Given $\nabla \cdot u = 0$ and $w = \nabla \times u$ equation 9 of has the identity $u = - \nabla \times (\nabla^{-1} w)$.
What is $\nabla^{-1}$ exactly and how was this identity derived?
Thanks a lot.
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$\begingroup$The fact that $u$ is divergence free does mean that $u$ is the curl of something, locally at least. The fact that we have, for some $v,$ that $u = \nabla \times v$ amounts to little more than the fact that mixed partial derivatives commute, and in general is called Poincare's Lemma. Furthermore, one can replace any such $v$ by $v + \nabla \cdot f$ for some function $f.$ As in the comments, there is therefore little reason to talk about an operator $\nabla^{-1},$ such a thing is not going to be well defined. If you want to give an exact reference and convince us that a responsible person wrote the material you are quoting, things might be different.
$\endgroup$ 1 $\begingroup$There is no such thing as $\nabla^{-1}$. What is true is that $\nabla \times (\nabla \times u) = \nabla(\nabla \cdot u) - \nabla^2 u$.
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