Does the problem $\operatorname{div}(\vec{p}) = f$ on $\Omega$ with $\vec{p}\cdot\vec{\nu}=0$ on $\partial \Omega$ admits non-conservative solutions?

Question

Define $V:=\{\vec{p} \in L^2(\Omega;\mathbb{R}^n) \mid \operatorname{div}(\vec{p}) \in L^2(\Omega)\}$ where $\Omega$ is a non-empty connected bounded open subset of $\mathbb{R}^n$ with $C^2$ boundary and $\operatorname{div}$ is the distributional divergence defined by $$\forall \vec{p} \in L^2(\Omega;\mathbb{R}^n), \forall \varphi \in C^1_c(\Omega), \operatorname{div}(\vec{p})(\varphi):=- \int_\Omega \vec{p}\cdot \nabla\varphi \operatorname{d}x.$$ It is known that this space is a Banach space and its elements admit normal traces (see the answer to this question). Let $f \in L^2(\Omega)$ and consider the problem of finding $\vec{p} \in V$ such that $$\forall \varphi\in H^1(\Omega), - \int_\Omega \vec{p}\cdot \nabla\varphi \operatorname{d}x=\int_\Omega f\varphi\operatorname{d}x.$$ This is clearly the weak formulation of the problem \begin{equation} \begin{cases} \operatorname{div}(\vec{p})=f, & \text{on}\ \Omega \\ \vec{p}\cdot \vec{\nu}=0, & \text{on}\ \partial\Omega \end{cases} \end{equation} where $\vec{p}\cdot \vec{\nu}$ is the normal trace of $\vec{p}$ on $\partial\Omega$.

It is a consequence of the Fredholm theory about compact operators that this problem admits an irrotational solution $\vec{p} = \nabla u$ with $u \in H^1(\Omega)$, if and only if $\int_\Omega f \operatorname{d}x = 0$, and that in this case the solution is unique.

However, I never have seen treated the general case, and so the question:

Are there non-conservative solutions to this problem, i.e. solutions that don't admit potentials $u \in H^1(\Omega), \nabla u = \vec{p}$?

Answer 1

Assume that $n = 2$ and that $\int_\Omega f \operatorname{d} x=0$. Let $u \in H^1(\Omega)$ such that $\vec{p} := \nabla u$ is a solution to our problem. Get any non-trivial $g \in C^2_c(\Omega)$ and define $\vec{q}_g:=(\partial_yg,-\partial_xg).$ Then by Schwarz theorem we have that $\operatorname{div}(\vec{q}_g) = 0$ and, by the fact that $\operatorname{supp}(g) \subset\subset\Omega$, we also have that the normal trace of $\vec{q}_g$ on $\partial\Omega$ is zero. Then $\vec{p}+\vec{q}_g$ is another (non-conservative) solution to the problem.