The other day, I was helping some friend to study maths for an exam when we came across with this exercise:

Let $\Omega\subset\mathbb R^3$ be a connected bounded subset with differentiable boundary $\partial\Omega$. If the divergence of $F:\mathbb R^3\rightarrow \mathbb R^3$ is zero in $\Omega$, then

$$ \int_\Omega F^TJ^T F dx = 0 , $$

where $J$ means the jacobian of $F$ and $J^T$ the transpose matrix of $J$.

The exercise also suggests use integration by parts.

To be honest, I have never seen some identity like such a one. I have looked at the classical divergence theorem, but I don't know how to apply it here. Also it seems that the identities for the curl and the divergence can't work here.

We have develop the integral expression, which can be written compactly as follows:

$$F^TJ^TF= \sum_i F_i^2\frac{\partial F_i}{\partial x_i} + \sum_{i,j\\i<j} F_iF_j\left(\frac{\partial F_i}{\partial x_j} + \frac{\partial F_j}{\partial x_i}\right) $$

The first sum is ($1$ over $3$ times) the divergence of the vector field $F^3=(F_1^3,F_2^3,F_3^3)$, so in this case we can use the divergence theorem. But I can't see how to use the fact that the divergence of $F$ is zero (clearly it doesn't imply that the divergence of $F^3$ is also $0$).

May you give us some clue please? To be honest I'm very lost with that (and it is supposed I'm the savant of both...)

PD: There is no tag for homework or something similar :(

  • $\begingroup$ What are $J$ and "$J^T$"? $\endgroup$
    – Mark Viola
    May 15, 2018 at 16:29
  • $\begingroup$ The jacobian of $F$ and its transpose. Sorry. I add it. $\endgroup$
    – Dog_69
    May 15, 2018 at 16:39
  • $\begingroup$ I thought that it was the case. $\endgroup$
    – Mark Viola
    May 15, 2018 at 16:41
  • $\begingroup$ Yes but, anyway. It is worth writing the definition just in case. $\endgroup$
    – Dog_69
    May 15, 2018 at 16:43
  • $\begingroup$ Does $\vec F\cdot \hat n$ vanish on $\partial \Omega$? $\endgroup$
    – Mark Viola
    May 15, 2018 at 16:57

1 Answer 1


First, we note that

$$\require\cancel \begin{align} \vec F\cdot \overleftrightarrow J^T\cdot\vec F&=\sum_{i=1}^3\sum_{j=1}^3F_i\frac{\partial F_j}{\partial x_i}F_j\\\\ &=\frac12 \vec (F\cdot \nabla)(\vec F\cdot \vec F)\\\\ &=\frac12 \nabla \cdot (\vec F(\vec F\cdot \vec F))-\frac12 (\vec F\cdot \vec F) \color{red}{\cancelto{=0}{\nabla \cdot \vec F}}\tag1 \end{align}$$

Using $(1)$ reveals

$$\begin{align} \int_\Omega \vec F\cdot \overleftrightarrow J^T\cdot\vec F\,dV=\int_\Omega \frac12 \nabla \cdot (\vec F(\vec F\cdot \vec F))\,dV\tag 2 \end{align}$$

Applying the Divergence Theorem to the right-hand side of $(2)$, we find that

$$\int_\Omega \vec F\cdot \overleftrightarrow J^T\cdot\vec F\,dV=\frac12\oint_{\partial \Omega}(\vec F\cdot\vec F)(\hat n\cdot \vec F)\,dS$$

If $(\hat n\cdot \vec F)=0$ on $\partial \Omega$, then we have

$$\int_\Omega \vec F\cdot \overleftrightarrow J^T\cdot\vec F\,dV=0$$

as was to be shown!

  • $\begingroup$ Fantastic. I can't say anymore. The key of the problem is (1). I don't realize of that relation. Brilliant. $\endgroup$
    – Dog_69
    May 15, 2018 at 17:58
  • $\begingroup$ @Dog_69 Thank you. It was a pleasure. We do need to assume that $\hat n\cdot \vec F$ vanishes on the boundary $\partial \Omega$. $\endgroup$
    – Mark Viola
    May 15, 2018 at 17:59
  • $\begingroup$ I'm convince it does it. The boundary is C^1$ and F is free-divergence. I mean, things should be simply. $\endgroup$
    – Dog_69
    May 15, 2018 at 18:08
  • $\begingroup$ Suppose $\vec F = \hat xx-\hat yy$ on the box $[-1,1]\times[-1,1]\times[-1,1]$. Then, clearly $\nabla \cdot \vec F=0$ and of course $\int_{\Omega}\hat n\cdot\vec F\,dS=0$. But, $\hat n\cdot \vec F$ does not vanish on $\partial \Omega$. But we do have $\oint_{\partial \Omega }(\vec F \cdot\vec F)(\hat n\cdot\vec F)\,dS=0$ also. $\endgroup$
    – Mark Viola
    May 15, 2018 at 18:17
  • $\begingroup$ Oh, nice counterexample. However the boundary of the box isn't $C^1$ on the edges. On the other hand, it is $C^1$ almost everywhere... Hummm, I don't know. $\endgroup$
    – Dog_69
    May 15, 2018 at 18:25

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