On Lyapunov numbers

118 Views Asked by At

I'm working on a paper , I almost understood the whole paper , I just didn't understand some examples of that. I try to explain a little about some definitions.

$(X,f)$ is a dynamical system , where $X$ is a compact metric space with metric $d$ and $ f : X \rightarrow X $ is a continuous map.

I try to define some Lyapunov numbers :

The first Lyapunov number is defined the way like this :
$ L_{r} := \sup \lbrace \varepsilon : \forall x \in X , \forall U_{x} , \exists y \in U_{x} , \exists n \in \mathbb{Z}_{+} \text{such that}: d(f^{n}(x),f^{n}(y)) > \varepsilon \rbrace $
The third Lyapunov number is defined the way like this :
$ \overline{L_r} := \sup \lbrace \varepsilon : \forall x \in X , \forall U_{x} \quad \exists y \in U_{x} \quad \text{such that}: \limsup_{n \to \infty} d(f^{n}(x) , f^{n}(y) ) > \varepsilon \rbrace $

And the orbit of a map is defined :

$\lbrace f^{n}(x) \quad : n\geq 0 \rbrace $

Now some examples were solved in the paper , but I can not understand how they were solved:

Consider the following map :
$ g : [0,1] \rightarrow [0,1] $ and $ g(x) = 3((x-\frac{1}{3}) - |x-\frac{1}{3}| + |x-\frac{2}{3}|)$

In the paper , it is said that $\frac{1}{2}$ is a fixed point of the map , so clearly that $L_{r} =\frac{1}{2}$.

I can't understand how the $L_{r}$ was computed.
This is the link of the paper : On the Lyapunov numbers

Could you please help me ?

1

There are 1 best solutions below

2
On BEST ANSWER

Since $x_0=1/2$ is a fixed point then

$$d(f^n(x_0), f^n(y))=d(x_0, f^n(y))=d(1/2, f^n(y))\leq 1/2$$

since we are inside $[0,1]$. Therefore $L_r\leq 1/2$.

On the other hand the crucial thing you missed from the paper is that $g$ is topologically transitive. And so in this scenario there is $y_0$ such that the orbit of $y_0$ is dense. Therefore $d(1/2, f^n(y_0))$ is arbitrarly close (from below) to $1/2$. Meaning $d(f^n(x_0), f^n(y_0))$ is arbitrarly close to $1/2$ and therefore $L_r\geq 1/2$.