Lipschitz Continuity and Uniqueness

[Kajsa_Stina]

Dear all,

If a differential equation is Lipschitz continuous, then the solution is unique. But what about the implication in the other direction? I know that uniqueness does not imply Lipschitz continuity. But is there a counterexample? A differential equation that is not L-continuous, still the solution is unique?
I would really like to know the answer to this question.
Thank you all for your help,

Hendrik

 

[Kajsa_Stina]

Now I have found an answer to my question:

http://archive.numdam.org/ARCHIVE/A..._1996_6_5_4_577_0/ AFST_1996_6_5_4_577_0.pdf

or, rather, Bernis and Qwang have found an answer. However, my differential equation is only of first degree, like

\dot K(t) = F(K(t))

One example would be F(K) = K^\alpha with 0 < \alpha < 1, then you get multiple solutions if you start from K = 0. But that's not surprising, because the differential equation is not Lipschitz continuous in K = 0. My question now would be: Is it possible to have a unique solution even if F were L-continuous in 0?

Thanks, many greetings,
Hendrik

 

[tacman]

Dear Hendrik,

Thank you for your question. You are correct that uniqueness of a solution does not necessarily imply Lipschitz continuity. In fact, there are examples of differential equations that are not Lipschitz continuous, yet still have a unique solution.

One such example is the equation y' = √|y|, which is known as the Van der Pol equation. This equation does not satisfy the Lipschitz condition, as the derivative of √|y| is unbounded near y = 0. However, it can be shown that this equation still has a unique solution for any given initial condition.

This example highlights the fact that Lipschitz continuity is a sufficient condition for uniqueness, but not a necessary one. There are other conditions, such as the Picard-Lindelöf theorem, that can also guarantee uniqueness of solutions without requiring Lipschitz continuity.

I hope this helps answer your question. If you have any further inquiries, please don't hesitate to ask.