1
$\begingroup$

I was wondering how could I prove such a property stated in [Angeli, 2004]. For instance, consider the system $\dot{x}=f(x)$, where $f:\mathbb{R}^n\to\mathbb{R}^n$ is Lipschitz continuous.

Claim. Suppose that the system is locally stable and globally attractive. Then, it is globally asymptotically stable.

Proof. I was trying to prove it as follows. From local stability definition, there exists neighborhood $N$ of the origin for which, $$\forall\varepsilon>0, \exists\delta>0:|\xi|<\delta\Rightarrow|\phi(t,\xi)|<\varepsilon,\forall t\geq0,\forall \xi\in N,$$ and \begin{equation*} \lim_{t\to\infty}\phi(t,\xi)=0,\quad\forall \xi\in N. \end{equation*} where $\phi$ is a solution for the system.

Let $\tilde{\xi}\notin N$ be an initial condition for a solution $\phi$ of the system. From the global attracvity, $$\lim_{t\to\infty}\phi(t,\tilde{\xi})=0.$$ From the local asymptotic stability of the origin, there exists $T<\infty$ such that, $\forall t>T$, $\phi(T,\tilde{\xi})\in N$. Moreover, \begin{equation*} \forall\varepsilon>0, \exists\delta>0:|\phi(T,\tilde{\xi})|<\delta\Rightarrow|\phi(t,\phi(T,\tilde{\xi}))|<\varepsilon,\forall t\geq0. \end{equation*} From the time-invariance property and unicity of solutions, $$\forall t\geq0,\quad\phi(t,\phi(T,\tilde{\xi}))=\phi(t+T,\tilde{\xi}).$$

Thus, $\forall\xi\in\mathbb{R}^n$, \begin{equation*} \forall\varepsilon>0,\exists\delta>0:|\xi|<\delta\Rightarrow|\phi(t,\xi)|<\varepsilon \end{equation*} and \begin{equation*} \lim_{t\to\infty}\phi(t,\xi)=0. \end{equation*} Hence, the origin of the system is globally asymptotically stable.$\square$

Is this proof correct?

References:

[Angeli, 2004] Angeli, D. (2004). An Almost Global Notion of Input-to-State Stability. IEEE Trans. Aut. Control, 49(6):866–874.

$\endgroup$
1
$\begingroup$

I think you misunderstood the paper. The property "global asymptotic stability" is defined to be the combination of local stability and global attractivity. There is nothing to prove here; it's a definition. Here is the relevant part:

definition

$\endgroup$
0
$\begingroup$

No, the proof is not correct. Because $\phi(t+T,\tilde \xi) < \epsilon \not \Rightarrow \phi(t,\xi) < \epsilon$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.