# locally asymptotically stable

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

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 $\mathcal{B}$).
• Hi. Thanks for that.. I guess I'm still slightly confused because Boyce and diPrima say the following: "If V is positive-definite and $\dot{V}$ is negative-definite on Some domain D containing the origin, then the origin is an asymptotically stable critical point, which is a much stronger result than just being stable. But, here they just use a domain "D", not all of R^n. – Thomas Moore Mar 23 '17 at 3:02
• Second point: $\dot V(x) <0$ tells us that "solutions flow downhill"--think of water running downhill and collecting in basins. If we show $\dot V (x) <0$ on a domain $D$, then solutions in $D$ must approach our equilibrium point, but there might be other solutions outside of $D$ that go elsewhere (drop of water fell on the other side of a ridge and flows towards a different lake). If we can show that $\dot V(x) <0$ on $D=\mathbb{R}^n$ (as well as other listed hypotheses), then every initial condition "runs downhill" to our globally stable equilibrium point. – erfink Mar 23 '17 at 3:18
• Ahh. I see. So, if you can show for a center for example that $\dot{V} < 0$ on a domain $D$, then would that center be asymptotically stable by this theorem in Boyce and di Prima? – Thomas Moore Mar 23 '17 at 3:22