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Let us adopt the following definition of stability and asymptotic stability of a dynamical system of the form:

$$ \dot{x}=f(x) $$

The trajectory of this system starting from the initial point $x_0$ at $t_0=0$ will be hereinafter denoted by $\varphi(t,x_0)$.

Definition 1 (Stability). Let $x^\star$ be an equilibrium point for $\dot{x}=f(x)$, i.e. $f(x^\star)=0$. This point is said to be stable if for every $\varepsilon>0$ there is a $\delta=\delta(\varepsilon)>0$ such that $$ \|x-x^\star\|<\delta \implies \|\varphi(t,x_0)-x^\star\|<\varepsilon $$ for all $t\geq 0$.

Definition 2 (Asymptotic Stability). Let $x^\star$ be an equilibrium point for $\dot{x}=f(x)$, i.e. $f(x^\star)=0$. This point is said to be asymptotically stable if it is stable and additionally there is an $\alpha>0$ such that $$\lim_{t\to\infty}\|\varphi(t,x_0)-x^\star\|=0$$ for all $x$ with $\|x-x^\star\|<\alpha$.

Definition 3 (Classes of functions). We define the following four classes of functions: A function $\alpha:\mathbb{R}^+\to\mathbb{R}^+$ is said to be

  1. $\mathcal{K}$-class and we denote $\alpha\in\mathcal{K}$ if it is continuous, strictly increasing and $\alpha(0)=0$.
  2. A function $\beta:\mathbb{R}^+\to\mathbb{R}^+$ is said to be $\mathcal{K}_\infty$-class and we denote $\beta\in\mathcal{K}_\infty$ if $\beta\in\mathcal{K}$ and it is unbounded.
  3. A $\mu:\mathbb{R}^+\to\mathbb{R}^+$ is said to be $\mathcal{L}$-class and we denote $\mu\in\mathcal{L}$ if it is continuous, strictly decreasing and $\lim_{\tau\to\infty}\mu(\tau)=0$
  4. A $\gamma:\mathbb{R}^+\times\mathbb{R}^+\to\mathbb{R}^+$ is said to be $\mathcal{KL}$-class and we denote $\gamma\in\mathcal{KL}$ if $\gamma(\cdot,t)\in\mathcal{K}$ for all $t\in\mathbb{R}^+$ and $\gamma(r,\cdot)\in\mathcal{L}$ for all $r\in\mathbb{R}^+$

Proposition 1. An equilibrium point $x^\star$ is asymptotically stable if and only if there is a $\gamma\in\mathcal{KL}$ and a constant $\kappa>0$ so that: $$ \|\varphi(t;x_0)-x^\star\|\leq \gamma(\|x_0-x_\star\|,t) $$ for all $x_0$ such that $\|x_0-x^\star\|<\kappa$ and for all $t\geq 0$.

I have some problem with the proof!

Proof: 1. [The first part looks easy] Let us assume that $x^\star$ satisfies the inequality mentioned above. Let us choose an $\varepsilon>0$; $\gamma(\cdot,t)$ is $\mathcal{K}$-class, so we may choose a $\delta$ such that $0<\delta<\kappa$ and $\gamma(\delta,0)\leq \varepsilon$. Now for $\|x_0-x^\star\|<\delta$ we have:

$$ \|\varphi(t;x_0)-x^\star\|\leq\gamma(\|x_0-x_\star\|,t)\leq\gamma(\|x_0-x_\star\|,0)\leq\gamma(\delta,0)\leq\varepsilon $$

So if the inequality state above holds then $x^\star$ is stable and more: $$ 0\leq\lim_{t\to\infty}\|\varphi(t;x_0)-x^\star\|\leq \lim_{t\to\infty}\gamma(\|x_0-x^\star\|,t)=0 $$ Therefore $$ \lim_{t\to\infty}\|\varphi(t;x_0)-x^\star\|=0\Leftrightarrow\lim_{t\to\infty}\varphi(t;x_0)=x^\star $$

What about the converse? How can one prove it?

We need to prove that if $x^\star$ is a (locally) asymptotically stable equilibrium point, then there is a $\gamma\in\mathcal{KL}$ and a constant $\kappa>0$ so that: $$ \|\varphi(t;x_0)-x^\star\|\leq \gamma(\|x_0-x^\star\|,t) $$ for all $x_0$ such that $\|x_0-x^\star\|<\kappa$ and for all $t\geq 0$.

I would appreciate a proof or at least a reference to some book or article where I can find it clearly explained.

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A proof can be found in Khalil, Nonlinear Systems, 3rd edition, appendix C.6.

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It would be nice if you can give the outline here – Shailesh Sep 15 '15 at 16:09
    
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. – Lord_Farin Sep 15 '15 at 16:34

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