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?
Best Answer
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.