Continuity of the period for a periodic dynamical system

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?

• Not an answer, but probably this indicates a negative response to your question. numdam.org/item/PMIHES_1976__46__5_0 In any case, it shows that your problem will depend on the fact that the vector field is defined in Rn with those properties. Jun 29 '21 at 17:32
• This seems implicit, but just to clarify: for this question, the vector field $v$ may have zeros (stationary points)? Jun 30 '21 at 12:04

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.

For $$n = 2$$, the answer is Yes. Topological considerations (as in the proof of the Poincare-Bendixson theorem) mean that the first-return map for a transversal to a point on a periodic orbit will be the identity locally, which implies continuity of $$T$$.

For sufficiently high $$n$$, the answer is No. The paper by Sullivan, mentioned by rpotrie, describes a compact smooth 5-manifold $$M$$ with a $$C^{\infty}$$ vector field $$f$$ on which every orbit is periodic, the periods are unbounded, and there are no stationary points. For this flow the period map $$T$$ can't be continuous.

Now consider $$n$$ large enough that $$M$$ smoothly embeds in $$\mathbb{R}^n$$. A sufficiently small tubular neighborhood of $$M$$ will be diffeomorphic to $$M \times D$$, where $$D$$ is the unit $$(n-5)$$-ball. You can extend the flow $$f$$ on $$M$$ to a flow $$g$$ on $$M \times D$$, slowing down to zero velocity as you reach the boundary $$M \times \partial D$$. Define $$g$$ to be zero outside this tubular neighborhood, and you have a counterexample.

As for intermediate values of $$n$$: there is an example like Sullivan's that works on a 4-manifold (Epstein and Vogt, 1978), so that gives a counterexample for $$n \geq 9$$, but I can't say much else.