No periodic solution for ODE

Question

My math class is over but I never managed to solve the following problem:

Assume we have a K periodic solution $\lambda(t)$ to

$\dot{x}=f(x).$

Furthermore, $f(x)$ is locally Lipschitz continuous. Prove: No nonconstant $K$ periodic solution $\lambda$ can be asymptotically stable.

What I tried: Since $\lambda$ is $K$ periodic, we know $\lambda (t)=\lambda (t+K)$.

If two solutions $\lambda,\mu$ that satisfy the ODE are asymptotically stable, we have $lim_{t\rightarrow \infty}|\lambda(t)-\mu(t)|=0.$

Inserting our periodicity for $\lambda$:

$lim_{t\rightarrow \infty}|\lambda(t)-\mu(t)|=lim_{t\rightarrow \infty}|\lambda(t+K)-\mu(t)|.$

Clarification: Locally lipschitz means, $f(x)$ is Lipschitz continuous in every neighborhood $x$.

But what can I do now?

Where does the assumption that $\lambda(t)$ is $K$-periodic come from? — Michael, Oct 19 '17 at 18:36
Can you also state the precise assumption for "locally Lipschitz" and "asymptotically stable"? — Michael, Oct 19 '17 at 18:37
Hello, Michael! You are completely right! We assume that $\lambda$ is $K$-periodic from the start. My professor used the term locally Lipschitz interchangeably with Lipschitz continuity. Basically the conditions to apply Picard Lindelöf uniqueness axioms. I edited the question for clarification. — Theodor Johnson, Oct 19 '17 at 18:40
What if you consider $\lambda(t)$ and $\mu(t)=\lambda(t+K/2)$? — Michael, Oct 19 '17 at 18:45
I unfortunately do not immediately understand how this would benefit us. — Theodor Johnson, Oct 19 '17 at 18:53
What do you mean? Can you tell me anything about $\mu(t)=\lambda(t +r)$ for some constant $r$? — Michael, Oct 19 '17 at 18:54
Well, if $K$ is the period then $\lambda (t+r)=\lambda (t+r+K)$ — Theodor Johnson, Oct 19 '17 at 18:56
So now you are holding in your hand two particular solutions to the ODE. So... [you can answer your own question below, which is standard practice based on hints.] — Michael, Oct 19 '17 at 19:05
I think I see! Because of translation invariance of ODE's, both, $\lambda$ and $\mu$ are valid solutions. Since they cannot intersect due to uniqueness and are both periodic with a slightly different period, the difference $\lambda-\mu$ will never converge, hence no asymptotic stability (but simple stability maybe). If this is correct then I will add an answer — Theodor Johnson, Oct 19 '17 at 19:09
I hope that you mean under "ODE" a vector ODE (in other words, an ODE system). Otherwise, the discussion is vacuous, see Prove that an autonomous ODE f(x)=x' has no nonconstant periodic solutions, or Proof by contradiction that there are no periodic solutions to x˙=f(x). Incidentally, in either case the Lipschitz assumption is not needed. — user539887, Apr 07 '18 at 21:51

No periodic solution for ODE

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