# $\nabla({\bf u}^2)=2({\bf u}\cdot \nabla){\bf u} - 2(\nabla \times {\bf u}) \times {\bf u}$

In (2.13), he used: $$\nabla({\bf u}^2)=2({\bf u}\cdot \nabla){\bf u} - 2(\nabla \times {\bf u}) \times {\bf u}.$$

Any help appreciated!

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See e.g. Wikipedia. –  Qmechanic Mar 9 '13 at 13:46
Look at pg 68 of Panasatasiou et al.This is a fluid mechanics book available for free. –  drN Mar 9 '13 at 14:17

This is more like a maths question to me. This is just an identity, which is true and facilitates the calculation and it is valid for any vector field.

The proof, using Einstein summation convention would be something like:

$$(\nabla \times \vec u )\times \vec u = \epsilon_{ijk}(\nabla \times u)_j u_k = \\ \epsilon_{ijk}\epsilon_{jlm}\partial_l (u_m) u_k = \epsilon_{jki}\epsilon_{jlm}\partial_l (u_m) u_k = \\ (\delta_{kl}\delta_{im} - \delta_{km}\delta_{il}) \partial_l (u_m) u_k = \\ \partial_k (u_i) u_k - \partial_i (u_k) u_k = (\vec u \cdot \vec \nabla) \vec u - \frac 1 2 \nabla(\vec u \cdot \vec u)$$

I hope that helps.

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