I want to solve this problem :
M is a manifold. Let $t\mapsto \gamma(t)$ be an integral curve of a vector field X on M. Suppose there exists $t_0$ such that $\gamma'(t_0)=0$. Prove that $\gamma(t)=\gamma(t_0)$ for all t.
I know that we have $X(\gamma(t))=\gamma'(t)$ and that we have to use uniqueness of a solution but i have some difficulties writing a "technical" solution. Thank you for any help.
If $t \mapsto \gamma(t)$ is an integral curve of a vector field $X$ on a manifold $M$ and we impose the initial conditions $$ \begin{cases} \gamma(t_0) = p,\\ \gamma'(t_0) = 0, \end{cases} $$ for some time $t_0$, then by definition $$ X(\gamma(t_0)) = \gamma'(t_0) = 0. $$ That is, $X(p) = 0$. Thus, $t \mapsto \zeta(t) \equiv p$ is also an integral curve of $X$ satisfying $\zeta(t_0) = p =\gamma(t_0)$. By uniqueness of the integral curve of $X$ passing through $p$ at the time $t_0$, we conclude that $\gamma(t) = \zeta(t) \equiv p $ for all times $t$.