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It seems to me that, in some derivations on fluid dynamic books I am reading, the identity $$\nabla \cdot (\nabla^2 u) = 0 $$ where $u$ is a vector field, is used.

Does this identity exist? Is it true?

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2 Answers 2

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Note: $\nabla^2 u= \nabla(\nabla \cdot u) - \nabla \times (\nabla \times u)$. Taking the divergence gives: $\nabla \cdot(\nabla^2 u)= \nabla^2 (\nabla \cdot u)$. The reason the statement holds in your case is likely that $u$ is assumed to be an incompressible flow field, i.e. $\nabla \cdot u=0$, which is the incompressible continuity equation. Else, it may not hold.

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    $\begingroup$ Why is the divergence of the RHS of the equation equals to that? Is it because "curl is always solenoidal"? $\endgroup$
    – toliveira
    Jan 6, 2016 at 1:07
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The definition of vector Laplacian is $$ \nabla^2 \mathbf{u} = \nabla(\nabla \cdot \mathbf{u}) - \nabla \times (\nabla \times \mathbf{u}) $$ Since curl is always solenoidal, the divergence of the second term is $\mathbf{0}$, so we are left with $$ \nabla \cdot (\nabla^2 \mathbf{u}) = \nabla^2(\nabla \cdot \mathbf{u}) $$ and it may not be $\mathbf{0}$.

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  • $\begingroup$ By "second term", you mean "div curl curl u", and not "curl curl u", right? $\endgroup$
    – toliveira
    Jan 6, 2016 at 1:06
  • $\begingroup$ Thanks, Henry W. The other answer was very similar and I can mark just one... :( $\endgroup$
    – toliveira
    Jan 6, 2016 at 1:22

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