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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?

happyle
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    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? – Rollen S. D'Souza Sep 27 '23 at 01:01
  • @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. – happyle Sep 27 '23 at 07:02
  • @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 $D\subset { x|V(x)\leq c }$. – happyle Sep 27 '23 at 07:07
  • 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. – SampleTime Sep 29 '23 at 05:38
  • @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. – happyle Sep 29 '23 at 07:25
  • @happyle Yes I think it is obvious that only the connected component containing the origin can be a subset of the roa but as far as I can see the definition in the linked MO post does not rule out the option of multiple disconnected regions within $F$, but I guess I am missing something here. – SampleTime Sep 29 '23 at 09:05

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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.

AVK
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