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Consider the differential equation $x'=f(t,x)$, where $f(t,x)$ is continuously differentiable in $t$ and $x$.

Suppose that $f(t+T,x)=f(t,x)$ for all $t$ . Suppose there are constants $p$, $q$ such that $f(t,p)>0,f(t,q)<0$ for all $t$. Prove that there is a periodic solution $x(t)$ for this equation with $p<x(0)<q$

What I have to show is that $ x(t+T) = x(t) $, that is $x(t)$ is periodic. I tried this

$\int_{0}^{t}{x'} = x(t)-x(0) = \int_{0}^{t}{f(s,x(s))}ds $

But I don't know how to continue...

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tnt235711
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  • It would be good if you told us what your knowledge of time-periodic DEs is. Do you know, for instance, that periodic solutions are in one-to-one correspondence with fixed points of the period map? Then the problem reduces to the Darboux property of the period map. – user539887 Jul 06 '18 at 08:54
  • if $p(x_{0}) = x_{0}$, that is $x_{0}$ is a fixed point of the Poincaré map, then the solution satisfying $x(0) = x_{0}$ is periodic. – tnt235711 Jul 06 '18 at 09:11
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    The symbol $p$ already means some point, so it's better to denote the Poincaré map by $P$. From the assumptions it follows that $P(p)>p$ and $P(q)<q$, By uniqueness, $P(p)\le P(q)$. So, $p<P(p)\le P(q)<q$, and by the Darboux property of the continuous function $P$ on $[p,q]$, there is $x_0\in(p,q)$ such that $P(x_0)=x_0$. (What I wrote here is only a sketch: to have a complete proof one has to show that, under the assumptions, $P$ is defined for all $x\in[p,q]$.) – user539887 Jul 06 '18 at 09:26
  • Maybe this answer will provide a useful illustration. Also, in principle there could be multiple fixed points of Poincaré map, but one of them is necessarily stable. – Evgeny Jul 06 '18 at 12:00
  • @Evgeny Now I can imagine what is happening. We know that there are constants $p, q$ such that $f(t,p)>0$ ,$f(t,q)<0$ but how do we know that $ p < q $?... – tnt235711 Jul 06 '18 at 17:38
  • @tnt235711 In case $ p > q$ you can consider the problem in backward time: all arrows would be reversed, and the picture would be the same as in my answer. – Evgeny Jul 06 '18 at 21:33

1 Answers1

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The Poincaré map is defined in the following way:

For $\xi \in \mathbb{R}$, $P(\xi)$ is defined as the value at time $T$ of the (unique) solution of the IVP $$\begin{cases} x'= f(t,x), \\ x(0) = \xi. \end{cases}$$

It is easy to see that fixed points of the Poincaré map are in a one-to-one correspondence with the values at time $t = 0$ of $T$-periodic solutions of the ODE $x' = f(t, x)$ (here $T$ need not be the prime period).

Let $\varphi$ (resp. $\psi$) stand for the nonextendible solution of the ODE satisfying $\varphi(0) = p$ (resp. $\psi(0) = q$). It follows from the uniqueness property that $$ \varphi(t) < \psi(t) $$ for all $t \ge 0$ for which both $\varphi(t)$ and $\psi(t)$ exist.

Denote by $(\alpha, \beta)$ the domain of $\varphi$. We claim that $\varphi(t) > p$ for all $t \in (0, \beta)$. Observe that $\varphi(t) > p$ for $t > 0$ sufficiently small. Suppose to the contrary that $\tau$ is the first moment in $(0, \beta)$ such that $\varphi(\tau) = p$. But $\varphi'(\tau) = f(\tau, p) > 0$, a contradiction. In a similar way we prove that $\psi(t) < q$ for all $t > 0$ for which $\psi(t)$ exists.

Note that, by uniqueness, for any $x \in [p, q]$ the solution of the ODE starting at $x$ takes, for positive times, the values in the compact set $[p, q]$ (as long as it is defined). Consequently, the domain of such a solution contains $[0, \infty)$.

To sum up, we have proved that the continuous (indeed, $C^1$) Poincaré map maps $[p, q]$ into $[P(p), P(q)] \subset (p, q)$. By the Darboux property of continuous functions applied to $$ [p, q] \ni x \mapsto P(x) - x $$ there exists $x_0 \in (p, q)$ such that $P(x_0) = x_0$. So, the solution of the ODE starting at $x_0$ has period $T$.

user539887
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  • How do we know that the Poincare map is continuous? Is it just because $f(t,x)$ is continuously differentiable? Or are there other requirements? Can you point me to a proof of that? Thanks! – SweepingBishops Jun 14 '23 at 14:19
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    @SweepingBishops Theorems on continuous dependence of solutions of an ODE on initial values are formulated in any good textbook on ODEs. Let’s try Gerald Teschl’s Ordinary Differential Equations and Dynamical Systems (accessible for personal use at https://www.mat.univie.ac.at/~gerald/ftp/book-ode/index.html). From our assumptions (indeed, they can be relaxed slightly) it follows that RHS is Lipschitz continuous in $x$, uniformly with respect to $t$. Then apply Theorem 2.8 on p. 43 to obtain continuity of the Poincaré map (or Theorem 2.9 to obtain Lipschitz continuity). – user539887 Jun 17 '23 at 11:23