# Divergence operator of higher order and intrinsic point of view

Let $\underline{u}$ be a $1$ - order tensor (say a column vector) I want to prove that :

$\underline{\operatorname{div}} \left( (\underline{\underline{\operatorname{grad}}} \, \underline{u})^T\right)= \underline{\operatorname{grad}} \, (\operatorname{div} \underline{u})$

where $\underline{\operatorname{grad}}$ is the one order gradient (the usual one) and $\underline{\underline{\operatorname{grad}}}$ is the second order gradient (it is the jacobian matrix)

I want a proof that those not involve any coordinates. Because it is easy to find a proof using for example cartesian coordinates.

Here is what I've done so far :

Since for any volume $V$ we have : $$\iiint _V \underline{\operatorname{div}} \left( (\underline{\underline{\operatorname{grad}}} \, \underline{u})^T\right) \; \mathrm{d}V = \iint_{S} (\underline{\underline{\operatorname{grad}}} \, \underline{u})^T \cdot \underline{n} \; \mathrm{d}S$$ (it it the definition of $\underline{\operatorname{div}}$) where $\underline{n}$ is the normal vector to the surface $S$ at the limit of the volume $V$.

And we can write : $$\iiint_V \underline{\operatorname{grad}} \, (\operatorname{div} \underline{u}) \; \mathrm{d}V = \iint_S (\operatorname{div} \underline{u} )\underline{n} \; \mathrm{d}S$$

Thus it is sufficient to prove that : $$\iint_{S} (\underline{\underline{\operatorname{grad}}} \, \underline{u})^T \cdot \underline{n} \; \mathrm{d}S =\iint_S (\operatorname{div} \underline{u} )\underline{n} \; \mathrm{d} S$$

The problem is that we don't have $(\underline{\underline{\operatorname{grad}}} \, \underline{u})^T \cdot \underline{n} = (\operatorname{div} \underline{u} )\underline{n}$

So how can I finish the proof ?

If you need details, please tell me.

Thank you.

• weird notation. what do you mean by underlined divergence? since you also use the non-underlined divergence operator there seems to be a difference... – Max Dec 7 '15 at 14:25
• Yes there is a difference, the notations are the one used in continuum mechanics (in France), the usual divergence is not underlined and is applicated on vectors (tensors of order $1$) but the underlined divergence is applicated on matrices (tensors of order $2$) and is the vector defined by the divergenc theorem (the first equality) – M.LTA Dec 7 '15 at 14:42
• Do you have a coordinate-free definition of the double-underlined grad and the underlined div? – Justpassingby Dec 18 '15 at 10:13
• @Justpassingby It is a really good question. A coordinate free definition of the underlined div is given by $\iiint_V \underline{\operatorname{div}} (\underline{u}) \, \mathrm{d}V = \iint_{S}\underline{u} \cdot \underline{n} \, \mathrm{d}S$ For the double underlined gradient you can see it as the differential of $\underline{u}$ seen as a linear map of $\mathbb{R}^3$ – M.LTA Dec 20 '15 at 19:29

This calculation is done component wise, so it isn't most elegant way to show the fact. I would like to see more general approach (This probably needs that the metric tensor is taken account. Or we could switch formalism to k-forms).

Let $A$ be matrix valued function. The i th component of $div(A)$ is given by

$$div(A)_i = \sum_j \partial_j A_{ij}$$

and the ij component of gradient of vector function $u$ is

$$grad(u)_{ij} = \partial_j u_i.$$

So the transpose of gradient is $$(grad(u)^{\top})_{ij} = \partial_i u_j.$$

Now, evaluating i th component the left hand side gives:

$$div(grad(u)^{\top})_i = \sum_j \partial_j \partial_i u_j = \partial_i( \sum_j \partial_j u_j ) = \partial_i(div(u)),$$ where we have assumed that $u$ is smooth enough. Now, the i th component of gradient of scalar function f is given by

$$grad(f)_i = \partial_i f.$$

And we notice that $\partial_i (div(u)) = grad(div(u))_i$ and thus

$$div(grad(u)^{\top})_i = grad(div(u))_i$$ so $$div(grad(u)^{\top}) = grad(div(u))$$.

• Well, I also proved it this way but this is component wise like you said so it does not really interest me, sorry. But still you wrote a proof... – M.LTA Dec 20 '15 at 19:23