Discrete Chaos: With Applications in Science and Engineering, Second Edition (Solutions, Instructor Solution Manual) [2 ed.] 9781584885924, 9781420069853, 1584885920

Citation preview

SOLUTIONS MANUAL FOR

Discrete Chaos, Second Edition: With Applications in Science and Engineering

by Saber N. Elaydi

C6985.indd 1

9/27/07 3:28:28 PM

C6985.indd 2

9/27/07 3:28:28 PM

SOLUTIONS MANUAL FOR Discrete Chaos, Second Edition: With Applications in Science and Engineering

by Saber N. Elaydi

Boca Raton London New York

Chapman & Hall/CRC is an imprint of the Taylor & Francis Group, an informa business

C6985.indd 3

9/27/07 3:28:29 PM

Chapman & Hall/CRC Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2008 by Taylor & Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-13: 978-1-4200-6985-3 (Softcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

T&F_LOC_C_Master.indd 1 C6985.indd 4

9/27/07 AM 9/27/07 11:33:03 3:28:29 PM

Contents

1 The Stability of One-Dimensional Exercises - (1.2–1.4) . . . . . . . . . . Exercises - (1.5 and 1.6) . . . . . . . Exercises - (1.7) . . . . . . . . . . . . Exercises - (1.8) . . . . . . . . . . . . Exercises - (1.9) . . . . . . . . . . . .

Maps . . . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

1 . 1 . 4 . 6 . 11 . 19

2 Attraction and Bifurcation Exercises - (2.2 and 2.3) . . . Exercises - (2.4) . . . . . . . . Exercises - (2.5) . . . . . . . . Exercises - (2.6) . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

23 23 25 27 28

3 Chaos in One Dimension Exercises - (3.2 and 3.3) . . Exercises - (3.4) . . . . . . . Exercises - (3.5) . . . . . . . Exercises - (3.6 and 3.7) . . Exercises - (3.8) . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

33 33 34 36 37 40

Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

43 43 45 47 50 53 54 56

6 Fractals Exercises - (6.1–6.3) . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises - (6.4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises - (6.5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

59 59 60 63

. . . . .

4 Stability of Two-Dimensional Exercises - (4.1 and 4.2) . . . . Exercises - (4.3 and 4.4) . . . . Exercises - (4.5–4.7) . . . . . . . Exercises - (4.8) . . . . . . . . . Exercises - (4.9) . . . . . . . . . Exercises - (4.10 and 4.11) . . . Exercises - (4.12) . . . . . . . .

7 The Julia and Mandelbrot Sets 65 Exercises - (7.2 and 7.3) . . . . . . . . . . . . . . . . . . . . . . . . 65 Exercises - (7.4 and 7.5) . . . . . . . . . . . . . . . . . . . . . . . . 67

i

1 The Stability of One-Dimensional Maps

Exercises - (1.2–1.4) 1. 1 x(n) + 2, x(0) = c 2   n   n 1 − 12 1 c+2 x(n) = 2 1 − 12  n  n 1 1 c+4−4 = 2 2  n 1 x(n) = 4 + (c − 4) 2

x(n + 1) =

3. (a) Letting x(n) =

1 z(n)

(b)

yields z(n + 1) = 

z(n) =

 β + α−1 1− c + nβ 1 αn c

1 α z(n) 1 αn



+ αβ . if α = 1, if α = 1.

Since z(0) = z0 = c, we have

 β z0 + (α−1) (αn − 1) /αn z(n) = z0 + nβ Now 1 = x(n) = z(n) This implies

 lim x(n) =

n→∞



x0 (α−1)αn (α−β−1)+βx0 αn x0 1+nx0 β

(α−1) β

0

if α = 1, if α = 1. if α = 1, if α = 1.

if |α| > 1, if α = 1 or |α| < 1.

If α = −1, every solution is periodic of period two and has two points {x0 , −x0 /(1 + βx0 )} for any initial point x0 .

1

2

Discrete Chaos 4. x(n + 1) = 2x(n)[1 − x(n)] Let x(n) = 12 [1 − y(n)]. Then we have   1 1 (1 − y(n + 1)) = (1 − y(n)) 1 − (1 − y(n)) 2 2 1 1 1 1 1 1 − y(n + 1) = 1 − + y(n) − y(n) + y(n) − y 2 (n). 2 2 2 2 2 2 (2n )

This yields y(n + 1) = y 2 (n). Then by iterations we have y(n) = y0 where y0 = y(0). Hence 1 (2n ) . 1 − y0 x(n) = 2 But y0 = 1 − 2x0 which implies that

n 1 1 − (1 − 2x0 )2 . x(n) = 2

5. Let x(n) = sin2 θ(n). Then θ(n) = sin−1 x(n). sin2 θ(n + 1) = 4 sin2 θ(n)(1 − sin2 θ(n)) = 4 sin2 θ cos2 θ = sin2 2θ(n) θ(n + 1) = 2θ(n) θ(n) = 2n θ(0)

sin−1 x(n) = 2n sin−1 x(0) √ x(n) = sin2 (2n sin−1 x0 ) 7. a(n + 1) = rq(1 − m)a(n) = 2a(n) (a) a(n + 1) = 2a(n) (b) a(1) = 20, a(2) = 40, . . . , a1 0 = 210 · 10 (c)

,

The Stability of One-Dimensional Maps

3

9. (a) a(n + 1) = rq(1 − m)a(n) − 9 (b) a(n + 1) = 2a(n) − 9

11. (a) y  = −y + 1,

y(0) = 0,

0 ≤ t ≤ 1,

h = 0.25

y(n + 1) = y(n) + 0.25(−y(n) + 1) 1 3 y(n + 1) = y(n) + 4 4 (b) y(n) .75

DE Δ E

.5 .25 1

2

3

4

h = .25

.5

.75

1

n

(c) 

 d (yet ) = et dt yet = et + c y = 1 + ce−t 0 = y(0) = 1 + c ⇒ c = −1 y(t) = 1 − e−t

DE

t 0 .25 .5 .75 1 y(t) 0 .22 .39 .53 .63

ΔE

n 01 2 3 4 y(n) 0 .25 .44 .5 .63

4

Discrete Chaos

Exercises - (1.5 and 1.6) 

1. (a) f (x) = |x − 1| =

x − 1 if x ≥ 1 1 − x if x < 1 f(x) y=x

x x*

There are no fixed points for x ≥ 1. 1 − x = x ⇒ x∗ = 12 is a fixed point. (b) The eventually fixed points are ± 23 , ± 25 , ± 2k+1 2 ,... 2. (b) There are two fixed points x∗1 = 0, x∗2 = μ=2:

x∗1 is unstable;

μ−1 . μ

x∗2 is asymptotically stable.

μ = 2.5 : x∗1 is unstable; x∗2 is asymptotically stable. μ = 3.2 : Both x∗1 x∗2 are unstable. 3. (a) Let f (x) = 2x(x − 1)(x − 2)(x + 1) + x. Then f has four unstable fixed points x∗1 = 0, x∗2 = 1, x∗3 = 2, x∗4 = −1. Notice that f  (0) = 5,

f  (1) = −6,

f  (2) = 8,

f  (−1) = −11.

All four fixed points are unstable. (b) f (x) = x2 + 1 has no fixed points since x2 + 1 = x or x2 − x + 1 = 0 has two complex roots. (c) Let f (x) = x2 − 12 . Then f (x) has two fixed points obtained by solving the equation 1 x2 − x − = 0. 2 √

√

The fixed point x∗1 = 1+2 3 , x∗2 = 1−2 3 is asymptotically stable √ since√f  (x) = 2x, f  (x∗1 ) = 1 + 3 > 1, x∗1 is unstable. |f2 (x∗2 )| = |1 − 3| < 1, x∗2 is asymptotically stable.

The Stability of One-Dimensional Maps

5

6 x To find the fixed points, let f (x) = x.

4. f (x) = 5 −

6 = x, x = 0 x x2 − 5x + 6 = (x − 3)(x − 2) = 0 x∗1 = 3, x∗2 = 2 5−

From the cob-web diagram, x∗1 = 3 is asymptotically stable and x∗2 = 2 is unstable. 5. (a) αx(n) , α > 1, 1 + βx(n) αx =x 1 + βx βx2 + x(1 − α) = 0

x(n + 1) =

β>0

Hence the positive equilibrium point is given by x∗ =

α−1 β .

7. Let x0 = x(0) = 0. Then (a) x(1) = f (x(0)) = f (0) = 0, x(2) = f (x(1)) = f (0) = 0, . . . , x(n) = 0. (b) If x(m) = 0 for some m, then x(m + 1) = f (x(m)) = f (0) = 0, which contradicts the graph. 8. Qc (x) = x2 + c (a) x2 + c = x x2 − x + c = 0 √ 1 ± 1 − 4c x= 2 √ √ 1 − 4c 1 + 1 − 1 − 4c ∗ ∗ x1 = , x2 = . 2 2 1 1 Both exist if 1 − 4c ≤ 0 ⇒ c ≤ . If c > , we have no fixed points. 4 4 From the cob-web diagram it is clear that x∗1 is unstable, while x∗2 3 is asymptotically stable if c > − . 4

6

Discrete Chaos 9. Let x0 = k/2n in its reduced form with 0 < k/2n ≤ 1. Then T (x0 ) =

 2n−1

(1.1)

whether x0 < 12 or x0 ≥ 12 . Now on the interval [0, 1] there are only finitely many points of the form 2r , and thus x0 is eventually periodic. By (1.1) no point repeats in the orbit of x0 and hence x0 must be eventually fixed. In fact x0 is an eventually fixed point of the fixed point x∗1 = 0, since the eventually fixed points of x∗2 = 23 all have denominators of multiples of 3. 11. x(n + 1) = 12 sin(πx(n)) There are three fixed points. x∗1 =

1 2

x∗2 = − x∗3 = 0

1 2

:

asymptotically stable

:

asymptotically stable

:

unstable

13. (a) 2xe−x = x ∗ We have two fixed points x∗1 = 0 and x∗2 with 2e−x2 = 1. Hence ∗ 1 −x∗ x ∗  e 2 = 2 or e 2 = 2. Thus x2 = ln 2. Now f (x) = 2e−x − 2xe−x . x∗1 = 0 is unstable. (b) f  (ln 2) = 1 − ln 2 < 1. Hence x∗2 = ln 2 is asymptotically stable.

Exercises - (1.7) 1. Fixed Points: x∗1 = 0, x∗2 = 1 Now f  (x) = 2x f  (0) = 0 ⇒ x∗1 is asymptotically stable f  (1) = 2 ⇒ x∗2 is unstable

2. x(n + 1) = 12 x3 (n) + 12 x(n) The equilibrium points are givn by 1 3 1 x + x=x 2 2 1 2 x(x − 1) = 0 2 x∗1 = 0, x∗2 = −1, x∗3 = 1.

The Stability of One-Dimensional Maps

7

To find their stability we evaluate f  . 3 1 f  (x) = x2 (n) + 2 2 1 f  (0) = implies by Theorem 1.3, that x∗1 is asymptotically stable. 2 f  (−1) = 2 implies by Theorem 1.3, that x∗2 is unstable and similarly for x∗3 . 3. x(n) = 3x(n)[1 − x(n)] There are two fixed points x∗1 = 0, x∗2 = 23 . Now f (x) = 3x − 3x2 , f  (x) = 3 − 6x. Since f  (0) = 3, it follows by Theorem 1.3, that x∗1 is unstable. However, since f  23 = −1 we appeal to Theorem 1.5. The Schwarzian derivative of f at x∗2 is given by  2 f  (x) 3 f  (x) − Sf (x) =  f (x) 2 f  (x) 3 Sf (x∗2 ) = − (36) < 0. 2 This implies by Theorem 1.5, that x∗2 is asymptotically stable. x∗1 = 0 is unstable 2 x∗2 = is asymptotically stable 3 4. Fixed Points: tan−1 (x) = x ⇒ tan x = x f  (x) =

1 , 1 + x2

f  (x) =

−2x , (1 + x2 )2

f  (x) =

6x2 − 2 (1 + x2 )3

Fixed Points: x∗1 = 0, f  (0) = 1, f  (0) = 0, f  (0) = −2 Hence by Theorem 1.11, x∗1 is asymptotically stable. 5. x(n + 1) = x(n) exp(1.5(1 − x(n))) f (x) = xe1.5(1−x) ,

f  (x) = e1.5(1−x) − 1.5x2 e1.5(1−x)

Fixed Points: x∗1 = 0, f  (0) = e1.5 > 1 ⇒ x∗1 = 0 is unstable.



e1.5(1−x) = 1 ⇒ x∗2 = 1

f (1) = 1 − 1.5 = −.5 ⇒ x∗2 = 1 is asymptotically stable. 6. There is only one fixed point x∗ = 0. Since f  (0) = 0.8, x∗ is asymptotically stable by Theorem 1.3.

8

Discrete Chaos y=x 0.4

0

1/2

1

7. x(n + 1) = −x3 (n) − x(n) f (x) = −x3 − x,

f  (x) = −3x2 − 1,

f  (x) = −6x,

f  (x) = −6

Fixed Points: −x3 − 2x = −x(x2 + 2) = 0 x∗1 = 0,

f  (0) = −1

3 Sf (0) = −f  (0) − [f  (0)]2 2 =6>0 Hence by Theorem 1.13, x∗1 = 0 is unstable. 8. There are two fixed points x∗1 = 0, x∗2 = 1. Since f  (0) = 2, x∗1 is unstable by Theorem 1.3. Now f  (1) = 2 implies that x∗2 is unstable by Theorem 1.3. f(x) 1

y=x

x 1/2

9. x(n + 1) =

αx(n) 1+βx(n) ,

f (x) = Fixed Points:

α > 1,

αx , 1 + βx αx 1+βx

1

β>0

f  (x) =

α(1 + βx) − αβx α = (1 + βx)2 (1 + βx)2

=x

βx2 + x(1 − α) = 0

x∗1 = 0, f  (0) = α > 1 ⇒ x∗1 = 0 is unstable   α−1 α−1 1 α ∗  x2 = , f = 0 such that that for  ∗ |f 1(x ∗)| M . Since xn → x∗ as n → ∞, it follows that lim |f  (xn )| = |f  (x∗ )|. This n→∞

implies that |f  (x∗ )| ≥ M , which is a contradiction. Hence there exists η > 0 such that x ∈ (x∗ − η, x∗ + ε) implies that |f  (x)| ≤ M . 11. (a) f  (x) = 2ax + b, f  (x) = 2a. By Theorem 1.5 x∗ is unstable since f  (x∗ ) = 0.  ∗ 2 ) (b) Notice that f  (x) = 0. Hence Sf (x∗ ) = − 32 ff (x < 0. By (x) ∗ Theorem 1.6, x is asymptotically stable. 12. g(x) g  (x) g(x∗ ) fN (x∗ ) = x∗ −  ∗ g (x ) g(x) g  (x∗ ) = lim∗  ∗ lim∗  x→x g (x) x→x g (x ) fN (x) = x −

which is defined. 13. There are two fixed points x∗1 = 0 and x∗2 = 1. f  (x) = 1 + α − 2αx, f  (x) = −2α, f  (x) = 0, x∗1 : f  (0) = 1 + α. Thus x∗1 = 0 is asymptotically stable if |1 + α| or −2 ≤ α < 0. (i) α = 0, f  (0) = 1. Since f  (0) = 0, by Theorem 1.5, x∗1 = 0 is unstable if a = 0. (ii) α = −2, f  (0) = −1, Sf (0) = − 32 (−2α)2 < 0. Hence by Theorem 1.6, x∗1 is asymptotically stable. Thus x∗1 is asymptotically stable for −2 ≤ α < 0. x∗2 : f  (1) = 1 − α. Thus x∗2 is asymptotically stable if |1 − α| < 1 or 0 < α < 2. (i) If α = 0, f  (1) = 1. Since f  (1) = 0, by Theorem 1.5, x∗2 is unstable. (ii) If α = 2, f  (1) = −1. Now Sf (1) = − 32 (−2α)2 < 0 and by Theorem 1.5, x∗2 is asymptotically stable. Therefore x∗2 is asymptotically stable if 0 < α ≤ 2. x∗2 = 1 is asymptotically stable if 0 < α ≤ 2 14. Suppose that |f  (x∗ )|> 1. Let |f  (x∗)| > M > 1. Assume that for each n there exists xn ∈ x∗ − n1 , x∗ + n1 such that |f  (xn )| < M . Since lim xn = x∗ , lim |f  (xn )| = |f  (x∗ )|. This implies that |f  (x∗ )| ≤ n→∞

n→∞

10

Discrete Chaos M , which is a contradiction. Hence there exists η > 0 such that x ∈ (x∗ − η, x∗ + η) = I implies |f  (x)| ≥ M > 1. Let ε = η. Now for x0 ∈ I, |f (x0 ) − x∗ | = |f  (0)||x0 − x∗ | where c is between x0 and x∗ and hence is in I. Thus |f (x0 ) − x∗ | ≥ M |x0 − x∗ |. If f n−1 (x0 ) ∈ I, |f n (x0 ) − x∗ | ≥ M n |x0 − x∗ |. Thus there exists a positive integer N such that f N (x0 ) ∈ I, and hence x∗ is unstable.

15. Part 2. Since f  (x∗ ) = 1, f  (x∗ ) = 0 and f  (x∗ ) > 0, it follows that x∗ is an inflection point of f . Furthermore, for a sufficiently small open interval I around x∗ , f  (x) < 0 to the left of x∗ and f  (x) > 0 to the right of x∗ . Thus on I, f  (x) ≥ M > 1. This implies that for x0 ∈ I |f (x0 ) − f (x∗ )| = |f (x0 ) − x∗ | ≥ M |x0 − x∗ |. By induction, we conclude that |f n (x0 ) − x∗ | ≥ M n |x0 − x∗ |. This implies that x∗ is unstable. 16. Let g = f 2 and assume that x∗ is asymptotically stable. (a) Since x∗ is stable with respect to g, given ε > 0, there exists δ > 0 such that |x0 − x∗ | < δ implies |g n (x0 ) − x∗ | < ε for all n ∈ Z+ . This is equivalent to saying that |f 2n (x0 ) − x∗ | < ε for all n ∈ Z+ . Suppose that for this x0 , there exists a least N such that |f 2N +1 (x0 ) − x∗ | ≥ ε. But then we have |f 2N (x0 ) − x∗ | < ε and |f 2N +2 (x0 ) − x∗ | < ε. Now there are two cases to consider. Case (i) f 2N +1 (x0 ) > x∗ + ε > f 2N (x0 ). Then f 2N +2 (x0 ) < f 2N +1 (x0 ). Hence the function g(x) = f (x) − x changes sign between f 2N (x0 ) and f 2N +1 (x0 ) and thus has a root there. This implies that f has a fixed point different from x∗ . Case (i) The second case where f 2N +1 (x0 ) < x∗ −ε < f 2N (x0 ) is analogous. Assuming that there are no other periodic or fixed points in the vicinity of x∗ we have a contradiction. To prove attractivity, it suffices to notice that if lim f 2n (x0 ) = x∗ , then by the continuity n→∞

of f it follows that lim f (f 2n (x0 )) = lim f 2n+1 (x0 ) = f (x∗ ) = n→∞ n→∞ x∗ . 18. The fixed points are given by x3 + x3 + x = x x2 (x + 1) = 0 Hence x∗ = 0 is a fixed point. Notice that f  (0) = 1, f  (0) = 2 > 0. Hence x∗ is semiasymptotically stable from the left.

The Stability of One-Dimensional Maps

11

19. x3 − x2 + x = x x2 (x − 1) = 0 ⇒ x∗1 = 0, f  (x) = 3x2 − 2x + 1,

x∗2 = 1

f  (x) = 6x − 2

x∗1 : f  (0) = 1, f  (0) = −2 < 0 ⇒ x∗1 = 0 is semiasymptotically stable from the right. x∗2 : f  (1) = 2. x∗2 = 1 is unstable.

Exercises - (1.8) 1. Q(x) = x2 − .85 Fixed Points: x2 − x − .85 = 0 √ 1 + 1 + 3.4 x∗1 = ≈ 1.5488, 2

x∗2 =

1−

√

4.4

2

≈ −.5488,

2-period points: Q2 (x) = x or (x2 − .85)2 − .85 = x x4 − 1.7x2 − x − .1275 = 0 (x2 − x − .85)(x2 + x + .15) = 0 Solve x2 + x + .15 = 0 √ −1 + .4 ≈ −.1838 x1 = 2√ −1 − .4 ≈ −.8162 x2 = 2 The 2-cycle is {x1 , x2 } |Q (x1 )Q (x2 )| = |(−.1838)(−.8162)| = .15 < 0. The 2-cycle is stable. 3. f (x) = xe2(1−x) (a) Fixed Points: xe2(1−x) = x. Hence there are two fixed points x∗1 = 0 and x∗2 = 1. (b) 2-cycles: f 2 (x) = x 2(1−x)

xe2(1−x) e2(1−xe

)

=x

12

Discrete Chaos Discarding the fixed point x∗1 = 0 we can divide by x to get 2(1−x)

e2(1−x) · e2(1−xe

) = 1.

 2(1 − x) + 2 1 − xe2(1−x) = 0

Hence or

2 − x − xe2(1−x) = 0.

Let h(x) = 2 − x − xe2(1−x) . Then h(1) = 0. h(x)

x 1

Moreover, h (x) = −1 − e2(1−x) + 2xe2(1−x) ,

h (1) = 0

and h (x) = 4e2(1−x) − 4xe2(1−x)  >0 h (x) = 4e2(1−x)[1 − x] 1.

Thus for x < 1, h (x) < 0 and hence h(x) > 0; for x > 1, h (x) > 0 and hence h(x) < 0. Hence 1 is the only zero of h. We conclude that there are no 2-cycles. 6 6 4. g(x) = 5 − , g  (x) = 2 x x There are two fixed points x∗1 = 2, x∗2 = 3. g 2 (x) = 5 −

6 5−

6 x

=5−

19x − 30 6x = 5x − 6 5x − 6

g 2 (x) = x implies 19x − 30 = 5x2 − 6x 5x2 − 25x + 30 = 0 or x2 − 5x + 6 = 0 x ¯1 = 2 = x∗1 ,

x ¯2 = 3 = x∗2

Hence there are no cycles of period 2. Since g  (2) = 32 and g  (3) = 23 , x∗1 = 2 is unstable while x∗2 = 3 is asymptotically stable.

The Stability of One-Dimensional Maps

13

5. Every point is a 2-periodic cycle under h, since h2 (x) = x. Moreover, every point is stable but not asymptotically stable. 6. f (x) = |x − 1|, f 2 (x) = ||x − 1| − 1|. To find the 2-periodic cycles let ||x − 1| − 1| = x.

y=x

|x−1|

1

Then either (i) |x − 1| − 1 = x or (ii) |x − 1| − 1 = −x. For (i) if x > 1, then there is no solution. On the other hand, if x ≤ 1, we have x ¯1 = 0. The 2-periodic cycle is given by {0, 1}. For (ii) if x ≥ 1, we have x ¯1 = 1 which leads to the same cycle. On the other hand if x < 1 we have no solution. Since f  (1) is undefined, we cannot use Theorem 1.6. Now  |x − 2| if x > 1, 2 f (x) = |x| if x < 1.

1 1 Thus  1 2 every point in the interval [0, 1] is in fact a 2-periodic point: 2 , 2 , 3 , 3 , . . ., {0, 1} and more generally {t, 1 − t}. Moreover, every other point is eventually periodic point. Thus every periodic point is stable but not asymptotically stable. 7. Q(x) = ax2 +bx+c,

a = 0,

Q (x) = 2ax+b,

[Q2 (x)] = Q (Q(x))Q (x)

(a) [Q2 (x0 )] = Q (x0 )Q (x1 ) = −1. We need to find SQ2 . Observe that h(x) = Q2 (x) = a[ax2 + bx + c]2 + b[ax2 + bx + c] + c.

14

Discrete Chaos Then h (x) = 2a(ax + b)[ax2 + bx + c] + b[ax + b] h (x) = 2a[ax + b]2 + 4a2 (ax2 + bx + c) + 2ab h (x0 ) = 2a[Q (x0 )]2 + 4a2 x1 + 2ab = 2a[(Q (x0 ))2 + Q (x1 )] h (x) = 8a2 [2ax + b] + 4a2 (2ax + b) h (x0 ) = 12a2 Q (x0 ). Hence 3 SQ2 (x0 ) = −12a2Q (x0 ) − [2a(Q (x0 ))2 + Q (x1 )]2 2  2 1 = −12a2Q (x0 ) − 6a2 (Q (x0 ))2 −  Q (x0 ) = −12a2Q (x0 ) − 6a2 (Q (x0 ))4 + 12a2 Q (x0 ) −

6a2 (Q (x0 ))2

< 0.

Hence by Theorem 1.14, the cycle {x0 , x1 } is asymptotically stable. (b) Here we need to apply Theorem 1.12 to Q2 .   1 h (x0 ) = 2a (Q (x0 ))2 +  . Q (x0 ) If h (x0 ) = 0, then (Q (x0 ))3 = −1 or Q (x0 ) = −1. Hence Q (x1 ) = −1. This implies that h (x0 ) = −12a2 < 0 and the 2-cycle {x0 , x1 } is then asymptotically stable by Theorem 1.12. On the other hand if Q (x0 ) = −1, then h (x0 ) = 0 and the 2-cycle {x0 , x1 } is unstable by Theorem 1.12. 8. g  (0) = −b, g  (1) = 3a − b |g  (0)g  (1)| = |b2 − 3ab| < 1 or −1 < b2 − 3ab < 1

(1.2)

Since g(1) = 0, a − b + 1 = 0 or a = b − 1. Substituting in (1.2) yields −1 < b2 − 3b(b − 1) < 1 −1 < −2b2 + 3b < 1

The Stability of One-Dimensional Maps 15   √ √   1  3 − 17 3 + 17 S1 = −∞, , ≈ (−.28, 1.78), (1, ∞), S2 = 2 4 4   1  b ∈ S1 ∩ S2 = −.28, (1, 1.78). 2  2x if 0 ≤ x ≤ 12 9. (a) B(x) = 2x − 1 if 12 < x ≤ 1 B(x)

x 1/2

1

Fixed Points:2x = x ⇒ x∗1 = 0 2x − 1 = x ⇒ x∗2 = 1 ⎧ 4x ⎪ ⎪ ⎪ ⎨4x − 1 B 2 (x) = ⎪4x − 2 ⎪ ⎪ ⎩ 4x − 3

if 0 ≤ x ≤ 14 if 14 < x ≤ 12 if 12 < x ≤ 34 if 34 < x ≤ 1 B 2 (x)

x 1/4

1/2

3/4

1

2-periodic points:x∗1 = 0, x∗2 = 1 1 2 x1 = , x = 3 3

is a 2-cycle

|B  (x1 )B  (x2 )| = |2 · 2| = 4 > 1 ⇒ the 2-cycle is unstable.

16

Discrete Chaos (b) There are 21 fixed points, 22 points of period 2 including the fixed points. There are 23 points of period 3 with two 3-cycles and two fixed points. Hence there are (2k ) k-periodic points.

10. Let x ¯=

i m,

where m = 2r + 1, 1 ≤ i ≤ m − 1.

(i) Suppose first that i ≤ r. Then mi < 12 . The Baker map can be written as B(x) = 2x mod 1 identifying 0 and 1. There exists least n0 such that   i 1 2n0 i B n0 > , = 2r + 1 2r + 1 2   i 1 2n0 −1 i B n0 −1 < , = 2r + 1 2r + 1 2   n0 2n0 +1 i − 2r − 1 i 2 i B n0 +1 −1= . = 2r + 1 2r + 1 2r + 1  i 1 2 2r are 2r+1 Hence the possible values of B n 2r+n , 2r+1 , · · · , 2r+1 .

n0 The value 2r+1 2r+1 is not possible since in this case 2 i would be either odd or not an integer. Hence the orbit of x ¯ has at most 2r elements and consequently, it is eventually periodic. Hence there exists a least nonnegative integer m and a least nonnegative integer x) = B s (¯ x), m < s. If m = 0, we are done and x ¯ is s such that B m (¯ m−1 s−1 s-periodic. But suppose that m > 0. Then B (¯ x ) and B (¯ x)     cannot be in the same interval  0, 12  or 12 , 1 . Assume without loss   of generality that B m−1 (¯ x) ∈ 0, 12 and B s−1 (¯ x) ∈ 12 , 1 . Then

B m (¯ x) = 2(B m−1 (¯ x)) = even B s (¯ x) = 2[B s−1 (¯ x)] − 1 = odd which is a contradiction. Another solution. k Let x ¯ = 2r+1 , r ≥ 1. Then in the binary system x ¯ cannot be represented by a finite string of 0’s and 1’s. For example 1 1 = 4 1 = 0.010101 . . . 3 1− 4   1 2 B = = 0.10101 . . . 3 3   2 1 B = 2x − 1 = 1.010101 − 1 = . 3 3

There exists a least positive integer s such that 2s − 1 = 2r + 1 or 2s − 1 = t(2r + 1), t is an odd integer < 2s . If 2s − 1 = 2r + 1, then 1 = 0.00 . . . 100 . . . 100 . . . 1 . . . 2r + 1

The Stability of One-Dimensional Maps

17

where 1 appears in the sth slot. This gives an s-periodic point. If 1 on the other hand 2s − 1 = t(2r + 1). Then 2st−1 = 2r+1 and we arrive at a similar conclusion. 11. x(n) = c0 + (−1)n c1 is a 2-periodic solution. So c0 + (−1)n+1 c1 = μ(c0 + (−1)n c1 )[1 − c0 − c1 (−1)n ] c0 − (−1)n c1 = μ[c0 (1 − c0 ) − c21 ] + μc1 (−1)n (1 − 2c0 ) which is equivalent to c0 = μ[c0 (1 − c0 ) − c21 ] c1 = −μc1 (1 − 2c0 ) whose solution is 1 c0 = 2

  1 1+ , μ

c21 =

(μ + 1)(μ − 3) . 4μ2

This solution is real if and only if μ ≥ 3. In this case we have

1 μ + 1 + (−1)n (μ + 1)(μ − 3) , if μ > 3; x(n) = 2μ   1 1 2 x(n) = 1+ = , if μ = 1 which is a fixed point. 2 μ 3 13. x(n + 1) =

αx(n) 1 + βx(n)

(1.3)

Let x(n) = c0 + (−1)n c1 . Then substitution into equation (1.3), c0 − (−1)n c1 =

α(c0 + (−1)n c1 ) 1 + β(c0 + (−1)n c1 )

c0 (1 + βc0 ) + (−1)n βc0 c1 − (−1)n c1 (1 + βc0 ) − βc21 = αc0 + (−1)n αc1 . (1.4) Comparing both sides of equation (1.4) we get c0 (1 + βc0 ) − βc21 − αc0 = 0 βc0 c1 − c1 (1 + βc0 ) = αc1 (1 + α)c1 = 0.

(1.5) (1.6)

If α = −1, then c1 = 0 which leads to the fixed point x(n) = c0 . In this case the equation has no nontrivial 2-periodic solution. If α = −1, then we have two cases: (i) If β = 0, then c0 = 0 and x(n) = (−1)n c1 , c1 arbitrary, is a 2-cycle.

18

Discrete Chaos (ii) If β = 0, then c21 =

c0 (2+βc0 ) . β

For any c0 for which c1 is real, the solution is  c0 (2 + βc0 ) . β

x(n) = c0 + (−1)n

(1.7)

If β < 0, there is a family of 2-periodic solutions: for any c0 < 0 or any c0 > − β2 . However, for β > 0 there is a family of 2-periodic solutions: for any c0 > 0 or any c0 < − β2 . The solution is given by equation (1.7). 14. x(n + 1) = μx(n)e−x(n) , x(n) ≥ 0, μ > 0. Solution 1 A periodic point of period 2 is of the form x(n) = c0 + (−1)n c1 . And a 2-cycle is given by {c0 − c1 , c0 + c1 }. Hence by substituting into the equation we obtain μ(c0 − c1 )ec1 −c0 = c0 + c1 μ(c0 + c1 )e−c1 −c0 = c0 − c1 μ2 (c0 − c1 )e−2c0 = c20 − c21 . If c20 = c21 , e−2c0 =

1 μ2

⇒ ec0 = μ c0 = ln μ

since both c0 − c1 and c0 + c1 are positive, c0 > 0 and c0 > c1 . Hence μ > 1. Solution 2

x(x + 2) = x(n) 2

μ xe

−x −μxe−x

e

−x

ex+μxe

=x = μ2

x + μxe−x = ln(μ2 ) since every term on the left side is positive, μ2 > 1 or μ > 1.

The Stability of One-Dimensional Maps

19

Exercises - (1.9) 1. Fμ (x) = μx(1 − x), Fμ (x) = μ − 2μx, Fμ (x) = −2μ, Fμ (x) = 0 [Fμ2 (x)] = Fμ (Fμ (x))Fμ (x) [Fμ2 (x)] = Fμ (Fμ (x))Fμ (x) + Fμ (Fμ (x))[Fμ (x)]2 √ √ [Fμ2 (¯ x1 )] = −2(1 + 6)Fμ (¯ x2 ) − 2(1 + 6)[Fμ (¯ x1 )]2 [Fμ2 (x)] = Fμ (Fμ (x))Fμ (x) + Fμ (Fμ (x))Fμ (x)Fμ (x) + 2Fμ (Fμ (x))Fμ (x)Fμ (x) + Fμ (Fμ (x))[Fμ (x)]3 [Fμ2 (¯ x1 )] = 0 + 4μ2 Fμ (¯ x1 ) + 8μ2 Fμ (¯ x1 ) + 0 √ 2  x1 ) = 12(1 + 6) Fμ (¯ √ √ 3 x1 ) = −12(1 + 6)2 Fμ (¯ x1 ) − 4(1 + 6)2 [Fμ (¯ x1 )]4 SFμ2 (¯ 2 √ √  − 8(1 + 6)2 Fμ (¯ x1 ) + 4(1 + 6)2 /[Fμ (¯ x1 )]2 √   = −6(1 + 6)2 [Fμ (¯ x1 )]4 + 1/[Fμ (¯ x1 )]2 < 0. ¯2 } is asymptotically stable. Hence {¯ x1 , x 2. Fμ (x) = μx(1 − x) (a) Fixed point: There are 21 fixed points (b) 2-cycles: Fμ (x) = x gives μ(μx(1 − x))[1 − μx(1 − x)] = x a polynomial of degree 22 . There are 22 roots but 21 of them are fixed points. Hence there are 22 −21 = 21 periodic points of prime period 2. Hence we have only one cycle of prime period 2. (c) 3-cycles: Fμ3 (x) = x μ(μx(1 − x))[1 − μx(1 − x)][1 − μx(1 − x)[1 − μx(1 − x)]] = x which is of degree 23 . Hence we have 23 − 21 = 2 · 3 = 6 points of prime period 3 or 2 cycles of prime period 3. 4-cycles: There are 24 − 22 = 22 · 3 = 4 · 3 periodic points of prime period 4 and thus we have 3 cycles of prime period 4. 5-cycles: There are 25 − 22 = 21 periodic points of prime period 5 (25 − 21 = 6 · 5) and consequenty we have 6 cycles of prime period 5. 6-cycles: There are 26 − 23 − 2 = 9 × 6 periodic points of prime period 6 and thus we have 9 cycles of primer period 6. 3. Use Maple, Mathematica, or Phaser.

20

Discrete Chaos √

√

(a) x∗1 = 1+ 21−4μ , x∗2 = 1− 21−4μ , x∗1 is unstable, x∗2 is asymptotically stable if − 43 < μ < 0.   √ √ (b) x1 = −1+ 2−3−4μ , x2 = −1− 2−3−4μ is a 2-cycle which is asymptotically stable if − 54 < μ < − 34 .

(c) μ1 = −0.75, μ2 = −1.25, μ3 = −1.35708, μ4 = −1.38 (d)

5. Gμ (x) = μ sin πx, 0 < μ ≤ 1, −1 ≤ x ≤ 1 Fixed points: Gμ (x) = x = μ sin πx (i) x∗1 = 0 is a fixed point. Gμ (x) = μπ cos πx |Gμ (0)| = |μπ| < 1 ⇒ μ < π1 x∗1 = 0 is asymptotically stable if 0 < μ < π1 . If μ = π1 , then 3 Gμ (0) = 1. Now Gμ (0) = 0, G μ (0) = −μπ < 0. By Theorem 1.5, 1 ∗ x1 = 0 is asymptotically stable for μ = π . For μ > π1 , x∗1 = 0 is unstable. (ii) The second fixed point x∗2 , x∗2 = μ sin πx∗ appears when μ > Gμ (x∗2 ) = μπ cos πx∗2 . 8. k = 2m, k = 4, m = 2 4-periodic points are given by x(n) = c0 + (−1)n c2 + c1 cos

 nπ 2

+ d1 sin

n = 0, x(0) = c0 + c2 + c1 n = 1, x(1) = c0 − c2 + d1 n = 2, x(2) = c0 + c2 − c1 n = 3, x(3) = c0 − c2 − d1

 nπ 2

1 π.

The Stability of One-Dimensional Maps

21

since x(4) = x(0), we get by substitution in the equation x(4) = x(0) = μx(3)[1 − x(3)] c0 + c2 + c1 = μ(c0 − c2 − d1 )[1 − c0 + c2 + d1 ] x(5) = x(1) = μx(4)[1 − x(4)] = μx0 (1 − x0 )

(1.8)

c0 − c2 + d1 = μ(c0 + c2 + c1 )[1 − c0 − c2 − c1 ] c0 + c2 − c1 = μ(c0 − c2 + d1 )[1 − c0 + c2 − d1 ]

(1.9) (1.10)

c0 − c2 − d1 = μ(c0 + c2 − c1 )[1 − c0 − c2 + c1 ].

(1.11)

We have 4 equations in 4 unknowns; using Maple we obtain c0 , c1 , c2 , d1 .

2 Attraction and Bifurcation

Exercises - (2.2 and 2.3) 1.

 f (x) =

x2 for −3 ≤ x ≤ 5, √ 4 x − 3 for 1 < x ≤ 9.

There are three fixed points x∗1 = 0, x∗2 = 1, x∗3 = 9. Now • f  (0) = 0 ⇒ x∗1 = 0 is asymptotically stable • f  (1) = 2 ⇒ x∗2 = 1 is unstable • f  (9) =

2 3

⇒ x∗3 = 9 is asymptotically stable

W s (0) = (−1, 1), W s (0) = [−3, −1) ∪ (1, 9]

3. Let y ∈ W s (f (¯ x)), and let g = f k . Then lim g n (f (¯ x)) = y. Now n→∞

g n (f (¯ x)) = f nk+1 (¯ x) = f (f nk (¯ x)). Suppost that f nk (¯ x) → z ∈ W s (¯ x), nk s then f (f (¯ x)) → f (z) = y. Thus y ∈ f (W (¯ x)). Conversely, let u ∈ f (W s (¯ x)). Then there exists v ∈ W s (¯ x) with f nk (¯ x) → v as nk x)) = f (f kn (¯ x)) → f (v) = u as n → ∞ and u = f (v). But f (f (¯ n → ∞. Thus u ∈ W s (f (¯ x)). This implies that W s (f (¯ x)) = f (W s (¯ x)).

23

24

Discrete Chaos 5.

⎧ x ⎪ ⎨2 f (x) = 3x − 12 ⎪ ⎩ 2 − 2x

0 ≤ x ≤ 0.2, 0.2 < x ≤ 12 1 2 < x ≤ 1.

There are three fixed: x∗1 = 0, x∗2 = 14 , x∗3 = 23 . ⎧ 1 ⎪ 0 ≤ x ≤ 0.2, ⎨2  f (x) = 3 0.2 < x ≤ 12 ⎪ ⎩ −2 12 < x ≤ 1. Hence the only asymptotically stable fixed point is x∗1 = 0 and the other   ∞ two fixed points are unstable. W s (0) = 0, 14 ∪ G, where G = ∪ Gi i=1

defined as follows. G1 (f −1 (0.25), 1] = (0.875, 1], G2 = (t1 , t2 ), t1 , t2 ∈ f −1 (0.875). We find t1 , t2 by solving 3x − 12 = 0.875, 2 − 2x = 0.875. We obtain t1 ≈ 0.4583, t2 ≈ 0.5625. G3 = (f −1 (t1 ), f −1 (t2 )), etc. 7. Let y ∈ Int(M ). Since f is continuous and one-to-one, f is either strictly increasing or strictly decreasing. Assume, without loss of generality, that f is increasing. Then M must contain a fixed point either as an interior point or a right end point. In either case, an interior point is mapped to an interior point. 10.

(i) For 0 < μ < 1, there is only one fixed point x∗1 = 0. μK(K + (μ − 1)p) − (μ − 1)μKp [K + (μ − 1)p]2 μK 2 = [K + (μ − 1)p]2

f  (p) =

f  (0) = μ. Hence x∗1 = 0 is asymptotically stable for 0 < μ < 1. For μ = 1, we have f  (0) = 1. Now f  (0) = 0 implies by Theorem 1.5 that for μ = 1, x∗1 = 0 is unstable. Moreover, for 0 < μ < 1, W s (0) = [0, ∞). (ii) For μ > 1, x∗1 = 0 is unstable and there is a new positive fixed point x∗2 = K. 1 f  (K) = < 1 for μ > 1. μ Hence x∗2 = K is asymptotically stable. Furthermore, W s (K) = (0, ∞). 11. g  (1) = 12 implies that x∗ = 1 is asymptotically stable. However, x∗ = 1 is not globally asymptotically stable since 12 → 32 → 0.

Attraction and Bifurcation

25

Exercises - (2.4) 1. f (x) = x + x3 has positive Schwarzian derivative for − √16 ≤ x ≤

√1 . 6

3. Consider the function f (x) = 2 cot−1 x. There are two asymptotically stable fixed points x∗1 ≈ −1.8, x∗2 ≈ 1.8 where 2 cot−1 x∗1 = x∗1 , μ cot−1 x∗2 = x∗2 .

2 cot −1 x

x

−2 1 + x2 2 |Lμ (x)| = < 1 ⇒ 1 < x2 ⇒ either x > 1 or x < −1. 1 + x2 Lμ (x) =

So we have two attracting fixed points with basins of attraction (−∞, −1) and (1, ∞), respectively. Another example: f (x) = 2 sin x on [−π, π].

26

Discrete Chaos 5. x

f (x) = xer(1− k ) x x r f  (x) = er(1− k ) − xer(1− k ) k rx

r (1− x ) k 1− =e k x rx r r(1− xk ) r r(1− k )  1− − e f (x) = − e k k k x rx r = − er(1− k ) 2 − k k r2 r(1− xk )  rx r2 r(1− xk )  + 2e f (x) = 2 e 2− k k k r2 r(1− xk ) rx

= 2e 3− k k  2 2 (3 − rx ) 3 r2 2 − rx r k k − Sf (x) = 2 k (1 − rx 2 k 2 1 − rx k ) k   rx 2 )(1 − ) − (2 − rx r2 (3 − rx 1 r2 2 − k k k ) = 2 − 2 k (1 − rx 2 k2 1 − k )   2 1 r2 1 r2 2 − rx k = 2 − 0. This implies that f˜6 (p) = p, a contradiction. Thus p, f˜(p), . . . , f˜5 (p) ∈ (5, 9). By simple computations, one may show ˜ that the only fixed point of f˜, f˜2 , . . . , f˜6 is p = 19 3 . Hence f has no points of prime period 6. 9. A map with 2 × 7 cycle but no 2 × 5 cycles. First we start with the map f in problem 5. Then define the double map f˜ : [1, 19] → [1, 19] as follows.  f (x) + 12, 1 ≤ x ≤ 7 ˜ f (x) = x − 12, 13 ≤ x ≤ 19

10. The general procedures for constructing a continuous map of prime period 2(2n + 1) but no points of prime period 2(2n − 1) may be explained as follows. We start with a map f : [1, 1 + h] → [1, 1 + h] with points of prime period (2n + 1) but no points of prime period (2n − 1). We define the double map f˜ : [1, 1 + 3h] → [1, 1 + 3h] as follows:  f (x) + 2h for 1 ≤ x ≤ 1 + h, f˜(x) = x − 2h for 1 + 2h ≤ x ≤ 1 + 3h and by linearity for 1 + h < x < 1 + 2h. 12. To construct a map with a prime period 8 but no points of prime period 16, we start with the map f : [1, 4] → [1, 4] defined as follows: f (1) = 3,

f (2) = 4,

f (3) = 2,

f (4) = 1,

and on each interval [n, n + 1] we assume f to be linear.

Attraction and Bifurcation

31 4 3 2 1 1

2

Now we use the double map  f (x) + 6 f˜(x) = x−6

3

4

for 1 ≤ x ≤ 4, for 7 ≤ x ≤ 10

f˜ : [1, 10] → [1, 10] Then f˜ has points of period 8 but no points of period 16. 10 9 8 7 6 5 4 3 2 1 1

2

3

4

5

6

7

8

9 10

16. Let f : I → I defined f (1) = 1, f (x) = fk (x)  as follows:  if x ∈ 2 1 1 + , k ∈ Z → 0, Ik = 1 − 31k , 1 − 3k+1 , where f (x) : 0, 0 3 3 defined  2 7 2 7 13 → defined by f = , , (x) = −x by f0 (x) = x, f1 (x) 1 3 9  8 25 3 9  8+739; 8 25 49 f2 (x) = 9 , 27 → 9 , 27 defined  by f2 (x) = −x + 27 for x ∈ 9 , 81 , 2 74 25 f2 (x) = x − 81 for x ∈ 81 , 27 and then connect linearly. In general,   k+1 1 we have fk (x) = −x + 2(33k+1)−5 for x ∈ 1 − 31k , 1 − 31k + 3k+2 and 2 fk (x) = x − 3k+2 and then connect linearly. 18. Let f : [a, b] → [a, b] be continuous such that there exists x0 ∈ [a, b] with either f 2 (x0 ) < x0 < f (x0 ) or f (x0 ) < x0 < f 2 (x0 ). Assume that f 2 (x0 ) < x0 < f (x0 ). Let w = min{x|x0 ≤ x ≤ f (x0 ) such that f (x) = x}. Now we have two cases to consider. (a) If f has no fixed points in [a, x0 ], then f 2 (a) ≥ a and f 2 (x0 ) < x0 implies that we have a periodic point with prime period 2 in [a, x0 ].

32

Discrete Chaos (b) If f has a fixed point in [a, x0 ], let t = max{x|a ≤ x ≤ x0 such that f (x) = x}. Then f has no fixed points in (t, x0 ]. Let u ∈ [t, x0 ] such that f (u) = x0 (this is true since f [t, x0 ] ⊃ [t, x0 ]). Since f 2 (u) > u and f 2 (x0 ) < x0 , we have a point of prime period 2 in [t, x0 ].

3 Chaos in One Dimension

Exercises - (3.2 and 3.3) 1. Let us wrtie each x ∈ [0, 1] in its binary expansion, x = 0.x1 x2 x3 . . . ,where each xi is either 0 or 1. There are two fixed points 0 = 0.0 and 1 = 0.1. The Baker map acts as a shift map, B(0.x1 x2 x3 x4 . . . ) = 0.x2 x3 x4 . . . . (a) A periodic point of period n is of the form x = 0.x1 x2 . . . xn =

Thus x =

r 2n −1

x2 x1 xn x1 x2 + 2 + · · · + n + n+1 + n+2 + . . . 2 2 2 2 2 xn + .... 22n

for 0 ≤ r ≤ 2n − 2.

(b) Let y ∈ (0, 1) be written in its binary form y = 0.y1 y2 y3 . . . . For a given ε > 0, 21k < ε for some k ∈ Z+ . Consider the periodic point x of period k, x = 0.x1 x2 . . . xk , where x1 = y1 , x2 = y2 , . . . , xk = yk . Then ∞ ∞   1 xi − yi 1 = k < ε. < |x − y| = i 2 2i 2 i=k+1

i=k+1

Hence PerB is dense [0,1]. 2. (a) Let a = rs be a periodic point of period n. Then B n (a) = a = 2n (a) mod 1. Hence a = 2n a − r. Consequently a = 2nr−1 , r = 0, 1, 2, . . . , 2n − 2. (b) Let J = (a, b) ⊂ [0, 1] with t = b − a. Choosen sufficiently large  n such that 2n −1 > 2t . Consider now the set A = 2n1−1 , 2n2−1 , 22n −2 −1 which consists of periodic points by Problem 8 for any two consecr+1 r 1 t utive numbers 2nr−1 , 2r+1 n −1 , 2n −1 − 2n −1 = 2n −1 < 2 . Hence the set J must contain one of the numbers in the set A. Hence the set of periodic points is dense in [0, 1]. 3. Write x ∈ [0, 1] in its ternary expansion x = 0.x1 x2 x3 x4 . . . , where xi is 0, 1, or 2. Then f (x) = 0.x2 x3 x4 . . . . Consider the point y ∈ (0, 1) of

33

34

Discrete Chaos the form y =

∞  j=1

yj 3j ,

where yj ’s appear as follows:

012 00 01 02 10 11 12 20 21 22 000 001 002 010 020 . . . . 1 − block 2 − block 3 − block We claim that the orbit O(y) is dence in [0, 1]. Let x = 1 2N

∞  i=1

xi 3i

be

an arbitrary point [0, 1], and let ε > 0 be given. Then < ε for some N ∈ Z+ . Now the string x1 , x2 , . . . , xN must appear as one of the N -blocks in the ternary expansion of y. Thus for some k, f k (y) = ∞  1 1 0 · x1 x2 . . . xN zN +1 zN +2 . . . . Now |f k (y) − x| ≤ 2j = 2n < ε. j=N +1

This proves the claim. Hence f is transitive on [0, 1]. 5. Let U = {θ : θ1 < θ < θ2 } and V = {θ : θ3 < θ < θ4 } be two open arcs in S 1 . Now f (U ) is an arc of length 2|θ2 − θ1 |, f 2 (U ) is an arc of length 4|θ2 − θ1 |, etc. There exists k ∈ Z+ such that f k (U ) ⊃ S 1 and consequently, f k (U ) ∩ V = ∅. Thus g is transitive. 7. Let us write θ as eiθ . Then h(eiθ ) = e3iθ . If x0 = eiθ is k-periodic, k then hk (eiθ ) = ei3 θ = eiθ . Thus 3k θ = θ + 2nπ, n = 0, 1, . . . , 3k − 2. 2(n+1)π 2nπ are two Hence θ = 32nπ k −1 . Now if θ(n) = 3k −1 and θ(n + 1) = 3k −1 consecutive angles in the periodic orbit of x0 , then θ(n+1)−θ(n) = 3k2π −1 . Let U = {θ | θ1 < θ < θ2 } be open arc in S 1 . Let d = (θ2 − θ1 )/2. Choose k sufficiently large such that 2π/(3k − 1) < d. Hence there exists n and k such that θ(n) = 2nπ/(3k − 1) ∈ U . Hence the set of periodic points is dense in S 1 .

Exercises - (3.4) 1. (a) n = 5, f 5 (0.1+ 0.01) = f 5 (0.11) = 0.97, f 5 (0.1) = 0.59 ≈ 0.73 λ(0.1) ≈ 15 ln 0.97−0.59 0.01 (b) n = 6, f 6 (0.11) = 0.12, f 6 (0.1) = 0.97 = 0.74 λ(0.1) ≈ 16 ln 0.12−0.97 0.01 (c) n = 7, f 7 (0.11) = 0.42,f 7 (0.1) = 0.11  0.42−0.11 1 = 0.49 λ(0.1) ≈ 7 ln 0.01 3. (a) n = 5, |f 5 (0.31) − f 5 (0.3)| = 0.0080057346, λ(x0 ) ≈ −0.044 (b) n = 6, λ(x0 ) = −0.044 (c) n = 7, λ(x0 ) ≈ −0.043

Chaos in One Dimension ⎧ ⎪ 0 ≤ x ≤ 13 ⎨3x, 5. f (x) = 2 − 3x, 13 < x ≤ 23 ⎪ ⎩ 3 − 3x, 23 < x ≤ 1  |f (x)| = 3

35

n−1 1  ln |f  (x(k))| n n→∞ k=0

= log 3 ≈ 1.0986

(a) λ(x0 ) = lim

(b) Let ε = 1. Let x0 ∈ [0, 1]. If |x0 − y| = δ > 0, then |f n (x0 ) − f n (y)| = |(f n (x0 )) − (f n (y)) ||x0 − y| = 3n δ. Choose n large enough such that 3n δ > 1. Hence f possesses sensitive dependence on initial conditions. 9. 1 < μ < 3, μ = 2. Show λ(x) = ln |2 − μ| for the logistic map n−1 1  ln |μ − 2μx(k)|. n→∞ n

λ(x0 ) = lim

K=0

 Since 1 < μ < 3, x(n) = Fμn (x0 ) → x∗ = μ−1 μ . Thus Fμ (x(n)) →  ∗  Fμ (x ) = 2 − μ and consequently, ln |Fμ (x(n))| → ln |2 − μ| as n → ∞. Let ε > 0. Then there exists N1 such that for n > N .

ln |2 − μ| − ε < ln |Fμ (x(n))| < ln |2 − μ| + ε. Hence for n > N n−N 1 (ln |2 − μ| − ε) = n n
0, there exists N such that |xn − x| < exists N2 > N1 such that n > N2 we have

x1 +x2 +···+xN1 n


N2 . Thus for

n

N

n 

i=1

i=1

i=N1 +1

<
N1 . There

xi − x

|xi − x|

i=N1 +1

ε ε + = ε. 2 2

Exercises - (3.5) 1. Let U = {z : |f i (z) − f i (p)| < ε, 0 ≤ i ≤ k − 1}. Then U = ∅ since it contains the point p. Now for each j, the set Uj = {z : |f j (z) − f j (p)| < k−1

ε} is an open interval of the form (aj , bj ) containing p. Since U = ∩ Uj , j=0

it follows that U is open. 4. Consider f (x) = −x, defined on R. Then every point is periodic of period 2. Clearly f does not possess sensitive dependence on initial conditions. 5. Let U = {θ ∈ X : θ1 < θ < θ2 }, V = {θ ∈ X : θ3 < θ < θ4 }. Then f (U ) is an arc, minus countably many points, with double the length of U , f 2 (U ) is an arc with four times the length of U , etc. Eventually, f k (V ) ⊃ X for some k. Hence f k (U ) ∩ V = ∅, and f is topologically transitive. Let us write a periodic point x0 as eiθ . If x0 is of period k, then 2nπ f k (θ) = e2ikθ = eiθ . Hence θ + 2nπ = 2kθ. This gives θ = 2k−1 ∈ X. Consequently, f has no periodic points in X.  7. f (x) =

3 2 x, 3 2 (1

− x),

 (a) f (x) =

0 ≤ x ≤ 12 1 3 2 < x≤ 4

3 4 3 4

− −

3 2 3 2

  on 0, 34

1  − x = 32 x 2   x − 12 = 32 (1 − x)

if 0 ≤ x ≤ 12 if 12 < x ≤ 1

Chaos in One Dimension

37

(b) |f  (x)| =

3 2

n−1 1 3 λ(x) = lim log n→∞ n 2 k=0

1 3 = lim · n log n→∞ n 2 3 = log > 0 2 Hence f possesses sensitive dependence on initial conditions.

.75 .5 .25

.25

.5

.75

  (c) Let x ∈ 0, 38 . Then there exists r > 0 such that f r (x) ≥ 12 . Claim that if z ≥ 12 , then f (z) ≥ 38 . Assume the contrary, that is, 3 3 3 3 3 implies that z > 34 , which is 2 (1 − z) < 8 . Then 2 −  2z <  8 3which  3 not possible since  f  0, 4 = 0, 4 . Hence f has no periodic points in the interval 0, 38 .

Exercises - (3.6 and 3.7) 1. (a)  3x h(x) = 3(1 − x) x∗1 = 0,

x∗2 =

if x ≤ if x > 3 = .75 4

1 2 1 2

38

Discrete Chaos (b) Algorithm: 0.75 3 × 0.75 = 2.25 3 × 0.25 = 0.75 3 × 0.75 = 2.25 3 × 0.25 = 0.75 In ternary expansion x∗1 = 0.0 x∗2 = 0.2020 Justification: x2 x3 x1 + + + ... 3 3 3 x3 x2 + + · · · ⇒ x1 = 2, etc. 2.25 = 3 × 0.75 = x1 + 3 3           3 h 3 9 1 2 h 1 2 h h , , → → 1, → − , 0 → − , 0 . . . (−∞, 0) 2. 9 9 3 3 2 2 2 .75 =

3. (a) Let x = ·x1 x2 x3 · · · ∈ E be in a ternary expansion. If x ∈ K, then for some i, xi = 1. There are two cases to consider. (i) x1 = 2. Then



x3 x2 xi h(x) = 3 − x1 + + 2 + · · · + i−1 + . . . 3 3 3 = ·¯2 − ·x2 x3 x4 . . . = ·y1 y2 y3 . . .

where xi = yi−1 = 1. (ii) x1 = 0. Then h(x) = ·x2 x3 . . . xi · · · = ·y1 y2 . . . yi−1 . . . with xi = yi−1 = 1. Hence in either case h(x) = ·xi xi+1 · · · = xi+1 1 i 3 + 32 + . . . which implies that h (x) = 1 · xi+1 xi+2 · · · > 1, a contradiction. Thus E ⊂ K. Conversely, let x = ·x1 x2 x3 · · · ∈ K in ternary expansion, xi ∈ {0, 2}. Then clearly hn (x) ∈ [0, 1] for all n ∈ Z+ , and consequently, K ⊂ E. This implies that E = K. (b) The map f (x) = 13 x is clearly a homeomorphism. ∞

4. Consider ∧ = ∩ An . n=1

(i) ∧ is totally disconnected. Given any open interval (a, b) with d = 1 b − a, then by Lemma there exists large n such that (1+ε) n < d. Thus (a, b) may not be contained in any subinterval of An .

Chaos in One Dimension

39

(ii) Perfect. Let p ∈ ∧. Then p ∈ An for each n. Hence p ∈ 1 [an−1 , bn−1 ] ⊂ An , with bn−1 − an−1 < (1+ε) It follows that n. 1 |an−1 − p| < (1+ε)n . Thus lim |an−1 − p| = 0 which implies that n→∞ lim an = p. n→∞

5.

˜ = ∩∞ S˜n is closed, being the intersection of closed sets. (i) Clearly K n=1

(ii) Totally Notice that the length of each subinterval n  disconnected.  in ˜ contains an interval of length d, then d > 2 n Sn is 25 . If K 5 for all large n, which is absurd. ˜ Then p ∈ S˜n for all n. Hence (iii) Perfect. Assume that p ∈ K.  n  n ˜ p ∈ [an , bn ] ⊂ Sn , with bn − an = 25 . Thus |an − p| < 25 . For  2 N ε > 0, let N be large enough such that 5 < ε. Then for n > N , |an − p| < ε. This implies that an → p as n → ∞.  7. Ω = {x ∈ 2 : x contains no consecutive zeros} (a) If x = {x0 x1 x2 x3 . . . } has no consecutive zeros, then σ(x){x0 x1 x2 x3 . . . } has no consecutive zeros either. Thus Ω is invariant. (b) Infinite fixed points 1

2-cycles 3-cycles 4-cycles . . . 3 4 7

This looks like the fibonacci sequence Fn+1 = Fn + Fn−1

F0 = 1,

F1 = 3.

Therefore  √ n √ n 1− 5 1+ 5 + c1 Fn = c1 2 2    √ n+1 √ n+1 1+ 5 1− 5 = + . 2 2 

(c) Yes. 1. Transitive: Modify z in the proof of Theorem 3.28 to include only blocks with no consecutive zeros. Then O(z) = Ω. 2. P = Ω. Let x = {x1 x1 x2 . . . } ∈ Ω. Define a sequence {yn } as y1 = {¯ x0 . . . } if x0 = 1 or {x0 x1 . . . } otherwise. In the first case let y2 = {x0 x1 . . . } and in the latter y2 = {x0 x1 x2 . . . } if x2 = 1 or {x0 x1 x2 x3 . . . } if x2 = 0, etc. Then lim yn = x. 8. h : K →

n→∞

 2

40

Discrete Chaos • For x = ·x1 x2 x3 . . . , h(x) = y = {y0 y1 . . . }, yi = x12+1 , h−1 (y) = x = ·x1 x2 x3 . . . with xi+1 = 2yi . Hence h is one-to-one and onto. • h is continuous: Let ε > 0 be given, x ∈ C, y = h(x). If ε < 21N let δ = 31N . Then if d(x, x˜) < δ = 31N , xi = x ˜i for i = 1, 2, . . . , N . Thus h(xi ) = h(˜ xi ), i = 0, 1, 2, . . . , N . Hence d(y, h(˜ xi )) ≤ 21N < ε. To show that h is open, let Bδ (x) be an open ball in C, with δ < 31N , and y = h(x). Suppose that y˜ ∈ h(Bδ (x)). Then a = d(˜ y , y) < 21N . 1 Let b = 2N − a. Put ε = min(a, b). Then claim that Bε (˜ y) ⊂ y ). Then h(Bδ (x)). To show this let z ∈ Bε (˜ d(z, y) ≤ d(z, y˜) + d(˜ y , y) 1 1 < N − a + a = n. 2 2 Hence z ∈ h(Bδ (x)). Thus Bε (˜ y ) ⊂ h((Bδ (x)) and the proof is now complete.

13. In Lemma 3.5, it was shown that Fμ (x) > 1 + ε for all x ∈ A1 . Now for x ∈ A2 , we have Fμ (x) ∈ A1 . It follows that  2  Fμ (x) = Fμ (Fμ (x)) Fμ (x) > (1 + ε)2 since both Fμ (x) and x are in A1 . Inductively, for x ∈ An  n  Fμ (x) > (1 + ε)n .

(3.1)

Let [an−1 , bn−1 ] = Is0 s1 ...sn−1 ⊂ An . Then either Fμn (an−1 ) = 0 or 1. Assume Fμn (an−1 ) = 0, then Fμn (bn−1 ) = 1. By the Mean Value Theorem, Fμn (bn−1 ) − Fμn (an−1 )  n  = Fμ (c) bn−1 − an−1 1 > (1 + ε)n . bn−1 − an−1 Hence bn−1 − an−1
0. Then there exists N such that 21N < ε. Since √ μ > 2 + 5 there exists γ > 0 such that the length of each subinterval  1 1 in AN is less than (1+γ) N . Put δ = (1+γ)N . Let a = a0 a1 a2 · · · ∈ 2, h−1 (a) = x. Then x ∈ Ia0 a1 ...an for all n ∈ Z+ . If d(x, x˜) < δ, then x, x ˜ ∈ Ia0 a1 ...aN . If h(x) = a ˜, then a1 = a ˜i , i = 0, 1, 2, . . . , N . Hence d(a, a ˜) < 21N < ε.

Similarly, one may show that h−1 is continuous. 7. Gλ (x) = 1 − λx2

(a) Let h(x) = to use h

−1

a = −λ,

b = 0,

c=1

μ μ2 − 2μ μ x− , c = = 1, λ = λ 2λ 4λ

μ2 −2μ . 4

It is convenient

,

1 λ x+ μ 2 μ2 + 2μ 1 μ λ 1 + =1− h(−1) = − + = − μ 2 4μ 2 4 2 μ μ − 2μ 1 + = h(1) = 4μ 2 4   Therefore h−1 : [−1, 1] → 1 − μ4 , μ4 h−1 (x) =

(b) Now λ = 2 corresponds to μ = 4. Since F4 ≈ Q2 and F4 is chaotic on [0, 1], Q2 is chaotic on [−1, 1]. 11. Q−2 (x) = x2 − 2, F4 (x) = 4x(1 − x). Using (3.17), we let h(x) = −4x + 2. Then h is a conjugation map which takes [0, 1] onto [−2, 2]. Moreover, h(F4 (x)) = h(4x(1 − x)) = −16x(1 − x) + 2 = 16x2 − 16x + 2. Q−2 (h(x)) = Q−2 (−4x + 2) = (4x + 2)2 − 2 = 16x2 − 16x + 2. This shows that F4 is conjugate to Q−2 . Since F4 is chaotic, it follows by Theorem 3.9, that Q−2 is chaotic.

4 Stability of Two-Dimensional Maps

Exercises - (4.1 and 4.2) 1. A =

  −4.5 5 −7.5 8

Let |A − λI| = 0 so

−4.5 − λ 5 = (−4.5 − λ)(8 − λ) + 37 = −7.5 8 − λ √ 3.5± 3.52 −4·1.5 2

= 3.5±2.5 3, 2 .Then   λ1 =  −7.5 5 v11 0 1 λ2 = 2 . For λ1 = 3, (A − λI)v1 = 0. So = . −7.5 5 v21 0   1 Then v11 = 23 v21 . Let v1 = 3 . For λ2 = 12 , (A − λI)v2 = 0. So 2        1 0 −5 5 v12 . Let P = = . Then v12 = v22 . Let v2 = 1 v22 0 −7.5 7.5      n  3  0 11 −2 2 , so An = P , then P −1 = P −1 = (v1 v2 ) = 3 n 3 −2 1 0 12 2     1 −2 · 3n + 23n 2 · 3n − 2n−1 1 . Thus eigenvector is associated with 1 3 −3n+1 + 23n 3n+1 − 2n−1 2     1 1 −2 · 3n + 23n 2 · 3n − 2n−1 1 n is associated with λ2 = 2 , A = λ1 = 3, 1 1 −3n+1 + 23n 3n+1 − 2n−1  8 1 3 3 3. A = − 34 34 8    −λ 1 = 83 − λ 43 − λ + 49 =. Then Let |A − λI| = 0, so 3 4 4 3 −3 3 − λ 2 = 0. So λ = (3λ − 6) 1   λ2 = 2. For λ1 = 2, (A − λI)V1 =0, so  2 1   1 V 0 11 3 3 . To = . Then V11 = − 21 V21 . Let V1 = −2 − 34 − 23 V21 0  2 1     V21 1 3 3 = . Then find V2 , let (A − λI)V2 = V1 . So 4 2 V −2 − − 22 3 3     1 1 1 . So P = (V1 V2 ) = , then P −1 = 2V21 + V22 = 3. Let V2 = 1 −2 1  n   n n−1  1 1 2 (n+3) n·2n−1 −1 3 − 3 . So An = P 2 n2 3 3 = . The P n+1 2 1 0 2n − n23 2n(3n−1) 3 3 3 λ2 − 3.5λ + 1.5 = 0. So λ =

43

44

Discrete Chaos  n  n−1



   2 (n+3) n·2 1 1 3n+1 3 eigenVector is associated with λ = 2, An = . −2 1 − n23 2n(3n−1) 3   −2 −3 5. A = 1 1 −2 − λ −3 Let |A − λI| = 0, so = (2 + λ)(λ − 1) + 3 = λ2 + λ + 1 = 0. 1 1−λ √ −1± 1−4 2

√ −1± 3i . 2

√

√

Then λ1 = −1+2 3i , λ2 = −1−2 3i .  √   −2 − −1+2 3i λ −3 √ Since(A − λI)V = 0, therefore = −1+ 3i y 1 1− 2 √ √ √   3        −2 0 − 23 23 − 23 + 23 i − 23 = . Then P = , . So V = +i 0 1 1 0 1 0 !   √  −1 2  √3 2 0 1 so P −1 = 2√3 √ . Since λ1 = −1+2 3i , therefore |λ1 | = + 2 = 2 2 √ 3 3 √ 1, θ = arctan −21 = arctan(− 3) = 120◦ = 23 π. Then So λ =

=

2

  cos nω sin nω An = P · |λ1 |n P −1 − sin nω cos nω  √  √ 2nπ − 3 sin + cos 2nπ −2 3 sin 2nπ 3 3 3 √ √ = . 2nπ 2 3 cos 2nπ 3 sin 2nπ 3 sin 3 3 + 3  √ √ − 32 + −1+ 3i −1− 3i Thus eigenValues are λ1 = , λ = , V = 2 1 2 2 1 √   3 3 −2 − 2 i associated with λ1 , V2 = is associated with λ2 . 1  √  √ 2nπ 2nπ 2nπ 3 sin + cos −2 3 sin − 3 √ 3 √ 3 2nπ . An = 2 3 2nπ cos 2nπ 3 sin 3 3 sin 3 3 +  8 1   1 3 3 7. Solve x(n + 1) = Ax(n), A = and x(0) = . 1 − 43 34  n  n−1 2 (n+3) n·2 3 3 n −n·2n+1 2 (3−n) 3 3

From Problem 3, An = therefore

 n

x(n) = A x(0) =  = X(n) =



3 2 i

is

. Since x(n + 1) = Ax(n),

2n (n+3) n·2n−1 3 3 n −n·2n+1 2 (3−n) 3 3



2n−1 (n+2n+6) 3 2n (3−3n) 3   n−1

2

√

(n + 2) . 2n (1 − n)

  1 1

Stability of Two-Dimensional Maps

45

    −2 3 2 2 n 0 9. Let f : R → R be defined by f (X) = AX, , find f . 1 1 1 From Problem 5,  √  √ 2nπ + cos 2nπ −2 3 sin 2nπ − 3 sin n 3 3 3 √ √ A = . 2 3 2nπ cos 2nπ 3 sin 2nπ 3 sin 3 3 + 3 √       2nπ 0 −2 3 sin n 0 3 √ =A = . f 1 1 cos 2nπ 3 sin 2nπ 3 + 3

Then

n

Exercises - (4.3 and 4.4)

  −1 1 . The eigenvalues of A are λ1 = −1 and λ2 = 2. The 0 2     1 1 corresponding eigenvectors are V1 = and V2 = . Thus 0 3

1. A =

    1 1 X(n) = k1 (−1)n + k2 2n 0 3       1 1 1 + k2 X(0) = = k1 0 3 2 2 1 3k2 = 2 ⇒ k2 = , k1 + k2 = 1 ⇒ k1 = . 3 3 Hence     1 2 1 1 (−1)n + (2n ) 3 0 3 3   n n+1 1 (−1) + 2 = . 2n+1 3

X(n) =



 1 1 . The eigenvalues of A are λ1 = 2 and λ2 = 3. The −2 4     1 1 corresponding eigenvectors are V1 = and V2 = . Hence 1 2

3. A =

    1 1 X(n) = k1 2n + k2 3n 1 2   n n k1 2 + k2 3 = . k1 2n + 2k2 3n

46

Discrete Chaos

  31 5. A = 03 X(n) = An X0  n n−1    x1 (0) 3 3 = x2 (0) 0 3n   n 3 x1 (0) + 3n−1 x2 (0) = 3n x2 (0)

7. F (n + 2) − F (n + 1) − F (n) = 0. The associated characteristic equation is given by λ2 − λ − 1 = 0 √ √ 1+ 5 1− 5 , λ2 = . λ1 = 2 2 Hence  √ n √ n 1+ 5 1− 5 F (n) = k1 + k2 2 2   √  √  1+ 5 1− 5 + k2 F (1) = 1 = k1 2 2   √ n √ n 1+ 5 1− 5 F (2) = 1 = k1 + k2 . 2 2 

Solving equations (4.1) and (4.2) yields 1 k1 = √ , 5 Hence 1 F (n) = √ 5

"

1 k2 = − √ . 5

√ n  √ n # 1+ 5 1− 5 − . 2 2

9. x(n + 2) + 16x(n) = 0. The characteristic equation is given by λ2 + 16 = 0 λ1 = 4i, λ2 = −4i. Hence x(n) = k1 cos nθ + k2 sin nθ where θ = tan−1 (4) ≈ 76◦ .

(4.1)

(4.2)

Stability of Two-Dimensional Maps

47

15.

Y (n) = An Y (0) +

n−1 

An−K−1 g(K)

K=0

 n n−1    n−1  2n−K−1 (n − K − 1)2n−K−2  k  2 n2 1 = + 0 2n 0 2n−K−1 0 1 K=0  n  n−1  k2n−K−1 + (n − K − 1)2n−K−2  2 = + 0 2n−K−1 K=0 ⎛ n−1 ⎞ n−1  (n−1)  −k 1 −k  n k2 + 2 4 ⎜4 ⎟ 2 K=0 K=0 ⎟ = + 2n ⎜ n−1 ⎝ ⎠  0 1 −k 2 2 K=0

Using the formula

n−1 

n−1 

(1 − an ) , a = 1−a k

K=0

K=0

kak =

a(1 − an ) − nan+1 (1 − a) . (1 − a)2

We obtain 

  n−1 3  2n − 4n n2 + 2n − 1 0  n  2 + n2n−1 − 34 n = . 2n − 1

Y (n) =

Exercises - (4.5–4.7) 1 1. A =

2

0



0 12 The origin is asymptotically stable since ρ(A) =

1 2

< 1 (Theorem 4.13).

48

Discrete Chaos y

x

λ1 = λ2 = 12 Phase portrait: origin is asymptotically stable.   21 02 Since ρ(A) = 2 > 1, it follows by Theorem 4.13 that the origin is asymptotically stable.

3. A =

y

x

λ1 = λ2 = 2 Phase portrait: origin is unstable. 

 0.5 0.25 5. A = −0.25 0.5 Eigenvalues of A are λ1 =

1 2

+ 14 i, λ2 = !

|λ1 | = |λ2 | +

1 2

− 14 i.

√ 1 1 5 + = 1. y

x

Y1 Y2

λ1 = 2, λ2 = 3 Unstable node. 

 1 2 −1 −1 The eigenvalues of A are λ1 = i, λ2 = −i. The origin is stable but not asymptotically stable.

9. A =

50

Discrete Chaos y

x

λ1 = i, λ2 = −i A stable center.

Exercises - (4.8) 

1. A =

 2 3 1+a 4

trA = 6,

det A = 5 − 3a

The point (T, D) = (6, 5 − 3a) moves vertically along the line T = 6 in the trace-determinant plane, as a varies. Note det A < −trA − 1 if a > 4, where we have an oscillatory source. If a = 4, det A = −trA − 1 and consequently, λ2 = −1 and λ1 = − det A = −6. Hence the origin is unstable. If a decreases below 4, we have a saddle and thus a = 4 is a bifurcation value. However, when the line T = 6 intersects the parabola T 2 = 4D when a = − 43 , we have two repeated eigenvales λ1 = λ2 = 12 trA = 3. In this case, we have an unstable origin and a = − 43 is a bifurcation value. For a < − 34 , we have spiral source. (i) (ii) (iii) (iv) (v)

a > 4 : oscillatory source a = 4 : unstable origin (bifurcation value) a < 4 : a saddle a = − 43 : unstable origin a < − 43 : spiral sources   2+a 1 3. A = 0 2 trA = 4 + a, det A = 4 + 2a Hence det A = 2trA − 4. a. Notice that tr2 A − 4 det A = tr2 A − 8trA + 16 = (trA − 4)2 ≥ 0. Hence all eigenvalues of A are real.

Stability of Two-Dimensional Maps

51

b. Our line det A = 2trA − 4 intersects the line det A = trA − 1 at the point (T, D) = (3, 2) when a = −1, while it intersects the line det A = −trA − 1 at the point (T, D) = (1, −2) when a = −3. c. The bifurcation value of a is -3, for a < −3 we have an oscillatory source and for a > −3 we have a saddle. For a = −3, we have the eigenvalues λ2 = −1 and λ1 = − det A = 2 and hence we have an unstable (oscillatory) origin. Bifurcation value: a = −3 a < −3 : a saddle a = −3 : unstable oscillatory   a + 1 a2 + a 5. A = 1 a+1 trA = 2a + 2, det A = a2 + 2a + 1 − a2 − a = a + 1 Hence det A = 12 trA. Our line intersects   (a) the line det A = −tr − 1 at (T, D) = − 31 , − 23 , when a = − 67   (b) the line det A = 1 at 12 , 1 , when a = 0 (c) the det A = 4tr2 A at the points (0, 0) when a = −1 and  1 parabola  2 , 1 , when a = 0. Hence there are three bifurcation values. (i) a >= − 67 where the eigenvalues are λ2 = −1 and λ1 = − det A = 23 and we have an “oscillatory” stable origin, for a < − 76 , we have an oscillatory saddle. (ii) a = −1, the eigenvalues are λ1 = λ2 = 0 and every point is a fixed point, for − 67 < a < −1, we have a sink, and for −1 < a < 0, we have a spiral sink. (iii) a = 0, where the eigenvalues are λ1 = 1, λ2 = 1, where we have eigher a stable or unstable origin, for a > 0, we have a saddle.   a+1 1 7. A = b 2 trA = a + 3, det A = 2a + 2 − b Region 5: det A < trA − 1 2a + 2 − b < a + 2 a −trA − 1 and 2a + 2 − b > −a − 4 and

3a + 6 > b

52

Discrete Chaos Region 6: det A < trA − 1 a trA − 1 and det A < −trA − 1 a>b and 3a + 6 < b

Regions 1 and 2: det A > trA − 1 b −trA − 1 and det A < 1 and

b < 3a + 6

and

b > 2a + 1

(i) Region 1: tr2 A > 4 det A a2 + 6a + 9 > 8a + 8 − 4b 4b > −a2 + 2a − 1 1 b > − (a − 1)2 4 Region 8: det A > −trA − 1 and tr2 A > 4 det A and 1 b < 3a + 6 and b > − (a − 1)2 4 b < 2a + 1

det A > 1

Region 4: det A > trA − 1 and tr2 A > 4 det A 1 b − (a − 1)2 4

and det A > 1 and

b < 2a + 1

Stability of Two-Dimensional Maps

53

Exercises - (4.9) 1. Let V (x1 , x2 ) = x21 + x22 . Then ΔV (x1 (n), x2 (n)) = x22 (n)[1 − (x21 (n) + x22 (n))]2 + x21 (n)[1 − (x21 (n) + x22 (n))]2 − x21 (n) + x22 (n) = x22 (n)[x21 (n) + x22 (n)]2 − 2x22 (n)(x21 (n) + x22 (n)) + x21 (n)[x21 (n) + x22 (n)]2 − 2x21 (n)(x21 (n) + x22 (n)) = [x21 (n) + x22 (n)]2 [x21 (n) + x22 (n) − 2]. If x21 + x22 < 2, ΔV < 0 and the origin is asymptotically stable by Thoerem 4.13. 3. x1 (n + 1) = g1 (x1 (n), x2 (n)), x2 (n + 1) = g2 (x1 (n), x2 (n)) Let V = x1 x2 . Then ΔV (x1 (n), x2 (n)) = g1 (x1 (n), x2 (n))g2 (x1 (n), x2 (n)) − x1 (n)x2 (n) > 0.   Let an = n1 , n1 , n ≥ 1. Then an → (0, 0), and 0 < V (an ) = n12 → 0 as n → ∞. Hence by Theorem 4.18, the origin is unstable.     x 2y − 2yx2 5. f = 1 2 y 2 x + xy 2 Let V (x, y) = x + 4y 2 . Then ΔV = 4y 2 (1 − x2 )2 + 4x2



1 + y2 2

2

− x2 − 4y 2

= 4y 2 + 4x4 y 2 − 8x2 y 2 + x2 + 4x2 y 4 + 4x2 y 2 − x2 − 4y 2 = 4x2 y 2 [x2 + y 2 ] − 4x2 y 2 = 4x2 y 2 [x2 + y 2 − 1]. If x2 + y 2 < 1, then the origin is asymptotically stable.     x y 7. F = , β > 0. y αx/(1 + βy 2 ) Let V (x, y) = x2 + y 2 . Then ΔV (x(n), y(n)) = x2 (n + 1) + y 2 (n + 1) − x2 (n) − y 2 (n) = y 2 (n) +

α2 x2 − x2 (n) − y 2 (n) [1 + βy 2 (n)]2

≤ (α2 − 1)x2 (n).

54

Discrete Chaos Case (i) If α2 < 1, then ΔV < 0. In this case X ∗ = 0 is the only fixed point and by Theorem 4.13, the origin is asymptotically stable. Case (ii) If α2 = 1, then ΔV ≤ 0 and ΔV = 0 on the x and y-axes. Hence E = {(x, 0), (0, y) : x, y ∈ R}. There are two subcases:         x 0 x x (i) α = 1; then F = , F2 = . Thus E = M , and 0 x 0 0 the origin is not asymptotically stable.             x 0 −x 0 2 x 3 x (ii) α = −1; F = , F = , F = , 0 −x 0 0 0 x     0 x F4 = . Hence E = M and the origin is not asymptotix 0 cally stable. (iii) α2 > 1 may not be determined by either Theorem 4.13 or Theorem 4.16.

11. Let V (x, y) = xy. Then ΔV (x(n), y(n)) = y(n)[x(n) + f (x(n))] − x(n)y(n) = y(n)f (x(n)) 2

Δ V (x(n), y(n)) = Δ[y(n)f (x(n))] > 0. The conclusion now follows from Problem 8.

Exercises - (4.10 and 4.11)     0 01 3. Df = 0 α0

  √ √ 0 The eigenvalues of Df are λ1 = α, λ2 = − α, and ρ(A) = |λ1 | = 0 √ √ | α|. Hence by Theorem 4.22, if ρ(A) = | α| < 1 or |α| < 1, then the origin is aysmptotically stable.       0 0 00 ∗ : Df = (a) X1 = 0 0 40 Since ρ(A) = 0, it follows from Theorem 4.22 that X1∗ is asymptotically stable.       6 6 12 0 ∗ (b) X2 = : A = Df = 3 3 4 −6 The eigenvalues of the matrix A are given by λ1 = 12, λ2 = −6. Hence ρ(A) > 1. By virtue of Theorem 4.22, X2∗ is unstable.

Stability of Two-Dimensional Maps     0 α 0 5. A = Df = 0 −1 β

55

By Theorem 4.11, ρ(A) < 1 if and only if |α + β| < 1 + αβ < 2. Hence the zero solution is asymptotically stable by virtue of Theorem 4.22. αβ < 1 6. To find the fixed points we solve the system 1 y2 − x = x 2 1 1 x + y = y. 4 2

(4.3) (4.4)

From Eq. (4.4) we have y = 12 x. Substituting in Eq. (4.3) yields   1 1 1 2 3 x − x= x x − 3 = 0. 4 2 2 2     0 6 ∗ ∗ Hence x = 0 or x = 6. Hence the fixed points are X1 = , X2 = . 0 3 The Jacobian  1     1  − 2 2y 0 −2 0 J= , J = 1 1 1 . 1 0 4 2 4 2   0 is asymptotically stable (a) X ∗ = 0   6 is unstable (b) X2∗ = 3     0 01 7. A = Df = 0 ab By Theorem 4.11, ρ(A) < 1 if and only if |b| < 1 − a < 2. Hence by Theorem 4.22, the origin is asymptotically stable if |b| < 1 − a < 2. 11. Y (n + 1) = AY (n) + g(Y (n)) If V (Y ) = Y T BY , then ΔV (Y ) = V (Y (n + 1)) − V (Y (n)) = (Y T (n)AT + g T (Y (n))B(AY (n) + g(Y (n))) − Y T (n)BY = Y T [AT BA − B]Y + Y T AT Bg(y) + g T (Y )BAY + g T (Y )Bg(Y ). Observe that AT BA − B = −C, where C is a positive definite matrix. Furthermore, since B is symmetric Y T AT Bg(Y ) = g T (Y )BAY . Hence ΔV (Y ) = −Y T CY + 2Y T AT Bg(Y ) + V (g(Y )).

56

Discrete Chaos

13. Suppose that f is conjugate to g. Then there exists a homeomorphism h such that h ◦ f = g ◦ h. (a) Assume that p is heteroclinic with respect to f . Then lim f n (P ) = n→∞

X1∗ and lim f −n (P ) = X2∗ . Hence lim g n (h(P )) = lim h(f n (P )) = n→∞

n→∞

n→∞

h(X1∗ ). Furthermore, lim g −n (h(P )) = lim h(f −n (P )) = h(X2∗ ). n→∞

n→∞

This proves that h(P ) is heteroclinic with respect to g. Similarly, one may show that if P is a homoclinic point of f , then h(P ) is homoclinic with respect to g.

Exercises - (4.12)

  ⎞ ⎛  ∗ 1−γ1 γ2 1 γ2 a1 − a1 1−γ x1 ⎝1 − b1 − a1 1+γ2 1+γ  2 ⎠ 1. A = Df x∗2 1 − b2 − a2 1−γ1 γ2 a2 − a2 1−γ1 γ2 1+γ2

1+γ2

0 < ai < 1, 0 < bi < 1, i = 1, 2 3. (a) To find the fixed points we solve rN e−aP = N N (1 − e−aP ) = P. Hence re−aP =1 or −aP = − ln r. Thus P ∗ = a1 ln r. This implies r that N ∗ 1 − 1r = a1 ln r. Hence N ∗ = a(r−1) ln r, r > 1, a > 0.  ∗   −r 1 ln r N (b) A = Df = r−1 r−1 −1 P∗ r r−1 ln r We need to show that |tr A| < 1 + det A < 2. But this is evident since ln r ln r 34 (1 − b)2 , then

*

1 1−b − (1 − b) − [1 − b − (1 − b)2 + 4a]2 + 4b 2 * 2

1 b 1 [1 − b − (1 − b)2 + 4a]2 + 4b. 0

1 + 1 − b − (1 − b)2 + 4a − b = 2(1 − b) − (1 − b)2 + 4a > 0 tr A1 = 1 − b −

4(1 − b)2 > (1 − b)2 + 4a 7. Use the Liapunov function       p q −1 ∗ ∗ −1 ∗ ∗ p − p − p ln + c(1 − β) q − q − q ln . V (p, q) = (1−α) ∗ p q∗

6 Fractals

Exercises - (6.1–6.3) 1. (b) Dt = 1 (c) h 1

N 1 1 8  142 2 8 4 .. ..  1.n .n 8 4 ln N (h) ln 8n  1  = lim = 1.5 n→0 ln 4n h→0 ln h

Df = lim 3. (b) Dt = 1 ln 5n ln 5 = h→0 ln 3n ln 3

(c) Df = lim 5. (b) Dt = 1

ln 3 ln(3n+1 ) = n→∞ ln(2n ) ln 2

(c) Df = lim 7. (b) Dt = 2 (c)

h 1 1 3

..  1.n 3

N 1 20 .. . 20n .. .

.. . ln 20n ln 20 ≈ 2.73 = Df = lim n→∞ ln 3n ln 3

59

60

Discrete Chaos 9. (b) Dt = 1 ln 4n =2 n→∞ ln 2n

(c) Df = lim

19. Let us find Df (C5 ). h 1

N 1 2 2  252 2 2 5 .. ..  2.n .n 2 5 ln 2 ln 2n  n =  5  Df (C5 ) = lim n→∞ ln 5 ln 2 2 Dt (C5 ) = 0 18. Let N1 (h) and N2 (h) be the smallest number of sets of diameter atmost h that covers A and B, respectively. If A ⊂ B, then N1 (h) ≤ N2 (h). Hence ln N1 (h) ≤ ln N2 (h). Thus ln N2 (h) ln N1 (h) 1 ≤   = Df (B). h→0 ln ln h1 h

Df (A) = lim

20. Dt (C2n+1 ) = 0, Df (C2n+1 ) =

ln 2 ln( 2n+1 n )

Exercises - (6.4)

    x −x = y y    1  x −2x (b) F = y 2y

1. (a) F

(c)      x cos π4 − sin π6 x F = sin π4 cos π6 y y √   2 −1 x = √22 √32 y 2

2

Fractals 3.

61   1 x = 3 y 0   1 x F2 = 3 0 y   1 x F3 = 3 0 y   1 x F4 = 3 0 y   1 x F5 = 3 y 0 F1

5.

  x 1 y 3     0 x 0 + 2 1 y 3     31  0 x + 31 1 y 3     32  0 x + 3 1 0 y 3    2 x 0 + 32 1 y 3 3 0

  1    0 x x F1 = 3 1 0 3 y y   1    2 0 x x = 3 1 + 3 F2 0 3 0 y y

7.

  1    0 x x F1 = 2 1 0 2 y y   1    1 0 x x F2 = 2 1 + 2 0 2 0 y y   1      0 x x 0 F2 = 2 1 + 1 y y 0 2 2

10. First Iteration

Second Iteration

62 11. Koch Snowflake

12. Weed

13. Leaning Bush

14. Island

15. Chain

Discrete Chaos

Fractals

63

Exercises - (6.5)

√ 1. (a) d(A, B) = d(B, A) = D(A, B) = 5 √ (b) d(A, B) = 2 d(B, A) = 2 2 − 1 ≈ 1.8 Therefore D(A, B) = 2.

3.

A

d(A,C)

B

d(B,E)

C

d(C,A)

E

d(E,B)

D(A ∪ B, C ∪ E) = max{d(A ∪ B, C ∪ E), d(C ∪ E, A ∪ B)} Now d(a, C ∪ E) = inf{d(a, z) : z ∈ C ∪ E}. Thus d(A, C ∪ E) ≤ d(A, C) d(B, C ∪ E) ≤ d(B, E). Hence d(A ∪ B, C ∪ E) ≤ d(A, C), d(B, E) and d(C ∪ E, A ∪ B) ≤ d(C, A), d(E, B). Hence D(A ∪ B, C ∪ A) ≤ max{d(A, C), d(C, A), d(B, E), d(E, B)} ≤ max{max{d(A, C), d(C, A)}, max{d(B, E), d(E, B)} ≤ max{D(A, C), D(B, E)}.

64

Discrete Chaos 5. By Lemma 8.16, F1 , F2 , . . . , FN are contractions on H. This implies by Lemma 8.17 that F = ∪N i=1 Fi is also a contraction on H. The conclusion now follows by applying the contraction mapping principle (Theorem 8.15). 7. Let F : X → X be a contraction with a contraction factor α ∈ (0, 1). Now d(x(n + 1), x(n)) = d(F (x(n)), F (x(n − 1)) ≤ αd(x(n), x(n − 1)) .. . ≤ αn d(x(1), x(0)). Since lim x(i) = x∗ , for a given n there exists a sufficiently large m > n i→∞

such that

d(x(n), x∗ ) ≤ d(x(m), x(n)).

(6.1)

But d(x(m), x(n)) ≤ d(x(m), x(m − 1)) + d(x(m − 1), x(m − 2)) + · · · + d(x(n + 1), x(n)) ≤ αn (1 + α + α2 + · · · + αm−n )d(x(1), x(0)) αn ≤ d(x(1), x(0)). 1−α From (6.1) and (6.2) we conclude d(x(1), x∗ ) ≤

αn d(x(1), x(0)). 1−α

(6.2)

7 The Julia and Mandelbrot Sets

Exercises - (7.2 and 7.3) 1. (a) i = cos π2 + i sin π2 square roots of i: z1 = cos π4 + i sin π4 , z2 = − cos π4 − i sin π4 Z

Z

1

Z 2

√ √ (c) −1 + 3i, θ = tan−1 (− 3) = 2π ,r=2 √  √  3  roots: z1 = 2 cos π3 + i sin π3 , z2 = − 2 cos π3 + i sin π3 , Z Z 1

Z 2

√ √ (e) −2 + 2i, θ = tan−1 (−1), θ = 3π 8=2 2 4 ,r = √ √    4 3π 3π 3π roots: z1 = 4 8 cos 3π 8 + i sin 8 , z2 = − 8 cos 8 + i sin 8 , Z Z 1

Z 2

65

66

Discrete Chaos 3. (a) f (z) = az + b, a, b ∈ C −b az + b = z ⇒ z = a−1 if a = 1 (b) Let h(z) = a +

b a−1 .

Then

ab b = az + . a−1 a−1  b cb If g(z) = cz, then g(h(z)) = g z + a−1 = cz + a−1 put c = a, then we have conjugacy. h(f (z)) = h(az + b) = az + b +

5. Q1/4 (z) = z 2 + (a) z 2 − z +

1 4

1 4

= 0, z ∗ = 12 , Q1/4 (z ∗ ) = 1. Now Q1/4 (z) = |2z| < 1 if

and only if |z| < 12 . Observe that |z| < 12 is the interior of a circle passing through z ∗ = 12 . Hence Q1/4 (z) > 1 for |z| > 12 . Thus z ∗ = 12 is unstable. (b) 2  1 1 + =2 z2 + 4 4 1 5 z4 + z2 − z + =0 2 16  2   1 5 2 z− z +z+ =0 2 4   The two cycles are: − 12 − i, − 21 + i .

Q1/4 (z1 )Q1/4 (z2 ) = 5 Thus the 2-cycles are repelling. 7. f (z) = eiθ z

(a) Let z0 = r0 eiθ0 . Then f n (z0 ) = r0 ei(nθ+θ0 ) . If z0 is n-periodic, then f n (z0 ) = z0 . Hence nθ + θ0 = 2kπ + θ0 . Solving for θ yields θ = 2k n π. Hence z0 is periodic if and only if θ is a rational multiple of π. (b) Suppose that θ is not a rational multiple of π. Then f m (z0 ) = z0 for m = 1, 2, 3, . . . . Let ε > 0, r0 = |z0 |. Then for m > 2πr ε , none of the first m iterates of z0 are the same and there must be two between which the distance is less than ε. Suppose that |f m2 (z0 ) − f m1 (z0 )| < ε, m2 > m1 . Hence r0 ei(θ0 +m2 θ) − r0 ei(θ0 +m1 θ) = ei(θ0 +m1 θ) r0 ei(m2 −m1 )θ − r0 = r0 ei(m2 −m1 )θ − r0 < ε.

The Julia and Mandelbrot Sets Thus

67

f m2 −m1 (z0 ) − z0 = r0 ei(m2 −m1 )θ − r0 < ε.

Thus the sequence {z0 , f m2 −m1 (z0 ), f 2(m2 −m1 ) (z0 ), . . . } covers the circle with radius |z0 | = r0 by points that are separated by a distance less than ε. Hence given any point z on the circle of radius r0 , then the neighborhood of z with radius ε must contain an interate of z0 . So ε is arbitrary, O(z0 ) is dense in the circle with radius r0 and center at the origin. 9. Q2 (z) = z 2 + 2

Exercises - (7.4 and 7.5) −1 3. K0 = {z ∈ C : |z| ≤ r(c)}, K−1 = Q−1 c (K0 ), . . . , K−n−1 = Qc (K−n ). Let z ∈ Kc . Then |z| ≤ r(c) and z ∈ K0 . Suppose that z ∈ K−N for some N ∈ Z+ . Then Qc (z) ∈ K−N +1 , QN c (z) ∈ K0 , a contradiction. Conversely, let z ∈ ∩K−n . Then for each n ∈ Z+ , Qnc (z) ∈ K0 . Hence O(z) is bounded, and, consequently, z ∈ Kc .

5. Let z ∈ Kc , then O(z) is bounded. Now Qc (−z) = (−z)2 + c = z 2 + c = Qc (z). Hence O(−z) is bounded and, consequently, −z ∈ Kc . 7. (a) |Fμ (z)| = |μ| |z| |1 − z| ≥ |μ| |z| (|z| − 1) 1 ≥ |μ| |z| |μ| = |z|. Now if |z| >

1 |μ|

+ 1, then |z| − 1 = |Fμ (z)| ≥

1 |μ|−δ

for some δ > 0. Then

|μ| |z| = M |z| |μ| − δ

68

Discrete Chaos where M > 1. By mathematical induction, one may show that |Fμn (z)| ≥ M n |z| → ∞ as n → ∞. (b) Conclusion: If |z| >

1 |μ| 1 2

9. (a) Suppose that |z|