Chapter 6
HARVESTING AND HOPF BIFURCATION IN A PREYPREDATOR MODEL WITH HOLLING TYPE IV FUNCTIONAL
RESPONSE
INTRODUCTION
In chapter 5, we saw the effect of harvesting on dynamics of prey-predator
model with Holling type-III functional response. The aim of this chapter is to study
the effect of harvesting and time delay on predator species in a prey-predator model
with Holling type-IV functional response. The prey-predator model is a kind of
“pursuit and evasion” system in which the prey try to evade the predator and the
predator try to catch the prey if they interact.
Two species continuous time models of prey (pest) – predator (natural enemy)
type of interaction with functional response have been extensively discussed in the
literature (Baek, 2010; Cantrell and Consner, 2001). Andrews (2006) suggested a
form of functional response as f ( X , Y ) 
mX
, which is called as the Monoda  bX  Y 2
Haldane function, and also called a Holling type- IV function. Further this form is
simplified by taking b  0 and then it becomes f ( X , Y ) 
mX
.
a X2
Li and Wei (2009) have proposed a prey-predator system with BeddingtonDeAngelis functional response and harvesting of predator species. This system also
obeys Holling type-IV functional response. So in this chapter we have modified the
model for those species by considering Holling type IV functional response.
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Dynamics of the system has been studied in terms of local and Hopf bifurcation
analysis. Finally, numerical simulation is done to support the analytical findings.
6.1 MATHEMATICAL MODEL
Let us consider the prey-predator model with Holling type-IV functional
response and harvesting of predator (natural enemy) species.
dX
X
XY

 rX 1   
,
dt
K  1  e1 X 2

dY
X
 b0Y  (d 0 
)Y  qE 0Y ,
dt
1  e2 X 2
X (0)  0,
(6.1.1)
Y (0)  0.
Here X (t ) and Y (t ) denote the population densities of prey (pest) and predator (natural
enemy) at time t , respectively. r and K
represent the intrinsic growth rate and
carrying capacity of prey in the absence of predation, respectively. ei , i  1,2 are the
half-saturation constants. The predator consumes the prey with functional
response
X
1  e1 X 2
its growth rate
, known as Holling type-IV functional response and contributes to
XY
1  e2 X 2
. b0 is a constant birth rate and d 0 is the maximum death rate
in the absence of food. Moreover, the catch rate function qE 0Y is based on the catchper-unit-effort (CPUE) hypothesis. Here, q is the catchability coefficient of the
predator species and E 0 is the harvesting effort.
To analyze the model (6.1.1), we need the bounds of dependent variables involved.
For this we find the region of attraction in the following lemma.
2rK 

Lemma6.1.1: The set   ( X , Y ) : 0  X  Y 
, where
m 

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m  min r , d 0  qE0  b0  is the region of attraction for all solutions initiating in the
interior of the positive octant.
Proof: Let ( X (t ), Y (t )) be any solution with positive initial conditions ( X 0 , Y0 ).
From first equation of model (6.1.1), we get
dX
r
 rX  X 2 .
dt
K
Now using comparison principle, we obtain
lim sup X t   K for all t  0,
t 
X max  K .
Now define a function
W (t )  X (t )  Y (t ).
Computing the time derivative of W (t ) along solutions of system (6.1.1), we get
dW
X
XY
X

 rX 1   
 b0Y  (d 0 
)Y  qE0Y ,
2
dt
K  1  e1 X
1  e2 X 2

 2rK  mW ,
where
m  min r , d 0  qE0  b0 .
Thus
dW
 mW (t )  2rK .
dt
Applying a theorem of differential inequalities, we obtain
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153
0  W ( X ,Y ) 
2rK
2rK W ( X 0 , Y0 )
. Therefore all solutions

, and for t  , 0  W 
m
m
exp( mt)
of system (6.1.1) enter into the region
2rK 

  ( X , Y ) : 0  X  Y 
.
m 

This completes the proof of lemma.
6.2 EXISTENCE OF EQUILIBRIA
Equating the derivatives on the left hand sides to zero and solving the resulting
possible equilibria E1 0,0, E2 K ,0 and
algebraic equations we find three
E3  X *, Y * . The trivial equilibrium point E1 0,0 always exists. The equilibrium
point E2 K ,0. exists on the boundary of the first octant. We show the existence of
interior equilibrium as follows
Existence of E3  X *, Y *
Here X * and Y * are the positive solutions of the system of algebraic equations given
below
X *
Y *

r 1 
 0,

K  1  e1 X *2

b0  (d 0 
X *
1  e2 X *2
(6.2.1)
)  qE 0  0.
(6.2.2)
From equations (6.2.1) and (6.2.2) we get
X* 
   2  4e2 d 0  qE0  b0 2
2e2 d 0  qE0  b0 
, Y* 
r K  X *1  e1 X *2 
.
K
And for X * and Y * are the positive, we must have
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154
d 0  qE0  b0  0.
(6.2.3)
Hence the equilibrium E3  X *, Y * exists under the above condition.
6.3 STABILITY ANALYSIS
To discuss the local stability of system (6.1.1), we compute the variational
matrix of system (6.1.1). The entries of general variational matrix are given by
differentiating the right side of system (6.1.1) with respect to X and Y , i.e.



2rX 1  e1 X 2 Y
r



2
K
1  e1 X 2

M ( X ,Y ) 

1  e2 X 2 Y

2
1  e2 X 2









1  e1 X
.
X 
b0  d 0  qE 0 

1  e 2 X 2 

X
2
Using analogous notations to the equilibria, i.e. M ( E1 ) is the variational matrix
corresponding to E1 , M ( E2 ) is the variational matrix corresponding to E2 and
M ( E 3 ) is the variational matrix corresponding to E 3 , we get
0
r

M ( E1 )  
.
0 b0  d 0  qE 0 
The eigenvalues of M ( E1 ) are 1  r and 2  b0  d 0  qE0 . Equilibrium point E1
is stable in Y  direction with condition (6.2.3) and unstable in X  direction . So E1
is a saddle point of the system.
K



2
 r

1  e1 K
.
M (E2 )  
 0 b  d  qE   K 
0
0
0

1  e2 K 2 
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The eigenvalues of M ( E2 ) are 1  r and
2  b0  d 0  qE 0 
K
1  e2 K 2
.
Equilibrium point E2 is locally stable in X  direction and unstable in Y  direction.
So E2 has stable manifold in X  direction and unstable manifold in Y  direction.
Now we discuss the stability of the interior equilibrium point



2rX * 1  e1 X *2  Y *
X * 


r 

2
2
K
1  e1 X *2 
1  e1 X *

M ( E3 ) 
.


1  e2 X *2  Y *
0


2 2
1

e
X
*
2








The characteristic equation corresponding to variational matrix M ( E 3 ) is given by
2  A1  A2  0,
where
 1  e2 X *2 X * Y *
rX * 2e1X *2 Y *
.
A1 

,A 
K
1  e1 X *2 2 2 1  e1 X *2 1  e2 X *2 2
By the Routh-Hurwitz’s condition, E 3 is locally asymptotically stable provided the
following conditions are satisfied
A1 
 1  e2 X *2 X * Y *
rX * 2e1X *2 Y *


0
A

 0.
,
2
K
1  e1 X *2 2
1  e1 X *2 1  e2 X *2 2
6.4 ANALYSIS OF THE MODEL WITH DELAY
In this section, we consider the following modification of the model (6.1.1) by
incorporating discrete time delay. We assume that the juveniles of prey and predator
take  1 and  2 units of time to mature, respectively. We get the desired delay-induced
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prey-predator model with Holling type-IV functional response and harvesting of
predator (natural enemy) species as follows
X t   1  
dX
XY

 rX 1 
,

2
dt
K

 1  e1 X
(6.4.1)
dY
X
 b0Y (t   2 )  (d 0 
)Y  qE 0Y .
dt
1  e2 X 2
with initial condition given by
X ( )   1 ( )  0, Y ( )   2 ( )  0,     i ,0, i  1,2.
6.5 STABILITY AND LOCAL HOPF BIFURCATION
To determine the local stability of interior point E3  X *, Y * we linearize the
system (6.4.1). The linearized form of the system (6.4.1) is given by
rX * xt   1 
yX *
dx  
X *
Y * 
 r  1 
x

,

2 
2
dt  
K  1  e1 X *  1  e1 X *
K
 X *

dy
xY *

 
 d 0  qE 0  y  b0 y (t   2 ) .
2
2
dt 1  e 2 X *  1  e 2 X *

(6.5.1)
Where x and y are small perturbations given to X and Y respectively. Such that
X t   X *  x(t ) , Y t   Y *  y(t ) .
The characteristic equation associated with system (6.5.1) is given by
2  A  ( B1  B2 )e    (C1  C2 )e    D1e   (  )  D2  0,
1
2
1
2
(6.5.2)
where
rX *
X *
 X *
, B1 
,
 r 1 
  d 0  qE0 
2
K
1  e1 X *
K 
1  e2 X *

 Y*
rX * 
 X* 
X * 

 d 0  qE 0 
 , C1  b0 , C 2  b0 
B2 
 r 1 
 ,
2 
2
K 
1  e2 X * 
K 

1  e1 X *
A
Y*
2
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D1 
  Y * (d 0  qE0 )
 b0 rX *
X *   X *

, D2  r 1 
 d 0  qE0  
.

2
2
K 1  e2 X *
1

e
X
*
K

1

Case 1:  1   2  0
In this case, Equation (6.5.2) becomes
2  ( A  B1  C1 )  B2  C 2  D1  D2  0 .
(6.5.3)
All roots of Equation (6.5.3) have negative real parts if and only if
(H1) A  B1  C1  0 and B2  C2  D1  D2  0 .
So the equilibrium point E3  X *, Y * is locally asymptotically stable when (H1) holds
and unstable if either of these conditions is not satisfied. Hence, in the absence of time
delay,
the
A  B1  C1  0
system
is
locally
asymptotically
stable
if
and
only
if
and B2  C 2  D1  D2  0 hold simultaneously.
Case 2:  1  0, 2  0
Here, Equation (6.5.2) becomes
2  ( A  B1 )  B2  D2  (C1  C 2  D1 )e   0 .
2
(6.5.4)
Let   iw be one such root. Substituting this in equation (6.5.4) and equating real
and imaginary parts, we get
(C 2  D1 ) cos w 2  C1 w sin w 2  w 2  ( B2  D2 ) ,
(6.5.4(a))
C1 w cos w 2  (C2  D1 ) sin w 2  w( A  B1 ) .
(6.5.4(b))
Squaring and adding the equations (6.5.4(a)) and (6.5.4(b)) we get
w 4  {( A  B1 ) 2  C12  2( B2  D2 )}w 2  ( B2  D2 ) 2  (C 2  D1 ) 2  0 .
(6.5.5)
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From (6.5.5), it follows that if
(H2) ( A  B1 ) 2  C12  2 (B2  D2 )  0 and ( B2  D2 ) 2  (C2  D1 ) 2  0
holds, then Equation (6.5.5) has no positive roots. Hence, all roots of (6.5.4) have
negative real parts when  2  (0, ) under (H1) and (H2).
If (H1) and (H3) ( B2  D2 ) 2  (C 2  D1 ) 2  0 hold, then (6.5.5) has a unique positive
root w 02 . Substituting w 02 into (6.5.4(a)) and (6.5.4(b)), we have
2
 2
 2n
1
1 w0 {C 2  D1  C1 ( A  B1 )}  (C 2  D1 )( B2  D2 )

cos 
, n  0,1,2,.....

2
2
2
w0
C1 w0  (C2  D1 )

 w0
n
Again, if
(H4) C12  2( B2  D2 )  ( A  B1 ) 2  0, ( B2  D2 ) 2  (C 2  D1 ) 2  0
C
2
1
and

 2( B2  D2 )  ( A  B1 ) 2  4( B2  D2 ) 2  (C 2  D1 ) 2 
2
hold, then there are two positive solutions w 2 . Substituting w 2 into (6.5.4(a)) and
(6.5.4(b)), we have
 2 
k
 w 2 C  D1  C1 ( A  B1 )  (C 2  D1 )( B2  D2 )  2k
1
cos 1   2
,

w
C12 w2  (C 2  D1 ) 2

 w
k  0,1,2,.....
(6.5.5(a))
Differentiating equation (6.5.4) with respect to  2 , we obtain
2  A  B
1
  2 e  2 (C1   C 2  D1 )  C1 e  2
 dd
 e  2 (C1   C 2  D1 )  0 . (6.5.6(a))
2
Therefore,
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 d

 d 2
1

2  A  B1
C1

 

 2,
3
2
    ( A  B1 )  ( B2  D2 )   (C1  C 2  D1 ) 

by using e
  2
(6.5.6(b))
 2   ( A  B1 )  B2  D2 
.

(C1  C2  D1 )

 d (Re  ) 
  d
Thus, sign 
 sign Re


 d 2   iw
  d 2



1


 ,

  iw
 ( A  B1 ) 2  2(1  B2  D2 )

C12
 sign 

. (6.5.6(c))
2
4 2
2
C1 w0  (C 2  D1 ) 
 ( A  B1 )  (1  B2  D2 )
Theorem 6.5.1. If (H1) and (H3) hold, then the equilibrium E3  X *, Y * is unstable
for  2   20 and asymptotically stable for  2   20 . Further, as  increases through  20 ,
E3  X *, Y * bifurcates into small amplitude periodic solutions where  20   2n as
n  0.
Proof. For  2  0 , E3  X *, Y * is unstable if condition (H1) holds. Hence, by Butler’s
lemma, E3  X *, Y * remains unstable for  2   20 . Now we show that
 d (Re  ) 
 0.


 d 2    0 , w w0
(6.5.7)
This signifies that there exists at least one eigenvalue with positive real part
for  2   20 . Moreover, the conditions of Hopf bifurcation (Hale and Lunel, (1993))
are satisfied yielding the required periodic solution.
From (6.5.6(c)), it follows that
 ( A  B1 ) 2  2(1  B2  D2 )

 d(Re  ) 
C12
sign 
 sign 
 4 2
.

2
2 
C1 w0  (C 2  D1 ) 
 d 2   iw 0
 ( A  B1 )  (1  B2  D2 )
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Therefore,
 d (Re  ) 
 0.


 d 2    0 , w w0
Therefore, the transversality condition holds and hence Hopf bifurcation occurs
at w  w0 , 2   20 . This completes the proof.
Theorem 6.5.2. Let  2k be defined in (6.5.5(a)). If (H1) and (H4) hold, then there
exists a positive integer m such that there are m switches from stability to instability
and instability to stability. In other words, when   [0, 0 )  ( 0 , 1 )  .. .  ( m 1 , m ) ,
the
equilibrium
point
E3  X *, Y *
is
unstable
and
when
  [ 0 , 0 ) 
( 1 , 1 )    ( m 1 , m ) , E3  X *, Y * is stable.
Therefore, there are bifurcations at E3  X *, Y * when    k , k  0,1,2,...
Proof. If conditions (H1) and (H4) hold, then to prove the theorem we need only to
verify the transversality conditions,
 d (Re  ) 
0


 d 2    2k
and
 d (Re  ) 
 0.


 d 2    2k
From (6.5.6(c)), it follows that
 d(Re  ) 
 ( A  B1 ) 2  2(1  B2  D2 )

C12
sign 
 sign 
 4 2
,

2
2 
C1 w  (C 2  D1 ) 
 d 2   iw 
 ( A  B1 )  (1  B2  D2 )
 d (Re  ) 
and therefore 
0.

 d 2  w w ,  2k
Again,
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 d(Re  ) 
 ( A  B1 ) 2  2(1  B2  D2 )

C12
sign 

sign

,


2
4
2
2 
d

(
A

B
)

(
1

B

D
)
C
w

(
C

D
)
2
1
2
2
1

2
1

  iw 


 d (Re  ) 
and therefore 
 0.

 d 2  w w ,  2k
Hence, the transversality conditions hold.
Case 3:  1  0, 2  0
We consider equation (6.5.2) with  2 in its stable interval and  1 as a
parameter. Without loss of generality, we consider system (6.5.1) under the
assumptions (H1) and (H3). Let   iw be one such root. Substituting this in equation
(6.5.2) and equating the real and imaginary parts we get
cos w1B2  D1 cos w 2  sin w1{B1w  D1 sin w 2}  w2  D2  C1wsin w 2  C2 cos w 2 , (6.5.8(a))
cos w 1 B1 w  D1 sin w 2  sin w 1{B2  D1 cos w 2 }  C2 sin w 2  C1 w cos w 2  Aw . (6.5.8(b))
Squaring and adding equation (6.5.8(a)) and (6.5.8(b)), we get
~
~
~
w 4  Aw 2  D22  C 22  D12  B22  2 B sin w 2  2C cos w 2  0 ,
(6.5.9)
where
~
~
A  C12  A 2  B12  2 D2 , B  C1 w 3  D2 C1 w  AwC2  B1 D1 w ,
~
C   w 2 C 2  D2 C 2  Aw 2 C1  B2 D1 .
~
~
~
We define F ( w)  w 4  Aw 2  D12  C 22  D12  B22  2 B sin w 2  2C cos w 2 and
assume that
(H5) {( D2  C2 ) 2  ( D1  B2 ) 2 }  0 holds.
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162
It is easy to check that F (0)  0 and F ()  0 . We can obtain that (6.5.9) has finite
positive roots w1 , w2 ,.....wk . For every fixed wi , i  1,2,......k , there exists a sequence

j
1i



| j  1,2,3,... , such that (6.5.9) holds. Let  10  min  1ji | i  1,2,......, k; j  1,2,3...... .
When  1   10 , equation (6.5.2) has a pair of purely imaginary roots  iw0 for
 2  [0, 20 ) .In the following we assume that
 d (Re  ) 
 0.
(H6) 

d

1

  iw0
Therefore, by the general Hopf bifurcation theorem (Hale,1993), we have the
following result on stability and bifurcation in system (6.4.1).
Theorem 6.5.3. For the system (6.4.1), suppose (H1), (H3) and (H5) are satisfied and
 2  [0, 20 ). Then the equilibrium point E3  X *, Y * is asymptotically stable when
  (0, ) and system (6.4.1) undergoes a Hopf bifurcation at E3  X *, Y * when
1
10
1  1 .
0
6.6 NUMERICAL SIMULATION AND DISCUSSION
In this section, we present numerical simulation to illustrate the results
obtained in the previous sections. The system (6.4.1) is solved using the MATLAB
software package under the following set of parameters.
(a) r  20, K  120,   10, e1  2, b0  2.6, d 0  2.5,   4, e2  2, q  0.4, E0  0.5
The interior equilibrium point of system (6.4.1) with data (a) is
X *  0.0250313 , Y *  2.00209 .
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Then, we can easily obtain that (H1) and (H4) to be satisfied. By computation,
we have w0  2.1415, 20  0.245831 . Here we show that the transversality condition
 d (Re  ) 
(6.5.7) is satisfied as 
 1.20339  0 .

 d 2   
0
Therefore, E3  X *, Y * is unstable for  2   20  0.245831 and asymptotically
stable for 2.5   2   2  0.245831 .When  2   20 , system (6.4.1) with data (a)
0
undergoes a Hopf bifurcation at E3  X *, Y * , that is there is a small amplitude
periodic solution around E3  X *, Y * when  1  0 and  2 is close to  2 0 which is shown
in Figure (1a-1c).
2.25
2.2
2.15
Predator
2.1
2.05
2
1.95
1.9
1.85
1.8
0
0.02
0.04
0.06
0.08
0.1
0.12
Prey
Figure 1a. When  1  0, 2  0.2 .
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164
0.04
0.035
Prey
0.03
0.025
0.02
0.015
0.01
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 1b. When  1  0, 2  0.3 , stable behavior of prey with time and other
parameter values are same as data (a).
2.08
2.06
Predator
2.04
2.02
2
1.98
1.96
1.94
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 1c. When  1  0, 2  0.3 , stable behavior of predator with time and other
parameter values are same as data (a).
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Regard  1 as a parameter and let  1  0.034 . From Figure (2a) and (2b) we can
easily see that when  1  0.034,  2  0.2   20 the system is unstable and when
 1  0.034,  2  0.3   20 the system becomes stable that is there is a small amplitude
periodic solution around E3  X *, Y * when  1  0.034, 2   20 . But when  2 reaches
2.5 the system becomes unstable, it means stability of the system lies
in [0.245831, 2.5] . Thus, delay plays an important role in Hopf bifurcation.
2.08
2.06
Predator
2.04
2.02
2
1.98
1.96
1.94
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
Prey
Figure 2a. When  1  0.034, 2  0.2 .
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166
2.4
2.3
Predator
2.2
2.1
2
1.9
1.8
1.7
0
0.02
0.04
0.06
0.08
0.1
0.12
Prey
Figure 2b. When  1  0.034, 2  0.3 .
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
120.15
-1
2.5
120.1
120.05
2
120
1.5
1
0.5
Predator
119.95
0
119.9
Prey
Figure 2c. When  1  0, 2  0 .
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167
When there is no delay, the system (6.4.1) becomes system (6.1.1). From
figure (2c), system shows chaotic behavior that means the system is unstable. Thus,
from numerical simulation we have observed that when population of prey is very
small, maturation delay in predator plays an important role in stability of the system.
When maturation time of predator lies between 0.245831 and 2.5 , the system remains
stable and the stability of the system is lost when  2  2.5 , refer Figure 3. Here, we can
show ecological interpretation of maturation delay. Some insects, such as ladybirds
are very helpful to farmers, because they eat pest species such as greenfly, whitefly
and mealy bug. Ladybirds are natural enemies of pests. If maturation delay of natural
enemies is  2  2.5 , the population of pests will increase and this is harmful for crops.
For good production, it is necessary maturation time of natural enemies (predator) lies
between 0.245831 and 2.5 .
2.04
2.03
Predator
2.02
2.01
2
1.99
1.98
0.015
0.02
0.025
0.03
0.035
0.04
Prey
Figure 3. When  1  0.034, 2  2.5 .
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168
Now, we see effect of harvesting on the system. Harvesting has a strong
impact on the dynamic evolution of a population. To a certain extent, it can control
the long-term stationary density of population efficiently. However, it can also lead to
the incorporation of a positive extinction probability and therefore to potential
extinction in finite time. We take harvesting of predator species. If the population of
ladybirds increases out of limit, it will create problem for the human beings. So we
had taken harvesting of ladybirds (predator). Figure (4) is the plots of prey population
and predator population against time for different value of E 0 at  2  0.2 . Figure 4
shows that predator population decreases with E0 increases and becomes extinct
if E 0  0.5 . Here, we can also note that the prey population increases with the increase
of E0 and reaches to the peak level.
140
120
100
E0=0.4
E0=0.5
Prey
80
E0=0.6
60
E0=0.7
40
20
0
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 4a. When  1  0, 2  0.2 , Variation of the prey population with time for
different values of harvesting effort and other parameter values are same as data (a).
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169
3
2.5
Predator
2
E0=0.4
1.5
E0=0.5
E0=0.6
1
E0=0.7
0.5
0
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 4b. When  1  0, 2  0.2 , Variation of the predator population with time for
different values of harvesting effort and other parameter values are same as data (a).
0.045
0.04
E0=0.4
0.035
E0=0.5
E0=0.6
Prey
0.03
0.025
0.02
0.015
0.01
0.005
0
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 5a. When  1  0, 2  0.3 , Variation of the prey population with time for
different values of harvesting effort and other parameter values are same as data (a).
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Predator
2.1
2.08
E0=0.4
2.06
E0=0.5
2.04
E0=0.6
2.02
2
1.98
1.96
1.94
1.92
1.9
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 5b. When  1  0, 2  0.3 , Variation of the predator population with time for
different values of harvesting effort and other parameter values are same as data (a).
2.5
E0=0.4
Prey,Predator
2
1.5
Prey
Predator
1
0.5
0
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 6a.
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171
2.5
E0=0.46
Prey,Predator
2
1.5
Prey
Predator
1
0.5
0
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 6b.
140
Prey
Predator
120
Prey,Predator
100
E0=0.5
80
60
40
20
0
0
50
100
150
200
250
300
350
400
450
500
Time
Figure 6c.
Figure (6a-6c). When  1  0, 2  0.2 .
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From figure (4a), figure (4b) and figure (6a-6c) we can easily see that when
the system is unstable that is. when  2  0.2 , the range of Harvesting is (0.4,0.5) . Here
we can note that when E 0  0.4 the prey population becomes extinct and however for
E 0  0.5 the predator population becomes extinct. Both of these cases do not have
any biological meaning. Figure 5 is the plots of the prey population and predator
population against time for different value of E 0 at  2  0.3 . Figure (5a) and figure
(5b) show that the prey population increases with E 0 increases and predator
population decreases with E0 increases. We can also note that the prey and predator
population will not tend to extinction if 0.4  E0  0.7 . For good production it is
necessary E 0 lies between 0.4 and 0.7.
6.7 CONCLUSION
In this chapter, a nonlinear mathematical model is proposed and analyzed to
see the effect of harvesting on dynamics of prey- predator system with Holling type
IV functional response. We carried out simulations and found that when the system is
unstable i.e. when  2  0.2 , range of harvesting is (0.4,0.5) . We had taken ladybirds
as natural enemy (predator) and pests as prey. If the maturation delay of ladybirds
is  2  2.5 , the population of pests will increase and it will be harmful for crops. So the
maturation delay of
ladybirds should less than 2.5. We had taken harvesting of
ladybirds for controlling the population of them. If the population of ladybirds
increases out of limit, it will create problems for human beings. For good production
it is necessary E0 lies between 0.4 and 0.7. Hence, we conclude from our analysis
that the stability properties of the system could switch with the Harvesting with time
delay that is incorporated on different densities in the model.
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173