3
$\begingroup$

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.

  • 3
    @QED: it would be helpful if you told us where you got this from...2011-09-27
  • 0
    QED is working through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer.2011-09-27
  • 0
    @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
  • 0
    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
  • 0
    Oh. Well, you'll have to wait a bit more for the award, then!2011-09-27
  • 0
    Mariano, if there is money involved, maybe we can talk about this.2011-09-27
  • 0
    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
  • 0
    @Will: so far, the award amounts to a stack of StackExchange stickers to be used at will.2011-09-28

2 Answers 2