Whether a flow domain always admitted a flow?

Let $M$ be a non compact manifold, a flow domain $D\subset \mathbb R\times M$ is a open set such that its fibre at a point $D_p=\{t: (t,p)\in D\}$ is a open interval containing 0 (i.e. it's connected).

For every vector field, its maximal domain of the flow is a flow domain. I want to think about the converse. That is, given a flow domain, can we find a (smooth) vector field on $M$ such that the maximal domain of its flow is exactly $D$.

I need M noncompact just because it admits some flows that cannot be globally defined

Let $M = \mathbb R^2$, and let $D$ be the following flow domain: $$D = \{(t,p): |t|(1-|p|)^2<1\}.$$ Thus for each point $p$ on the unit circle, $D_p$ is all of $\mathbb R$, while for any other $p$, $D_p$ is the interval $(-1/(1-|p|)^2,\ +1/(1-|p|)^2)$.
Suppose $X$ is a vector field whose flow domain is $D$. The only points whose integral curves are defined for all time are those on the unit circle. If an integral curve starting anywhere intersects the unit circle at some time $t_0$, then it must be defined for all time, because the integral curve starting at $0$ is just a time-translation of the one starting at $t_0$. Therefore, integral curves starting in the interior of the unit disk must stay in the interior, because otherwise they'd have to cross the circle and thus be defined for all time.
• @yaoliding: Yes, something like that is the basic idea behind that example. You might think about whether there could be some assumptions on $D$ that would rule out such examples. – Jack Lee Nov 8 '17 at 20:08