Lipschitz Continuity and Uniqueness

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SUMMARY

A differential equation that is Lipschitz continuous guarantees a unique solution, but uniqueness does not imply Lipschitz continuity. Hendrik's inquiry revealed that a first-degree differential equation, specifically of the form \(\dot K(t) = F(K(t))\) with \(F(K) = K^\alpha\) (where \(0 < \alpha < 1\)), can yield multiple solutions at \(K = 0\) due to the lack of Lipschitz continuity. The discussion references a paper by Bernis and Qwang, which addresses the implications of Lipschitz continuity on solution uniqueness.

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Kajsa_Stina
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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
 
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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
 
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