Assume that $$\nabla H_1\times\nabla H_2 \:=\: V\quad\text{and}\quad V \:=\: \big(σy, x(r − z), xy\big)\,.$$ My question: If $V$ is given, is there any way to find out what $\,\nabla H_1\,$ and $\,\nabla H_2\,$ are?
The answer for the given $V$ should be $$ V\:=\:\big(σy, x(r − z), xy\big) \;=\;\underbrace{(0, y, z − r)}_{\qquad =\,\nabla H_1}\:\times\:\underbrace{(−x, 0, σ)}_{\qquad =\,\nabla H_2}\,. $$ From a mathematical point of view, it is impossible to solve since we have six variables and only three equations. But there is a paper in which it is mentioned that it can be decomposed. I am not getting any idea how this is possible.
I am attaching the link to the paper https://arxiv.org/abs/0910.3881 and refer to equation $(3.17)$ in it.
A bit of searching for the phrase “Clebsch–Monge potentials” (which they mention in the paper) revealed that it's a known but nontrivial fact that any (nice) divergence-free vector field in $\mathbf{R}^3$ can be written as the cross product of two gradients (at least locally, and away from points where the vector field is zero). I didn't find much material about this online, but there are proofs in some old vector analysis books (L. Brand, Vector and Tensor Analysis, §104, and R. Aris, Vectors, Tensors, and the Basic Equations of Fluid Mechanics, §3.43), and also a proof of a more general result in $\mathbf{R}^n$ in the paper Representation of divergence-free vector fields by C. Barbarosie.