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Quick question about Picard–Lindelöf theorem.

We know the existence and uniqueness of a local solution around $[t_0-\epsilon,t_0+\epsilon ]$. But what if we have Lipschitz continuity everywhere, does this mean that a unique solution exists say on the whole of $\mathbb{R}$.

For example $\frac{d}{dx}(x(t))=x(t)^3$ and $x(0)=1$ can we conclude without solving this that the solution will exist everywhere and be unique? I don't know how to apply the Picard–Lindelöf theorem here.

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You need a global Lipschitz condition with the same constant to apply a global version of the theorem.

Your example $\dot x = x^3$ does not possess a global Lipschitz constant since its derivative $3x^2$ is unbounded. And indeed, the solution $x^{-2}=C-2t$ has a couple limitations on its domain.

However, there is a theorem of Cauchy that tells you that if the ODE function is continuously differentiable (and thus locally Lipschitz), then any solution can be uniquely continued to the boundary of its domain, which may be the $x$ infinity at finite time.

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  • $\begingroup$ So in out case the theorem of Cauchy would imply global existence of the solution but not necessarily the uniqueness? $\endgroup$ – Caps May 18 '16 at 9:44
  • $\begingroup$ No, it says that the solution is unique as long as it exists, and in this example there is a blow-up in finite time, $x(t)=\frac{x_0}{\sqrt{1-2x_0^2·t}}$. But it does not guarantee a global solution. That you only get with a global Lipschitz constant. $\endgroup$ – Lutz Lehmann May 18 '16 at 9:54

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