Henon map's sink fixed point and saddle fixed point.

Question

It is noticeable that the sinking and saddling points of the Henon map are appropriately satisfied in both $a = 0.27, -0.09$, and $b=0.4$. As written in this fig:

My question is, why the Henson map has such properties for $a$ and $b$, what is the relationship between these two parameters?

I also noticed that when a = 0.27 the loss of stability of the sink happens, is this regarding to parameter $b$ or not? For example, for $0.27 < a < 0.85$, the map has a period-two sink.

And if the problem above is irregular, then if there exists the largest magnitude eigenvalue of $Df^2$, the Jacobian of $f^2$ at the period-two orbit, when $a = 0.85$.

clc
clear all

%% PARAMETERS
iterations = 1000;
x = zeros(1, iterations);
y = zeros(1, iterations);
a = 1.4;
b = 0.3;
dt = 0.001;

%% FLAGS
drawing = 0; % set to 1 to plot at each step
bifurcation = 1; % set to 1 or 2 to plot birfurcation (x & a, x & b)
plotter = 0; % set to 1 for normal plot (x vs y)

%% PLOTTING
% Normal plot
if plotter==1
    % set initial value of x
    x(1,1) = 0.1;
    for i=1:iterations
        % apply update
        x(1,i+1) = 1-(a*x(1,i)^2) + y(1,i);
        y(1,i+1) = b*x(1,i);
        % plotting
        plot(x(5:iterations),y(5:iterations), '.k','MarkerSize',3)
        line1 = sprintf('Henon Map with %.0f iterations', iterations);
        line2 = sprintf('a = %.2f, b = %.4f', a, b);
        title({line1, line2});
        xlabel('x')
        ylabel('y')
        % if model is set to update after each iteration
        if drawing==1
            drawnow
        end
    end
end

% to demonstrate system dynamics with varying 'a' parameter
if bifurcation==1
    minAlpha = 0.1;
    for alpha = minAlpha:dt:(a)
        % set initial values
        x(1,1) = 0.1;
        y(1,1) = 0;
        % start iterating
        for i=1:iterations
            % apply Henon Map update
            x(1,i+1) = 1-(alpha*x(1,i)^2) + y(1,i);
            y(1,i+1) = b*x(1,i);
        end
        % plotting
        plot(alpha*ones(1,10), x(1,iterations-9:iterations),'.k','MarkerSize',1)
        line1 = sprintf('Bifurcation diagram for the Henon Map with %.0f iterations', iterations);
        line2 = sprintf('b = %.2f', b);
        title({line1, line2});
        xlabel('a')
        ylabel('x')
        if drawing==1
            drawnow
        end
        hold on
    end
end

% to demonstrate system dynamics with varying 'b' parameter
if bifurcation==2
    minBeta = -0.6;
    for beta1 = minBeta:dt:(b+0.1)
        % set initial values
        x(1,1) = 0.1;
        y(1,1) = 0;
        % start iterating
        for i=1:iterations
            % apply Henon Map update
            x(1,i+1) = 1-(a*x(1,i)^2) + y(1,i);
            y(1,i+1) = beta1*x(1,i);
        end
        % plotting
        plot(beta1*ones(1,10),x(1,iterations-9:iterations),'.k','MarkerSize',1)
        line1 = sprintf('Bifurcation diagram for the Henon Map with %.0f iterations', iterations);
        title({line1});
        xlabel('a')
        ylabel('x')
        if drawing==1
            drawnow
        end
        hold on
    end
end

Answer 1

To prove that for $0.09 < a < 0.27$, the Hénon map $f$ has one sink fixed point and one saddle fixed point, we need to analyze the stability of the fixed points for different values of $a$. The Hénon map is defined by the following equations:

\begin{align*} x_{n+1} &= 1 - a x_n^2 + y_n \\ y_{n+1} &= 0.4 x_n \end{align*}

The fixed points of the map are found by setting $x_{n+1} = x_n$ and $y_{n+1} = y_n$. In other words, we are looking for values $(x, y)$ such that:

\begin{align*} x &= 1 - a x^2 + 0.4 y \\ y &= 0.4 x \end{align*}

Now, we can solve for $x$ and $y$ simultaneously:

From the second equation, we get $y = 0.4x$, and we can substitute this into the first equation:

$$x = 1 - a x^2 + 0.4(0.4x) = 1 - a x^2 + 0.16x$$

Now, we have a single equation in one variable, $x$:

$$x = 1 - a x^2 + 0.16x$$

Rearrange it to a quadratic equation:

$$a x^2 - 0.16x + x - 1 = 0$$

Now, we have a quadratic equation in $x$, which has two solutions. To determine their stability, we need to compute the derivative of the map with respect to $x$ and evaluate it at these fixed points. We use the following derivatives:

\begin{align*} \frac{df}{dx} &= -2ax + 1 \\ \frac{df}{dy} &= 0.4 \end{align*}

Now, we will find the values of $x$ that satisfy the quadratic equation for each $a$:

1. For $x_1$, let's denote the smaller solution:

$$a x_1^2 - 0.16x_1 + x_1 - 1 = 0$$

2. For $x_2$, let's denote the larger solution:

$$a x_2^2 - 0.16x_2 + x_2 - 1 = 0$$

Now, for each $a$ in the range $0.09 < a < 0.27$, we need to compute $x_1$ and $x_2$, and then calculate the derivatives $\frac{df}{dx}$ at these points.

The stability of a fixed point depends on the magnitude of the derivative at that point, as we can also see from the code and figure:

• If $\left|\frac{df}{dx}\right| < 1$, the fixed point is a stable sink.
• If $\left|\frac{df}{dx}\right| > 1$, the fixed point is a saddle point.

\begin{lstlisting}
clc
clear all

%% PARAMETERS
iterations = 1000;
x = zeros(1, iterations);
y = zeros(1, iterations);
a = 1.4;
b = 0.3;
dt = 0.001;

%% FLAGS
drawing = 0; % set to 1 to plot at each step
bifurcation = 1; % set to 1 or 2 to plot birfurcation (x & a, x & b)
plotter = 0; % set to 1 for normal plot (x vs y)

%% PLOTTING
% Normal plot
if plotter==1
    % set initial value of x
    x(1,1) = 0.1;
    for i=1:iterations
        % apply update
        x(1,i+1) = 1-(a*x(1,i)^2) + y(1,i);
        y(1,i+1) = b*x(1,i);
        % plotting
        plot(x(5:iterations),y(5:iterations), '.k','MarkerSize',3)
        line1 = sprintf('Henon Map with %.0f iterations', iterations);
        line2 = sprintf('a = %.2f, b = %.4f', a, b);
        title({line1, line2});
        xlabel('x')
        ylabel('y')
        % if model is set to update after each iteration
        if drawing==1
            drawnow
        end
    end
end

% to demonstrate system dynamics with varying 'a' parameter
if bifurcation==1
    minAlpha = 0.1;
    for alpha = minAlpha:dt:(a)
        % set initial values
        x(1,1) = 0.1;
        y(1,1) = 0;
        % start iterating
        for i=1:iterations
            % apply Henon Map update
            x(1,i+1) = 1-(alpha*x(1,i)^2) + y(1,i);
            y(1,i+1) = b*x(1,i);
        end
        % plotting
        plot(alpha*ones(1,10), x(1,iterations-9:iterations),'.k','MarkerSize',1)
        line1 = sprintf('Bifurcation 
        
        diagram for the Henon Map with %.0f iterations', iterations);
        
        line2 = sprintf('b = %.2f', b);
        title({line1, line2});
        xlabel('a')
        ylabel('x')
        if drawing==1
            drawnow
        end
        hold on
    end
end

% to demonstrate system dynamics with varying 'b' parameter
if bifurcation==2
    minBeta = -0.6;
    for beta1 = minBeta:dt:(b+0.1)
        % set initial values
        x(1,1) = 0.1;
        y(1,1) = 0;
        % start iterating
        for i=1:iterations
            % apply Henon Map update
            x(1,i+1) = 1-(a*x(1,i)^2) + y(1,i);
            y(1,i+1) = beta1*x(1,i);
        end
        % plotting
        plot(beta1*ones(1,10),x(1,iterations-9:iterations),'.k','MarkerSize',1)
        line1 = sprintf('Bifurcation diagram for the Henon Map with %.0f iterations', iterations);
        title({line1});
        xlabel('a')
        ylabel('x')
        if drawing==1
            drawnow
        end
        hold on
    end
end

\end{lstlisting}

For your second question,To analyze the stability of the fixed point at a specific value of $a = 0.27$ for the Hénon map, We can calculate the Jacobian matrix and find its largest magnitude eigenvalue. The Jacobian matrix is given by:

\begin{equation} J = \begin{bmatrix} \frac{\partial f_x}{\partial x} & \frac{\partial f_x}{\partial y} \\ \frac{\partial f_y}{\partial x} & \frac{\partial f_y}{\partial y} \end{bmatrix} \end{equation}

Where:

\begin{align*} f_x &= 1 - a x^2 + 0.4 y \\ f_y &= 0.4 x \end{align*}

Now, let's calculate the partial derivatives:

\begin{align*} \frac{\partial f_x}{\partial x} &= -2a x \\ \frac{\partial f_x}{\partial y} &= 0.4 \\ \frac{\partial f_y}{\partial x} &= 0.4 \\ \frac{\partial f_y}{\partial y} &= 0 \end{align*}

Now, plug these values into the Jacobian matrix:

\begin{equation} J = \begin{bmatrix} -2a x & 0.4 \\ 0.4 & 0 \end{bmatrix} \end{equation}

At the fixed point, we have $x = 1 - a x^2 + 0.4 y$ and $y = 0.4x$. For the fixed point, We should solve these equations to find the values of $x$ and $y$. These values will give We the specific fixed point We want to analyze.

Let's denote the fixed point as $(x^*, y^*)$.

Next, find the Jacobian matrix at this fixed point:

\begin{equation} J^* = \begin{bmatrix} -2a x^* & 0.4 \\ 0.4 & 0 \end{bmatrix} \end{equation}

Now, find the eigenvalues of this matrix $J^*$, and specifically, find the largest magnitude eigenvalue.

We can use MATLAB or a similar tool to calculate the eigenvalues, or We can do it analytically. The following is to solve the eigenvalues out:

\begin{lstlisting}
% Define the fixed point (x*, y*)
a = 0.27;
x_star = 1 - a * x_star^2 + 0.4 * y_star;
y_star = 0.4 * x_star;

% Define the Jacobian matrix J at the fixed point
J = [-2*a*x_star, 0.4; 0.4, 0];

% Calculate the eigenvalues
eigenvalues = eig(J);

% Find the largest magnitude eigenvalue
largest_magnitude_eigenvalue = max(abs(eigenvalues));

% Display the eigenvalues and the largest magnitude eigenvalue
disp('Eigenvalues:');
disp(eigenvalues);
fprintf('Largest magnitude eigenvalue: %.6f\n', largest_magnitude_eigenvalue);

The analytical solution is to find the eigenvalues of the Jacobian matrix $J^*$ at the fixed point $(x^*, y^*)$ for $a = 0.27$ analytically, we'll use the Jacobian matrix $J$ given by:

\begin{equation} J^* = \begin{bmatrix} -2a x^* & 0.4 \\ 0.4 & 0 \end{bmatrix} \end{equation}

Now, let's calculate the eigenvalues:

The characteristic equation for a 2x2 matrix $A$ with eigenvalues $\lambda_1$ and $\lambda_2$ is given by:

$$\text{det}(A - \lambda I) = 0$$

Where $I$ is the identity matrix. For our Jacobian matrix $J^*$, this becomes:

$$\text{det}\left(\begin{bmatrix} -2a x^* - \lambda & 0.4 \\ 0.4 & -\lambda \end{bmatrix}\right) = 0$$

Now, calculate the determinant: $$(-2a x^* - \lambda)(-\lambda) - (0.4)(0.4) = 0$$

Now, factor out $\lambda ^2$:

$$\lambda^2(2a x^* - 1) - 0.16 = 0$$

Now, solve for $\lambda$: $$\lambda^2 = \frac{0.16}{2a x ^* - 1}$$

Take the square root: $$\lambda = \pm \sqrt{\frac{0.16}{2a x^* - 1}}$$

The eigenvalues depend on $a$ and $x^*$. To determine the stability, we need to find the values of $x^*$ and $a$ for the fixed point and then calculate the eigenvalues based on those values.

It's important to note that for a sink, we typically expect the eigenvalues to have a magnitude less than 1, indicating stability. If the magnitude is greater than 1, it would suggest an unstable behavior. You should substitute the specific values of $a$ and $x^*$ at the fixed point to get the numerical eigenvalues and determine the stability accordingly.

As we can see, the largest magnitude eigenvalue will determine the stability of the fixed point. If the largest magnitude eigenvalue is greater than 1 in magnitude, the fixed point is unstable. If it's less than 1, the fixed point is stable. If it's equal to 1, further analysis is needed.

Once We find the largest magnitude eigenvalue, We can explain the loss of stability of the sink. If the magnitude is greater than 1, it means the system has moved from a stable (attractive) state to an unstable state, indicating the loss of stability.

For your next question, of course this is not related with $b$, to prove that for $0.27 \leq a \leq 0.85$, the Hénon map $f$ has a period-two sink, we seek values of $a$ within this range for which the map exhibits a period-two orbit. A period-two orbit consists of two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ such that $f(f(x_1, y_1)) = (x_1, y_1)$ and $f(f(x_2, y_2)) = (x_2, y_2)$.

The Hénon map is defined by the following equations:

\begin{align*} x_{n+1} &= 1 - a x_n^2 + y_n \\ y_{n+1} &= 0.4 x_n \end{align*}

For a period-two orbit, we are looking for solutions $(x_1, y_1)$ and $(x_2, y_2)$ that satisfy:

\begin{align*} f(x_1, y_1) &= (x_2, y_2) \\ f(x_2, y_2) &= (x_1, y_1) \end{align*}

Let's assume $(x_1, y_1)$ is the first point:

\begin{align*} x_2 &= 1 - a x_1^2 + y_1 \\ y_2 &= 0.4 x_1 \end{align*}

Now, plug $(x_2, y_2)$ into the second equation:

$$y_2 = 0.4 x_2$$

This results in two equations with two unknowns $(x_1, y_1)$:

\begin{align*} 1 - a x_1^2 + y_1 &= x_2 \\ 0.4 x_1 &= y_2 \end{align*}

To find solutions for $(x_1, y_1)$ and $(x_2, y_2)$ that satisfy these equations for $0.27 \leq a \leq 0.85$, numerical methods or software may be used. Once such values are found, it can be demonstrated that the Hénon map has a period-two sink for $0.27 \leq a \leq 0.85$.

And based on this irregular, to find the largest magnitude eigenvalue of $Df^2$ analytically, we need to calculate $Df$ at the points $(x_1, y_1)$ and $(x_2, y_2)$ corresponding to the period-two orbit and then multiply these matrices. The largest magnitude eigenvalue of the resulting matrix $Df^2$ will determine the stability.

Let's calculate $Df$ at the fixed points $(x_1, y_1)$ and $(x_2, y_2)$:

1. For $(x_1, y_1)$, we have:

$$Df_1 = \begin{bmatrix} \frac{\partial f_x}{\partial x} & \frac{\partial f_x}{\partial y} \\ \frac{\partial f_y}{\partial x} & \frac{\partial f_y}{\partial y} \end{bmatrix}_{(x_1, y_1)} = \begin{bmatrix} -2ax_1 & 0.4 \\ 0.4 & 0 \end{bmatrix}$$

2. For $(x_2, y_2)$, we have:

$$Df_2 = \begin{bmatrix} \frac{\partial f_x}{\partial x} & \frac{\partial f_x}{\partial y} \\ \frac{\partial f_y}{\partial x} & \frac{\partial f_y}{\partial y} \end{bmatrix}_{(x_2, y_2)} = \begin{bmatrix} -2ax_2 & 0.4 \\ 0.4 & 0 \end{bmatrix}$$

Now, let's multiply these matrices to get $Df^2$. The multiplication of two 2x2 matrices is performed by multiplying corresponding elements and summing them:

$$Df^2 = Df_2 \cdot Df_1 = \begin{bmatrix} -2ax_2 & 0.4 \\ 0.4 & 0 \end{bmatrix} \cdot \begin{bmatrix} -2ax_1 & 0.4 \\ 0.4 & 0 \end{bmatrix} $$

Perform the matrix multiplication to obtain $Df^2$:

$$Df^2 = \begin{bmatrix} (4a^2 x_1 x_2 - 0.16) & 0.4(-2ax_2) \\ 0.4(-2ax_1) & 0.16 \end{bmatrix}$$

Now, to find the eigenvalues of $Df^2$, we solve for $\lambda$ in the characteristic equation:

$$\text{det}(Df^2 - \lambda I) = 0$$

Where $I$ is the identity matrix:

$$\begin{vmatrix} (4a^2 x_1 x_2 - 0.16 - \lambda) & 0.4(-2ax_2) \\ 0.4(-2ax_1) & (0.16 - \lambda) \end{vmatrix} = 0$$

Now, solve this characteristic equation for $\lambda$. You will obtain two eigenvalues. Calculate the magnitude of both eigenvalues and identify the largest one to determine the stability of the period-two orbit when a = 0.85.

The Jacobian matrix of $f^2$ at the period-two orbit is obtained by applying the Jacobian matrix $Df$ twice. We've already calculated $Df$ at the fixed points $(x_1, y_1)$ and $(x_2, y_2)$ for the Hénon map when $a = 0.85$.

Now, to find the Jacobian matrix of $f^2$, you can simply multiply the Jacobian matrix $Df$ at each point in the period-two orbit. Let's denote $Df_1$ as the Jacobian matrix at $(x_1, y_1)$ and $Df_2$ as the Jacobian matrix at $(x_2, y_2)$.

The Jacobian matrix of $f^2$ at the period-two orbit is given by:

$$D(f^2) = Df_2 \cdot Df_1$$

We've already calculated $Df_1$ and $Df_2$ earlier:

1. For $(x_1, y_1)$, we have $Df_1 = \begin{bmatrix} -2ax_1 & 0.4 \\ 0.4 & 0 \end{bmatrix}$.

2. For $(x_2, y_2)$, we have $Df_2 = \begin{bmatrix} -2ax_2 & 0.4 \\ 0.4 & 0 \end{bmatrix}$.

Now, multiply these matrices to find $D(f^2)$. You will obtain a 2x2 matrix representing the Jacobian of $f^2$ at the period-two orbit when $a = 0.85$.

The result should look like:

$$D(f^2) = \begin{bmatrix} \text{element 11} & \text{element 12} \\ \text{element 21} & \text{element 22} \end{bmatrix}$$

Then to calculate the product of $Df_2$ and $Df_1$ to obtain the specific elements of the Jacobian matrix $D(f^2)$ for $a = 0.85$ and the period-two orbit.