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Let $I\subseteq \mathbb{R}^{n}$ be an arbitrary (not necessarily closed) interval and $f:I\times \mathbb{R}^{n}\to \mathbb{R}^{n}$ a continuous function such that in $I\times \mathbb{R}^{n}$ satisfies a global Lipschitz condition on its second variable. Then is it true that:

For every point $(a,b)\in I\times \mathbb{R}^{n}$ there exists a a solution to the equation $y^{\prime}=f(x,y)$ defined over the entire $I$

Picard-Lindelöf theorem states that there exists a solution in a local closed neighborhood $[a-\epsilon,a+\epsilon]$. One could try to glue the local solutions to get a global one but then there will be a problem with the boundary of the resulting (possibly) open interval.

Does such a globally defined solution always exist? Are there references in the literature on this?

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    $\begingroup$ Yes, this implies global existence because you have an a priori lower bound on the maximal interval of existence that you can iterate. Any moderately advanced basic ODE textbook will discuss this. $\endgroup$ Commented Jul 8, 2014 at 20:54
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    $\begingroup$ I'm not using that $I$ is compact. Another easy argument is to use the fact that I have a comparison principle for first order ODEs, so $y$ can't grow faster than the solutions of $u'=Lu+g(x)$, and you can't have blow up in finite time. $\endgroup$ Commented Jul 8, 2014 at 22:47
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    $\begingroup$ Look up the (easy) proof that you have global existence for linear ODEs. The same argument will answer your question. $\endgroup$ Commented Jul 8, 2014 at 22:49
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    $\begingroup$ There is a global solution in the interior of $I$. For any $t$ in the interior of $I$ you just define $y(t)$ to be the value of the unique solution of your equation. If $I$ has a boundary, you can define $y(b)$ by a limit from the interior. $\endgroup$ Commented Jul 8, 2014 at 23:07
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    $\begingroup$ @DariusMath This cannot happen: since the function $y$ is a bounded solution of the differential equation, the function $x \mapsto y' (x) = f (x, y (x))$ is bounded in a neighbourhood of the point $b$, and thus the function $y$ is uniformly continuous, and hence has a limit at the point $b$. $\endgroup$ Commented Jul 9, 2014 at 8:11

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You have existence on open intervals $(a-\epsilon,a+\epsilon)$, whose union covers entire closed interval $I$. Therefore you can choose a finite subcover, call it $I_n, 0<n<N$. You can also contract them so that only consecutively numbered ones would intersect, $I_n\cap I_m=\emptyset, |m-n|>1$. That would allow you to extend solution consecutively from one subinterval to the next.

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  • $\begingroup$ But you are assuming that $I$ is closed. Of course if I is compact it will be a simple generalization of PL. I meant for an arbitrary -possibly open-interval. $\endgroup$ Commented Jul 8, 2014 at 22:37

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