Proof of the vector Calculus identity $\vec{u}\times(\nabla\times\vec{u}) = \nabla(\frac{1}{2}\vert{\vec{u}}\vert^{2})-\vec{u}\cdot\nabla\vec{u}$

Question

So far I have,

$$=\varepsilon_{ijk}u_{j}\varepsilon_{klm}\partial_{l}u_{m}$$ $$=\varepsilon_{kij}\varepsilon_{klm}u_{j}\partial_{l}u_{m}$$ $$=\delta_{il}\delta_{jm}u_{j}\partial_{l}u_{m} - \delta_{im}\delta_{jl}u_{j}\partial_{l}u_{m}$$ After this step I am unsure of where to go, many thanks in advance.

Answer 1

You're almost there. Note that $\partial_i(|\vec{u}|^2) =\partial_i(u_j u_j)= 2u_j \partial_i u_j$. We also know that the Kronecker delta acts as $\delta_{ij} v_j = v_i$. So your last line reads

$$ u_m \partial_i u_m - u_j \partial_j u_i = \partial_i (|\vec{u}|^2/2) - u_j \partial_j u_i\,, $$ which is precisely the $i$th component of the right hand side.