Suppose $v$ is a vector field on a manifold $X$ with flow $\phi^t$. Suppose $v$ carries a first integral $f:X \rightarrow \mathbb{R}$ (i.e. $f$ is constant on the orbits of $v$). Suppose $\gamma:[0,T]\rightarrow X$ is a closed orbit of $v$, that is, $\gamma(x)=\phi^t(x)$ with $\phi^T(x)=x$. Suppose $f(\gamma)\equiv c$. Suppose $U \subset f^{-1}(c)$ is a compact neighborhood of $\mathrm{Im}(\gamma)$ in $f^{-1}(c)$ such that $U$ contains no other closed orbits of $v$ other than $\gamma$ (or it's translates).

What conditions are sufficient to impose on $v$ so that there exists some $t_* \ge 0$ such that no flow line of $v$ apart from $\gamma$ can live in $U$ for time longer than $t_*$. More precisely, if $\gamma_1$ is another flow line of $v$ such that $\mathrm{Im}(\gamma_1) \cap U \ne \emptyset$, and $[a,b] \subset \mathbb{R}$ is any interval such that $\gamma_1([a,b]) \subset U$ then $|b-a| \le t_*$.

**Edit**

(Thanks to Willie Wong)

Assume $X$ has dimension $\ge 3$. The question is perhaps better phrased as "are there well known conditions that always guarantee such a bound?".

**Edit**

(Thanks to Pietro Majer)

Sorry the question is badly phrased. I've tried to improve it. I'm most interested in the case where $v$ carries a first integral $f:X \rightarrow \mathbb{R}$ (i.e. $f$ is constant on the orbits of $v$), and we only care about orbits $\gamma_1$ such that $f(\gamma_1)=f(\gamma)$. I have thus rewritten the question. But the alternative question where the orbit $\gamma_1$ is required to have the same period as $\gamma$ is also interesting to me.

anyperiod, or just: contains no other closed of $v$ of the same period $T$ as $\gamma$? Thanks. – Pietro Majer Mar 7 '11 at 18:27