Student Solutions Manual for Non Linear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering [3 ed.] 9780429293979, 9780367265663

This official Student Solutions Manual includes solutions to the odd-numbered exercises featured in the third edition of

178 117 34MB

English Pages 399 [400] Year 2024

Report DMCA / Copyright

DOWNLOAD FILE

Polecaj historie

Student Solutions Manual for Non Linear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering [3 ed.]
 9780429293979, 9780367265663

  • Commentary
  • Publisher PDF | Published: March 15, 2024

Table of contents :
Cover
Title Page
Copyright Page
Table of Contents
Part I One-Dimensional Flows
2 Flows on the Line
2.1 A Geometric Way of Thinking
2.2 Fixed Points and Stability
2.3 Population Growth
2.4 Linear Stability Analysis
2.5 Existence and Uniqueness
2.6 Impossibility of Oscillations
2.7 Potentials
2.8 Solving Equations on the Computer
3 Bifurcations
3.1 Saddle-Node Bifurcation
3.2 Transcritical Bifurcation
3.3 Laser Threshold
3.4 Pitchfork Bifurcation
3.5 Overdamped Bead on a Rotating Hoop
3.6 Imperfect Bifurcations and Catastrophes
3.7 Insect Outbreak
4 Flows on the Circle
4.1 Examples and Definitions
4.2 Uniform Oscillator
4.3 Nonuniform Oscillator
4.4 Overdamped Pendulum
4.5 Fireflies
4.6 Superconducting Josephson Junctions
Part II Two-Dimensional Flows
5 Linear Systems
5.1 Definitions and Examples
5.2 Classification of Linear Systems
5.3 Love Affairs
6 Phase Plane
6.1 Phase Portraits
6.2 Existence, Uniqueness, and Topological Consequences
6.3 Fixed Points and Linearization
6.4 Rabbits versus Sheep
6.5 Conservative Systems
6.6 Reversible Systems
6.7 Pendulum
6.8 Index Theory
7 Limit Cycles
7.1 Examples
7.2 Ruling Out Closed Orbits
7.3 Poincaré-Bendixson Theorem
7.4 Liénard Systems
7.5 Relaxation Oscillations
7.6 Weakly Nonlinear Oscillators
8 Bifurcations Revisited
8.1 Saddle-Node, Transcritical, and Pitchfork Bifurcations
8.2 Hopf Bifurcations
8.3 Oscillating Chemical Reactions
8.4 Global Bifurcations of Cycles
8.5 Hysteresis in the Driven Pendulum and Josephson Junction
8.6 Coupled Oscillators and Quasiperiodicity
8.7 Poincaré Maps
Part III Chaos
9 Lorenz Equations
9.1 A Chaotic Waterwheel
9.2 Simple Properties of the Lorenz Equations
9.3 Chaos on a Strange Attractor
9.4 Lorenz Map
9.5 Exploring Parameter Space
9.6 Using Chaos to Send Secret Messages
10 One-Dimensional Maps
10.1 Fixed Points and Cobwebs
10.2 Logistic Map: Numerics
10.3 Logistic Map: Analysis
10.4 Periodic Windows
10.5 Liapunov Exponent
10.6 Universality and Experiments
10.7 Renormalization
11 Fractals
11.1 Countable and Uncountable Sets
11.2 Cantor Set
11.3 Dimension of Self-Similar Fractals
11.4 Box Dimension
11.5 Pointwise and Correlation Dimensions
12 Strange Attractors
12.1 The Simplest Examples
12.2 Hénon Map
12.3 Rössler System
12.4 Chemical Chaos and Attractor Reconstruction
12.5 Forced Double-Well Oscillator
Part IV Collective Behavior
13 Kuramoto Model
13.1 Governing Equations
13.2 Visualization and the Order Parameter
13.3 Mean-Field Coupling and Rotating Frame
13.4 Steady State
13.5 Self-Consistency
13.6 Remaining Questions

Citation preview

Student Solutions Manual for Nonlinear Dynamics and Chaos, Third Edition With Applications to Physics, Biology, Chemistry and Engineering Mitchal Dichter

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

A CHAPMAN & HALL BOOK

Third edition published 2024 by CRC Press 2385 Executive Center Drive, Suite 320, Boca Raton, FL 33487-2742 and by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN CRC Press is an imprint of the Taylor & Francis Group, LLC © 2024 Mitchal Dichter First edition published by Westview Press 2017 Second edition published by Taylor and Francis 2018 Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. 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. For works that are not available on on CCC please contact [email protected] Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Every effort has been made to secure required permissions for all text, images, maps, and other art reprinted in this volume.

ISBN 13: 978-0-367-26566-3 (pbk) ISBN 13: 978-0-429-29397-9 (ebk) DOI: 10.1201/9780429293979

CONTENTS Part I One-Dimensional Flows 2 Flows on the Line 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

3 Bifurcations 3.1 3.2 3.3 3.4 3.5 3.6 3.7

1

A Geometric Way of Thinking 1 Fixed Points and Stability 2 Population Growth 6 Linear Stability Analysis 8 Existence and Uniqueness 10 Impossibility of Oscillations 11 Potentials 12 Solving Equations on the Computer

17

Saddle-Node Bifurcation 17 Transcritical Bifurcation 25 Laser Threshold 28 Pitchfork Bifurcation 30 Overdamped Bead on a Rotating Hoop Imperfect Bifurcations and Catastrophes Insect Outbreak 51

4 Flows on the Circle 4.1 4.2 4.3 4.4 4.5 4.6

13

40 42

61

Examples and Definitions 61 Uniform Oscillator 62 Nonuniform Oscillator 62 Overdamped Pendulum 70 Fireflies 73 Superconducting Josephson Junctions

75

Part II Two-Dimensional Flows 5 Linear Systems 5.1 5.2 5.3

6 Phase Plane 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

81

Definitions and Examples 81 Classification of Linear Systems Love Affairs 94

86

96

Phase Portraits 96 Existence, Uniqueness, and Topological Consequences Fixed Points and Linearization 103 Rabbits versus Sheep 111 Conservative Systems 123 Reversible Systems 137 Pendulum 152 Index Theory 154

102

Contents

7 Limit Cycles 7.1 7.2 7.3 7.4 7.5 7.6

162

Examples 162 Ruling Out Closed Orbits 169 Poincar´e-Bendixson Theorem 178 Li´enard Systems 187 Relaxation Oscillations 187 Weakly Nonlinear Oscillators 192

8 Bifurcations Revisited 8.1 8.2 8.3 8.4 8.5 8.6 8.7

207

Saddle-Node, Transcritical, and Pitchfork Bifurcations 207 Hopf Bifurcations 214 Oscillating Chemical Reactions 224 Global Bifurcations of Cycles 227 Hysteresis in the Driven Pendulum and Josephson Junction 235 240 Coupled Oscillators and Quasiperiodicity Poincar´e Maps 254

Part III Chaos 9 Lorenz Equations 9.1 9.2 9.3 9.4 9.5 9.6

261

A Chaotic Waterwheel 261 Simple Properties of the Lorenz Equations 264 Chaos on a Strange Attractor 267 Lorenz Map 279 Exploring Parameter Space 279 Using Chaos to Send Secret Messages 290

10 One-Dimensional Maps 10.1 10.2 10.3 10.4 10.5 10.6 10.7

11 Fractals 11.1 11.2 11.3 11.4 11.5

342

Countable and Uncountable Sets 342 Cantor Set 343 Dimension of Self-Similar Fractals 344 Box Dimension 348 Pointwise and Correlation Dimensions 351

12 Strange Attractors 12.1 12.2 12.3 12.4 12.5

294

Fixed Points and Cobwebs 294 Logistic Map: Numerics 305 Logistic Map: Analysis 309 Periodic Windows 316 Liapunov Exponent 323 Universality and Experiments 326 Renormalization 336

352

The Simplest Examples 352 H´enon Map 361 R¨ossler System 367 Chemical Chaos and Attractor Reconstruction Forced Double-Well Oscillator 371

369

Part IV Collective Behavior 13 Kuramoto Model

379

13.1 Governing Equations 379 13.2 Visualization and the Order Parameter 13.3 Mean-Field Coupling and Rotating Frame 13.4 Steady State 386 13.5 Self-Consistency 388 13.6 Remaining Questions 392

381 384

Chapter 2 2.1.1 The fixed points of the flow x˙ = sin(x) occur when x˙ = 0 ⇒ sin(x) = 0 ⇒ x = zπ

z∈Z

2.1.3 a) We can find the flow’s acceleration x ¨ by first deriving an equation containing x ¨ by taking the time derivative of the differential equation. d d x˙ = sin(x) ⇒ x ¨ = cos(x)x˙ dt dt We can obtain x ¨ solely as a function of x by plugging in our previous equation for x. ˙ x ¨ = cos(x) sin(x) b) We can find what values of x gives the acceleration x ¨ maximum positive values by using the trig identity 1 sin(2x) = sin(x) cos(x) 2 which can be used to rewrite x ¨ as x ¨= which has maximums when

1 sin(2x) 2

ã 1 π x= z+ 4 Å

z∈Z

2.1.5 a) A pendulum submerged in honey with the pendulum at the 12 o’clock position corresponding to x = 0 is qualitatively similar to x˙ = sin(x). The force near the 12 o’clock position is small, is greatest at the 3 o’clock position, and is again small at the the 6 o’clock position. b) x = 0 and x = π being unstable and stable fixed points respectively is consistent with our intuitive understanding of gravity.

1

Chapter 2: Flows on the Line

2.2.1 x˙ = 4x2 − 16 x˙

x −2

x(t)

2

t ∈ [0, 2]

2

−2  2 x0 e16t + x0 − 2e16t + 2 x(t) = −x0 e16t + x0 + 2e16t + 2 2.2.3 x˙ = x − x3 x˙

x −1

0

2

1

Chapter 2: Flows on the Line

t ∈ [0, 1]

x(t) 1 0 −1

x(t) = »

1 x20

±et

depending on the sign of the initial condition.

+ e2t − 1

2.2.5 x˙ = 1 +

1 cos(x) 2



x

There are no fixed points, but the rate increase for x(t) does vary. x(t)

t

Ç x(t) = 2 arctan



Ç

Ç

3 tan arctan

3

tan x20 √ 3



√ åå 3t + 4

Chapter 2: Flows on the Line

2.2.7 x˙ = ex − cos(x) We can’t solve for the fixed points analytically, but we can find the fixed points approximately by looking at the intersections of ex and cos(x), and determine the stability of the fixed points from which graph is greater than the other nearby.



x 0

We could also plot the graph using a computer.



x 0 Å There are fixed points at x ≈ π x(t)

ã 1 − n , n ∈ N, and x = 0. 2 t ∈ [0, 4]

0 −1.57 −4.71 −7.85 Unable to find an analytic solution. 4

Chapter 2: Flows on the Line

2.2.9 f (x) = x(x − 1) x˙

0

x 1

2.2.11 RC Circuit Q 1 V0 V0 − ⇒ Q˙ + Q= Q˙ = R RC RC R

Q(0) = 0

t

Multiply by an integrating factor e RC to both sides. ä V0 t t t t 1 V0 t d Ä ˙ RC Qe e RC Q = e RC ⇒ Qe RC = e RC + RC R dt R Apply an indefinite integral to both sides with respect to t. Z Z ä t V0 t d Ä Qe RC dt = e RC dt dt R t

−t

t

Qe RC = V0 Ce RC + D ⇒ Q(t) = V0 C + De RC And using the initial condition. ä Ä −t Q(0) = V0 C + D = 0 ⇒ D = −V0 C ⇒ Q(t) = V0 C 1 − e RC 2.2.13 Terminal Velocity The velocity v(t) of a skydiver falling follows the equation mv˙ = mg − kv 2 with m the mass of the skydiver, g is the acceleration due to gravity, and k > 0 the the drag coefficient. a)

… v(t) =

mg tanh k 5

Ç…

gk t m

å

Chapter 2: Flows on the Line

b) t→∞

v(t) −−−→



mg k

c) The terminal velocity should occur at a fixed point. The fixed point occurs when … mg v˙ = 0 ⇒ mg − kv 2 = 0 ⇒ v = k d) vavg =

31400ft − 2100ft ft ≈ 253 116sec sec

e) s(t) = which can be rewritten using V =

»

mg k

Ç Ç… åå m gk ln cosh t k m

into s(t) =

 g  V2  ln cosh t g V

ft Using s(116sec) = 29300ft and g = 32.2 sec 2 gives

Ç Ç åå ft 32.2 sec V2 2 29300ft = ln cosh t ft V 32.2 sec 2 ft which can be solved numerically to give V ≈ 266 sec .

2.3.1 Logistic Equation

ã Å N N˙ = rN 1 − K

N (0) = N0

a) Solve by separating the variables and integrating using partial fractions Å ã  N rN 2 rKN − rN 2 r rN 1 − = rN − = = KN − N 2 K K K K  dN r r dN N˙ = = = dt KN − rN 2 ⇒ 2 dt K KN − N K 1 1 1 = − KN − N 2 KN K(N − K) Z Z Z dN dN dN 1 1 = − = ln(N ) − ln(N − K) 2 KN − N KN K(N − K) K K Z r r = dt = t + C K K 1 1 r ln(N ) − ln(N − K) = t + C K K K ln(N ) − ln(N − K) = rt + KC Å ã N ln = rt + KC N −K 6

Chapter 2: Flows on the Line

Å − ln

N N −K

ã

Å = ln

N −K N

ã

Å ã K = ln 1 − = −rt − KC N

K = e−rt−KC = e−KC e−rt N K 1 − e−KC e−rt = N K N (t) = 1 − e−KC e−rt K K N (0) = = N0 ⇒ e−KC = 1 − 1 − e−KC N0 K ä Ä N (t) = 1 − 1 − NK0 e−rt

1−

b) Making the change of variables x =

1 N

Å ã x˙ 1 1 1 ⇒ N˙ = − 2 = r 1− x x x xK r x˙ = − rx K 1 1 KCe−rt + 1 1 ⇒ = Ce−rt + = x(t) = Ce−rt + K N (t) K K K ⇒ N (t) = KCe−rt + 1 K Ä ä N (0) = N0 ⇒ N (t) = 1 − 1 − NK0 e−rt N=

2.3.3 Tumor growth N˙ = −aN ln(bN ) a) a can be interpreted as specifying how fast the tumor grows, and

1 specifies the stable size of the tumor. b

b) N˙

N 1 b

0

7

Chapter 2: Flows on the Line

t

N (t) 1 b 0

2.3.5 Dominance of the fittest X˙ = aX

Y˙ = bY

x(t) =

X(t) X(t) + Y (t)

X0 , Y 0 > 0

a>b>0

a) Show as t → ∞, x(t) → 1 by solving for X(t) and Y (t). X(t) = eat

Y (t) = ebt ⇒ x(t) =

a > b > 0 ⇒ 0 > b − a ⇒ x(t) →

1 eat = + ebt 1 + e(b−a)t

eat

1 = 1 as t → ∞ 1+0

b) Show as t → ∞, x(t) → 1 by deriving an ODE for x(t).

Å ã ˙ − X Y˙ X aXY − bXY X XY 1− = (a − b)x(1 − x) = = (a − b) x˙ = (X + Y )2 (X + Y )2 X +Y X +Y

This is the logistic equation  x x˙ = rx 1 − K X0 , Y 0 > 0 ⇒ 0 < x 0 =

r =a−b>0

K=1

X0 < 1 ⇒ x(t) → 1 = K X0 + Y0

And x(t) increases monotonically as t → ∞ from our previous analysis of the logistic equation. 2.4.1 x˙ = x(1 − x)

dx˙ = 1 − 2x dx dx˙ (0) = 1 ⇒ unstable fixed point dx dx˙ (1) = −1 ⇒ stable fixed point dx

2.4.3 x˙ = tan(x) dx˙ = sec2 (x) dx 8

Chapter 2: Flows on the Line

dx˙ (πz) = 1 ⇒ unstable fixed point z ∈ Z dx π There are also stable fixed-like points at πz + because x˙ is positive on the left and negative on the right. 2 However, these points aren’t true fixed points because tan(x) has infinite discontinuities at these points and hence is not defined there. 2.4.5 x˙ = 1 − e−x

2

2 dx˙ = 2xe−x dx dx˙ (0) = 0 ⇒ inconclusive dx



x 0

The fixed point 0 is stable from the left and unstable from the right. 2.4.7 √ √ x˙ = ax − x3 = x( a − x)( a + x) dx˙ = a − 3x2 dx √ Assuming a ≥ 0 in order for a to be a real root.    unstable :a 0 if r < 0 dx2 √ −1 d2 V d2 V (x2 ) = (x4 ) = 4r + 4r + 1 + 1 > 0 if 0 if 0 < r 2 2 dx dx so the three candidate wells occur at x1 , x2 , and x4 with

−1 4

< r < 0.

V (x0 ) = C, so for simplicity we’ll set C = 0 which leaves V (x2 ) = V (x4 ) = V (x0 ) ä Ä 3 −3 1 −(4r + 1) 2 − 6r − 1 = 0 ⇒ r = 24 16

39

Chapter 3: Bifurcations

V −3 16

r=

C=0

x

So rc =

−3 16

and the three minima x1 , x2 , and x4 occur at the value of the integrating constant C.

3.5.1 There shouldn’t be an equilibrium between

π 2

and π because the rotation would force the bead outward and

down, and gravity would force the bead down. The forces wouldn’t cancel, so the bead would move. 3.5.3

 A Taylor Series expansion of f (ϕ) = sin(ϕ) γ cos(ϕ) − 1 centered at ϕ = 0 gives Å ã  1 2γ (γ − 1)ϕ + − ϕ 3 + O ϕ5 6 3

3.5.5 a)

Starting with the dimensionless equation ϵ

d2 ϕ dϕ + = f (ϕ) dτ 2 dτ

t br = from the text, but we don’t know what order of magnitude Tfast is in τ mg relation to Tslow , so we’ll guess a new time scale τ = ϵk ξ. The dimensionless equation is now We already have Tslow =

ϵ1−2k

dϕ d2 ϕ + ϵ−k = f (ϕ) 2 dξ dξ

We still need to be able to solve the equation, meaning each of the terms must be of the same order or negligible compared to all the other terms. To have all the terms of the same order, we need ϵ1−2k = ϵ−k = 1 which is impossible.

To have two of the terms of the same order and one term negligible, we need ϵ1−2k = ϵ−k ≫ 1 ⇒ k = 1

or ϵ1−2k = 1 ≫ ϵ−k ⇒ k =

Choosing k = 0 gives the equation we started with, and k =

1 gives 2

d2 ϕ 1 dϕ +√ = f (ϕ) dξ 2 ϵ dξ 40

1 2

or ϵ−k = 1 ≫ ϵ1−2k ⇒ k = 0

Chapter 3: Bifurcations

1 which is bad because ϵ ≪ 1 makes the √ term enormous, but the k = 1 choice looks good. ϵ If we choose τ = ϵξ for the new time scale, then we can find Tfast in terms of m, g, r, ω, and b by using t br = From the text τ mg Å ã t τ br t τ br br m2 g br m = = = =ϵ = 2 = ξ ξ t mg ξ mg mg b r mg b

Tslow = Tfast b)

Rescaling with the fast time scale −br dϕ rω 2 r d2 ϕ = − sin(ϕ) + sin(ϕ) cos(ϕ) 2 gTfast dξ 2 mgTfast dξ g −b2 r dϕ rω 2 b2 r d2 ϕ = 2 − sin(ϕ) + sin(ϕ) cos(ϕ) 2 2 m g dτ m g dτ g b2 r d2 ϕ b2 r dϕ rω 2 + 2 = − sin(ϕ) + sin(ϕ) cos(ϕ) 2 2 m g dτ m g dτ g Å ã rω 2 m2 g d2 ϕ dϕ − sin(ϕ) + + = sin(ϕ) cos(ϕ) dτ 2 dτ b2 r g = ϵf (ϕ) Now the ϵf (ϕ) is the negligible term. c) Tfast = ϵ

br = ϵTslow mg

So if ϵ ≪ 1 then Tfast ≪ Tslow . 3.5.7

Å ã N N˙ = rN 1 − K

N (0) = N0

a) r∼

1 time

K ∼ population N0 ∼ population

b) ã Å N ˙ N = rN 1 − K x=

N K

N (0) = N0

K x˙ = rKx(1 − x) x(0) = x˙ = rx(1 − x) x(0) = x0

τ = rt

N0 = x0 K

dx = rx(1 − x) x(0) = x0 dτ dx = x(1 − x) x(0) = x0 dτ

r

41

Chapter 3: Bifurcations

c)

u=

N N0

τ = rt κ=

N0 K

Å ã N N˙ = rN 1 − N (0) = N0 K Å ã N0 N (0) N0 u˙ = rN0 u 1 − u u(0) = =1 K N0 Å ã N0 1 u˙ = u 1 − u r K Å ã du N0 =u 1− u dτ K du = u (1 − κu) u(0) = 1 dτ

d) The x nondimensionalization would be useful for investigating the effect of the initial condition relative to carrying capacity on the solution, whereas the u nondimensionalization would be useful for studying the rate of population growth relative to carrying capacity.

By the way, scaling in such way that all solutions have the same initial condition is mathematically valid, but runs counter to the geometric approach we are using in this book. Our whole approach is based on visualizing different solutions emanating from different initial conditions. 3.6.1 x˙ = h + rx − x3 This x versus r bifurcation diagram. x

r

has a saddle-node bifurcation at a negative x value. Vertically shifting the cubic equation and then varying r will achieve this.

42

Chapter 3: Bifurcations



x˙ r=1 h=0

r=1 h = 0.25 x

x

So h is positive. 3.6.3 x˙ = rx + ax2 − x3 a) a = −1

x

a=0

x

r

a=1

r

x

r

b)

43

Chapter 3: Bifurcations

a

3 fi

xe

d

po

int s

de

transcritical

sa dd leno

1 fixed point

3 fixed points

r

3.6.5 a) The spring has spring constant k and rest length L0 but only the force from the spring that is parallel, and not perpendicular, with the wire will affect the bead. Äp

ä x x2 + a 2 − L0 √ x2 + a2 Å ã L0 = kx 1 − √ x2 + a2

k(L − L0 ) = k

The force of gravity will also move the bead along the wire, but only the fraction of the force that is parallel with the wire. mg sin(θ) At equilibrium, the force of the spring and the force from gravity have to be equal and opposite, which gives Å ã L0 mg sin(θ) = kx 1 − √ x2 + a2 b) ã Å L0 mg sin(θ) = kx 1 − √ x2 + a2 L0 mg sin(θ) = kx − kx √ 2 x + a2 L0 mg sin(θ) − kx = −kx » 2 a xa2 + 1 44

Chapter 3: Bifurcations

−mg sin(θ) L0 +1= » 2 kx a xa2 + 1 1−

mg sin(θ) L0 = » kx a 1+

x L0 R= a a R h 1− = √ u 1 + u2

u=

x2 a2

h=

mg sin(θ) ka

c) Both h and R are strictly positive quantities from the physical constants they’re made of. The following graphs show the different types and number of possible intersections. In each graph, h shapes the discontinuous graph and R shapes the bell shaped graph with u as the independent variable.

h=1 R = 0.5

u

45

Chapter 3: Bifurcations

h=1 R ≈ 2.8

u

h=1 R=5

u

For 0 < R < 1 there is only one intersection since the other curve asymptotes to height 1 from above. As R increases, there is a point when the two curves just touch, making another fixed point in a saddle-node bifurcation. As R increase further there are three fixed points. d)

1−

h R =√ u 1 + u2

r =R−1 1−

r+1 h =√ u 1 + u2 46

Chapter 3: Bifurcations

Taylor series expansion

p p u 1 + u2 − h 1 + u2 = (r + 1)u p p h 1 + u2 + (r + 1)u − u 1 + u2 = 0

Å ã Å ã   1 2 1 2 4 4 h 1+ u +O u + (r + 1)u − u 1 + u + O u =0 2 2  1 h h + u2 + ru + u − u − u3 = O u4 2 2  h 2 1 3 h + ru + u − u = O u4 2 2 h 1 h + ru + u2 − u3 ≈ 0 2 2 e) The saddle-node bifurcations will approximately occur at the local max of the approximate equation when the slopes of these equations are equal at an equal height. h 1 3 u − ru = h + u2 2 2

h=1 r = 1.95

u

h 1 3 u − ru = h + u2 2 2 Å Å ã ã d 1 3 h 3 2 d h + u2 u − ru = u − r = hu = du 2 2 du 2 3 2 u − hu = r 2 Å ã 1 3 3 2 1 3 h u − ru = u − u − hu u = h + u2 2 2 2 2 1 3 3 3 h u − u + hu2 = h + u2 2 2 2 47

Chapter 3: Bifurcations

h 2 u − h = u3 2 ã Å 1 2 u − 1 = u3 h 2 2u3 h= 2 u −2 u4 + 6u2 3 2 2u3 3 2 u= r = u − hu = u − 2 2 2 u −2 2(2 − u2 ) f) Here we use the same procedure as in part (e). The saddle-node bifurcations will occur at the local max of the equation when the slopes of these equations are equal at an equal height. R h =√ u 1 + u2 Å ã Å ã d R −Ru h h d √ 1− = 2 = 3 = du u u du 1 + u2 (1 + u2 ) 2 1−

3

h(1 + u2 ) 2 = −Ru3

3

−h(1 + u2 ) 2 u3 3 h R −h(1 + u2 ) 2 −h(1 + u2 ) √ √ 1− = = = u u3 1 + u2 u3 1 + u2 R=

u3 − hu2 = −h(1 + u2 ) h = −u3

3

3

3 −h(1 + u2 ) 2 −(−u3 )(1 + u2 ) 2 R= = = (1 + u2 ) 2 u3 u3

The approximate result and the exact result agree in the limit of u → 0. h = −u3 ≈

2u3 2u3 ≈ 2 0−2 u −2 3

3

R = (1 + u2 ) 2 ≈ (1 + 02 ) 2 ≈

u4 + 6u2 04 + 3(0)2 +1≈ +1=r+1 2 2(1 − 0 ) 2(2 − u2 )

g) h(u) = −u3

R(u) = 1 + u2

48

 32

⇒ r(u) = 1 + u2

 32

−1

Chapter 3: Bifurcations

h 1 fixed point

e od n dle ad

s

3 fixed points

r

h) 3 h(u) = −u3 R(u) = 1 + u2 2 x u= a Å  x 3  x 2 ã 32 L0 mg sin(θ) h=− = = R= 1+ a ka a a Solving for x and equating the two equations gives the conditions that the catastrophe will occur and the position x of its occurrence. Since many of the parameters are fixed, we could find the catastrophe angle θ in terms of the physical constants of the experiment.

49

Chapter 3: Bifurcations

3.6.7 a) h = T arctanh(m) − Jnm We can nondimensionalize to α + βm = arctanh(m)

α=

h T

β=

Jn >0 T

0 β





(ii) δK − α =

β β ⇒ h(K) = 2 2

p

p α δ

α δ 55

K

α+β δ

Chapter 3: Bifurcations

(iii) δK − α < 0 ⇒ h(K) < 0



p α+β δ c) We’ll assume α ≥ 0 and β > 0 due to being biological measurements, as well as δK > β from the problem statement. If we start α = 0 and increase α, we’ll get a saddle-node bifurcation at p ≈ K when the line h(p) touches the top corner of g(p). Then we get another saddle-node bifurcation when h(p) touches the bottom corner of g(p). α ≈ δK − β



α ≈ δK



p

p α δ

K

K

In summary, the fixed points are 0 < α < δK − β



α ≈ δK − β



δK − β < α < δK



α ≈ δK



δK < α



α stable δ α p ≈ , K stable, semistable δ α α+β p ≈ , K, stable, unstable, stable δ δ α+β semistable, stable p ≈ K, δ α+β p≈ stable δ p≈

and now we can plot the bifurcation diagram.

56

α+β δ

Chapter 3: Bifurcations

p

K

α δK − β

δK

d) Hysteresis is possible for this system, in the path shown through parameter space below for example. p

K

α δK − β

δK

Starting at α = 0, p = 0. The system stays at the lower fixed point until α increased past δK and the system jumps to the upper fixed point. The system exhibits hysteresis because the system does not jump back to the lower fixed point once α decreases past δK, despite the lower fixed point existing. Instead, we have to decrease α past δK − β in order for the system to jump back to the lower fixed point. Here is a similar path viewed on the phase line where we have combined the graph of both functions as g(p) − h(p) with a very large n value, K = 3, δ = 1, and β = 2.

57

Chapter 3: Bifurcations



p˙ α=

α=0

1 2

p

p



p˙ α=

α=1

3 2

p

p



p˙ α=

α=2

p

5 2

p

58

Chapter 3: Bifurcations



p˙ α=

α=3

7 2

p

p





α=

α=3

5 2

p

p





α=

α=2

3 2

p

p

59

Chapter 3: Bifurcations





α=

α=1

1 2

p

p

p˙ α=0

p

The systems exhibits hysteresis when p starts at the left fixed point, but moves to the right fixed point once 3 = δK < α, but then α must be decreased to α < δK − β = 1 instead of α < δK = 3 for p to go back to the original left fixed point.

60

Chapter 4 4.1.1 For what real values of a does the equation θ˙ = sin(aθ) give a well-defined vector field on the circle? The value of a must ensure that f (θ) = sin(aθ) is 2π periodic. f (θ) = f (θ + 2π)

4.1.3

 sin(aθ) = sin a(θ + 2π) = sin(aθ + 2πa) ⇒ 2πa = 2πz, z ∈ Z ⇒ a ∈ Z θ˙ = sin(2θ) π 3π θ˙ = 0 ⇒ sin(2θ) = 0 ⇒ θ = 0, , π, , 2π 2 2 θ˙ θ

4.1.5 θ˙ = sin(θ) + cos(θ) Å ã √ pi 3π 7π =0⇒θ= , θ˙ = 0 ⇒ sin(θ) + cos(θ) = 2 sin θ + 4 4 4 θ˙ θ

4.1.7 θ˙ = sin(kθ) where k is a positive integer. p θ˙ = 0 ⇒ sin(kθ) = 0 ⇒ θ = π , p = 0, 1, 2, ...2k k Example: k = 3 θ˙ θ

61

Chapter 4: Flows on the Circle

4.1.9 Exercise 2.6.2 and 2.7.7 do not carry over to vector fields on the circle because of the different interpretations of points. Put simply, the definition of periodic for vector fields on the line is fundamentally different from vector fields on the circle.

A vector field on the line is an interval with each value in the interval corresponding to a value of x that is interpreted as a unique point. Periodicity is defined as the particle returning to the same location after a time T , which can’t occur.

A vector field on a circle with interval [0, 2π) interprets θ(t) = 1 and θ(t + T ) = 1 + 2π as the same point. Periodicity is defined as the particle returning to the same location modulo 2π, which can occur. The particle isn’t really where it started, but unlike vector fields on the line, the two locations are considered identical. 4.2.1 Using common sense, the two bells will ring again in 12 seconds.

Using the method of Example 4.2.1 Å ã−1 Å ã ã Å 1 1 1 −1 1 −1 1 = = 12 = − − T1 T2 3 4 12 4.2.3 Using the methods of this section Å ã−1 ã Å ã Å 11 12 1 −1 1 −1 1 1 = = − hours − = T1 T2 1 12 12 11 So the hour and minutes hands will be aligned again 1 hour and 5 and 27 seconds, to the nearest second, later.

We also know that the hour and minute hands will be aligned again in 12 hours. During this time, the minute hand will pass the hour hand 11 times. The time between passes is equal so the minute hand will lap the hour hand every

1 11

of the perimeter, which is 5 minutes and 27 seconds, to the nearest second.

4.3.1 Z



√ √ dx x = r tan(θ) dx = r sec2 (θ)dθ 2 r + x −∞ √ Z π2 √ Z π2 r sec2 (θ) r sec2 (θ) = dθ = dθ 2 r + r tan (θ) r(1 + tan2 (θ)) −π −π 2 2 Z π2 Z π2 sec2 (θ) 1 √ √ dθ dθ = = 2 r sec (θ) r −π −π 2 2 π =√ r

Tbottleneck =

62

Chapter 4: Flows on the Circle

4.3.3 θ˙ = µ sin(θ) − sin(2θ) µ = −2 subcritical pitchfork bifurcation at θ = π µ = 2 subcritical pitchfork bifurcation at θ = 0

µ = −3

θ˙

θ

µ = −2

θ˙

θ

µ = −1

θ˙

θ

63

Chapter 4: Flows on the Circle

θ˙

µ=0

θ

θ˙

µ=1

θ

θ˙

µ=2

θ

θ˙

µ=3

θ

64

Chapter 4: Flows on the Circle

4.3.5 θ˙ = µ + cos(θ) + cos(2θ) µ = −2 saddle-node bifurcation at θ = 0 µ = 0 saddle-node bifurcation at θ = π 9 8

saddle-node bifurcation at »  5 θ = 2 arctan 3 »  5 θ = 2π − 2 arctan 3

µ=

µ = −3

θ˙

θ

µ = −2

θ˙

θ

µ = −1

θ˙

θ

65

Chapter 4: Flows on the Circle

θ˙

µ=0

θ

θ˙

µ=

9 16

θ

θ˙

µ=

9 8

θ

θ˙

µ=2

θ

66

Chapter 4: Flows on the Circle

4.3.7 θ˙ = µ = −1 saddle-node bifurcation at θ =

sin(θ) µ + sin(θ)

π 2

µ = 0 weird The function hugs closer and closer to the θ˙ = 1 line and the two vertical asymptotes, forming right angles there while simultaneously disappearing. µ = 1 saddle-node bifurcation at θ =

3π 2

Note: Some of the fixed points on the circles are not actually fixed points, but correspond to infinite discontinuities. However, the discontinuities act like stable or unstable fixed points since the point of discontinuity is either attracting or repelling respectively. µ = −2

θ˙

θ

θ˙

µ=

−3 2

θ

µ = −1

θ˙

θ

67

Chapter 4: Flows on the Circle

θ˙

µ=

−1 2

θ

θ˙

µ=0

θ

θ˙

µ=

θ

68

1 2

Chapter 4: Flows on the Circle

θ˙

µ=1

θ

θ˙

µ=

3 2

θ

θ˙

µ=2

θ

4.3.9 a) x˙ = r + x2

x = ra u t = rb τ

dx du ra du 2 = b = ra−b = r + x2 = r + (ra u) = r + r2a u2 dt r dτ dτ b) r2a u2 = ru2 ⇒ a = ra−b

1 2

du du 1 −1 =r ⇒a−b= −b=1⇒b= dτ dτ 2 2

69

Chapter 4: Flows on the Circle

4.4.1 mL2 θ¨ + bL2 θ˙ + mgL sin(θ) = Γ dθ d2 θ + bL2 + mgL sin(θ) = Γ dt2 dt 1 d2 θ 1 dθ d2 θ dθ = = t = Tτ dt T dτ dt2 T 2 dτ 2 2 2 2 mL d θ bL dθ + + mgL sin(θ) = Γ T 2 dτ 2 T dτ bL dθ Γ L d2 θ + + sin(θ) = 2 2 gT dτ mgT dτ mgL mL2

We want the

dθ dτ

to be O(1) so that all the terms except the

d2 θ dτ 2

term are of the same order.

bL bL =1⇒T = mgT mg m2 g d2 θ dθ Γ + + sin(θ) = b2 L dτ 2 dτ mgL Then we want the

d2 θ dτ 2

coefficient to be very small so the system relaxes quickly and we can neglect the

d2 θ dτ 2

term.

m2 g m2 g ≪ 1 ⇒ ≪ b2 b2 L L So as long as the above condition is satisfied, then the second order differential equation can be well approximated by the first order differential equation after an initial transient. 4.4.3 The length of the simulation increases in each graph in order to compare the oscillations. θ˙

γ = 100

t

The oscillations are almost indistinguishable due to the vertical scaling for 1 ≪ γ because the derivative is approximately 100 all the time. θ˙

γ=2

t

70

Chapter 4: Flows on the Circle

The oscillations look irregular as γ → 1 because the pendulum is slowing down significantly when going through the horizontal portion of the swing. θ˙

γ = 1.1

t

These oscillations continue to sharpen and lengthen as γ → 1 and the period will become infinity if γ = 1 because the pendulum never passes through the horizontal portion of the swing. θ˙

γ=1

t

θ˙

γ = 0.9

t

Now the torque isn’t strong enough to continue the oscillations, so the pendulum comes to rest where the torque balances the force of gravity.

71

Chapter 4: Flows on the Circle

θ˙

γ = 0.1

t

θ˙

γ=0

t

Now there is no torque and the pendulum settles to the bottom of the swing.

72

Chapter 4: Flows on the Circle

4.5.1 ϕ˙ = Ω − ω − Af (ϕ)

f (ϕ) =

a)

  ϕ

−π 2

 π − ϕ

π 2

≤ϕ≤

≤ϕ≤

π 2

3π 2

f (ϕ)

ϕ

b) Entrainment occurs when there are still fixed points of the ϕ˙ equation. The max and min of f (ϕ) are ± π2 so ϕ˙ = 0 = Ω − ω − Af (ϕ) Ω − ω = Af (ϕ) π −π A≤Ω−ω ≤ A 2 2 π |Ω − ω| ≤ A 2 c) The phase difference is simply the ϕ value of the stable fixed point. ϕ˙ = 0 = Ω − ω − Af (ϕ∗ ) = Ω − ω − A(π − ϕ∗ ) ϕ∗ =

ω−Ω +π A

|Ω − ω| ≤

π A 2

d) Tdrift

Z

Z 3π 2 dϕ dϕ = = −π Ω − ω − Af (ϕ) Ω − ω − Af (ϕ) 0 0 2 Z 3π Z π2 2 dϕ dϕ + = −π Ω − ω − Af (ϕ) π Ω − ω − Af (ϕ) 2 2 Z π2 Z 3π 2 dϕ dϕ = + −π Ω − ω − Aϕ π Ω − ω − A(π − ϕ) 2 2 ò π2 ò 3π ï ï   2 1 −1 ln Ω − ω − Aϕ ln Ω − ω − A(π − ϕ) + = −π π A A 2 2 ï ò π2 ò −π ï 2   −1 1 = ln Ω − ω − Aϕ ln Ω − ω − Aϕ + −π π A A Z



dϕ = ϕ˙



2

73

2

Chapter 4: Flows on the Circle

ò −π ï ò −π  2  2 1 1 + ln Ω − ω − Aϕ ln Ω − ω − Aϕ π π A A 2 2 ò −π ï 2  2 ln Ω − ω − Aϕ = π A 2 Å Å ã Å ãã 2 −π π = ln Ω − ω − A − ln Ω − ω − A A 2 2 Å Å ã Å ãã π π 2 ln Ω − ω + A − ln Ω − ω − A = A 2 2 ï

=

4.5.3 θ˙ = µ + sin(θ) a) θ˙ µ = 0.9

θ

The system is periodic, so the unique globally attracting “rest state” and “threshold” are the stable and unstable fixed point respectively. Also, a large enough stimulus will shift the system past the unstable fixed point “threshold” and will only return to the stable fixed point “rest state” after going through most of the 2π length periodicity. b)  cos θ(t)

t

 cos θ(t)

t

74

Chapter 4: Flows on the Circle

 cos θ(t)

t

 cos θ(t)

t

4.6.1 a) I ϕ˙ = − sin(ϕ) Ic I Ic

ϕ

t

75

= 1.1

Chapter 4: Flows on the Circle

I Ic

ϕ

= 10

t

The vertical axis in the second case has been scaled for a reasonable size. b)

ℏ ℏ V (t) = ϕ˙ = 2e 2e

Å

ã I − sin(ϕ) Ic I Ic

V (t)

= 1.1

ϕ

I Ic

V (t)

= 10

t

The vertical axis in the second case has been scaled for a reasonable size. 4.6.3 a) I − sin(ϕ) ϕ˙ = Ic dV I −I − = − sin(ϕ) ⇒ V (ϕ) = ϕ − cos(ϕ) dϕ Ic Ic 76

Chapter 4: Flows on the Circle

V (ϕ) ̸= V (ϕ + 2π) So V (ϕ) a not a single-valued function on the circle. b) I Ic

V (ϕ)

= 0.1

ϕ

I Ic

V (ϕ)

= 0.9

ϕ

I Ic

V (ϕ)

=1

ϕ

I Ic

V (ϕ)

ϕ

77

= 1.1

Chapter 4: Flows on the Circle

I Ic

V (ϕ)

= 10

ϕ

I Ic

V (ϕ)

= 1.1 ϕ

c) Increasing I brings the potential well and hill closer together, which eventually collide and annihilate. 4.6.5 a) Ib = Ia + IR b) The current running through the middle is Ia , which has to be the same current running through all sections because they are in series. The current in a section is split between the current through the Josephson junction Ic sin(ϕi ) and the current running through the resistor because they are in parallel. This gives Ia = Ic sin(ϕk ) + 78

Vk r

Chapter 4: Flows on the Circle

c) Ia = Ic sin(ϕk ) +

Vk r

Vk = rIa − rIc sin(ϕk )

I ℏ ˙ ϕ ϕ˙ = − sin(ϕ) 2e Ic ã Å ℏ Ia − sin(ϕk ) = ϕ˙k Vk = rIc Ic 2e

V =

d) Using Kirchhoff’s voltage law, the voltage from the top to the bottom of the circuit is the same going through the Josephson’s junctions and the resistor. N X

Vi = VR

i=1

N

ℏ X ˙ ϕi = IR R = (Ib − Ia ) R 2e i=1 N

ℏ X ˙ ϕi Ib = Ia + 2eR i=1 = Ic sin(ϕk ) +

N

Vk ℏ X ˙ ϕi + r 2eR i=1 N

= Ic sin(ϕk ) +

ℏ X ˙ ℏ ˙ ϕi ϕk + 2er 2eR i=1

e) N

Ib = Ic sin(ϕk ) + N Ib = Ic

N X

ℏ ˙ ℏ X ˙ ϕk + ϕi 2er 2eR i=1

sin(ϕi ) +

i=1

Ib =

Ic N

N X

sin(ϕi ) +

i=1

= Ic sin(ϕk ) +

N N Nℏ X ˙ ℏ X ˙ ϕi + ϕi 2er i=1 2eR i=1

N N ℏ X ˙ ℏ X ˙ ϕi + ϕi 2N er i=1 2eR i=1 N

ℏ ˙ ℏ X ˙ ϕi ϕk + 2er 2eR i=1

N N ℏ X ˙ ℏ ˙ Ic X sin(ϕi ) + ϕk ϕi = Ic sin(ϕk ) + N i=1 2N er i=1 2er

N N ℏ X ˙ Ic X ℏ ˙ sin(ϕi ) ϕk − ϕi = Ic sin(ϕk ) + 2N er i=1 2er N i=1

N N ℏ X ˙ Nℏ ˙ r X Nr sin(ϕi ) Ic sin(ϕk ) + ϕk − Ic ϕi = 2eR i=1 R 2eR R i=1

79

Chapter 4: Flows on the Circle

N

Ib = Ic sin(ϕk ) +

ℏ X ˙ ℏ ˙ ϕk + ϕi 2er 2eR i=1

N Nr Nℏ ˙ r X ℏ ˙ sin(ϕi ) ϕk + Ic sin(ϕk ) + ϕk − I c 2er R 2eR R i=1 ã Å ã Å 2 1 N r X Nr Ic sin(ϕk ) + ℏ + ϕ˙k − Ic sin(ϕj ) = 1+ R 2er 2eR R j=1

= Ic sin(ϕk ) +

Å

Å ã ã 2 1 Nr N r X ˙ ℏ ϕk = Ib − 1 + + Ic sin(ϕk ) + Ic sin(ϕj ) 2er 2eR R R j=1 2 R + Nr r X ℏ(R + N r) ˙ sin(ϕj ) ϕk = Ib − Ic sin(ϕk ) + Ic 2erR R R j=1

2 ℏ(R + N r) ˙ R + Nr 1 X RIb sin(ϕj ) − sin(ϕ ) + ϕ = k k 2N er2 Ic N rIc Nr N j=1 2 dϕk 1 X = Ω + a sin(ϕk ) + sin(ϕj ) dτ N j=1

Ω=

RIb N rIc

a=

−(R + N r) Nr

τ=

80

2N er2 Ic t ℏ(R + N r)

Chapter 5 5.1.1 Harmonic Oscillator v˙ = −ω 2 x

x˙ = v a)

Show that the orbits are given by ω 2 x2 + v 2 = C where C is any non-negative constant.

Divide the first equation by the second equation. v x˙ = ⇒ −ω 2 xx˙ = v v˙ ⇒ ω 2 xx˙ + v v˙ = 0 v˙ −ω 2 x Integrate

ω2 2 1 2 x + v = D ⇒ ω 2 x2 + v 2 = C 2 2

b) Show that this is equivalent to conservation of energy.

The harmonic oscillator with mass m and spring constant k stores a constant amount of energy as the sum of kinetic and potential energy.

Potential Energy PE =

Z

− kxdx =

−k 2 x +D 2

Kinetic Energy KE =

−m 2 v 2

Total Energy E = P E + KE =

−m 2 −2 2D k −k 2 x +D+ v ⇒ E+ = x2 + v 2 2 2 m m m

If we set C=

−2 2D E+ m m

ω2 =

k m

then we get the same equation. 5.1.3 x˙ = −y y˙ = −x Ñ é Ñ éÑ é x˙ 0 −1 x = y˙ −1 0 y

81

Chapter 5: Linear Systems

5.1.5 x˙ = 0 y˙ Ñ é Ñ x˙ 0 = y˙ 1

=x+y éÑ é 0 x 1

y

5.1.7 x˙ = x

y˙ = x + y y

x

82

Chapter 5: Linear Systems

5.1.9 x˙ = −y

y˙ = −x

a) y

x

b) xx˙ − y y˙ = x(−y) − y(−x) = −xy + xy = 0 Z Z xx˙ − y y˙ dt = 0 dt

x2 − y 2 = C c)

From the vector field, it looks like y = x and y = −x are the stable and unstable manifold respectively. We can check by plugging y = x and y = −x into the system. y = x gives x˙ = −x and y˙ = −y, which has a stable fixed point at the origin. y = −x gives x˙ = x and y˙ = y, which has an unstable fixed point at the origin. d) v =x−y

u=x+y

u˙ = x˙ + y˙ = −y − x = −u u˙ = −u ⇒ u(t) = Ce−t 83

Chapter 5: Linear Systems

u(0) = u0 ⇒ C = u0 u(t) = u0 e−t

v˙ = x˙ − y˙ = −y + x = v v˙ = v ⇒ v(t) = Cet v(0) = v0 ⇒ C = v0 v(t) = v0 et e) The stable and unstable manifold occur when u = 0 and v = 0 respectively. f) u+v u0 e−t + v0 et (x0 + y0 )e−t + (x0 − y0 )et = = 2 2 2 u0 e−t − v0 et (x0 + y0 )e−t − (x0 − y0 )et u−v = = y= 2 2 2

x=

5.1.11 a) x˙ = y

y˙ = −4x

x ¨ = y˙ = −4x x(t) = x0 cos(2t) +

Liapunov stable

y0 sin(2t) 2

y(t) = −2x0 sin(2t) + y0 cos(2t)   (x0 , y0 < δ ⇒ x(t), y(t) < 2δ = ϵ

b) x˙ = 2y

y˙ = x

x ¨ = 2y˙ = 2x Ä√ ä Ä√ ä √ 2t + 2y0 sinh 2t x = x0 cosh ä Ä ä Ä √ √ x0 y = √ sinh 2t + y0 cosh 2t 2 Trajectories go away from the origin None of the above c) x˙ = 0

y˙ = x 84

Chapter 5: Linear Systems

x = x0

y = x0 t + y0

Trajectories go away from the origin None of the above d) y˙ = −y

x˙ = 0

Liapunov stable

x = x0 y = y0 e−t    δ = x0 , y0 ≥ x(t), y(t) ≥ x0 , 0 for t ≥ 0

e) x˙ = −x

Asymptotically stable

y˙ = −5y

x = x0 e−t y = y0 e−5t   δ = x0 , y0 ≥ x(t), y(t) for t ≥ 0

f) x˙ = x

y˙ = y

x = x0 et

y = y0 et

Trajectories go away from the origin None of the above 5.1.13 Going back to multivariable calculus, a saddle point on a graph is neither a local minimum or local maximum, but the first partial derivatives of x and y both equal zero at the saddle point. The typical example

85

Chapter 5: Linear Systems

looks just like a horse saddle, and the level sets of graph make this pattern. The phase plane of a saddle point for an ODE looks the same, so we call these fixed points saddle points. 5.2.1 x˙ = 4x − y a)

Ñ é x˙

=

y˙ = 2x + y

Ñ éÑ é 4 −1 x



4 − λ 2

2 1 y −1 = (4 − λ)(1 − λ) + 2 = λ2 − 5λ + 6 = (λ − 2)(λ − 3) = 0 1 − λ λ1 = 2

Ñ 4 − λ1 2

Ñ 4 − λ2 2

é −1 1 − λ1 −1 1 − λ2

⃗v1 = é ⃗v2 =

λ2 = 3

Ñ é 2 −1 2 −1 Ñ é 1 −1 2 −2 86

⃗v1 = ⃗0



⃗v1 =

⃗v2 = ⃗0



⃗v2 =

Ñ é 1 2 Ñ é 1 1

Chapter 5: Linear Systems

b)

Ñ é x y

= c1 eλ1 t⃗v1 + c2 eλ2 t⃗v2 = c1 e

Ñ é 1 2t 2

+ c2 e

Ñ é 1 3t 1

c) Unstable node d)

é Ñ x(0) y(0)

=

Ñ é 3 4

= c1

Ñ é 1

Ñ é x

2 =e

+ c2

Ñ é 1

Ñ é 1 2t 2

y

1 +e



c1 = 1 and c2 = 2

Ñ é 2 3t 2

5.2.3 x˙ = y y˙ = −2x − 3y Ñ é Ñ éÑ é x˙ 0 1 x = y˙ −2 −3 y Ñ é Ñ é 1 1 λ1 = −2 ⃗v1 = λ2 = −1 ⃗v2 = −2 −1 Stable node y

x

87

Chapter 5: Linear Systems

5.2.5 x˙ = 3x − 4y y˙ = x − y Ñ é Ñ éÑ é x˙ 3 −4 x = y˙ 1 −1 y Ñ é 2 λ1 = 1 ⃗v1 = 1 Unstable degenerate node y

x

88

Chapter 5: Linear Systems

5.2.7 x˙ = 5x + 2y y˙ = −17x − 5y Ñ é Ñ éÑ é x˙ 5 2 x = y˙ −17 −5 y Ñ é Ñ é 5 + 3i 5 − 3i λ1 = 3i ⃗v1 = λ2 = −3i ⃗v2 = −17 −17 Center y

x

89

Chapter 5: Linear Systems

5.2.9 x˙ = 4x − 3y y˙ = 8x − 6y Ñ é Ñ éÑ é x˙ 4 −3 x = y˙ 8 −6 y Ñ é Ñ é 1 3 λ1 = −2 ⃗v1 = λ2 = 0 ⃗v2 = 2 4 Non-isolated fixed point y

x

5.2.11

Ñ é x˙

=

Ñ a



b

éÑ é x

0 a y a − λ b = (a − λ)2 = 0 ⇒ λ = a 0 a − λ é Ñ é é Ñ Ñ 1 0 b a−λ b ⃗v = ⃗0 ⇒ ⃗v = ⃗v = 0 0 0 0 a−λ y˙ = ay ⇒ y = c2 eat 90

Chapter 5: Linear Systems

x˙ = ax + by = ax + c2 beat x˙ − ax = c2 beat

d −at  e x = c2 b e−at x˙ − ae−at x = dt Z e−at x = c2 b dt = c1 + c2 bt x = c1 eat + c2 bteat

Ñ é x y

Ñ é Ñ Ñ é Ñ éé 1 b 0 = c1 eat + c2 eat t + 0 0 1

Stable degenerate node y

x

5.2.13 m¨ x + bx˙ + kx = 0

b>0

a) x˙ = y

−k b −k b y˙ = x ¨= x − x˙ = x− y m m m m éÑ é Ñ é Ñ x x˙ 0 1 = −b −k y y˙ m m

91

Chapter 5: Linear Systems

b)

0 − λ −k m

ã Å = −λ −b − λ + k = 1 (mλ2 + bλ + k) = 0 m m m − λ √ −b ± b2 − 4mk λ= 2m

1 −b m

If the inside of the square root is positive, then the origin is stable because the square root can’t exceed −b and both roots will be nonnegative. If the inside of the square root is negative, then the −b term still determines the stability. b > 0 ensures the origin is asymptotically stable. y

x

m = 21 , b = 32 , k = 1 ⇒ b2 − 4mk > 0 ⇒ Stable node

92

Chapter 5: Linear Systems

y

x

m = 1, b = 1, k = 1 ⇒ b2 − 4mk < 0 ⇒ Stable spiral

93

Chapter 5: Linear Systems

y

x

m = 1, b = 2, k = 1 ⇒ b2 − 4mk = 0 ⇒ Stable degenerate node c) Stable node ←→ overdamped Stable degenerate node ←→ critically damped Stable spiral ←→ underdamped 5.3.1 See the Answers to Selected Exercises Section in the textbook. 5.3.3

ab < 0 ⇒ Center

R˙ = aJ J˙ = bR Ñ é Ñ éÑ é 0 a R˙ R = ˙ J b 0 J 0 − λ a = λ2 − ab = 0 b 0 − λ

0 < ab ⇒ Saddle point

94

Chapter 5: Linear Systems

5.3.5

a < b ⇒ Saddle point

R˙ = aR + bJ J˙ = bR + aJ éÑ é Ñ é Ñ a b R˙ R = J˙ b a J a − λ b = (a − λ)2 − b2 = 0 ⇒ λ = a ± b b a − λ

a < b < 0 ⇒ Stable node 0 < b < a ⇒ Unstable node

95

Chapter 6 6.1.1 y˙ = 1 − ex

x˙ = x − y (x, y) = (0, 0) is the fixed point. y

x

96

Chapter 6: Phase Plane

6.1.3 x˙ = x(x − y)

y˙ = y(2x − y)

(x, y) = (0, 0) is the fixed point. y

x

97

Chapter 6: Phase Plane

6.1.5 x˙ = x(2 − x − y)

y˙ = x − y

(x, y) = (0, 0) and (1, 1) are the fixed points. y

x

98

Chapter 6: Phase Plane

6.1.7 x˙ = x + e− y = c2 e−t If we pick c1 = 2 and c1 =

e−2 −1 2

y˙ = −y

x = c1 et −

et−c2 e et−c2 + c2 c2

−t

then the growing exponentials destructively interfere with each other,

leaving the stable manifold. y = 2e−t

x=

ä et Ä −2e−t −1 e 2

y

x

99

Chapter 6: Phase Plane

6.1.9 Dipole fixed point y˙ = y 2 − x2

x˙ = 2xy (x, y) = (0, 0) is the fixed point. y

x

100

Chapter 6: Phase Plane

6.1.11 Parrot x˙ = y + y 2

6 1 y˙ = −x + y − xy + y 2 5 5

(x, y) = (0, 0) is the fixed point. y

x

101

Chapter 6: Phase Plane

6.1.13 y

x

6.2.1 Trajectories in a phase portrait never intersect, despite trajectories that approach a stable fixed point. Trajectories never actually reach the stable fixed point because they decay like an exponential function, never vanishing but always decreasing.

However, less well behaved ODEs such as having non-unique solutions can reach a fixed point in finite time.

102

Chapter 6: Phase Plane

6.3.1 y˙ = x2 − 4

x˙ = x − y (x, y) = (−2, −2) and (2, 2) are the fixed points. y

x

103

Chapter 6: Phase Plane

6.3.3 x˙ = 1 + y − e−x

y˙ = x3 − y

(x, y) = (0, 0) is the fixed point. y

x

104

Chapter 6: Phase Plane

6.3.5 x˙ = sin(y) (x, y) =

n1 −

1 2



 π, n2 π are the fixed points.

y˙ = cos(x)

y

x

6.3.7 See graphs for Exercises 6.3.1 through 6.3.6 6.3.9 x˙ = y 3 − 4x

y˙ = y 3 − y − 3x

a) (x, y) = (−2, −2), (0, 0), and (2, 2) are the fixed points. Ñ é −4 3y 2 A= −3 3y 2 − 1 é Ñ −4 12 ∆ = −8 ⇒ Saddle point A(−2,−2) = −3 11 é Ñ −4 0 ∆ = 4 τ = −5 ⇒ Unstable spiral A(0,0) = −3 −1 Ñ é −4 12 A(2,2) = ∆ = −8 ⇒ Saddle point −3 11 105

Chapter 6: Phase Plane

b) x = y ⇒ x˙ = x3 − 4x = y 3 − 4y = y˙ So as long as the initial condition satisfies x(0) = y(0) then x(t) = y(t) since the two variables and equations are indistinguishable. c) u=x−y u˙ = x˙ − y˙ = y 3 − 4x − (y 3 − y − 3x) = −x + y = −u u(t) = ce−t |x(t) − y(t)| = |u(t)| = ce−t lim |x(t) − y(t)| = lim ce−t → 0 t→∞

t→∞

This analysis implies u → 0 as t → ∞, meaning all trajectories approach the line x = y. d) Sketching the phase portrait with some accuracy around the fixed points will be a little difficult and require computing the eigenvectors. Some creativity and guessing of the stable and unstable manifolds will be required to fill in the phase portrait away from the fixed points. The main feature should be that all trajectories will approach the line x = y. e) y

x

106

Chapter 6: Phase Plane

The hand drawn phase portrait from part (d) might look similar to this numerically generated phase portrait in many ways, except the trajectories are not exponentially decaying to the line x = y. Instead, the trajectories seem to run parallel to the line x = y. This does not agree with part (c) at all.

Due to a very interesting and subtle assumption in the exercise, there is something wrong with the reasoning above. Finding u(t) this way assumes that x(t) and y(t) are defined for all time t, which isn’t true. Turns out the y 3 terms in the x˙ and y˙ equations cause x(t) and y(t) to go to infinity in finite time for most initial conditions. (Solve y˙ = y 3 and you’ll see why solutions blow up in finite time.) This is what is happening in the numerically generated phase portrait.

So in fact what happens is the trajectories that escape to infinity asymptote to lines of the form y = x + c, since y(t) − x(t) will decay exponentially to some constant c as t approaches the finite blow up time. 6.3.11 a) r˙ = −r θ˙ =

r(t) = r0 e−t

1 1 1 1 = = = −t −t ln(r) ln(r0 e ) ln(r0 ) + ln(e ) ln(r0 ) − t

θ(t) = − ln | ln(r0 ) − t| + ln(ln(r0 )) + θ0 Å ã ln(r0 ) = ln + θ0 | ln(r0 ) − t| b) lim r(t) = lim r0 e−t → 0 ⇒ Stable t→∞ Å ã ln(r0 ) lim |θ(t)| = lim ln + θ0 → ∞ ⇒ Spiral t→∞ t→∞ | ln(r0 ) − t| t→∞

c) r=

p x2 + y 2

θ = arctan

y

x p x x ˙ + y y ˙ dp 2 x + y2 = p r˙ = = −r = − x2 + y 2 dt x2 + y 2   d 2 y xy˙ − y x˙ 1 1 p θ˙ = arctan = 2 = = = 2 2 2 2 dt x x +y ln(r) ln(x + y 2 ) ln x +y p xx˙ + y y˙ p = − x2 + y 2 x2 + y 2 2 xy˙ − y x˙ = 2 2 2 x +y ln(x + y 2 )

xx˙ + y y˙ = −(x2 + y 2 ) xy˙ − y x˙ =

2(x2 + y 2 ) ln(x2 + y 2 )

x(xx˙ + y y) ˙ − y(xy˙ − y x) ˙ = (x2 + y 2 )x˙ = −x(x2 + y 2 ) − y 107

2(x2 + y 2 ) ln(x2 + y 2 )

Chapter 6: Phase Plane

x˙ = −x −

2y + y2 )

ln(x2

y(xx˙ + y y) ˙ + x(xy˙ − y x) ˙ = (x2 + y 2 )y˙ = −y(x2 + y 2 ) + x y˙ = −y +

2(x2 + y 2 ) ln(x2 + y 2 )

2x ln(x2 + y 2 )

d) á

4y 2 2 − 2 2 2 + y2 ) 2 2 2 ln(a (x + y ) ln (a + y )

4xy −1 2 2 (x + y ) ln2 (x2 + y 2 )

A=

4x2 −4xy 2 − 2 −1 2 2 + y ) (x + y ) ln2 (x2 + y 2 ) (x2 + y 2 ) ln2 (x2 + y 2 ) Ñ é −1 0 = x˙ = −x y˙ = −y ⇒ Stable star 0 −1 ln(x2

A(0,0) 6.3.13

x˙ = −y − x3 r 2 = x2 + y 2 r˙ =

y˙ = x

2rr˙ = 2xx˙ + 2y y˙

xx˙ + y y˙ −x4 x(−y − x3 ) + yx p =p < 0 for (x, y) ̸= (0, 0) = r x2 + y 2 x2 + y 2 A=

A(0,0) =

Ñ −3x2

é −1

1 0 é Ñ 0 −1 1

0

∆ > 0 τ = 0 ⇒ Center

108

ë

Chapter 6: Phase Plane

6.3.15 θ˙ = 1 − cos(θ)

r˙ = r(1 − r2 ) (r, θ) = (0, 0) and (1, 0) are the fixed points. y

x

(1,0) is attracting, but all trajectories except those starting on the positive x-axis have to traverse near the unit circle before reaching the fixed point.

109

Chapter 6: Phase Plane

6.3.17 x˙ = xy − x2 y + y 3

y˙ = y 2 + x3 − xy 2

(x, y) = (0, 0) is the fixed point. y

x

110

Chapter 6: Phase Plane

6.4.1 x˙ = x(3 − x − y)

y˙ = y(2 − x − y)

(x, y) = (0, 0), (0, 2), and (3, 0) are the fixed points. y

x

111

Chapter 6: Phase Plane

6.4.3 x˙ = x(3 − 2x − 2y)

y˙ = y(2 − x − y)

(x, y) = (0, 0), (0, 2), and ( 32 , 0) are the fixed points. y

x

6.4.5

Å ã N1 N˙ 1 = r1 N1 1 − − b1 N1 N2 K1 Å ã N1 dN1 = r1 N1 1 − − b1 N1 N2 dt K1 t = uτ

N˙ 2 = r2 N2 − b2 N1 N2 dN2 = r2 N2 − b2 N1 N2 dt

N1 = vx N2 = wy Å ã v dx vx w dy = r1 vx 1 − = r2 wy − b2 vwxy − b1 vwxy u dτ K1 u dτ ã Å b1 w vx 1 dy r2 1 dx − =x 1− xy = y − xy r1 u dτ K1 r1 b2 uv dτ b2 v 1 1 b1 w r1 r1 r1 1 =1⇒u= =1⇒w= = =1⇒v= r1 u r1 r b1 b2 uv b2 v b2 Å ã 1 r1 x r2 dy dx =x 1− = y − xy − xy dτ b2 K1 dτ r1 r2 r1 ρ= κ= r1 b2 K1 112

Chapter 6: Phase Plane

dx = x(1 − κx − y) dτ ρ=3 κ=

dy = y(ρ − x) dτ

1 2

(x, y) = (0, 0), (0, 1), (2, 0), and (3, 0) are the fixed points. y

x

ρ=3 κ=

1 3

(x, y) = (0, 0), (0, 1), and (3, 0) are the fixed points. y

x

ρ=3 κ=

1 4

 (x, y) = (0, 0), (0, 1), (3, 0), 3, 41 , and (4, 0) are the fixed points.

113

Chapter 6: Phase Plane

y

x

The top and bottom graphs have a different number of fixed points, with the middle graph being the borderline case. Except by starting with an initial condition of zero population, the top two graphs grow infinitely in y and x decays to zero. Depending on the initial condition, the bottom graph can grow infinitely in y and x decays to zero, but it can also go towards a stable fixed point where x becomes finite and y decays to zero. 6.4.7 n˙1 = G1 (N0 − α1 n1 − α2 n2 )n1 − K1 n1

n˙2 = G2 (N0 − α1 n1 − α2 n2 )n2 − K2 n2

a) A=

A(0,0) =

é

Ñ G1 (N0 − 2α1 n1 − α2 n2 ) − K1 −G2 α1 n1 Ñ G1 N0 − K1 0 0

λ1 = G1 N0 − K1

é

−G1 α2 n2 G2 (N0 − α1 n1 − 2α2 n2 ) − K2

G2 N0 − K2 λ2 = G2 N0 − K2

So the origin in unstable if G1 N0 − K1 > 0 or G2 N0 − K2 > 0 b)

Ä ä Ä ä G1 N0 −K1 2 The other two fixed points that exist are 0, G2αN20G−K and , 0 α1 G1 2 é Ñ G1 K2 G1 K 2 − K −G N + 1 1 0 G2 G2 AÄ0, G2 N0 −K2 ä = α 2 G2 0 −G2 N0 + K2 é Ñ −G N + K 0 1 0 1 AÄ G1 N0 −K1 ,0ä = G2 K 1 G2 K 1 α1 G1 −G2 N0 + G1 G1 − K2 Since these are triangular matrices, the eigenvalues are on the diagonals. Notice that two of the eigenvalues G1 K2 G2

− K1 and

G2 K 1 G1

− K2 will have opposite sign or both zero, meaning at most one of the fixed points can

be stable.

114

Chapter 6: Phase Plane

The other pair of eigenvalues doesn’t have that relationship, and can be positive, negative, or zero depending on how big N0 is.

If you look closely, you’ll see that the eigenvalues of a fixed point and its coordinate are somewhat simply related. This relationship makes for some interesting graphs. c) N0 = 5 G1 = 1 G2 = 1 α1 = 1 α2 = 1 K1 = 7.5 K2 = 7.5 y

x

N0 = 5 G1 = 1 G2 = 1 α1 = 1 α2 = 1 K1 = 7.5 K2 = 2.5

115

Chapter 6: Phase Plane

y

x

N0 = 5 G1 = 1 G2 = 1 a1 = 1 a2 = 1 K1 = 2.5 K2 = 7.5 y

x

N0 = 5 G1 = 1 G2 = 1 α1 = 1 α2 = 1 K1 = 2.5 K2 = 2.5

116

Chapter 6: Phase Plane

y

x

There are four qualitatively different phase portraits. n1 and n2 decay to the origin, n1 axis, n2 axis, or an infinite number of fixed points. No other phase portraits are possible because the nullclines are the axes and parallel lines. 6.4.9 I˙ = I − αC a) (x, y) =

Ä

αG G α−1 , α−1

ä

C˙ = β(I − C − G)

I, C, G ≥ 0,

1 < α < ∞,

1≤β 1 then the real part is negative making the fixed point stable. The sign of the expression inside the square root determines the type of stable behavior. α> α= α
kc and β > 1 then λ2 < 0 < λ1 Ñ v1 =

β+λ1 β(1−k)

é

Ñ v2 =

1

β+λ2 β(1−k)

é

1

The attracting eigendirection is v1 which has positive slope and is present in the positive quadrant. The economy will follow this line, with I and C always increasing. c)

Å

G = G0 + kI 2 ⇒ (x, y) = G0 < G0 = G0 >

(α−1)2 4α2 k (α−1)2 4α2 k (α−1)2 4α2 k



α−1±



(α−1)2 −4α2 G0 k α−1± , 2αk

(α−1)2 −4α2 G0 k 2α2 k

ã

then there are two fixed points in the positive quadrant. then there is only one fixed point in the positive quadrant. then there are no real fixed points.

α = 2 β = 2 k = 0.05 G0 = 0.75

118

are the fixed points.

Chapter 6: Phase Plane

C

I

α = 2 β = 2 k = 0.01 G0 = 1.25

119

Chapter 6: Phase Plane

C

I

α = 2 β = 2 k = 0.01 G0 = 1.75

120

Chapter 6: Phase Plane

C

I

In all cases, the trajectories can result in the national income I going negative, but the amount of consumer spending C is still positive. Also, it looks like the government spending lots of money is a good thing. Too little spending and the economy can get stuck at a fixed point, but a lot of government spending with the right initial condition will result in the national income growing continually with consumer spending not increasing much. This is the the unstable manifold of the saddle point. 6.4.11 x˙ = rxz

z˙ = −rxz − ryz

y˙ = ryz

a) x+y+z =1

(The total sum of leftists, centrists, and rightists should add to 1)

x˙ + y˙ + z˙ = rxz + ryz − rxz − ryz = 0 So the sum never changes, also known as invariant. b) (x, y, z) = (x, y, 0) and (0, 0, z) are the fixed points. á rz 0 rx A=

ë

0

rz

ry

−rz

−rz

−rx − ry 121

Chapter 6: Phase Plane

A(x,

,0)

=

ë

á 0

0

rx

0

0

ry

λ1 = 0 λ2 = 0 λ3 = −rx − ry

0 0 −rx − ry ë á rz 0 0 A(0,0,z) =

0

rz

0

−rz

−rz

0

λ1 = rz

λ2 = rz

λ3 = 0

c) For r > 0, a purely centrist government z = 1 is unstable. If there are any leftists or rightists, then the centrists are converted to each side with the limiting number of leftists and rightists determined by the initial values. For r < 0, a purely centrist government z = 1 is stable. If there are no centrists z = 0 then the leftists and rightists will stay at their initial value because neither side can ever convert opposite members.

122

Chapter 6: Phase Plane

6.5.1 x ¨ = x3 − x a) x˙ = y

y˙ = x3 − x

(x, y) = (−1, 0), (0, 0), and (1, 0) are the fixed points. b)

E=

1 2 x˙ + 2

Z

−(x3 − x) dx =

1 2 1 2 1 4 x˙ + x − x 2 2 4

c) y

x

123

Chapter 6: Phase Plane

6.5.3

E=

1 2 x˙ + 2

Z

x ¨ = a − ex −(a − ex ) dx =

x˙ = y

1 2 x˙ − ax + ex 2

y˙ = a − ex

a0

125

Chapter 6: Phase Plane

y

x

6.5.5 x ¨ = (x − a)(x2 − a)

√ (x, x) ˙ = (a, 0) and (± a, 0) are the fixed points.

y˙ = (x − a)(x2 − a)

x˙ = y Ñ A=

A(a,0) =

A(√a,0) =

A(−√a,0) =

é 1

0 −2ax − a + 3x2 é Ñ 0 1 a2 − a 0 Ñ 0

é 1

3

−2a 2 + 2a 0 é Ñ 0 1 3

2a 2 + 2a

0

0 p λ1,2 = ± a2 − a

» 3 λ1,2 = ± −2a 2 + 2a

» 3 λ1,2 = ± 2a 2 + 2a

The eigenvalues tell us that the fixed points, if they are real, are saddle points. The only ones to be unsure about are when a = −1, 0, 1 which causes some eigenvalues to be zero.

126

Chapter 6: Phase Plane

We can determine what happens at these troublesome values of a by finding a conserved quantity E and classifying the critical points. Z 1 1 E = x˙ 2 − (x − a)(x2 − a) dx = y 2 − 2 2 Exx Ex −3x2 + 2ax − a = E x E y − x3 + ax2 − ax + a2 a = −1 Exx Ex E x E a=0 Exx E x

Ex E

a=1 Exx E x Exx E x

(1,0)

(0,0)

1 4 a 3 a 2 x + x − x + a2 x 4 3 2 3 2 2 y − x + ax − ax + a 1

−4 0 = −4 < 0 ⇒ Saddle point = 0 1 0 0 = 0 < 0 ⇒ Inconclusive = 0 1

−2 Ex = 0 E (1,0) −6 Ex = 4 E (−1,0)

0 = −2 < 0 ⇒ Saddle point 1 4 = −22 < 0 ⇒ Saddle point 1

So everything was a saddle point, except possibly (x, y) = (0, 0) when a = 0, but we can figure out by looking at the conserved quantity. E=

1 2 1 4 y − x 2 4

which is most definitely a saddle since the x-axis cross section is concave down and the y-axis cross section is concave up. 6.5.7

d2 u + u = α + ϵu2 dθ2

u=

a) du =v dθ dv = α + ϵu2 − u dθ b) (u, v) =

Ä

√ ä 1+ 1−4αϵ ,0 2ϵ

and

Ä

√ ä 1− 1−4αϵ ,0 2ϵ

are the fixed points.

c) é

Ñ A=

0

1

2ϵu − 1

0 127

1 r

Chapter 6: Phase Plane

Ñ AÄ 1+√1−4αϵ ,0ä = 2ϵ

√ ∆ = − 1 − 4αϵ AÄ 1−√1−4αϵ ,0ä 2ϵ

∆=



é 0

1

1 − 4αϵ 0

τ = 0 ⇒ Saddle point Ñ é 0 1 = √ − 1 − 4αϵ 0

√ 1 − 4αϵ

τ = 0 ⇒ Linear Center

This is also a nonlinear center by Theorem 6.5.1 with the only difficult condition to check is that the fixed point is a local minimum of a conserved quantity. Z 1 2 1 ϵ 1 E = x˙ + −(α + ϵx2 − x) dx = x˙ 2 − αx − x3 + x2 2 2 3 2 ∇E = ⟨α + ϵx2 − x, x⟩ ˙ Ç å √ 1 − 1 − 4αϵ ∇E , 0 = (0, 0) 2ϵ Exx Exx ˙

Exx˙ −2ϵx + 1 x˙ − α − ϵx2 + x = Ex˙ x˙ x˙ − α − ϵx2 + x 1 = −2ϵx + 1 − (x˙ − α − ϵx2 + x)2

∇E = (0, 0) at the fixed fixed point. Both Exx , Ex˙ x˙ > 0 at the fixed point for sufficiently small ϵ, as well as the evaluation of the Jacobian at the fixed point, Ç å √ Exx Exx˙ √ 1 − 4αϵ 1 − = 1 − 4αϵ > 0 ⇒ , 0 is a local minimum. 2ϵ Exx Ex˙ x˙ Ä 1−√1−4αϵ ä ˙ 2ϵ

,0

d)

If the fixed point corresponds to a circular orbit then the radius is constant. √ 1 − 1 − 4αϵ 2ϵ 1 √ =u= ⇒r= r 2ϵ 1 − 1 − 4αϵ which is a constant. 6.5.9 If H(x, p) is a conserved quantity, then H is constant in time. ∂H ∂H d H(x, p) = x˙ + p˙ = p˙x˙ − x˙ p˙ = 0 dt ∂x ∂p So H˙ = 0 and H is a conserved quantity. 6.5.11 x˙ = y

y˙ = −by + x − x3 128

00 = = Exx x˙ + x + ϵx3 0 1 Ex˙ x˙ 1 ˙ (0,0)

(0,0)

So along with Exx (0, 0) = 1 > 0 means the fixed point is a local minimum and therefore a nonlinear center of the system. b) The origin will still be a nonlinear center for any value of ϵ because the origin is always a local minimum of E and therefore will be surrounded by closed trajectories. However, we don’t know how far away those closed trajectories exist for any value of ϵ, or even if the value of ϵ matters at all.

To find out, we can look at the level sets of E for arbitrary ϵ. E=

1 2 1 2 ϵ 4 x˙ + x + x 2 2 4

4Eϵ = 2ϵx˙ 2 + 2ϵx2 + ϵ2 x4 = 2ϵx˙ 2 + (ϵ2 x4 + 2ϵx2 + 1) − 1 = 2ϵx˙ 2 + (ϵx2 + 1)2 − 1 4Eϵ + 1 = 2ϵx˙ 2 + (ϵx2 + 1)2 This equation looks a lot like, but not quite due to the x exponent, the equations for the conic sections (hyperbola, parabola, ellipse, circle). Even more interesting is that by varying ϵ we can get all the different conic-like sections as level sets. ϵ > 0 ⇒ ellipse like trajectories ϵ = 0 ⇒ (x, x) ˙ = (0, 0) −1 4E

< ϵ < 0 ⇒ hyperbola like trajectories ä Ä ä Ä √ −1 ⇒ (x, x) ˙ = √±1 ϵ = 4E , 0 = ±2 E, 0 −ϵ

ϵ
0 and all trajectories loop around the origin no matter how far away we are.

130

Chapter 6: Phase Plane



x

ϵ 0 leading into a stable node or spiral, we know that the trajectory has to always travel towards the fixed point sufficiently close to the fixed point. However, the twin trajectory with y < 0 will necessarily travel away from the fixed point because the arrows are reversed. This is a contradiction because the trajectory has to travel toward the fixed point whether or not y is positive or negative.

So a reversible system will never contain stable nodes or spirals. 6.6.7 x ¨ + xx˙ + x = 0 This system’s symmetry is across the x-axis. ˙ As a consequence, we’ll check that f (x, y) and g(x, y) are even and odd in x respectively. x˙ = y = f (x, y)

f (−x, y) = y = f (x, y)

y˙ = −xy − x = g(x, y)

g(−x, y) = xy + x = −g(x, y)

140

Chapter 6: Phase Plane

y

x

6.6.9

N dϕk 1 X sin(ϕj ) = Ω + a sin(ϕk ) + dτ N j=1

for k = 1, 2

a) θ k = ϕk −

π 2

N   π π 1 X dϕk sin θj + = Ω + a sin θk + + dτ 2 N j=1 2

N dθk 1 X cos(θj ) = Ω + a cos(θk ) + dτ N j=1

Applying the transformation θk → −θk

τ → −τ

N 1 X −1 dθk cos(−θj ) = Ω + a cos(−θk ) + −1 dτ N j=1 N dθk 1 X cos(θj ) = Ω + a cos(θk ) + dτ N j=1

results in the same equations, so the system is reversible.

141

Chapter 6: Phase Plane

b) N dθk 1 X cos(θj ) = 0 = Ω + a cos(θk ) + dτ N j=1

Ω + a cos(θk ) +

 1 cos(θ1 ) + cos(θ2 ) = 0 2

2a cos(θk ) + cos(θ1 ) + cos(θ2 ) = −2Ω (2a + 1) cos(θ1 ) + cos(θ2 ) = −2Ω

cos(θ1 ) + (2a + 1) cos(θ2 ) = −2Ω (2a + 1)2 cos(θ1 ) + (2a + 1) cos(θ2 ) = −2(2a + 1)Ω − cos(θ1 ) − (2a + 1) cos(θ2 ) = 2Ω  (2a + 1)2 − 1 cos(θ1 ) = 2Ω − 2(2a + 1)Ω

(4a2 + 4a + 1 − 1) cos(θ1 ) = (2 − 4a − 2)Ω 4a(a + 1) cos(θ1 ) = −4aΩ cos(θ1 ) =

−Ω a+1

Ω 1 and zero solutions if a + 1 (2a + 1) cos(θ1 ) + cos(θ2 ) = −2Ω cos(θ1 ) + (2a + 1) cos(θ2 ) = −2Ω − cos(θ2 ) − (2a + 1) cos(θ1 ) = 2Ω (2a + 1)2 cos(θ2 ) + (2a + 1) cos(θ1 ) = −2(2a + 1)Ω .. .

cos(θ2 ) =

−Ω a+1

which has the same conditions on the number of solutions. Ω 1 There are zero (θ1 , θ2 ) fixed points if a + 1 c)

Ω=0 θ2

θ1

Ω=1

143

Chapter 6: Phase Plane

θ2

θ1

Ω = 1.75

144

Chapter 6: Phase Plane

θ2

θ1

Ω=2

145

Chapter 6: Phase Plane

θ2

θ1

Ω = 2.5

146

Chapter 6: Phase Plane

θ2

θ1

Ω=3

147

Chapter 6: Phase Plane

θ2

θ1

6.6.11 θ˙ = cot(ϕ) cos(θ) a) t → −t

t → −t

 ϕ˙ = cos2 (ϕ) + A sin2 (ϕ) sin(θ)

θ → −θ −1 dθ = cot(ϕ) cos(−θ) −1 dt dθ = cot(ϕ) cos(−θ) dt

 1 dϕ = cos2 (ϕ) + A sin2 (ϕ) sin(−θ) −1 dt   dϕ − = cos2 (ϕ) + A sin2 (ϕ) − sin(θ) dt  dϕ = cos2 (ϕ) + A sin2 (ϕ) sin(θ) dt

1 dθ = cot(−ϕ) cos(θ) −1 dt

 −1 dϕ = cos2 (−ϕ) + A sin2 (−ϕ) sin(θ) −1 dt

ϕ → −ϕ

148

Chapter 6: Phase Plane



b)

dθ = − cot(ϕ) cos(θ) dt dθ = cot(ϕ) cos(θ) dt

2  dϕ = cos2 (ϕ) + A − sin(ϕ) sin(θ) dt  dϕ = cos2 (ϕ) + A sin2 (ϕ) sin(θ) dt

A = −1 φ

θ

A=0 149

Chapter 6: Phase Plane

φ

θ

A=1

150

Chapter 6: Phase Plane

φ

θ

c) From the phase portraits, it seems that almost all trajectories approach one of two stable fixed points when A = −1 or A = 0. This means that the object settles down to a fixed orientation in the shear flow. In contrast, when A = 1, the trajectories trace out neutrally stable periodic orbits starting from almost all initial conditions. This means the object tumbles periodically in the shear flow without ever converging to a fixed orientation. 151

Chapter 6: Phase Plane

6.7.1 θ¨ + bθ˙ + sin θ = 0

b>0

y˙ = − sin(x) − by

x˙ = y (x, y) = (zπ, 0) are the fixed points, z ∈ Z.

é

Ñ 0

A=

1

− cos(x) −b

Ñ

é 0

A(2zπ,0) =

1

−1 −b

∆ = 1, τ = −b < 0 ⇒ stable fixed point √ −b ± b2 − 4 λ1,2 = 2 0 < b < 2 ⇒ stable spiral b = 2 ⇒ stable degenerate node 2 < b ⇒ stable node

A

=

Ñ 0

(2z+1)π,0

é 1

1 −b

∆ = −1, τ = −b < 0 ⇒ saddle point b=1 y

x

b=2

152

Chapter 6: Phase Plane

y

x

b=4 y

x

6.7.3

 θ¨ + 1 + a cos(θ) θ˙ + sin θ = 0 x˙ = y

(x, y) = (zπ, 0) are the fixed points, z ∈ Z. Ñ A=

a≥0

 y˙ = − sin(x) − 1 + a cos(x) y

é 0

1

− cos(x) + ay sin(x) −1 − a cos(x) é

Ñ A(2zπ,0) =

0

1

−1 −1 − a 153

Chapter 6: Phase Plane

∆ = 1, τ = −1 − a < 0 ⇒ stable fixed point

A

=

Ñ 0

(2z+1)π,0

é 1

1 a−1

∆ = −1 ⇒ saddle point So as expected, the low point of the swing is stable and the very top of the swing is unstable for this damped pendulum. 6.7.5 α

2π 1 +

1 2 16 α



Quadrature

0 18 π



1 18 π

6.29514

6.29517

2 18 π

6.3310

6.3313

3 18 π

6.3908

6.3925

4 18 π

6.4746

6.4801

5 18 π

6.5822

6.5960

6 18 π

6.7138

6.7430

7 18 π

6.8693

6.9250

8 18 π

7.0488

7.1471

9 18 π

7.2521

7.4163

10 18 π

7.4794

7.7423

11 18 π

7.7306

8.1389

12 18 π

8.0058

8.6261

13 18 π

8.3048

9.2352

14 18 π

8.6278

10.0182

15 18 π

8.9747

11.0723

16 18 π

9.3455

12.6135

17 18 π

9.7403

15.3270

18 18 π

10.1590

82.0530

6.8.1 a) Stable Spiral

154



Chapter 6: Phase Plane

y 3 4

2

1,9

8

2 1,9

5

7

x 3

6 4

6

5

8 7

b) Unstable Spiral y 3 4

2 5

4

6 1,9

5

3

x 7

2 8

6

8 7

c) Center

155

1,9

Chapter 6: Phase Plane

y 3 4

2 5 6 1,9

5

x

7

3 8

6

4

1,9

2

8 7

d) Star y 3 2 4

3 4 1,9

x

2 1,9

5

5 6 8 6 7

e) Degenerate Node

156

8 7

Chapter 6: Phase Plane

y

x

The vector field on this circle is pretty finicky near the point when the trajectory turns around and starts going the opposite direction. Due to this, we’ll be graphing the path of the vector heads instead of the individual vectors themselves since they really bunch around θ = π and the side directly opposite. y

x

And the degenerate node has an index of +1 as expected. 6.8.3 x˙ = y − x

157

y˙ = x2

Chapter 6: Phase Plane

y

x

Despite looking complicated, we can see this curve has an index of 0 because the arrows never point down so can’t make a full rotation. 6.8.5 x˙ = xy

y˙ = x + y

y 5

4

3

1,4,13

2,3

6 5,12

2 1,13

7

12 8 9

10

x 6,11 7,10

11

8,9

The arrows almost make a full counterclockwise rotation but then turn around, so the index is 0. 6.8.7 x˙ = x(4 − y − x2 )

y˙ = y(x − 1)

(x, y) = (0, 0), (2, 0), (1, 3), and (−2, 0) are the fixed points. Ñ é 4 − y − 3x2 −x A= y x−1

A(0,0) =

Ñ 4

é 0

0 −1 158

Chapter 6: Phase Plane

λ1 = 4 λ2 = −1 ⇒ Saddle Point Ñ é Ñ é 1 0 ⃗v1 = ⃗v2 = 0 1

A(2,0) =

Ñ é −8 −2 0

1

λ1 = −8 λ2 = 1 ⇒ Saddle Point Ñ é Ñ é 1 −2 ⃗v1 = ⃗v2 = 0 9

A(1,3) = λ1,2

é Ñ −2 −1

3 0 √ = −1 ± 2i ⇒ Stable Spiral

A(−2,0) =

Ñ −8

é 2 −3

0

λ1 = −8 λ2 = −3 ⇒ Stable Node Also notice from the equations that x(0) = 0 ⇒ x(t) = 0 and y(0) = 0 ⇒ y(t) = 0. This means the x-axis and y-axis are invariant. This means a closed orbit can’t cross the axes.

A closed orbit must enclose a fixed point, and the only candidate is the stable spiral. However, this is also impossible since the unstable manifold of the saddle point goes into the stable spiral. We didn’t actually prove this, but use numerical results as evidence. y

x

An intuitive argument is the upper branch of the unstable manifold of the saddle point (2,0) must approach the spiral. It has to go somewhere. If it does not go to the stable spiral, where else could it go? 159

Chapter 6: Phase Plane

i) Both the x or y axis are invariant, so the unstable manifold can’t get out of the first quadrant. Said another way, both of the axes contain trajectories and trajectories cannot cross by the uniqueness theorem for solutions of ODEs.

ii) The unstable manifold can’t approach the saddle point at (0,0) since the only trajectories that do lie on the y-axis, which is the stable manifold of the saddle point at (0,0).

iii) Conceivably the upper branch of the unstable manifold could escape to infinity. However, the flow in the first quadrant pushes the unstable manifold upward and to the left until it crosses the vertical line x = 1. Then the unstable manifold moves left because x˙ < 0 when x lies to the right of the nullcline given the parabola 4 − y − x2 = 0 and moves upward because y˙ > 0 when x > 1, which is the case initially since the unstable manifold starts at (2,0). Once the manifold crosses the line x = 1 it has to move downward. And after that, there’s no way for the trajectory to escape out to infinity.

Clearly this is only a plausibility argument, not a proof. To really prove that the unstable manifold cannot escape to infinity, one could construct a trapping region, a technique discussed in Section 7.3. 6.8.9 False. A counter example is r˙ = r(r − 1)(r − 3) θ˙ = r2 − 4 There is only one fixed point at the origin. The inner closed orbit is attracting and θ˙ < 0, so rotation is in the clockwise direction. The outer closed orbit is repelling with θ˙ > 0, so rotation is in the counterclockwise direction. 6.8.11 a) k=1 z˙ = z = x + iy = reiθ

r=

z˙ = z¯ = x − iy = re−iθ

p x2 + y 2

tan(θ) =

k=2 z˙ = z 2 = (x + iy)2 = x2 − y 2 + 2ixy 2 = reiθ = r2 e2iθ

z˙ = (¯ z )2 = (x − iy)2 = x2 − y 2 − 2ixy 160

y x

Chapter 6: Phase Plane

= re−iθ k=3

2

= r2 e−2iθ

z˙ = z 3 = (x + iy)3 = x3 − 3xy 2 + i(3x2 y − y 3 ) 3 = reiθ = r3 e3iθ

z˙ = (¯ z )3 = (x − iy)3 = x3 + i(y 3 − 3x2 y) − 3xy 2 3 = re−iθ = r3 e−3iθ

bc)

It’s not so easy to see that the origin is the only stable fixed point in the Cartesian coordinates, but in the polar coordinates it’s clear that the only way to satisfy z˙ = 0 is r = 0 in both cases, which is the origin. The index, also not easy to see from the Cartesian coordinates, is ±k in the z k and (¯ z )k case respectively. The arrows will do k rotations on a loop enclosing the origin when θ goes through a 2π rotation due to the k in e±kiθ with (+) causing counterclockwise and (−) causing clockwise rotations. 6.8.13 a) Å ã Å ã y˙ g(x, y) = arctan x˙ f (x, y) ã Å ã Å d g(x, ) d g d 1 g(x, y) d dt f (x, ) = = ϕ= arctan Ä g(x, ) ä2 2 dt dt f (x, y) dt f 1 + fg 2 1 + f (x, ) ϕ = arctan

dϕ =

f dg − gdf f dg − gdf 1 2 = 2 g f f 2 + g2 1 + f2

b) IC =

1 2π

I

C

dϕ =

1 2π

161

I

C

f dg − gdf f 2 + g2

Chapter 7 7.1.1 r˙ = r3 − 4r

θ˙ = 1

y

x

162

Chapter 7: Limit Cycles

7.1.3 r˙ = r(1 − r2 )(4 − r2 )

θ˙ = 2 − r2

y

x

163

Chapter 7: Limit Cycles

7.1.5 θ˙ = 1

r˙ = r(1 − r2 ) x˙ =

 d r cos(θ) dt

= r˙ cos(θ) − r sin(θ)θ˙ = r(1 − r2 ) cos(θ) − r sin(θ) = (1 − r2 )r cos(θ) − r sin(θ) = (1 − x2 − y 2 )x − y = x − y − x(x2 + y 2 ) y˙ =

 d r sin(θ) dt

= r˙ sin(θ) + r cos(θ)θ˙ = r(1 − r2 ) sin(θ) + r cos(θ) = (1 − r2 )r sin(θ) + r cos(θ) = (1 − x2 − y 2 )y + x = x + y − y(x2 + y 2 ) 7.1.7 θ˙ = 1

r˙ = r(4 − r2 )

The initial condition of (r, θ) = (0.1, 0) in Cartesian is (x, y) = (0.1, 0). The system will approach the r = 2 limit cycle, which would make x(t) = 2 cos(t) the limiting waveform for any initial condition except the origin.

We can see this in the following graph. y

x

164

Chapter 7: Limit Cycles

7.1.9 a) Since the dog and the duck are moving at the same speed, then they will both move the same amount in the time it takes θ to change by ∆θ. Thus the problem can be done using ∆θ instead of ∆t in the derivation.

During a small amount of time ∆t, the length of the segment R decreases from the dog swimming a length of ∆θ and increases from the duck moving along the perimeter of the circle by length also ∆θ. We’ll eventually be taking a limit, so we can make our lives easier with some approximations.

∆θ

ϕ

R ∆θ c θ

The line segment along the perimeter of the circle is approximately the length of the arc the line segment spans, and the angle bordering ϕ is approximately a right angle. Using these approximations, the Law of Cosines, and a trick for taking a limit, we can arrive at a differential equation for R. ∆R ≈ −∆θ + c − R …  π −R = −∆θ + (∆θ)2 + R2 − 2∆θR cos ϕ + 2  (∆θ)2 + R2 − 2∆θR cos ϕ + π2 − R2 = −∆θ + »  (∆θ)2 + R2 − 2∆θR cos ϕ + π2 + R

(∆θ)2 + 2∆θR sin(ϕ) = −∆θ + p (∆θ)2 + R2 + 2∆θR sin(ϕ) + R ∆θ + 2R sin(ϕ) ∆R = −1 + p 2 ∆θ (∆θ) + R2 + 2∆θR sin(ϕ) + R 165

Chapter 7: Limit Cycles

lim

∆θ→0

∆R ∆θ + 2R sin(ϕ) = −1 + p ∆θ (∆θ)2 + R2 + 2∆θR sin(ϕ) + R 2R sin(ϕ) = −1 + sin(ϕ) R′ = −1 + √ R2 + R

For the ϕ′ equation we can use some basic geometry.

a b ϕ + ∆ϕ

ϕ

R c d e

θ + ∆θ θ

∆θ

Next we have to remember that △acb is similar to △dce (angles subtended by the same arc are equal) meaning the ratio between ab and de is equal to the ratio of ac and ce.

166

Chapter 7: Limit Cycles

a b

ϕ

R c 1

π − 2ϕ

d e 1

Ù = ∆θ and de as de. Ù Approximating ab as ab Ù≈ ab : de = ac : cd ⇒ de

2 cos(ϕ) − R ∆θ R

Next we have to remember that the arc made by the inscribed angle ϕ is half of the arc of the central angle (Inscribed Angle Theorem).

167

Chapter 7: Limit Cycles

a ϕ d



Ù and a decrease Now putting it all together, the change in the central angle 2∆ϕ is from an increase of arc de of arc ∆θ. Ù − ∆θ ≈ 2 cos(ϕ) − R ∆θ − ∆θ = 2 cos(ϕ) − 2R ∆θ 2∆ϕ = de R R ∆ϕ ≈ cos(ϕ) − R R ∆θ ∆ϕ = Rϕ′ = cos(ϕ) − R lim R ∆θ→0 ∆θ As for whether or not the dog catches the duck, the answer is no. The proof is difficult enough that it is the subject of a published paper. Please take a look at Applications of Center Manifolds by Keith Promislow if you’re interested. b) Repeating the derivation now with the speed of the dog as k∆θ R′ = −k + sin(ϕ) Rϕ′ = cos(ϕ) − R c) Assuming there is a limit cycle R′ = −k + sin(ϕ) = 0 ⇒ ϕ = arcsin(k) p Rϕ′ = cos(ϕ) − R = 0 ⇒ R = cos(ϕ) = 1 − k 2 168

Chapter 7: Limit Cycles

So for k =

1 2

the dog will endlessly chase the duck while asymptotically approaching a circle of radius

a limit cycle. 7.2.1 V = x2 + y 2 y

x

169

√ 3 2

as

Chapter 7: Limit Cycles

7.2.3 V = ex sin(y) y

x

7.2.5 Let x˙ = f (x, y) and y˙ = g(x, y) be a smooth vector field defined on the phase plane. a) If this is a gradient system, then ã Å ∂V ∂V = (f (x, y), g(x, y)) ,− −∇V = − ∂x ∂y −

∂V ∂V ∂f ∂2V ∂2V ∂g = f and − =g⇒ =− =− = ∂x ∂y ∂y ∂y∂x ∂x∂y ∂x

b) Yes. If f and g are smooth on R2 and

∂f ∂

=

∂g ∂x

then the system is a gradient system.

f : R2 → R is smooth and the Fundamental Theorem of Calculus together imply there exists V : R2 → R R such that V (x, y) is continuous and V (x, y) = f (x, y)dx R R R ∂g ∂ Leibniz integral rule ⇒ ∂V f (x, y)dx = ∂f ∂ = ∂ ∂ (x, y)dx = ∂x (x, y)dx = g

Hence there exists V (x, y) such that

∂V ∂x

= f and

∂V ∂

=g

7.2.7 x˙ = y + 2xy a)

y˙ = x + x2 − y 2

  ∂ ∂ y + 2xy = 1 + 2x = x + x2 − y 2 ∂y ∂x

b) −

∂V = x˙ = y + 2xy ∂x Z ⇒ −V =

y + 2xy dx = xy + x2 y + g(y) 170

Chapter 7: Limit Cycles

∂V = y˙ = x + x2 − y 2 ∂y  ∂ xy + x2 y + g(y) = x + x2 + g ′ (y) = ∂y 1 ⇒ g ′ (y) = −y 2 ⇒ g(y) = − y 3 + C (set to 0 for convenience) 3 1 3 2 V (x, y) = −xy − x y + y 3 −

c) y

x

7.2.9 Recall that a gradient system will have ⟨x, ˙ y⟩ ˙ = ⟨f (x, y), g(x, y)⟩ = ⟨−Vx , −V ⟩ ⇒ f = −Vx = −V

x

= gx

a) x˙ = y + x2 y f =

y˙ = −x + 2xy

∂ ∂ (y + x2 y) = 1 + x2 ̸= −1 + 2y = (−x + 2xy) = gx ∂y ∂x

So this is not a gradient system. b) x˙ = 2x

y˙ = 8y 171

Chapter 7: Limit Cycles

∂ ∂ (2x) = 0 = (8y) = gx ∂y ∂x So this is a gradient system with V (x, y) = −x2 − 4y 2 . f =

y

y

x

x

c) 2

2

2

2

y˙ = −2yex + x˙ = −2xex +   2 2 2 2 2 2 ∂  ∂  − 2xex + = −4xyex + = − 2yex + = gx f = ∂y ∂x

So this is a gradient system with V (x, y) = ex

2

+

2

.

y

y

x

x

7.2.11 V = ax2 + 2bxy + cy 2 The shape is paraboloid like, but the middle term can make the origin into a saddle point for the surface. This problem is exactly like a multivariable calculus problem to find an absolute minimum, so we will be using the second derivative test. Vxx V

− Vx

2

(0,0)

= (2a)(2c) − (2b)2

(0,0)

172

= 4 ac − b2



Chapter 7: Limit Cycles

The second derivative tests states the origin is a local extrema if ac − b2 > 0 and a local minimum if Vx (0, 0) = V (0, 0) = 0 for the origin to be a critical point and Vxx (0, 0) = 2a > 0. (We could have used V instead of Vxx but we simply had to use one of them.) Hence, V (x, y) is positive definite if and only if a > 0 and ac − b2 > 0. 7.2.13

ã Å N1 − b1 N1 N2 N˙ 1 = r1 N1 1 − K1

ã Å N2 N˙ 2 = r2 N2 1 − − b2 N1 N2 K2

Using Dulac’s criterion with weighting function g(N1 , N2 ) =

1 N1 N2

 ∇ · g(N˙ 1 , N˙ 2 )⟨N1 , N2 ⟩ ≠ Å ã ã Å ∑ 1 1 N1 1 N2 1 = ∇ · r1 − − − b1 , r2 − b2 N2 K1 N2 N1 K2 N1 r1 1 r2 1 =− − 0

a) The −R and −S terms signify that the more love Rhett and Scarlett feel for the other, the faster their love for the other decreases.

The AS and AR is the baseline amount of love one feels for the other. For instance, if Scarlett love for Rhett is fixed at 0, then Rhett will still love Scarlett an amount AS , and similarly for Scarlett when the roles are reversed. The kSe−S and kRe−R are 0 for R, S = 0 and approximately 0 when R, S are large, and maximum somewhere in between. The function having a maximum implies too little love and too little love for their partner have effectively no effect, and there is an optimal amount of love for their partner to cause the greatest increase in their partner’s love. b) Making normal vectors pointing inward on the boundary of the first quadrant.

176

Chapter 7: Limit Cycles

S

R

and then dot product with the vector field on the boundary ˙ S⟩ ˙ ⟨1, 0⟩ · ⟨R,

R=0

˙ S⟩ ˙ ⟨0, 1⟩ · ⟨R,

= ⟨1, 0⟩ · ⟨−R + AS + kSe−S , −S + AR + kRe−R ⟩

= ⟨1, 0⟩ · ⟨AS + kSe

−S

R=0

, −S + AR ⟩

= AS + kSe−S > 0 S=0

= ⟨0, 1⟩ · ⟨−R + AS + kSe−S , −S + AR + kRe−R ⟩

= ⟨0, 1⟩ · ⟨−R + AS , AR + kRe

−R

S=0



= AR + kRe−R > 0 meaning the vector field always points inward on the positive axes. Thus the trajectory can never escape the first quadrant. c) Picking g(R, S) = 1 ˙ S⟩ ˙ = ∇ · ⟨−R + AS + kSe−S , −S + AR + kRe−R ⟩ ∇ · ⟨R, = (−1) + (−1) = −2 then by Dulac’s criterion there are no periodic solutions, and not just in the first quadrant. d) AS = 1.2 AR = 1 k = 15

R(0) = S(0) = 0

177

Chapter 7: Limit Cycles

S

R

7.3.1 x˙ = x − y − x(x2 + 5y 2 )

y˙ = x + y − y(x2 + y 2 )

a) Classify the fixed point at the origin.

A=

Ñ é 1 −1 1

1

(1 − λ)2 + 1 = 0 ⇒ λ2 − 2λ + 2 = (λ − (1 + i)) (λ − (1 − i)) = 0 λ1 = 1 + i

λ2 = 1 − i ⇒ unstable spiral

b) Rewrite the system in polar coordinates using xy˙ − y x˙ θ˙ = r2

rr˙ = xx˙ + y y˙ xx˙ = x2 − xy − x2 (x2 + 5y 2 )

y y˙ = xy + y 2 − y 2 (x2 + y 2 )

xx˙ + y y˙ = x2 − xy − x2 (x2 + 5y 2 ) + xy + y 2 − y 2 (x2 + y 2 ) = x2 + y 2 − x2 (x2 + 5y 2 ) − y 2 (x2 + y 2 ) = x2 + y 2 − x4 − y 4 − 6x2 y 2 = r2 − r4 cos4 (θ) − r4 sin4 (θ) − 6r4 cos2 (θ) sin2 (θ)   = r2 − r4 cos4 (θ) + r4 cos2 (θ) sin2 (θ) − r4 sin4 (θ) + r4 cos2 (θ) sin2 (θ) − 4r4 cos2 (θ) sin2 (θ)   = r2 − r4 cos2 (θ) cos2 (θ) + sin2 (θ) − r4 sin2 (θ) sin2 (θ) + cos2 (θ) − 4r4 cos2 (θ) sin2 (θ)

= r2 − r4 cos2 (θ) − r4 sin2 (θ) − 4r4 cos2 (θ) sin2 (θ) 178

Chapter 7: Limit Cycles

 = r2 − r4 cos2 (θ) + sin2 (θ) − 4r4 cos2 (θ) sin2 (θ)

= r2 − r4 − 4r4 cos2 (θ) sin2 (θ)

2

= r2 − r4 − r4 (2 cos(θ) sin(θ)) = r2 − r4 − r4 sin2 (2θ)

rr˙ = xx˙ + y y˙ = r2 − r4 − r4 sin2 (2θ) ⇒ r˙ = r − r3 − r3 sin2 (2θ) = r 1 − r2 − r2 sin2 (2θ)



 xy˙ = x x + y − y(x2 + y 2 ) = x2 + xy − xy(x2 + y 2 )

 y x˙ = y x − y − x(x2 + 5y 2 ) = xy − y 2 − xy(x2 + 5y 2 ) xy˙ − y x˙ = x2 + y 2 − xy(x2 + y 2 ) − xy − y 2 − xy(x2 + 5y 2 ) = x2 + y 2 + xy(−x2 − y 2 + x2 + 5y 2 )



= x2 + y 2 + 4xy 3 = r2 + 4r4 cos(θ) sin3 (θ) xy˙ − y x˙ = 1 + 4r2 cos(θ) sin3 (θ) θ˙ = r2 c) Determine the circle of maximum radius, r1 , centered on the origin such that all trajectories have a radially o tward component on it.

This is equivalent to 0 ≤ r˙ for all θ.  0 ≤ r 1 − r2 − r2 sin2 (2θ) ⇒ 0 ≤ 1 − r2 − r2 sin2 (2θ) 1 1 + sin2 (2θ)

⇒0≤r≤

1 ⇒ 0 ≤ r ≤ √ = r1 2 d) Determine the circle of minimum radius, r2 , centered on the origin such that all trajectories have a radially inward component on it.  r 1 − r2 − r2 sin2 (2θ) ≤ 0 ⇒ 1 − r2 − r2 sin2 (2θ) ≤ 0



1 ≤ r ⇒ r2 = 1 ≤ r 1 + sin2 (2θ)

e) 1 We know that there is at least one limit cycle in the region √ ≤ r ≤ 1 by the Poincar´e-Bendixson Theorem. 2 The trapping region is a closed and bounded subset of the plane. The vector field is continuously differentiable, easily seen in the Cartesian representation, on an open set 179

Chapter 7: Limit Cycles

containing the trapping region. The trapping region does not contain any fixed points.  r˙ = 0 ⇒ r 1 − r2 − r2 sin2 (2θ) = 0 ⇒ r = 0 or r2 = θ˙ = 0 ⇒ 1 + 4r2 cos(θ) sin3 (θ) = 0 ⇒ r2 =

1 1 + sin2 (2θ)

−1 4 cos(θ) sin3 (θ)

−1 1 = ⇒ 4 cos(θ) sin3 (θ) = −1 − sin2 (2θ) 2 1 + sin (2θ) 4 cos(θ) sin3 (θ) Determining the solutions of this equation is difficult analytically, but graphically it’s pretty clear that there are no real solutions to the equation. 7.3.3 x˙ = x − y − x3

y˙ = x + y − y 3

The only fixed point of the system is the origin, and linearization predicts the origin is a unstable spiral.

We’ll make a trapping region by converting to polar coordinates. r 2 = x2 + y 2 2rr˙ = 2xx˙ + 2y y˙ x(x − y − x3 ) + y(x + y − y 3 ) r x2 + y 2 − x4 − y 4 = r r2 − r4 cos4 (θ) − r4 sin4 (θ) = r

r˙ =

We can bound the θ term with

1 2

= r − r3 cos4 (θ) − r3 sin4 (θ)  = r − r3 cos4 (θ) + sin4 (θ)

≤ cos4 (θ) + sin4 (θ) ≤ 1 which gives r − r3 ≤ r˙ ≤ r −

r3 2

So a suitable trapping region is bounded by a circle of radius less then 1 which has a positive radial compo√ nent and a circle of radius greater than 2 which has a negative radial component.

Then by the Poincar´e-Bendixson Theorem, there is at least one periodic solution in the trapping region. 7.3.5 x˙ = −x − y + x(x2 + 2y 2 )

y˙ = x − y + y(x2 + 2y 2 )

This system has a fixed point at (x, y) = (0, 0) which is best seen using a graph of the nullclines. 180

Chapter 7: Limit Cycles

y

x

A=

Ñ 3x2 + 2y 2 − 1 2xy + 1

A(0,0) =

é 4xy − 1

x2 + 6y 2 − 1

Ñ é −1 −1 1

−1

λ1,2 = −1 ± i ⇒ Stable Spiral The Poincar´e-Bendixson Theorem can still be applied here, despite the fixed point being unstable. In this case the trapping region will be more of a repulsion region as the vector field will point outward on the boundaries. Thus there will be at least one unstable limit cycle in the region. (You can see this by transforming t → −t which runs time backwards for the system. Then we’re showing existence for the usual stable limit cycle.)

Next we’ll transform the system into polar coordinates to find the boundaries of the trapping region. r 2 = x2 + y 2 181

Chapter 7: Limit Cycles

2rr˙ = 2xx˙ + 2y y˙ xx˙ + y y˙ r   x − x − y + x(x2 + 2y 2 ) + y x − y + y(x2 + 2y 2 ) = r x4 + 3x2 y 2 − x2 + 2y 4 − y 2 = r (x2 + y 2 )(x2 + 2y 2 − 1) = r r2 (r2 + y 2 − 1) = r  2 = r r + r2 sin2 (θ) − 1

r˙ =

From here we can upper and lower bound the θ term to bound r. ˙

r(r2 − 1) ≤ r˙ ≤ r(2r2 − 1) So a suitable trapping region is bounded by a circle of radius less than

√1 2

which has a negative radial

component and a circle of radius greater than 1 which has a positive radial component.

Then by the Poincar´e-Bendixson Theorem, there is at least one periodic solution in the trapping region. 7.3.7 x˙ = y + ax(1 − 2b − r2 )

y˙ = −x + ay(1 − r2 )

0 0 will have all vectors pointing inward on the boundary. Therefore, by the Poincar´eBendixson Theorem, there is at least one limit cycle in the trapping region. We also know that all limit cycles have the same period because θ˙ = 2ab sin(2θ) − 1 is independent of r, meaning the rotation rate is the same at every θ value no matter which limit cycle is being traversed. c) Setting b = 0 simplifies the equations to r˙ = ar 1 − r2



θ˙ = −1

for which r < 1 ⇒ r˙ > 0 and 1 < r ⇒ r˙ > 0 making r = 1 the only limit cycle. 7.3.9 r˙ = r(1 − r2 ) + µr cos(θ) θ˙ = 1 a) dr dr dθ dr = = = r(1 − r2 ) + µr cos(θ) dt dθ dt dθ  r(θ) = 1 + µr1 (θ) + O µ2 dr = µr1′ (θ) dθ

r˙ = r(1 − r2 ) + µr cos(θ)     2  µr1′ (θ) = 1 + µr1 (θ) 1 − 1 + µr1 (θ) + µ 1 + µr1 (θ) cos(θ)

r1′ = −2r1 + cos(θ) r1 =

1 2 sin(θ) + cos(θ) + Ce−2θ 5 5

The exponential term eventually dies off, leaving only the periodic trajectory. Å ã  1 1 2 2 r1 = sin(θ) + cos(θ) r(θ) = 1 + µ sin(θ) + cos(θ) + O µ2 5 5 5 5

b)

Å ã dr 1 2 =µ cos(θ) − sin(θ) = 0 dθ 5 5 Å ã Å ã 1 1 1 2 , arctan +π cos(θ) − sin(θ) = 0 ⇒ θ = arctan 5 5 2 2 183

Chapter 7: Limit Cycles

c)

Å Å Å ãã ã   1 1 1 2 2 µ √ + √ =1+µ r arctan + O µ2 = 1 + √ + O µ2 2 5 5 5 5 5 ã Å Å Å ã ã   1 −1 2 −2 1 µ √ + √ +π =1+µ r arctan + O µ2 = 1 − √ + O µ2 2 5 5 5 5 5 p p µ µ 1−µ> 1, show that the positive branch of the cubic nullcline begins at xA = 2 and ends at xB = 1.

The cubic nullcline is from x˙ = µ[y − F (x)]

F (x) =

187

1 3 x −x 3

Chapter 7: Limit Cycles

y

(xA , yA ) x

(xB , yB )

xB occurs at the local minimum F ′ (x) = x2 − 1 = 0 ⇒ x = ±1 ⇒ xB = 1 xA occurs at the intersection of the tangent line of the local maximum with F (x) F (−1) = yA =

2 3

= F (x) ⇒ x = −1, 2 ⇒ xA = 2

7.5.3 x ¨ + k(x2 − 4)x˙ + x = 1 We can do a substitution to transform the equation into the form of Example 7.5.1. x→z+1

 x ¨ + k(x2 − 4)x˙ + x = 1 → z¨ + k (z + 1)2 − 4 z˙ + (z + 1) = 1

Å Å ãã  1 d z˙ + k (z + 1)3 − 4z 0 = z¨ + k (z + 1)2 − 4 z˙ = dt 3 1 F (z) = (z + 1)3 − 4z 3  w = z˙ + kF (z) w˙ = z¨ + k (z + 1)2 − 4 z˙ = −z   w z y= z˙ = k y − F (z) y˙ = − k k

Now that we transformed the equation into something similar to Example 7.5.1, we can perform the integral over the branches since F (z) is not symmetric. Z tB Z T ≈ dt + tA

tD

dt

tC

 dz z dy dz = ≈ F ′ (z) = (z + 1)2 − 4 k dt dt dt −k (z + 1)2 − 4 dz dt ≈ z −

Next we have to find the turning points of the limit cycle for the integral bounds for zA → zB → zC → zD → zA . F ′ (z) = (z + 1)2 − 4 = z 2 + 2z − 3 = (z + 3)(z − 1) 188

Chapter 7: Limit Cycles

zB = 1 zD = −3 F (zB ) = F (zC ) ⇒ zC = −5 and F (zD ) = F (zA ) ⇒ zA = 3 And now we can compute the integrals for the approximate period. ÇZ 1 å Z tB Z tD Z −3  (z + 1)2 − 4 (z + 1)2 − 4 T ≈ dt + dt = −k dz + dz = k 12 − 3 ln(5) z z 3 tA tC −5 7.5.5

 x ¨ + µ |x| − 1 x˙ + x = 0

Define

 x˙ = µ y − F (x)

F (x) =

  − x2 − x x ≤ 0 2   x2 − x 2

0 0 implies unstable. We can actually see from the graph that unstable only occurs

between the local maximum and local minimum. v

u

The u˙ has a positive increase to the rightward direction, either starting positive and becoming more positive above the nullcline, or starting negative and increasing towards zero below the nullcline.

191

Chapter 7: Limit Cycles

Therefore

∂ u˙ ∂u

> 0 when c lies on the middle branch, and thus the fixed point is unstable and there is a

stable limit cycle when 2α ≈ c1 < c < c2 ≈

1 − 3αb b

c) v

u

The fixed point is stable when c < c1 and all trajectories eventually approach the fixed point, but the route can be very long depending the initial condition. Consequently a bit of noise in the system can have varying amounts of effect depending on which way the noise nudges the system. 7.6.1

√ sin(t 1 − ϵ2 ) √ eϵt 1 − ϵ2 ϵ2 ∂ 2 x ∂x + 0 1 and a + b > 0 ⇒ Origin is a stable node and other fixed point is a saddle point. 0 < ab < 1 and a + b > 0 ⇒ Origin is a saddle point and other fixed point is a stable node. ab > 1 and a + b < 0 ⇒ Origin is an unstable node and other fixed point is a saddle point. 0 < ab < 1 and a + b < 0 ⇒ Origin is a saddle point and other fixed point is an unstable node. ab < 0 ⇒ Origin and the other fixed point are saddle points. This last case completes all possible cases. From this we can conclude that transcritical bifurcations occur along the boundary a =

1 in parameter b

space. 8.1.9 x ¨ + bx˙ − kx + x3 = 0 First we transform into a 2-D system. y˙ = kx − by − x3

x˙ = y

Next we find the fixed points, which are the intersections of the nullclines. y=0

y=

√ kx − x3 ⇒ (x, y) = (0, 0) and (± k, 0) b

Now we find the linearization at the fixed points. Ñ A= Ñ A(0,0) =

A(±√k,0)

=

0

é 0

1

k − 3x2 é

−b

1

λ1,2 =

−b

k Ñ

−b ±

√ b2 + 4k 2

é

0

1

−2k

−b

λ1,2 =

−b ±



b2 − 8k 2

The second pair of fixed points come into existence and split off from the origin when k > 0, and the stability of all three fixed points is determined by the sign of b. If b < 0 and k becomes positive, then there’s a subcritical pitchfork bifurcation. If b > 0 and k becomes positive, then there’s a supercritical pitchfork bifurcation. When b = 0 there is a bifurcation when k becomes positive too. The origin changes from a center to a saddle, and the two new fixed points are centers.

Note that the new fixed points can be spirals or nodes depending on k and b, but we only care about the stability for the stability diagram. 209

Chapter 8: Bifurcations Revisited

k

1 saddle 2 centers

1 saddle 2 unstable

subcritical pitchfork

1 saddle 2 stable

supercritical pitchfork

1 center

1 unstable

b

1 stable

8.1.11 u˙ = a(1 − u) − uv 2

v˙ = uv 2 − (a + k)v

a, k > 0

The fixed points of the system are (u, v) = (1, 0), which always exists, and å Ç p p a ± a2 − 4(a + k)2 a ∓ a2 − 4(a + k)2 , (u, v) = 2a 2(a + k) if the inside of the square root is positive.

We for sure have a saddle-node bifurcation when the inside of the square root is negative with zero fixed points, the inside of the square root is zero with one fixed point, and then the inside of the square root is positive with two fixed points.

The inside of the square root is zero when a2 − 4(a + k)2 = 0 ⇒ (a + k)2 =

√ a a ⇒ k = −a ± 4 2

and thus saddle-node bifurcations occur on this curve in a − k parameter space. 8.1.13 n˙ = GnN − kn

N˙ = −GnN − f N + p

210

G, k, f > 0

Chapter 8: Bifurcations Revisited

a) dN dn = GnN − kn = −GnN − f N + p dt dt G dn Gn G dN f GN pG G2 nN G2 nN − − + 2 = = − k 2 dt k2 k k 2 dt k2 k2 k GN f pG Gn y= a= b= 2 τ = kt x = k k k k dy dx = x(y − 1) = −xy − ay + b dτ dτ

 b) (x, y) = 0, ab and (b − a, 1) are the fixed points. é Ñ y−1 x A= −y −x − a é Ñ b −1 0 b A(0, b ) = a ∆=a−b τ = −1−a a a −a − ab Ñ é 0 b−a A(b−a,1) = ∆ = b − a τ = −b −1 −b c) y

y

a=2 b=1

y

a=2 b=2

x

x

x

d)

an s

cr

iti

ca lb

ifu

rc

at

io n

b=

a

 0, ab unstable (b − a, 1) stable

tr

b

a

 0, ab stable (b − a, 1) unstable 211

a=2 b=3

Chapter 8: Bifurcations Revisited

The picture only includes positive values of a due to the sign restrictions of the nondimensionalized variables. 8.1.15 The power of true believers n˙ A = (p + nA )nAB − nA nB

n˙ B = nB nAB − (p + nA )nB

nAB = 1 − (p + nA ) − nB

a) The first equation is the rate of change of the A-believers, which is the sum of gaining A-believers and losing A-believers. Gain comes from AB-believers converting at a rate proportional to the fraction of Abelievers, and loss comes from A-believers converting to AB-believers at a rate proportional to the fraction of B-believers.

The second equation is the rate of change of the B-believers, which is the sum of gaining B-believers and losing B-believers. Gain comes from AB-believers converting at a rate proportional to the fraction of Bbelievers, and loss comes from B-believers converting to AB-believers at a rate proportional to the fraction of A-believers.

While there is a fixed fraction of unconvertable A-believers, p, there is no analogous group for the Bbelievers. So the total fraction of A-believers is (p + nA ), but the total number of B-believers is nB .

An equation for n˙ AB is unnecessary because nAB can be found by conservation of believers. b)

212

Chapter 8: Bifurcations Revisited

nB p = 0.1 p = 0.15 p = 0.2

p = 0.5

nA An initial condition of p = 0.15 successfully converts everyone to A-believers but an initial condition of p = 0.1 is unsuccessful. Then there should be a 0.1 < pc < 0.15 that is a critical point for the change in behavior. c) n˙ B = 0 ⇒ nB = 0 or nB = −2nA − 2p + 1 n˙ A = 0 and nB = 0 ⇒ nA = 1 − p n˙ A = 0 and nB = −2nA − 2p + 1 ⇒ nA =

−4p + 1 ±

p

4p2 − 8p + 1 6

Then the fixed points in terms of p are Ç å p p −4p + 1 ± 4p2 − 8p + 1 4p − 1 ∓ 4p2 − 8p + 1 (1 − p, 0) , − 2p + 1 6 3 The first fixed point is always there, but the other two fixed points will collide when √ 3 2 4p − 8p + 1 = 0 ⇒ p = 1 ± 2 out of which we have pc = 1 −

√ 3 2

to be in the [0,1] interval.

The last two fixed points are complex if pc < p and therefore to check if the remaining fixed point (1 − p, 0) is globally attracting we only need to look at its linearization.

213

Chapter 8: Bifurcations Revisited

Ñ A=

d ˙A dnA n

d ˙A dnB n

d ˙B dnA n

d ˙B dnB n

é

A(1−p,0) =

=

Ñ −2nA − 2nB − 2p + 1

Ñ −1 0

é

−2nB

−1

é −2nA − 2p −2nA − 2nB − p + 1

λ1 = −1 λ2 = p − 1

p−1

The eigenvalues are always negative. Therefore (1 − p, 0) is a stable node and when pc < p all trajectories will converge to it. d) A saddle-node bifurcation occurs at p = pc as two of the fixed points approach, collide, and then disappear. 8.2.1 x ¨ + µ(x2 − 1)x˙ + x = a First convert the ODE into a system. y˙ = a − µ(x2 − 1)y − x

x˙ = y

Now we solve for fixed points, which gives (x, y) = (a, 0), and then compute the Jacobian Ñ é 0 1 A= −2µxy − 1 −µ(x2 − 1) é

Ñ

A(a,0) =

0

1

∆ = 1 τ = −µ(a2 − 1)

2

−1 −µ(a − 1)

Hopf bifurcations occur when the trace changes from negative to zero to positive, and vice versa. The trace changes sign when crossing the lines µ = 0 and a = ±1 in parameter space. 8.2.3 x˙ = −y + µx + xy 2

y˙ = x + µy − x2

y

y

µ = −0.04

µ = −0.02

x

214

x

Chapter 8: Bifurcations Revisited

y

y

µ=0

µ = 0.02

x

x

There’s an unstable limit cycle for µ negative and the limit cycle constricts towards the origin as µ increase to zero. (Only a portion of the spiral trajectory is shown for µ = −0.02 because the individual spiral trajectories are very densely packed.) The limit cycle disappears when µ = 0 with a very slight unstable spiral now at the origin. The unstable spiral’s radial growth increases as µ increases past zero. This is a subcritical Hopf bifurcation. 8.2.5 x˙ = y + µx

y˙ = −x + µy − x2 y

y

y

µ = −0.1

µ = 0.1

x

x

Supercritical Hopf bifurcation 8.2.7 x˙ = µx + y − x2

y˙ = −x + µy + 2x2

These graphs have bounds of (x, y) ∈ [−0.4, 0.8] × [−0.4, 0.8].

215

Chapter 8: Bifurcations Revisited

µ = −0.06

y

y

µ = 0.06

x

x

This is a subcritical Hopf bifurcation. For µ < 0 the origin is stable and there is an unstable limit cycle, but for µ > 0 the origin is unstable and there is no limit cycle. 8.2.9

Å x˙ = x b − x −

y 1+x

ã

Å

x y˙ = y − ay 1+x

ã x, y ≥ 0 a, b > 0

a) The x and y nullclines are x = 0 y = (1 + x)(b − x)

y=0 y=

x a(1 + x)

y

x

The fixed points are (x, y) = (0, 0), (b, 0), and the intersection of the curves, which is difficult to find because we need to solve a cubic equation.

Next, we check stability. A=

Ñ b−

(x+1)2

b

é 0

0

0

Ñ A(0,0) =

− 2x

(x+1)2

−x 1+x x 1+x

é

− 2ay

b > 0 ⇒ Unstable

216

Chapter 8: Bifurcations Revisited

A(b,0) =

Ñ −b 0

−b 1+b b 1+b

é λ1 = −b < 0
0 so we have to have exactly one strictly positive intersection, or exactly three strictly positive intersections. We can’t have three intersections graphically, which means the strictly positive fixed point is unique. c) A Hopf bifurcation can only occur if the trace of the linearized system is zero. So we’ll plug in the strictly positive fixed point (x∗ , y ∗ ) and find conditions on a and b. y x − 2x + − 2ay 2 (x + 1) 1+x x∗ y∗ ∗ − 2x + − 2ay ∗ τ |(x∗ , ∗ ) = b − ∗ (x + 1)2 1 + x∗ x∗ y ∗ = (1 + x∗ )(b − x∗ ) y∗ = a(1 + x∗ ) ∗ ∗ (1 + x )(b − x ) x∗ 2x∗ τ |(x∗ , ∗ ) = b − − 2x∗ + − ∗ 2 ∗ (x + 1) 1+x 1 + x∗ b − x∗ x∗ 2x∗ =b− − 2x∗ + − ∗ ∗ 1+x 1+x 1 + x∗ b−2 b = 0 ⇒ x∗ = = b − 2x∗ − 1 + x∗ 2 τ =b−

(1 + x∗ )(b − x∗ ) = a=

x∗ a(1 + x∗ )

x∗ = (1 + x∗ )2 (b − x∗ ) 1+ 217

b−2 2

 b−2 2 2

b−

b−2 2



Chapter 8: Bifurcations Revisited

=

b−2 2 b+2 2 2

 b 2

=

4(b − 2) = ac b2 (b + 2)

We should also check the determinant is positive here to show there is a Hopf bifurcation. Ñ é ∗ −x∗ b − (x∗ +1)2 − 2x∗ 1+x∗ A(x∗ , ∗ ) = ∗ x∗ ∗ (x∗ +1)2 1+x∗ − 2ay y ∗ = (1 + x∗ )(b − x∗ )

A(x∗ ,

∗)

=

Ñ 2x∗ + 2 −

=

Ñ 2− Ñ

=

∆=

y∗ =

x∗ a(1 + x∗ ) b−x∗ 1+x∗

b−x∗ 1+x∗

b−x∗ 1+x∗

b−x∗ 1+x∗

−x∗ 1+x∗

x∗ =

=

−x∗ 1+x∗

2(1+x∗ )−(x∗ +2) 1+x∗

−x∗ 1+x∗

x∗ +2 1+x∗

−x∗ 1+x∗

é

−x∗ 1+x∗

− 2x∗ é

b−2 ⇒ b = 2x∗ + 2 2

x∗ 1+x∗

Ñ 2− é



x − 2 1+x ∗ −x∗ 1+x∗

x∗ +2 1+x∗

x∗ +2 1+x∗

Ñ

=

é

−x∗ 1+x∗

x∗ 1+x∗

−x∗ 1+x∗

x∗ +2 1+x∗

−x∗ 1+x∗

é

−x∗ −x∗ x∗ + 2 2x∗ x∗ − = >0 1 + x∗ 1 + x∗ 1 + x∗ 1 + x∗ (1 + x∗ )2

With the last inequality following from x∗ being strictly positive.

Hence we have a whole curve of Hopf bifurcations in (a, b) parameter space. d) y

a = 0.08 b = 4

y

x

218

a=

1 12

b=4

x

Chapter 8: Bifurcations Revisited

a = 0.1 b = 4

y

x

8.2.11 x ¨ + µx˙ + x − x3 = 0 a) x˙ = y A=

A(0,0) =

Ñ

y˙ = −µy − x + x3 é 0 1

−1 + 3x2 Ñ é 0 1 −1 −µ

−µ ∆ = 1 τ = −µ

µ negative and close to zero implies an unstable spiral µ positive and close to zero implies a stable spiral b) µ0

x

The origin becomes a center when µ = 0 which is a degenerate case of a Hopf bifurcation because there are an infinite number of closed orbits instead of a single closed orbit. 8.2.13 y˙ = −x + µy − x2 y

x˙ = y + µx

µ=0

ω = −1 f (x, y) = 0 g(x, y) = −x2 y fx = 0

f =0

fxx = 0

fx = 0

fxxx = 0 fx

f

=0

gx = −2xy

g = −x2

gxx = −2y

gx = −2x

gxx = −2

=0

g

g

=0

=0

16a = fxxx + fx + gxx + g i 1h + fx (fxx + f ) − gx (gxx + g ) − fxx gxx + f g ω −1 16a = −2 ⇒ a = −1 − b(u − u2 ) − = λ1 + λ2 T τ = −1 − b(u − u2 ) −

Therefore λ1 and λ2 are both negative. d) However for the A − B matrix

Å ã 1 1 1 + (g − b)(u − u2 ) ∆ = − 1 + b(u − u ) − − − g(u − u2 ) = T T T 2



223

Chapter 8: Bifurcations Revisited

1 T

τ = −1 + b(u − u2 ) −

The determinant is positive or negative if g > b or b > g respectively, and the trace can be positive or negative depending on the relative sizes of b and T .

From these facts about the eigenvalues, we can cause either a pitchfork bifurcation or a Hopf bifurcation at (u, u, u, u) by varying g and T respectively. If we start with negative trace, then by changing g the system will undergo a pitchfork bifurcation at (u, u, u, u) when the determinant switches sign. If we start with positive determinant, then by changing T the system will undergo a Hopf bifurcation at (u, u, u, u) when the trace switches sign. e) For small T , the variables settle to the stable fixed point (u, u, u, u). z t Increasing T causes the stable limit cycle to appear. z t And increasing T further enlarges the size of the limit cycle. z t We can’t plot the limit cycle since it’s 4-dimensional, but we can plot the individual variables in time. We used z for the horizontal label because all four of the variables x1 , y2 , x3 , and y4 have indistinguishable graphs. So not only does the system tend toward a symmetric fixed point, the variables even follow the limit cycle symmetrically. 8.3.1 x˙ = 1 − (b + 1)x + ax2 y

y˙ = bx − ax2 y

a, b > 0 x, y ≥ 0

a)

  x, y = 1, ab is the fixed point. A=

A

b 1, a

=

Ñ −(b + 1) + 2axy b − 2axy

Ñ b−1 −b

é

ax2

é

−ax2

a

∆=a

−a

τ = b − (1 + a)

224

Chapter 8: Bifurcations Revisited

The determinant is positive since a is positive and the trace can vary greatly depending on b and a. Therefore all we really know is the fixed point can’t be a saddle point. b) Finding a trapping region implies that we’re looking for a limit cycle and b > 1 + a for an unstable fixed point. A trapezoid with a circle around the fixed point will suffice. y 2−

b(b + 1) a

1 b+1

Å y =− x−2+

1 b+1

ã +

b(b + 1) a

x 1 b+1 First, we should make sure that the fixed point is always inside the trapezoid. This is actually very easy with the choice of trapezoid. x:

1 1 0 ⟨0, −1⟩ · ⟨x, ˙ y⟩ ˙ b(b+1) = bx − ax2 a = a ⟨−1, −1⟩ · ⟨x, ˙ y⟩ ˙ = 1 − (b + 1)x + ax2 y − (bx − ax2 y) = −1 + x > 0 1 ⟨0, 1⟩ · ⟨x, ˙ y⟩ ˙

x>2− b+1

=0

= bx − ax2 (0) = bx > 0

Lastly, since the linearization predicts an unstable fixed point, we can always draw a sufficiently small circle enclosing the fixed point with the vector field pointing outward. Therefore we have our trapping region. 225

Chapter 8: Bifurcations Revisited

c) The Hopf bifurcation has to occur at b = 1 + a = bc where the sign of the trace changes and there are no other fixed points. d) The limit cycle exists for b > bc since we were able to construct a trapping region and apply the Poincar´eBendixson Theorem. (However we should keep in mind that there could be more than one limit cycle inside the trapping region, but a stable limit cycle will appear after the Hopf bifurcation.) e) b=1+a+µ 0 0 b ≪ 1 y ˙ = bx 1 − x˙ = a − x − 1 + x2 1 + x2 λ1,2 =

µ−

The solution for the problem done in the text is applicable here as well. The only portion that needs to be redone is estimating the period of the limit cycle since b ≪ bc now. y a = 10 b ≪ 1 B

C

A

D

x

The entire period of the relaxation oscillation is the time spent on all four branches, but the time spent on the BC and DA branches are negligible compared to the time spent on the AB and CD branches since 226

Chapter 8: Bifurcations Revisited

y˙ ≪ x, ˙ so we can neglect the fast branches. Z tB Z tC Z T = dt + dt + tA

tB

tD

dt +

tC

Z

tA

dt ≈

tD

Z

tB

dt +

tA

Z

tD

dt

tC

Next we change the integrals from dt to dx Z xB Z xB Z tB Z xB 1 dy 1 dy dy dt ä dx Ä dx = dx = dt = xA dx y˙ xA dx bx 1 − 1+x2 tA xA dx dy From here we can make a substitution from the x˙ nullcline equation making the integrand a function of x only.

(a − x)(1 + x2 ) dy a−x y ax2 − a − 2x3 = ⇒ = 4x 1 + x2 4x dx 4x2 Z tB Z xB Z xB dy ax2 − a − 2x3 1 1  ä Ä dt = dx ≈ a−x dx 2 dx 4x bx 1 − bx 1 − 1+x2 tA xA xA 4x y=

We had to go this very roundabout way because we couldn’t use x˙ for the approximation, our first of which is in the previous step. We know that the trajectory is close to the x nullcline, so we approximated the (x, y) coordinates as exactly on the x nullcline. The problem is by definition x˙ = 0 on the x nullcline, which would make an undefined integral. Z tB

dt =

tA

Now back to integrating Z tB tA

Z

Z

xB

B

A

1 dx = x˙

Z

xB

xA

1 a−x−

ax2 − a − 2x3  dx = 4bx3 1 − a−x 4x

Z

4x 1+x2

dx =

Z

xB

xA

1 dx 0

xB

ax2 − a − 2x3 dx 3 2 xA xA 4bx − bx (a − x) xB (3a2 − 125)x ln(5x − a) − 5a(2x2 + 5) + 125x ln(x) = 25abx

dt ≈

xA

Next we can us a bit of basic calculus to find the turn around points xA , xB , xC , and xD from the local minimum and maximum of the x nullcline. Unfortunately, this requires solving a cubic equation and plugging the results back in gets rather messy. We don’t show the details here, but the approximate period would be Z tB Z tD T ≈ dt + dt tA

tC

xB (3a − 125)x ln(5x − a) − 5a(2x2 + 5) + 125x ln(x) = 25abx 2

xA

xD (3a − 125)x ln(5x − a) − 5a(2x + 5) + 125x ln(x) + 25abx 2

2

xC

8.4.1

θ˙ = µ − sin(θ)

r˙ = r(1 − r2 ) x, y ∈ [−1.25, 1.25] t ∈ [0, 100] 227

Chapter 8: Bifurcations Revisited

µ = 1.1

x

t

y t

µ = 1.01

x

t

y t

8.4.3 x˙ = µx + y − x2

y˙ = −x + µy + 2x2

Homoclinic bifurcation at µ ≈ 0.066 These graphs have bounds of (x, y) ∈ [−0.4, 0.8] × [−0.4, 0.8] µ = 0.060

y

µ = 0.072

x

y

x

In each case, the gray trajectory starts near the origin. The left graph traps the trajectory in a stable limit cycle, but the the stable limit cycle eventually makes a homoclinic orbit with the saddle point and then 228

Chapter 8: Bifurcations Revisited

disappears in a homoclinic bifurcation. The right graph is after the homoclinic bifurcation. 8.4.5

 x ¨ + x + ϵ bx3 + k x˙ − ax − F cos(t) = 0

0 < ϵ 0

 h(x, x) ˙ = bx3 + k x˙ − ax − F cos(t) = bx3 + k x˙ − ax − F cos(θ − ϕ)h r cos(θ), −r sin(θ) = br3 cos3 (θ) − kr sin(θ) − ar cos(θ) − F cos(θ − ϕ)

  − kr + F sin(ϕ) h r cos(θ), −r sin(θ) sin(θ)dθ = 2 0  Z 2π  − 4ar − 3br3 + 4F cos(ϕ) 1 = ⟨h cos(θ)⟩ = h r cos(θ), −r sin(θ) cos(θ)dθ = 2π 0 8  3 − 4ar − 3br + 4F cos(ϕ) = 8r

dr 1 = ⟨h sin(θ)⟩ = dT 2π r

dϕ dT dϕ dT

Z



8.4.7 x′ = (r′ , rϕ′ )

g(rϕ) ≡ 1

The reason for the (r′ , rϕ′ ) instead of (r′ , ϕ′ ) is due to using the polar coordinate unit vectors. x′ = (r′ , rϕ′ ) = r′ rˆ + rϕ′ ϕˆ and the polar coordinate unit vectors, unlike Cartesian coordinates, are dependent on the angle of the point.

 − kr + F sin(ϕ) r = 2 ′

− 4ar − 3br3 + 4F cos(ϕ) ϕ = 8r ′



1 ∂ 1 ∂ (rr′ ) + (rϕ′ ) r ∂r r ∂ϕ Å ã Å ã 1 ∂ 4ar − 3br3 + 4F cos(ϕ) 1 ∂ kr2 + F r sin(ϕ) − =− r ∂r 2 r ∂ϕ 8 Å ã Å ã 1 −4F sin(ϕ) 1 2kr + F sin(ϕ) − = −k =− r 2 r 8

∇ · x′ =

Thus, the value of ∇ · (gx′ ) < 0 since k > 0 and there are no closed orbits for the averaged system by Dulac’s criterion. 8.4.9

 − kr + F sin(ϕ) r = 2 ′

− 4ar − 3br3 + 4F cos(ϕ) ϕ = 8r ′



a)  − kr + F sin(ϕ) kr =0⇒− = sin ϕ r = 2 F k2 r2 sin2 (ϕ) + cos2 (ϕ) = 1 ⇒ cos2 (ϕ) = 1 − 2 F  3 − 4ar − 3br + 4F cos(ϕ) ϕ′ = = 0 ⇒ 4ar − 3br3 + 4F cos(ϕ) = 0 8r ′

3br3 − 4ar = 4F cos(ϕ)

229

Chapter 8: Bifurcations Revisited

ã Å k2 r2 (3br3 − 4ar)2 = 16F 2 cos2 (ϕ) = 16F 2 1 − 2 = 16F 2 − 16k 2 r2 F 16k 2 r2 + (3br3 − 4ar)2 = 16F 2 " Å ã2 # 3 2 2 2 r k + = F2 br − a 4 b) a

b=0

r

c) Increasing b slightly positive, and then a bit more gives the following two graphs.

230

Chapter 8: Bifurcations Revisited

a

a

0 < b < bc

bc < b

r

r

The tip of the b = 0 graph continually moves up until the range is eventually overlapped. Now there are three different r values in a range of a values where the graph doubles back.

We can solve for a in terms of r somewhat easily and then find when the solutions overlap. … F2 3 2 − k2 a = br ± 4 r2 The graph will double back when there is zero slope with respect to r. da 3 F2 = br ± p =0 dr 2 r2 (F − kr)(F + kr)

We throw out the plus from the ± because we need the terms to cancel.

3 F2 da =0 = br − p dr 2 r2 (F − kr)(F + kr)

We also know that the second derivative will equal zero at b = bc when the graph is cubic shaped and about to develop a local minima.

d2 a 3 2F 4 − 3F 2 k 2 r2 = b − 3 = 0 dr2 2 r3 (F − kr)(F + kr) 2

We can find the conditions under which there is a real value of r as a solution to these two equations. The results are

√ 32 3k 3 = bc b≥ 27F 2

d)

231

Chapter 8: Bifurcations Revisited

r

b

a

8.4.11 k=1 b=

4 3

F =2

a)

Averaged system

 x ¨ + x + ϵ bx3 + k x˙ − ax − F cos(t) = 0  − kr + F sin(ϕ) r = 2 ′

r

0 < ϵ 0

− 4ar − 3br3 + 4F cos(ϕ) ϕ = 8r ′



a = −1

φ

232

Chapter 8: Bifurcations Revisited

r

a=0

φ

r

a=1

φ

r

a=2

φ

b)

233

Chapter 8: Bifurcations Revisited

r

a = 2.8

φ

c) k = 1, b = 34 , F = 2, ϵ = 0.01 The simulation is run from t = 0 to t = 1200. a starts at -1, linearly increases to 5 at t = 300, then linear decreases back to -1 at t = 600. x

t

234

Chapter 8: Bifurcations Revisited

8.4.13 a) x˙ = a − bxy 2 x(t) =

c2 u(τ ) ab

y(t) =

u′ = 1 − uv 2 b) A=

αv

a2 b t c2 c3 α= 2 a b

a v(τ ) c

τ=

v ′ = αv(uv − 1)

é

Ñ −v 2 2

y˙ = bxy 2 − cy

−2uv

A(1,1) =

α(2uv − 1)

é Ñ −1 −2 α

∆=α

α

τ =α−1

Since a, c, b > 0 then α > 0, and transitioning across α = 1 will change the linearization from an unstable to stable spiral with only one fixed point, so a Hopf bifurcation occurs. c) v 3

2

1

u 1

2

3

def) As α is increased, the limit cycle keeps expanding into the first quadrant, with some critical value αc where the limit cycle expands to infinity. There’s several research papers analyzing this behavior in more detail and reviewing the literature would make for a good research project. 8.5.1 1 ln(I − Ic )

f (I) = Taking the first few derivatives d dI d2 dI 2

1 ln(I − Ic )

!

1 ln(I − Ic ) =

!

= 2

(I − Ic

)2

−1

2 (I − Ic ) ln(I − Ic )

1 3 + 2 2 (I − Ic ) ln(I − Ic ) ln(I − Ic ) 235

Chapter 8: Bifurcations Revisited

d3 dI 2

1 ln(I − Ic )

!

=

(I − Ic )3

−6

−6 −2 4 + 3 + 2 ln(I − Ic ) (I − Ic )3 ln(I − Ic ) (I − Ic )3 ln(I − Ic )

we can rewrite them all in terms of f (I) = f

f (1) = f (2) = f (3) =

−f 2 I − Ic

f2 2f 3 + f 2 2f 3 + = (I − Ic )2 (I − Ic )2 (I − Ic )2

−6f 4 −6f 3 −2f 2 −6f 4 − 6f 3 − 2f 2 + + = 3 3 3 (I − Ic ) (I − Ic ) (I − Ic ) (I − Ic )3

Maybe we can rewrite all the derivatives in terms of f Pn+1 f (n) =

f

(n+1)

=

Pn+1 p=2

ap pf p−1

f

p=2

ap f p

(I − Ic )n

(1)

−n

Pn+1 p=2

ap f p

(I − Ic )n (I − Ic )n+1 Pn+1 Pn+1 Å ã p−1 p −f 2 p=2 ap pf p=2 ap f = − n (I − Ic )n I − Ic (I − Ic )n+1 Pn+1 Pn+1 p p+1 − p=2 ap pf p=2 ap f − n = (I − Ic )n+1 (I − Ic )n+1 Pn+2 Pn+1 P n+1 ˜p f p − p=2 ap pf p+1 − n p=2 ap f p p=2 a = = (I − Ic )n+1 (I − Ic )n+1

The coefficients are a bit confusing to figure out, but we have a pattern. Now we just need to make sure that there is at least one term in the numerator for all derivatives in case everything cancels to zero somehow. Looking a little closer at the recurrence relation, the coefficient of the highest power of f in the numerator is Pn+1 p p=2 ap f (n) f = an+1 = (−1)n n! (I − Ic )n So the highest power of f in f (n) has an infinite limit at Ic . (In fact, all terms have an infinite limit as long as the coefficient doesn’t cancel out.) Therefore f (I) has infinite derivatives of all orders at Ic . 8.5.3

ã Å N ˙ N = rN 1 − K(t)

a) K(t) is positive, smooth, and T -periodic in t implies K(t) has a maximum and minimum Kmax and Kmin on the cylindrical phase space. Notice that N > Kmax ⇒ N˙ < 0 and N < Kmin ⇒ N˙ > 0 from the differential equation.  Next, we’ll define a Poincar´e map P N (t) = N (t + T ) as successive iterates of trajectories around the

cylinder. Just as in the text, we know P (N ) is continuous and monotonic, and there must be a fixed point of the Poincar´e map P (N ∗ ) = N ∗ corresponding to a limit cycle of the differential equation. 236

Chapter 8: Bifurcations Revisited

P (N )

P (Nj ) P (Ni )

N N∗

Ni

Nj

We also know that Kmin < N ∗ < Kmax since N = Kmin and N = Kmax form a trapping region for the stable limit orbit. b) We don’t know the stable limit cycle is unique from part (a) since there could be multiple crossings. We’ll need to prove it.

First, we do a change of variables. x=

ã Å ã Å r 1 x˙ N 1 = 1− ⇒ − 2 = N˙ = rN 1 − N x K(t) x xK(t) r ⇒ x˙ + rx = K(t)

This is a first order linear ODE, which can be solved with an integrating factor. x˙ + rx =

r K(t)

ert x˙ + rert x =

rert K(t)

rert d rt  e x = dt K(t) Z rert ert x = dt + C K(t) ! Z rert −rt dt + C x(t) = e K(t) We know the integral has a unique solution because K(t) being positive and smooth is well behaved, and that all solutions converge to this unique solution. lim x(t) = lim e

t→∞

t→∞

−rt

Z

rert dt + C K(t)

Therefore the stable limit cycle is unique. 237

!

−rt

−→ e

Z

rert dt K(t)

Chapter 8: Bifurcations Revisited

8.5.5 ˙ θ| ˙ + sin(θ) = F θ¨ + αθ|

α, F > 0

a) θ˙ = ν

ν˙ = F − αν|ν| − sin(θ)

   The fixed points are θ, ν = arcsin(F ), 0 and π − arcsin(F ), 0 F < 1 é Ñ 0 1 A= − cos(θ) −2α|ν| Ñ

A

=

é 1

0

√ − 1 − F2 0 Ñ é 0 1 = √ A π−arcsin(F ),0 1 − F2 0 arcsin(F ),0

1

λ1,2 = ±i(1 − F 2 ) 4 Center 1

λ1,2 = ±(1 − F 2 ) 4 Saddle Point

Now we aren’t sure if the predicted center is really a center. A quick numerical simulation will show it’s actually a stable spiral. Now we look for a suitable Liapunov function. With a little bit of experimentation with an “energy” relative to the fixed point gives p 1 2 ν − F θ − cos(θ) + F arcsin(F ) + 1 − F 2 2  V˙ = ν ν˙ − F θ˙ + sin(θ)θ˙ = ν F − αν|ν| − sin(θ) − F ν + sin(θ)ν = −αν 2 |ν| V (θ, ν) =

Our V is not strictly Liapunov because V˙ (θ, 0) = 0 meaning V˙ is not negative on the line ν = 0. However, this suffices to prove the fixed point is a stable spiral. That’s because the trajectories always move inward to lower and lower contours of V , except at instants when they cross the line ν = 0. At each such instant they move tangent to the V -contour, but an instant later they resume moving inward. Also, the V contours approach ellipses as we approach the fixed point, so the geometry of this situation is clear.

Thus the Liapunov “energy” V is almost always decreasing (but never increasing) towards the fixed point after each intersection with ν = 0 and the “energy” of the fixed point is zero. Therefore the fixed point is a stable spiral. b) We’ll construct a trapping region to prove the existence of the stable limit cycle. ν˙ = F − αν|ν| − sin(θ) ≥ 0 F − sin(θ) ≥ F − 1 ≥ αν|ν| … F −1 ⇒ ν˙ ≥ 0 ν= α ν˙ = F − αν|ν| − sin(θ) ≤ 0 238

Chapter 8: Bifurcations Revisited

F − sin(θ) ≤ F + 1 ≤ αν|ν| … F +1 ν= ⇒ ν˙ ≤ 0 α There are no fixed points in this region. Therefore there must be at least one stable limit cycle within … … F −1 F +1 ≤ν≤ 0 ≤ θ < 2π α α Next, we can use the total energy of the system after each cycle to prove there is a unique stable limit cycle within the bounds. Even though the system is losing energy to friction and gaining energy from the externally applied torque, the system will repeat the same velocity at the same angle each cycle. Therefore the total energy at that position will always be the same.

The total energy for a pendulum is given by E=

 1 2 ν + 1 − cos(θ) 2

Now we can repeat the reasoning in Section 8.5 Uniqueness of the limit cycle with some small changes. E is different, and also y ⇒ ν and ϕ → θ. Starting with Equation (7)

 dE d 1 2 dν = ν + 1 − cos(θ) = ν + sin(θ) dθ dθ 2 dθ dν F − αν|ν| − sin(θ) ν˙ = = ˙ dθ ν θ

Hence dE F − αν|ν| − sin(θ) =ν + sin(θ) = F − αν|ν| dθ ν Going back to the change in energy integral Z 2π Z 2π Z 2π  dE 2πF 0= dθ = F − αν|ν| dθ ⇒ ν(θ) ν(θ) dθ = dθ α 0 0 0

To prove uniqueness, suppose there were two distinct limit cycles. Following the notation and reasoning in Section 8.5 of the text, denote these by νU (θ) and νL (θ). Hence νU (θ) > νL (θ) implies Z 2π Z 2π νL (θ) νL (θ) dθ νU (θ) νU (θ) dθ > 0

0

which is a contradiction. Therefore the limit cycle is unique. c)

 1 2 ν t(θ) 2   1 d 2 dν dt 1 d du = ν t(θ) = ν ν t(θ) = ν = ν ν˙ = ν˙ = θ¨ dθ 2 dθ dθ dt dθ ν

u(θ) =

d)

ν > 0 −→ θ˙ > 0 239

Chapter 8: Bifurcations Revisited

˙ θ| ˙ + sin(θ) = F −→ θ¨ + αθ˙2 + sin(θ) = F θ¨ + αθ| 1 du = θ¨ u = ν 2 dθ 2

du θ¨ + αθ˙2 + sin(θ) = F −→ + 2αu + sin(θ) = F dθ This is an integrating factor problem with solution u(θ) =

2α 1 F cos(θ) − sin(θ) + Ce−2αθ + 2α 1 + 4α2 1 + 4α2

And the limit cycle is when the transient decays to zero, leaving u(θ) =

F 2α 1 cos(θ) − sin(θ) + 2α 1 + 4α2 1 + 4α2

e) We can find F at the homoclinic bifurcation by constraining u(θ) to go through one of the fixed points, which has to be the saddle point because homoclinic bifurcations occur with saddles.

The saddle point is

  (θ, ν) = π − arcsin(F ), 0 −→ (θ, u) = π − arcsin(F ), 0 F 1 2α + cos(θ) − sin(θ) 2α 1 + 4α2 1 + 4α2 Ä p ä 1 F 2α 2 − − + F 0= 1 − F 2α 1 + 4α2 1 + 4α2 2α −1 p F 1 − F2 = F− 2 1 + 4α 2α 1 + 4α2 p 4α2 F − F (1 + 4α2 ) = −2α 1 − F 2 p − F = −2α 1 − F 2

u(θ) =

F 2 = 4α2 (1 − F 2 ) = 4α2 − 4α2 F 2 F 2 (1 − 4α2 ) = 4α2 F = Fc (α) = √

2α 1 + 4α2

And that’s where the homoclinic bifurcation occurs. 8.6.1 θ˙1 = ω1 + sin(θ1 ) cos(θ2 )

θ˙2 = ω2 + sin(θ2 ) cos(θ1 )

ω1 , ω2 ≥ 0

a) Let ϕ1 = θ1 − θ2 which corresponds to the phase difference of the two oscillators. ϕ˙ 1 = θ˙1 − θ˙2 = ω1 + sin(θ1 ) cos(θ2 ) − ω2 − sin(θ2 ) cos(θ1 ) = ω1 − ω2 + sin(θ1 − θ2 ) = ω1 − ω2 + sin(ϕ1 ) 240

Chapter 8: Bifurcations Revisited

The fixed points correspond to phase-locked solutions of the original system, i.e. periodic trajectories on the torus. ϕ˙1 = ω1 − ω2 + sin(ϕ1 ) = 0 ϕ1 = 2π − arcsin(ω1 − ω2 ) π + arcsin(ω1 − ω2 )

|ω1 − ω2 | ≤ 1

A graphical approach for ω1 − ω2 ≥ 0 (and similarly for ω1 − ω2 ≤ 0) shows ϕ˙ 1

ω1 − ω2 ϕ1 ϕ˙ 1

ω1 − ω2 ϕ1 ϕ˙ 1

ω1 − ω2 ϕ1 ϕ˙ 1 ϕ1

ω1 − ω2

Next, we’ll look at ϕ2 = θ1 + θ2 , the fixed points of which corresponds to periodic trajectories on the torus running in opposite directions, in contrast to the fixed points of ϕ1 . (θ˙1 + θ˙2 = 0 ⇒ θ˙1 = −θ˙2 ) ϕ˙ 2 = θ˙1 + θ˙2 = ω1 + sin(θ1 ) cos(θ2 ) + ω2 + sin(θ2 ) cos(θ1 ) = ω1 + ω2 + sin(θ1 + θ2 ) = ω1 + ω2 + sin(ϕ2 ) 241

Chapter 8: Bifurcations Revisited

The fixed points are ϕ˙2 = ω1 + ω2 + sin(ϕ2 ) = 0 ϕ2 = 2π − arcsin(ω1 + ω2 )

π + arcsin(ω1 + ω2 )

ω1 + ω2 ≤ 1

A graphical approach for ω1 − ω2 ≥ 0 is the same as previously, except this time there is no ω1 + ω2 ≤ 0 case to consider because ω1 , ω2 ≥ 0. Now we have to go through all the possible cases for ω1 and ω2 and the existence of fixed points in the (ϕ1 , ϕ2 ) system and see what happens in the (θ1 , θ2 ) system. (Actually we only show graphs from about half the regions due to symmetry.) θ2

ω1 =

1 4

ω2 =

1 4

θ1

Fixed points Å (θ1 , θ2 ) =

ã Å ã Å ã Å ã Å ã Å ã Å ã 5π 17π 7π 7π 11π 11π 13π π 17π 5π 19π 19π π 13π , , , , , , , , , , , , , 12 12 12 12 12 12 12 12 12 12 12 12 12 12

242

Chapter 8: Bifurcations Revisited

θ2

ω1 =

1 2

ω2 = 2

θ1 θ2

ω1 = 1 ω2 = 1

θ1

Stable and unstable limit cycles present. 243

Chapter 8: Bifurcations Revisited

ω1 =

θ2

1 2

ω2 =

1 2

θ1

Fixed points

Å (θ1 , θ2 ) =

ã Å ã Å ã Å ã 3π 3π 5π π 7π 7π π 5π , , , , , , , 4 4 4 4 4 4 4 4

In this phase portrait, the fixed points are about to disappear in an infinite-period bifurcation if we view the phase portrait as on a torus. In their places will be limit cycles.

244

Chapter 8: Bifurcations Revisited

θ2

ω1 = 1 ω2 = 2

θ1

θ2

ω1 = 0 ω2 = 0

θ1

245

Chapter 8: Bifurcations Revisited

Fixed points Å ã Å ã Å ã Å ã Å ã Å ã 0π 0π 0π 2π 0π 4π 1π 1π 1π 3π 2π 0π (θ1 , θ2 ) = , , , , , , , , , , , , 2 2 2 2 2 2 2 2 2 2 2 2 ã Å ã Å ã Å ã Å ã Å ã Å ã Å 2π 4π 3π 1π 3π 3π 4π 0π 4π 2π 4π 4π 2π 2π , , , , , , , , , , , , , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ω1 = 0 ω2 = 1

θ2

θ1

Fixed points Å (θ1 , θ2 ) =

ã Å ã Å ã 0π 3π 2π π 4π 3π , , , , , 2 2 2 2 2 2

b) We have saddle-node bifurcation of cycles occurring along the |ω1 + ω2 | = 1 lines and infinite-period bifurcations occurring along the ω1 − ω2 = 1 line. c)

246

2|

=

1

Chapter 8: Bifurcations Revisited

1 2|

=

of c

s

Limit cycles

+

cy cl e

ω1

Sa dd

le -n o

de

bi



1



ω

fu rc at io n

No fixed points No limit cycles

yc le s



1



ω

ω2

Sa

n io

at

rc

fu bi

dd

d

le -

no

io er

de

-p ite

bi fu

rc

fin In

at

1

io n

=

of

ω2

Fixed points

No fixed points No limit cycles ω1

The oscillator death region is the lower left triangle. Once we cross the infinite-period bifurcation ω1 +ω2 = 1 line, the limit cycles are broken by fixed points and the oscillations stops. 8.6.3 This problem is actually quite common and there are multiple proofs. Here is one without too much prerequisite knowledge. ω1 < 1 ω since exchanging ω1 with ω2 is an equivalent problem. Also, we only need to prove that the irrational orbit Without loss of generality, assume that q = (0, 0) as we can translate to any point and ω =

is dense in {(θ1 , θ2 ) : θ2 = 0} (the bottom of the phase space) since a dense set at a cross section of the torus will also be dense everywhere when translated by the irrational flow. So now we a have a somewhat simpler problem to work on.

We’ll be using Dirichlet’s Approximation Theorem which implies that for every irrational number ω there exists integers a and b such that

a 1 ω − < 2 b b a The orbits that start at (θ1 , θ2 ) = (0, 0) with slope intersect the bottom of the phase space at b ß ™ 1 2 b−1 0, , , · · · b b b 247

Chapter 8: Bifurcations Revisited

So the rational orbit passes within distance

1 of every point at the bottom of the phase space. 2b

Now applying the implication of Dirichlet’s Approximation Theorem n 1 a n = 0, 1, 2, . . . (b − 1) n ω − < 2 < b b b

n 1 distance of and As long as 0 ≤ n ≤ b − 1, then the n’th intersection of the irrational orbit is within b b 1 therefore the first b − 1 iterations of the irrational orbit are with distance of every point at the bottom of b the phase space. Hence the irrational orbit can be made arbitrarily close to any point at the bottom of the phase space, and consequently any point on the entire torus, by picking b large enough. 8.6.5 a)

b)

y

c)

y

x

x

ω=3 d)

e)

ω= f)

y

x

ω=

m¨ r=

5 3

y

x

√ 2

8.6.7

x

2 3

ω=

y

y

x

ω=π

h2 −k mr3

θ˙ =

ω=

h mr2

√ 1+ 5 2

m, k, h > 0

a) r(t) = r0 ⇒ r¨ = 0 h2 h2 − k −→ 0 = − k ⇒ r0 = m¨ r= mr3 mr03 Å 2 ã 31 Å ã 32 k ˙θ = h = h mk = = ωθ 2 2 mr0 m h mh

Å

h2 mk

ã 13

b) First, we’ll transform the r¨ equation to a system and then linearize about (r, r) ˙ = (r0 , 0) x=r

y = r˙

x˙ = y 248

y˙ =

k h2 − m2 x3 m

Chapter 8: Bifurcations Revisited

Ñ A=

0 −3h2 m 2 x4

é 1 0

Ñ A(r0 ,0) =

é 0

1

−3h2 m2 r04

0

Then the frequency of small radial oscillations is ωr =

√ h √ 3h2 = 3 2 = 3ωθ m2 r04 mr0

c) The winding number ωr = ωθ

√ √ 3ωθ = 3 ωθ

is irrational and hence the small radial oscillations are quasiperiodic. d) The linearization from part (c) predicts a center, but we didn’t actually check that it’s a center. Turns out it really is a center and you can find a conserved quantity, OR if we look closely the r equation is a nonlinear spring with no damping and r = r0 rest position. The solution to r(t) is periodic for any initial conditions. The periodicity of r(t) implies that θ˙ is periodic. The θ˙ equation is also strictly positive. Both these facts together imply θ(t) has constant finite period, which is determined by the r(t) initial conditions. The period can vary for different r(t) initial conditions, but the period is constant for each choice of r(t) initial conditions. More precisely  r(0), r(0) ˙ = (a, b) θ(Ta,b ) − θ(0) = 2π ⇒ θ(t + Ta,b ) − θ t) = 2π  r(0), r(0) ˙ = (α, β) θ(Tα,β ) − θ(0) = 2π ⇒ θ(t + Tα,β ) − θ t) = 2π

However, if (a, b) ̸= (α, β) then Ta,b may or may not equal Tα,β because they correspond to different initial conditions.

Now to the geometric argument. We just proved that ωr and consequently ωθ are constant for any amωr plitude of radial oscillation. Then the winding number is also constant. The winding number is either a ωθ rational or irrational number, which correspond to periodic and quasiperiodic orbits on a torus with dimensions corresponding to the r(t) and θ(t) periods respectively. There is no chaos. e) Two masses are connected by a string of fixed length. The first mass plays the role of the particle; it moves on a frictionless, horizontal “air table.” It is connected to the second mass by a string that passes through a hole in the center of the table. This second mass hangs below the table, bobbing up and down and supplying the constant force of its weight. This mechanical system obeys the equations given in the text, after some rescaling.

249

Chapter 8: Bifurcations Revisited

8.6.9 θ˙1 = ω + H(θ2 − θ1 ) θ˙2 = ω + H(θ1 − θ2 ) θ˙1 = ω + H(θ2 − θ1 ) + H(θ3 − θ1 ) θ˙2 = ω + H(θ1 − θ2 ) + H(θ3 − θ2 ) θ˙3 = ω + H(θ1 − θ3 ) + H(θ2 − θ3 ) a) ϕ˙ = θ˙1 − θ˙2 = H(θ2 − θ1 ) − H(θ1 − θ2 ) = H(−ϕ) − H(ϕ) ϕ˙ = θ˙1 − θ˙2 = H(θ2 − θ1 ) + H(θ3 − θ1 ) − H(θ1 − θ2 ) − H(θ3 − θ2 ) = H(−ϕ) + H(−ϕ − ψ) − H(ϕ) − H(−ψ) ψ˙ = θ˙2 − θ˙3 = H(θ1 − θ2 ) + H(θ3 − θ2 ) − H(θ1 − θ3 ) − H(θ2 − θ3 ) = H(ϕ) + H(−ψ) − H(ϕ + ψ) − H(ψ) b) H(x) = a sin(x) For the two frog system ϕ˙ = H(−ϕ) − H(ϕ) = a sin(−ϕ) − a sin(ϕ) = −2a sin(ϕ) = 0 ⇒ ϕ = 0, π dϕ˙ dϕ˙ dϕ˙ = −2a = 2a = −2a cos(ϕ) dϕ dϕ dϕ ϕ=0

ϕ=π

If a < 0 then the ϕ = π antiphase synchronization state will be stable.

For the three frog system ϕ˙ = H(−ϕ) + H(−ϕ − ψ) − H(ϕ) − H(−ψ) = a sin(−ϕ) + a sin(−ϕ − ψ) − a sin(ϕ) − a sin(−ψ) = −2a sin(ϕ) + a sin(ψ) − a sin(ϕ + ψ) ψ˙ = H(ϕ) + H(−ψ) − H(ϕ + ψ) − H(ψ) = a sin(ϕ) + a sin(−ψ) − a sin(ϕ + ψ) − a sin(ψ) = −2a sin(ψ) + a sin(ϕ) − a sin(ϕ + ψ) ˙ π) = ψ(0, ˙ π) = 0 ϕ(0, Å ã Å ã 2π 2π 2π 2π ˙ ˙ ϕ , =ψ , =0 3 3 3 3 250

Chapter 8: Bifurcations Revisited

The two experimentally stable states are fixed points according to the model.

Next we linearize A=

Ñ −2a cos(ϕ) − a cos(ϕ + ψ)

é a cos(ψ) − a cos(ϕ + ψ)

a cos(ϕ) − a cos(ϕ + ψ) −2a cos(ψ) − a cos(ϕ + ψ) Ñ é −a 0 ∆ = −3a2 τ = 2a A(0,π) = 2a 3a ∆ < 0 ⇒ Saddle point Ñ é 2 3a ∆ = 9a4 0 τ = 3a 2 A( 2π , 2π ) = 3 3 3a 0 a < 0 ⇒ τ < 0 Stable 2

So the experimental fixed points exist, but the pair of frogs in unison and the third out of phase is unstable. c) H(x) = a sin(x) + b sin(2x)

a = −5 b = 1

ϕ 2π

π

t

0

251

Chapter 8: Bifurcations Revisited

ψ

ϕ

d) H(x) = a sin(x) + b sin(2x) + c cos(x)

a = −5 b = 1 c = 0.1

252

Chapter 8: Bifurcations Revisited

ϕ 2π

π

t

0 ψ

ϕ

So the fixed points and their stability remained even after adding a small even component to H. 253

Chapter 8: Bifurcations Revisited

8.7.1

Z

r1

r0

dr = 2π r(1 − r2 )

1 A B C 1 = = + + r(1 − r2 ) r(1 + r)(1 − r) r 1+r 1−r 1 = A(1 + r)(1 − r) + Br(1 − r) + Cr(1 + r) = A(1 − r2 ) + B(−r2 + r) + C(r2 + r) = (−A − B + C)r2 + (B + C)r + A ⇒A=1 B= Z

r1

r0

−1 2

C=

1 2

Z r1 Å ã ã 1 1 1 1 1 1 dr = dr − + − − r 2(1 + r) 2(1 − r) r 2(r + 1) 2(r − 1) r0 r òr1 ï 0 1 1 = ln |r| − ln |r + 1| − ln |r − 1| 2 2 r0 Å ã Å ã 1 1 1 1 = ln |r1 | − ln |r1 + 1| − ln |r1 − 1| − ln |r0 | − ln |r0 + 1| − ln |r0 − 1| 2 2 2 2 p r r r2 − 1 1 r2 (r2 − 1) r 1 0 1 = 2π = ln p 2 − ln p 2 = ln p 02 = ln 1 0 r0 − 1 r0 r1 − 1 2 r02 (r12 − 1) r1 − 1

dr = r(1 − r2 )

Z

r1

Å

We can drop the absolute value because r = 1 is a limit cycle, so (r1 − 1) and (r0 − 1) have to have the same sign. r12 (r02 − 1) = e4π r02 (r12 − 1) r12 (r02 − 1) = r02 (r12 − 1)e4π r12 (r02 − 1) − r02 (r12 − 1)e4π = 0 r12 (r02 − 1 − r02 e4π ) + r02 e4π = 0 r12 =

−r02

1 r02 e4π 1 = = + 1 + r02 e4π −e−4π + r0−2 e−4π + 1 1 + e−4π (r0−2 − 1) 1

r1 = » 1 + e−4π (r0−2 − 1)

1 rn+1 = P (rn ) = » 1 + e−4π (rn−2 − 1) d e−4π r−3 P (r) = p dr 1 + e−4π (r−2 − 1) d P (1) = e−4π dr

8.7.3 x˙ + x = F (t)

F (t) =

254

  +A 0 < t <  −A

T 2

T 2