# What if the level set of Lyapunov function is disconnected? - when estimating region of attration

Consider $$\frac{dx}{dt}=f(x)$$, where $$x\in\mathbb{R}^n$$. Suppose $$x=0$$ is a stable equilibrium.

It is classical way to estimate region of attraction of $$0$$ by finding a $$C^1$$ function $$V(x)$$ such that when $$x\in D\subset \mathbb{R}^n$$

(1) V(x) is lower bounded

(2) $$\frac{dV(x)}{dx}\frac{dx}{dt}<0$$, $$V(x)=0$$ if and only if $$x=0$$

(3) solution $$x$$ cannot leave the region $$D$$. This can be guaranteed by imposing compactness and positive invariance on region $$D$$.

The level set $$\{ x|V(x)\leq c \}$$ satisfies the condition (3). Thus, level set of $$V(x)$$ is an estimate of region of attraction of $$x=0$$. Please see Section 8.2 - H. K. Khalil, Nonlinear systems.

I am a bit confused on this, once I start to think of the level set of $$V(x)$$ might be disconnected. In this case, for the connected components that do not contain $$x=0$$, they cannot be part of region of attraction for sure.

Do we need to change the conclusion to: The connected component who contain $$x=0$$ is an estimate of region of attraction? Or the level set should be connected such that it becomes an estimation.

I didn't see any literature noticing the connectedness of level set. Am I missing something or the literature are not rigorous?

• A domain (the word Khalil uses) is usually interpreted as a (non-empty) open, connected set. Also did you see the footnote 11 on pg 317? Commented Sep 27, 2023 at 1:01
• @RollenS.D'Souza "The set $\{ V(z)\leq c \}$ may have more than one component, but there can be only one bounded component in D, and that is the component we work with". I see! But I didn't get why there can be only one bounded component? I think there can be a few. Commented Sep 27, 2023 at 7:02
• @RollenS.D'Souza I am not sure whether this is true: I have a Lyapunov function $V(x)$ that satisfies all requirements. Now I want to verify whether a subset $D$ is the estimation of DOA (of the stable equilibrium $x=0$) or not. How could we do that? I think I need verify these 3 conditions: (1) The set $D$ is connected; (2) The set $D$ contains $x=0$; (3) The set $D$ is a subset of sublevel set of $V(x)$, i.e. there exists a constant $c$ such that $D\subset \{ x|V(x)\leq c \}$. Commented Sep 27, 2023 at 7:07
• I posted a similar question 2,5 years ago here on MathOverflow, unfortunatly with no response so far (even though I offered a bounty back then). So I would be very interested in an answer too. Commented Sep 29, 2023 at 5:38
• @SampleTime I feel this is a very natural question But a bit strange that seems no where address the problem explicitly. maybe intuitively only the connected component who contains the equilibrium can be considered as an estimation of roa. Commented Sep 29, 2023 at 7:25

If $$\forall x\ne 0, x \in\Omega_C\; \dot V(x)<0$$ (where $$\Omega_C=\{ x:V(x)\leq C \}$$), then there can be no bounded connected components of $$\Omega_C$$ other than the one containing $$0$$.
Indeed, let $$\Omega_C^1$$ be the bounded connected component of $$\Omega_C$$, $$0\notin \Omega_C^1$$. Since $$\dot V(x)$$ is continuous on the compact set $$\Omega_C^1$$, it reaches on this set some maximum value $$M$$, $$M<0$$. Thus, $$\forall x\in \Omega_C^1\; \dot V(x)\le M<0$$. Let us recall that any trajectory that enters $$\Omega_C^1$$ remains in it. But then the function $$V(x)$$ cannot be bounded below, because for any trajectory in $$\Omega_C^1$$ and for any initial moment $$t_0$$, $$x(t_0)\in \Omega_C^1$$ $$V(x(t))=V(x(t_0))+\int_{t_0}^t \dot V(x(t))\, dt\le V(x(t_0))+M(t-t_0),$$ i.e. $$\lim_{t\to\infty} V(x(t))=-\infty$$. We got a contradiction.
• It should be highlighted that the fact that $0 \notin \Omega_C^1$ is being used in the claim that $M < 0.$ Commented Sep 30, 2023 at 22:09