Nonlinear Dynamics and Chaos
Nonlinear Dynamics and Chaos
2nd Edition
ISBN: 9780813349107
Author: Steven H. Strogatz
Publisher: PERSEUS D
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Question
Chapter 10.7, Problem 1E
Interpretation Introduction

Interpretation:

For the functional equation g(x) = αg2(xα), arise in renormalization analysis of period doubling.

  1. If g(x)1+c2x2 for small x, then solve for c2 and α

  2. For g(x)1+c2x2+c4x4, solve equation for c2, c4 and α, and compare the results with exact values c2-1.527..., c40.1048... and α-2.5029...

Concept Introduction:

  • The power series expansion of the g(xα) is

    g(xα) = 1+c2(xα)2+c4(xα)4+...

Expert Solution & Answer
Check Mark

Answer to Problem 1E

Solution:

  1. α=2.73205 and c2=1.36603

  2. α=2.534...   c2=1.52224...   c4=0.12761... and these are compared with exact values of α,c2 and c4

Explanation of Solution

  1. The function equation arose in renormalization equation is,

    g(x) = αg2(xa)

    The power series expansion of the above equation is

    g(xα) = 1+c2(xα)2+c4(xα)4+...

    g2(xα)=g(g(xα))=g(1+c2(xα)2)=1+c2+2c22(xα)2+O(x4)

    αg2(xa)=α(1+c2+2c22(xα)2+O(x4))=α(1+c2)+2c22x2α+αO(x4)

    The given approximation is

    g(x)=1+c2x2

    Comparing this with αg2(xa)

    1 + c2x2 = α(1+c2)+2c22x2α + O(x3)

    Comparing the coefficients of x0 and x2

    1= α(1+c2)

    c2=2c22α

    Solving above two equations

    (α,c2)={(1,0)(13,132)(2.73205,1.36603)(1+3,1+32)(0.732051,0.366025)

    Here, take the second solution because in that solution α is closer to Feigenbaum α=2.5...

  2. g(x) = αg2(xa)=α(1+c2+c4+2(c22+2c2c4)(xα)2+(c23+6c22c4+2c2c4+c42)(xα)4+O(x5))

    The given approximation is

    g(x)=1+c2x2+c4x4

    Equating this with αg2(xa)1+c2x2+c4x4(1+c2+c4)+2(c22+2c2c4)x2α+(c23+6c22c4+2c2c4+c42)x4α3+O(x5)

    Comparing the coefficients of x0,x2and x4, we get

    1= α(1+c2+c4)

    c2=2(c22+2c2c4)α

    c4=(c23+6c22c4+2c2c4+c42)α3

    Using first and second equations for solving c2 and c4 in terms of α

    c2=2+2αα2

    c4=11α+α2

    Substituting these values of c2 and c4 in third equation, then it becomes

    4α6+3α560α4104α3+168α2+240α384+128α=0

    The root of above equation closest to Feigenbaum is α=2.534...

    Substituting this value of α in c2=2+2αα2 and c4=11α+α2

    α=2.534...   c2=1.52224...   c4=0.12761...

    Comparing the results with exact values c2-1.527..., c40.1048... and α-2.5029... they are nearly equal.

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