Given parameters $(a,k,A) \in \mathbb{R}^3$, we consider on $\mathbb{S}^1$ the $2\pi$-periodic ODE $$ \dot{\theta} \ = \ - a\sin(\theta) + k + A\cos(t) \hspace{4mm} \mathrm{mod} \ 2\pi. $$ Identifying $\mathbb{S}^1$ with the one-point compactification $\hat{\mathbb{R}}$ of the reals via the stereographic projection $\tan(\frac{\cdot}{2})$, this equation becomes the Riccati equation $$ \dot{y} \ = \ -ay + \tfrac{1}{2}(k+A\cos(t))(1+y^2). $$ As shown in this question, the time-$t$ map is a Möbius transformation for every $t$, and so we have the following: Either

(a) all solutions are neutrally stable,

(b) there are exactly two periodic solutions, one stable and one unstable, or

(c) there is one periodic solution, and this is globally attractive but not stable;

moreover, in cases (b) and (c), the periodic solutions must be $2\pi$-periodic; in case (a), if there is a $2\pi$-periodic solution then all solutions are $2\pi$-periodic.

Is it the case that for every $(a,k,A)$ for which there exists at least one $2\pi$-periodic solution, there are parameters $(\tilde{a},\tilde{k},\tilde{A})$ arbitrarily close to $(a,k,A)$ for which (b) holds?

Remark. If I understand correctly, the set $M$ of orientation-preserving real Möbius transformations of $\hat{\mathbb{R}}$ is a 3-dimensional connected noncompact Lie group (which is naturally embedded into the 3-dimensional complex Lie group of all Möbius transformations of $\hat{\mathbb{R}}$), with each $f \in M$ admitting the local chart $$ \frac{f(\cdot) + \alpha}{\beta f(\cdot) + \gamma} \ \mapsto \ (\alpha,\beta,\gamma). $$ [This chart is defined on the set of all $\tilde{f} \in M$ for which $\tilde{f}{}^{-1}(0) \neq f^{-1}(\infty)$.]

The set $M$ can be partitioned as \begin{align*} M \ &= \ U_1 \cup U_2 \cup \bar{S} \\ &\textrm{with } \ \bar{S} \ = \ \partial U_1 \ = \ \partial U_2 \ = \ S \cup \{e\} \end{align*} where [using notation in which an expression involving $x$ is intended to mean that expression as a function of $x$]: $$ U_1 \ = \ \left\{ \lambda - \frac{\mu}{x+\nu} : 0 < \mu < \left( \frac{\lambda+\nu}{2} \right)^{\!2} \right\} \ \cup \ \big\{\mu x + \nu : \mu \in (0,\infty) \setminus \{1\} \big\} $$ is the open set consisting of all maps with a stable and an unstable fixed point, $$ U_2 \ = \ \left\{ \lambda - \frac{\mu}{x+\nu} : \mu > \left( \frac{\lambda+\nu}{2} \right)^{\!2} \right\}$$ is the open set consisting of all maps that are topologically conjugate to a nontrivial circle rotation, $$ S \ = \ \left\{ \lambda - \frac{\frac{1}{4}(\lambda+\nu)^2}{x+\nu} : (\lambda,\nu) \neq (0,0) \right\} \ \cup \ \big\{ x + \nu : \nu \neq 0 \big\} $$ is the set of all maps admitting a unique fixed point (in which case all trajectories are homoclinic), and $e$ is the identity function.

Now let $\Phi \colon \mathbb{R}^3 \to M$ be the map sending a tripet $(a,k,A)$ onto the time-$2\pi$ mapping of the Riccati equation above. As in Anthony's idea, this map will (I believe) be an analytic map. The question then becomes:

Is it the case that $\Phi^{-1}(\bar{U}_1) \subset \overline{\Phi^{-1}(U_1)}\,$?

[Equivalently, since $\partial U_1 = \bar{S}$: Is it the case that $\Phi^{-1}(\bar{S}) \subset \overline{\Phi^{-1}(U_1)}\,$?]

I had hoped that perhaps $\Phi$ is a local diffeomorphism, in which case the result would be clear. But $\Phi$ is clearly not a local diffeomorphism: indeed, it only has rank $1$ at points $(0,k,0)$, and has rank at most $2$ at points $(0,k,A)$.


2 Answers 2


The answer to your question is YES, and you can look up the proof in our article with A. Klimenko https://arxiv.org/pdf/1305.6746.pdf for example. The key idea is to mix monotonicity with Moebius property, this will suffice. First this was proven by Foote in a completely different setting (for Prytz planimeter!!) and then re-proven by Buchstaber, Karpov and Tertychnyj when they came up with the study of this equation because they were motivated by the understanding of the physics of supraconductivity.

  • $\begingroup$ Thanks for this. I guess the point is that (for each $\mu$ and $k$) the functions $a_{0,k}$ and $a_{\pi,k}$ introduced near the top of p4 are analytic, and therefore they cannot be equal on an interval? But why are $a_{0,k}$ and $a_{\pi,k}$ analytic functions? I can see that they are analytic about every point where they do not coincide (by the analytic implicit function theorem applied to the derivative of $\tilde{P}_{a,b,\mu}$), but that would obviously be useless for concluding that there is no interval on which they coincide. $\endgroup$ Dec 20, 2017 at 20:29
  • $\begingroup$ Okay, I think I was getting confused. Analyticity of $a_{0,k}$ will follow from the analytic implicit function theorem applied to the map $(b,a) \mapsto \tilde{P}_{a,b,\mu}(0)-k$, provided $\partial_a \tilde{P}_{a_{0,k}(b),b,\mu}(0)>0$ for all $b$. In other words, to formulate this requirement in the notation of my question: For fixed values of $a$ and $A$, and taking $\theta(0)=\frac{\pi}{2}$, the map $k \mapsto \theta(2\pi)$ should have non-zero derivative at any $k$-value for which $\theta(2\pi)=\frac{\pi}{2}$. Is this statement true? Or have I completely misunderstood things? $\endgroup$ Dec 21, 2017 at 2:08

Yes, I think so. Once you know that the time $t$ map of your differential equation for $y$ look like Möbius transformations, you can set $$ y_t=\frac{\alpha_t y_0+\beta_t}{y_0+\gamma_t}. $$ By equating constant, $y$ and $y^2$ coefficients, you obtain differential equations for $\alpha$, $\beta$ and $\gamma$. In particular, the solutions are analytic functions of $a$, $k$ and $A$. Clearly there are values of $a$, $k$ and $A$ for which the solution does have the property you want; indeed the Möbius transformations with a period $2\pi$-periodic solution that is not of type (b) form a nowhere dense set. Since the parameters of the Möbius transformation vary analytically, there should be nearby parameter values of $a$, $k$ and $A$ such that the time $2\pi$ map satisfies property (b).

  • $\begingroup$ Thanks for this! Presumably your reasoning relies on the map $(a,k,A) \mapsto (\alpha_t,\beta_t,\gamma_t)$ being nonsingular everywhere? I think it's possible to come up with other fairly simple 3-parameter families of Riccati equations for which my desired conclusion does not hold. But I guess it should be possible to manually verify (or disprove!) the nonsingularity in our case. $\endgroup$ Nov 21, 2017 at 6:55
  • $\begingroup$ And I guess that if we do need the nonsingularity of this map, then analyticity isn't really so relevant (since being a local homeomorphism would be enough)? Or have I completely misunderstood things? $\endgroup$ Nov 21, 2017 at 7:06
  • $\begingroup$ I would argue as follows: when you look for $2\pi$-periodic points, you are solving a quadratic equation whose coefficients are analytic functions of $k,a,A$ and hence the discriminant is an analytic function of $k,a,A$ also. For cases (a) and (c), the discriminant vanishes. If an analytic function vanishes on a set with a limit point, then it's identically 0. $\endgroup$ Nov 21, 2017 at 22:40
  • $\begingroup$ In case (a), the discriminant only vanishes when the solutions are $2\pi$-periodic, otherwise the discriminant is negative. I am not presupposing that there is a neighbourhood of $(a,k,A)$ on which at least one $2\pi$-periodic solution exists. (Or am I still misunderstanding something?) $\endgroup$ Nov 22, 2017 at 12:05
  • $\begingroup$ I see. I will think some more. $\endgroup$ Nov 22, 2017 at 17:16

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