Let $S\subset\mathbb R^d$ be a subset. There can be done all the hypotheses one wants on the set $S$. But, at the moment, suppose that $S$ is closed and $\partial S$ is compact. Let us define
$$ t_x:=\inf\{t\geq0:\Phi(x, 0, t)\in \partial S\}, $$ where $\Phi(x, 0, t)$ is the flow solving the following ODE system $$ \begin{cases} \partial_s\Phi(x, s)=b(\Phi(x, s), s)\\ \Phi(x, 0)=x \end{cases}. $$ The field $b$ is assumed to be Lipschitz continuous in space-time and bounded. Now, for $s\in[0, T]$ ($T>0$ is fixed) let us define the set $$ \mathcal B(s):=\{x\in\mathbb R^d:t_x\leq s\} $$ which is such that $$ s_1\leq s_2\quad\Longrightarrow\quad\mathcal B(s_1)\subseteq\mathcal B(s_2).\qquad(\star) $$ I proved that:
- the map $s\longmapsto\mathcal L^d(\mathcal B(s))$, where $\mathcal L^d$ is the $d$-dimensional Lebesgue measure, is continuous. Moreover, thanks to $(\star)$ it is increasing too.
- denoted by $\rho$ the Hausdorff distance, $\rho(\mathcal B(s'), \mathcal B(s))\leq L|s'-s|$ for some $L>0$.
I want to prove that the map $s\longmapsto\mathcal L^d(\mathcal B(s))$ is Lipschitz continuous, that is there exists $L>0$ such that $$ \mathcal L^d(\mathcal B(s')\setminus\mathcal B(s))\leq L|s'-s|. $$
I expect that it is true but I cannot prove it. This would be true if $$ \mathcal L^d(\mathcal B(s')\setminus\mathcal B(s))\leq C\rho(\mathcal B(s'), \mathcal B(s))\ \ \text{for some}\ \ C>0 $$ but this is not true in general (think to two concentric balls for example). Some helps?
Thank You