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Consider two smooth vector fields $v$ and $u$ in $\mathbb{R}^n$, and a smooth set $\Omega$. Consider the flow of $\Omega$ via $v$ and $u$ for a time $T$, namely let

$$ \Omega_v =\{x(T, x_0) | x \text{ is the flow of $x_0 \in \Omega$ via v} \}$$

and

$$ \Omega_u =\{x(T, x_0) | x \text{ is the flow of $x_0 \in \Omega$ via u} \} .$$

My question is: intuitively, if $v$ is close to $u$, then $\Omega_u \approx \Omega_v$. So, is there a way to quantify the measure of the symmetric difference of $\Omega_u$ and $\Omega_v$ in terms of some norm of the difference of $u$ and $v$? Namely:

$$ |\Omega_v \Delta \Omega_v| \lesssim ||u- v|| ?$$

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Not a full answer but too long for a comment.

I'd first bound the Hausdorff distance between the two (eg through Gronwall as suggested in an answer), and then bound the volume of the symmetric difference using the Hausdorff estimate and regularity conditions on the input domain.

For the second step you might wanna assume that the domain has positive reach as well as its complement, and then argue using the tube formula.

The two steps (Hausdorff and volume) seem pretty much independant from each other in terms of arguments.

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  • $\begingroup$ Could you elaborate a little? In particular bounding the volume using the Hausdorff distance and the argument using the tube formula and the reach (never heard of neither of them to be honest) $\endgroup$
    – tommy1996q
    Commented Jul 2 at 16:57
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    $\begingroup$ Just to have my understanding verified, not because I think I can explain @alesia's nice idea better: the measure of all points which have distance less than $\epsilon$ to $\Omega_u$ (but are not included in $\Omega_u$) is $\lesssim \epsilon \vert \partial \Omega_u \vert \lesssim \epsilon \vert \partial \Omega \vert $. Therefore the measure of all points which are in $\Omega_v$ but not in $\Omega_u$ is $\lesssim d_H(\Omega_u, \Omega_v) \vert \partial \Omega \vert$. By symmetry, $\vert \Omega_u \Delta \Omega_v \vert \lesssim d_H(\Omega_u, \Omega_v) \vert \partial \Omega \vert$. $\endgroup$
    – unwissen
    Commented Jul 2 at 17:43
  • $\begingroup$ yes the idea is essentially as @unwissen describes. The tube formula provides an exact formula refining the above area based first order expansion, which is useful if you want an upper bound and not a first order behavior $\endgroup$
    – alesia
    Commented Jul 2 at 21:21
  • $\begingroup$ also, if you don't want to assume regularity, you can bound the volume of tubular neighborhoods using covering numbers of the boundary (consider union of balls) $\endgroup$
    – alesia
    Commented Jul 2 at 21:37

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