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I read the book A Linear Systems Primer by Antsaklis, Panos J., and Anthony N. Michel (Vol. 1. Boston: Birkhäuser, 2007) and I am confused by the definition of stable. It is defined as:

\begin{equation} \dot{x}=f(x), f(0)=0\tag{4.8} \end{equation} Definition 4.6. The equilibrium $x = 0$ of (4.8) is said to be stable if for every $\epsilon > 0$, there exists a $\delta(\epsilon)>0$, such that \begin{equation} ||φ(t, x_0 )|| < \epsilon ~for ~all~ t ≥ 0 \end{equation} whenever $||x_0||<\delta(\epsilon) $

In this definition, does it mean that $ \lim_{t\to\infty} φ(t, x_0 ) = 0$ because $\epsilon$ could be any arbitrarily small positive number? If this is true, then the definition of stable equals to that of asymptotically stable which should NOT be correct. So what is the wrong part in my thought?

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  • $\begingroup$ This definition says "if you start close enough then you will stay within any epsilon of the equilibrium point, where close enough could depend on epsilon". Asymptotic stability would mean that delta does not go to zero as epsilon goes to zero. $\endgroup$
    – Ian
    Mar 22, 2017 at 10:49
  • $\begingroup$ @Ian Do you mean that in this case, the equilibrium point is zero, so $\lim_{t \to \infty}\phi(t,x_0)=0$. But if the equilibrium point is nonzero, say $\alpha$, then $\lim_{t \to \infty}\phi(t,x_0)=\alpha$? $\endgroup$
    – winston
    Mar 22, 2017 at 12:12
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    $\begingroup$ If the equilibrium point is nonzero then several things must change (for example the condition at the end becomes $\| x_0-\alpha \|<\delta(\epsilon)$). Anyway, mere stability does not tell you anything about limits, even about their existence. $\endgroup$
    – Ian
    Mar 22, 2017 at 12:15

2 Answers 2

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Stability means that the solution of the differential equation will not leave the $\epsilon$-ball. But asymptotic stability means that the solution does not leave the $\epsilon$-ball and goes to the origin.

Decreasing $\epsilon$ will force the initial condition to approach the zero in the stable case and not the solution at infinity.

Asymptotic stability implies stability but the converse is not true in general.

Example of stable but not asymptotic stable system: $\dot{x}=0$, The solution stays at its initial condition for any sufficiently small $\epsilon$ but will not go to zero if the initial condition is not zero.

Asymptotic example $\dot{x}=-x$ The solution $x=x_0e^{-t}$ goes to zero for any initial condition $x_0$.

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  • $\begingroup$ Thank you for your intuitive explanation. $\endgroup$
    – winston
    Mar 22, 2017 at 13:14
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    $\begingroup$ Your example is really good for understanding. For $\dot{x}=0$, if we choose $\delta(\epsilon)=\epsilon$, then this proves the system is stable. However , $\lim_{t \to \infty}\phi(t,x_0)=\alpha \ne 0$, where $\alpha$ is a positive number that is smaller than $\epsilon$. $\endgroup$
    – winston
    Mar 22, 2017 at 13:40
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    $\begingroup$ Not necessarily the origin, but rather the equilibrium point. Great answer though $\endgroup$ Feb 5, 2019 at 17:52
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Generically, for a stable system your $\delta$ can be chosen to be a strictly increasing function, with $\delta(0)=0$. If $\lim_{\epsilon \to 0} \delta(\epsilon)=0$ and $x_0 \neq 0$, then the condition $\| x_0 \|<\delta(\epsilon)$ can only be satisfied for $\epsilon>\epsilon_0$ for some fixed number $\epsilon_0$. Thus for such an $x_0$ your definition only gives you that $\| \varphi(t,x_0) \|<\epsilon_0$ which is now fixed, not arbitrary anymore.

To modify this definition to get asymptotic stability, you would need to change $\delta(\epsilon)$ to be defined by "if $\| x_0 \|<\delta(\epsilon)$ then $\| \varphi(t,x_0) \|<\epsilon$ for all $t \geq t_0(\epsilon)$". Moreover you would need this modified $\delta(\epsilon)$ to not go to zero (so that this condition actually continues to apply to some nonzero values of $x_0$ even as $\epsilon \to 0$).

More intuitively speaking, stability says that a system starting in some $\delta$-ball around the equilibrium will not leave a $\epsilon$-ball around the equilibrium (where perhaps $\epsilon>\delta$). Asymptotic stability says that a system starting is some $\delta$-ball around the equilibrium will converge to the equilibrium.

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  • $\begingroup$ Some formulations in your post are misleading since "stability of $0$" plus "$\delta$ uniformly bounded", would mean $$\exists\delta>0\quad\forall\epsilon>0\quad\forall x\quad\|x\|<\delta\implies\sup_{t>0}\|\varphi(t,x)\|<\epsilon$$ which of course, never holds. Asymptotic stability is concerned with $\delta$ being uniformly bounded below, yes, but also with limits when $t\to\infty$ instead of supremum over every $t>0$, that is, asymptotic stability at $0$ is stability at $0$ plus the property that $$\exists\delta>0\quad\forall x\quad\|x\|<\delta\implies\lim_{t\to\infty}\|\varphi(t,x)\|=0$$ $\endgroup$
    – Did
    Mar 22, 2017 at 12:06
  • $\begingroup$ @Did Ah, yes, I skimmed and didn't catch their exact definition of $\delta$. I'll fix it. $\endgroup$
    – Ian
    Mar 22, 2017 at 12:10
  • $\begingroup$ @Ian In your modified version of definition, why $t \ge t_0(\epsilon)$ matters? Does it mean that when $\epsilon \to 0$, it is possible that $t_0(\epsilon) \to \infty $ so the limit could be achieved? However, without the $\epsilon$ as its independent variable, $t$ could not go to inifinity? $\endgroup$
    – winston
    Mar 22, 2017 at 13:40
  • $\begingroup$ @winston Yes, we should have $t_0 \to \infty$ as $\epsilon \to 0$, unless $x_0=0$. Without introducing this variable $t_0$, the definition again cannot apply to any nonzero value of $x_0$, because by definition $\varphi(0,x_0)=x_0$. (This was the content of Did's comment.) $\endgroup$
    – Ian
    Mar 22, 2017 at 13:44

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