Chapter 10.4, Problem 5E

Interpretation Introduction

Interpretation:

To show numerically that the period-doubling bifurcations of the $\text{3}$-cycle for the logistic map accumulate near $\text{r }=\text{3}.\text{8495 }\dots$, to form three small chaotic bands. Also, show that these chaotic bands merge near $\text{r }=\text{3}.\text{8495 }\dots$ to form a much larger attractor that nearly fills an interval.

Concept Introduction:

• Determine the value of $\text{r}$ for the third cycle.

• Plot the cobweb for $\text{r}$ and check the merging of chaotic bands.

Answer to Problem 5E

Solution:

It is numerically shown that the period-doubling bifurcations of the $\text{3}$-cycle for the logistic map accumulate near $\text{r }=\text{3}.\text{8495 }\dots$ The cobweb plots are shown and verified similarly.

Explanation of Solution

Consider the logistic map ${\text{x}}_{n+1}\text{ = f(x)}$ and function ${\text{f(x) = rx}}_{\text{n}}{\text{(1- x}}_{\text{n}}\text{)}$ for the period cycle $\text{3}$ that become unstable and period-doubling bifurcation occurs. The second derivative of the function is defined by $\text{ f}\left(\text{f(x)}\right)={\text{f}}^2\text{(x)}$ this is followed in higher derivatives. The third-iterate map ${\text{f}}^{\text{3}}\text{(x)}$ will give the period-$\text{3}$ cycle such that it should satisfy ${\text{p = f}}^{\text{3}}\text{(p)}$ and are the fixed points of the third-iterate map. Unfortunately, since ${\text{f}}^{\text{3}}\text{(x)}$ is an eighth-degree polynomial, we cannot solve for the fixed points explicitly. But a graph gives sufficient insight.

Now for computation of third iterate method of the function is ${\text{f}}^{\text{3}}\text{(x)}$ and parameter $\text{r}$ must be larger than the start value in the orbit.

${\text{f(x) = rx}}_{\text{n}}{\text{(1- x}}_{\text{n}}\text{)}$

For the third iteration method, the function is written in the form of,

$\frac{\text{d}}{\text{dx}}\left({\text{f}}^3\text{(x)}\right)\text{= }\frac{\text{d}}{\text{dx}}{\left(\text{f(x)}\right)|}_{x=c}\frac{\text{d}}{\text{dx}}{\left(\text{f(x)}\right)|}_{x=b}\frac{\text{d}}{\text{dx}}{\left(\text{f(x)}\right)|}_{x=a}$

By substituting ${\text{f(x) = rx}}_{\text{n}}{\text{(1- x}}_{\text{n}}\text{)}$ in the above equation

$\frac{\text{d}}{\text{dx}}\left({\text{f}}^3\text{(x)}\right)\text{= }\frac{\text{d}}{\text{dx}}{\left({\text{rx}}_{\text{n}}{\text{(1- x}}_{\text{n}}\text{)}\right)|}_{x=c}\frac{\text{d}}{\text{dx}}{\left({\text{rx}}_{\text{n}}{\text{(1- x}}_{\text{n}}\text{)}\right)|}_{x=b}\frac{\text{d}}{\text{dx}}{\left({\text{rx}}_{\text{n}}{\text{(1- x}}_{\text{n}}\text{)}\right)|}_{x=a}$

$\frac{\text{d}}{\text{dx}}\left({\text{f}}^3\text{(x)}\right)\text{= }{\left({\text{r- 2rx}}_{\text{n}}\right)|}_{x=c}{\left({\text{r- 2rx}}_{\text{n}}\right)|}_{x=b}{\left({\text{r- 2rx}}_{\text{n}}\right)|}_{x=a}$

$\frac{\text{d}}{\text{dx}}\left({\text{f}}^3\text{(x)}\right){\text{= r}}^3\left(\text{1- 2c}\right)\left(\text{1- 2b}\right)\left(\text{1- 2a}\right)$

$\frac{\text{d}}{\text{dx}}\left({\text{f}}^3\text{(x)}\right)\text{= 1}$

Hence

${\text{r}}^3\left(\text{1- 2c}\right)\left(\text{1- 2b}\right)\left(\text{1- 2a}\right)=1$

By solving the above equation

${\text{r}}^{\text{3}}\left(\text{1- 2}\left(\text{a + b + c}\right)\text{+4}\left(\text{ab + bc +ac}\right)\text{ - 8abc}\right)\text{=1}$

Let the above equation in the form of $\text{α = a + b + c}$, $\text{β = ab + bc +ac}$ and $\text{γ = abc}$.

${\text{r}}^{\text{3}}\left(\text{1- 2α + 4β - 8λ}\right)\text{=1}$

Now consider another function $\text{g(x)}$, it has a double root at different three points at a fixed point $\text{x = 0}$ and $\text{x = }\frac{\text{r -1}}{\text{r}}$. So

$\text{g(x)αx}\left(\text{x -1 +}\frac{\text{1}}{\text{r}}\right){\left[\left(\text{x - a}\right)\left(\text{x - b}\right)\left(\text{x - c}\right)\right]}^{\text{2}}$

After solving this,

$\text{g(x)αx}\left(\text{x -1 +}\frac{\text{1}}{\text{r}}\right){\left(\text{x - a}\right)}^2{\left(\text{x - b}\right)}^2{\left(\text{x - c}\right)}^2=$

$\text{g(x)αx}\left(\text{x -1 +}\frac{\text{1}}{\text{r}}\right)\left({\text{x}}^{\text{2}}{\text{- 2xa + a}}^{\text{2}}\right)\left({\text{x}}^{\text{2}}{\text{- 2xb + b}}^{\text{2}}\right)\left({\text{x}}^{\text{2}}{\text{- 2xc + c}}^{\text{2}}\right)=$

$\text{g(x)αx}\left(\text{x -1 +}\frac{\text{1}}{\text{r}}\right)\left(\begin{array}{l}\left({\text{x}}^{\text{6}}{\text{- 2x}}^{\text{5}}{\text{b + x}}^{\text{4}}{\text{b}}^{\text{2}}{\text{- 2ax}}^{\text{5}}{\text{+4abx}}^{\text{4}}{\text{- 2ab}}^{\text{2}}{\text{x}}^{\text{3}}{\text{+ a}}^{\text{2}}{\text{x}}^{\text{4}}{\text{- 2a}}^{\text{2}}{\text{bx}}^{\text{3}}{\text{ + a}}^{\text{2}}{\text{b}}^{\text{2}}{\text{x}}^{\text{2}}\right)\\ +\left({\text{- 2cx}}^5{\text{+ 4bcx}}^{\text{4}}{\text{- 2b}}^{\text{2}}{\text{cx}}^{\text{3}}{\text{+4acx}}^{\text{4}}{\text{- 8abcx}}^{\text{3}}{\text{+4ab}}^{\text{2}}{\text{cx}}^2{\text{- 2a}}^{\text{2}}{\text{cx}}^{\text{3}}{\text{+ 4a}}^{\text{2}}{\text{bcx}}^2{\text{- 2a}}^{\text{2}}{\text{b}}^{\text{2}}\text{cx}\right)+\\ \left({\text{c}}^{\text{2}}{\text{x}}^{\text{4}}{\text{- 2x}}^{\text{3}}{\text{bc}}^{\text{2}}{\text{ + x}}^{\text{2}}{\text{b}}^{\text{2}}{\text{c}}^{\text{2}}{\text{- 2ac}}^{\text{2}}{\text{x}}^{\text{3}}{\text{+4abc}}^{\text{2}}{\text{x}}^{\text{2}}{\text{- 2ab}}^{\text{2}}{\text{c}}^{\text{2}}{\text{x + a}}^{\text{2}}{\text{c}}^{\text{2}}{\text{x}}^{\text{2}}{\text{- 2a}}^{\text{2}}{\text{bc}}^{\text{2}}{\text{x + a}}^{\text{2}}{\text{b}}^{\text{2}}{\text{c}}^{\text{2}}\right)\end{array}\right)$

After solving the above expression,

${\text{g(x)αx}}^{\text{8}}\text{- }\left(\text{2α + 1-}\frac{\text{1}}{\text{r}}\right){\text{x}}^{\text{7}}\text{+ }\left({\text{2β + α}}^{\text{2}}\text{+ 2}\left(\text{1-}\frac{\text{1}}{\text{r}}\right)\text{α}\right){\text{x}}^{\text{6}}\text{- }\left(\text{2γ + 2αβ + 2}\left(\text{1-}\frac{\text{1}}{\text{r}}\right)\text{β +}\left(\text{1-}\frac{\text{1}}{\text{r}}\right){\text{α}}^{\text{2}}\right){\text{x}}^{\text{5}}\mathrm{......}$

${\text{g(x) = r}}^{\text{8}}\text{x}\left(\text{1- x}\right)\left[\text{1- rx}\left(\text{1- x}\right)\right]\left[{\text{1- r}}^{\text{2}}\text{x}\left(\text{1- x}\right)\left(\text{1- rx}\left(\text{1- x}\right)\right)\right]\text{- x}$

${\text{g(x) = r}}^{\text{7}}\left[{\text{x}}^{\text{8}}{\text{- 4x}}^{\text{7}}\text{+}\left(\text{6+}\frac{\text{2}}{\text{r}}\right){\text{x}}^{\text{6}}\text{-}\left(\text{4+}\frac{\text{6}}{\text{r}}\right){\text{x}}^{\text{5}}\text{+}\mathrm{.....}\right]$

Now comparing the terms of the above equation with ${\text{r}}^{\text{3}}\left(\text{1- 2α + 4β - 8λ}\right)\text{=1}$ and the values of $\text{α , β and γ}$ are as follows,

$\text{2α + 1-}\frac{\text{1}}{\text{r}}=4$, Hence $\text{2α = 3+}\frac{\text{1}}{\text{r}}$

${\text{2β + α}}^{\text{2}}\text{+ 2}\left(\text{1-}\frac{\text{1}}{\text{r}}\right)\text{α = 6+}\frac{\text{2}}{\text{r}}$ Hence $\text{4β = }\frac{\text{3}}{\text{2}}\text{+}\frac{\text{5}}{\text{r}}\text{+}\frac{\text{3}}{{\text{2r}}^{\text{2}}}$

Similarly $\text{2γ + 2αβ + 2}\left(\text{1-}\frac{\text{1}}{\text{r}}\right)\text{β +}\left(\text{1-}\frac{\text{1}}{\text{r}}\right){\text{α}}^{\text{2}}=\text{4+}\frac{\text{6}}{\text{r}}$ Hence $\text{8γ = -}\frac{\text{1}}{\text{2}}\text{+}\frac{\text{7}}{\text{2r}}\text{+}\frac{\text{5}}{{\text{2r}}^{\text{2}}}\text{+}\frac{\text{5}}{{\text{2r}}^{\text{3}}}$

By substituting above values in the equation ${\text{r}}^{\text{3}}\left(\text{1- 2α + 4β - 8λ}\right)\text{=1}$

${\text{r}}^{\text{3}}\left(\text{1- 2α + 4β - 8λ}\right)\text{=1}$

${\text{r}}^{\text{3}}\left(\text{1- }\left(\text{3+}\frac{\text{1}}{\text{r}}\right)\text{ + }\left(\frac{\text{3}}{\text{2}}\text{+}\frac{\text{5}}{\text{r}}\text{+}\frac{\text{3}}{{\text{2r}}^{\text{2}}}\right)\text{ - }\left(\text{-}\frac{\text{1}}{\text{2}}\text{+}\frac{\text{7}}{\text{2r}}\text{+}\frac{\text{5}}{{\text{2r}}^{\text{2}}}\text{+}\frac{\text{5}}{{\text{2r}}^{\text{3}}}\right)\right)\text{=1}$

${\text{r}}^{\text{3}}\left(\text{1 -3-}\frac{\text{1}}{\text{r}}\text{ + }\frac{\text{3}}{\text{2}}\text{+}\frac{\text{5}}{\text{r}}\text{+}\frac{\text{3}}{{\text{2r}}^{\text{2}}}+\frac{\text{1}}{\text{2}}-\frac{\text{7}}{\text{2r}}-\frac{\text{5}}{{\text{2r}}^{\text{2}}}-\frac{\text{5}}{{\text{2r}}^{\text{3}}}\right)\text{=1}$

${\text{r}}^{\text{3}}\left(\frac{1}{\text{2r}}-\frac{\text{1}}{{\text{r}}^{\text{2}}}\text{-}\frac{\text{5}}{{\text{2r}}^{\text{3}}}\right)\text{=1}$

$\left(\frac{{\text{r}}^{\text{2}}}{\text{2}}\text{- r}-\frac{5}{\text{2}}\right)\text{=1}$

$\left(\frac{{\text{r}}^{\text{2}}}{\text{2}}\text{- }\frac{2\text{r}}{\text{2}}-\frac{5}{\text{2}}\right)\text{=1}$

$\left({\text{ r}}^{\text{2}}\text{- 2r}-5\right)\text{= 2}$

${\text{r}}^{\text{2}}\text{- 2r}-7\text{= 0}$

$\text{r =}\frac{2\pm \sqrt{4+28}}{2}$

$\text{r =}\frac{2\pm \sqrt{32}}{2}$

$\text{r =}\frac{2\pm 4\sqrt{2}}{2}$

$\text{r =}1\pm 2\sqrt{2}$

Hence, the logistic map of the period cycle-3 has a tangent at parameter $\text{r = 1+2}\sqrt{\text{2}}$ when the map has a maximum value at $\text{x = 0}$ and $\text{x = }\frac{\text{1}}{\text{2}}$.

There is a value of the parameter $\text{r = 1+2}\sqrt{\text{2}}\text{=3}\text{.828}$ for period cycle-$3$ for the period-doubling bifurcation of the period cycle-$3$. If the accumulated parameter value $\text{r =3}\text{.8495}$, then it indicates that the chaotic bands are merged to each other and the minor change in the parameter value $\text{r = 1+}\sqrt{\text{8}\text{.12}}$ for period cycle-$3$. The logistic function may loseits stability and it only consists of the period-doubling. The distance between slots will increase and have three small chaotic bands.

If the chaotic bands are merging, then the value of the parameter $\text{r}$ is near about $\text{3}\text{.857}$. As we change the parameter $\text{r}$ of the logistic map for the period cycle and the attractor are formed in the system, this is the last point of the chaotic band of the period cycle-$3$. Hence the function is ready to enter in the period cycle-$6$ and due to the large attractor bands gap. The stability of the system at a different fixed point is loss and its stability depends on the period-doubling bifurcation and ready to jump in the next period cycle.

Hence, the logistic mapping for the parameter value of the period cycle-$3$ is followed the period-doubling process for any change in the parameter values in the orbit. And at this point, the chaotic bands are full all the interval due to the merging in the interval bands are full, loses its stability and is ready to enter into the period cycle-$6$.

Conclusion

The period-doubling bifurcations of the $\text{3}$-cycle for the logistic map accumulate near $\text{r }=\text{3}.\text{8495 }\dots$