# Vector Identity Question

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:

Prove:

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

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 $$(u \cdot \nabla)u + u \times \nabla \times u = \nabla\left(\frac12\lvert u \rvert^2 \right).$$
• @user434180, because $\nabla \times u = (w_y -v_z, u_z - w_x, v_x - u_y)$ – abel Jun 14 '15 at 10:38
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 $\vec u =\vec v$