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The following seemingly-simple problem came up when working on a problem in the fluid theory of plasmas.

Given a vector field $\mathbf{A}$, find a symmetric tensor $\mathbf{P}$ such that $\boldsymbol{\nabla}\times\mathbf{A} = \boldsymbol{\nabla}\cdot\mathbf{P}$.

This isn't very hard if you don't require $\mathbf{P}$ to be symmetric (e.g. $P_{ij} = \epsilon_{ijk}A_k$ works), but the symmetry requirement is proving harder than I thought. I tried this over on Math Stack Exchange and got no responses, so I'm hoping you guys might be able to help.

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More generally, you can consider a given vector field v, not necessarily a curl. To solve $\nabla\cdot P=v$, set $P=\nabla u+(\nabla u)^T$. This results in the equation $\Delta u+\nabla(\nabla\cdot u)=v$, which is an elliptic system for u.

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  • $\begingroup$ Actually I think that does it. $\nabla^2 \mathbf{u} = \boldsymbol \nabla( \boldsymbol \nabla\cdot\mathbf{u}) - \boldsymbol\nabla \times (\boldsymbol\nabla \times \mathbf{u}) $. If we assume $\boldsymbol \nabla\cdot \mathbf{u} = 0$, then $\mathbf{u}$ simply has to satisfy $\boldsymbol \nabla\times\mathbf{u} = - \mathbf{A}$, which is easily solved. $\endgroup$ – eyeballfrog Feb 22 '17 at 6:04

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