Chapter 10.3, Problem 6E

Interpretation Introduction

Interpretation:

Consider the cubic map ${\text{x}}_{\text{n+1}}{\text{ = f (x}}_{\text{n}}\text{)}$ where ${\text{f(x}}_{\text{n}}{\text{) = rx}}_{\text{n}}{\text{ - x}}_{\text{n}}^{\text{3}}$

• a) To find the fixed points and to find the values of r for which they do exist. Also, find the values for which they are stable.

• b) To show that $\text{p}$ and $\text{q}$ are the roots of the equation ${\text{x(x}}^{\text{2}}{\text{- r+1)(x}}^{\text{2}}{\text{- r -1)(x}}^4{\text{-rx}}^2\text{+1)=0}$ and use these roots to find all the $\text{2-}$ cycles.

• c) To determine the stability of the $\text{2-}$ cycle as a function of $\text{r}$.

• d) To plot the partial bifurcation diagram based on the information obtained.

Concept Introduction:

• ➢ The logistic mapping ${\text{x}}_{\text{n+1}}{\text{ = f (x}}_{\text{n}}\text{)}$ the function f is differential function and it is able to solve. The fixed points of the function are those points where the curve of the function meets the green line.

Answer to Problem 6E

Solution:

• a) The fixed points ${\text{x}}_{\text{*}}\text{ = 0}$ have $\left|{\text{f '(x}}_{*}\text{)}\right|\text{ > 1}$ since they are unstable and for the next two fixed points ${\text{x}}_{\text{*}}\text{ = ± }\sqrt{\text{r - 1}}$ are stable.

• b) $\text{p}$ and $\text{q}$ are the roots of the equation ${\text{x(x}}^{\text{2}}{\text{- r+1)(x}}^{\text{2}}{\text{- r -1)(x}}^4{\text{-rx}}^2\text{+1)=0}$ is proved.

• c) The stability of the 2-cycle as a function of r is determined.

• d) The partial bifurcation diagram plotted using the calculated information.

Explanation of Solution

• a)

  The cubic map equation is ${\text{x}}_{\text{n+1}}{\text{ = rx}}_{\text{n}}{\text{ - x}}_{\text{n}}^{\text{3}}$,

  The difference equation ${\text{x}}_{\text{n+1}}{\text{ = f (x}}_{\text{n}}\text{)}$.

  Let ${\text{x}}_{\text{n}}{\text{ = x}}_{\text{*}}$ then the given quadratic map can be written as

  ${\text{f(x}}_{*}{\text{) = rx}}_{*}{\text{ - x}}_{*}^2$.

  To determine the root of the function using the quadratic formula to find the value of the variable.

  $\begin{array}{l}{\text{f(x}}_{\text{*}}{\text{) = rx}}_{\text{*}}{\text{ - x}}_{*}^3\\ {\text{x}}_{\text{*}}{\text{ = x}}_{\text{*}}{\text{(r - x}}_{\text{*}}^{\text{2}}\text{)}\\ {\text{(r - x}}_{\text{*}}^{\text{2}}\text{) = 1}\end{array}$

  The first solution of the mapping is ${\text{x}}_{\text{*}}\text{ = 0}$ this is the first fixed point.

  Also;

  $\begin{array}{l}{\text{r - x}}_{\text{*}}^{\text{2}}\text{ = 1}\\ {\text{x}}_{\text{*}}^{\text{2}}\text{ = r - 1}\\ {\text{x}}_{\text{*}}\text{ = ± }\sqrt{\text{r - 1}}\end{array}$

  The next two fixed points are $\text{± }\sqrt{\text{r - 1}}$. Now consider the stability of these two fixed points the mapping function ${\text{f(x}}_{\text{*}}{\text{) = rx}}_{\text{*}}{\text{ - x}}_{*}^3$ taking the first derivative of the mapping.

  So,

  ${\text{f '(x}}_{\text{*}}{\text{) = r - 3x}}_{*}^2$

  At the first fixed point put ${\text{x}}_{\text{*}}\text{ = 0}$.

  Then, $\text{f '(0) = r}$

  At the next two fixed point.

  $\begin{array}{l}\text{f '(±}\sqrt{\text{r - 1}}\text{)= r - 3 (±}\sqrt{\text{r - 1}}{\text{)}}^{\text{2}}\\ \text{f '(±}\sqrt{\text{r - 1}}\text{)= r - 3(r - 1)}\\ \text{f '(±}\sqrt{\text{r - 1}}\text{)= r - 3r + 3}\\ \text{f '(±}\sqrt{\text{r - 1}}\text{)= 3 - 2r}\end{array}$

  Hence, all of these fixed points ${\text{x}}_{\text{*}}\text{ = 0}$ have $\left|{\text{f '(x}}_{*}\text{)}\right|\text{ > 1}$ since they are unstable and for the next two fixed points, ${\text{x}}_{\text{*}}\text{ = ± }\sqrt{\text{r - 1}}$ are stable.

• b)

  Consider the cubic map ${\text{x}}_{\text{n+1}}{\text{ = f (x}}_{\text{n}}\text{)}$, where ${\text{f(x}}_{\text{n}}{\text{) = rx}}_{\text{n}}{\text{ - x}}_{\text{n}}^{\text{3}}$ of the $\text{2-cycle}$ orbit. Suppose that p and q are the two fixed point of the cubic mapping and $\text{f(p) = q and f(q) = p}$ these are the roots of the polynomial equation,

  Here consider the $\text{2-cycle}$ orbit so let the value of the roots be:

  $\begin{array}{l}{\text{f(x}}_{\text{*}}{\text{) = rx}}_{\text{*}}{\text{ - x}}_{*}^3\\ {\text{x}}_{\text{*}}{\text{ = x}}_{\text{*}}{\text{(r - x}}_{\text{*}}^{\text{2}}\text{)}\\ {\text{(r - x}}_{\text{*}}^{\text{2}}\text{) = 1}\end{array}$

  The first solution of the mapping is ${\text{x}}_{\text{*}}\text{ = 0}$ this is the first fixed point. Consider also:

  $\begin{array}{l}{\text{r - x}}_{\text{*}}^{\text{2}}\text{ = 1}\\ {\text{x}}_{\text{*}}^{\text{2}}\text{ = r - 1}\\ {\text{x}}_{\text{*}}\text{ = ± }\sqrt{\text{r - 1}}\end{array}$

  Hence, the first root is $\text{p, q = 0}$ then the second roots are $\text{p,q =}\pm \sqrt{\text{r - 1}}$ $\text{x}$.

  If ${\text{f(x}}_{\text{*}}{\text{) = -x}}_{*}$ for negative fixed root so,

  $\begin{array}{l}{\text{f(x}}_{\text{*}}{\text{) = rx}}_{\text{*}}{\text{ - x}}_{\text{*}}^{\text{3}}\\ {\text{-x}}_{\text{*}}{\text{=x}}_{\text{*}}{\text{(r - x}}_{\text{*}}^{\text{2}}\text{)}\\ {\text{(r - x}}_{\text{*}}^{\text{2}}\text{) = -1}\\ {\text{x}}_{\text{*}}\text{= ±}\sqrt{\text{r + 1}}\end{array}$

  Hence, the root of the third polynomial equation is ${\text{x}}_{\text{*}}\text{ = ± }\sqrt{\text{r + 1}}$.

  Now taking the rational fixed point at ${\text{f(x}}_{\text{*}}\text{) = }\frac{\text{1}}{{\text{x}}_{\text{*}}}$ so,

  $\begin{array}{l}{\text{f(x}}_{\text{*}}{\text{) = rx}}_{\text{*}}{\text{ - x}}_{\text{*}}^{\text{3}}\\ \frac{\text{1}}{{\text{x}}_{\text{*}}}{\text{ = x}}_{\text{*}}{\text{(r - x}}_{\text{*}}^{\text{2}}\text{)}\\ {\text{(rx}}_{\text{*}}^{\text{2}}{\text{ - x}}_{\text{*}}^{\text{4}}\text{) = 1}\\ {\text{x}}_{\text{*}}^{\text{4}}{\text{ - rx}}_{\text{*}}^{\text{2}}\text{ + 1 = 0}\end{array}$

  Hence, the root p, q are the roots of the given equation.

• c)

  Now consider the stability of $\text{cycle-2}$ orbit as a function of r. Consider the cubic equation of a $\text{period-2}$ orbit,

  The difference equation ${\text{x}}_{\text{n+1}}{\text{ = f (x}}_{\text{n}}\text{)}$ period is $\text{2}$ orbit so the equation ${\text{x}}_{\text{p}}{\text{=f}}^{\text{2}}{\text{(x}}_{\text{p}}\text{)}$

  So,

  $\begin{array}{l}{\text{x}}_{\text{p}}{\text{= f}}^{\text{2}}{\text{(x}}_{\text{p}}\text{)}\\ {\text{x}}_{\text{p}}{\text{= f(f(x}}_{\text{p}}\text{))}\\ {\text{x}}_{\text{p}}{\text{= f(x}}_{\text{p}}{\text{(r-x}}_{\text{p}}^2\text{))}\\ {\text{x}}_{\text{p}}\text{= f(y)}\end{array}$

  Here, the quadratic equation ${\text{f(x) = rx - x}}^{\text{3}}$

  Then,

  The equation is written as:

  $\begin{array}{l}{\text{f(y) = ry - y}}^{\text{3}}\\ {\text{f(y) = r(rx}}_{\text{p}}{\text{ - x}}_{\text{p}}^{\text{3}}{\text{) - (rx}}_{\text{p}}{\text{ - x}}_{\text{p}}^{\text{3}}{\text{)}}^{\text{3}}\\ {\text{f(y) = (rx}}_{\text{p}}{\text{ - x}}_{\text{p}}^{\text{3}}{\text{)(r - (rx}}_{\text{p}}{\text{ - x}}_{\text{p}}^{\text{3}}{\text{)}}^{\text{2}}\text{)}\end{array}$

  Let the fixed point be ${\text{x}}_{\text{\&*\#x00B1;}}$ so, $\left({\text{x - x}}_{\text{x+}}\right)\left({\text{x - x}}_{\text{x-}}\right)$ it can be written as ${\text{x}}^{\text{2}}\text{ + 1 - r}$. Now the quadratic equation is written as:

  ${\text{(rx}}_{\text{p}}{\text{ - x}}_{\text{p}}^{\text{3}}{\text{)(r - (rx}}_{\text{p}}{\text{ - x}}_{\text{p}}^{\text{3}}{\text{)}}^{\text{2}}{\text{) = x}}_{\text{p}}{\text{(r - x}}_{\text{p}}^{\text{2}}{\text{)(r - x}}_{\text{p}}^{\text{2}}{{\text{(r - x}}_{\text{p}}^{\text{2}}\text{)}}^{\text{2}}\text{)}$

  Compare with ${\text{(x}}_{\text{p}}^3{\text{+αx}}_{\text{p}}^2{\text{+βx}}_{\text{p}}{\text{+γ)(x}}_{\text{p}}^2\text{+1- r)}$ the coefficient of the cubic degree equation.

  $\begin{array}{l}{\text{(x}}^{\text{3}}{\text{+ αx}}^{\text{2}}{\text{+ βx + γ)(x}}^{\text{2}}{\text{+1- r)=x}}^{\text{5}}{\text{+ x}}^{\text{3}}{\text{- rx}}^{\text{3}}{\text{+ αx}}^{\text{4}}{\text{+ αx}}^{\text{2}}{\text{-rαx}}^{\text{2}}{\text{+βx}}^{\text{3}}{\text{+βx-rβx+γx}}^{\text{2}}\text{+γ-γr}\\ {\text{(x}}^{\text{3}}{\text{+ αx}}^{\text{2}}{\text{+ βx + γ)(x}}^{\text{2}}{\text{+1- r)=x}}^{\text{5}}{\text{+ αx}}^{\text{4}}+(1+\text{β}-\text{r}){\text{x}}^3+(\text{α}-\text{rα}+\text{γ}){\text{x}}^{\text{2}}+(\text{β}-\text{rβ})\text{x+γ}-\text{γr}\mathrm{............}\text{1)}\end{array}$

  And

  $\begin{array}{l}{\text{x(r - x}}^{\text{2}}{\text{)( r- (rx - x}}^{\text{3}}{\text{)}}^{\text{2}}{\text{) = (rx-x}}^{\text{3}}{\text{)(r - x}}^{\text{6}}{\text{- rx}}^{\text{2}}{\text{+ 2rx}}^{\text{4}}\text{)}\\ {\text{x(r - x}}^{\text{2}}{\text{)( r- (rx - x}}^{\text{3}}{\text{)}}^{\text{2}}{\text{) = r}}^{\text{2}}{\text{x - r}}^{\text{2}}{\text{x}}^{\text{3}}{\text{+2r}}^{\text{2}}{\text{x}}^{\text{5}}{\text{- rx}}^{\text{3}}{\text{+x}}^{\text{9}}{\text{+rx}}^{\text{5}}{\text{-3rx}}^{\text{7}}\\ {\text{x(r - x}}^{\text{2}}{\text{)( r- (rx - x}}^{\text{3}}{\text{)}}^{\text{2}}{\text{) = x}}^{\text{9}}{\text{-3rx}}^{\text{7}}{\text{+(2r}}^{\text{2}}+{\text{r)x}}^{\text{5}}{\text{-(r}}^{\text{2}}{\text{+r)x}}^3{\text{ + r}}^2\text{x}\mathrm{.........}\text{2)}\end{array}$

  Comparing equations $\text{1) and 2)}$

  $\begin{array}{l}{\text{1= 2r}}^{\text{2}}+\text{r}\\ \text{α = 0}\\ {\text{β = -r}}^2-1\\ \text{α}-\text{rα + γ=0}\end{array}$

  And

  $\begin{array}{l}\text{β}-\text{γβ}={\text{r}}^2\\ \text{γ}-\text{γr}=0\end{array}$

  Now the values are $\text{α}=0$ and $\text{r=1}$.

  If $\text{r=-2}$ then $\text{β}=-5,\text{γ}=0$.

  Hence, the period of the second orbit ${\text{x = x}}_{\text{p}}$ satisfies the cubic equation,

  $\begin{array}{l}{\text{x}}^{\text{3}}\text{- 5x = 0}\\ \text{x =}\pm \sqrt{5},0\end{array}$

  Since there are three fixed points of the cubic equation, where two points are positive and one is negative. So, the value of the constant term computes:

  ${\text{b}}^{\text{2}}\text{- 4ac = 0}$

  $\begin{array}{l}{\text{b}}^{\text{2}}\text{- 4ac = 0}\\ {\text{x}}^2-\text{r}+1=0\\ 4(\text{r}-1)=0\\ \text{r =1}\end{array}$

  Hence, the value of r is positive at $\text{r = -2}$ point is negative. So the fixed points lie in the lower and upper branch at ${\text{x}}_{\text{*-}}{\text{ , x}}_{\text{*+}}$ is stable.

• d)

  The bifurcation diagram is as shown below,