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Recall that given a time-invariant dynamical system

$$\dot x = f(x)$$

We say that an equilibrium point at the origin, $x_e \in \mathbb{R}^n$, of the above system is stable (in the sense of Lyapunov) if:

$$\forall \epsilon > 0, \exists \delta > 0, \text{ s.t. } \|x(t_0) - x_e\| < \delta \implies \|x(t) - x_e\|< \epsilon, \forall t \geq t_0$$

Suppose that the above condition holds for all $x_0 = x(t_0) \in \mathbb{R}^n$, then we can say that the equilibrium point is globally stable.

Suppose that $x_e$ is globally stable, then is it the unique equilibrium of $\dot x = f(x)$? enter image description here

Note that global asymptotic stability implies uniqueness because every trajectory has to converge to that point.

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  • $\begingroup$ There's no $x_0$ in the condition, what do you mean when you say that it holds for all $x_0\in\mathbb{R}^n$? $\endgroup$ Commented Jul 22, 2017 at 0:07

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Consider $\dot x =0$: every point is a globally stable equilibrium.

For a less ridiculous example, $\dot x = (-x_1, 0)$ has a line of globally stable equilibria.

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  • $\begingroup$ What happens if the set of equilibria not connected? Does this imply uniqueness? $\endgroup$ Commented Dec 31, 2020 at 22:14

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