I'm trying to do the following 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

$$ \int_\Omega F^TJ^TFdx=0, $$

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

I have asked a solution for it here and @MarkViola has proved that the integral is equal to

$$\int_{\partial\Omega} (F\cdot F)(F\cdot n) \, dS , $$

where $n$ is the normal vector to the surface defined by the boundary. Now, why is this integral also zero??? I can see that if $\operatorname{div}F=0$ on $\Omega$ then

$$\int_{\partial \Omega} F\cdot n \, dS = 0. $$

But, does it imply that $F\cdot n =0$ on the boundary? If not, why the integral

$$\int_{\partial\Omega} (F\cdot F)(F\cdot n) \, dS $$

is equal to zero?

Remark I should remark that @Mark also gave an example for which $F\cdot n$ is not zero on the boundary and the integral is. The only problem I see in that case is that the boundary is $C^1$ a.e. but not actually $C^1$.

  • $\begingroup$ Is $F\in C^1({\mit Ω})$ or $F\in C^1(\overline{\mit Ω})$? $\endgroup$
    – Ѕааԁ
    May 19, 2018 at 16:29
  • $\begingroup$ The statement doesn't say anything about that, so suppose what you prefer. The strogest condition. $\endgroup$
    – Dog_69
    May 19, 2018 at 16:38
  • $\begingroup$ @AlexFrancisco: In fact I would suppose $C^1(\mathbb R^3)$. $\endgroup$
    – Dog_69
    May 19, 2018 at 16:52
  • $\begingroup$ @ChristianBlatter: $J$ is the Jacobian of $F$ and $T$ denotes the transpose matrix. Far than that... Wll Mark Viola proved that $FJ^TF$ is equal to the material derivative of $\|F\|^2$: $FJ^TF=(F\cdot\nabla)(F\cdot F)$. $\endgroup$
    – Dog_69
    May 19, 2018 at 18:32
  • $\begingroup$ @ChristianBlatter: Sure, but they've got it. $F$ is a vector $(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z))$, where each $F_i$ is a function from $\mathbb R^3$ into $\mathbb R$. And $J$ is the Jacobian of $F$ and hence a matrix $3\times 3$. So $FJ^TF$ is a function from $\mathbb R^3$ into $\mathbb R$. Furthermore, it turns out to be the function given by $(F\cdot\nabla)(F\cdot F)$. $\endgroup$
    – Dog_69
    May 19, 2018 at 18:44

1 Answer 1


It is not necessary that $$ F\cdot n=0. $$ For example, you may consider

  • $F(x,y,z)=\left(y,x,0\right)$, and
  • $\Omega=\left\{x\in\mathbb{R}^3:\left\|x\right\|\le 1\right\}$.

In this case, $\partial\Omega$ is the unit sphere, with its outward unit normal being $$ n=\left(x,y,z\right). $$ Then on $\partial\Omega$, we have $$ F\cdot n=\left(y,x,0\right)\cdot\left(x,y,z\right)=2xy\not\equiv 0. $$

In general, it is not necessary that $$ \int_{\Omega}FJ^{\top}F{\rm d}V=0, $$ even if $F$ is divergence-free. For example, consider

  • $F(x,y,z)=\left(x,y,-2z\right)$, so that $F$ is divergence-free, and
  • $\Omega=\left\{x\in\mathbb{R}^3:\left\|x\right\|\le 1\right\}$.

In this case, $\partial\Omega$ is still the unit sphere, with its outward unit normal being $$ n=\left(x,y,z\right). $$ Then $\partial\Omega$, we have \begin{align} F\cdot F&=x^2+y^2+4z^2=1+3z^2,\\ F\cdot n&=x^2+y^2-2z^2=1-3z^2. \end{align}

We will make use of this standard spherical coordinate transform \begin{align} x&=r\sin\theta\cos\varphi,\\ y&=r\sin\theta\sin\varphi,\\ z&=r\cos\theta \end{align} with $r=1$ on $\partial\Omega$, $\theta\in\left[0,\pi\right)$, and $\varphi\in\left[0,2\pi\right)$. Then \begin{align} \int_{\Omega}FJ^{\top}F{\rm d}V&=\int_{\partial\Omega}\left(F\cdot F\right)\left(F\cdot n\right){\rm d}S\\ &=\int_{\partial\Omega}\left(1+3z^2\right)\left(1-3z^2\right){\rm d}S\\ &=\int_{\partial\Omega}\left(1-9z^4\right){\rm d}S\\ &=\int_0^{\pi}\sin\theta{\rm d}\theta\int_0^{2\pi}\left(1-9z^4\right){\rm d}\varphi\\ &=\int_0^{\pi}\sin\theta{\rm d}\theta\int_0^{2\pi}\left(1-9\cos^4\theta\right){\rm d}\varphi\\ &=\int_0^{\pi}\sin\theta\left(1-9\cos^4\theta\right){\rm d}\theta\int_0^{2\pi}{\rm d}\varphi\\ &=2\pi\int_0^{\pi}\left(9\cos^4\theta-1\right){\rm d}\cos\theta\\ &=-\frac{16\pi}{5}\\ &\ne 0. \end{align}

  • $\begingroup$ Yes, good point. But the integral is still zero. Why? There must be a reason.(In this case is because you integrate $\sin(2\phi)$ from $0$ to $2\pi$. but I mean in general) $\endgroup$
    – Dog_69
    May 19, 2018 at 16:51
  • $\begingroup$ @Dog_69: Excuse me. I misunderstood your question. Well, I am afraid that the expected integral $\int_{\Omega}FJ^{\top}F{\rm d}V$ is not always $0$, even if $F$ is divergence free. I just updated my answer, and you may check the second example I provided there. $\endgroup$
    – hypernova
    May 19, 2018 at 17:49
  • $\begingroup$ Oh my God. I can't believe it. This problem was an exercise demanded by a teacher. Is he wrong? Well, it seems... But... I don't know what to say. In the problem he says ''regular boundary'', but I have understood this as $C^1$ frontier. I'm going to check some notes. Anyway, tmillions of thanks for your counterexample. $\endgroup$
    – Dog_69
    May 19, 2018 at 17:55
  • $\begingroup$ @Dog_69: Never mind. Maybe your teacher has provided some additional conditions as well? Or maybe he just wanted you to check how to transform the volume integral into the surface integral using the divergence theorem? Anyway, good luck to you :-) $\endgroup$
    – hypernova
    May 19, 2018 at 17:57
  • $\begingroup$ I don't know... One question @hypernova: If I say you ''regular boundary'', what do you understand? $\endgroup$
    – Dog_69
    May 19, 2018 at 18:04

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.