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Theorem: Suppose ${\bf g} :\mathbb{R}^n \mapsto \mathbb{R}^n$ is continuously differentiable, there exists a set $\mathcal{A} \subset \mathbb{R}^n$ such that $\bf g$ is uniformly Lipschitz on $\mathcal{A}$, and for all points ${\bf x}_0\in\mathcal{A}$ and $t>0$ there exists a unique solution ${\bf x} (t) \in \mathcal{A}$ to the ODE $\dot{{\bf x}}(t)= {\bf g} ( {\bf x} (t))$ such that ${\bf x} (t) = {\bf x}_0$, then the set of initial points that converge to an unstable equilibrium point has measure zero.


I need the aforementioned result for some work that I am doing. I have been unable to find any similar result in the the continuous case. I have a proof, but would like to try a bit harder to see if someone has done it already before making part of the manuscript when I could reference it. The core of the proof is the center-stable manifold theorem.

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  • $\begingroup$ For example a saddle point. In particular, an equilibrium point for which the derivative of g has at least eigenvalue with real part > 0 $\endgroup$
    – RLip2
    Commented Jan 21, 2022 at 17:48
  • $\begingroup$ Through a related question, generated by posting this question, I found the answer. $\endgroup$
    – RLip2
    Commented Jan 21, 2022 at 21:29
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    $\begingroup$ As it is written it is false, unless you have in mind a special definition of “unstable equilibrium” (you should include it). Note that A has no role. $\endgroup$ Commented Jan 21, 2022 at 23:11
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    $\begingroup$ To support the comment of @PietroMajer: We know examples of fixed points that are attractive but not stable (in the sense of Lyapunov). Here attractive means that a neighborhood of the fixed point is attracted to it. So the claim of the theorem cannot be true. A famous example is due to Vinograd (1957), explained in Hahn, Stability of Motion, p.191. $\endgroup$ Commented Jan 22, 2022 at 3:16
  • $\begingroup$ It is easy to make a smooth field in $\mathbb R^n$ whose flow has a (degenerate) rest point x, such that the unstable manifold of x is e.g. a line and the the basin of attraction of x has non-empty interior. So maybe the question needs clarification on the definitions. $\endgroup$ Commented Jan 22, 2022 at 5:49

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