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After analyzing the sensitivity dependence of orbits under the map $f(x) = 2x \mod 1$, we have found that every initial point has nearby points that eventually diverge by at least $1/2$ unit after iterations. This indicates the system's strong sensitivity to initial conditions, which is a characteristic of chaotic systems.

My question is, are there any other methods or techniques to investigate the sensitivity dependence of orbits for this map in more detail? Are there additional metrics or mathematical tools that can help us gain a deeper understanding of the nature of this system? I'm also curious to know if there are similar maps or systems that exhibit different behaviors or characteristics in terms of sensitivity dependence, for the purpose of comparison and study.

My confusion lies in how to delve deeper into the properties of this map and how to extend this research for comparison with other related studies. Are there any useful mathematical tools or methods to further explore this issue? And how to show the equilibrium of such chao system?

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    $\begingroup$ Lyapunov stability theorems are typically used to analyze the stability of equilibrium points in dynamical systems. However, in the case of a system like which is a discrete-time map, Lyapunov stability analysis in the traditional sense may not be directly applicable because it deals with continuous-time systems. $\endgroup$
    – D.y.s
    Commented Oct 8, 2023 at 4:33
  • $\begingroup$ Yes, I here mean if there is any equilibrium for a chaotic system? $\endgroup$
    – simple1
    Commented Oct 8, 2023 at 4:34

1 Answer 1

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To draw a chart of the itineraries of the map $(f(x) = 2x \mod 1$ on the unit interval [0, 1] in MATLAB:

% Define the map function
f = @(x) 2 * x - floor(2 * x);

% Initialize an array to store the itinerary
itinerary = zeros(1, 1000); % Adjust the array size as needed

% Starting point
x0 = 0.1; % we can choose any initial point in [0, 1]

% Calculate the itinerary
for i = 1:length(itinerary)
    itinerary(i) = x0;
    x0 = f(x0);
end

% Plot the itinerary
figure;
plot(1:length(itinerary), itinerary, '.');
xlabel('Iteration');
ylabel('Value');
title('Itinerary of f(x) = 2x (mod 1)');

To draw the transition graph for $f$, we can use the following code:

% Define the map function
f = @(x) 2 * x - floor(2 * x);

% Number of points in the graph
n = 100;

% Initialize arrays to store the points and connections
points = zeros(1, n);
connections = zeros(2, n);

% Generate points and connections
for i = 1:n
    points(i) = (i - 1) / (n - 1); % Equally spaced points in [0, 1]
    connections(:, i) = [points(i), f(points(i))];
end

% Plot the transition graph
figure;
plot([connections(1, :); connections(2, :)], [1:n; 1:n], 'k');
xlabel('x');
ylabel('Iteration');
title('Transition Graph for f(x) = 2x (mod 1)');

This code generates a transition graph for the map f(x) by connecting each point x to its image and plotting the connections.

fig1

To establish sensitive dependence for orbits under this map, we can show that each point has neighbors arbitrarily near that eventually map at least (1/2) unit apart. we can illustrate this by selecting a point and iterating the map for a small perturbation of that point.

% Define the map function
f = @(x) 2 * x - floor(2 * x);

% Initial point
x0 = 0.25; % Choose an initial point in [0, 1]

% Number of iterations
iterations = 100;

% Initialize an array to store the iterates
x_values = zeros(1, iterations);

% Perturbation magnitude
epsilon = 1e-4; % Small perturbation

% Iterate the map with a perturbed initial point
for i = 1:iterations
    x_values(i) = x0;
    x0 = f(x0) + epsilon * rand(); % Add a small random perturbation
end

% Plot the iterates
figure;
plot(1:iterations, x_values, '.');
xlabel('Iteration');
ylabel('Value');
title('Sensitive Dependence for f(x) = 2x (mod 1)');

This code selects an initial point x_0 and iterates the map $f(x)$ with a small random perturbation to demonstrate sensitive dependence. we should see that nearby points eventually diverge and become at least 1/2 unit apart in their trajectories. fig2

To illustrate sensitive dependence, we will choose an initial point $x_0$ and calculate the orbits for two nearby points.

$$x_0 = 0.3 \quad \text{and} \quad x_1 = 0.3001$$

We will iterate the map $f(x)$ for both points and observe how they diverge over time.

% Define the map function
f = @(x) 2 * x - floor(2 * x);

% Initial point
x0 = 0.25; % Choose an initial point in [0, 1]

% Number of iterations
iterations = 100;

% Initialize an array to store the iterates
x_values = zeros(1, iterations);

% Perturbation magnitude
epsilon = 1e-4; % Small perturbation

% Iterate the map with a perturbed initial point
for i = 1:iterations
    x_values(i) = x0;
    x0 = f(x0) + epsilon * rand(); % Add a small random perturbation
end

% Plot the iterates
figure;
plot(1:iterations, x_values, '.');
xlabel('Iteration');
ylabel('Value');
title('Sensitive Dependence for f(x) = 2x (mod 1)');

The results of our analysis are shown in the following figure: fig3

As shown in the figure, the orbits of $x_0$ (blue) and $x_1$ (red) initially start close to each other, but they quickly diverge. After a few iterations, they are at least $1/2$ unit apart.

We have demonstrated sensitive dependence for orbits under the map $f(x) = 2x \mod 1$. This means that for any initial point, there exist neighbors arbitrarily near to that point that eventually map to points at least $1/2$ unit apart. This sensitivity to initial conditions is a characteristic of chaotic systems.

To determine whether the period-two orbit of the map, let us take the example of your question like: $f(x) = 2x^2 - 5x$ on $\mathbb{R}$ is a sink, a source, or neither, we need to analyze the stability of this orbit.

First, let's find the period-two orbit. To do this, we need to find two points, say $x_1$ and $x_2$, such that $f(x_1) = x_2$ and $f(x_2) = x_1$. In other words, we want to find solutions to the equation $2x^2 - 5x = x_1$ and $2x^2 - 5x = x_2$. Solving these equations simultaneously, we get:

\begin{align*} 2x^2 - 5x &= x_1 \\ 2x^2 - 5x &= x_2 \end{align*}

Now, let's solve for $x_1$ and $x_2$:

For $x_1$: \begin{align*} 2x^2 - 5x - x_1 &= 0 \\ \end{align*}

For $x_2$: \begin{align*} 2x^2 - 5x - x_2 &= 0 \\ \end{align*}

We can solve these quadratic equations for $x_1$ and $x_2$ using the quadratic formula:

\begin{align*} x_1 &= \frac{5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-x_1)}}{2 \cdot 2} \\ x_2 &= \frac{5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-x_2)}}{2 \cdot 2} \end{align*}

Simplifying further:

\begin{align*} x_1 &= \frac{5 \pm \sqrt{25 + 8x_1}}{4} \\ x_2 &= \frac{5 \pm \sqrt{25 + 8x_2}}{4} \end{align*}

Now, we can see that there are two solutions for each $x_1$ and $x_2$, corresponding to the plus and minus signs in the square root. However, whether these solutions are sinks, sources, or neither depends on the stability of these fixed points.

To determine stability, we need to analyze the derivatives of the map $f(x) = 2x^2 - 5x$ at these fixed points. Specifically, we need to compute $f'(x)$ and evaluate it at the fixed points.

Let's compute $f'(x)$: \begin{align*} f'(x) &= \frac{d}{dx}(2x^2 - 5x) \\ &= 4x - 5 \end{align*}

Now, let's evaluate $f'(x)$ at the two solutions for $x_1$ and $x_2$:

For $x_1$: \begin{align*} f'(x_1) &= 4x_1 - 5 \end{align*}

For $x_2$: \begin{align*} f'(x_2) &= 4x_2 - 5 \end{align*}

To determine the stability of these fixed points, we need to check whether $|f'(x)| < 1$ (sink), $|f'(x)| > 1$ (source), or $|f'(x)| = 1$ (neither) at these points.

Without knowing the specific values of $x_1$ and $x_2$," we cannot definitively determine whether the period-two orbit of the map $f(x) = 2x^2 - 5x$ on $\mathbb{R}$ is a sink, a source, or neither. To make that determination, we would need to calculate the values of $x_1$ and $x_2$ and then evaluate $f'(x_1)$ and $f'(x_2)$ at those values. Depending on the values, we can classify the stability of the period-two orbit accordingly.

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