Let $\mathcal{O}$ be a bounded open subset of $\mathbb{R}^n$ $(n\geq 1)$ and $v\in\mathcal{C}^1(\mathcal{O},\mathbb{R}^n)$. 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 $V\subset\mathcal{O}$ a submanifold of dimension $n-1$, such that $x\_0\notin V$ and, for all $x\in V$, $v(x)\notin T\_x(V)$ (where $T\_x(V)$ is the tangent space at $x$ of $V$). Denote by $\tau^V(x\_0)$ the reaching time of $V$ by the trajectory $\big(x(t)\big)\_{t\geq 0}$:


\begin{align\*}
\tau^V(x\_0)=\inf\big\{t\geq 0,\,x(t)\in V\big\}
\end{align\*}


**Question:** If $\tau^V(x\_0)<+\infty$, is $\tau^V$ continuous on a neighborhood of $x\_0$?



Continuity of a reaching time of a submanifold
If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x\_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).


To begin with, let $t\_0=\tau^V(x\_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t\_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t\_0))\not\in T\_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t\_0-\varepsilon,t\_0)$ and $x^1(x(t))>0$ for $t\in(t\_0,t\_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $W\times(t\_0-\delta,t\_0+\delta)$ of $(x\_0,t\_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in W\times(t\_0-\delta,t\_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t\_0+\delta))>0$ and $x^1(x(t\_0-\delta))<0$, shrinking the neighborhood $W$ if necessary we get that $x^1(y(t\_0+\delta))>0$ and $x^1(y(t\_0-\delta))<0$ for all $y\in W$. So by the intermediate value theorem, for every $y\in W$ there is some $s\in(t\_0-\delta,t\_0+\delta)$ such that $y(s)\in W$.


As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)<t\_0+\delta\;\forall y\in W$. (We have still not used that $V$ is closed)


Now we just need to prove that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)>t\_0-\delta\;\forall y\in W$. But if this was false for some $\delta$, there would be a sequence of points $p\_n\to x$ and a sequence of times $t\_n<t\_0-\delta$ such that $p\_n(t\_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq t\_0-\delta$ of the sequence $t\_n$ we have $x(t)\in V$, a contradiction.


