0
$\begingroup$

Consider a discrete-time system $$x_{t+1} = f(x_t),$$ where $t \in \mathbb{N}_0$ and $x_t \in \mathbb{R}^n$. This system has only one equilibrium $x^* = 0$.

Is there any example for $f: \mathbb{R}^n \to \mathbb{R}^n$ such that the system is asymptotically stable but not uniformly stable?

Thanks!

PS: We need two definitions to define asymptotic stability and uniform asymptotic stability.

Definition 1 (Lyapunov Stable and Uniformly Stable) The equilibrium $x^* = 0$ is (Lyapunov) stable at $t = t_0$ if for any $\epsilon > 0$, there exists a $\delta(t_0,\epsilon) > 0$ such that $$\|x(t_0)\| \leq \delta(t_0,\epsilon) \rightarrow \|x(t)\| < \epsilon, \forall t \geq t_0.$$ Furthermore, if $\delta(t_0,\epsilon)$ is not dependent on $t_0$, then the equilibrium $x^* = 0$ is uniformly stable.

Definition 2 (Asymptotic Stability and Uniform Asymptotic Stability) The equilibrium $x^* = 0$ is asymptotically stable if:

  1. $x^* = 0$ is stable;

  2. There exists $\delta(t_0)$ such that $$\|x(t_0)\| \leq \delta(t_0) \rightarrow \lim_{t\to\infty}\|x(t)\| = 0.$$ Furthermore, if $\delta(t_0)$ is not dependent on $t_0$, then the equilibrium $x^* = 0$ is uniformly asymptotically stable.

$\endgroup$
  • $\begingroup$ Perhaps you can explain your definitions. Depending on your definitions, I suspect there are easy examples $f:\mathbb{R}\rightarrow\mathbb{R}$. Like when $x_t\rightarrow 0$ when $x_0$ is near zero, but $x_t\rightarrow\infty$ if $x_0$ is too far from $0$. You can make a simple $f$ that does this. $\endgroup$ – Michael Jul 10 '18 at 5:27
  • $\begingroup$ @Michael Thanks for your comments, and I have added the definitions for the asymptotic stability and uniform asymptotic stability. For your example, sorry I think did not get the point. It seems that this example is both asymptotically stable and uniformly asymptotically stable. Can you provide more details. Thanks very much! $\endgroup$ – Ryan Jul 10 '18 at 5:52
  • $\begingroup$ I don't get what $t_0$ means in def 2. $\endgroup$ – Michael Jul 10 '18 at 6:05
  • $\begingroup$ @Michael Sorry, I have revised Definition 2 by replacing $x_0$ with $x(t_0)$. $\endgroup$ – Ryan Jul 10 '18 at 6:18
  • 1
    $\begingroup$ Your system $x_{t+1} = f(x_t)$ is time invariant, so the $t_0$ business can be removed. $\endgroup$ – Michael Jul 10 '18 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.