# Harvesting Model for Fishery Resource with Reserve Area and Bird Predator

Authors: Amit Sharma; Bhanu Gupta
Journal: Journal of Marine Biology
 (2014)

---

## Abstract

The aim of this paper is to study the dynamics of fishery resource with reserve area in the presence of bird predator. The aquatic region under investigation is divided into two zones: one free for fishing and another restricted for any kind of fishery. The criteria of biological and bionomic equilibrium of system are established. The points of local stability, global stability, and instability are obtained for the proposed model. An optimal harvesting policy is established using Pontryagin’s maximum principle. At last the theoretical results obtained are illustrated with the help of numerical simulation.


---

## Body

## 1. Introduction

Renewable resources like fishery, forestry, and oil exploration are important sources of food and materials which play an important role for survival and growth of biological population. Extensive and unregulated harvesting of marine fishes can lead to the depletion of several fish species. One potential solution of these problems is the creation of marine resources where fishing and other exploration activities are strictly prohibited. Sometimes, though marine reserves protect species inside the reserve area, they lead to increasing fish abundance in adjacent areas. So this aspect should also be considered to make effective use of reserves. Restrictions on gear and/or effort may also be considered as other ways to protect species from extinction.Economic and biological aspects of renewable resources management have been considered by Clark [1]. Chaudhuri [2] studied the problem of combined harvesting of two competing fish species. Ganguly and Chaudhuri [3] proposed a model to study the regulation of single species fishery by taxation. Mesterton-Gibbons [4] also described a technique to find an optimal harvesting policy for Lotka-Voltra ecosystem to two interdependent populations. Kar and Misra [5] explained the possibilities of the existence of bionomic equilibrium in prey-predator model. Kar and Chaudhuri [6] derived the condition for global stability of system using a Lyapunov function. Kar [7] discussed the optimal harvesting policy using Pontryagin’s maximal principle of prey-predator system. Dubey et al. [8] proposed a dynamic model for single species fishery which depends partially on a logistically growing resource. They showed that both the equilibrium density of the fish population and the maximum sustainable yield increase as the resource biomass density increases. Dubey [9] proposed and analyzed the dynamics of a prey-predator model. The role of reserved zone is investigated and it is shown that the reserve zone has a stabilizing effect on predator-prey interactions. Zhang et al. [10] discussed the dynamics of prey-predator fishery model with harvesting of both prey and predator. Kar and Pahar [11] studied the dynamical behavior and harvesting problem of a prey-predator fishery.From the literature discussed above and to the best of our knowledge, in the models considered by different authors [6, 7, 9–11] the predator is feeding in unreserved area only. In this paper, we have proposed a model in which predator birds are feeding in both reserved and unreserved areas. Moreover the predators are also being harvested from unreserved zone. The criteria of biological and bionomic equilibrium of system are established. The points of local stability, global stability, and instability are obtained for the proposed model. An optimal harvesting policy is also discussed using Pontryagin’s maximum principle. The paper is concluded with numerical simulation.


## 2. The Model

Consider a fishery resource system consisting of two zones: a free fishery and a reserve zone where fishing is not allowed. Each zone is supposed to be homogeneous. There is a bird predator feeding on both of them, that is, fishes of reserved as well as unreserved zones. It is assumed that the predator population is also harvested in unreserved zone. We suppose that the prey species migrate between the two zones randomly. The growth of prey in each zone in the absence of predator is assumed to be logistic. Keeping these in view, the model becomes(1)
d
x
d
t
=
r
x
(
1
-
x
K
)
-
σ
1
x
+
σ
2
y
-
m
1
x
z
-
q
1
E
1
x
,
d
y
d
t
=
s
y
(
1
-
y
L
)
+
σ
1
x
-
σ
2
y
-
m
2
y
z
,
d
z
d
t
=
-
d
z
+
k
1
m
1
x
z
+
k
2
m
2
y
z
-
q
2
E
2
z
.Herex
(
t
) and y
(
t
) are the respective biomass densities of the prey species inside the unreserved and reserved areas, respectively, at a time t
;
z
(
t
) is the biomass density of predator at time t; σ
1 and σ
2 are migration rates from the unreserved area to reserved area and the reserved area to the unreserved area, respectively; E
1 and E
2 are the efforts applied to harvest the fish population and predator in unreserved zone, respectively; r and s are intrinsic growth rates of prey species inside the unreserved and reserved zones, respectively; K and L are the carrying capacities of prey species in the unreserved and reserved zones, respectively; q
1 and q
2 are the catchability coefficient of prey and predator in unreserved zone, respectively; d is death rate of predator; m
1 and m
2 are the capturing rates and k
1 and k
2 are the conversion rates of prey in unreserved and reserved zones, respectively.All the parameters are assumed to be positive. Here we observe that if there is no migration of fish population from the reserved area to the unreserved area (i.e.,σ
2
=
0) and r
-
σ
1
-
q
1
E
1
<
0, then x
°
<
0. Similarly, if there is no migration of fish population from the unreserved area to reserved area (i.e., σ
1
=
0) and s
-
σ
2
<
0, then y
°
<
0. Hence, throughout our analysis, we assume that
(2)
r
-
σ
1
-
q
1
E
1
>
0
,
s
-
σ
2
>
0
.In order to simplify the model proposed in (1), we assume that the capturing rates are the same from both reserves, that is, m
1
=
m
2
=
m and conversion rates of prey in unreserved and reserved zones are the same, that is, k
1
=
k
2
=
k. After incorporating these assumptions, the system (1) becomes
(3)
d
x
d
t
=
r
x
(
1
-
x
K
)
-
σ
1
x
+
σ
2
y
-
m
x
z
-
q
1
E
1
x
,
d
y
d
t
=
s
y
(
1
-
y
L
)
+
σ
1
x
-
σ
2
y
-
m
y
z
,
d
z
d
t
=
-
d
z
+
α
(
x
z
+
y
z
)
-
q
2
E
2
z
,
where α
=
k
m.Lemma 1.
All the solutions of the system (3) which initiate in R
3
+ are uniformly bounded.Proof.
Letw
=
x
(
t
)
+
y
(
t
)
+
z
(
t
) and η
>
0 be a constant. Then
(4)
d
w
d
t
+
η
w
=
d
x
d
t
+
d
y
d
t
+
d
z
d
t
+
η
x
+
η
y
+
η
z
=
-
r
K
(
x
-
K
2
r
(
r
-
q
1
E
1
+
η
)
)
2
-
s
L
(
y
-
L
2
s
(
s
+
η
)
)
2
-
z
(
(
α
-
m
)
(
x
+
y
)
+
d
+
q
2
E
2
-
η
)
+
K
4
r
(
r
-
q
1
E
1
+
η
)
2
+
L
4
s
(
s
+
η
)
2
≤
K
4
r
(
r
-
q
1
E
1
+
η
)
2
+
L
4
s
(
s
+
η
)
2
=
μ
.
By the theory of differential inequality [12], we have
(5)
0
<
w
(
x
(
t
)
,
y
(
t
)
,
z
(
t
)
)
≤
μ
η
(
1
-
e
-
η
t
)
+
w
(
x
(
0
)
,
y
(
0
)
,
z
(
0
)
)
e
-
η
t
and for t
→
∞
,
0
<
w
≤
μ
/
η. This proves the lemma.


## 3. Existence of Equilibria

We find the steady states of (3) by equating the derivatives on the left hand sides to zero and solving the resulting algebraic equations. This gives three possible steady states, namely, P
1
(
0,0
,
0
)
,
P
2
(
x
-
,
y
-
,
0
), and P
3
(
x
*
,
y
*
,
z
*
).AtP
1
(
0,0
,
0
) the population is extinct and this equilibrium point always exists.Now, consider the equilibrium pointP
2
(
x
-
,
y
-
,
0
), where the predator is not present. Here x
- and y
- are the positive solutions of
(6)
r
x
-
(
1
-
x
-
K
)
-
σ
1
x
-
+
σ
2
y
-
-
q
1
E
1
x
-
=
0
,
s
y
-
(
1
-
y
-
L
)
+
σ
1
x
-
-
σ
2
y
-
=
0
.
This system (6) is already solved by Dubey et. al. [8] and local and global stability results for the system at P
2
(
x
-
,
y
-
,
0
) are discussed there.Now assume here that the interior equilibrium pointP
3
(
x
*
,
y
*
,
z
*
) exists and is a solution of
(7)
r
x
*
(
1
-
x
*
K
)
-
σ
1
x
*
+
σ
2
y
*
-
m
x
*
z
*
-
q
1
E
1
x
*
=
0
,
s
y
*
(
1
-
y
*
L
)
+
σ
1
x
*
-
σ
2
y
*
-
m
y
*
z
*
=
0
,
-
d
z
*
+
α
(
x
*
z
*
+
y
*
z
*
)
-
q
2
E
2
z
*
=
0
.
Next, we discuss the local and global stability results at these equilibrium points.


## 4. Stability Analysis

The variational matrix of system (3) is(8)
V
(
x
,
y
,
z
)
=
[
r
-
2
r
K
x
-
σ
1
-
m
z
-
q
1
E
1
σ
2
-
m
x
σ
1
s
-
2
s
L
y
-
σ
2
-
m
z
-
m
y
α
z
α
z
-
d
+
α
(
x
+
y
)
-
q
2
E
2
]
.At P
2
(
x
-
,
y
-
,
0
) the characteristic equation is λ
2
+
a
1
λ
+
a
2
=
0, where
(9)
a
1
=
r
K
x
-
+
σ
2
y
-
x
-
+
s
L
y
-
+
σ
1
x
-
y
-
,
a
2
=
(
r
K
x
-
+
σ
2
y
-
x
-
)
(
s
L
y
-
+
σ
1
x
-
y
-
)
-
σ
1
σ
2
.
Since
(10)
λ
1
+
λ
2
=
-
(
r
K
x
-
+
σ
2
y
-
x
-
+
s
L
y
-
+
σ
1
x
-
y
-
)
<
0
,
λ
1
λ
2
=
(
(
r
K
x
-
+
σ
2
y
-
x
-
)
(
s
L
y
-
+
σ
1
x
-
y
-
)
-
σ
1
σ
2
)
>
0
.
Therefore, all eigenvalues are negative and hence P
2 is locally asymptotically stable. Let us now suppose that system (3) has a unique positive equilibrium P
3
(
x
*
,
y
*
,
z
*
). The variational matrix of (3) at P
3 is
(11)
V
(
x
,
y
,
z
)
=
[
-
r
K
x
*
-
σ
2
y
*
x
*
σ
2
-
m
x
*
σ
1
-
s
L
y
*
-
σ
1
x
*
y
*
-
m
y
*
α
z
*
α
z
*
0
]
.
The characteristic equation of variational matrix of system (3) at P
3 is given by λ
3
+
b
1
λ
2
+
b
2
λ
+
b
3
=
0, where
(12)
b
1
=
r
K
x
*
+
σ
1
x
*
y
*
+
s
L
y
*
+
σ
2
y
*
x
*
,
b
2
=
r
σ
1
K
y
*
x
*
2
+
r
s
x
*
y
*
K
L
+
α
m
(
x
*
z
*
+
y
*
z
*
)
+
s
σ
2
y
*
2
L
x
*
,
b
3
=
α
m
s
x
*
y
*
z
*
L
+
α
m
r
x
*
y
*
z
*
K
+
α
m
σ
1
x
*
2
z
*
y
*
+
α
m
σ
2
y
*
2
z
*
x
*
+
α
m
σ
1
x
*
z
*
+
α
m
σ
2
y
*
z
*
.According to Routh-Hurwitz criteria, the necessary and sufficient conditions for local stability of equilibrium pointP
3 are b
1
>
0, b
3
>
0, and b
1
b
2
-
b
3
>
0.It is evident thatb
1
>
0 and b
3
>
0. Thus the stability of P
3 is determined by the sign of b
1
b
2
-
b
3. By direct calculation, we obtain
(13)
b
1
b
2
-
b
3
=
(
r
K
x
*
+
σ
1
x
*
y
*
+
s
L
y
*
+
σ
2
y
*
x
*
)
×
(
r
σ
1
K
x
*
2
y
*
+
r
s
x
*
y
*
K
L
+
s
σ
2
L
y
*
2
x
*
)
+
α
m
s
L
z
*
y
*
2
+
α
m
r
K
z
*
x
*
2
>
0
and hence P
3
(
x
*
,
y
*
,
z
*
) is locally asymptotically stable.Now we will discuss the global stability of the endemic equilibrium pointP
3
(
x
*
,
y
*
,
z
*
) of the system (3).Theorem 2.
The equilibrium pointP
3
(
x
*
,
y
*
,
z
*
) of system (3) is globally asymptotically stable if (
1
-
y
*
σ
2
/
x
*
σ
1
)
(
y
-
y
*
)
(
z
-
z
*
)
<
0.Proof.
Let us consider the following Lyapunov function:(14)
V
(
x
,
y
,
z
)
=
(
x
-
x
*
-
x
*
ln
⁡
x
x
*
)
+
l
1
(
y
-
y
*
-
y
*
ln
⁡
y
y
*
)
+
l
2
(
z
-
z
*
-
z
*
ln
⁡
z
z
*
)
,
where l
1 and l
2 are positive constants to be chosen later on.
DifferentiatingV
(
x
,
y
,
z
) with respect to time t, we get
(15)
d
V
d
t
=
(
x
-
x
*
)
x
d
x
d
t
+
l
1
(
y
-
y
*
)
y
d
y
d
t
+
l
2
(
z
-
z
*
)
z
d
z
d
t
.
Choosing l
1
=
(
y
*
/
x
*
)
(
σ
2
/
σ
1
) and l
2
=
m
/
α, a little algebraic manipulation yields
(16)
d
V
d
t
=
-
r
K
(
x
-
x
*
)
2
-
s
L
y
*
σ
2
x
*
σ
1
(
y
-
y
*
)
2
-
σ
2
x
y
x
*
(
y
x
*
-
x
y
*
)
2
+
m
(
1
-
y
*
σ
2
x
*
σ
1
)
(
y
-
y
*
)
(
z
-
z
*
)
<
0
.
Clearly d
V
/
d
t
<
0 if and only if (
1
-
y
*
σ
2
/
x
*
σ
1
)
(
y
-
y
*
)
(
z
-
z
*
)
<
0.
Therefore,P
3
(
x
*
,
y
*
,
z
*
) is globally asymptotically stable provided (
1
-
y
*
σ
2/x
*
σ
1
)
(
y
-
y
*
)
(
z
-
z
*
)
<
0.


## 5. Bionomic Equilibrium

In this section, we study the bionomical equilibrium of the model system (3). Let c
1 be the fishing cost per unit effort for prey species, let c
2 be the harvesting cost per unit effort for predator species, let p
1 be the price per unit biomass of the prey, and let p
2 be the price per unit biomass of the predator.Therefore, the net economic revenue at any timet is given by
(17)
π
=
(
p
1
q
1
x
-
c
1
)
E
1
+
(
p
2
q
2
z
-
c
2
)
E
2
=
π
1
+
π
2
,
where π
1
=
(
p
1
q
1
x
-
c
1
)
E
1 and π
2
=
(
p
2
q
2
z
-
c
2
)
E
2; that is, π
1 and π
2 represent the net revenues for the prey and predator species, respectively.The bionomical equilibrium(
x
∞
,
y
∞
,
z
∞
,
E
1
∞
,
E
2
∞
) is given by the following simultaneous equations:
(18)
r
x
(
1
-
x
K
)
-
σ
1
x
+
σ
2
y
-
m
x
z
-
q
1
E
1
x
=
0
,
(19)
s
y
(
1
-
y
L
)
+
σ
1
x
-
σ
2
y
-
m
y
z
=
0
,
(20)
-
d
z
+
α
(
x
z
+
y
z
)
-
q
2
E
2
z
=
0
,
(21)
(
p
1
q
1
x
-
c
1
)
E
1
+
(
p
2
q
2
z
-
c
2
)
E
2
=
0
.
In order to determine the bionomic equilibrium, we now consider the following cases.Case 1.
Ifc
2
>
p
2
q
2
z and c
1
<
p
1
q
1
x, then the cost of harvesting is greater than the revenue for the predator and cost of harvesting of prey is less than revenue. Here the harvesting of predator will be stopped, that is, (
E
2
=
0
), and only the prey fishing remains operational.
We then havex
∞
=
c
1
/
p
1
q
1. Substituting x
∞ into (20), we get y
∞
=
(
1
/
α
)
(
d
-
α
c
1
/
p
1
q
1
). Now substituting x
∞ and y
∞ into (19), we get z
∞
=
(
1
/
m
)
(
s
-
(
s
/
L
α
)
(
d
-
α
c
1
/
p
1
q
1
)
+
σ
1
c
1
α
/
(
d
p
1
q
1
-
α
c
1
)
-
σ
2
).
Now, substitutingx
∞, y
∞, and z
∞ into (18), we get
(22)
E
1
∞
=
1
m
(
s
-
s
L
α
(
d
-
α
c
1
p
1
q
1
)
+
σ
1
c
1
α
d
p
1
q
1
-
α
c
1
-
σ
2
)
.Case 2.
Ifc
1
>
p
1
q
1
x and c
2
<
p
2
q
2
z, then the cost of fishing is more than the revenue and cost of harvesting of predator is less than revenue. Here harvesting of prey will be closed (i.e., E
1
=
0) and only predator harvesting will remain operational.
We then havez
∞
=
c
2
/
p
2
q
2. Substituting z
∞ into (17) we get y
∞
=
(
x
∞
/
σ
2
)
(
-
r
(
1
-
x
∞
/
K
)
+
σ
1
+
m
c
2
/
p
2
q
2
).
Now, substitutingz
∞ into (18), we get x
∞
=
(
y
∞
/
σ
1
)
(
-
s
(
1
-
y
∞
/
L
)
+
σ
2
+
m
c
2
/
p
2
q
2
).
From these two equationsx
∞ and y
∞ can be found. Substituting x
∞ and y
∞ into (20), we get
(23)
E
2
∞
=
-
d
q
2
+
α
q
2
(
x
∞
+
y
∞
)
,
whereE
2
∞
>
0, provided α
(
x
∞
+
y
∞
)
>
d.Case 3.
Ifc
1
>
p
1
q
1
x and c
2
>
p
2
q
2
z, then the cost is greater than revenues for both the species and so the whole fishing and harvesting of predator will be closed.Case 4.
Ifc
1
<
p
1
q
1
x and c
2
<
p
2
q
2
z, then the revenues for both the species are being positive, and then the whole fishing and harvesting of predator will be in operation.
In this case,x
∞
=
c
1
/
p
1
q
1 and z
∞
=
c
2
/
p
2
q
2.
Now substitutingx
∞ and y
∞ into (18), (19), and (20), we get
(24)
y
∞
=
L
2
s
(
(
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
2
+
4
s
L
σ
1
c
1
p
1
q
1
)
1
/
2
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
s
s
k
s
s
s
s
+
(
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
2
+
4
s
L
σ
1
c
1
p
1
q
1
)
1
/
2
)
,
E
1
∞
=
1
q
1
(
r
+
σ
2
y
∞
x
∞
-
(
r
K
x
∞
+
σ
1
+
m
z
∞
)
)
,
E
2
∞
=
1
q
2
(
-
d
+
α
(
c
1
p
1
q
1
+
y
∞
)
)
.
Now,
(25)
E
1
∞
>
0
,
if
r
+
σ
2
y
∞
x
∞
>
(
r
K
x
∞
+
σ
1
+
m
z
∞
)
,
E
2
∞
>
0
,
if
α
(
c
1
p
1
q
1
+
y
∞
)
>
d
.
Thus the nontrivial bionomic equilibrium point (
x
∞
,
y
∞
,
z
∞
,
E
1
∞
,
E
2
∞
) exists if conditions (25) hold.


## 6. Optimal Harvesting Policy

In this section, the optimal management of a fishery resource in the presence of predator is discussed. Here, our objective is to maximize the present valueJ of a continuous time stream of revenues given by
(26)
J
=
∫
0
∞
e
-
δ
t
(
(
p
1
q
1
x
-
c
1
)
E
1
(
t
)
s
s
s
s
s
s
s
s
s
s
s
s
s
+
(
p
2
q
2
z
-
c
2
)
E
2
(
t
)
)
d
t
,
where δ denotes the instantaneous annual rate of discount. We intend to maximize (26) subject to the state equations (3) by invoking Pontryagin’s maximal principle (Clark [1]). The control variable E
i
(
t
)
(
i
=
1,2
) is subjected to the constraints 0
≤
E
i
(
t
)
≤
(
E
i
)
max
⁡.The Hamiltonian function for the problem is given by(27)
H
=
e
-
δ
t
[
(
(
p
1
q
1
x
-
c
1
)
E
1
(
t
)
+
(
p
2
q
2
z
-
c
2
)
E
2
(
t
)
)
]
+
λ
1
[
r
x
(
1
-
x
K
)
-
σ
1
x
+
σ
2
y
-
m
1
x
z
-
q
1
E
1
x
]
+
λ
2
[
s
y
(
1
-
y
L
)
+
σ
1
x
-
σ
2
y
-
m
y
z
]
+
λ
3
[
-
d
z
+
α
(
x
z
+
y
z
)
-
q
2
E
2
z
]
,
where λ
1
,
λ
2, and λ
3 are the adjoint variables.The control variablesE
1 and E
2 appear linearly in the Hamiltonian function H.Assuming that the control constraints are not binding, that is, the optimal solution does not occur at(
E
i
)
max
⁡, we have singular control.According to Pontryagin’s maximum principle(28)
∂
H
∂
E
1
=
0
;
∂
H
∂
E
2
=
0
;
∂
λ
1
∂
t
=
-
∂
H
∂
x
;
∂
λ
2
∂
t
=
-
∂
H
∂
y
;
∂
λ
3
∂
t
=
-
∂
H
∂
z
.
Substitution and simplification yield
(29)
∂
H
∂
E
1
=
0
⟹
λ
1
=
e
-
δ
t
(
p
1
-
c
1
q
1
x
)
,
(30)
∂
H
∂
E
2
=
0
⟹
λ
3
=
e
-
δ
t
(
p
2
-
c
2
q
2
z
)
,
(31)
∂
λ
1
∂
t
=
-
[
e
-
δ
t
p
1
q
1
E
1
+
λ
1
(
r
-
2
r
x
K
-
σ
1
-
m
z
-
q
1
E
1
)
s
s
s
s
s
x
K
+
λ
2
σ
1
+
λ
3
α
z
]
,
(32)
∂
λ
2
∂
t
=
-
[
λ
1
σ
2
+
λ
2
(
s
-
2
s
y
L
-
σ
2
-
m
z
)
+
λ
3
α
z
]
,
(33)
∂
λ
3
∂
t
=
-
[
e
-
δ
t
p
2
q
2
E
2
-
λ
1
m
x
-
λ
2
m
y
s
s
s
s
s
e
-
δ
t
+
λ
3
(
-
d
+
α
(
x
+
y
)
-
q
2
E
2
)
]
.
Now, substituting λ
1 and λ
3 into (33) and using equilibrium equations, we get
(34)
z
*
=
(
δ
c
2
q
1
)
×
(
q
1
q
2
p
2
(
δ
-
q
2
E
2
)
+
q
2
m
(
p
1
q
1
x
*
-
c
1
)
e
δ
t
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
+
e
δ
t
q
1
q
2
y
*
λ
2
m
)
-
1
.
From (32), we get ∂
λ
2
/
∂
t
-
A
1
λ
2
=
-
A
2
e
-
δ
t, whose solution is given by
(35)
λ
2
(
t
)
=
A
2
A
1
+
δ
e
-
δ
t
,
where A
1
=
(
s
/
L
)
y
*
+
σ
1
(
x
*
/
y
*
) and A
2
=
(
p
1
-
c
1
/
q
1
x
*
)
σ
2
+
(
p
2
z
*
-
c
2
/
q
2
)
α.From (31), we get ∂
λ
1
/
∂
t
-
A
3
λ
1
=
-
A
4
e
-
δ
t, whose solution is given by
(36)
λ
1
(
t
)
=
A
4
A
3
+
δ
e
-
δ
t
,
where A
3
=
(
r
/
K
)
x
*
-
σ
2
(
y
*
/
x
*
)
,
A
4
=
p
1
q
1
E
1
+
(
A
2
σ
2
)
/
(
A
1
+
δ
)
+
(
p
2
-
c
2
/
q
2
y
*
)
α
z
*.From (29) and (36), we get the singular path
(37)
(
p
1
-
c
1
q
1
x
*
)
=
A
4
A
3
+
δ
e
-
δ
t
.
Using
(38)
y
*
=
L
2
s
(
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
(
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
2
+
4
s
L
σ
1
c
1
p
1
q
1
)
1
/
2
s
s
s
s
k
s
s
s
+
(
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
2
+
4
s
L
σ
1
c
1
p
1
q
1
)
1
/
2
)
,
(39)
z
*

=

δ
c
2
q
1
q
1
q
2
p
2
(
δ
-
q
2
E
2
)

+

q
2
m
(
p
1
q
1
x
*
-
c
1
)

+

e
δ
t
q
1
q
2
y
*
λ
2
m
,
A
1, A
3, and A
4 can be written as
(40)
A
1
=
s
L
y
*
+
σ
1
x
*
y
*
,
A
3
=
r
K
x
*
-
σ
2
y
*
x
*
,
A
4
=
p
1
q
1
E
1
+
A
2
σ
2
A
1
+
δ
+
(
p
2
-
c
2
q
2
y
*
)
α
z
*
.
Thus (37) can be written as
(41)
F
(
x
*
)
=
(
p
1
-
c
1
q
1
x
*
)
-
A
4
A
3
+
δ
=
0
.
There exists a unique positive root x
*
=
x
δ of F
(
x
*
)
=
0 in the interval 0
<
x
*
<
K, if the following inequalities hold:
(42)
F
(
0
)
<
0
,
F
(
K
)
>
0
,
F
′
(
x
*
)
>
0
for
x
*
>
0
.
For x
*
=
x
δ, we get z
*
=
z
δ from (34).We then have(43)
y
δ
=
L
2
s
(
(
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
2
+
4
s
L
σ
1
c
1
p
1
q
1
)
1
/
2
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
s
s
k
s
s
s
s
s
+
(
(
s
-
σ
2
-
m
c
1
p
1
q
1
)
2
+
4
s
L
σ
1
c
1
p
1
q
1
)
1
/
2
)
,
E
1
δ
=
1
q
1
(
r
+
σ
2
y
δ
x
δ
-
(
r
K
x
δ
+
σ
1
+
m
z
δ
)
)
,
E
2
δ
=
1
q
2
(
-
d
+
α
(
c
1
p
1
q
1
+
y
δ
)
)
.
Hence once the optimal equilibrium (
x
δ
,
y
δ
,
z
δ
) is determined, the optimal harvesting efforts E
1
δ and E
2
δ can be determined.From (30), (35), and (36), we observe that λ
i
(
t
)
(
i
=
1,2
,
3
) do not vary with time in optimal equilibrium. Hence they remain bounded as t
→
∞.


## 7. Simulation

In order to investigate the dynamics of the system (3) with help of computer simulation, we choose the following set of values of parameters:
(44)
r
=
2.1
,
s
=
1.7
,
K
=
100
,
L
=
100
,
σ
1
=
0.7
,
σ
2
=
0.1
,
q
1
=
0.02
,
q
2
=
0.01
,
E
1
=
50
,
E
1
=
20
,
m
=
0.02
,
α
=
0.003
,
d
=
0.0
2
in appropriate units with initial conditions [
30
30
30
].From Figure1, it is clear that the biomass density of prey species in unreserved area increases with respect to time and then decreases slightly and settles down at its equilibrium level.Figure 1
Plot of prey populationx in unreserved area versus t.From Figure2, it is clear that biomass density of prey population in reserved zone increases sharply near to its carrying capacity and then settles down at its equilibrium level near the carrying capacity of this zone.Figure 2
Plot of prey population in reserved areay versus time t.Figure3 shows that biomass density of predator increases approximately linearly with respect to time and tries to settle at their equilibrium level. Figures 1, 2, and 3 also show that the endemic equilibrium point P
3
(
x
*
,
y
*
,
z
*
) is locally asymptotically stable for the assumed values of parameters.Figure 3
Plot of predator populationz versus time t.From Figure4, we may conclude that the steady state P
3
(
x
*
,
y
*
,
z
*
) is globally asymptomatically stable. All the solutions with different initial conditions and set of parameters in (44) satisfying the conditions of Theorem 2 converge to the same equilibrium point. Hence the theory established earlier is verified.Figure 4
Plot of global stability ofP
3
(
x
*
,
y
*
,
z
*
).



---

*Source: 218451-2014-09-22.xml*