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Following the steps of derivation, everything is clear just for one small argument which is:

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Why is the divergence of the transpose of gradient equal to gradient of the divergence, and why does it vanish? By the way, $u$ is the velocity vector field.

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Using index notation (Einstein convention is used)

$$ div(u) = \partial_k u^k $$

$$ grad(u) = \partial_a u^b $$

Therefore, the statement in the question is just the fact that partial derivatives commute:

$$ \partial_j \partial_i u^j = \partial_i \partial_j u^j $$

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  • $\begingroup$ Thanks, but what about the transpose and vanishing of the divergence of $u$ ? $\endgroup$
    – user134613
    Commented Feb 10, 2022 at 13:43
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    $\begingroup$ the transpose is just a "useless" symbol used to recreate the "row-column" product. Once you go to index notation there is no need for it. In your case, the result is zero because your are considering an incompressible fluid (it is written in the first line of the picture you posted) en.wikipedia.org/wiki/Incompressible_flow $\endgroup$
    – Quillo
    Commented Feb 10, 2022 at 13:55
  • $\begingroup$ Thanks! I understand now $\endgroup$
    – user134613
    Commented Feb 10, 2022 at 14:28
  • $\begingroup$ see also this question, it may help you to understand the formal aspects of Navier Stokes and index notation: physics.stackexchange.com/q/142588/226902 $\endgroup$
    – Quillo
    Commented Feb 10, 2022 at 16:53

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