Let $M$ be a smooth manifold.
Let $X:M \to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=\overline{\{p\in M: X_p\not=0\}}$. Suppose $K$ a is compact subset of $M$.
Let $p\notin K$, and let $\theta^p$ be the maximal integral curve of $X$ starting at $p$.
Why $\theta^p$ is costant?
I know that $\theta^p:U^p\to M$ is a smooth map, with $U^p$ open interval in $\mathbb{R}$, and for each $t \in U^p$ I have $(\theta^p)'(t)=X_{\theta^p(t)}$.
The definition of $(\theta^p)'(t)$ is $(\theta^p)'(t):C^\infty(M)\to \mathbb{R}$ defined by $(\theta^p)'(t)(f)=\frac{d}{dt}(f\circ\theta^p)(t)$.
I know I have to use that $X_q=0$ for each $q\notin K$ but I can't see how.
For example, I don't know if $\theta^p(t)$ is in $K$ or not, so when I have $(\theta^p)'(t)=X_{\theta^p(t)}$ I don't know if $X_{\theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(\theta^p)'(t)$ is $0$ for each $t\in U^p$, then I don't know how to deduce from this that $\theta^p $ is in fact costant.
Since
$p \notin K, \tag 1$
$X(p) = 0; \tag 2$
let
$\theta^p(t) \tag 3$
be the unique solution of
${\theta^p}'(t) = X(\theta^p(t)) \tag 4$
such that
$\theta^p(0) = p; \tag 5$
we note that the constant curve
$\phi^p(t) = p, \; \forall t \in \Bbb R, \tag 6$
satisfies
${\phi^p}´(t) = 0 = X(p) = X(\phi^p(t)), \; \forall t\in \Bbb R; \tag 7$
we note that $\theta^p(t)$ and $\phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have
$\theta^p(t) = \phi^p(t) = p, \; \forall t \in \Bbb R. \tag 8$