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Let $n\geq 1$ and $v\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$. For $x_0\in\mathcal{O}$, let $\big(x(t)\big)_{t\geq 0}$ be the solution of

\begin{align*} & x(0)=x_0 \\ & \dot{x}=v(x). \end{align*}

Let $K\subset\mathbb{R}^n$ a non-empty compact set, and for $x_0\in K$ denote by $\tau^K(x_0)$ the exit time of $K$ by the trajectory $\big(x(t)\big)_{t\geq 0}$:

\begin{align*} \tau^K(x_0)=\inf\big\{t\geq 0,x(t)\notin K\big\} \end{align*}

Question: If for all $x_0\in K$, $\tau^K(x_0)<+\infty$, do we have $\sup_{x_0\in K}\tau^K(x_0)<+\infty$?

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For any $x_0\in K$ we have the trajectry $x(t)$ which go out of $K$ ,say, at time $t_0=\tau^K(x_0)+\varepsilon$. Hence, by the continuous dependence on its initial value, the solution starts from an neighborbood $U(x_0)$ of $x_0$ also goes out of $K$ at $t=t_0$. Covering the compact set $K$ by finite number of such $U(x_0)$'s you have the uniform bound of the exit time.

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  • $\begingroup$ This argument needs more precision. If the hypothesis would allow to vary the compact set $K$, this would be okay (if a trajectory leaves a neighbourhood of $K$ then continuity w.r.t. initial conditions can be applied). But a trajectory could maybe stay in the boundary of a fixed compact set $K$. $\endgroup$
    – Jochen Wengenroth
    Commented Dec 31, 2023 at 8:11
  • $\begingroup$ I may misunderstand your point, but his definition of exit time allows the trajectory actually leaves K, out of the boundary, within finite time. $\endgroup$
    – anything
    Commented Dec 31, 2023 at 17:30
  • $\begingroup$ Okay, I agree with your argument. $\endgroup$
    – Jochen Wengenroth
    Commented Jan 1 at 10:06

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