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Let $$f(x, y, z) = \langle f_1(x,y,z), f_2(x,y,z), f_3(x,y,z)\rangle$$ be sufficiently differentiable and $\nabla \cdot f = 0$. Can anything be said about the expression $$(\nabla \times f) \cdot \nabla (f_i) - f \cdot \nabla \big((\nabla \times f)_i\big)$$ for $i = 1, 2, 3$?

I have tried computing this through brute force but it doesn't seem to simplify at all.

CBBAM
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  • any context you can provide? – Calvin Khor Jul 14 '23 at 04:44
  • @CalvinKhor This is part of a larger problem that I have been working on (not homework related). I have a suspicion that the above should simplify to zero, but I am having trouble proving it (if it is even true). To keep the question focused I have omitted as many unnecessary details as possible. Should I add anything more to the original post? – CBBAM Jul 14 '23 at 04:48
  • The standard method to show such an equation is to mechanically write out all differential operators. Index notation and a Levi-Civita symbol might help. There are plenty of other but similar examples here on MSE. – Kurt G. Jul 14 '23 at 07:24
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    I would assume that the terms do not cancel (product of 1st order derivs minus f times 2nd order derivs...) so for them to be zero you would need the individual terms to be zero. And the first term is not zero for arbitrary divergence free fields f (otherwise, it would not appear as a term in the vorticity formulation of 3D Euler/Navier--Stokes). So if instead you only want an inequality to be true, instead of perfect cancellation, that would be good to add to your post. – Calvin Khor Jul 14 '23 at 09:55
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    The linear shear flow $f(x,y,z) = (-y,x,0)$ is a special solution. This is because (A) $\nabla f_i$ has no $z$-component while $\nabla\times f$ is pure $z$-component, and (B) linear functions have zero second derivatives – Calvin Khor Jul 14 '23 at 11:04
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    @CalvinKhor Thank you for your comments! – CBBAM Jul 14 '23 at 19:19

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That equation is not true. Consider \begin{align} f=\begin{pmatrix}-y^2\\x^2\\0\end{pmatrix}\,. \end{align} This satisfies $\nabla\cdot f=0\,.$ Clearly, \begin{align} \nabla\times f&=\begin{pmatrix}0\\0\\2x+2y\end{pmatrix}\,,&\nabla f&=\begin{pmatrix}0&-2y&0\\2x&0&0\\0&0&0\end{pmatrix}\,,&\nabla(\nabla\times f)=\begin{pmatrix}0&0&0\\0&0&0\\2&2&0\end{pmatrix}\,. \end{align} The two terms in your equation are

\begin{align} (\nabla\times f)^\top(\nabla f)&=\begin{pmatrix}0\\0\\0 \end{pmatrix}\,,& f^\top(\nabla(\nabla\times f))&=\begin{pmatrix}0\\0\\\color{red}{2x^2} \end{pmatrix}\,. \end{align}

Kurt G.
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