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Motivated by this question, and my own related research interest, I'm thinking to see under what conditions/hypotheses, the initial point in the Cauchy-Picard-Lindelöf theorem lies on the boundary point of the open set where the solution is supposed to lie in for a short time. The precise details of what I'm seeking are:

Consider the initial value problem (IVP): $$\frac{dx}{dt} = f(x(t),t), \\ x(0) = x_0. $$

Let $U \subset \mathbb{R}^n$ be an open set. Let $x_0\in \partial{U}.$ ADD EXTRA CONDITIONS HERE. Let $ f: U \times [0,T] \to \mathbb{R}^n $ be a continuous function which satisfies the Lipschitz condition $$ |f(x_1,t) - f(x_2,t)| \leq M |x_1 - x_2|, \quad \forall (x_1,t), (x_2,t) \in U \times [0,T], $$ where $M$ is a given constant. Then for some positive $ \delta $, there is a unique solution $ x: [0,\delta] \to U $ of the above initial value problem.

The above theorem is wrong in full generality without extra conditions: just solve the geodesic flow equation $dx_i/dt=v_i; dv_i/dt=0, x_i(0)=0, v_i(0)=c_i \in \mathbb{R}, i=1,2. $ Take $U\subset \mathbb{R}^2$ to be the domain bounded by $y=x,y=x^2.$ Note that $0\in \partial{U}.$ The solutions are straight lines passing through the origin. But no straight line is contained inside $U$, since the boundary curves got the same tangents at $0.$ So this means that we may need more assumption on $U$ for the proposed theorem to hold. See the figure below.

enter image description here

To circumvent the previous issue, we can make $U$ to be locally convex at $x_0,$ i.e. $\exists \delta >0$ so that $U\cap B(x_0;\delta)$ is convex. But this seems to only prevent the straight lines emanating from $x_0$ going outside $U$ at a small enough time. But this doesn't seem to necessarily prevent the solutions to the above equation to leave $U$ at an arbitrary short time.

So my question is: is there a modified version of the above theorem where the initial point $x(0)=:x_0$ lies on the boundary of $U,$ yet the solution $x(t)$ stay within $U$ for $t\in [0,T].$

If yes, I'd love to see a reference(s). Thank you!

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    $\begingroup$ "... the above initial value problem." Where is it? I think you forgot the initial value problem. $\endgroup$ Commented Feb 4 at 14:43
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    $\begingroup$ You did not say what the "above initial value problem" is. The natural assumption would be that the vector of initial conditions $f(x(0),0)$ points points from $x(0)$ to the interior of $U$. $\endgroup$ Commented Feb 4 at 14:44
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    $\begingroup$ For example in dimension $1$, if your open set is $(0, \infty)$, your initial point is $0$ and your vector field $f=f(x)$ is such that $f(0)<0$, then your problem is ill-posed: your solution immediately exits the allowed open set. $\endgroup$ Commented Feb 4 at 16:02
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    $\begingroup$ I'm pretty sure the following version is true. (1) $\partial U$ is Lipschitz (2) $f$ extends continuously to $\bar{U}$ and that $f|_{\partial U}$ is transverse to the boundary and points inwards. (3) standard Picard-Lindelof assumptions. Then local existence holds. $\endgroup$ Commented Feb 4 at 16:42
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    $\begingroup$ Instead of another existece theorem, you should consider the notion of (positively) invariant set and corresponding sufficient/necessary conditions $\endgroup$ Commented Feb 5 at 0:54

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