Is $\vec{F}\cdot \hat{n}$ zero on the boundary?

Question

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 $C^1$.

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 $$ F\cdot n=\left(y,x,0\right)\cdot\left(x,y,z\right)=2xy\not\equiv 0. $$

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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}