Does "$\exists \delta >0$ S.T $||x(0)-x_e||<\delta\Rightarrow \displaystyle \lim_{t\rightarrow \infty}||x(t)-x_e||=0$" imply stability?

Question

Recall the definition of stable and Asymptotically stable:

1. A fixed point $x_e$ of a vector field is called (Lyapunov) stable if $\forall \varepsilon>0,\exists \delta(\varepsilon)$ such that $\forall ||x(0)-x_e||<\delta \Rightarrow ||x(t)-x_e||<\varepsilon,\forall t\geq 0$
2. A fixed point $x_e$ of a vector field is called asymptotically stable if it's stable and the condition mentioned in the title "$\exists \delta >0$ S.T $||x(0)-x_e||<\delta\Rightarrow \displaystyle \lim_{t\rightarrow \infty}=||x(t)-x_e||=0$" is true.

So, my question is the that if the second condition in asymptotically stable implies stable? If not, what counter example can we take?

I guess the second condition doesn't imply stable. "stable" means I start my flow close to the fixed point $x_e$ and it will be always close but not necessarily convergent to $x_e$. And "asymptotically stable" means, besides the stability, the flow will not only be just close to $x_e$ but also converges to it in the end.

Thus, from this point of view, the second condition cannot imply "stable", because there definitely exists a flow that it will finally converge to $x_e$ but it won't necessarily be close to $x_e$ all the time, which means it might detour for a certain period and come back. However, I have hard time finding such flow.

I would really appreciate if you can help me find an example of such flow.

Answer 1

Asymptotic stability is defined as Lyapunov stability plus the condition you cite; see, for example, this.

Under these definitions, asymptotic stability certainly implies Lyapunov stability, but not conversely. As a simple example, think of a trajectory that oscillates around $x_e$ with amplitude $\varepsilon/2$ for any $\varepsilon>0$.

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Edit: Ok, maybe my first reading didn't focus on the main issue of your question. Just because a trajectory limits to $x_e$ doesn't mean that it must "start close" (with a $\delta$-neighborhood of $x_e$) and "stay close" (in an $\varepsilon$-neighborhood of $x_e$) for all $t\ge 0$.

Think of $f(t)=te^{-t}$ with $\varepsilon={1\over 10}$.

Answer 2

No, it does not. Because stability means that we are always close to the equilibrium. It is a simple matter to present an example when it is true that $$ \|x(t)-x_e\|\to 0 $$ for any initial conditions close enough to $x_e$, such that there are orbits that leave any small neighborhood of $x_e$.