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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...2011-09-27
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    QED is working through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer.2011-09-27
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    @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. :)2011-09-27
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    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.2011-09-27
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    Oh. Well, you'll have to wait a bit more for the award, then!2011-09-27
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    Mariano, if there is money involved, maybe we can talk about this.2011-09-27
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    Whatever they are doing, they must be using more than just $\nabla\cdot u=0$ and $\omega=\nabla\times u$. Consider for example $u=(x,y,-2z)$ and $u=(-2x,y,z)$. In both cases, $\nabla\cdot u=0$ and $\omega=(0,0,0)$, so $u$ can't be recovered from $\omega$.2011-09-27
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    @Will: so far, the award amounts to a stack of StackExchange stickers to be used at will.2011-09-28

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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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    It is http://projecteuclid.org/euclid.cmp/1103941230 equation 9 that I want to understand.2011-09-27
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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)2011-09-27
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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.2011-09-28
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    @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.2011-09-28
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    Actually using [these identities](http://en.wikipedia.org/wiki/Vector_calculus_identities) I proved that $- \nabla \times f$ has the same divergence and curl as $u$. Does this prove that $u = - \nabla \times f$?2011-09-28
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    @QED: You need some assumptions about decay at infinity. See http://math.stackexchange.com/questions/19412/is-any-divergence-free-curl-free-vector-field-necessarily-constant for links to more info.2011-09-29
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    @Hans, Wonderful! Thanks a lot!2011-09-29