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$\textbf{Theorem}:$ Let $D=[a,b]\times\mathbb{R}(or~D=\mathbb{R}\times\mathbb{R})$ and $f$ is globally Lipschitz on $D$ then the I.V.P $$\frac{\mathrm dy}{\mathrm dx}=f(x,y),y(x_{0})=y_{0}$$ has unique solution on $[a,b](or ~on ~\mathbb{R}).$

I have this theorem. Now my question is that is there any similar theorem for our domain of the type $[a,b]\times [c,d] $ i.e. can the same theorem hold for bounded domain also so that we can directly say up to which interval the I.V.P. has unique solution. Please suggest me. Thanks a lot.

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It does not exist, unless you have an appropriate bound for the Lipscitz constant: take a smaller interval than the image of a given solution.

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  • $\begingroup$ If it is globally Lipschitz then clearly it has Lipschitz constant.. $\endgroup$ – neelkanth Jan 5 '17 at 5:38
  • $\begingroup$ Sorry, but for your question it is irrelevant whether the dynamics enters or not a region where the map is no longer Lipschitz. It is as I replied: take $[c,d]$ strictly inside the interval $y([a,b])$ for some solution $y$. Surely you see that this is a trivial counterexample. $\endgroup$ – John B Jan 5 '17 at 10:50
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The problem is that the dynamics of $y$ may enter a region where $f$ is no longer Lipschitz. At this time a problem can arise. Conversely, until then no problem can arise. As an example, look at $f(x,y)=-|y|^{1/2},y_0=1$, with $0<c<1<d$. Here a problem arises when $y$ hits zero (in particular, $y$ can stay zero for any length of time before it begins to fall again, so that there are continuum-many solutions to this IVP), but until then you have existence/uniqueness. It is sometimes possible to use some bounds on $f$ in order to get existence/uniqueness on $[a,b] \times [c,d]$ "on short times", i.e. when $b$ is sufficiently close to $a$.

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