locally asymptotically stable

Question

So, I am slightly confused about something that is written on Wikipedia regarding Lyapunov stability, namely: it says that if some function $V(x)$ is such that $\dot{V}(x) < 0$ $\forall \ x \in B$, where $B$ is some set excluding the origin, then the equilibrium point is locally asymptotically stable. (https://en.wikipedia.org/wiki/Lyapunov_function#Locally_asymptotically_stable_equilibrium)

Am I to understand this that if you can only show that $\dot{V(x)} < 0$ within some set containing the equilibrium point, then this point is locally stable, but not globally stable. For global stability, one needs to show that $B = \mathbb{R}^{n}$?

Thanks. Thomas

Answer 1

Correct. Looking down two headings on the wikipedia page, you have:

If the Lyapunov-candidate-function $ V$ is globally positive definite, radially unbounded and the time derivative of the Lyapunov-candidate-function is globally negative definite: $\dot {V}(x)<0 ~\forall x\in \mathbb {R} ^{n}\setminus \{0\}$, then the equilibrium is proven to be globally asymptotically stable. The Lyapunov-candidate function $V(x)$ is radially unbounded if $\|x\|\to \infty \implies V(x)\to \infty $. (This is also referred to as norm-coercivity.)

Roughly speaking, the maximal set $\mathcal{B}$ (w.r.t. inclusion) on which you can show $\dot V (x) < 0$ would be the basin of attraction for the equilibrium. Note that this isn't quite precise due to some topological concerns (certainly everything within $\mathcal{B}$ approaches the equilibrium, but there could be other points in the basin of attraction for which there is no such $V$ and $\mathcal{B}$).