# An example for dynamical systems and its Lyapunov number

I already asked a question about Lyapunov numbers , as I said , I'm working on a paper in which we introduce Lyapunov numbers and find relations between them, as I said in the previous post, the first and the third Lyapunov number are defined i, the following way:

The first :

$$L_{r} := \sup \lbrace \varepsilon : \forall x \in X , \forall U_{x} , \exists y \in U_{x} , \exists n \in \mathbb{Z}_{+} \text{such that}: d(f^{n}(x),f^{n}(y)) > \varepsilon \rbrace$$

The third :

$$\overline{L_r} := \sup \lbrace \varepsilon : \forall x \in X , \forall U_{x} \quad \exists y \in U_{x} \quad \text{such that}: \limsup_{n \to \infty} d(f^{n}(x) , f^{n}(y) ) > \varepsilon \rbrace$$

And a dynamical system $$(X,f)$$ is defined the way like this :

$$(X,f)$$ is a dynamical system , where $$X$$ is a compact metric space with metric $$d$$ and $$f : X \rightarrow X$$ is a continuous map and $$O_{f} = \lbrace f^{n}(x) \quad : n\geq 0 \rbrace$$ is the orbit of the map $$f$$.

I have problem understanding the proof of this proposition :

There exists a topological dynamical system $$(X, f)$$ for which $$L_{r}=2\overline{L_{r}}$$.

Its proof is this :

Proof : We define the space $$X$$ as a compact surface in $$R^{3}$$ which is homeomorphic to a two-dimensional disk in $$R^{2}$$.
More precisely, the cylindrical coordinates of a point $$(x, y, z) \in X$$ have the form $$(r,\varphi, z)$$, where $$r = \sqrt{x^{2} + y^{2}}$$ and $$\varphi$$ is an angle, for which $$x = r\cos \varphi$$ and $$y = r\sin \varphi$$.
In other words, $$(r,\varphi)$$ are the polar coordinates of $$(x, y)$$, and $$z$$ remains unchanged.
Let $$h(r) = 8r(1 − r)$$.
Now, define $$X$$ as a set of points with cylindric coordinates $$(r,\varphi, h(r))$$, where $$0 ≤ r ≤ 1$$,$$\varphi ∈ R$$, and let the Euclidean metric (in $$R^{3}$$) $$d$$ be the metric on $$X$$.
Now we define a continuous map $$f$$ from $$X$$ to itself as follows $$f : (r,\varphi, h(r)) → (g(r), 2\varphi, h(g(r)))$$, where $$g(x)$$ is a continuous map $$[0, 1] → [0, 1]$$ with $$g(0) = 0$$, $$g(1) = 1$$ and $$g(x) > x$$ for all $$x ∈ (0, 1)$$.
From this properties one can easily deduce that $$\lim_{n \to \infty} g^{n}(x) = 1$$ for any $$x ∈ (0, 1]$$.
For example, let $$g(x) = 2x − x2$$.
Let $$p ∈ X$$ and $$U$$ be a neighborhood of $$p$$. If $$p \neq (0, 0, 0)$$, then for any $$\delta > 0$$ there are $$n ∈ N$$ and $$q ∈ U$$ such that $$d(f^{n}(p), f^{n}(q)) > 2 − \delta$$.
If $$p = (0, 0, 0)$$, then there are $$n ∈ N$$ and $$q ∈ U$$, for which $$f^{n}(q)$$ lies on a circumference of $$X$$ with the center $$(0, 0, 2)$$ (in $$R^{3}$$) and the radius $$\dfrac{1}{2}$$ .
For these $$n$$ and $$q$$ we have $$d(f^{n}(p), f^{n}(q)) > 2$$ and so $$L_{r} ≥ 2$$.
Now, let $$p = (0, 0, 0)$$. The equality $$\lim_{n \to \infty} d(f^{n}(p), f^{n}(q)) = 1$$ holds for any $$q \neq p$$.
So $$Lr ≤ 1$$.
Since $$L_{r} ≤ 2\overline{L_{r}}$$ (by Proposition 2.1), it gives $$Lr = 2\overline{L_{r}}$$.

My questions :

1- How did they find out when $$p \neq (0,0,0)$$ then for every $$\delta > 0$$ there exists $$n \in \mathbb{N}$$ and $$q \in U$$ which $$d(f^{n}(p) , f^{n}(q)) > 2 - \delta$$ ?
2- How did they find out when $$p=(0,0,0)$$ then those things in the proof of the theorem happen ?

Here is the link of the paper On Lyapunov Numbers

• I guess near the end of the proof should be $\overline{L_{r}}≥ 2$ instead of $L_{r} ≥ 2$ and $L_r\le 1$ instead of $Lr\le 1$, also $g(x)$ should be $2x−x^2$ instead of $2x−x2$. – Alex Ravsky Jul 27 at 19:02

Since $$f(0,0,0)=(0,0,0)$$, we have $$f^n(0,0,0)=(0,0,0)$$ for each $$n\in\Bbb N$$.
2- How did they find out when $$p=(0,0,0)$$ then those things in the proof of the theorem happen ?
Let $$q=(r,\varphi,h(r))$$, where $$0\le r\le 1$$ and $$\varphi\in\Bbb R$$. Since $$q\ne p$$, $$r>0$$. Then $$f^n(q)=(g^n(r),n\varphi,h(g^n(r))$$ for any $$n\in\Bbb N$$. Since $$r>0$$, $$g^n(r)$$ tends to $$0$$ when $$n$$ tends to infinity. Then $$h(g^n(r)$$ tends to zero. Thus for any $$\delta>0$$ there exists $$N$$ such that for each $$n>N$$, $$f^n(q)$$ belongs to $$\delta$$-neighborhood of the circle $$C=\{(0, \psi,0): \psi\in \Bbb R\}$$. Since the distance from $$p$$ to any point of $$C$$ equals $$1$$, the distance $$d(p,f^n(q))$$ tends to $$1$$.
1- How did they find out when $$p \neq (0,0,0)$$ then for every $$\delta > 0$$ there exists $$n \in \mathbb{N}$$ and $$q \in U$$ which $$d(f^{n}(p) , f^{n}(q)) > 2 - \delta$$ ?
Let $$p=(r,\psi,h(r))$$, where $$0 and $$\psi\in\Bbb R$$. There exists a number $$m\in\Bbb N$$ such that a point $$q=(r,\psi+\pi/m,h(r))\in U$$. Similarly to the answer to Question 2, we can show that for any $$\varepsilon>0$$ there exists a number $$N$$ such that for each odd $$k>N$$, $$f^{km}(p)$$ and $$f^{km}(q)$$ belong to $$\delta/2$$-neigborhood of the circle $$C$$. Since $$km(\psi+\pi/m)-km\psi=k\pi$$ and $$k$$ is odd, points $$f^{km}(p)$$ and $$f^{km}(q)$$ are in $$\delta/2$$-neigborhoods of endpoints of a diameter of the circle $$C$$. Then the triangle inequality implies that $$d(f^{km}(p), f^{km}(q))> 2 - \delta$$.