Nonlinear Dynamics and Chaos
Nonlinear Dynamics and Chaos
2nd Edition
ISBN: 9780429972195
Author: Steven H. Strogatz
Publisher: Taylor & Francis
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Chapter 10.3, Problem 6E
Interpretation Introduction

Interpretation:

Consider the cubic map xn+1 = f (xn) where f(xn) = rxn - xn3

  • 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 p and q are the roots of the equation x(x2- r+1)(x2- r -1)(x4-rx2+1)=0 and use these roots to find all the 2- cycles.

  • c) To determine the stability of the 2- cycle as a function of r.

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

Concept Introduction:

  • ➢ The logistic mapping xn+1 = f (xn) 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.

Expert Solution & Answer
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Answer to Problem 6E

Solution:

  • a) The fixed points x* = 0 have |f '(x*)| > 1 since they are unstable and for the next two fixed points x* = ± r - 1 are stable.

  • b) p and q are the roots of the equation x(x2- r+1)(x2- r -1)(x4-rx2+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 xn+1 = rxn - xn3,

    The difference equation xn+1 = f (xn).

    Let xn = x* then the given quadratic map can be written as

    f(x*) = rx* - x*2.

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

    f(x*) = rx* - x*3x* = x*(r - x*2)(r - x*2) = 1

    The first solution of the mapping is x* = 0 this is the first fixed point.

    Also;

    r - x*2 = 1x*2 = r - 1x* = ± r - 1

    The next two fixed points are ± r - 1. Now consider the stability of these two fixed points the mapping function f(x*) = rx* - x*3 taking the first derivative of the mapping.

    So,

    f '(x*) = r - 3x*2

    At the first fixed point put x* = 0.

    Then, f '(0) = r

    At the next two fixed point.

    f '(±r - 1)= r - 3 (±r - 1)2f '(±r - 1)= r - 3(r - 1)f '(±r - 1)= r - 3r + 3f '(±r - 1)= 3 - 2r

    Hence, all of these fixed points x* = 0 have |f '(x*)| > 1 since they are unstable and for the next two fixed points, x* = ± r - 1 are stable.

  • b)

    Consider the cubic map xn+1 = f (xn), where f(xn) = rxn - xn3 of the 2-cycle orbit. Suppose that p and q are the two fixed point of the cubic mapping and f(p) = q and f(q) = p these are the roots of the polynomial equation,

    Here consider the 2-cycle orbit so let the value of the roots be:

    f(x*) = rx* - x*3x* = x*(r - x*2)(r - x*2) = 1

    The first solution of the mapping is x* = 0 this is the first fixed point. Consider also:

    r - x*2 = 1x*2 = r - 1x* = ± r - 1

    Hence, the first root is p, q = 0 then the second roots are p,q =±r - 1 x.

    If f(x*) = -x* for negative fixed root so,

    f(x*) = rx* - x*3-x*=x*(r - x*2)(r - x*2) = -1x*= ±r + 1

    Hence, the root of the third polynomial equation is x* = ± r + 1.

    Now taking the rational fixed point at f(x*) = 1x* so,

    f(x*) = rx* - x*31x* = x*(r - x*2)(rx*2 - x*4) = 1x*4 - rx*2 + 1 = 0

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

  • c)

    Now consider the stability of cycle-2 orbit as a function of r. Consider the cubic equation of a period-2 orbit,

    The difference equation xn+1 = f (xn) period is 2 orbit so the equation xp=f2(xp)

    So,

    xp= f2(xp)xp= f(f(xp))xp= f(xp(r-xp2))xp= f(y)

    Here, the quadratic equation f(x) = rx - x3

    Then,

    The equation is written as:

    f(y) = ry - y3f(y) = r(rxp - xp3) - (rxp - xp3)3f(y) = (rxp - xp3)(r - (rxp - xp3)2)

    Let the fixed point be x&*#x00B1; so, (x - xx+)(x - xx-) it can be written as x2 + 1 - r. Now the quadratic equation is written as:

    (rxp - xp3)(r - (rxp - xp3)2) = xp(r - xp2)(r - xp2(r - xp2)2)

    Compare with (xp3+αxp2+βxp+γ)(xp2+1- r) the coefficient of the cubic degree equation.

    (x3+ αx2+ βx + γ)(x2+1- r)=x5+ x3- rx3+ αx4+ αx2-rαx2+βx3+βx-rβx+γx2+γ-γr(x3+ αx2+ βx + γ)(x2+1- r)=x5+ αx4+(1+βr)x3+(α+γ)x2+(β)x+γγr............1)

    And

    x(r - x2)( r- (rx - x3)2) = (rx-x3)(r - x6- rx2+ 2rx4)x(r - x2)( r- (rx - x3)2) = r2x - r2x3+2r2x5- rx3+x9+rx5-3rx7x(r - x2)( r- (rx - x3)2) = x9-3rx7+(2r2+r)x5-(r2+r)x3 + r2x.........2)

    Comparing equations 1) and 2)

    1= 2r2+rα = 0β = -r21αrα + γ=0

    And

    βγβ=r2γγr=0

    Now the values are α=0 and r=1.

    If r=-2 then β=5,γ=0.

    Hence, the period of the second orbit x = xp satisfies the cubic equation,

    x3- 5x = 0x =±5,0

    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:

    b2- 4ac = 0

    b2- 4ac = 0x2r+1=04(r1)=0r =1

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

  • d)

    The bifurcation diagram is as shown below,

    Nonlinear Dynamics and Chaos, Chapter 10.3, Problem 6E

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