Cfrgtkky

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).