Kumar upadhyay solutions manual for introduction to mathematical modeling and chaotic dynamics CRC (2014)

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

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Introduction to Mathematical Modeling
and
Chaotic Dynamics

Solution Manual

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

1

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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.
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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⎦ ⎣

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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.
Choose
25. x1′ = −2 x1 + 4 x 23 − 4 x 24 ,
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

∫

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

0

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

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.
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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

0

0

x

50

100

150

Time

Fig.1.1. Phase plot for Problem 43.

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])
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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

15

0

10

-10

5

-10

0
-5
-20

y

-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.

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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.

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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

⎛r⎞
⎛P⎞
= ⎜ ⎟ P[1 − t ( 2 + α ) + t 2 (1 + α )],
where t = ⎜ ⎟ .
⎝α ⎠
⎝K⎠
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.
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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.
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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
For P = ,
= ⎜ − ⎟ exp( r − 1) < 0.
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.

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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
3. We have F ( X , Y ) = a1 − b1 X −
, G ( X , Y ) = −a2 + 1

.
X +D
X + D1
Conditions (i), (ii), (iv), (v), (vi), (vii) are satisfied. Equality condition in (iii) is satisfied.
wC
a D
(viii) G (C , 0) = −a2 + 1
= 0, gives C = 2 1 .
C + D1
w1 − a 2
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

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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

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⎛ a D ⎞
⎛ a D ⎞
(A)
2⎜⎜ 2 1 ⎟⎟ + D − K > 0, or 2b1 ⎜⎜ 2 1 ⎟⎟ + b1 D − a1 > 0.
⎝ 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

2 fvh Z2
bu
.
=
−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
a 21 =

=

⎡
⎤
hP v
v

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

bhP v
2 fvhZ2
bu
fv( v 2 − hZ2 )
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
+
> 0,
B = a11a 22 − a12 a 21 = u ⎢1 −
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.

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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
α + βY + γX w1 X
wY
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 − αa 2 ].
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

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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 )
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a11 + a 22 =

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

⎡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,
⎤⎡ a β
⎡ γa
⎤
( 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.
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⎛ ∂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).
⎞
⎛
X⎞
BZ
w3
⎛
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 ⎞
+Z
(iv) X ⎜
⎟ > 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

.
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(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
.
2
( X + D3 )
( X + D3 )
If X m is the maximum value of X, we can choose c < w3 /( X m + D3 ).
(b) B > 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,
K
α1 w4 + β1cZ + w4γ 1Z

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.
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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
> 0, if w3 > Aβ1 .
(vi) F (0, A*) = 0, gives A* =
( 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.

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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

1
0.6
0.4

0.2
y

(a)

0.2

0.8

0.4
0

0.2
0

x

0
4000

4500

5000

5500

t

(b)

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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

4

5

6

7

-6

b

(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).
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