# Solutions to $\nabla \times\nabla \times\vec{F}=0$?

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.

• I replaced your \curl by \nabla \times assuming this is what you intended Apr 15, 2023 at 11:30
• 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 ... Apr 15, 2023 at 11:32
• Note that this problem translates into the mere vectorial equation $\vec{k} \times (\vec{k} \times \vec{F}) = \vec{0}$ in the Fourier space. Apr 15, 2023 at 11:36
• So $\vec{k}||\vec{F}$? Whatever that means in real space. Apr 15, 2023 at 11:43

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 $$a\in \mathbb R^3$$.

• 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 $-\nabla q+C\,.$ Apr 16, 2023 at 6:53
• 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? Apr 16, 2023 at 7:34
• @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 Apr 16, 2023 at 7:39
• @RaphaelJ.F.Berger Ah yes thanks. That was a typo, it should be constant vector Apr 16, 2023 at 7:39
• 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 Apr 16, 2023 at 7:50