Omega) $ such that $v=\nabla \phi$. This condition on $v$ will imply $curl (v)=0$ but not the other way around. This is in fact quite a general result and would work for $v$ in more general Sobolev spaces or in the space of distributions (each component of $v$ is a distribution).


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