Consider $u$ a function in $\mathbb R^n$ with $div u = 0$, i.e. divergence free, then why is $(u \cdot \nabla)u = \nabla \cdot (uu)$?

Question

Consider $u$ a function in $\mathbb R^n$ with $div u = 0$, i.e. divergence free, then why is $(u \cdot \nabla)u = \nabla \cdot (uu)$?

My understanding is to use integration by parts and the above is actually $u_i \partial_i u_j = \partial_i u_i u_j$,but I can't see why it's true explicitly.

Answer 1

We have the 'product rule' $$\let\del\partial ∇· (u\otimes v)=\del_i (u_i v_j) = (\del_iu_i)v_j + u_i\del_iv_j = (∇· u)v + (u·∇)v $$ If $u=v$ and $∇ · u = 0$ then the wanted equality holds.