2
$\begingroup$

I am interested in the following statement:

Let $F$ be a vector field in $\mathbb{R}^n$ that is $C^1$-smooth in a domain $U$, continuous up to the boundary $\partial U$, and vanishing on $\partial U$. Then $\int_U div(F)=0$.

Naively, if $U_n$ is a smooth domain approximating $U$ from inside and $|F|\leq 1/n$ on $\partial U_n$, then the regular divergence theorem gives $|\int_{\partial U_n} F\cdot N| \leq Area(\partial U_n)/n$ which might very well diverge. So it looks like one needs a finite perimeter condition. However, is there a known example of a construction in a case of non-finite perimeter where the integral of the divergence is not zero?

$\endgroup$
3
  • $\begingroup$ do you allow unbounded sets? $\endgroup$
    – username
    Nov 29, 2013 at 20:08
  • $\begingroup$ Sure, but $U$ bounded would be better. $\endgroup$ Nov 29, 2013 at 23:11
  • 1
    $\begingroup$ If you let $F(x,y)=(xy,0)$ and $U=\{(x,y):y>0\}$,then that would be an easy counterexample when $U$ is unbounded, but in that case I would want $F$ to "vanish at infinity" as well. $\endgroup$ Nov 30, 2013 at 5:06

1 Answer 1

3
$\begingroup$

I will show that if $U$ is open and bounded, $f \in C^1 (U) \cap C(\bar{U})$, $f=0$ on $\partial U$ and $v \in \mathbb{R}^n$, then $$ \int_U \partial_v f = 0, $$ where $\partial_v f$ is the directional derivative of $f$ in the direction $v$.

The proof goes as follows. First, the statement can be proved in the one-dimensional case by writing the open set $U$ as a countable collection of intervals, applying the fundamental theorem on each interval.

The higher dimensional case follows from the one-dimensional case and Fubini's theorem.

If $U$ is unbounded, the statement still holds if the directional derivative $\partial_v f$ is absolutely integrable: $\int_U \lvert \partial_v f\rvert < \infty$.

In fact it can be proved under these conditions the extension of $f$ to the whole space $\mathbb{R}^n$ by $0$ is weakly differentiable.

$\endgroup$
2
  • $\begingroup$ Yes. It would also work under the assumption that $f \in W^{1, 1}_0 (U)$, where $W^{1, 1}_0 (U)$ is the completion of $C^1$ functions with compact support with respect to the Sobolev norm $\int_U \vert \nabla u\vert$. $\endgroup$ Dec 11, 2013 at 11:16
  • $\begingroup$ Actually, this might not work after all. Because Fubini's Theorem might not hold for this $f$. $\endgroup$ Apr 2, 2014 at 20:34

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.