What is the value of $\Delta(|u|^q)$?

Question

Suppose $u$ is a vector in $\mathbb R^n$ and for $q>2$, I know $\nabla (|u|^q)=q|u|^{q-2}u\nabla u$, then what is $\Delta(|u|^q)$?

Update: What I got is $|u|^{q-2}(\nabla \cdot u \nabla u + u \Delta u)$.

Answer 1

$\let\del\partial$ $$ \del_i (|u|^q) = \del_i ((u_j u_j)^{q/2}) = q (u_ju_j)^{q/2-1} \del_iu_ku_k = q|u|^{q-2}(u\nabla u)_i$$

where $(\nabla u)_{ij} = \del_j u_i$. Then $$ \del_i\del_i (|u|^q) = q\del_i(|u|^{q-2}(u\nabla u)_i) = q(\del_i(|u|^{q-2}))\del_iu_ku_k + q|u|^{q-2}\del_i((u\nabla u)_i) $$ and $$ \del_i(|u|^{q-2}) = (q-2)|u|^{q-4} (u\nabla u)_i,\quad \del_i((u\nabla u)_i) = \Delta u\cdot u + |\nabla u|^2$$ so $$ \del_i\del_i (|u|^q) = q(q-2)|u|^{q-4} (u\nabla u)_i(u\nabla u)_i + q|u|^{q-2}(\Delta u\cdot u + |\nabla u|^2) \\ = q|u|^{q-4} ( (q-2)|u\nabla u|^2 + |u|^2(\Delta u\cdot u + |\nabla u|^2)) $$