On a (simply connected) domain $\Omega$ for a smooth vector field $F\colon \Omega \to \mathbb{R}^3$, when does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$. I know that $n\cdot(\nabla\times F)=0$ on $\partial\Omega$ is sufficient, and also $t\cdot(\nabla\times F)=0$ is sufficient ($t$ the tangential). Is there a weaker condition?

$\begingroup$ I think (perhaps naively) the answer should be "always". Evidence: (1) On all of $\mathbb{R}^3$, switch to the Fourier transform. Your vector field $V(x)$ is $\int_{\omega \in \mathbb{R}^3} \hat{V}(\omega) \exp(i \omega \cdot x)$. The Fourier transform of the curl is $\omega \times \hat{V}(\omega)$, and the nested curl is $\omega \times (\omega \times \hat{V}(\omega))$. But, for ordinary vectors, it is true that $\omega \times (\omega \times V)=0$ implies $\omega \times V=0$. So, as long as the Fourier transform is defined, this covers the case of $\mathbb{R}^3$. $\endgroup$– David E SpeyerDec 18, 2015 at 13:59

1$\begingroup$ @DavidSpeyer: the question is (almost) equivalent to asking that for a differential form $\omega$, when does $\mathrm{d}\omega = 0 \wedge \mathrm{d}*\omega = 0 \implies \omega = 0$. This is a linear elliptic system and on noncompact domains and domains with boundary is necessarily tied to boundary conditions. The question, at least as written, is poor. Rather than "are there other conditions" I would prefer to see the asker give some indications about what types of conditions he/she is looking for. $\endgroup$– Willie WongDec 18, 2015 at 14:45

3$\begingroup$ The main problem and the pity of this site is that many participants are too fast and not objective with their "ON HOLD" decision and demonstrating "WANT TO BE CLEVER THEN OTHERS" position. Say for not simply connected domains this is an interesting question, I do not know is it solved or not. $\endgroup$– SergeiDec 18, 2015 at 14:48

3$\begingroup$ @Sergei While I can see why you are unhappy about the situation there is quite a bit on the side of an asker to be done. Terse question often get closed, detailed questions with context and motivation much less so. This is not in itself "not objective," indeed it is perfectly in line with the official documentation of the site that requests that such information is given. $\endgroup$– user9072Dec 18, 2015 at 15:10

1$\begingroup$ @WillieWong Here is one way I could imagine making the question precise: Suppose that $X \subset \mathbb{R}^3$ is bounded with boundary a $2$fold. Let $b$ be a function on $\partial X$ with $\int_{\partial X} b=0$. Is there a vector field $F$ on $X$ such that $\nabla \times F$ has flux $b$ across $\partial X$, and $\nabla \times (\nabla \times F)=0$? That would basically say that no weaker condition on $(\nabla \times F)_{\partial X}$ could suffice. It would also be interesting to think about unbounded regions, but I don't have a precise formulation to propose there. $\endgroup$– David E SpeyerDec 18, 2015 at 15:17
2 Answers
Here are some basic thoughts. Let $G$ be a vector field which is of the form $\nabla \times F$, and also obeys $\nabla \times G = 0$.
Since $\nabla \times G = 0$, the vector field $G$ is locally of the form $\nabla h$ for some scalar valued function $h$. The condition that $G = \nabla \times F$ imples that $\nabla \cdot G=0$ or, in other words, $\nabla^2(h)=0$. This says that $h$ is harmonic. So, locally, the condition is equivalent to $G$ being the gradient of a harmonic function. Globally, if $\Omega$ is not simply connected, then traveling around a loop in $\Omega$ may change $h$ by a constant.
From this, we can see that, if $G$ is compactly supported, it is zero: If $G$ is $0$ outside a ball of radius $R$, then $h$ is constant outside that ball, and thus $h$ is constant everywhere.
We can also trot out our favorite conditions to ensure that a harmonic function is constant. For example, if $\Omega = \mathbb{R}^3$, and $G(x) \to 0$ as $x \to \infty$, then $h(x) = o(x)$ and a variant of Liouville's theorem tells us that $h$ is constant and $G=0$.
The OP seems to be interested in conditions on the flux of $G$ across $\partial \Omega$. In order to make this make sense, I am going to assume that the set up is that $\overline{\Omega}$ is a compact manifold with boundary.
Let $R$ be the flux of $G$ across $\partial \Omega$. As the OP already knows, if $R=0$ then $G=0$. And the assignment of $G \mapsto R$ is linear, so this shows that $R$ determines $G$. Let $\mathcal{V}$ be the vector space of functions on $\partial \Omega$ which can occur as such an $R$.
The OP asks for conditions which force $G$ to be zero. In other words, he wants a vector space $\mathcal{W}$ of functions on $\partial \Omega$ which is transverse to $\mathcal{V}$. I find that an odd way to think about the question  better to just characterize $\mathcal{V}$!
Now, the obvious observation is that $\int_{\partial \Omega} R=0$ for any $R$ in $\mathcal{V}$, since $\nabla \cdot G=0$. It would be really cool if that were a precise characterization of $\mathcal{V}$. But I don't think it is. Let $\Omega$ be a solid cylinder, and let $R$ be zero on the sides of $\Omega$ and a hat function on top and bottom, dying out smoothly as it approaches the sides. Is $R$ the flux of some $G$? I couldn't figure out how to make it be.
For some scalar function g: $ rot\ F=grad\ g$. Necessary and sufficient!

$\begingroup$ I am confused  don't {functions}grad>{vector fields}rot>{vector fields}div>{functions} form a complex, so if rot$f=$grad$g$ then for sure rot rot$f=0$, even if rot$f\ne 0$? See the example in David's comment above. $\endgroup$ Dec 18, 2015 at 16:54


$\begingroup$ Exactly. So this does not tell you anything about rot$f=$grad$g$ itself. $\endgroup$ Dec 18, 2015 at 19:43

$\begingroup$ Is not the question we consider then equivalent that $grad g =O$? $\endgroup$– SergeiDec 18, 2015 at 20:36

$\begingroup$ Is not the question we consider then equivalent that $grad g =O$? More: $rot(rot F)=O => rot F =grad g $ for some g. Question: when this $rot F=O$ ? Answer: only if g=const. $\endgroup$– SergeiDec 18, 2015 at 20:43