Question

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.

Answer 1

First a heuristic argument that either $v$ must be a very bad vector field, or $U$ cannot contain $\gamma$ strictly in its interior.

Fix a coordinate system on $U$. Suppose the vector field $v$ were Lipschitz continuous in the set $U$. Then as a consequence of the usual existence and uniqueness theorem of ODEs, you have that $|\gamma_1(t) - \gamma(t)| \leq e^{Kt}|\gamma_1(0) - \gamma(0)|$ for two integral curves, where $K$ is essentially determined by the Lipschitz constant of $v$. So if we assume that $U$ contains an $\epsilon$ neighborhood of $\gamma$, then for any $t_0$ we can pick $\gamma_1(0)$ sufficiently close to $\gamma$ such that for $t\in [0,t_0]$ $\gamma_1(t)\in U$. In particular, a bound of the type you want is impossible. This is completely independent of any statement about existence of closed orbits.

On the other hand, if you do not pre-suppose that $v$ is Lipschitz, you may in general lose uniqueness of integral curves, so the flow "$\phi^t$"may not be well defined.

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Now, since $\gamma$ is closed, consider the local $\epsilon$ neighborhood of $Im(\gamma)$ with axial coordinates along $\gamma$ and cross-sectional coordinates $\in B^{n-1}_\epsilon$, the $n-1$ dimensional ball. Now suppose $\gamma(0) = \gamma(T)$. Take $U$ to be something like $Im(\gamma) \cup Im(\gamma([0,T-1]))\times B^{n-1}_\epsilon$, then for sufficiently small $\epsilon$, using the existence and uniqueness theorem again, you can show that any $\gamma_1\neq\gamma$ must leave $U$ in finite time: that assuming that $T_0$ minimizes $|\gamma(T_0) - \gamma_1(0)|$, then at time $T- T_0 -1/2$, $\gamma_1$ must be (assuming $\epsilon$ sufficiently small) within $1/4$ of $\gamma(T-1/2)$, but there are no such points in $U$ other than $\gamma$.