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

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


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?

  • $\begingroup$ Where does the assumption that $\lambda(t)$ is $K$-periodic come from? $\endgroup$ – Michael Oct 19 '17 at 18:36
  • 1
    $\begingroup$ Can you also state the precise assumption for "locally Lipschitz" and "asymptotically stable"? $\endgroup$ – Michael Oct 19 '17 at 18:37
  • $\begingroup$ 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. $\endgroup$ – Theodor Johnson Oct 19 '17 at 18:40
  • $\begingroup$ What if you consider $\lambda(t)$ and $\mu(t)=\lambda(t+K/2)$? $\endgroup$ – Michael Oct 19 '17 at 18:45
  • 1
    $\begingroup$ 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.] $\endgroup$ – Michael Oct 19 '17 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.