Chapter 12.2, Problem 7E

Interpretation Introduction

Interpretation:

Considering Henon map with $-1<b<1$, show the map has the 2-cycle for $a>a_1=\frac{3}{4}{\left(1-b\right)}^2$, and find the values of ‘a’ for which the $\text{2-}$ cycle is stable.

Concept Introduction:

• Henon map is given by

  ${\text{x}}_{\text{n+1}}{\text{= y}}_{\text{n}}+1-{\text{ax}}_{\text{n}}^{\text{2}},{\text{ y}}_{\text{n+1}}{\text{= bx}}_{\text{n}}$

  Here, a and b are adjustable parameters.

• The first transformation is

  $\text{T': x' = x, y' =1+y}-{\text{ax}}^{\text{2}}$

  The second transformation is

  $\text{T'': x'' = bx', y'' = y'}$

  The third transformation is

  $\text{T''': x''' = y'', y''' = x''}$

  The composite transformation $\text{T = T''' T'' T'}$ gives the Henon mapping using notation ${\text{(x}}_{\text{n}}{\text{,y}}_{\text{n}}\text{) for (x,y)}$ and ${\text{(x}}_{\text{n+1}}{\text{,y}}_{\text{n+1}}\text{) for (x''',y''')}$

• The properties of Henon map are as follows:

  1. It is invertible

  2. It is dissipative

  3. For certain parameter values, it has a trapping region.

  4. Some trajectories of it may escape to infinity.

Answer to Problem 7E

Solution:

It is shown that the map has $\text{2-}$ cycle for $a>a_1=\frac{3}{4}{\left(1-b\right)}^2$; 2-cycle is stable for the range $\frac{3}{4}{\left(b-1\right)}^2<a<\frac{1}{4}\left({\text{5b}}^{\text{2}}-\text{6b}+5\right)$

Explanation of Solution

From the Henon map equations,

$\begin{array}{l}{\text{x}}_{\text{n+2}}{\text{= bx}}_{\text{n}}+1-\text{a}{\left({\text{y}}_{\text{n}}\text{+1}-{\text{ax}}_{\text{n}}^{\text{2}}\right)}^2{\text{ y}}_{\text{n+2}}\text{=b}\left({\text{y}}_{\text{n}}\text{+1}-{\text{ax}}_{\text{n}}^{\text{2}}\right)\\ \text{x = bx+1}-\text{a}{\left(\text{y+1}-{\text{ax}}^2\right)}^2\text{ y = b}\left(\text{y}+1-{\text{ax}}^2\right)\end{array}$

Rearranging $\text{y = b}\left(\text{y}+1-{\text{ax}}^2\right)$ to solve for y

$\begin{array}{l}\text{y = by}+\text{b}-{\text{bax}}^2\\ y-\text{b}y=\text{b}-{\text{bax}}^2\\ y(1-\text{b})=b\left(1-{\text{ax}}^2\right)\\ y=\frac{\text{b}\left({\text{ax}}^2-1\right)}{\left(\text{b}-1\right)}\end{array}$

Substituting the above value of y in $\text{x = bx+1}-\text{a}{\left(\text{y+1}-{\text{ax}}^2\right)}^2$

$\begin{array}{l}\text{x = bx+1}-\text{a}{\left(\text{y+1}-{\text{ax}}^2\right)}^2=\text{bx+1}-a{\left(\frac{y}{b}\right)}^2\\ \text{x}=\text{bx+1}-\frac{\text{a}}{{\text{b}}^{\text{2}}}{\left(\frac{\text{b}\left({\text{ax}}^{\text{2}}-1\right)}{\left(\text{b}-1\right)}\right)}^2\\ \text{x}=\text{bx+1}-\text{a}{\left(\frac{\left({\text{ax}}^{\text{2}}-1\right)}{\left(\text{b}-1\right)}\right)}^2\end{array}$

Further, rearranging the above equation

$0={\left(\text{b}-1\right)}^3\text{x}+{\left(\text{b}-1\right)}^2-\text{a}{\left({\text{ax}}^{\text{2}}-1\right)}^2$

The roots of the above equation are the coordinates of the $\text{2-}$ cycle; from these four roots two roots of the equation are fixed points of the Henon map $\text{x = y+}1-{\text{ax}}^{\text{2}}=\text{bx+1}-{\text{ax}}^{\text{2}}$

Thus, the roots can be divided out

$0=\frac{{\left(\text{b}-1\right)}^3\text{x}+{\left(\text{b}-1\right)}^2-\text{a}{\left({\text{ax}}^{\text{2}}-1\right)}^2}{\text{bx}+1-{\text{ax}}^{\text{2}}-\text{x}}={\text{a}}^{\text{2}}{\text{x}}^{\text{2}}+\text{a}\left(\text{b}-1\right)\text{x}-\text{a}+{\left(\text{b}-1\right)}^2$

The solutions of the equations are ${\text{a}}^{\text{2}}{\text{x}}^{\text{2}}+\text{a}\left(\text{b}-1\right)\text{x}-\text{a}+{\left(\text{b}-1\right)}^2=0$

$\text{x}=\frac{1-\text{b}\pm \sqrt{4\text{a}-3{\left(\text{b}-1\right)}^2}}{2\text{a}}\Rightarrow \text{y}=\frac{b\left(1-\text{b}\mp \sqrt{\text{4a}-3{\left(\text{b}-1\right)}^2}\right)}{\text{2a}}\text{ a}\ne \text{0}$

The case $\text{a = 0}$ can be neglected because it is not a $\text{2-}$ cycle.

For $\text{2-}$ cycle to exist, the coordinates must be real,

$\text{4a}-3{\left(\text{b}-1\right)}^2>0\Rightarrow \text{a}>{\text{a}}_{\text{1}}=\frac{3}{4}{\left(1-\text{b}\right)}^2$

To determine the stability of the fixed points, use Jacobian matrix,

$J=\left(\begin{array}{cc}\frac{\partial x_{n+2}}{\partial x_n}& \frac{\partial x_{n+2}}{\partial y_n}\\ \frac{\partial y_{n+2}}{\partial x_n}& \frac{\partial y_{n+2}}{\partial y_n}\end{array}\right)$

$J=\left(\begin{array}{cc}\frac{\partial \left({\text{bx}}_{\text{n}}+1-\text{a}{\left({\text{y}}_{\text{n}}\text{+1}-{\text{ax}}_{\text{n}}^{\text{2}}\right)}^2\right)}{\partial x_n}& \frac{\partial \left({\text{bx}}_{\text{n}}+1-\text{a}{\left({\text{y}}_{\text{n}}\text{+1}-{\text{ax}}_{\text{n}}^{\text{2}}\right)}^2\right)}{\partial y_n}\\ \frac{\partial \left(\text{b}\left({\text{y}}_{\text{n}}\text{+1}-{\text{ax}}_{\text{n}}^{\text{2}}\right)\right)}{\partial x_n}& \frac{\partial \left(\text{b}\left({\text{y}}_{\text{n}}\text{+1}-{\text{ax}}_{\text{n}}^{\text{2}}\right)\right)}{\partial y_n}\end{array}\right)$

$J=\left(\begin{array}{cc}{\text{b+4a}}^{\text{2}}\text{x}\left(\text{y+1}-{\text{ax}}^{\text{2}}\right)& -\text{2a}\left(\text{y+1}-{\text{ax}}^{\text{2}}\right)\\ -\text{2abx}& \text{b}\end{array}\right)$

$J_{\left(x^{*},y^{*}\right)}=\left(\begin{array}{cc}\text{b+}\frac{{\text{4a}}^{\text{2}}{\text{x}}^{*}{y}^{*}}{b}& -\frac{{\text{2ay}}^{*}}{b}\\ -{\text{2abx}}^{*}& \text{b}\end{array}\right)$

The eigenvalues of the above matrix are

$\begin{array}{l}{\text{λ}}_{\text{±}}\text{=}\frac{{\text{2a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}{\text{+b}}^{\text{2}}\text{±}\sqrt{{\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}\left({\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}{\text{+b}}^{\text{2}}\right)}}{\text{b}}\\ {\text{λ}}_{\text{±}}\text{=}\frac{\text{2}\left({\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}\right){\text{+b}}^{\text{2}}\text{±2}\sqrt{{\left({\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}\right)}^{\text{2}}{\text{+b}}^{\text{2}}\left({\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}\right)}}{\text{b}}\end{array}$

From $y=\frac{\text{b}\left({\text{ax}}^2-1\right)}{\left(\text{b}-1\right)}$

$y^{*}=\frac{\text{b}\left(\text{a}{\left({\text{x}}^{*}\right)}^2-1\right)}{\left(\text{b}-1\right)}$

Substituting $y^{*}=\frac{\text{b}\left(\text{a}{\left({\text{x}}^{*}\right)}^2-1\right)}{\left(\text{b}-1\right)}$ in ${\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}$

${\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}={\text{a}}^{\text{2}}{\text{x}}^{\text{*}}\frac{\text{b}\left(\text{a}{\left({\text{x}}^{*}\right)}^2-1\right)}{\left(\text{b}-1\right)}=\frac{{\text{abx}}^{\text{*}}\left({\text{a}}^2{\left({\text{x}}^{\text{*}}\right)}^2-a\right)}{\left(\text{b}-1\right)}$

Using ${\text{a}}^{\text{2}}{\left({\text{x}}^{*}\right)}^{\text{2}}+\text{a}\left(\text{b}-1\right){\text{x}}^{*}-\text{a}+{\left(\text{b}-1\right)}^2=0$ to simplify ${\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}$

Rearranging above equation,

${\text{a}}^{\text{2}}{\left({\text{x}}^{*}\right)}^{\text{2}}-\text{a}+{\left(\text{b}-1\right)}^2=-\left(\text{b}-1\right)\left({\text{ax}}^{*}+\left(\text{b}-1\right)\right)$

$\begin{array}{l}{\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}=-{\text{abx}}^{\text{*}}\left({\text{ax}}^{\text{*}}+\left(\text{b}-1\right)\right)\\ {\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}=-\text{b}\left({\text{a}}^{\text{2}}{\left({\text{x}}^{\text{*}}\right)}^2+\text{a}\left(\text{b}-1\right){\text{x}}^{\text{*}}\right)\end{array}$

$\begin{array}{l}{\text{a}}^{\text{2}}{\left({\text{x}}^{*}\right)}^{\text{2}}+\text{a}\left(\text{b}-1\right){\text{x}}^{*}-\text{a}+{\left(\text{b}-1\right)}^2=0\\ {\text{a}}^{\text{2}}{\left({\text{x}}^{*}\right)}^{\text{2}}+\text{a}\left(\text{b}-1\right){\text{x}}^{*}=\text{a}-{\left(\text{b}-1\right)}^2\end{array}$

${\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}=-\text{b}\left(\text{a}-{\left(\text{b}-1\right)}^2\right)=\text{b}\left({\left(\text{b}-1\right)}^2-\text{a}\right)$

$\therefore {\text{λ}}_{\text{±}}\text{=}\frac{\text{2}\left({\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}\right){\text{+b}}^{\text{2}}\text{±2}\sqrt{{\left({\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}\right)}^{\text{2}}{\text{+b}}^{\text{2}}\left({\text{a}}^{\text{2}}{\text{x}}^{\text{*}}{\text{y}}^{\text{*}}\right)}}{\text{b}}$ becomes

${\text{λ}}_{\text{±}}\text{=}\frac{\text{2b}\left({\left(b-1\right)}^2-a\right){\text{+b}}^{\text{2}}\text{±2}\sqrt{b^2{\left({\left(b-1\right)}^2-a\right)}^{\text{2}}{\text{+b}}^{\text{3}}\left({\left(b-1\right)}^2-a\right)}}{\text{b}}$

${\text{λ}}_{\text{±}}\text{=2}\left({\left(b-1\right)}^2-a\right)\text{+b±2}\sqrt{{\left({\left(b-1\right)}^2-a\right)}^{\text{2}}\text{+b}\left({\left(b-1\right)}^2-a\right)}$

The fixed points are stable for $\left|{\text{λ}}_{\text{±}}\right|\text{<1}$ and stability of the fixed points also dependson whether the eigenvalues are complex or not.

By assuming a positive argument to the square root, a negative argument to the square root and the inequality from the existence of the $\text{2-}$ cycle.

$a<\frac{1}{4}\left({\text{5b}}^{\text{2}}-\text{6b}+5\right)$

$-1<b<1$

$a>\frac{3}{4}{\left(b-1\right)}^2$

Following is a plot of intersection of the above three inequalities