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Chaos and Fractals

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Chaos and Fractals. Chapter 6: Bifurcations. 2. Chapter 6: Family of ... Last example page 65: 26. F?(x)= ?x-x3. FP: F?(0)=0 for all ? -1 ? 1: FP attracting ... – PowerPoint PPT presentation

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Title: Chaos and Fractals


1
Chaos and Fractals
  • Chapter 6
  • Bifurcations

2
Chapter 6
  • Family of functions Qc(x) x2 c (with c
    constant)
  • It looks simple but..
  • Complicated dynamics
  • Behavior not understood for some c
  • C is a parameter each c ? different system

3
Goal
  • Understand how dynamics of
  • Qc(x) x2 c change when c varies
  • ?
  • introduce 2 important bifurcations

4
Fixed points (FP) solve x2 c x
FP should be real (1-4c) 0 ? c ¼ 1) c gt
¼ no FP 2) c ¼ 1 FP (p p-) 3) c lt ¼ 2 FP
(distinct)
5
Case c gt ¼ (does not meet yx) Case c
¼ 1 FP (at x ½) x2
¼ is tangent to line yx Case c lt ¼ 2 FP p
and p- attracting/repelling/neutral ??
6
  • Look at Qc(x) in the FP p and p-
  • Qc(x) 2x

Case for p
  • If c ¼ Qc(p) 1 ? neutral FP
  • If c lt ¼ Qc(p) gt 1 ? repelling FP

7
Case for p- slightly more complicated
  • If c ¼ Qc(p-) 1 ? one neutral FP
  • If c lt ¼ Qc(p-) lt 1 ? attracting FP
  • For which values of c is p- attracting
    ??Condition Qc(p-) lt 1, solving this yields

8
  • This phenomenon is called a bifurcation
  • Saddle-node (or tangent bifurcation)
  • Summarized in Proposition on page 54
  • For the family Qc(x)x2c
  • All orbits tend to infinity if c gt ¼.
  • When c ¼, Qc has a single neutral FP at pp-
    ½.
  • For c lt ¼, Qc has 2 FP at p and p-. FP p is
    always repelling,a. If -¾ lt c lt ¼, p- is
    attracting,b. If c -¾, p- is neutral,c. If c
    lt -¾ , p- is repelling.

9
Second part 6.1
  • For c ¼
  • interesting dynamics occur in p x p
  • Orbits for x lt -p and x gt p tend to infinity
  • Qc(-p) p eventually FP !

10
  • For ¾ lt c lt ¼ orbits in (-p,p) tend to
    attracting FP p-.
  • Case c gt ¼, c ¼ and -¾ lt c lt ¼ known, but what
    about c lt -¾ ??

11
  • When c lt -¾1) p- becomes repelling2) and a
    cycle of period 2 occurs !!To see this solve
    Qc2(x) x
  • (x2c)2 c x ? x4 2cx2 c2 c x
    04th order difficult ! But we know 2 solutions
    !(x-p) and (x-p-)

12
Solving x2 x (c 1) gives
  • FP q and q- of Qc2(x) are FP of
  • Qc(x) of period 2 !

Note that (-4c-3) should be 0 c -¾. As
c decreases below ¾, p- changes from attracting
to repelling and a 2-cycle occurs at q and q-.
If c -¾, q q- p- -½.
13
  • Is this 2-cylce attracting/neutral/repelling
    ?Condition if (Qc2)(q) lt 1, then
    attracting2 ways to solve this (Qc2)
    4x34xc, substitute q 44c lt 1
    (Qc2)(q) lt 1 ? Qc(q)Qc(q-) lt 1 ?

    44c lt 1Both cases if -1¼ ltclt -¾, attracting
    2-cycle !And if clt-1¼ repelling 2 cycle.

14
  • This phenomenon is also called a bifurcation
  • Period doubling bifurcation.
  • Summarized in Proposition on page 56
  • For the family Qc(x)x2c
  • For ¾ lt c lt ¼, Qc has attracting FP at p- and
    no 2-cycles.
  • For c -¾, Qc has neutral FP at p-q and no
    2-cycles.
  • For -1¼ lt c lt -¾, Qc has repelling FP at p and
    an attracting 2 cycle at q.

15
Section 6.2
  • Look at more general family of functions F? to
    study bifurcations.
  • For each ? F?(x) is function of x
  • Examples ?x(1-x), ? sin(x), ex ?
  • Bifurcations occur in F? when there is change in
    fixed or period point structure as ? passes
    through some particular value ?0.

16
Saddle-Node bifurcation
  • DefinitionA one parameter family of functions
    F? undergoes a saddle-node (or tangent)
    bifurcation at the parameter value ?0, if there
    is open interval I and e gt 0 st1) For ?0-e lt ?
    lt ?0, F? has no FP in I2) For ??0, F? has one
    neutral FP in I.3) For ?0 lt ? lt ?0e, F? has 2
    FP in I, one attracting, one repelling.

17
Remarks
  • Reversed direction also bifurcation possible
  • Also saddle-node bifurcation of periodic points.
  • Saddle-node bifurcation typically occurs
    whenF?0(x0) 1 and F?0(x0) ? 0 (concave and
    one FP x0).

18
Typical saddle-node bifurcation
19
Bifurcation diagram
  • 2 dimensional plot
  • X-axis value of ?
  • Y-axis fixed points xExample Q?(x) x2 ?,
    saddle-node bifur. at ?¼ ? gt ¼ 0
    FP ? ¼ 1 FP ? lt ¼ 2 FP

20
  • Another example
  • family E?(x)ex? saddle-node bifurcation when
    ?-1E-1 (0)0E-1 (0)1E-1(0)1

21
6.3 Period-Doubling bifurcation
  • Definition A one-parameter family of functions
    F? undergoes a period-doubling bifurcation at
    ??0 if there is open interval I and e gt 0
    st1. For each ? in ?0- e, ?0 e, there is
    unique FP p? for F? in I.2. For ?0- e lt ?
    ?0, F? has no 2-cycles in I and p is
    attracting (resp. repelling).3. For ?0 lt ? lt
    ?0e, there is unique 2-cycle q1?, q2? in I
    with F?(q1?) q2?.This 2-cycle is attracting
    (resp. repelling). Meanwhile the FP p? is
    repelling (resp. attracting).4. As ? ? ?0,
    we have qi? ? p?0

22
Remarks
  • 2 cases for period-doubling when ? changesFP
    changes attracting?repelling attracting 2-cycle
    orFP changes repelling? attracting repelling
    2-cycle
  • Direction of ? may be reversed.
  • Cycle can period-double too n-cycle ? 2n-cycle
  • Period-doubling occurs when F? - diagonal yx
    F?(x) -1

23
  • Example Qc(x) x2 c, period-doubling at
    c -¾. Look at p-

24
saddle-node
period-doubling
25
Last example page 65
26
  • F?(x) ?x-x3
  • FP F?(0)0 for all ?
  • -1 lt ? lt 1 FP attracting ? -1 FP neutral ?
    lt -1 FP repellingWhy ? F?(x) ? 3x2 and
    F?(0) ? Obvious !
  • Odd function F?(-x) -F?(x) for all x.F(x0)
    -x0 after 1 applyF(-x0) -F(x0) F2(x0) x0,
    so solve F?(x0) -x
  • ?x-x3x0, x0 or x1,2 Sqrt(?1)
  • Repelling or attracting ?(F2?)(x1) (F?)(x1)
    (F?)(x2) (?-3x2)(?-3x2) 4?2 12? 9For ?
    -1 value 1 neutralFor ? -1/2 value 4
    repelling
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