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Show that any periodic points of a one-dimensional ODE is a fixed point

My attempt:

If $x_0$ is not a fixed point then there exists some $\epsilon>0$ and $T > 0 $ so that if $|x-x_0|<\epsilon$, we have $|\phi(t,x)-x_0|>\epsilon$ for any $t>T$. But if $x_0$ is a periodic point, then ...

Don't know what else to do. I only know the definitions of fixed points, periodic points, etc.

sucksatmath
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    Well, if there is a periodic solution of one-dimensional ODE, then you have some period $T > 0$ and (at least) $x(T) = x(0)$. What can you say about behaviour of derivative $x'(t)$ in the interval $\lbrack 0, T \rbrack$ ? – Evgeny Dec 03 '16 at 09:55
  • @Evgeny. I'm not sure, though $x'(0)=0$? – sucksatmath Dec 03 '16 at 14:36
  • Not necessarily $x'(0) = 0$, but there is at least one moment of time $\tau$ at which $x'(\tau) = 0$. If you have a nice ODE $x' = f(x)$ with well-behaving rhs, what can you say about point $x(\tau)$? – Evgeny Dec 03 '16 at 18:01
  • @Evgeny. the point is fixed? – sucksatmath Dec 03 '16 at 18:28
  • Right. Do you see an upcoming contradiction? – Evgeny Dec 03 '16 at 20:13
  • @Evgeny. So is time $\tau$ in the interval $[0,T]$? Is the contradiction that at some time $t>\tau$ we have $ |\phi(t,x)-x_0|>\epsilon|$? – sucksatmath Dec 03 '16 at 21:16
  • Yes, $\tau \in \lbrack 0, T \rbrack$ by Rolle's theorem; 2) No, this isn't the contradiction you are looking for. Your periodic solution passes through the equilibrium point $x(\tau)$ and is not equilibrium itself. Are you okay with that?
    • – Evgeny Dec 04 '16 at 06:18
  • @Evgeny. Oh, I see now, the periodic solution has to be an equilibrium point if it passes through one. Thank you! – sucksatmath Dec 04 '16 at 06:34
  • You are welcome. It was still possible to prove the absence of periodic trajectory using your idea, but I like this way more :) – Evgeny Dec 04 '16 at 07:56