# No periodic solution for ODE

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
• 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