I've been analyzing one PDE and got tangled in notation. If in a diffusion-reaction PDE one has the term
$$\nabla(D(u)\nabla u),$$
where $u=u(x,y)$, does this mean that
$\nabla(D(u)\nabla u) = (\partial_x D(u)\nabla u, \partial_y D(u)\nabla u)=\left(\frac{dD}{du}\frac{\partial u}{\partial x}\nabla u+\nabla^2 u, \frac{dD}{du}\frac{\partial u}{\partial y}\nabla u+\nabla^2 u\right)$
$=(D'(u)\partial_x u, D'(u)\partial_y u) + \nabla^2 u(1,1)$ (which is a vector)?
Or is it $\nabla(D(u)\nabla u)=\nabla D(u) \cdot \nabla u + D(u) \nabla^2 u$ (which is a scalar)?
And, in the second expression, what is $\nabla D(u)$ exactly, isn't it just the derivative of $D$ w.r.t. $u$? I.e., is it just $(D'(u) \partial_x u, D'(u)\partial_y u)=D'(u)\nabla u$?
If the second expression is correct, then the term would come out to be
$$\nabla(D(u)\nabla u)=D'(u)\nabla u \cdot \nabla u + D(u)\Delta u=D'(u)\Delta u + D(u)\Delta u = (D'(u)+D(u))\Delta u$$
May seem like a lame question, but just wanted to make sure.
Best Answer
As rafa11111 noted in his comment, the usual diffusion term is the divergence, and not the gradient, of $D(u)\nabla u$. If that was what you meant, than the result is a scalar:
$\nabla\cdot(D(u)\nabla u)) =\\ =\nabla(D(u))\cdot \nabla (u)+D(u)\Delta(u)=\\ =D'(u)||\nabla(u)||^2+D(u)\Delta(u)$
If, instead, you really meant
$\nabla(D(u)\nabla u)$
the result is:
$\nabla(D(u))\otimes \nabla u+D(u)H(u)=\\D'(u)\nabla u\otimes \nabla u+ D(u)H(u)$
(Where $\otimes$ indicates dyadic product or tensor product and $H(u)$ is the Hessian matrix of $u$)
The big difference between the two possibilities is generated by the fact that, while $\text{div}$ lowers your tensor degree by one, $\text{grad}$ increases it by one.