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Which differentiable and smooth vector fields $\vec{F}: \mathbb{R}^3 \to \mathbb{R}^3 $ satisfy

$$\nabla \times\nabla \times \vec{F}=\vec{0}$$

Here one can apply the identity

$$\nabla \times(\nabla \times(\vec{F}))=\nabla(\nabla \cdot \vec{F}) - \nabla^2 \vec{F}$$

where $\nabla^2$ is the vector Laplacian, then it would reduce to

$$\nabla(\nabla \cdot \vec{F}) = \nabla^2 \vec{F}$$

which looks like a rather complicated vector PDE. I wonder if this is "another well known problem" like the Laplace-equations or solenoidal or conservative vector fields and so on.

I came across it in the context of this vector field decomposition, where I was wondering if these solutions are unique or up to what they are unique.

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    $\begingroup$ I replaced your \curl by \nabla \times assuming this is what you intended $\endgroup$
    – Henry
    Apr 15, 2023 at 11:30
  • $\begingroup$ Thank you! I did the same immediately :-D, strange it didn't conflict. In the editing mode it was working not in the final. Also in edit mode my browser was incredibly slow in parsing, strange things ... $\endgroup$ Apr 15, 2023 at 11:32
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    $\begingroup$ Note that this problem translates into the mere vectorial equation $\vec{k} \times (\vec{k} \times \vec{F}) = \vec{0}$ in the Fourier space. $\endgroup$
    – Abezhiko
    Apr 15, 2023 at 11:36
  • $\begingroup$ So $\vec{k}||\vec{F}$? Whatever that means in real space. $\endgroup$ Apr 15, 2023 at 11:43

1 Answer 1

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Partial Answer: If $\mathbf F \in L^2(\mathbb R^3 ; \mathbb R^3) \cap C^2(\mathbb R^3;\mathbb R^3)$ then all solutions are of the form $$ \mathbf F = - \nabla q +a \tag{$\ast$}$$ where $q \in (H^1(\mathbb R^3))^3$ with $a\in \mathbb R^3$ is a constant vector. Here $H^1(\mathbb R^3)$ is a Sobolev space. If you are not familiar with these spaces just think of $q\in C^3(\mathbb R)$ since the assumption $\mathbf F \in C^2(\mathbb R^3;\mathbb R^3)$ will force this to be true anyways.

Indeed, on one hand, if $\mathbf F \in L^2(\mathbb R^3 ; \mathbb R^3) \cap C^2(\mathbb R^3;\mathbb R^3)$ is of the form ($\ast$) then one can check that $\mathbf F$ satisfies $\operatorname{curl}(\operatorname{curl}\mathbf F)=0$.

On the other hand, suppose that $\mathbf F \in C^2(\mathbb R^3 ; \mathbb R^3) \cap L^2(\mathbb R^3;\mathbb R^3) $ is a solution to $\operatorname{curl} (\operatorname{curl} \mathbf F)=0$ in $\mathbb R^3$. By the Helmholtz decomposition, there exists $ q\in H^1(\mathbb R) $ and $ \mathbf w \in (H^1( \mathbb R^3 ))^3$ such that $$ \mathbf F = - \nabla q + \operatorname{curl} \mathbf w .$$

Then, since $\operatorname{curl}(\nabla q) =0$ and $\operatorname{div}(\operatorname{curl} \mathbf w)=0$,\begin{align*} 0 &= \operatorname{curl}(\operatorname{curl} \mathbf F) = \operatorname{curl}(\operatorname{curl} (\operatorname{curl} \mathbf w)) = \nabla (\operatorname{div} (\operatorname{curl} \mathbf w))- \Delta (\operatorname{curl} \mathbf w)=- \Delta (\operatorname{curl} \mathbf w). \end{align*} Thus, $\operatorname{curl} \mathbf w$ is harmonic in $ \mathbb R^3$. But $\operatorname{curl} \mathbf w$ is in $L^2(\mathbb R^3 ; \mathbb R^3)$, so it is a constant (this is a property of harmonic functions similar to Liouville's theorem - harmonic functions in $L^2$ are necessarily constant). Thus, $\mathbf F = -\nabla q + a$ for some constant $a\in \mathbb R^3$.

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    $\begingroup$ I am a bit confused. The vector field $\mathbf{F}=\left(\begin{smallmatrix} -y\\0\\0\end{smallmatrix}\right)$ has $\operatorname{curl}\mathbf{F}=\left(\begin{smallmatrix}0\\0\\1\end{smallmatrix}\right)$ and $\operatorname{curl \,(curl}\mathbf{F})=\left(\begin{smallmatrix}0\\0\\0\end{smallmatrix}\right)\,.$ But I do not see how $\mathbf{F}$ can be of the form $-\nabla q+C\,.$ $\endgroup$
    – Kurt G.
    Apr 16, 2023 at 6:53
  • $\begingroup$ In addition to @KurtG. s comment to which I agree, I do not see how one can add a scalar $C$ to a vector $-\nabla q$ but this is is only a minor issue of the initial definitions (quote "$C \in \mathbb{R}$ is a constant"), I think. And also if $C$ would be a constant vector, wouldn't it be expressible as a gradient as well? $\endgroup$ Apr 16, 2023 at 7:34
  • $\begingroup$ @KurtG. Yes, I agree with you. The issue is that that particular $\mathbf F$ is not $L^2$ which is essential to use the Helmholtz decomposition. I will update my answer to say that it is incomplete $\endgroup$
    – JackT
    Apr 16, 2023 at 7:39
  • $\begingroup$ @RaphaelJ.F.Berger Ah yes thanks. That was a typo, it should be constant vector $\endgroup$
    – JackT
    Apr 16, 2023 at 7:39
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    $\begingroup$ Though I should be careful because if I replace $q$ with $q+ a_1x+a_2y+a_3z$ then my function is no longer in $H^1(\mathbb R^3)$ since functions in this space also have to satisfy that they are $L^2$. But if you only care about $q\in C^3$ then you can do that $\endgroup$
    – JackT
    Apr 16, 2023 at 7:50

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