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I have been thinking about this question for a while. I haven't found a definite answer, but I am led to believe that there can be a unique solution to an IVP outside of interval of validity. I just fail to prove it.

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  • $\begingroup$ This is a good question. Often existence and uniqueness theorems are not formed at the maximum level of generality because the maximum level of generality is difficult to identify and does not arise too often in applications people care about. It may help to spell out more hypotheses of the theorem you are referring to if you want an example that relates to those hypotheses. I have seen different sets of hypotheses in different books. $\endgroup$ Commented Feb 18, 2021 at 4:14
  • $\begingroup$ Uh, this seems to have been edited into an entirely different question. See if it can be posted as a different question. $\endgroup$ Commented Feb 18, 2021 at 4:32
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    $\begingroup$ Why are you editing the question into a completely different one? If you sincerely want to ask the question in the edit, please just start a new question. If not, please do not vandalize your own question. $\endgroup$
    – Elliot Yu
    Commented Feb 18, 2021 at 4:33

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For example, let $$ f(x,y) = \cases{1 & if $xy < 0$\cr -1 & if $xy > 0$\cr 0 & if $xy = 0$\cr}$$ and consider the initial value problem $$ \eqalign{ \dfrac{dy}{dx} &= f(x,y) \cr y(0) &= 0\cr} $$ The hypotheses of the Existence and Uniqueness Theorem are not satisfied because $f(x,y)$ is not continuous in a neighbourhood of $(0,0)$, but there is a unique solution, namely $y=0$.

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