# Inverse of the curl

Given $\nabla \cdot u = 0$ and $w = \nabla \times u$ equation 9 of http://projecteuclid.org/euclid.cmp/1103941230 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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@QED: it would be helpful if you told us where you got this from... – Mariano Suárez-Alvarez♦ Sep 27 '11 at 2:39
QED is working through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer. – Will Jagy Sep 27 '11 at 4:14
@Will: either your knowledge of the literature is simpy outstanding or you have great changes of getting the site's Mind-Reader of the Year Award. :) – Mariano Suárez-Alvarez♦ Sep 27 '11 at 4:21
Mariano, I am a pretty good mind-reader, but my first indication was the OP putting a comment with a link to the reference under my answer. – Will Jagy Sep 27 '11 at 4:29
Oh. Well, you'll have to wait a bit more for the award, then! – Mariano Suárez-Alvarez♦ Sep 27 '11 at 4:41
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## 2 Answers

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.

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– user16697 Sep 27 '11 at 3:11

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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QED is working his way through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer. Yesterday's question was about equation (7), today equation (9) – Will Jagy Sep 27 '11 at 3:37

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So presumably this was a misprint and they meant $(\nabla^2)^{-1} w$, i.e. some vector field $f$ such that $\nabla^2 f = w$, instead of $\nabla^{-1} w$. This is nonunique because you could add any vector field whose components are harmonic functions. – Robert Israel Sep 28 '11 at 22:21
@Robert, That could be it but if so how do I prove $u = - \nabla \times f$? I tried to show the divergence and curl of $- \nabla \times f$ match that of $u$ but could only do divergence.. and I'm not sure if both those are enough to prove it. – user16697 Sep 28 '11 at 22:47
– user16697 Sep 28 '11 at 22:59

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– Hans Lundmark Sep 29 '11 at 8:57
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