Let $W$ be an open subset of $\mathbb R^n$ and $f: W \to \mathbb R^n$ a locally Lipschitz field.
Suppose I know for some reason (say I have one or more first integrals) that a maximal solution $x$ of the ODE $\dot x = f(x)$, $x(t_0) = x_0$ stays inside the set $M \subseteq W$. Furthermore, suppose I know that $M$ is diffeomorphic to $S^1$ and $f \neq 0$ on all of $M$.
Is this enough to tell that $x$ is a peroidic solution defined for all times with image $M$ and period \begin{align*} p = \int_M \lVert f\rVert^{-1}\,\quad ? \end{align*}
Here is a sketch of my attempt to prove it:
I do not know a lot about differentiable manifolds. So $M \cong S^1$ means for me that there is some diffeomorphism $\varphi^{-1}: \mathbb R^n \to \mathbb R^n$ which sends $M$ bijectively to the circle $S^1$ embedded in the $x_1$-$x_2$-plane. Then $y = \varphi^{-1}\circ x$ is a maximal solution of the ODE $\dot y = (\varphi^*f)(y)$, $y(t_0) = \varphi^{-1} (x_0)$, where $\varphi^*f$ is the pullback of $f$ under $\varphi$. That is \begin{align*} (\varphi^*f) (y) = (\varphi'(y))^{-1} f(\varphi(y))\,. \end{align*} It suffices to show that $y$ is surjective and periodic and has period \begin{align*} p = \int _{S^1} \lVert \varphi^*f \rVert^{-1}\, . \end{align*} Choose three points $a_0 = \varphi^{-1}(x_0)$, $a_1$ and $a_2$ on$S_1$ and three diffeomorphisms $\psi_j^{-1}$ which map the circle segment between $a_j$ and $a_{j+1}$ respectively to a line segment on the $x_1$-axis. The transformations are sketched in the picture. This effectively reduces the roblem to a one dimensional one. The result should follow from the one-dimensional version:
Suppose $g: J \to \mathbb R$, $J$ open in $\mathbb R$, is locally Lipschitz and $[a,b] \subseteq J$ and $g > 0$ on $[a,b]$ and $x$ is a maximal solution of the ODE $\dot x = g(x)$, $x(0) = a$. Then $x$ is strictly monotone increasing, $[a,b] \subseteq im(x)$ and the unique $t$ such that $x(t) = b$ is given by
\begin{align*}
t = \int_0^b 1/g(s)\, ds\,.
\end{align*}
The one-dimensional version can be shown with help of the inverse function theorem. 