Does taking $\nabla\times$ infinity times from an arbitrary vector exists?

Question

Is it possible to get the value of:

\begin{equation} \underbrace{\left[\nabla\times\left[\nabla\times\left[\ldots\nabla\times\right.\right.\right.}_{\infty\text{-times taking curl operator}}\mathbf{V}\left.\left.\left.\right]\right]\ldots\right] = ? \end{equation}

For any possible values of vector $\mathbf{V}$.

Answer 1

For a general vector field $\mathbf{V}$, this sequence need not converge. Consider for example $\mathbf{V} = (e^{x-y}, e^{x-y}, 0)$. We have $\nabla \times \mathbf{V} = \mathbf{W} = (0,0,2 e^{x-y})$, and $\nabla \times \mathbf{W} = -2 \mathbf{V}$, and the cycle then repeats with a factor of $-2$.

It's worth noting that this operation is "unnatural" from the point of view of differential forms; curl is really the $d$ operator taking 1-forms to 2-forms; it's just that in $\mathbb{R}^3$, both 1-forms and 2-forms can be viewed as vector fields. But the result of the curl operator is a 2-form, which is not something that it makes sense to take the curl of. The other relations you cite in your comment don't suffer from this problem: grad takes 0-forms to 1-forms (so curl grad makes sense) and div takes 2-forms to 3-forms (so div curl makes sense), and they are both special cases of the fact that $d^2 = 0$.

Answer 2

Two applications of $\nabla$ yield $\nabla \times (\nabla \times F) = -\nabla^2 F + \nabla(\nabla \cdot F)$. Why? Well, setting $F = \sum_i F_i e_i$ where $e_i$ is the standard cartesian frame of $\mathbb{R}^3$ allows the formula: $$ (\nabla \times F)_k = \sum_{ij} \epsilon_{ijk} \partial_i F_j $$ Curling once more, $$ [\nabla \times (\nabla \times F)]_m = \sum_{kl}\epsilon_{klm}\partial_k\sum_{ij} \epsilon_{ijl} \partial_i F_j $$ But, the antismmetric symbol is constant and we can write this as $$ [\nabla \times (\nabla \times F)]_m = \sum_{ijkl}\epsilon_{klm}\epsilon_{ijl} \partial_k \partial_i F_j $$ A beautiful identity states:
$$ \sum_{l}\epsilon_{klm}\epsilon_{ijl} = -\sum_{l}\epsilon_{kml}\epsilon_{ijl} = -\delta_{ki}\delta_{mj}+\delta_{kj}\delta_{mi}$$ Hence, $$ [\nabla \times (\nabla \times F)]_m = \sum_{ijk}(-\delta_{ki}\delta_{mj}+\delta_{kj}\delta_{mi}) \partial_k \partial_i F_j = \sum_i [-\partial_i^2F_m+\partial_m(\partial_iF_i)]$$ and the claim follows since $m$ is arbitrary. Now, let's try for 3: $$ \nabla \times (\nabla \times (\nabla \times F)) = \nabla \times \bigl[-\nabla^2 F + \nabla(\nabla \cdot F)\bigr] =\nabla \times (-\nabla^2 F)$$ I used the curl of a gradient is zero. This need not be trivial, take $F = <x^2z,0,0>$ as an example. I suppose I could have shot for a four-folded curl by doubly applying the identity. $$ \nabla \times (\nabla \times (\nabla \times (\nabla \times F))) =?$$ Set $G = -\nabla^2 F $ since we know the gradient term vanishes, $$ \nabla \times (\nabla \times G) = -\nabla^2 G + \nabla \cdot G = \nabla^2(\nabla^2 F)-\nabla [\nabla \cdot (\nabla^2 F)]$$ So, there's the four-folded curl. Well, I see no reason this terminates. I guess you can give it a name. I propose we call (ordered as the edit indicates) $ \nabla \times \nabla \times \cdots \times \nabla = \top$

Answer 3

The identity $$ \nabla \times (\nabla \times \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})-\nabla^2 \mathbf{A} $$ is standard (where $\nabla^2$ denotes the component-wise Laplacian). Applying it repeatedly to $\nabla \times \mathbf{A}$ and using the fact that $\nabla \times (\nabla f)$ vanishes, we can see inductively that $$ (\nabla \times)^{2n+1}\mathbf{A}=(-1)^n\nabla \times (\nabla^2)^n \mathbf{A} \, . $$

So, when there's any hope of your thing converging, it'll be because the iterations of $-\nabla^2$ converge component-wise. Roughly speaking, this will happen when the Fourier transforms of your components are supported within some appropriate sphere, but getting the details right could be tricky (especially in cases where it includes some piece of the sphere's boundary).

A little more precisely, the equivalent of $-\nabla^2$ in the frequency domain is multiplication by $4\pi^2|\xi|^2$, so if $\operatorname{supp} \hat f \subset \{|\xi|<1/(2\pi)\}$ everything will converge to $0$. If $\operatorname{supp} \hat f \subset \{|\xi|\leq 1/(2\pi)\}$ everything should still converge; if $\hat f$ is a distribution with positive mass on the boundary of the sphere it should converge to that, but the analysis is icky enough that I don't want to say anything too definite...

You then have to worry about the even iterates, which you can get by replacing $\mathbf{A}$ with $\nabla \times\mathbf{A}$ everywhere in the above identity; there's no real reason why the even and odd iterates ought to have much to do with each other. On the other hand, in the nice case everything will converge to $0$ anyway, so this won't really matter.