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.

  • $\begingroup$ Is $D(u)$ a scalar function? $\endgroup$ – Gabriele Cassese Feb 11 at 23:27
  • $\begingroup$ @gabrielecassese Yes. $\endgroup$ – sequence Feb 11 at 23:31
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    $\begingroup$ Then I would say that, since you are taking the gradient of a vector function, you result should be a second rank tensor $\endgroup$ – Gabriele Cassese Feb 11 at 23:33
  • $\begingroup$ To be clearer: you should get something like an Hessian, is that what you mean with $\nabla^2u$? It seemed to me you assumed is as the scalar laplacian instead $\endgroup$ – Gabriele Cassese Feb 11 at 23:36
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    $\begingroup$ The diffusion term should be $$\nabla \cdot (D(u) \nabla u)$$ $\endgroup$ – rafa11111 Feb 11 at 23:47

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.

  • $\begingroup$ Thank you, this makes sense now. I think the divergence was meant, not the gradient of a vector. So what it was, then, is an abuse of notation. $\endgroup$ – sequence Feb 12 at 3:13

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