1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is $D(u)$ a scalar function? $\endgroup$ – Gabriele Cassese Feb 11 at 23:27
  • $\begingroup$ @gabrielecassese Yes. $\endgroup$ – sequence Feb 11 at 23:31
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ The diffusion term should be $$\nabla \cdot (D(u) \nabla u)$$ $\endgroup$ – rafa11111 Feb 11 at 23:47
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.