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According to Piccard-Lindelof Theorem, there exists a solution for every Initial Value Problem (under some conditions like Lipschitz, etc.) and that solution is unique. However, when we say that the solution is unique, we mean that it is unique GIVEN some interval, right?

Because if not suppose f: (1,2) --> R is a solution, then g: (1.5,2) ---> R is also a solution and so the solution would not be unique.

Agustin G.
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    If you look at the full specification of the theorem, it states that there exists an interval on which there is a unique solution, so yes it does also specify a domain. – ConMan Oct 09 '24 at 22:53
  • Maybe a counter-example of uniqueness would improve the concept: Norton's Dome example – Joako Oct 10 '24 at 02:49
  • An IVP $y'=f(x,y)$, $y(x_0)=y_0$ is said to be uniquely solvable if 1. it has a solution and 2. if $y_1:I_1 \to \mathbb{R}$ and $y_2:I_2 \to \mathbb{R}$ are solutions then $y_1=y_2$ on $I_1\cap I_2$. – Gerd Oct 11 '24 at 13:17

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