I am having some trouble with this question regarding vector diffiriential operators. It seems easy and I am not sure what I am missing.

The question:


$$ \mathbf{(u\cdot\nabla)u+u\times(\nabla\times u)=}\frac{1}{2}\nabla(|\mathbf{u}|^{2}). $$

My attempt:

\begin{align*} LHS= & \mathbf{(u\cdot\nabla)u+u\times(\nabla\times u)}\\ = & \mathbf{(u\cdot\nabla)u+\nabla(u\cdot u)-}\mathbf{u(u\cdot\nabla)}\text{ by vector triple product}\\ = & \mathbf{\nabla(u\cdot u)}\text{ and now I would like to write:}\\ = & \nabla(|\mathbf{u}|^{2}). \end{align*}

I think perhaps my cancellation on the second line is incorrect? I tried using index notation but since I am probably using an incorrect identity somewhere this just obscured what was going on.

Any help would be appreciated. Please try and use as few unproved identities as possible (unless they are really fundamental).


why don't you look at the $x$-component of the lhs? you have have $$\begin{align}\left( (u \cdot \nabla)u + u \times (\nabla \times u ) \right).i &= uu_x +vu_y + wu_z + v(v_x-u_y) -w(u_z-w_x)\\ &= uu_x+vv_x+ww_x\\ &= \nabla\left(\frac12\left(u^2 + v^2 + w^2\right) \right).i\\ &=\nabla\left(\frac12\lvert u \rvert^2 \right).i\end{align}$$

therefore $$(u \cdot \nabla)u + u \times \nabla \times u = \nabla\left(\frac12\lvert u \rvert^2 \right). $$

  • $\begingroup$ Hi, thanks for the answer. Could I ask you to clarify how the u X nabla X u turned into the last two terms on your first line? How is this consistent with the identity given by the other answerer? Thanks $\endgroup$ – user434180 Jun 14 '15 at 10:11
  • $\begingroup$ @user434180, because $\nabla \times u = (w_y -v_z, u_z - w_x, v_x - u_y)$ $\endgroup$ – abel Jun 14 '15 at 10:38
  • $\begingroup$ Yes thanks I'm with you now. $\endgroup$ – user434180 Jun 14 '15 at 13:35

Hint: Use the identity: $$ \nabla(\vec u \cdot \vec v)=\vec u \times (\nabla \times \vec v)+\vec v \times (\nabla \times \vec u)+(\vec u \cdot \nabla ) \vec v +(\vec v \cdot \nabla ) \vec u $$ with $\vec u =\vec v$

  • $\begingroup$ Could you give me some hints as how to prove this? Where do the first two terms come from? I can see that the last two are the standard sort of product rule. Thanks $\endgroup$ – user434180 Jun 14 '15 at 10:03
  • $\begingroup$ loock at: math.stackexchange.com/questions/170199/…. $\endgroup$ – Emilio Novati Jun 14 '15 at 12:31

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