Let $v\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$ $(n\geq 1)$ a velocity field such that every solution $(x\_t)\_{t\geq 0}$ of $(d/dt)x\_t=v(x\_t)$ is periodic. Denote, for a non-stationary point $x\in\mathbb{R}^n$ (meaning $v(x)\neq0$), by $T(x)$ the period of such a solution $(x\_t)\_{t\geq0}$ such that $x(0)=x$.


**Quetion:** Is $T$ continuous over the set of non-stationnary points?



Continuity of the period for a periodic dynamical system
The answer is trivially negative already for $n=3$: Start with the flow following along the "long" lines of a thickened (say with disc-shaped cut) Möbius strip. You can imagine this like a thick “rope” bent to a loop, making half a rotation on its way, and the flow follows the fibres of the rope (with a fixed speed).


Obviously, the fibre in the "center" of the rope has half of the period than the nearby fibres.


To extend the flow to $\mathbb R^3$, decrease the speed smoothly near the "boundary" of the rope to $0$, and then extend the field by letting it $0$ outside of the rope.


However, this extension works only, because the question explicitly admits stationary points. If one wants to exclude stationary points, there is a topological obstacle in the extension of this particular flow to $\mathbb R^3$. After an embedding into $\mathbb R^4$ this obstacle trivially vanishes, as the Möbius strip can be "unentangled" in $\mathbb R^4$, so in $\mathbb R^4$ there is obiously a non-singular counterexample.


I conjecture that for $n=3$ there is no non-singular counterexample at all, but I do not know how to prove this.


