Knowing when the Logistic map sequence converges for all initial guesses

Question

Let $4 \geq \alpha \geq 0$ and $T_{\alpha}:[0,1] \rightarrow [0,1]$ be the function defined by $$T_{\alpha}(x) =\alpha x(1-x)$$ for all $x\in [0,1]$. For which values of $\alpha$ the sequences given by $y^{(x,\alpha)}_n = {T_{\alpha}^{n}(x)}$ converges for all $x \in (0,1)$?

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Notation:

Inductively define $T^{n+1} = T^{n} \circ T $ and $T^{0} = I$, the identity function.

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Observations:

It looks that this is the Logistic map. I think I need only find a proof that these sequences converges for $1-1/\alpha$ for $1 \leq \alpha \leq 3$ and diverges for some $x \in (0,1)$ with $ \alpha >3.$

Answer 1

Partial answer

The sequence $y_n$ is constructed as follows \begin{cases} y_{n+1} = \alpha y_n(1-y_n) \\ y_0=x \in(0,1) \end{cases}

Case 1: $0 \le \alpha \le 1$ It's easy to notice that $y_n >0$ for all $n$ and the sequence is decreasing as

$$y_{n+1}-y_n = (\alpha-1) y_n - \alpha y_n^2 <0$$

So, the sequence converges for all values of $y_0 = x \in(0,1)$

$$$$ Case 2: $1 < \alpha < 2$ Let's denote $r$ the solution of $r = T_{\alpha}(r)$ then $r = 1-\frac{1}{\alpha}$. We notive that $$T'_{\alpha}(r) = \alpha(1-2r) = 2-\alpha<1$$ According to this result, as $x_{n+1} = T_{\alpha}(x_n)$ and the point $r = 1-\frac{1}{\alpha}$ is stable, the sequence converges to $r$.

Case 3: $2 \le \alpha \le 3$

Not yet

$$$$ Case 4: $3 < \alpha \le 4$

$\qquad$ First, we will find $(x,y)$ such that $x \ne y$, $x \in (0,1)$ and \begin{cases} y = \alpha x(1-x) \\\tag{1} x = \alpha y(1-y) \end{cases} From the 2 equations $(1)$, we deduce that $x+y=1+\frac{1}{\alpha}$. And then $x$ is the solution of $$1+\frac{1}{\alpha}-r=\alpha r(1-r) \tag{2}$$ The equation $(2)$ has 2 distinct real roots in $(0,1)$ $$r_{1,2} = \frac{(\alpha+1) \color{red}{\pm} \sqrt{(\alpha-1)^2-4}}{2\alpha}$$

$\qquad$ Second, take $y_0 = r1_1$, we will have $y_1 = r_2$ and in general \begin{cases} y_{2n} = r_1 \\ y_{2n+1} = r_2 \end{cases} So, with $3 < \alpha \le 4$, the sequence $y_n$ doesn't converge for all $y_0 \in (0,1)$ $$$$

Answer 2

$\textbf{Case 2:}$ $1 < \alpha \leq 2$

Let, first, $x\in (0,1/2]$, we have that

\begin{align*} T_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) = & \alpha x (1-x)-\left(1-\dfrac{1}{\alpha}\right) \\ = & \alpha x \left( 1 - \dfrac{1}{\alpha} -x + \dfrac{1}{\alpha}\right)-\left(1-\dfrac{1}{\alpha} - x + x\right) \\ = & \left(\alpha x- 1 \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right). \end{align*}

Now, since $| \alpha x- 1 | < 1, $ for $0<x\leq 1/2$, we get
\begin{equation*} \left| T_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) \right|<\left| x-\left(1-\dfrac{1}{\alpha}\right) \right|, \tag{1} \end{equation*} whenever $x \not = 1-\dfrac{1}{\alpha} $. Noting that $T_{\alpha} \left( 1- \dfrac{1}{\alpha}\right) = \left( 1- \dfrac{1}{\alpha}\right)$ and (1) holds, $\inf_{k\in\mathbb{N}} \left| x^{k} -\left( 1- \dfrac{1}{\alpha}\right) \right|$ cannot be greater than zero. Once the sequence $\left\lbrace \left| x^{k} -\left( 1- \dfrac{1}{\alpha}\right) \right|\right\rbrace_{k\in\mathbb{N}} $ strictly decreasing, $x^{k}$ converges to $\left(1- \dfrac{1}{\alpha}\right).$

On the other hand, if $ 1/2 < x \leq 1$, $0<T_{\alpha}(x) \leq 1/2$ and the argument below applies.

$\textbf{Case 3:}$ $2<\alpha \leq 3$

This case is slightly harder and its proof works for the other, but since is more complicated I have decided to split both cases. We need to prove:

1. The sequence defined by $T^{2 k}_{\alpha} (y)$ converges for all initial guess $y \in \left(0,\dfrac{\alpha}{4}\right]$;

2. Take into account that for $1 > x > \dfrac{\alpha}{4}$ its true $ 0 < T(x)\leq \dfrac{\alpha}{4}$;

3. Conclude that $x_{2k}$ converges to $1-1/\alpha$ and $x_{2k+1}$ converges to $1-1/\alpha$ as well and get the desired result;

The same argument applied to the $\textbf{Case 1}$, implies that for all $x \in (0,1),$ \begin{align*} T_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) = \left(\alpha x- 1 \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right), \end{align*} then \begin{align*} T^{2}_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) = & \left(\alpha T_{\alpha}(x) - 1 \right) \left(\left(1-\dfrac{1}{\alpha}\right) - T_{\alpha}(x) \right) \\ & \left(\alpha T_{\alpha}(x) - 1 \right) \left( 1 - \alpha x \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right) \\ = & \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right). \\ \end{align*}

Since problem

\begin{equation*} \begin{array}{c c} \text{maximize}_{(x,\alpha) \in \mathbb{R}^{2}} & \left| \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right)\right| \\ \text{subject to} & 0 \leq x \leq \dfrac{\alpha}{4}\\ & 1 \leq \alpha \leq 3 \end{array} \end{equation*} have maximum less or igual than $1$ and additionally $$ \left| \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right)\right| \not = 1$$ for all $ 0 < x \leq \dfrac{\alpha}{4}$ and $x \not = 1 - \dfrac{1}{\alpha},$ we have
\begin{equation*} \left| T^{2}_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) \right| < \left| \left(\left(1-\dfrac{1}{\alpha}\right) - x\right)\right|, \tag{2} \end{equation*} for all $ 0 < x \leq \dfrac{\alpha}{4}$ and $x \not = 1 - \dfrac{1}{\alpha}. $

Since $0 <T_{\alpha} (y) \leq \dfrac{\alpha}{4}$ for all $ 0 < y < 1$, given $0 < y < 1$ we have
$$ \left| T^{2(k+1)}_{\alpha}(y) - \left(1-\dfrac{1}{\alpha} \right) \right| = \left| T^{2}_{\alpha}( T_{\alpha}^{2k} (y)) - \left(1-\dfrac{1}{\alpha} \right) \right| < \left| T_{\alpha}^{2k} (y) - \left(1-\dfrac{1}{\alpha} \right) \right|. $$ Hence, the sequence $\left\lbrace \left| T^{2k}_{\alpha}(y) - \left( 1-\dfrac{1}{\alpha}\right) \right| \right\rbrace_{k\in\mathbb{N}}$ is decreasing and \begin{equation} \inf_{k\in\mathbb{N}} \left| T^{2k}_{\alpha}(y) - \left(1-\dfrac{1}{\alpha} \right) \right| \end{equation} cannot be greater than zero for all $ 0 < y <1 $, which means that \begin{equation*} \lim_{k\in\mathbb{N}} T^{2k}_{\alpha}(y) = 1-\dfrac{1}{\alpha} \tag{3} \end{equation*} for all $ 0 < y < 1$. With the aim to prove it, is needed to put all the sequence into a compact that does not contain $0$, $1-\dfrac{1}{\alpha}$ and $1$ and note that the inequality (2) would remain with a sufficiently small positive constant added in the LHS for all $x$ in that compact, which would would lead to a contradiction.

Now, for any $x \in \left(0,1\right)$, we have that $0 < T_{\alpha}(x) \leq \dfrac{\alpha}{4}$, hence \begin{align*} \left| T^{2k + 3}_{\alpha}(x) - \left(1-\dfrac{1}{\alpha} \right) \right| = & \left| T^{2(k+1)}_{\alpha}( T_{\alpha} (x) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \\ = & \left| T^{2}_{\alpha}( T_{\alpha}^{2k} ( T_{\alpha} (x) )) - \left(1-\dfrac{1}{\alpha} \right) \right| \\ < &\left| T_{\alpha}^{2k} ( T_{\alpha} (x) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \rightarrow 0. \end{align*}

On the other hand, the same argument implies that \begin{align*} \left| T^{2k+4}_{\alpha}(x) - \left(1-\dfrac{1}{\alpha} \right) \right| = &\left| T^{2(k +1)}_{\alpha}( T^{2}_{\alpha} (x) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \\ = & \left| T^{2}_{\alpha}( T_{\alpha}^{2(k+1)} ( x )) - \left(1-\dfrac{1}{\alpha} \right) \right| \\ < &\left| T_{\alpha}^{2(k+1)} ( x ) - \left(1-\dfrac{1}{\alpha} \right) \right| \\ = &\left| T_{\alpha}^{2k} ( T_{\alpha}^2 ( x ) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \rightarrow 0. \end{align*} Hence the sequence $\left\lbrace T^{k}(x)\right\rbrace_{k\in\mathbb{N}}$ converges to the same number for odd and even numbers. Finally, it must be that $\left\lbrace T^{k}(x)\right\rbrace_{k\in\mathbb{N}}$ converges.

Answer 3

I have conjectured that the problem \begin{equation*} \begin{array}{c c} \text{maximize}_{(x,\alpha) \in \mathbb{R}^{2}} & \left| \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right)\right| \\ \text{subject to} & \epsilon \leq x \leq \dfrac{\alpha}{4}\\ & 1 \leq \alpha \leq 3 \\ & \epsilon^2 \leq \left(x-\left(1-\dfrac{1}{\alpha}\right) \right)^{2} \\ \end{array} \end{equation*} has minimum strictly less than 1 for all $\epsilon>0$. If someone prove it, the prove above will be complete. =)