# Diffusion-reaction PDE gradient

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.

• Is $D(u)$ a scalar function? – Gabriele Cassese Feb 11 at 23:27
• @gabrielecassese Yes. – sequence Feb 11 at 23:31
• Then I would say that, since you are taking the gradient of a vector function, you result should be a second rank tensor – Gabriele Cassese Feb 11 at 23:33
• 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 – Gabriele Cassese Feb 11 at 23:36
• The diffusion term should be $$\nabla \cdot (D(u) \nabla u)$$ – rafa11111 Feb 11 at 23:47

## 1 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.

• 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. – sequence Feb 12 at 3:13