Introduction to Mathematical Modeling and Chaotic Dynamics (Instructor Solution Manual, Solutions) [1 ed.] 9781482224535, 9781439898864, 1439898863

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solutions MANUAL FOR

Introduction to Mathematical Modeling and Chaotic Dynamics by

Ranjit Kumar Upadhyay Satteluri R. K. Iyengar

solutionS MANUAL FOR

Introduction to Mathematical Modeling and Chaotic Dynamics by

Ranjit Kumar Upadhyay Satteluri R. K. Iyengar

Boca Raton London New York

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

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20130709 International Standard Book Number-13: 978-1-4822-2453-5 (Ancillary) This book contains information obtained from authentic and highly regarded sources. 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. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Introduction to Mathematical Modeling and Chaotic Dynamics

Solution Manual

Dr. Ranjit Kumar Upadhyay Dr. Satteluri R. K. Iyengar

1

Chapter 1 Exercise 1.1 1. Let the dimensions of the box be: width = l, length = 2l, height = h. Volume = 2l 2 h = 20. Cost of the total surface area of the open box is C = {30(l 3 + 10) / l}. Setting ( dC / dl ) = 0, we get l 3 = 5. Substituting in C, we get the minimum cost as 90(52 / 3 )$. Since ( d 2C / dl 2 ) > 0 when l 3 = 5, the cost is a minimum. 2. Let x denote the number of Electronic Music Systems expected to be sold per week after giving a rebate. The weekly increase in sale is ( x − 2000). The price function (demand function) is given by p ( x ) = 3500 − {100( x − 2000) / 200} = 4500 − 0.5 x. The revenue

function is R( x ) = xp ( x ) = 4500 x − 0.5 x 2 . Setting R ′( x ) = 0, we get x = 4500. Since R ′′( x ) < 0, the revenue is maximized. The corresponding price per unit should be p ( x = 4500) = 2250. The rebate that should be offered is 3500 − 2250 = 1250$. 3. By Newton’s law of cooling, we obtain T − 20 = ( 90 − 20)e − kt = 70e − kt . The temperature is reduced to 600 C in 10 minutes. Hence, 4 = 7e −10k , or k = − ln( 4 / 7) / 10. To cool to 35 0 C , we require the total time t = [10 ln(15 / 70) / ln(4 / 7)] ≈ 27.5 minutes, that is another 17.5 minutes. 4. x1′ = −2 x 2 + r x1 , x2′ = 3x1 + r x 2 , r = ( x12 + x22 )1 / 2 .

Choose V ( x ) = 3x12 + 2 x22 . V * ( x ) = 2 rV . Unstable. 5. x1′ = −4 x1 + 8 x1 x22 ,

x2′ = −6 x2 − 12 x12 x2 .

Choose V ( x ) = [( 3x12 + 2 x22 ) / 24]. V * ( x ) = −( x12 + x22 ). Asymptotically stable. 6. x1′ = − x1 − x22 , x2′ = 2 x1 x2 − x23 . Choose V ( x ) = ax12 + bx22 and obtain a = 2b. Let b = 1. V ( x ) = 2 x12 + x22 . V * ( x ) = −2( 2 x12 + x22 ). Asymptotically stable. 7. x1′ = − x13 + x14 , x2′ = x14 − x23 .

Choose V ( x) = x12 + x22 . We get V * ( x ) = −2{ x14 + x 24 − x14 ( x1 + x 2 )}. neighborhood of origin (0, 0), V * ( x ) < 0. Asymptotically stable.

In

the

8. x1′ = − x1 x22 p , x2′ = x12 q x2 , (p, q are positive integers). Choose V ( x ) = x12 q + x22 p . V * ( x ) < 0, for p < q. Asymptotically stable in this case. V * ( x ) = 0, for p = q. Stable. 9. x1′ = −4 x1 − x1 x22 ,

x2′ = −6 x2 − x12 x2 .

⎡− 2( 4 + x22 ) − 4 x1 x 2 ⎤ M ( x1 , x2 ) = ⎢ . det ( M 1 ) = −2( 4 + x22 ) < 0. 2 ⎥ − 2( 6 + x1 )⎦ ⎣ − 4 x1 x 2

det ( M 2 ) = 4( 24 + 4 x12 + 6 x22 − 3 x12 x22 ) > 0, in the neighborhood of origin. M 1 , M 2 are the leading minors of M. M is negative definite. Origin (0, 0) is asymptotically stable. 2

10. x1′ = −2 x1 − 3x 2 − 5 x15 ,

x2′ = 3x1 − 2 x2 − 2 x25 .

⎤ ⎡− 2( 2 + 25 x14 ) 0 . M ( x1 , x2 ) = ⎢ 4 ⎥ 0 4 ( 1 5 ) − + x 2 ⎦ ⎣ M is negative definite. Origin (0, 0) is asymptotically stable. 11. z ′′′ + 9 z ′′ + 26 z ′ + 24 z = 0. Eigen values of A are −2, −3, −4. B = diag (1 / 4, 1 / 6, 1 / 8).

V ( y ) = (1 / 4) y12 + (1 / 6) y22 + (1 / 8) y 32 . Asymptotically stable. 12. z ′′′ − 4 z ′′ − 4 z ′ + 16 z = 0. Eigen values of A are −2, 2, 4. B = diag(1 / 4, − 1 / 4, − 1 / 8). V(y) is not positive definite. Method cannot be applied. 13. P (λ ) = λ3 + 9λ2 + 6λ + 1. ai > 0 for all i and a1a 2 − a0 a 3 > 0. The roots of P(λ ) = 0 are negative or have negative real parts. 14. P(λ ) = λ3 + 3λ2 + λ + 8. ai > 0 for all i and a1a 2 − a0 a3 < 0. One or more roots of P(λ ) = 0 have positive real parts. 15.

P (λ ) = ( 2 − 3 p )λ3 + ( 4 + p )λ2 + 2(1 + p )λ + p. For P(λ ) = 0 are negative or have negative real parts.

0 < p < ( 2 / 3) , the roots of

16. x1′ = − x1 + 3e x 2 − 3 cos x1 , x 2′ = −1 + e x 2 − 6 x 2 − sin x1. 3⎤ ⎡− 1 A=⎢ ⎥. Eigen values of A are −2, −4. A is stable. R(x) satisfies the condition ⎣− 1 − 5⎦ (1.16). By Theorem 1.5, zero solution is asymptotically stable. 17. x1′ = −1 − x2 + e x1 , x2′ = 4 x1 − 2 sin x2 . − 1⎤ ⎡1 A=⎢ ⎥. Eigenvalues of A are [{−1 ± i 7 } / 2] and have negative real parts. A is ⎣4 − 2 ⎦ stable. R(x) satisfies the condition (1.16). By Theorem 1.5, zero solution is asymptotically stable.

18. x1′ = − x1 + 3e x 2 − 3 cos x1 , x 2′ = 1 + 6 x 2 − e x 2 − sin x1 . 3⎤ ⎡ 1 A=⎢ ⎥. Eigenvalues of A are 2, 4. By Theorem 1.6, zero solution is unstable. ⎣− 1 5 ⎦ 19. x1′ = − x2 − 4 x13 , x2′ = 9 x1 − 2 x23 . ⎡0 − 1⎤ . Eigenvalues of A are λ = ±3i. Critical case. Choose A= ⎢ 0 ⎥⎦ ⎣9 V ( x1 , x2 ) = 9 x12 + x22 . V * = −4(18 x14 + x24 ). Zero solution is asymptotically stable. 20. The unique positive equilibrium point ( x1* , x2* ) is the solution of the equations

3 ln x1 + 2 ln x2 = 11,

4 ln x1 + 3 ln x2 = 13. We obtain x1* = e 7 , x2* = e −5 . Using the

transformation y1 = ln( x1 / x1* ) = ln x1 − 7, and y 2 = ln( x 2 / x 2* ) = ln x 2 + 5, we obtain the linear system ⎡ y1′ ⎤ ⎡ − 3 − 2⎤ ⎡ y1 ⎤ ⎢ y ′ ⎥ = ⎢− 4 − 3⎥ ⎢ y ⎥. ⎦ ⎣ 2⎦ ⎣ 2⎦ ⎣ 3

The characteristic equation associated with the above linear system is λ2 + 6λ + 1 = 0. The roots of the characteristic equation have negative real parts. ( x1* , x2* ) is globally asymptotically stable. 21. x1′ = − x2 + 6 x1 ( x12 + x22 − 3), x2′ = x1 + 6 x2 ( x12 + x22 − 3). Period: 2π . C : x1 = 3 cos t , x2 = 3 sin t . Limit cycle is unstable. 22. x1′ = −3 x2 + 2 x1 (1 − x12 − x22 ), x 2′ = 3 x1 + 2 x 2 (1 − x12 − x 22 ). Period: ( 2π / 3) . C : x1 = cos 3t , x2 = sin 3t. Limit cycle is Stable. 23. x1′ = − x2 + x1 (1 − x12 − x22 ) 2 , x2′ = x1 − x2 (1 − x12 − x22 ). Period: 2π . C : x1 = cos t , x2 = sin t. I = 0. Study of the nonlinear terms is required to decide whether C is semistable. 24. x1′ = −2 x1 + 3 x2 , x2′ = − x1 − 6 x23 . Choose V ( x) = x12 + 3 x 22 . Now, V ( x) > 0 and V * ( x) < 0 for all ( x, y ) ≠ (0, 0). Hence, there are no closed orbits. 25. x1′ = −2 x1 + 4 x 23 − 4 x 24 , Choose x2′ = − x1 − x2 + x1 x2 . V ( x) = x12 + 2 x24 . Now, V ( x) > 0 and V * ( x) < 0 for all ( x, y) ≠ (0, 0). Hence, there are no closed orbits. 26. x + μ ( x 4 − 1) x + x = 0, μ > 0 . We have f ( x ) = μ ( x 4 − 1), g ( x ) = x, F ( x) = ( μ x / 5)( x 4 − 5), α = 5 . By Theorem 1.9 ( Lie ′nard' s Theorem), the system has a unique limit cycle. 27. x + μ ( x 2 − 1) x + x 3 = 0, μ > 0. We have f ( x ) = μ ( x 2 − 1), g ( x ) = x 3 , F ( x) = ( μ x / 3)( x 2 − 3), α = 3 . By Theorem 1.9 ( Lie ′nard' s Theorem), the system has a unique limit cycle. 28. Writing it as a system, we get x = y, y = −[ε ( x 6 − 1) y + x ]. Here, h ( x, y ) = ( x 6 − 1) y. On the limit cycle, we have the approximation, x( t ) ≈ a cos t, y ( t ) ≈ −a sin t , T ≈ 2π . The energy balance equation gives 2π

∫ 0

[ a 6 (cos6 t − cos8 t ) + cos2 t − 1)]dt =

π (5a 6 − 64) = 0. 64

The positive solution of this equation is a 0 = amplitude = 2 / 6 5. Also, πε g ′( a0 ) = − ( 40a 07 − 128a0 ) = −6πεa0 < 0. 64 The limit cycle is stable when ε > 0 and unstable when ε < 0. 29. x ′ = ( x / 2) − y , y ′ = x + ( y / 2 ). − 1⎞ ⎛1 / 2 ⎟ . The eigen values are λ = (1 / 2) ± i . We have A = ⎜⎜ 1 / 2 ⎟⎠ ⎝1

p = 1 > 0,

q = 5 / 4 > 0, Δ = p 2 − 4q = −4 < 0. Origin (0, 0) is an unstable focus. 30. x ′ = 3 x − 2 y , y ′ = 9 x − 3 y. ⎛3 − 2 ⎞ ⎟ . The eigen values are λ = ±3i . Here p = 0, q = 9 > 0. Therefore, (0, 0) is A = ⎜⎜ − 3 ⎟⎠ ⎝9 a centre and is stable. 4

31. x = −5 y − x ( x 2 + y 2 ) , r (0) = r0 , y = 5 x − y ( x 2 + y 2 ) , θ (0) = θ 0 . x 2 + y 2 ≠ 0. Define f(0) = 0. We obtain r = −r 2 , and θ = 5. Origin is an equilibrium point of the system. Integrating and using the initial conditions, we obtain r = r0 /(1 + t r0 ), for t > 0. θ = 5t + θ 0 . We find that r → 0 , and θ → ∞ as t → ∞ . Therefore, origin is a stable focus. 2y 2x 32. x = −3x − , r (0) = r0 , y = −3 y + , θ ( 0) = θ 0 . x 2 + y 2 ≠ 0. 2 2 5 ln( x + y ) 5 ln( x 2 + y 2 ) Define f(0) = 0. We obtain r = −3r, r = r0e−3t ,

θ =

1 1 1 ⎡ 3t ⎤ = , θ = θ 0 − ln ⎢1 − . 5 ln r 5(ln r0 − 3t ) 15 ⎣ ln r0 ⎥⎦

For, r0 < 1, r( t ) → 0 and θ → ∞ , as t → ∞. Origin is a stable focus in this case. 33. van der Pol oscillator x − ε (1 − x 2 ) x + x = 0, where ε > 0, is a damping constant. div( F ) = ε (1 − x 2 ) < 0 for

x > 1. System is dissipative for

x > 1. div( F ) > 0 for

x < 1. The system is an area expanding dynamical system for x < 1. 34. Lorenz system (a simplified model of atmospheric convection). dx dy dz = σ ( y − x ); = − xz + rx − y; = xy − bz. b and σ are positive constants. dt dt dt div( F ) = −(σ + 1 + b) < 0. System is dissipative. 35. Rossler equation : x = −( y + z ), y = x + ay, z = b + xz − cz, with a = b = 0.2 and c = 5.7. div( F ) = a + x − c = −5.5 + x < 0 for 0 < x < 5.5. System is dissipative for 0 < x < 5.5. Area expanding for x > 5.5. 36. The Jacobian matrix of the system is given by ⎡ ∂x n + 1 ∂x n + 1 ⎤ ⎢ ∂x − 2 yn ⎤ ∂y n ⎥ ⎡1 − a J(X ) = ⎢ n ⎥=⎢ ⎥. ⎢ ∂y n +1 ∂y n +1 ⎥ ⎣ y n 1 + b + x n ⎦ ⎢⎣ ∂x n ∂y n ⎥⎦

Now, det( J ( X ) = (1 − a )(1 + b + x n ) + 2 y n2 > 1 for a ≤ 0 and b ≥ 0. Therefore, Burgers map is area expanding for these cases. 37. The Jacobian matrix of the system is given by ⎡ ∂xn +1 ∂xn +1 ⎤ ⎢ ∂x 0 ⎤ ∂yn ⎥ ⎡− 2( a + c) xn J(X ) = ⎢ n . ⎥=⎢ 2b − 2( a − c ) y n − 2b⎥⎦ ⎢ ∂y n +1 ∂y n +1 ⎥ ⎣ ⎢⎣ ∂xn ∂yn ⎥⎦ Now, det( J ( X ) = 4( a 2 − c 2 ) x n y n + 4b( a + c ) xn . For a = c > 0, b > 0, det( J ( X ) = 8abxn < 8ab, in the unit square, 0 < xn < 1, 0 < yn < 1 . Therefore, the map is dissipative when 8ab < 1 , and area expanding when 8ab > 1.

38. x = y − x , y = 6 + y + x (4 − 2 x − x 2 ). Hamiltonian system. 5

H ( x, y ) = −6 x − xy − 2 x 2 + ( 2 / 3) x 3 + (1 / 4) x 4 . Equilibrium points are (−1, −1), (2, 2), (−3, −3). All points are saddle points. 39. x = y ( 4 − x 2 ), y = x ( x 2 + y 2 + 5). Hamiltonian system. H ( x, y ) = (8 y 2 − 2 x 2 y 2 − 10 x 2 − x 4 ) / 4. Equilibrium point (0, 0) is a saddle point.

40. x = 4 − x 2 − y 2 , y = x( x 2 + y 2 + 3). Not a Hamiltonian system. 41. x = y ( 4 + 5 x + x 2 ), y = [ x(1 − 2 y 2 ) − 5 y 2 ] / 2. Hamiltonian system. H ( x, y ) = [(8 + 10 x + 2 x 2 ) y 2 − x 2 ] / 4. Equilibrium points are (0, 0), ( −4, ± 2 / 3 ) . All are saddle points. 42. x = y ( 4 − 5 x − x 2 ), y = [(5 + 2 x ) y 2 − x ] / 2. Hamiltonian system. H ( x, y ) = [(8 − 10 x − 2 x 2 ) y 2 + x 2 ] / 4. Equilibrium points are (0, 0), ( x0 , ± y0 ), where

x0 = ( −5 + 41) / 2, y0 = ( 41 − 5 41) / 82 ) . (0, 0) is a centre. The remaining two points are saddle points. 43. Lotka-Volterra population model governing the growth/ decay of species x = ax − bxy, y = cxy − dy, for a = 1.0 , b = 0.2, c =0.4, d =0.5. % Matlab code for Lotka-Volterra model in Problem 43 g=inline('[1.*x(1)- (0.2.*x(1).*x(2));0.5.*x(2)+(0.04.*x(1).*x(2))]','t','x'); [t xa]=ode45(g,[0 150],[10 10]) plot(xa(:,1),xa(:,2)) %plot3(xa(:,1),xa(:,2),xa(:,3)); %comet3(xa(:,1),xa(:,2),xa(:,3)); hold on; figure; plot(t,xa(:,1),'r'); hold on; plot(t,xa(:,2),'g'); The phase plot and Time series are given in Figs. 1.1 and 1.2. 11 10

35

9

30

x y

8 25 Population density

y

7 6 5 4

20

15

10

3 5

2 1

0

5

10

15

20

25

30

35

x

Fig.1.1. Phase plot for Problem 43.

0

0

50

100

150

Time

Fig.1.2. Time series for Problem 43.

44. R o ssler model x = −( y + x ), y = x + ay, z = b + z ( x − c), when a = 0.2, b = 0.2 , c = 5.7 . % Matlab code for the R o ssler model g=inline('[-(x(2)+x(3));x(1)+0.2.*x(2);0.2+x(3).*(x(1)-5.7)]','t','x'); [t xa]=ode45(g,[0 500],[0.5 0.5 0.5]) 6

plot3(xa(:,1),xa(:,2),xa(:,3)) hold on; figure; plot(t,xa(:,1),'r'); hold on; plot(t,xa(:,2),'g'); plot(t,xa(:,3),'b'); The phase plot and Time series are given in Figs. 1.3 and 1.4. Chaotic Attractor

x y z

25 20

25

15

15

10

z

State Variables

20

10 5 0 10

5 0 -5

0 -10 -20

y

-5 -10

15

10

5

0

-10 -15 400

x

450

500

550

600

650

Time

Fig.1.3. Phase plot for Problem 44.

Fig.1.4. Time series for Problem 44.

45. Phase plot and the corresponding time series using MATLAB for the given model. when a1 = 2.5, b1 = 0.05, w = 0.85, α = 0.45, β = 0.2, γ = 0.6, a 2 = 0.95, and w1 = 1.65. % Matlab code for Problem 45 g= inline('[2.5.*u(1)−0.05.*u(1).*u(1) − (0.85.*u(1).*u(2))./(0.45+0.2.*u(2)+0.6.*u(1)); −0.95.*u(2)+(1.65.*u(1).*u(2))./(0.45+0.2.*u(2)+0.6.*u(1))]','t','u'); [t ua]=ode45(g,[400 550],[5.9157 35.6279]) figure; plot(t,ua(:,1),'b'); hold on; plot(t,ua(:,2),'r'); xlabel('time'); ylabel('state variable'); legend('x','y'); figure; plot(ua(:,1),ua(:,2)); xlabel('x'); ylabel('y'); The phase plot and Time series are given in Figs. 1.5 and 1.6. 40 35 30

y

25 20 15 10 5 0

0

2

4

6

8

10 x

12

14

16

18

20

Fig.1.5. Phase plot for Problem 45.

Fig.1.6. Time series for Problem 45. 7

46. Basin boundary for the limit cycle and chaos for the food-chain model (Rai and Upadhyay [24]) when a1 = 1.75, b1 = 0.05, a 2 = 1.0, c = 0.7, w = 1.0, w1 = 2.0, w2 = 1.5, w3 = 3.75, D = 10.0, D1 = 10.0, D2 = 10.0, D3 = 20.

Basin boundaries for the chaotic attractor are plotted in Fig. 1.7.

Fig.1.7. Basin boundaries for the chaotic attractor for Problem 46. (From Rai, V., Upadhyay, R. K., Chaotic population dynamics and biology of the top-predator. Chaos, Solitons Fractals, 21, 1195–1204, Copyright 2004, Elsevier. Reprinted with permission).

CHAPTER 2 Exercise 2.1 1. Without loss of generality, let the initial time be t = 0. The solution is given by P(t ) = P0 e rt . At time t = 0, we have P = P0 . Let at time t = T , the population is tripled, that is P(T ) = 3P0 . Hence, 3P0 = P0 e rT , or T = ln 3 / r. The time taken for the population to triple its size is T = ln 3 / r. 2. We have r = 0.09, K = 900, t0 = 0, P0 = 90. The value of the constant A is obtained as A = [( P0 − K ) / P0 ] = −9. The solution is given by K 900 900 = . P(90) = ≈ 897. P (t ) = − rt − 0.09t 1 − Ae 1 + 9e 1 + 9e − 8.1 3. (i) Equilibrium solutions are P1 = 0 and P2 = 500. Now, P ⎞ ⎛ f ′( P ) = 0.5⎜1 − f ′( P1 ) = 0.5 > 0, f ′( P2 ) = −0.5 < 0. ⎟, ⎝ 250 ⎠ Hence, the equilibrium point P1 = 0 is unstable and the equilibrium point P2 = 500 is stable. Thus, solutions initiating in a neighbourhood of K = 500 approach K as t → ∞ while no solution starting in a neighbourhood of P = 0 remains close to zero in the future. (ii) P (t ) = 500 /[1 + 9e −0.5t ]. K 100 dP P ⎞ ⎛ 4. = , = (0.0025) P⎜1 − ⎟. The solution is P(t ) = − rt dt ⎝ 100 ⎠ 1 − Ae 1 − Ae − 0.0025t where A = −11.5 e 4.9875. The estimated populations in the years 2005 and 2100 are 8.18 and 10.15 billions respectively.

8

5. Separating the variables and solving the differential equation, we get R /( R − 1) = ceαt , a, c are arbitrary constants to be determined. Applying the conditions that at 8 hours, R = 0.08, and at 12 hours R = 0.05, we obtain a = 0.25 ln(11.5), and c = − e −12a ≈ − 1 / 1520.9. Now, when R = 0.9, we obtain t = 15.5985 or approximately 3.36 P.M. 2 α ⎤ ⎡ ⎛ P ⎞α ⎤ d 2P ⎛ r ⎞ ⎡ ⎛ P ⎞ 6. We have α P 1 ( 1 ) = − + ⎥ ⎢1 − ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ dt 2 ⎝ α ⎠ ⎢⎣ ⎝ K ⎠ ⎥⎦ ⎢⎣ ⎝ K ⎠ ⎥⎦ α

2

⎛P⎞ ⎛r⎞ = ⎜ ⎟ P[1 − t ( 2 + α ) + t 2 (1 + α )], where t = ⎜ ⎟ . ⎝K⎠ ⎝α ⎠ Setting P ′′( t ) = 0, we get P = 0, t = 1, and t = 1 /(1 + α ), that is, P = 0, P = K, and P = K (1 + α ) −1 / α . The point of inflection is P = K (1 + α ) −1 / α . For α = 1, the point of inflection is P = K/2. Taking the limit as α → 0, we find that the point of inflection moves to P = K/e.

7. Comparing with (2.12), we have r = 10 −3 , γ = 10 −9 , P0 = 105. The solution is given by P (t ) =

rP0

γ P0 + ( r − γ P0 ) e − rt

=

106 . 1 + 9 exp( −0.001t )

The limiting value of the population is 106. 8. Comparing with the model for the density dependent growth (2.12), we have r = 0.09, γ = 0.0009, P0 = 2500. The solution of the model is obtained as rP0 P (t ) = γ P0 + ( r − γ P0 )e − rt (0.09)( 2500) 225 = = . − 0.09t (0.0009)( 2500) + {0.09 − 0.0009( 2500)}e 2.25 − 2.16e − 0.09t The limiting value of the population is 100. The equilibrium points are 0 and 100. The point 0 is unstable and the point 100 is stable. All other solutions move away from P = 0, towards P =100. 9. P(t ) = P0 exp[( r0 / α )(1 − e −αt )]. 10. Assume that no student leaves the campus throughout the duration of viral fever. Now, Number of students infected at a point of time = p, Number of students who are not infected at the same point of time = 2500 − p. Since, the rate at which viral fever spreads is proportional not only to the number of people affected, but also to the number of people who are not yet exposed to it, we obtain the model as dp = ap( 2500 − p ), p(0) = 1, a is an arbitrary constant. dt Separating the variables and solving, we get p (t ) = 2500 /(1 + 2499e −2500at ). Since, p (5) = 50, we obtain a = ln(51) / 12500. Hence, p (15) ≈ 2454 students. 11. The equilibrium points are the solutions of the equation P(6 − P ) − h = 0. The equilibrium points are given by P1 = 3 − 9 − h , P2 = 3 + 9 − h , h < 9. We have

f ′( P ) = 6 − 2 P,

f ′( P1 ) = 2 9 − h > 0,

f ′( P2 ) = −2 9 − h < 0. 9

The equilibrium point P1 is unstable whereas the equilibrium point P2 is asymptotically stable. The two equilibrium points coincide when h = 9. The critical value of h is hc = 9. When h > 9 , there are no equilibrium points and the population tends to extinction. If P > 3, f ′( P) < 0. In this case, we get stable solutions. Hence, the initial value of the population should satisfy P0 > 3. That is, the population may be extinct if P0 < 3 , even though the condition h < hc is satisfied. 12. The equation is a harvesting model with constant harvesting. (i) The equilibrium points are K 4h 1 N = [ rK ± r 2 K 2 − 4 rhK ] = [1 ± 1 − p ], where p = . 2r 2 rK K K Let, N 1 = [1 − 1 − p ], and N 2 = [1 + 1 − p ]. 2 2 Real solutions are obtained only when p < 1, or h < (rK / 4). We find

f ′( N ) = r[1 − (2 / K ) N ], f ′( N 1 ) = r 1 − p > 0, f ′( N 2 ) = − r 1 − p < 0. The equilibrium point N 1 is unstable whereas the equilibrium point N 2 is asymptotically stable. The two equilibrium points coincide when p = 1, that is when 4h = rK , or h = rK / 4 and N = K / 2. The critical value of h is hc = rK / 4. When h > rK / 4 , there are no equilibrium points and the population tends to extinction. (ii) If 2 N > K , f ′( N ) < 0. In this case, we get stable solutions. Hence, the initial value of the population should satisfy n0 > ( K / 2). That is, the population may become extinct if n0 < ( K / 2 ) , even though the condition h < hc is satisfied. 13. Equilibrium point is N 1 = Ke − qE / α . N = 0 can not be taken as an equilibrium point even though the limit of the right hand side exists. Now, f ′( N ) = α [ln( K / N ) − 1] − qE , and f ′( N 1 ) = −α < 0. Therefore, N 1 is asymptotically stable. Note that f ′( N = 0) is not defined. The sustainable yield is given by Y ( qE ) = qEN1 = qEKe − qE / α = α Kue − u = Y (u ), where u = qE / α .

Also, Y ′(u ) = αK [1 − u ]e − u , Y ′′(u ) = αK [u − 2]e − u . Setting Y ′(u ) = 0, we get the stationary point as u = 1. When u = 1, Y ′′(u ) < 0. The maximum occurs for u = 1, or qE = α . Maximum sustainable yield = Y (u = 1) = αK / e. 14. The location of the steady state varies with the length of the delay and the form of the characteristic equation changes due to the direct inclusion of the delay in the parameters and the indirect changes resulting from the varying location of the steady state. The model has the trivial steady state (0, 0) and the nontrivial steady state is given by 1 ⎛ be − μτ ⎞⎟ or P = ln⎜ . be − μτ e − aP = d , a ⎜⎝ d ⎟⎠ In particular, if τ > (1 / μ ) ln(b / d ), there is no positive steady state. In this case, given that the initial value is positive, we have dP ≤ be − μτ P (t − τ ) − dP(t ), with be − μτ < d . dt The solution goes to zero and the trivial steady state is globally stable. 10

15. P* = F ( P*), gives P* = K (1 + r ) which depends on r. At P = P*, F ′ = [1 /(1 + r )] < 1 for all r > 0. Asymptotically stable for all r. 16. We find the solution of the equation P* = F ( P*), that is of P * ⎞⎤ P * ⎞⎤ ⎡ ⎛ ⎡ ⎛ P* = P * exp ⎢λ ⎜1 − ⎟⎥, or exp ⎢λ ⎜1 − ⎟ = 1. K ⎠⎦ K ⎠⎥⎦ ⎣ ⎝ ⎣ ⎝ The solution is given by P* = K, for all λ . P ⎞⎤ dF P ⎞⎤ ⎡ λ P ⎤ ⎡ ⎛ ⎡ ⎛ Now, F ( P ) = P exp ⎢λ ⎜1 − ⎟⎥, = exp ⎢λ ⎜1 − ⎟⎥ ⎢1 − . K ⎠⎦ dP K ⎠⎦ ⎣ K ⎥⎦ ⎣ ⎝ ⎣ ⎝ dF At P = P* = K , we get = [1 − λ ]. dP Equilibrium is stable for 1 − λ < 1, or 0 < λ < 2. It is unstable for λ < 0 and λ > 2. Neighboring trajectories approach the equilibrium point asymptotically for 0 < λ < 1, and with damped oscillations for 1 < λ < 2. dF K d 2F ⎛ r ⎞ P ⎞⎤ ⎡ rP ⎤ ⎡ ⎛ = 0, when P = . Now, = ⎜ − ⎟ exp ⎢ r⎜1 − ⎟⎥ ⎢2 − ⎥. 2 dP r K ⎠⎦ ⎣ K⎦ dP ⎝ K⎠ ⎣ ⎝ d 2F ⎛ r ⎞ K = ⎜ − ⎟ exp( r − 1) < 0. For P = , r dP 2 ⎝ K ⎠ Maximum of the trajectory occurs at P = K / r. The maximum value = ( K / r ) exp( r − 1). (i) For K = 500, r = 1, P0 = 50, we have the following sequence of values: 122.98, 261.40, 421.26, 493.11, 499.95, 500, (see Fig.2.1). (ii) For K = 500, r = 1, P0 = 660, we get the sequence of values: 479.26, 499.56, 500, (see Fig. 2.1).

Fig.2.1. Discrete solution values for the data sets (i) and (ii). 17. Choose Vi = ( N i − K ) 2 . Vi ≥ 0 and has a minimum V = 0 at N i = K . The increment ΔVi on the trajectory is given by

ΔVi = K 2 ni [ e r (1− ni ) − 1][ ni e r (1− ni ) + ni − 2], where N i = Kni . ΔVi ≤ 0 when (i) e r (1− n i ) < 1, and [e r (1− ni ) + 1] ≥ [2 / ni ];

or (ii) e r (1− ni ) ≥ 1, and [e r (1− ni ) + 1] < [2 / ni ]. We obtain ΔVi ≤ 0 for 0 < r < 2 and for all n i . ΔVi = 0 only for ni = N * / K . Hence, equilibrium is globally asymptotically stable. 11

18. Equilibrium point is x* = 1. F ′(1) = 1 − r < 1 gives 0 < r < 2. Asymptotically stable for 0 < r < 2. Maximum occurs at x = 1/r and the maximum value is exp( r − 1) / r . 19. For steady state, we solve P* = F ( P*) = [( λP*) /(1 + bP *2 )] , to get P* = 0, and P* = (λ − 1) / b , λ > 1. We have

dF λ (1 − bPn2 ) . = dP (1 + bPn2 ) 2

Hence, the eigen values corresponding to P* = 0 and P* = (λ − 1) / b , are λ and ( 2 − λ ) / λ respectively. For λ =1, we have only a trivial equilibrium point. Setting dF / dP = 0, we get P = 1 / b . For this value of P, F ′′( P ) < 0. Hence, we obtain the maximum at P = 1 / b . The maximum value is F (1 / b ) = λ /(2 b ). Exercise 2.2 1. Setting F ( X , Y ) = 0, G ( X , Y ) = 0, we obtain the equilibrium point as X * = 212 / 3, Y * = 2332 / 15. Note that (0, 0) is not an equilibrium point. The elements of the Jacobian matrix of the system evaluated at an equilibrium point (X*, Y*) are X 8.0Y 2 1.6 X 2 a11 = 1 − − a = − , , 12 50 ( X + 5Y ) 2 ( X + 5Y ) 2 3Y 2 0.6 X 2 a = − 0.05. , 22 ( X + 5Y ) 2 ( X + 5Y ) 2 At the equilibrium point ( X *, Y *) : We obtain a11 = −0.682222, a12 = −0.011111, a21 =

a 21 = 0.100833, a 22 = −0.045833. The characteristic equation is λ2 + 0.72806λ + 0.032389 = 0. The coefficients in the equation are positive. By Routh-Hurwitz criterion, the eigen values are negative or have negative real parts. The equilibrium point ( X *, Y *) is asymptotically stable. 2. We have F ( P, Z ) = ( a1 − b1 P − c1Z ), and G ( P, Z ) = [a 2 − c2 ( Z / P )]. Conditions (i), (ii), (iii), (v), (vi) and (vii) are satisfied. Equality condition in (iv) is satisfied. The requirement (viii), G (C, 0) = 0 gives a 2 = 0, which violates the assumption that a 2 is positive. Hence, Kolmogorov theorem cannot be applied. wX wY , G ( X , Y ) = −a 2 + 1 . 3. We have F ( X , Y ) = a1 − b1 X − X +D X + D1 Conditions (i), (ii), (iv), (v), (vi), (vii) are satisfied. Equality condition in (iii) is satisfied. a D wC (viii) G (C , 0) = −a2 + 1 = 0, gives C = 2 1 . w1 − a 2 C + D1 C > 0 gives the condition w1 > a 2 . a D1a 2 Da (ix) B > C gives the condition 1 > , or w1 − a 2 > 1 2 . b1 ( w1 − a 2 ) K The two species system (2.80), (2.81) qualifies as a Kolmogorov system when the conditions w1 > a2 , K ( w1 − a 2 ) > D1a 2 are satisfied. 12

4. The elements of the Jacobian matrix of the system are wDY wX a11 = a1 − 2b1 X − , a12 = − , 2 X +D ( X + D) w1 X w1 D1Y a 21 = a = − a + , . 22 2 X + D1 ( X + D1 ) 2 At the equilibrium point E0 (0, 0) : We obtain a11 = a1 , a12 = 0, a 21 = 0, a22 = −a 2 . The eigen values are λ1 = a1 > 0, and λ2 = −a 2 < 0. The equilibrium point is unstable. Since Re(λ ) ≠ 0 for both eigen values, the fixed point is hyperbolic. Since the eigen values are real and are of opposite signs, we find that E0 is a hyperbolic saddle point which repels in the x-direction and attracts in y-direction. At the equilibrium point E1 ( K , 0) : We obtain wK a wK , a 21 = 0, a 22 = −a 2 + 1 a11 = a1 − 2b1 K , a12 = − , K = 1. ( K + D) K + D1 b1 The eigen values are λ1 = a11 = −a1 < 0, and w1a1 a ( w − a 2 ) − a 2 b1D1 λ2 = a 22 = −a 2 + = 1 1 >0 b1 ( K + D1 ) a1 + b1 D1 using the result from Kolmogorov condition (ix). The equilibrium point is unstable. The fixed point E1 ( K , 0) is also a hyperbolic saddle point which attracts in the x-direction and repels in y-direction. At the equilibrium point E*(X*, Y*): We obtain wDY * wX * a11 = a1 − 2b1 X * − a12 = − , , 2 X * +D ( X * +D) wX w1 D1Y * a 21 = , a 22 = G ( X , Y ) = −a 2 + 1 = 0. 2 X + D1 ( X * + D1 ) The eigen values of J are the roots of λ2 − a11λ − a12 a 21 = 0. By Routh-Hurwitz theorem, the necessary and sufficient conditions for the eigen values to be negative or have negative real parts are ( −a11 ) > 0, ( −a12 a 21 ) > 0 . That is

⎛ wX * ⎞⎛⎜ w1 D1Y * ⎞⎟ ( −a12 a 21 ) = ⎜ > 0, which is true. ⎟ ⎝ X * + D ⎠⎜⎝ ( X * + D1 ) 2 ⎟⎠ ⎡ wDY * ⎤ wY * D ( −a11 ) = − ⎢a1 − 2b1 X * − = . − ( a1 − 2b1 X *) 2⎥ ( X * + D) ⎦ ( X * + D) ( X * + D) ⎣ D = ( a1 − b1 X *) − (a1 − 2b1 X *) (from F ( X , Y ) = 0 ) ( X * +D)

= −(a1 − b1 X *)

⎡ X* X * −K ⎤ + b1 X * = b1 X * ⎢1 + ⎥ ( X * + D) ⎣ ( X * + D) ⎦

⎡2X * +D − K ⎤ = b1 X * ⎢ ⎥. ⎣ ( X * + D) ⎦ Now, ( −a11 ) > 0, if 2 X * + D − K > 0. Substituting the expression for X*, we get 13

⎛ a D ⎞ ⎛ a D ⎞ 2⎜⎜ 2 1 ⎟⎟ + D − K > 0, or 2b1 ⎜⎜ 2 1 ⎟⎟ + b1 D − a1 > 0. (A) ⎝ w1 − a2 ⎠ ⎝ w1 − a 2 ⎠ The equilibrium point E * ( X *, Y *) is locally asymptotically stable if the condition (A) is satisfied. 5. The elements of the Jacobian matrix of the system are u hP v a = − , , a11 = 1 − 2u − 12 hP + u ( hP + u ) 2

bhP v

a 21 =

bu 2 fvh Z2 = −c− 2 . (h P + u ) (hZ + v 2 ) 2

, a 22 ( hP + u ) 2 At the equilibrium point E0 (0, 0) : We obtain a11 = 1, a12 = 0, a 21 = 0, a 22 = −c. The eigen values are λ1 = 1, and λ2 = −c < 0. The equilibrium point is a hyperbolic saddle point. At the equilibrium point E1 (1, 0) : We obtain 1 b a11 = −1, a12 = − , a 21 = 0, a 22 = − c. (1 + hP ) hP + 1 (b − c) − ch p b The eigen values are λ1 = a11 = −1 < 0, and λ2 = a 22 = −c = . hP + 1 hP + 1 But (from text), b > c, 0 < chP < b − c, 0 < hP < 1. Therefore, the equilibrium E1 is a saddle point with stable manifold locally in the u -direction and with unstable manifold locally in the v -direction. At the equilibrium point E * (u*, v*) : We have hP v hP v =1− u − u − , a11 = 1 − 2u − 2 (hP + u ) ( hP + u ) 2 =

⎡ ⎤ hP v v v 1 −u− = u − ⎥. ⎢ 2 ( hP + u ) ( hP + u ) 2 ⎢⎣ ( hP + u ) ⎥⎦

bhP v bu fv( v 2 − hZ2 ) 2 fvhZ2 u a = − c − = , a 21 = , , 22 hP + u (h P + u ) (hZ2 + v 2 ) 2 ( hZ2 + v 2 ) 2 (hP + u ) 2 where all the quantities are evaluated at ( u*, v*). The eigen values of J are the roots of

a12 = −

λ 2 − (a11 + a22 )λ + (a11a22 − a12 a21 ) = 0. Using the Routh-Hurwitz theorem, we find the conditions for local stability as ⎡ ⎤ fv(hZ2 − v 2 ) v A = −( a11 + a 22 ) = u ⎢1 − + 2 > 0, 2⎥ 2 2 ⎣⎢ ( hP + u ) ⎦⎥ ( hZ + v ) ⎡ ⎤ ⎡ fv( hZ2 − v 2 ) ⎤ v bhP uv B = a11a 22 − a12 a 21 = u ⎢1 − + > 0, 2 ⎥⎢ 2 2 2 ⎥ 3 ⎢⎣ ( hP + u ) ⎥⎦ ⎢⎣ ( hZ + v ) ⎥⎦ (h P + u ) where all the quantities are evaluated at ( u*, v*). Sufficient conditions are v* < ( hP + u*) 2 , and ( v*) 2 < hZ2 . If the above conditions are satisfied, then both predator and prey species coexist, and they settle down at its equilibrium point.

14

6. The non-zero equilibrium points are the solution of the equations wY w1 X a1 − b1 X − = 0, − a2 + = 0. (α + βY + γX ) (α + βY + γX ) a1 − b1 X a 1 We have = = 2 . Solving the right equality, we obtain wY α + βY + γX w1 X 1 (2.1) Y = [( w1 − γa 2 ) X − αa 2 ]. a2 β Using the first and third terms in the equality and substituting the expression for Y, we w obtain ( a1 − b1 X ) w1 X = [( w1 − γa 2 ) X − αa2 ]. β Simplifying, we obtain β w1 b1 X

2

+ [ w ( w1 − γ a 2 ) − β w1 a 1 ] X − w α a 2 = 0 .

The roots of this equation are X = [ p ± p 2 + q ] /(2 βw1b1 ), where p = βw1a1 + wγa 2 − ww1 , q = 4 βw1b1wαa 2 . Irrespective of the sign of p, the root in the first quadrant is X * = [ p + p 2 + q ] /(2 βw1b1 ). The value of Y* is given by (2.1) (the chosen parameter values should satisfy Y* > 0). w1 X wY 7. We have F ( X , Y ) = a1 − b1 X − , and G ( X , Y ) = −a2 + . (α + β Y + γ X ) (α + βY + γX ) Conditions (i), (ii), (iii), (iv), (v), (vii) are satisfied. a1α (vi) F (0, A) = 0, gives A= > 0 if w > a1 β . ( w − a1 β ) a 2α (viii) G (C , 0) = 0, gives C = > 0 if w1 > a 2γ . ( w1 − a2γ ) a a2α (ix) B > C gives 1 > . b1 ( w1 − a2 γ ) a a2α Summarizing, we get the conditions as w > a1 β , w1 > a2 γ and 1 > . b1 ( w1 − a2 γ ) An oscillatory predator-prey dynamics (time series) exhibited by the model system for the given set of parameter values, a1 = 2.5, b1 = 0.05, w = 0.85, α = 0.45, β = 0.2, γ = 0.6, a 2 = 0.95, and w1 = 1.65, is presented in Fig. 2.2. 8. The equilibrium points are (0, 0), (a1 / b1 , 0) and (X*, Y*) (see Problem 6). The elements of the Jacobian matrix of the system are w(α + β Y )Y w(α + γ X ) X a11 = a1 − 2b1 X − , a12 = − , 2 (α + β Y + γ X ) (α + β Y + γ X ) 2 w1 (α + β Y )Y w (α + γ X ) X a21 = , a22 = −a2 + 1 . 2 (α + β Y + γ X ) (α + β Y + γ X )2

15

Fig.2.2. Time-series displaying oscillatory dynamics in the model (2.71)-(2.72). At the equilibrium point ( 0, 0) : We obtain a11 = a1 , a12 = 0, a21 = 0, a22 = −a2 . The eigen values are λ1 = a1 , and λ2 = −a2 < 0. The equilibrium point (0, 0) is unstable. Since, Re(λ ) ≠ 0 for both eigen values, the fixed point is hyperbolic. Since, the eigen values are real and are of opposite signs, we find that (0, 0) is a hyperbolic saddle point which repels in the x-direction and attracts in y-direction. At the equilibrium point (K, 0), where K = a1 / b1. wK wK We obtain a11 = −a1 , a12 = − , a 21 = 0, a 22 = 1 − a2 . α + γK α + γK The eigen values are λ1 = − a1 , and λ2 = a22 = ( w1 K /(α + γ K )) − a2 . If λ2 < 0, that is [ w1K /(α + Kγ )] < a 2 , that is, [ w1a1 /(b1α + a1γ )] < a2 , then the equilibrium point (K , 0) is asymptotically stable. Otherwise, (K , 0) is unstable. It depends on the values of the parameters w1 , a1 , a 2 , b1 , α , γ . If λ2 > 0, that is, [ w1a1 /(b1α + a1γ )] > a 2 , then the eigen values are real and are of opposite signs and the fixed point (K , 0) is a hyperbolic saddle point. At the equilibrium point (X*, Y*): The expressions for X*, Y* are X* = [p +

p 2 + q ] /(2 βw1b1 ); p = βw1a1 + wγa 2 − ww1 , q = 4 βw1b1wαa 2 .

a 2 β Y * = [( w1 − γa 2 ) X * −αa2 ]. The eigen values of J are the roots of λ2 − ( a11 + a 22 )λ + ( a11a 22 − a12 a 21 ) = 0. Using the Routh-Hurwitz theorem, the necessary and sufficient conditions are given by − ( a11 + a 22 ) > 0, and ( a11a 22 − a12 a 21 ) > 0. We have (dropping *) a1 − b1 X a 1 = = 2 . (see Problem 6) wY α + βY + γX w1 X 1 1 α + βY = α + [( w1 − γa 2 ) X − αa 2 ] = ( w1 − γa 2 ) X .. a2 a2 w(α + βY )Y a11 = a1 − 2b1 X − (α + βY + γX )2

a 22

⎡ w( w1 − γa 2 ) X ⎤ ⎡ ( a1 − b1 X )a 2 ⎤ γa 2 = a1 − b1 X − b1 X − ⎢ ⎥ ⎢ ww X ⎥ = w ( a1 − b1 X ) − b1 X . a2 1 1 ⎣ ⎦⎣ ⎦ a β w (α + γX ) X a (α + γX ) = − 2 ( a1 − b1 X ). = −a2 + 1 = −a2 + 2 2 (α + βY + γX ) w (α + βY + γX ) 16

a11 + a 22 =

a β γ a2 ( a1 − b1 X ) − b1 X − 2 ( a1 − b1 X ) w1 w

⎡γ ⎡ γa β⎤ βa ⎤ = a1a 2 ⎢ − ⎥ − b1 X ⎢ 2 + 1 − 2 ⎥ w ⎦ ⎣ w1 w ⎦ ⎣ w1 ⎡ γw − β w1 ⎤ b1 X [a2 ( βw1 − γw) − ww1 ]. = a1a 2 ⎢ ⎥+ ⎣ ww1 ⎦ ww1

Sufficient conditions for [ −( a11 + a 22 )] > 0 are γw − βw1 < 0, and a 2 ( βw1 − γw) − ww1 < 0. Now, ⎡ γa ⎤⎡ a β ⎤ ( a11a 22 − a12 a 21 ) = ⎢ 2 (a1 − b1 X ) − b1 X ⎥ ⎢− 2 ( a1 − b1 X )⎥ w w ⎦ ⎣ 1 ⎦⎣

wa 22

(α + γX )( w1 − γa2 )( a1 − b1 X ) w13 X a β = 2 ( a1 − b1 X )[b1 X ( w1 + γa 2 ) − γa1a2 ] + second term. ww1 Sufficient conditions for ( a11a 22 − a12 a 21 ) > 0 are ( w1 − γa2 ) > 0, ( a1 − b1 X ) > 0, and b1 X ( w1 + γa2 ) − γa1a2 > 0, γa1a2 that is, ( w1 − γa 2 ) > 0, and < b1 X < a1 . ( w1 + γa 2 ) We require the above conditions to be satisfied for asymptotic stability. However, it is possible to derive alternate conditions by simplifying in a different way. +

9. The equilibrium points are the solutions of the equations ⎤ ⎤ ⎡⎛ ⎡ βu u⎞ v −γ ⎥ = 0. u ⎢⎜ 1 − ⎟ − 2 ⎥ = 0, v ⎢ 2 ⎦ ⎣ (u / α ) + u + 1 ⎣⎝ K ⎠ ( u / α ) + u + 1 ⎦ Two of the equilibrium points are (0, 0) and (K, 0). From the second equation, we get [1 /{(u 2 / α ) + u + 1}] = (γ / β u ). Using this result in the first equation, we get v* = ( β / γK )( K − u*)u * . Simplifying the equations ( K − u )( u 2 + α u + α ) − vKα = 0, αβ u − γ (u 2 + α u + α ) = 0 , we obtain ( K − u ) [u 2 + α u + α − (αβ / γ )u ] = 0. The first root gives u = K, which gives the equilibrium point (K, 0). Setting S = α [1 − ( β / γ )], we obtain the solutions of u 2 + Su + α = 0 as u* = [ − S ± S 2 − 4α ] / 2. Hence, v* = ( β / γK )[ K − u*]u * .

The non-trivial solutions exist if S 2 > 4α , and u* < K, that is if [1 − ( β / γ )]2 > ( 4 / α ), and u* < K. If S < 0, that is β > γ , we obtain two positive equilibrium points. If S > 0, that is β < γ , we have u* < 0, and there are no positive equilibrium points. u⎞ v βu ⎛ 10. We have F (u, v ) = ⎜1 − ⎟ − 2 , and G (u, v ) = 2 −γ . ⎝ K ⎠ (u / α ) + u + 1 (u / α ) + u + 1 (i) ( ∂F / ∂v ) < 0 , (v) F ( 0, 0) > 0 , (vi) F (0, A) = 0, A > 0 , (vii) F ( B, 0) = 0, B = K > 0, are satisfied. Equality in condition (iii), ( ∂G / ∂v ) = 0 is satisfied. 17

⎛ ∂G ⎞ ⎛ ∂G ⎞ 2 2 (iv) u ⎜ ⎟ + v⎜ ⎟ > 0, gives uβ [1 − (u / α )] > 0. Hence, we get the condition u < α . ⎝ ∂u ⎠ ⎝ ∂v ⎠ ⎛ ∂F ⎞ ⎛ ∂F ⎞ (ii) u ⎜ ⎟+v⎜ ⎟ < 0, gives (after simplification) the condition ⎝ ∂u ⎠ ⎝ ∂v ⎠ vK [( u 2 / α ) − 1] < u[( u 2 / α ) + u + 1]2 .

Since u 2 < α , the left hand side is negative and the inequality is satisfied as all other quantities are positive. (viii) G (C , 0) = 0, C > 0, gives the quadratic equation for C as γ C 2 + (γ − β )α C + αγ

= 0. The roots of the equation are 2C = [{( β / γ ) − 1} ± α {( β / γ ) − 1}2 − ( 4 / α ) ]. Both the roots are real and positive if β > γ , and {( β / γ ) − 1}2 > ( 4 / α ). (ix) B > C gives K > C, that is K > (larger root of C). Summarizing, we get the conditions as u 2 < α , β > γ , {( β / γ ) − 1}2 > ( 4 / α ), K > 0.5 [{( β / γ ) − 1} + α {( β / γ ) − 1}2 − ( 4 / α ) ]. 11. A stable equilibrium solution for the given model system exhibited for a typical set of parameter values, K = 1, α = 3, β = 2.3 and γ = 0.3 is presented in Fig.2.3. We obtain the non-zero equilibrium solution as (u*, v*) = (0.1511, 0.9836). ⎛ ⎞ w3 X⎞ BZ ⎛ ⎟. 12. We have F ( X , Z ) = A ⎜1 − ⎟ − , G ( X , Z ) = Z ⎜⎜ c − ( X + D3 ) ⎟⎠ ⎝ K ⎠ ( D + dX + Z ) ⎝ Conditions (i), (ii), (v), (vii), (viii) are satisfied. w3 w − cD3 ∂G (iii) < 0, gives c − < 0, or X < 3 . X + D3 c ∂Z Since X > 0, we obtain the condition c < ( w3 / D3 ).

Fig.2.3. Phase plot and time series for the model system (2.75)-(2.76) for K = 1, α = 3, β = 2.3 and γ = 0.3. ⎡ w3 Z ⎤ ⎛ ∂G ⎞ ⎛ ∂G ⎞ (iv) X ⎜ +Z ⎟ > 0, gives X ⎢ ⎟ + Z⎜ 2⎥ ⎝ ∂Z ⎠ ⎝ ∂X ⎠ ⎢⎣ ( X + D3 ) ⎥⎦ ⎡ w3 D3 ⎤ A sufficient condition is Z ⎢c − > 0, 2⎥ ⎢⎣ ( X + D3 ) ⎥⎦

⎡ w3 ⎤ ⎢c − ⎥ >0. X + D3 ⎦ ⎣

or c >

w3 D3 ( X + D3 ) 2

. 18

(vi) From F (0, A*) = 0, we obtain A* = AD /( B − A). The condition A* > 0, gives the requirement B > A. (ix) The condition B* > C *, gives K > C * . From (iii), we have C* < [( w3 − cD3 ) / c] . We may choose K > [( w3 − cD3 ) / c] . Summarizing the results, Kolmogorov theorem gives the following conditions. w3 D3 w3 (a) Combining (iii), and (iv), we get . A, (c) K > [( w3 − cD3 ) / c] . ⎛ w3U Z ⎞ wU ⎟⎟ − 13. We have F ( Z ,U ) = A ⎜⎜1 − , and G ( Z , U ) = c − 4 . Z ⎝ K1 ⎠ ( Z + D3 ) Conditions (i), (ii), (iii), (v), (vi) are satisfied. Equality condition in (iv) is satisfied. The requirement (vii) gives B* = K1 > 0. The requirement (ix) gives K1 > C * . But condition (viii) is violated. We obtain G (C*, 0) = c. The condition G (C , 0) = 0, C > 0 is violated since c ≠ 0. Hence, Kolmogorov theorem cannot be applied. 14. The oscillatory predator-prey dynamics exhibited by Holling-Tanner model (2.84), (2.85) for the given set of parameter values is given in Fig.2.4.

Fig.2.4. Oscillatory predator-prey dynamics exhibited by Holling-Tanner model. 15. Note that (0, 0) is not an equilibrium point. (K, 0) is an equilibrium point. The second equation gives U = cZ / w4 . Substituting in the first equation, we get w3cZ A (K − Z ) − = 0, α1 w4 + β1cZ + w4γ 1Z K

AKα1w4 + Z [ AK ( β1c + w4γ 1 ) − Aα1 w4 − w3 Kc] − AZ 2 ( β1c + w4γ 1 ) = 0, Kα1 w4 α w + ( w3 Kc / A) or Z 2 + ( p − K ) Z − q = 0, where p = 1 4 ,q = . ( β1c + w4γ 1 ) ( β1c + w4γ 1 ) Irrespective of the sign of ( p − K ), the positive root is given by or

2 Z * = ⎡− ( p − K ) + ( p − K ) 2 + 4q ⎤. We have U * = cZ * / w4 . ⎢⎣ ⎥⎦ An oscillatory predator-prey dynamics exhibited by the model system for the given set of parameter values, A = 2, K = 100, w3 = 2.1, α1 = 0.45, β1 = 0.2, γ 1 = 0.6, c = 0.95 and w4 = 1.65, is presented in Fig. 2.5. 19

wU w3U Z⎞ ⎛ , and G ( Z ,U ) = c − 4 . F ( Z , U ) = A ⎜1 − ⎟ − Z K ⎠ (α1 + β1U + γ 1Z ) ⎝ Conditions (i), (ii), (iii), (v), (vii) are satisfied. Equality condition in (iv) is satisfied. Aα1 (vi) F (0, A*) = 0, gives A* = > 0, if w3 > Aβ1 . ( w3 − Aβ1 ) (viii) G(C ,0) = c ≠ 0. The condition is not satisfied. (ix) B > C is also not satisfied . Hence, Kolmogorov theorem cannot be applied. 17. The equilibrium point is X 1* = ( K1 − K 2 b1 ) /(1 − b1b2 ), X 2* = ( K 2 − K1b2 ) /(1 − b1b2 ).

16. We have

Since X 1* > 0, and X 2* > 0, we obtain the conditions b1 < ( K1 / K 2 ) < (1 / b2 ), and b1b2 < 1. The second condition is implied in the first condition. (Positivity holds also when the inequalities are reversed). The elements of the Jacobian matrix are (dropping the superfix *) a11 = ( r1 / K1 )[ K1 − 2 X 1 − b1 X 2 ] = −( r1 X 1 / K1 ), a12 = −( r1b1 X 1 ) / K1 , a 21 = −( r2 b2 X 2 ) / K 2 , a 22 = ( r2 / K 2 )[ K 2 − b2 X 1 − 2 X 2 ] = −( r2 X 2 / K 2 ). The characteristic equation is

λ2 + λ [( r1 X 1 / K1 ) + ( r2 X 2 ) / K 2 ] + [( r1 X 1 / K1 )( r2 X 2 ) / K 2 ](1 − b1b2 ) = 0. Applying the Routh-Hurwitz criterion, we find that the positive equilibrium point is asymptotically stable when b1b2 < 1. The required condition is b1 < ( K1 / K 2 ) < (1 / b2 ).

Fig.2.5. Time-series displaying oscillatory predator-prey dynamics exhibited by the modified HT model (2.90)-(2.91). 18. The positive equilibrium point E * ( X 1* , X 2* ) is the solution of the equations a1 − b1t1 − c1t 2 = 0, and a 2 − b2 t1 − c2 t2 = 0, where t1 = ln X 1 , t 2 = ln X 2 . a c − a 2 c1 b a − b2 a1 We obtain t1 = 1 2 , t2 = 1 2 , X 1 = e t1 , X 2 = et 2 . b1c2 − b2 c1 b1c2 − b2 c1 The elements of the Jacobian matrix are a11 = a1 − b1 (1 + ln X 1 ) − c1 ln X 2 = −b1 , a12 = −c1 X 1 / X 2 , a 21 = −b2 X 2 / X 1 , a 22 = a 2 − b2 ln X 1 − c2 (1 + ln X 2 ) = −c2 . The characteristic equation is

λ2 + λ (b1 + c2 ) + (b1c2 − c1b2 ) = 0.

20

Routh-Hurwitz criterion gives the necessary and sufficient conditions for the roots to be negative or have negative real parts, as b1 + c2 > 0, and b1c2 − c1b2 > 0. The positive equilibrium point is asymptotically stable when ( b1 / b2 ) > (c1 / c2 ). For the given set of parameter values ( b1 / b2 ) = 3 / 4, and ( c1 / c2 ) = 2 / 3. The condition is satisfied and the equilibrium point is asymptotically stable. The equilibrium point is ( X 1* , X 2* ) = ( e 7 , e −5 ). 19. For r = 1.5, α = 3, we get (1 / 3) < u* < 1. u* is a solution of r (u * −1) 1 − e r ( u * −1) = 1.5 − − e1.5( u *−1) = 0. f (u*) = 1 + αu * 2u * Newton-Raphson’s method applied with the initial approximation taken as 0.5, gives the sequence of iterates as 0.478602777, 0.480756737, 0.47971041, 0.480048641, 0.48004895. With u* = 0.48004895, we get v* = r (1 − u*) = 0.779926575. We obtain p = 4.5u * −1.5 = 0.660220275, A + B + C = r (1 − u * p) = 1.0245929 > 0, A − B + C = 2 − r + u * ( 2α − rp ) = 2.904886625 > 0, A − C = 1 − u * (α − rp ) = 0.035260224 > 0. By Miller’s theorem or Jury test, the equilibrium point is asymptotically stable. 20. The equilibrium points are obtained as (0, 0), and (4/9, 5/3). The elements of the Jacobian matrix J are a11 = 2.5 − 3N − 0.5P, a12 = −0.5 N , a 21 = 1.8 P, a 22 = 0.2 + 1.8 N . At (0, 0), the eigen values of J are 2.5 and 0.2. The system is unstable. At (4/9, 5/3), the eigen values of J are 1.0 and 1/3. The system is unstable.

Chapter 3 Exercise 3.1 1. MATLAB 7.0 is used to compute the phase plane diagram to generate the chaotic attractor and time series. Chaotic attractor and the temporal evolution for (i) t vs x, (ii) t vs y, (iii) t vs z are plotted in Figs. 3.1 (a), (b), (c) and (d). 1 11

0.8

0.6

9

x

z

10

8

0.4

7 0.8 0.6 0.4 0.2 y

(a)

0

0

0.2

0.4 x

0.6

0.8

1

0.2

0 4000

4500

5000

5500

t

(b)

21

0.5

11

0.4

10

0.3 y

z

9

0.2 8

0.1 0 4000

4500

5000

7 4000

5500

4500

5000

t

5500

t

(c) (d) Fig.3.1. (a) Chaotic attractor. (b) Temporal evolution (t vs x), (c) Temporal evolution (t vs y), (d) Temporal evolution (t vs z), in Problem 1. (From Upadhyay, R. K., Rai, V., Complex dynamics and synchronization in two non-identical chaotic ecological systems. Chaos, Solitons Fractals, 40, 2233–2241, Copyright 2009, Elsevier. Reprinted with permission.)

120

120

100

100

80

80

60

60

Z

Z

2. MATLAB 7.0 is used to generate the chaotic attractor in three different phase planes. The projections of the chaotic attractor are drawn in Figs. 3.2. (a), (b), (c).

40

40

20

20

0

0

5

10

15

20 X

25

30

35

0

40

0

5

10

15

20

25

30

35

40

45

Y

(a) (b) (c) Fig.3.2. Chaotic attractor: (a) (x-y) plane, (b) (x-z) plane, (c) (y-z) plane. Problem 2 1

b

0 0

1

2

3

4

5

6

7

-1

0.8

-2

f

log d

0.6

0.4

-3 -4

0.2

-5 0 0

1

2

3

b

4

5

6

7

-6

(a) (b) Fig.3.3. Points in the 2D parameter spaces (a) (b, f ) , (b) (b, log d ). Problem 3. (From Upadhyay, R. K., Kumari, N., Rai, V., Exploring dynamical complexity in diffusion driven predator–prey systems: Effect of toxin production by phytoplankton and spatial heterogeneities. Chaos, Solitons Fractals, 42(1), 584–594. Copyright 2009, Elsevier. Reprinted with permission).

3. MATLAB 7.0 is used to compute the discrete points at which chaos was observed and Microsoft office Excel 2007 to draw the 2D Scan diagram. All the points in the two dimensional parameter spaces where the model system exhibits chaotic dynamics are shown in Figs. 3.3(a), (b). 22

w1 xy dx wxy dy = a1 x − b1 x 2 − , = −a2 y + . dt (α + β y + γ x) dt (α + β y + γ x) The local stability conditions are γw − βw1 < 0, a2 ( β w1 − γ w) − ww1 < 0, ( w1 − γa 2 ) > 0, γ a1 a2 and < b1 x < a1 . ( w1 + γ a2 ) a a2α The Kolmogorov conditions give w > a1 β , w1 > a2 γ and 1 > . b1 ( w1 − a2 γ ) A set of parametric values satisfying the Kolmogorov conditions and violating the local stability condition are a1 = 2, b1 = 0.06, w = 0.85, α = 0.45, β = 0.2, γ = 0.6, a2 = 1.05, and w1 = 1.65. Now, assume that there exists a prey x, in addition to y whose growth rate is governed by  dx x ⎞ Bxz ⎛ , = Ax ⎜1 − ⎟ − dt K ⎠ (α + β z + γ x ) ⎝ where A is the rate of self-reproduction for this prey and K is the carrying capacity of its environment. In the Leslie-Gower scheme, the growth rate equations for the two populations will give the subsystem II  w z2 dx x ⎞ Bxz dz ⎛ = Ax ⎜1 − ⎟ − = cz 2 − 3 , . dt K ⎠ (α + β z + γ x ) dt x + D3 ⎝

4. The subsystem I is

For this subsystem to be a K-system, the following conditions are to be satisfied

w3 D3 w3 cD3 , (iii) K > [( w3 − cD3 ) / c] . Combining the above two conditions, we obtain D3 w3 (3.1) < c( x + D3 ) < w3 < c( K + D3 ); cD3 < w3 ( x + D3 ) The subsystem II exhibits a stable equilibrium when the following condition is satisfied: (i)

KBγ z * < A(α + β z* + γ x * ) 2 , x * =

w3 A(1 − x * / K )(α + γ x * ) − D3 , z * = c {B − Aβ (1 − x* / K )}

(3.2)

When the values of the parameters of this subsystem II are chosen in such a way that the constraints (3.1) are satisfied and the inequality (3.2) is violated, the subsystem II admits a limit cycle solution. For example, for the set of parametric values, A = 2.0, K = 130, α = 0.45, β = 0.2, γ = 0.6, B = 0.74, c = 0.0295, w3 = 2.1, D3 = 20 , the subsystem II has limit cycle solutions. The linking scheme would depend on how the individual populations of the two subsystems are related with each other. The link scheme can be mathematically represented by adding the term {−[( w2 yz ) /(α + β y + γx )]} to the second equation of subsystem I. The selection of values for w2 is based on the fact that it plays a role similar to that of w. w2 can be varied from 0.1 to 1. The parametric values for the original system are selected by omitting those appearing in the growth equation of pseudo prey ( x ). A set of parameter values for which the system admits limit cycle 23

solution is a1 = 2, b1 = 0.06, w =0.85, a2 = 1.05, w1 = 1.65, w2 = 0.3, α = 0.45, β = 0.2, γ = 0.6, c = 0.0295, w3 =2.1, D3 = 20. There exist other sets of parameter values which satisfy the above criteria. 5. BAS routine from Dynamics: Numerical Explorations software is used to compute the basin boundary structure, which bring changes in the dynamical behavior when the system parameters are varied . Basin boundary structure is plotted in Fig.3.4.

Fig.3.4. Basin boundary structure for the model given in Problem 5. (From Upadhyay, R. K., Rao, V. S. H., Short term recurrent chaos and role of toxin producing phytoplankton on chaotic dynamics in aquatic systems. Chaos, Solitons Fractals, 39, 1550–1564. Copyright 2009, Elsevier. Reprinted with permission).

6. MATLAB 7.0 is used to generate the bifurcation diagrams. The bifurcation diagrams are given in Fig. 3.5. From the figures, transition from chaos to order through a sequence of period halving bifurcation and hence a period-doubling cascade can be observed. 7.MATLAB 7.0 is used to generate the bifurcation diagrams. The successive maxima of y as a function of w5 for the given parameter values with w11 = 0.03 and 0.15 < w5 < 0.5, are plotted in Fig.3.6 (i); and with w11 = 0.06 and 0.25 < w5 < 0.5, are plotted in Fig.3.6 (ii). 8. We have f (0, 0) = 0, ( ∂f / ∂x )(0, 0) = 0, ( ∂f / ∂μ )(0, 0) = 1 / 4, ( ∂ 2 f / ∂x 2 ) ( 0, 0) = −2. For, μ < 0, the model has no fixed points and the vector field is decreasing in x . For,

μ > 0, the model has two fixed points x = ± μ / 2. The fixed point x = μ / 2 is stable, and the second fixed point x = − μ / 2 is unstable. 9. The nullclines intersect when x[ μax 2 − x + 9aμ ] = 0. The fixed points are (0, 0), and

x* = x1,2 = [1 ± 1 − 36a 2 μ 2 ] /(2aμ ). These two solutions coincide when 6aμ = 1 . Hence, the critical value is μc = 1/ 6a. At the bifurcation, the value of the fixed point is x* = 3. The Jacobian matrix is 1 ⎤ −μ ⎡ 2 2 J =⎢ 2 2 ⎥. Trace(J) = − ( a + μ ). J = aμ − [18 x /(9 + x ) ]. 18 /[( 9 ) ] x + x − a ⎦ ⎣ All the fixed points are either sinks or saddle points. At (0, 0), J = aμ > 0. Hence, (0, 0) is always a stable fixed point. At the other two fixed points, we have

24

Fig.3.5. Bifurcation diagrams taking θ as a control parameter for the model given in Problem 6. (From Upadhyay, R. K., Niji, R. K., Nitu Kumari. Dynamical complicity in some ecological models: Effect of touin production by phytoplankton. Nonlinear Andy.: Model. control 12(1), 123–138. Copyright 2007, Lithuanian Association of Nonlinear Analysts (LANA). Reprinted with permission).

(i)

(ii)

Fig.3.6. Bifurcation diagram of the model system. Problem 7. (From Upadhyay, R. K., Raw, S. N., Rai, V. 2010. Dynamical complexities in a tri-trophic hybrid food chain model with Holling type II and Crawley–Martin functional responses. Nonlinear Anal.: Model. Control 15(3), 361–375. Copyright © 2010, Lithuanian Association of Nonlinear Analysts, (LANA). Reprinted with permission).

⎡ x2 − 9 ⎤ μ = a ⎢ 2 ⎥. (9 + x 2 ) 2 (9 + x 2 )( x / μa ) 9 x + ⎣ ⎦ For 0 < x* < 3, J < 0. The fixed point is a saddle point. For x* > 3, J > 0. The fixed point is a stable node. J = aμ −

10. We have

18 x

= aμ −

18 x

f (0, 0) = 0, ( ∂f / ∂x )(0, 0) = 0, ( ∂f / ∂μ )(0, 0) = 0, ( ∂ 2 f / ∂x 2 )

( 0, 0) = 8,

( ∂ 2 f / ∂x∂μ )( 0, 0) = 1. Conditions (3.25) are satisfied. The system under-goes transcritical 25

bifurcation at (0, 0). Fixed points are x = 0 and x = 2 ± 4 + μ . Now, ( ∂f / ∂x )( 0, μ ) = μ , which changes sign as μ goes through zero. The system changes its stability at (0, 0). For x2 = 2 − 4 + μ , we have x2 > 0 for μ < 0, and x2 < 0 for μ > 0, and ( ∂f / ∂x )( x2 , μ ) = −4 x2 − 2 μ changes its sign from positive to negative as μ goes through zero.

11. (i) We have f (0, 0) = 0, ( ∂f / ∂x )(0, 0) = 0, ( ∂f / ∂μ )(0, 0) = 0, ( ∂ 2 f / ∂x 2 )(0, 0) = 0, ( ∂ 2 f / ∂x∂μ )(0, 0) = 1, ( ∂ 3 f / ∂x 3 )( 0, 0) = −6. The fixed points are x = 0 and x = ± μ . For μ < 0 , the fixed point x = 0 is stable. For μ > 0 , all the three fixed points exist. x = 0 is unstable and the other two fixed points are stable giving rise to supercritical pitchfork bifurcation.

(ii) We have

f (0, 0) = 0, ( ∂f / ∂x )( 0, 0) = 0,

( ∂f / ∂μ )(0, 0) = 0,

( ∂ 2 f / ∂x 2 )(0, 0) = 0,

( ∂ 2 f / ∂x∂μ )(0, 0) = 1, ( ∂ 3 f / ∂x 3 )( 0, 0) = 6. The fixed points are x = 0 and x = ± − μ .

For μ < 0 , the fixed point x = 0 is stable and x = ± − μ are unstable. The stationary solution (node) becomes an unstable saddle (a saddle-node, a saddle-focus), and together with it, the other two unstable stationary solutions disappear. The bifurcation is a crisis giving rise to subcritical pitchfork bifurcation. 12. We have f (0, 0) = 0, ( ∂f / ∂x )(0, 0) = 1, ( ∂f / ∂μ )( 0, 0) = 1 / 4, ( ∂ 2 f / ∂x 2 ) ( 0, 0) = −2. The map satisfies the conditions (3.24) and Remark 3.3. The sign of [ −(∂ 2 f / ∂x 2 )( 0, 0) / ( ∂f / ∂μ )( 0, 0)] = 8, tells us on which side of μ = 0 , the curve of fixed points is located. The map has no fixed points for μ < 0. For μ > 0, the map has two fixed points x = ± μ / 2 . The fixed point x = μ / 2 is stable and x = − μ / 2 is unstable. There are no fixed points on one side of μ = 0 and two fixed points on the other side. Saddle node bifurcation occurs at ( x, μ ) = (0, 0).

13. We have f (0, 0) = 0, ( ∂f / ∂x )(0, 0) = 1, ( ∂f / ∂μ )(0, 0) = 0, ( ∂ 2 f / ∂x∂μ ) ( 0, 0) = 1 / 2, ( ∂ 2 f / ∂x 2 ) ( 0, 0) = −2. Fixed points are x = 0 and x = μ / 2 , For μ < 0, x = 0 is stable and x = μ / 2 is unstable. For μ > 0, x = 0 is unstable and x = μ / 2 is stable. Thus, an exchange of stability occurs at μ = 0 with respect to both the fixed points. Transcritical bifurcation occurs at ( x, μ ) = (0, 0).

14. We have f (0, 0) = 0, ( ∂f / ∂x )(0, 0) = 1, ( ∂f / ∂μ )(0, 0) = 0, ( ∂ 2 f / ∂x∂μ ) ( 0, 0) = 1 / 4, ( ∂ 2 f / ∂x 2 )(0, 0) = 0, ( ∂ 3 f / ∂x 3 )( 0, 0) = −6. The fixed points of the map are x = 0 and

x = ± μ / 2. For μ < 0, the fixed point x = 0 is stable. For μ > 0, the fixed point x = 0 is unstable and other two fixed points are stable. Pitchfork bifurcation occurs at ( x, μ ) = (0, 0). 15. The fixed point x0 = [ −1 + 1 + 4 μ ] / 2 > 0 is called a period-1 cycle. It is stable for μ ∈ ( 0, 3 / 4] and its multiplier is − 2 x0 . At μ = μ1 = 3 / 4, the multiplier is −1 and the first period doubling bifurcation takes place giving rise to a stable period-2 cycle. The 2cycle consists of two points x1,2 = [1 ± 4 μ − 3 ] / 2. The 2-cycle is stable in the interval

μ ∈ [3 / 4, 5 / 4] and has the multiplier f x ( x1 ) f x ( x2 ) = 4(1 − μ ). At μ = μ2 = 5 / 4, the 26

second period doubling bifurcation takes place giving rise to a stable period- 2 2 cycle. The accumulation point of the bifurcation points is μ c = 1.40115... ⎡ μ − 2⎤ 16. We have at (0, 0), J = ⎢ . det( J ) = μ 2 + 1 > 0. Trace( J ) = 2μ . Eigenvalues are ⎥ ⎣1 / 2 μ ⎦ λ = μ ± i. As μ increases through zero, origin changes from a stable spiral to an unstable spiral. Hopf bifurcation occurs at μ = 0. Since an unstable limit cycle surrounds the stable fixed point, it is subcritical.

17. The variational matrix for E * = ( X ∗ , Y ∗ , Z ∗ ) is given by 0 ⎤ ⎡ a11 a12 ⎢ V4 = ⎢a 21 a 22 a 23 ⎥⎥ I have not checked as it is a paper. Is it OK. ⎢⎣ 0 a32 0 ⎥⎦

⎛ γ wY * ⎞⎟ w1 ( α + β Y * )Y ∗ wX * (α + γ X *) , where a11 = X * ⎜ − b1 + a , = − , a = 12 21 2 ⎟ 2 2 ⎜ S S S ⎝ ⎠ 1 1 1 ∗ ∗ ∗ ∗ ⎛w β X w Y (α 2 + γ 2Y ) w γ Z ⎞ a 22 = − Y * ⎜ 1 2 − 2 22 ⎟, a 23 = − 2 , 2 ⎜ S ⎟ S S 1 2 2 ⎝ ⎠

a32 = c 2 Z *2 / w3 ,

S1 = (α + βY * +γ X *),

S 2 = (α 2 + β 2 Z * +γ 2Y *).

Note that a12 < 0, a 21 > 0, a 23 < 0, a32 > 0. Also, a11 < 0, if and a 22 < 0, if

w2 γ 2 Z * S 22


0, D1 ( μ0 ) = p1 ( μ0 ) > 0, D2 ( μ0 ) = p1 ( μ0 ) p2 ( μ0 ) − p0 ( μ0 ) = 0,

and [ dD2 ( μ0 ) / dμ ] ≠ 0. Let, c, the growth rate of the generalist predator be a bifurcation parameter. Now, p2 is independent of c. From the above analysis, if a 22 ≤ 0 , along with the other conditions, then p1 p2 − p0 > 0 . Therefore, we assume that a 22 > 0. That is,

w2γ 2 Z * w1 β X * > . 2 (α 2 + β 2 Z + γ 2Y ) (α + β Y + γ X ) 2 Now, p2 = −(a11 + a22 ) > 0 , if ⎤ w γ2Z∗ Y * ⎡ ( wγ − w1 β ) X ∗ + < b1 . ⎢ 2 2 ⎥ X * ⎣ (α + β Y + γ X ) (α 2 + β 2 Z + γ 2Y ) ⎦

(3.3)

(3.4) 27

γ wY * < b1 is satisfied. (α + β Y + γ X ) 2 Solving p1 p2 − p0 = ( a11 + a22 ) (a12 a21 − a11 a22 ) + a22 a23 a32 = 0 for c, we obtain p0 = a11 a23 a32 > 0 , if a11 < 0. Now, a11 < 0, if

a 22 (− a 23 )(c 2 Z * 2 / w3 ) = ( a11 + a 22 )( a12 a 21 − a11a 22 ). The left hand side is positive. We require that the right hand side be positive so that we have a positive solution for c. Since (a11 + a 22 ) < 0, a sufficient condition is ( a12 a 21 − a11a 22 ) < 0. We require

⎞ ⎛ ⎞⎛ w2 γ 2 Z ∗ w1 β X ∗ γ wY * − − b ⎜ 1 2 2 ⎟ 2 ⎟ ⎜ (α + β Y + γ X ) ⎠ ⎝ (α 2 + β 2 Z + γ 2Y ) (α + β Y + γ X ) ⎠ ⎝ * ww1 (α + γ X *)( α + β Y ) . (3.5) < (α + β Y + γ X ) 4 Therefore, c exists and (dD2 / dc) ≠ 0. The conditions to be satisfied are (3.3), (3.4), (3.5). 18. The Lyapunov exponent bifurcation diagrams are drawn using MATLAB 7.0. The diagrams are plotted in Fig.3.7 for (i) ω12 = 0.25, 0.01 ≤ ω8 ≤ 0.6; and (ii) ω8 = 0.21,

and 0.15 ≤ ω12 ≤ 0.3.

Fig.3.7. Lyapunov exponent bifurcation diagram as a function of (i) ω8 , and (ii) ω12 , Problem 18. 19. We obtain f ′( x ) = 2. We may expect that all orbits will be unstable and chaos sets in. We obtain n n 1 1 λ = lim log ∏ f ′( xi ) = lim log ∏ ( 2) = log 2. n→∞ n n→∞ n i =1 i =1 20. We solve the system of equations using MATHEMATICA. The solutions are given in Table 3.1. The corresponding variational equations are dp dq = (4.5 − 0.25 F ) p − 0.25 Rq, = 0.5 Fp + q (0.5 R − 4). dt dt Computation of Lyapunov Exponents of order 1 The variational equations are also solved using MATHEMATICA. Assume the initial vectors as ω1 (0) = (1/ 2, 1/ 2)T , and

ω2 (0) = (1, 0)T , for computing λ1, λ2 respectively. At every time step, denote the normalized ω by ωˆ . The numerical solutions obtained for computing λ1 are given in Table 3.2 and for computing λ2 are given in Table 3.3. 28

Table 3.1. Solutions of Lotka-Volterra model. Problem 20. t R(t) F(t) t R(t) F(t) 0.1 12.0008 11.6259 0.6 7.00788 30.1885 0.2 13.573 14.8196 0.7 5.32247 27.4408 0.3 13.8507 19.854 0.8 4.41693 23.3969 0.4 12.2791 25.7597 0.9 4.06403 19.3467 0.5 9.54427 29.8573 1.0 4.11015 15.8837 Table 3.2. Solutions for computing λ1 , Problem 20. t

ω1 (t ) t

ω1 (t )

0.1 ⎛ 0.609011 ⎞ ⎜1.1332 ⎟ ⎝ ⎠

0.2 ⎛ 0.377234 ⎞ ⎜1.51901 ⎟ ⎝ ⎠

0.3 ⎛ 0.001498 ⎞ ⎜1.78594 ⎟ ⎝ ⎠

0.4 ⎛ −0.505948 ⎞ ⎜1.84843 ⎟ ⎝ ⎠

0.5 ⎛ −1.10702 ⎞ ⎜1.62439 ⎟ ⎝ ⎠

0.6

0.7

0.8

0.9

1.0

⎛ −1.73357 ⎞ ⎜1.05079 ⎟ ⎝ ⎠

⎛ −2.28991 ⎞ ⎜ 0.101106 ⎟ ⎝ ⎠

⎛ −2.6586 ⎞ ⎜ −1.19816 ⎟ ⎝ ⎠

⎛ −2.71242 ⎞ ⎜ −2.75219 ⎟ ⎝ ⎠

⎛ −2.3322 ⎞ ⎜ −4.39071 ⎟ ⎝ ⎠

Table 3.3. Solutions for computing λ2 , Problem 20. t

ω1 (t ) t

ω1 (t )

0.1 1.14687 ⎛ ⎞ ⎜ 0.57001 ⎟ ⎝ ⎠

0.2 1.155 ⎛ ⎞ ⎜1.24213 ⎟ ⎝ ⎠

0.3 0.973051 ⎛ ⎞ ⎜1.94191 ⎟ ⎝ ⎠

0.4 0.564971 ⎛ ⎞ ⎜ 2.5627 ⎟ ⎝ ⎠

0.6

0.7

0.8

0.9

⎛ −0.93794 ⎞ ⎜ 3.03038 ⎟ ⎝ ⎠

⎛ −1.93921 ⎞ ⎜ 2.60177 ⎟ ⎝ ⎠

⎛ −2.96714 ⎞ ⎜ 1.58855 ⎟ ⎝ ⎠

⎛ −3.85943 ⎞ ⎜ −0.0446107 ⎟ ⎝ ⎠

0.5 0.0803505 − ⎛ ⎞ ⎜ 2.97288 ⎟ ⎝ ⎠ 1.0 ⎛ −4.41958 ⎞ ⎜ −2.24203 ⎟ ⎝ ⎠

Using (3.31), at t = 1 , we obtain λ1 = 8.16805, and λ2 = 10.1024. The maximal Lyaponov exponent is 10.1024 > 0. The plot of λ1 , λ2 is given in Fig.3.8.

Fig. 3.8. Lyaponov exponents (LCE of order 1).

29

Chapter 4 Exercise 4.1 1. The non-dimensionalized form is ⎡ ∂p ∂2 p az ⎤ = p ⎢rx (1 − p ) − , + d ∂t 1 + bp ⎥⎦ ∂x 2 ⎣

⎡ np ∂z ∂2z f z ⎤ = z⎢ −m− + d , ⎥ ∂t ∂x 2 1 + g 2z2 ⎦ ⎣1 + bp

where a = K /( R H P ), b = K / H P , d = D /( L2 R ), m = δ / R , n = eυa = eυK /( R H P ), f = FK

/(υR H Z2 ), g = K /(υH Z ), rx = ( R x / R ) = s + lx. Let ( p * , z * ) be the positive equilibrium solution of the system in the absence of space and diffusion. Writing p = p * + U , z = z * + V , we obtain the linearized form of the model system about the positive equilibrium point E * ( p* , z * ) as

∂U ∂ 2U ∂V ∂ 2V = a11U + a12V + d 2 , = a21U + a 22V + d 2 , ∂t ∂t ∂x ∂x * * * az abp z ap * * where a11 = rx (1 − 2 p * ) − = − p r + a = − , , 12 x (1 + bp* ) 2 (1 + bp* ) 2 1 + bp* a 21 =

nz* (1 + bp * ) 2

, a 22 =

np * (1 + bp* )

−m−

2 fz * (1 + g 2 z *2 ) 2

=

fz* ( g 2 z *2 − 1) (1 + g 2 z *2 ) 2

.

Write the solution of the equations in the form U = se λt + ikx , V = we λt + ikx , where λ and k are the frequency and wave number respectively. Substitute the expressions for U, V in the equations The homogeneous equations in s and w have solution if determinant of the coefficient matrix is zero. We get ( λ − a11 + dk 2 )( λ − a 22 + dk 2 ) − a12 a21 = 0, or λ2 + p1λ + q1 = 0,

where p1 = 2dk 2 − ( a11 + a 22 ), q1 = d 2 k 4 − dk 2 ( a11 + a 22 ) + ( a11a 22 − a12 a 21 ). By Routh-Hurwitz criterion, the roots of the characteristic equation are negative or have negative real parts if p1 > 0, and q1 > 0. A sufficient condition for p1 > 0 is

a11 + a 22 < 0. This condition is satisfied when abz* < rx (1 + bp* ) 2 , and g 2 z*2 < 1. For these values, a11 < 0, a 22 < 0. Also, a12 < 0 and a 21 > 0. Hence, q1 > 0. Therefore, the positive equilibrium E * is locally asymptotically stable in the presence of diffusion if the above two sufficient conditions are satisfied. The positive equilibrium point E * is also locally asymptotically stable in the absence of diffusion. However, irrespective of the sign of p1 , the system is unstable if q1 < 0, as the characteristic equation has a positive root. This implies that there exist parameter values and a range for d, for which the system may be unstable. 2. Numerical simulations are done using MATLAB 7.0. For the numerical integration, Runge-Kutta fourth order method was used. Using the given values of the parameters, we have rx = 2 − 1.4 x. Now, ( p * , z * ) is the solution of the equations s(1 − p ) − [az /(1 + bp)]

= 0, [np /(1 + bp )] − m − [ fz /(1 + g 2 z 2 ] = 0. With the given set of values, we obtain using Newton’s iteration, p * = 0.1714390348, z * = 0.473 4767465. Computations are done 30

with the two given sets of initial conditions to study the sensitivity of the solutions to changes in the initial conditions. The results show that the prey density of model system is sensitive to initial condition at x = 0.85 , ( rx = 0.81) (see Fig. 4.1). The irregular spatial and temporal behaviors of the prey and predator densities of the model system are plotted in Figs. 4.2. Space-time plots of the prey and predator densities display irregular and complex dynamics in the model system. These spatiotemporal patterns reflect the effect of environmental heterogeneity at both spatial and temporal scales. Highly disordered oscillations are observed in both space and time. We observe that the fish predation stabilizes the dynamics of model system.

Fig. 4.1. Sensitivity of prey density to initial conditions (time series).

Fig.4.2. Complex spatiotemporal patterns of prey density and predator density. (From Upadhyay, R. K., Kumari, N., Rai, V. Wave of chaos in a diffusive systems: Generarting realistic patterns of patchiness in plankton-fish dynamics. Chaos, Solitons Fractals, 40(1), 262– 276. Copyright 2009, Elsevier. Reprinted with permission). 3. Perturb the solution of (4.38), (4.39) about the steady state ( P* , H * ) to ( P * + U , H * + V ), where U, V are small. The linearized system is given by

∂U ∂ 2U = a11U + a12V + d1 2 , ∂t ∂x

∂H ∂ 2V = a21U + a22V + d 2 2 . ∂t ∂x

(4.1)

31

where

a11 = − r +

a21 =

2 B2 P*3 H * ( P *2 + D 2 )

H *2C2 , P *2

, 2

a12 = −

a22 = −C1 +

B2 P *2 P *2 + D 2

2 FH *3 ( H *2 + D12 ) 2

,

,

(the equations governing the equilibrium point are used). Assume the solution of the above equations in the form U ( x, t ) = seλt cos(kx ), V ( x, t ) = we λt sin(kx), where k takes the discrete values, k = ( nπ L), n is an integer and λ is the eigen value determining the temporal growth (frequency). Here, k denotes the wave number. Substitute the expressions for U and V in (4.1). The homogeneous equations in s and w have a solution if the determinant of the coefficient matrix is zero. We obtain the characteristic equation as λ2 + pλ + q = 0, where p = ( d1 + d 2 )k 2 − (a11 + a 22 ), q = d1d 2 k 4 − ( a11d 2 + a 22 d1 ) k 2 + a11a 22 − a12 a 21 . Note that a12 < 0, a 21 > 0. Now, a sufficient condition for p > 0 is a11 + a22 < 0, that is, *⎡

⎤ (4.2) + r + C1 > 2 H ⎢ *2 ⎥. 2 2 ( H *2 + D12 ) 2 ⎦ ⎣(P + D ) If q > 0 is also satisfied, then both the eigenvalues are negative or have negative real parts. B2 P*3

FH *2

The steady state ( P* , H * ) is linearly stable both in the presence as well as in the absence of diffusion. For the parameter values, r = 1, B1 = 0.2, B2 = 0.91, D 2 = 0.3, C1 = 0.22, C 2 =

0.2, D1 = 0.1, F = 0.02, the equilibrium point is obtained as P* = 1.184 922937, and H * = 1.205819205. Without diffusion, we obtain the characteristic equation as λ2 − 0.07009λ + 0.10709 = 0. The eigenvalues are a complex pair with positive real parts. In the absence of diffusion, the system is unstable. In order to have Turing instability, we require that one or both the eigen values have positive real parts for some k ≠ 0 . Both the roots are positive or have positive real parts if p < 0, and q > 0. One of the roots is positive if p > 0, and q < 0; or when p < 0, and q < 0. Now, p > 0, for all d1 , d 2 , when the sufficient condition (4.2) is satisfied. Diffusive instability can arise if q < 0, that is if q( k 2 ) = Dk 4 − k 2C + B < 0, where D = d1d 2 , B = a11a22 − a12 a21, and C = (d 2a11 + d1a22 ). The roots of this equation in k 2 are real and positive when (i) B > 0, (ii) C > 0, and (iii) C 2 − 4 BD > 0. Then, q < 0 when k12 < k 2 < k22 , where k12 , k 22 = [C 

C 2 − 4 BD ] /(2 D ). Therefore,

diffusive instability occurs when these conditions and (4.2) are satisfied. Now, q(k 2 ) is a quadratic in k 2 and the graph of y = q( k 2 ), is a parabola opening upwards. The minimum occurs at the vertex of the parabola, that is for k 2 = km2 where k 2 = k m2 = C /(2 D). Consider again the above parameter values. The condition (4.2) is not satisfied. We obtain, B = 0.1070942, C > 0 when ( d 2 / d1 ) > 0.7276648; C 2 − 4 BD > 0 when ( d 2 / d1 ) > 7.8551. Let, d1 = 1, d 2 = 8. Then, p > 0 when k 2 > 0.00779. Now, q > 0 when k 2 > 0.134231. For all wave numbers k 2 > 0.134231, the system is stable under the action of diffusion. Note 32

that when there is no diffusion, the equilibrium point is unstable. When 0.09973 < k 2 < 0.13423, p > 0, and q < 0. For all these wave numbers, diffusive instability sets in. 4. For the one-dimensional case, the temporal dynamics is studied by observing the effect of time on space vs. density plot of prey populations. The spatiotemporal dynamics of the system depends to a large extent on the choice of initial conditions. In a real aquatic ecosystem, the initial spatial distribution of the species can be caused by spatially homogenous initial conditions. However, in this case, the distribution of species would stay homogenous for any time, and no spatial pattern can emerge. To get nontrivial spatiotemporal dynamics, the homogenous distribution is to be perturbed. We consider the constant-gradient distribution [54]: P( x, 0) = P* , H ( x, 0) = H * + εx + δ , where ε = −1.5 × 10−6 , δ = 10−6

are parameters and

( P * , H * ) ≈ (1.8789, 1.3984)

is the non-trivial

equilibrium point for the parameter values r = 1, B1 = 0.2, B2 = 0.91, D 2 = 0.3, C1 = 0.22, C2 = 0.2, D1 = 0.1, F = 0.1. Population distributions over space at t = 2000 are given in Fig.4.3a, and with ε = −1.5 × 10 −3 , δ = 10−6 are given in Fig.4.3b. In Fig.4.3a, the spatial distributions gradually vary in time, and the local temporal behavior of the dynamical variables P and H is strictly periodic following limit cycle of the non-spatial system, which is perhaps intuitively expected from the one dimensional spatial model system. When ε in increased in magnitude from − 1.5 × 10 −6 to − 1.5 × 10−3 , the dynamics of the system undergoes major qualitative changes (see Fig.4.4b). The size of the region occupied by this pattern steadily grows with time, so that finally irregular spatiotemporal oscillations invade the whole domain. Population distributions for ε = −1.5 × 10 −2 ,

δ = 10 −3 and over space at (a) t = 1000 and (b) t = 2000 are plotted in Figs.4.4a,b. (a) (b)

Fig.4.3. Population distributions over space at t = 2000. (From Upadhyay, R. K., Wang, W., Thakur, N. K. Spatiotemporal dynamics in a plankton system. Math. Model. Nat. Phenom. 5(5), 101–121. Copyright 2010, Cambridge University press(EDP Science). Reprinted with permission).

For the two-dimensional case, Turing patterns are obtained by performing numerical simulations with system size 200 × 200 for different values of F and time t. The values of the parameters are same as in the one dimensional case. The values of the other parameters are D1 = 0.01, d1 = 0.05, d 2 = 1. The uniform mesh size along both axis is taken as h = 1/3, and the time step is taken as Δt = 1 / 40. Initially, the entire system is placed in the stationary state ( P* , H * ) and the propagation velocity is of the order of 5 × 10 −4 space units per time unit. 33

(a)

(b)

Figs.4.4. Population distributions over space at (a) t = 1000 and (b) t = 2000. (From Upadhyay, R. K., Wang, W., Thakur, N. K. Spatiotemporal dynamics in a plankton system. Math. Model. Nat. Phenom. 5(5), 101–121. Copyright 2010, Cambridge University press(EDP Science). Reprinted with permission).

Then, the system is integrated for 105 or 2 × 105 time steps and some images are saved. After the initial period during which the perturbation spreads, the system goes either into a time dependent state or to an essentially steady state. As an illustration, the typical Turing patterns of phytoplankton P are plotted in Figs.4.5a-e. We observe that the patterns change as F is increased. For F = 0.0001, 0.01, 0.035, 0.065, 0.0768, we observe spots pattern, spotstripe mixtures pattern, stripes pattern, hole-stripe mixtures, and holes pattern respectively. For F = 0.035, the system is integrated for 105 time steps and for other cases for 2 × 105 time steps. For F > 0.077, wave pattern emerges.

Fig.4.5a-e. Typical Turing patterns of P in the two dimensional spatial model system. (From Upadhyay, R. K., Wang, W., Thakur, N. K. Spatiotemporal dynamics in a plankton system. Math. Model. Nat. Phenom. 5(5), 101–121. Copyright 2010, Cambridge University press(EDP Science). Reprinted with permission).

34

5. From the first equation, we get b1 x = a1 − [ wy /( x + D )]. Hence, x ≤ a1 / b1. Equilibrium points of the system are E0 (0, 0, 0), E1 ( a1 / b1 , 0, 0), E2 ( xˆ , yˆ , 0), where xˆ is the positive solution of ( xˆ + D1 )[ a 2 + θ f ( xˆ )] − w1 xˆ = 0, yˆ = ( a1 − b1 xˆ )( xˆ + D ) / w, xˆ < a1 / b1 . The positive equilibrium point is E3 ( x* , y * , z * ) , where y * = ( w3 − cD3 ) / c, w3 > cD3 , x* is the positive solution of the equation

b1 x 2 + ( Db1 − a1 ) x + ( wy * − a1 D ) = 0,

( y * + D2 )[ w1 x * − ( x* + D1 )( a 2 + θ f ( x* ))] /[ w2 ( x * + D1 )]. Since,

z* =

z * > 0, we get the

condition w1 x * > ( x * + D1 )( a 2 + θ f ( x * )). For E0 (0, 0, 0), the eigenvalues of the Jacobian matrix are 0, a1 , − a 2 , for f (0) = 0. E0 (0, 0, 0) is a non-hyperbolic fixed point. Furthermore, as one eigenvalue is positive and another one is negative, E0 is always non-stable. For E1 ( a1 / b1 , 0, 0), the eigenvalues of the Jacobian matrix are 0, − a1 , − a 2 − θ f ( a1 / b1 ) + [ w1a1 /(a1 + b1 D1 )]. If [ w1a1 /(a1 + b1 D1 )] < a 2 + θ f (a1 / b1 ), the third eigenvalue is negative. E1 is also a non-hyperbolic fixed point. Since two of the eigen values are negative, E1 has a stable manifold in at least two dimensions. To know the actual dimensions of the stable and unstable manifolds of E 1 we need to compute the center manifold corresponding to the eigenvalue zero [1]. For E2 ( xˆ , yˆ , 0), one eigenvalue of the Jacobian matrix is zero and the other eigenvalues are root of a quadratic equation. The Jacobian of the linearized systems about E0 , E1 , and E 2 has zero as one of the eigenvalues. These are non-hyperbolic points and the dynamical behavior near them can be stable, periodic, or even chaotic. At E3 ( x* , y * , z * ) , the characteristic equation of J is

λ3 + s1λ2 + s2 λ + s3 = 0, where s1 = −( a11 + a 22 ), s2 = ( a11a 22 − a12 a 21 − a 23a 32 ), s3 = a11a 23a 32 , a11 = x * [ −b1 + {wy * /( x * + D ) 2 }], a12 = − wx * /( x * + D ), a 21 = [{w1 D1 y * /( x* + D1 )2 } − θy * f ′( x* )], a 22 = w2 y * z * /( y * + D2 ) 2 , a23 = − w2 y * /( y * + D2 ), a 32 =

w3 z *2 /( y * + D3 ) 2 , a33 = 0. Note that a12 < 0, a 22 > 0, a 23 < 0, and a32 > 0. By RouthHurwitz criterion, the equilibrium point is asymptotically stable if the characteristic equation has negative roots, or has one negative root and a complex pair with negative real parts, that is if s1 > 0, s2 > 0, s3 > 0, and s1s2 − s3 > 0. For s3 > 0, we require a11 < 0. For s1 > 0, we require a11 + a 22 < 0. Now, s1s2 − s3 = ( a11 + a22 )( a12 a 21 − a11a 22 ) + a 22 a 23a32 . The required conditions are obtained from s2 > 0, a11 < 0, a11 + a 22 < 0, and s1s2 − s3 > 0. For the given set of parameter values, we get the characteristic equation as

λ3 + 1.05876λ2 − 0.041241λ + 0.148856 = 0. The equation has one negative root −1.197, and a complex pair with positive real parts. The equilibrium point is unstable. 6. Denote T1 = ( x 2 / i ) + x + a , and T2 = ( y 2 / i ) + y + a. From the first equation, we get ( dx / dt ) ≤ x (a1 − b1 x ), that is x(t ) ≤ [a1 /(b1 + ke − a1t )], where k is a constant depending on the initial condition. As t → ∞, x(t ) ≤ a1 / b1 . Equilibrium points of the system are E0 (0, 0, 0), E1 ( a1 / b1 , 0, 0), E2 ( xˆ , yˆ , 0), where xˆ is the solution of the quadratic equation a 2 x 2 + i ( a2 − w1 ) x + ia2 a = 0, and yˆ = (a1 − b1 xˆ )T1 ( xˆ ) / w, 0 < xˆ < ( a1 / b1 ). The positive

equilibrium point is E3 ( x* , y * , z * ) , where

y * is a positive root of the equation 35

cy 2 + i( c − w3 ) y + cia = 0. We require c < w3 , and i( w3 − c ) 2 > 4ac 2 . x* is a positive root f ( x) = b1 x 3 + (ib1 − a1 ) x 2 + i (b1a − a1 ) x + i ( wy * − a1a )

of the equation

= 0.

Now,

f (0) = i ( wy * − a1a ), and f ( a1 / b1 ) = iwy * > 0. If y * < a1a / w, then f (0) f ( a1 / b1 ) < 0,

and

a

positive

root

[ −a 2 + {w1 x* / T1 ( x* )}].

exists

in

( 0, a1 / b1 ).

z * is

given

by

z* = [T2 ( y* ) / w2 ]

Since z * > 0, we require {w1 x* / T1 ( x * )} > a 2 . The conditions

for the existence of a positive equilibrium point are c < w3 , i( w3 − c) 2 > 4ac 2 , y * < a1a

/ w, and {w1 x* / T1 ( x * )} > a 2 . At the equilibrium point E0 (0, 0, 0), the eigenvalues of the Jacobian matrix J are a1 , −c, − a 2 . E0 (0, 0, 0) is unstable. . It is a saddle point with a stable manifold in y-z direction ad unstable manifold in x direction. At the equilibrium point E1 ( a1 / b1 , 0, 0), the eigenvalues of the Jacobian matrix J are − c, − a1 , a 22 , where a 22 = −a 2 + [ w1a1 /{b1 T1 ( a1 / b1 )}]. The equilibrium point is asymptotically stable for {a 2 b1T1 ( a1 / b1 )} > w1a1. At the equilibrium point E2 ( xˆ , yˆ , 0), the characteristic equation of J is ( a33 − λ )(λ2 − a 11 λ − a12 a21 ) = 0. One eigenvalue is λ1 = a33 , which is negative when [ w3 yˆ / T2 ( yˆ )]

< c. Now, a12 = −[ wxˆ / T1 ( xˆ )] < 0 ; a 21 = [ w1 yˆ (ia − xˆ 2 )] /[iT12 ( xˆ )] > 0, when xˆ 2 < ia ; and a 11 < 0, when wyˆ (i + 2 xˆ ) < b1iT12 ( xˆ ). By Routh-Hurwitz criterion, the eigenvalues λ2 , λ3

are negative or has negative real parts when the above three conditions are satisfied. The equilibrium point E2 ( xˆ , yˆ , 0) is asymptotically stable. At E3 ( x* , y * , z * ) , the characteristic equation of J is λ3 + s1λ2 + s2 λ + s3 = 0, where s1 = −(a11 + a 22 ), s2 = ( a11a 22 − a12 a 21 − a23a32 ),

s3 =

a11a 23a 32 , a11 = x * [− b1 + {wy * ( 2 x * + i ) /(iT12

( x * )}],

a12 = wx * / T1

( x * ), a13 = 0, a 21 = [ w1 y * (ai − x *2 ) /{iT12 ( x * )}], a 22 = w2 y * z * ( 2 y * + i ) /{iT22 ( y * )}, a 23 = = − w2 y * / T2 ( y * ), a31 = 0, a32 = w3 z * ( ai − y *2 ) / T22 ( y * ), a 33 = 0. Note that a12 < 0, a22 > 0, and a 23 < 0. By Routh-Hurwitz criterion, the equilibrium point is asymptotically stable if the characteristic equation has negative roots, or has one negative root and a complex pair with negative real parts, that is if s1 > 0, s2 > 0, s3 > 0, and s1s2 − s3 > 0,

(sufficient conditions). For s3 > 0, we require a11 < 0. For s1 > 0, we require a11 + a 22 < 0. Now, s1s2 − s3 = ( a11 + a 22 )( a12 a 21 − a11a22 ) + a22a23a32 . The required conditions are obtained from s2 > 0, a11 < 0, a11 + a 22 < 0, and s1s2 − s3 > 0. Consider the given set of parameter values.

cy 2 + i( c − w3 ) y + cia = 0, y1*

= 0.213259 ,

y *2

that is,

y * is the positive solution of the equation 0.25 y 2 − 0.405 y +

0.075 = 0.

= 1.406741 . We find that corresponding to *

for the cubic equation for x . For

y1* ,

y 2*

The roots are

there is no positive root

the cubic equation in x has a positive root x1* = 0.9 *

11071, and a complex pair with negative real part. We obtain z * = 0.329525. For this value of the equilibrium point, we get the characteristic equation as λ3 + 0.528227λ2 − 36

0.107421λ + 0.084325 = 0. The equation has one negative root − 0.796175, and a complex pair with positive real parts. The equilibrium point is unstable. 7. Denote T1 = ( x 2 / i1 ) + x + a, and T2 = ( y 2 / i2 ) + y + a. From the first equation, we obtain 0 < x < ( a1 / b1 ). Equilibrium points of the system are E0 (0, 0, 0), E1 (a1 / b1 , 0, 0), E2 ( x , y , 0), where ~ x is the solution of the a 2 x 2 + i1 ( a2 − w1 ) x + aa2 i1 = 0, and ~ ~ y = ( a1 − b1 ~ x )T1 / w, 0 < x < a1 b1 . The positive equilibrium point is E3 ( x* , y* , z * ), where y * = ( w3 − cd ) / c, w3 > cd . x* is a positive root of the equation f ( x ) = b1 x 3 + (i1b1 − a1 )

x 2 + i1 ( ab1 − a1 ) x + [i1 ( wy * − aa1 )] = 0, 0 < x* < (a1 / b1 ). Now, f (0) = [i1 ( wy * − aa1 )] < 0,

when y * < aa1 / w; and f ( a1 / b1 ) = i1wy * > 0. Hence, there exists a positive root in

( 0, a1 / b1 ). From the second equation, we get z * = [{w1 x * − a 2T1 ( x * )}T2 ( y * )] /{w2T1 ( x * )}. Since z * > 0, we require w1 x* > a 2T1 ( x* ). In summary, the conditions for the existence of a positive equilibrium point are cd < w3 , y * < a1a / w, and w1 x * > a 2T1 ( x * ). At the equilibrium point E0 (0, 0, 0), the eigenvalues of the Jacobian matrix are a1 , − a2 , 0. E0 (0, 0, 0) is a non-hyperbolic fixed point. Furthermore, as one eigenvalue is positive, and another one is negative, E0 is always non-stable. At the equilibrium point E1 ( a1 / b1 , 0, 0), the eigenvalues of the Jacobian matrix J are 0, − a1 , a22 , where a22 = −a2 + [ w1a1 /{b1T1 ( a1 / b1 )}]. The equilibrium point is also a non-hyperbolic fixed point. At the equilibrium point E2 = ( x , y ,0) , one of the eigenvalues is zero and the other eigenvalues are roots of

λ2 − a11λ − a12 a 21 = 0. or a complex pair with eigenvalues is positive.

If a11 < 0, and a12 a21 < 0, then both the eigenvalues are negative negative real parts. If a11 > 0, and a12 a 21 > 0, then one of the E 2 is a non-hyperbolic point. The dynamical behavior near these

equilibrium points can be stable, periodic or even chaotic. At E3 ( x* , y * , z * ) , the characteristic equation of J s2 = ( a11a 22 − a12 a 21 − a 23a 32 ), ( x * )}],

a12 = − wx * / T1 ( x * ),

is λ3 + s1λ2 + s2 λ + s3 = 0, where s1 = −(a11 + a 22 ), s3 = a13 = 0,

a11a 23a 32 ,

a11 = x*[ −b1 + {wy * ( 2 x* + i1 ) /(i1T12

a 21 = [ w1 y * ( ai1 − x *2 ) /{i1T12 ( x * )}],

a 22 = w2

y * z* ( 2 y * + i2 ) /{i2T22 ( y* )}, a31 = 0, a23 = − w2 y * / T2 ( y * ), a32 = c 2 z * / w3 , a33 = 0. Note that a12 < 0, a 22 > 0, a 23 < 0, and a32 > 0. By Routh-Hurwitz criterion, the equilibrium point is asymptotically stable if the characteristic equation has negative roots, or has one negative root and a complex pair with negative real parts, that is if s1 > 0, s2 > 0, s3 > 0, and s1s2 − s3 > 0, (sufficient conditions). For s3 > 0, we require a11 < 0. For s1 > 0, we require a11 + a 22 < 0. Both the conditions are satisfied if [ wx * ( 2 x * + i1 ) /(i1T12 ( x * )] + [ w2 z * ( 2 y * + i2 ) /{i2T22 ( y * )} < (b1 x * / y * ). Now, s1s2 − s3 = ( a11 + a 22 )( a12 a 21 − a11a 22 ) + a 22 a 23a 32 . The required conditions are the above

condition, s2 > 0, and s1s2 − s3 > 0. Consider the given set of parameter values. (i) We obtain, y * = 625 / 27. x* is a positive root of the equation 0.075x 3 + 1.125x 2 − 73.575 x + 211.416666 = 0. We get two positive roots x1* = 22.718741, x2* = 3.043907. 37

Corresponding to x1* , we get z * = 70.790948. Corresponding to x2* , we get a negative value for z * . Therefore, the positive equilibrium point is (22.718741, 23.148148, 70.790 948). We get the characteristic equation as λ3 + 1.03824λ2 − 0.126348λ + 0.000776 = 0. The equation has one negative root − 1.14881, and the other roots are 0.104077, 0.00649. The equilibrium point is unstable. (ii) We obtain, y * = 330 / 13. x* is a positive root of the equation 0.088 x 3 + 1.71x 2 −

68.778 x + 312.057692 = 0. We get two positive roots x1* = 15.812703, x*2 = 5.503507. Corresponding to x1* , we obtain z * = 22.974668. Corresponding to x *2 , we find z * < 0. Therefore, we get one positive equilibrium point (15.812703, 25.384615, 22.974668). We get the characteristic equation as λ3 + 0.532106λ2 + 0.027195λ + 0.000417 = 0. The equation has one negative root − 0.47692, and the complex pair −0.027594 ± 0.01064i. The equilibrium point is asymptotically stable. The chaotic attractor for the first set of parameter values is given in Fig. 4.6a. Stable limit cycles for the second set of parameter values are given in Fig. 4.6b. 8 The model is numerically integrated to get the time series using MATLAB 7.5. Since every non-linear system has a finite amount of transients, the data points representing transient behavior were discarded. Phase portraits were drawn using this data to obtain the geometry of the attractors. The chaotic attractor and its corresponding time series are depicted in Figs.4.7a-b. We note that the solution of the model system is very sensitive to the death rate parameter c in the range 0.3 ≤ c ≤ 0.318 . The route of chaos is through the cascade of period doubling for decreasing values of the control parameter c. Other important changes in the chaotic set include an interior crisis in which a chaotic attractor undergoes a sudden increase in the size of the attractor. Closed curves marked as ‘A’ and ‘B’ in Fig.4.8 correspond to invariant KAM tori in the phase space. When the bifurcation parameter c is decreased further, these curves break and give rise to chaotic dynamics.

(a) (b) Fig.4.6a. Chaotic attractor the first set of parameter values. (b) Stable limit cycles for the second set of parameter values. (From Upadhyay, R. K., Rai, V., Raw, S. N. Challenges of living in the harsh environments: A mathematical modelling study. Appl. Math. Comp., 217, 10105–10117. Copyright 2011, Elsevier. Reprinted with permission).

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Fig.4.7.(a) Chaotic attractor with initial condition [0.8848, 0.2903, 0.0002]. (b) Temporal evolution of population densities x(t ), y (t ) and z (t ) for the chaotic attractor. (From Upadhyay, R. K., Banerjee, M., Parshad, R. D., Raw, S. N. Deterministic chaos versus stochastic oscillation in a prey-predator-top predator model. Math. Model. Anal.. 16(3), 343–364. Copyright 2011, Taylor & Francis. Reprinted with permission).

Fig.4.8. Bifurcation diagram of the model system with c as a control parameter. (From Upadhyay, R. K., Banerjee, M., Parshad, R. D., Raw, S. N. Deterministic chaos versus stochastic oscillation in a prey-predator-top predator model. Math. Model. Anal.. 16(3), 343–364. Copyright 2011, Taylor & Francis. Reprinted with permission).

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