# Stability of a fixed point of a discrete dynamical system

I'm stuck with studying the stability of one fixed point of a discrete dynamical system given in exercise (3) page 44 of Petr Kůrka's Topological and Symbolic Dynamics. Could you please help me?

I recall the definition given in page 5:

Definition 1.7 Let $(X,F)$ be a dynamical system.

(2)A fixed point $x\in X$ is stable if $$\forall\varepsilon>0,\exists\delta>0,\forall y\in B_\delta(x),\forall n\in\mathbb Z^+,d(F^n(y),x)<\varepsilon.$$

where $B_\delta(x)$ is the open ball centered at $x$ with radius $\delta$, and $F^n$ is the composition of $F$ with itself $n$ times.

The exercise is about determining the fixed points and their stabilities of the following dynamical system: $(I,F_a)$ where $I=[0,1]$, $a>0$ and \begin{align*} F:I&\to I\\ x&\mapsto x+\dfrac{x}{a+1}\sin(a\ln x). \end{align*} The set of fixed points of $F_a$ is $$\left\{\exp\left(k\frac{\pi}{a}\right)\mid k\in\mathbb{Z}^-\cup\{0\}\right\}\cup\{0\}.$$ Let $p_k:=\exp\left(k\frac{\pi}{a}\right).$ We have $F_a'(p_k)=1+(-1)^k\dfrac{a}{a+1}$ so $p_k$ is stable iff $k$ is odd.

I'm stuck with the stability of $0$. $F_a$ is not differentiable at $0$ and $F'_a$ takes values larger than $1$ in any neighborhood of $0$. Numerically, it seems that $0$ is stable, as when I take $x$ close to $0$, $F_a^n(x)$ seems to converge to a fixed point close to it.

So here's my attempt:

0 is the limit point of the other fixed points $(p_k)_{k\in\mathbb{Z}^-}$, with $p_{2k}$ not stable and $p_{2k+1}$ stable.

If we pick $\varepsilon>0$, $\delta>0$ small enough and $x\in (0,\delta)$, as $F$ is bounded, so is the sequence $(F_a^n(x))_{n\in\mathbb N}$. $I$ being compact, $F_a^n(x)$ would have a converging subsequence, which converges to a stable fixed point of $F_a$. But because this point is stable, for big enough $n$, all the $F_a^n(x)$ would stay close to that fixed point.

The problem is, I can't really tell a thing about the first values of $F_a^n(x)$, nor if that stable fixed point the sequence is gets close to will be in $(0,\varepsilon)$. I attempted to solve this problem as follows: we have the inequality $$|F_a(x)-x|\le\dfrac{x}{a+1}.$$

So, if $0<x<\delta$, we get

$$|F_a^n(x)-x|\le\dfrac{n(n+1)\delta}{2(a+1)}.$$

Fix $\varepsilon > 0$. We can take a negative integer $k$ such that $$x_0 := \exp\left(k\frac{\pi}{a}\right)$$ is closer to $0$ than $\varepsilon$.
We need to show now that the derivative of the mapping on $(0,1]$ is positive. Indeed, $$F_a'(x) = 1 + \frac{1}{a+1} \sin(a \ln{x}) + \frac{a}{a+1}\cos(a \ln{x}),$$ $\sin(a \ln{x}) \ge -1$, $\cos(a \ln{x}) \ge -1$, and they cannot be simultaneously equal to $-1$.
Consequently, $F_a$ preserves orientation. Therefore, for any $x \in (0,x_0)$ its iterates $F_a^n(x)$ are contained between the iterates of $0$ and the iterates of $F_a^n(x_0)$, that is, they belong to $(0, x_0)$.
So, a $\delta > 0$ required in the definition is just the distance between $0$ and $x_0$.