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