Chapter 10.7, Problem 6E

Interpretation Introduction

Interpretation:

Let $\text{f}\left(\text{x, r}\right){\text{ = r – x}}^{\text{2}}$, compare ${\text{αf}}^{\text{2}}\left({\text{x/α, R}}_{\text{1}}\right){\text{ to α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right)$ and match the coefficients of the lowest powers of x. determine value of $\text{α}$.

Concept Introduction:

• Obtain the expression for ${\text{αf}}^{\text{2}}\left({\text{x/α, R}}_{\text{1}}\right)$ from the given function.

• Obtain the expression for ${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right)$ from the given function.

• Compare the lowest power and determine $\text{α}$

Answer to Problem 6E

Solution:

By comparing ${\text{αf}}^{\text{2}}\left({\text{x/α, R}}_{\text{1}}\right){\text{ and α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right)$ the coefficients of the lowest powers of x is matched and value of $\text{α = -1}\text{.66499}$ is determined.

Explanation of Solution

First we introduce some notation. Let $\text{f}\left(\text{x, r}\right)$ denote a unimodal map that undergoes a period-doubling route to chaos as $\text{r}$ increases, and suppose that ${\text{x}}_{\text{m}}$ is the maximum. Let ${\text{r}}_{\text{n}}$ denote the value of $\text{r}$ at which a $\text{2n-}$ cycle is born, and let ${\text{R}}_{\text{n}}$ denote the value of $\text{r}$ at which the $\text{2n-}$ cycle is superstable.

For the given equation,

$\text{f}\left(\text{x, r}\right){\text{ = r – x}}^{\text{2}}$

For the expression provided for map, ${\text{R}}_{\text{0}}$ will be stable point. As ${\text{R}}_{\text{0}}$ has a super stable fixed point, therefore from the definition of the super stable fixed point,

${\text{x\&*\#x00A0;= R}}_{\text{0}}\text{ –}{\left(\text{x*}\right)}^{\text{2}}$

The superstability condition is

$\text{λ =}{\left(\frac{\partial \text{ƒ }}{\partial \text{x}}\right)}_{\text{x=x*}}\text{=0}$

Differentiating the given equation

$\left(\frac{\partial ƒ}{\partial \text{x}}\right)=\frac{\partial \left({\text{R}}_{\text{0}}\text{ –}{\left(\text{x}\right)}^{\text{2}}\right)}{\partial \text{x}}=\text{- 2x}$

Hence we must have fixed point $\text{x\&*\#x00A0;= 0}$.

Substituting $\text{x\&*\#x00A0;= 0}$ into the fixed point condition yields ${\text{R}}_{\text{0}}\text{= 0 }$.

Hence ${\text{R}}_{\text{0}}\text{= 0 }$.

At point ${\text{R}}_{\text{1}}$ the map has a superstable $\text{2-}$ cycle. Let $\text{p }$ and $\text{q}$ denote the points of the cycle.

The condition for super stability is depends on multiplier factor,

$\text{λ =}\left(\text{–2p}\right)\left(\text{–2q}\right)\text{ = 0 }$

$\text{λ = 0 }$

So the point $\text{x = 0}$ must be one of the points in the $\text{2-}$ cycle.

Form the given function $\text{f}\left(\text{x, r}\right){\text{ = r – x}}^{\text{2}}$, by substituting ${\text{x = 0 and r = R}}_{\text{1}}$. Then the period $\text{-2}$ condition to get the value of the ${\text{R}}_{\text{1}}$

${\text{f}}^{\text{2}}\left({\text{0,R}}_{\text{1}}\right)\text{ = 0 }$

$\text{f}\left(\text{f}\left({\text{0,R}}_{\text{1}}\right)\right)\text{ = 0 }$

$\text{f}\left(\left({\text{R}}_{\text{1}}\text{-}{\left(\text{0}\right)}^2\right),{\text{R}}_{\text{1}}\right)\text{ = 0 }$

Hence solving further,

${\text{R}}_{\text{1}}-{\left({\text{R}}_{\text{1}}\text{-}{\left(\text{0}\right)}^2\right)}^2\text{ = 0 }$

${\text{R}}_{\text{1}}-{\left({\text{R}}_{\text{1}}\right)}^2\text{ = 0 }$

${\text{R}}_{\text{1}}\left(1-{\text{R}}_{\text{1}}\right)\text{ = 0 }$

${\text{R}}_{\text{1}}{\text{= 0 and R}}_{\text{1}}\text{= 1 }$

${\text{R}}_{\text{1}}\text{= 0}$ is a fixed point and two cycle value is ${\text{R}}_{\text{1}}\text{= 1}$ .

Similarly, for 3-cycle, we will apply period 3 condition to get value of the value of ${\text{R}}_{\text{2}}$

${\text{f}}^{\text{4}}\left({\text{0,R}}_{\text{2}}\right)\text{ = 0 }$

$\text{f}\left(\text{f}\left(\text{f}\left(\text{f}\left({\text{0,R}}_{\text{1}}\right)\right)\right)\right)\text{ = 0 }$

$\text{f}\left(\text{f}\left(\text{f}\left({\text{R}}_{\text{2}}\text{-}{\left(\text{0}\right)}^2,{\text{R}}_{\text{2}}\right)\right)\right)\text{ = 0 }$

$\text{f}\left(\text{f}\left({\text{R}}_{\text{2}}\text{-}{\left({\text{R}}_{\text{2}}\text{-}{\left(\text{0}\right)}^{\text{2}}\right)}^{\text{2}}{\text{, R}}_{\text{2}}\right)\right)\text{ = 0 }$

$\text{f}\left({\text{R}}_{\text{2}}-{\left({\text{R}}_{\text{2}}\text{-}{\left({\text{R}}_{\text{2}}\text{-}{\left(\text{0}\right)}^{\text{2}}\right)}^{\text{2}}\right)}^2{\text{, R}}_{\text{2}}\right)\text{ = 0 }$

${\text{R}}_{\text{2}}\text{- }{\left({\text{R}}_{\text{2}}\text{- }{\left({\text{R}}_{\text{2}}\text{- }{\left({\text{R}}_{\text{2}}\text{-}{\left(\text{0}\right)}^{\text{2}}\right)}^{\text{2}}\right)}^{\text{2}}\right)}^{\text{2}}\text{ = 0 }$

Hence solving further,

${\text{R}}_{\text{2}}\text{= 0}$ is fixed point.

${\text{R}}_{\text{2}}\text{= 1}\text{.3107}$ is $\text{3-}$ cycle point

${\text{R}}_{\text{2}}\text{= 1}\text{.94079}$

and remaining are complex roots. Now from the given system equation $\text{f}\left(\text{x, r}\right){\text{ = r – x}}^{\text{2}}$, By substituting ${\text{r = R}}_{\text{1}}$

$\text{f}\left({\text{x, R}}_{\text{1}}\right)\text{ = }{\left({\text{R}}_{\text{1}}\right)}^2{\text{ – x}}^{\text{2}}$

By substituting ${\text{R}}_{\text{1}}=1$

$\text{f}\left({\text{x, R}}_{\text{1}}\right)\text{ = }{\left(\text{1}\right)}^2{\text{ – x}}^{\text{2}}$

Considering the given explicit function, ${\text{αf}}^{\text{2}}\left({\text{x/α, R}}_{\text{1}}\right)$

${\text{αf}}^{\text{2}}\left({\text{x/α, R}}_{\text{1}}\right)\text{ = αf}\left(\text{f}\left({\text{x/α, R}}_{\text{1}}\right)\right)$

By substituting ${\text{R}}_{\text{1}}=1$

${\text{αf}}^{\text{2}}\left(\text{x/α, 1}\right)\text{ = αf}\left(\text{f}\left(\text{x/α, 1}\right)\right)$

${\text{αf}}^{\text{2}}\left(\text{x/α, 1}\right)\text{ = αf}\left(\text{1-}{\left(\text{x/α}\right)}^2,\text{1}\right)$

${\text{αf}}^{\text{2}}\left(\text{x/α, 1}\right)\text{ = α}\left(1-{\left(\text{1-}{\left(\text{x/α}\right)}^2\right)}^2\right)$

Hence solving further,

${\text{αf}}^{\text{2}}\left(\text{x/α, 1}\right)\text{ = α}\left(1-{\left(\text{1-}{\left(\text{x/α}\right)}^2\right)}^2\right)$

${\text{αf}}^{\text{2}}\left(\frac{x}{\text{α}}\text{, 1}\right)\text{ = α}\left(\text{1-}\left(1+\frac{x^4}{{\text{α}}^4}-\frac{{\text{2 x}}^{\text{2}}}{{\text{α}}^{\text{2}}}\right)\right)$

${\text{αf}}^{\text{2}}\left(\frac{\text{x}}{\text{α}}\text{, 1}\right)\text{ = α}\left(\text{1 - 1- }\frac{{\text{x}}^{\text{4}}}{{\text{α}}^{\text{4}}}\text{+}\frac{{\text{2 x}}^{\text{2}}}{{\text{α}}^{\text{2}}}\right)$

${\text{αf}}^{\text{2}}\left(\frac{\text{x}}{\text{α}}\text{, 1}\right)\text{ = α}\left(\frac{{\text{2 x}}^{\text{2}}}{{\text{α}}^{\text{2}}}\text{- }\frac{{\text{x}}^{\text{4}}}{{\text{α}}^{\text{4}}}\right)$

${\text{αf}}^{\text{2}}\left(\frac{\text{x}}{\text{α}}\text{, 1}\right)\text{ = }\frac{{\text{2 x}}^{\text{2}}}{\text{α}}\text{- }\frac{{\text{x}}^{\text{4}}}{{\text{α}}^{\text{3}}}$

Similarly the given function for ${\text{R}}_2$, $\text{f}\left({\text{x, R}}_{\text{2}}\right)\text{ = }{\left({\text{R}}_{\text{2}}\right)}^2{\text{ – x}}^{\text{2}}$

By substituting ${\text{R}}_{\text{2}}\text{= 1}\text{.3107}$

$\text{f}\left({\text{x, R}}_{\text{2}}\right)\text{ = }{\left(\text{1}\text{.3107}\right)}^2{\text{ – x}}^{\text{2}}$

$\text{f}\left({\text{x, R}}_{\text{2}}\right)\text{ = 1}{\text{.71793449 – x}}^{\text{2}}$

$\text{f}\left({\text{x, R}}_{\text{2}}\right)\text{ = 1}{\text{.718 – x}}^{\text{2}}$

Considering the given explicit function, ${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right)$, By substituting ${\text{R}}_{\text{2}}\text{= 1}\text{.718}$

${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right){\text{ = α}}^{\text{2}}\text{f}\left(\text{f}\left(\text{f}\left(\text{1}\text{.78 - }{\left(\frac{\text{x}}{{\text{α}}^{\text{2}}}\right)}^{\text{2}}\text{,1}\text{.78}\right)\right)\right)$

${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right){\text{ = α}}^{\text{2}}\text{f}\left(\text{f}\left(\text{1}\text{.78}-{\left(\text{1}\text{.78 - }{\left(\frac{\text{x}}{{\text{α}}^{\text{2}}}\right)}^{\text{2}}\right)}^2,\text{1}\text{.78}\right)\right)$

${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right){\text{ = α}}^{\text{2}}\text{f}\left(\text{1}\text{.78-}\left(\text{1}\text{.78}-{\left({\left(\text{1}\text{.78 - }{\left(\frac{\text{x}}{{\text{α}}^{\text{2}}}\right)}^{\text{2}}\right)}^2\right)}^2,\text{1}\text{.78}\right)\right)$

${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right){\text{ = α}}^{\text{2}}\left(\text{1}\text{.78}-{\left(\text{1}\text{.78-}{\left(\text{1}\text{.78}-{\left(\text{1}\text{.78 - }{\left(\frac{\text{x}}{{\text{α}}^{\text{2}}}\right)}^{\text{2}}\right)}^2\right)}^2\right)}^2\right)$

Hence solving further,

${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right)\text{ = }\left(\begin{array}{l}1.67942{\text{α}}^{\text{2}}-\frac{x^{16}}{{\text{α}}^{\text{30}}}+\frac{\left(13.744\right)x^{14}}{{\text{α}}^{\text{26}}}-\frac{\left(75.7707\right)x^{16}}{{\text{α}}^{\text{22}}}+\frac{\left(213.124\right)x^{10}}{{\text{α}}^{18}}\\ -\frac{\left(319.835\right)x^8}{{\text{α}}^{\text{14}}}-\frac{\left(239.281\right)x^6}{{\text{α}}^{\text{10}}}+\frac{\left(66.2487\right)x^4}{{\text{α}}^{\text{6}}}-\frac{\left(3.32999\right)x^2}{{\text{α}}^{\text{2}}}\end{array}\right)$

Hence the given explicit function, ${\text{αf}}^{\text{2}}\left({\text{x/α, R}}_{\text{1}}\right){\text{ to α}}^{\text{2}}{\text{ f}}^{\text{4}}\left({\text{x/α}}^{\text{2}}{\text{, R}}_{\text{2}}\right)$ is

${\text{αf}}^{\text{2}}\left(\frac{\text{x}}{\text{α}}\text{, 1}\right)\text{ = }\frac{{\text{2 x}}^{\text{2}}}{\text{α}}\text{- }\frac{{\text{x}}^{\text{4}}}{{\text{α}}^{\text{3}}}$

${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left(\frac{\text{x}}{{\text{α}}^{\text{2}}}{\text{, R}}_{\text{2}}\right)\text{ = }\left(\begin{array}{l}1.67942{\text{α}}^{\text{2}}-\frac{x^{16}}{{\text{α}}^{\text{30}}}+\frac{\left(13.744\right)x^{14}}{{\text{α}}^{\text{26}}}-\frac{\left(75.7707\right)x^{16}}{{\text{α}}^{\text{22}}}+\frac{\left(213.124\right)x^{10}}{{\text{α}}^{18}}\\ -\frac{\left(319.835\right)x^8}{{\text{α}}^{\text{14}}}-\frac{\left(239.281\right)x^6}{{\text{α}}^{\text{10}}}+\frac{\left(66.2487\right)x^4}{{\text{α}}^{\text{6}}}-\frac{\left(3.32999\right)x^2}{{\text{α}}^{\text{2}}}\end{array}\right)$

To determine value of $\text{α}$

By comparing the lowest power of ${\text{αf}}^{\text{2}}\left(\frac{\text{x}}{\text{α}}\text{, 1}\right)$ and ${\text{α}}^{\text{2}}{\text{ f}}^{\text{4}}\left(\frac{\text{x}}{{\text{α}}^{\text{2}}}{\text{, R}}_{\text{2}}\right)$ is the term ${\text{x}}^2$,

$\frac{{\text{2 x}}^{\text{2}}}{\text{α}}\text{ = -}\frac{\left(\text{3}\text{.32999}\right){\text{x}}^{\text{2}}}{{\text{α}}^{\text{2}}}$

$\text{2 = -}\frac{\text{3}\text{.32999}}{\text{α}}$

$\text{α = -}\frac{\text{3}\text{.32999}}{\text{2}}$

$\text{α = -1}\text{.66499}$

Therefore the value of α from the given two function is determined as $\text{α = -1}\text{.66499}$.