Prove the vector calculus identity $\frac{1}{2}\mathbf{\nabla(\lVert u \rVert ^2) = (u \cdot \nabla)u+u \times (\nabla \times u )}$

Question

My attempt: Consider the first component of both sides. $$LHS=\frac{1}{2}\frac{\partial}{\partial x_1}(u_1^2+u_2^2+u_3^2)=u_1 \frac{\partial u_1}{\partial x_1}+u_2 \frac{\partial u_2}{\partial x_1} + u_3 \frac{\partial u_3}{\partial x_1}$$ $$RHS=u_1\left( \frac{\partial u_1}{\partial x_1}+\frac{\partial u_2}{\partial x_2}+\frac{\partial u_3}{\partial x_3}\right)+u_2(\partial_1u_2-\partial_2u_1)-u_3(\partial_3u_1-\partial_1u_3)$$

Then I got stuck as I couldn't cancel out some of the terms and don't know how to proceed. Any help will be appreciated!

Answer 1

I think you may have misinterpreted the notation $(\mathbf u . \nabla) \mathbf u$.

The first component of the RHS is actually this:

$$ \left(u_1 \partial_1 + u_2 \partial_2 + u_3 \partial_3 \right) u_1 + u_2 (\partial_1 u_2 - \partial_2 u_1) - u_3 (\partial_3 u_1 - \partial_1 u_3) $$

In other words, it is this:

$$ (u_1 \partial_1 u_1 + u_2 \partial_2 u_1 + u_3 \partial_3 u_1) + u_2 (\partial_1 u_2 - \partial_2 u_1) - u_3 (\partial_3 u_1 - \partial_1 u_3) $$

Answer 2

I'm going to use Einstein's notation. $$\begin{align} \left[(u \cdot \nabla)u+u \times(\nabla \times u)\right]_i &=u_j\partial_j u_i+\varepsilon_{ijk}u_j\varepsilon_{kpq}\partial_pu_q\\ &=u_j\partial_j u_i+\varepsilon_{kij}\varepsilon_{kpq}u_j\partial_pu_q\\ &=u_j\partial_j u_i+(\delta_{ip}\delta_{jq}-\delta_{iq}\delta_{jp})u_j\partial_pu_q\\ &=u_j\partial_j u_i+\delta_{ip}\delta_{jq}u_j\partial_pu_q-\delta_{iq}\delta_{jp}u_j\partial_pu_q\\ &=u_j\partial_j u_i+u_j\partial_i u_j-u_j\partial_ju_i\\ &=u_j\partial_i u_j\\ &=\frac{1}{2}\left(u_j\partial_i u_j+u_j\partial_i u_j\right)\\ &=\frac{1}{2}\left(\partial_i(u_j u_j)\right)\\ &=\frac{1}{2}\nabla(u \cdot u)_i \end{align}$$