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Chapter 8
INFLUENCE OF PREY RESERVE IN TWO PREYS AND ONE
PREDATOR SYSTEM
INTRODUCTION
Dynamics of interacting species is being studied since ages from various
prospective. It is well known that many species have already become extinct and
many others are at the verge of extinction due to several natural or manmade reasons
like over exploitation, indiscriminate harvesting, over predation, environmental
pollution, loss of habitat and mismanagement of natural resources etc. To save the
species from getting extinct we are taking measures like improving conditions of their
natural habitat, reduce the interaction of the species with external agents which tend to
decrease their numbers, impose restrictions on species harvesting, create natural
reserves, establish protected areas etc. so that the species grow in these protected
areas without any external disturbances and hence the protected population can
improve their numbers. While creating protected areas for a species, several factors
have to be taken into consideration like its capacity to house number of individuals of
the species to be protected, dynamics of the ecosystem supporting these species, the
number of the other important beings which depend on these protected species,
economics associated with the maintenance of the protected area, bio-economics of
the ecosystem and several others.
However, study of population dynamics in presence of a refuge/reserve for
one of the species is not new. In most of the works done in this area, the environment
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196
is assumed to be patchy, usually two patches, where one patch is assumed to be a
source and the other is assumed to be a sink which functions as a refuge for the prey
population and the source sink dynamics is employed to understand the dynamics of
these ecosystems. Dubey et al., 2002 analyzed a dynamic model for a single species
fishery which depends partially on a logistically growing resource in a two patch
environment. They showed that both the equilibrium density of the fish population as
well as the maximum sustainability yield increases as the resource biomass density
increases. Many works and research have been done related to stability and other
dynamical behaviour of the most fundamental Lotka- Volterra model, some of them
can be found in Freedman, 1980; Abbas et al., 2010; Strogatz, 2008. The main articles
and research related to teaming approach can be found in the Amy Sarver, 2006
Martin and Mitani, 2005. Our model is motivated by the two prey one predator model
given by M. F. Elettreby, 2009 in which he has taken the unity predation rates for
both the prey population, and Tripathi et al., 2012 in which they are considering two
important factors in their model:
(i.) different predation rates  1 and  2 for both the prey teams,
(ii.) the prey teams may interact in the absence of predator.
In this chapter, we have formulated and analyzed two preys one predator
model with one prey dispersal in a two patch environment. Each patch is supposed to
be homogeneous. Patch two constitutes a reserved area of prey and no predation is
permitted in this zone whereas patch one is an open access predation zone. The
growth of prey in each patch is assumed to be logistic. Biological equilibria of the
system are obtained and criteria for local stability and global stability of the system
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197
derived. We have also studied the effect of team approach on this model. Criteria for
local stability, instability and global stability of the non-negative equilibria are
obtained. Using differential inequality, we have obtained sufficient conditions that
ensure the persistence of the system. The key results developed in this article have
been illustrated using numerical simulation. These results can be interpreted in
different contexts like resource conservation, pest management, bio-economics of a
renewable resource etc.
8.1 MATHEMATICAL MODEL
In this Section, we study the model where one predator team interacts with the
two teams of preys. An interesting example of our study may be thought as the herds
of gazelles (prey) and zebras (prey) living in the same habitat and attacked by one
type of predator (lion). Such system can be described by the following set of
nonlinear differential equations
 x  αxz
dx1
 r1 x1 1  1   1 1  σ 1 x1  σ 2 x 2   x1 y  x1 yz,
dt
 k1  1  x1
x1 0  0,
 x 
dx2
 r2 x2 1  2   σ1 x1  σ 2 x2 ,
dt
 k2 
x2 0  0,

dy
y  β yz
 r3 y1    1
  x1 y  x1 yz,
dt
 k3  1  y
y0  0,
dz
α x z β yz
 cz 2  2 1  2 ,
dt
1  x1 1  y
z 0  0.
(8.1.1)
Here, x1 and y are biomass densities of prey species respectively interacting
with one team of predator with density z inside the unreserved area which is an open
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access predating zone at time t . x2 is the biomass density of prey species inside the
reserved area where no predation is permitted at this time t . All the parameters are
assumed to be positive. It is clear that the two teams of preys help each other e.g. in
foraging and in early warning against predation. Note that this help occurs only in the
presence of predator. This is presented by the term
x1 yz in the preys
equations. r1 , r2 and r3 are the intrinsic growth rates of prey species inside the
unreserved, reserved area and another competing prey respectively. k i i  1,2,3 are
the carrying capacities of prey species in the unreserved, reserved and another
competing prey species respectively.  1 and  2 are migration rates from the
unreserved area to the reserved area and the reserved area to the unreserved area,
respectively.  1 and  1 are the capturing rate of predators respectively.  2 and  2 are
the conversion rate of predators respectively. c is the death rate of predator and  is
the competing rate between preys in unreserved area in the absence of predator. In the
model (8.1.1),we suppose that the area under consideration is inhabited by two preys
(gazelles, zebras) and one common predator (lion). The area under consideration is
supposed to split into two subareas a reserve area where predation (for prey (gazelles)
species) is prohibited and an open area for predation. In the absence of any predation,
 x 
each team of preys grows logistically; this is r1 x1 1  1  and r3 y1  y  .The effect

k 3 
 k1 

of the predation is to reduce the prey growth rate by a term proportional to the prey
and predator populations; this is the 
β yz
α1 x1 z
and  1
terms.The teams of preys
1 y
1  x1
help each other against the predator, that is a x1 yz term exist. In the absence of any
prey for sustenance, the predator’s death rate results in inverse decay, that is the
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term  cz 2 .The preys’contribution to the predator growth rate are
α 2 x1 z β 2 yz
; that is
,
1  x1 1  y
proportional to the available prey as well as the size of the predator population.
8.2 EQUILIBRIUM ANALYSIS
Equating the derivatives on the left hand sides to zero and solving the resulting
algebraic equations. We can find seven possible equilibria
E0 0,0,0,0, E1 (0,0, k 3 ,0) , E2 xˆ1 , xˆ 2 ,0,0 , E3 0,0, y, z  , E4 x1 x2 , y,0 , E5 x1, x2 ,0, z  ,
and E6 ~
x1 , ~
x2 , ~
y, ~
z  . The equilibrium points E0 0,0,0,0 ,and E1 (0,0, k 3 ,0) obviously
exist. We show the existence of other equilibria as follows:
Existence of E2 xˆ1 , xˆ 2 ,0,0
Here xˆ1 , xˆ 2 are the positive solutions of the following algebraic equations
 xˆ 
r1 xˆ1 1  1    1 xˆ1   2 xˆ 2  0,
 k1 
(8.2.1)
 xˆ 
r2 xˆ 2 1  2    1 xˆ1   2 xˆ 2  0.
 k2 
(8.2.2)
From equation (8.2.1), we get
xˆ 2 
xˆ1 
r1 xˆ1 
 1  r1 
.
2 
k1 
Putting the value of x̂ 2 in equation (8.2.2), we get
A1 xˆ13  A2 xˆ12  A3 xˆ1  A4  0 ,
(8.2.3)
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where
r12 r2
2r r r   
r r   r r  2 
, A2   1 2 1 2 1 , A3  2 1 2 1  1 2
,
2
2
k 1 k 2 2
k 1 k 2 2
k 2 2
k1 2
r r   2 

A4  1 2
 r2 1 .
2
2
2
A1 
Equation (8.2.3) has unique positive solution x1  x̂1 , if the following inequalities hold
r2 r 1  1 
r r  2 
, r1 r2   2   r2 1 , r1   1 , r2   2 .
 1 2
k 2 2
k1
2
(8.2.4)
And for x̂ 2 to be positive, we must have
xˆ1 
r1   1 k1
r1
.
(8.2.5)
Hence the equilibrium E2 xˆ1 , xˆ 2 ,0,0 exists under the above conditions.
Existence of E3 0,0, y, z 
Here y, z are the positive solutions of the following algebraic equations

y  z
r3 1    1  0,
 k3  1  y
 cz 
2 y
1 y
 0.
(8.2.6)
(8.2.7)
From equation (8.2.7), we get
z
2 y
c1  y 
.
Putting the value of z in equation (8.2.6), we get
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B1 y 3  B2 y 2  B3 y  B4  0,
(8.2.8)
where
B1  r3 c, B2  cr3 k3  2, B3  r3c  1  2 k3  2cr3 k3 , B4  r3 k3 c.
Equation (8.2.8) has unique positive solution y  y , if the following inequalities hold
k 3  2, r3 c  1  2 k 3  2cr3 k 3 .
(8.2.9)
Hence the equilibrium E3 0,0, y, z  exists under the above conditions.
Existence of E4 x1 x2 , y,0
Here x1 , x2 and y  are the positive solutions of the system of algebraic equations
given below:
 x 
r1 x1 1  1    1 x1   2 x2   x1 y  0,
 k1 
(8.2.10)
 x 
r2 x2 1  2    1 x1   2 x2  0,
 k2 
(8.2.11)

y 
r3 1     x1  0.
 k3 
(8.2.12)
From equation (8.2.12), we get
y 
k 3 r3   x1 
.
r3
(8.2.13)
From equation (8.2.11), we get
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x1 
x2 
r x 
 2  r2  2 2  .
1 
k2 
(8.2.14)
Putting the value of x1 and y  in equation (8.2.10), we get
C1 x 2 3  C 2 x 2 2  C3 x 2  C 4  0,
(8.2.15)
where
r22  r1  2 k 3 
2 r2 r2   2   r1  2 k 3 


 
,


,
C


2
k 22 12  k1
r3 
 12 k 2  k1 r3 
2
   k3  r1 r2   2  .
r2  1   k 3  r1  r2   2   r1  2 k 3 
  
 C4   2  1
C3  

2
k 2 1
1
r3 
1
 k1
C1 
Let us consider a function of x2 ,
h( x 2 )  C1 x 2 3  C 2 x 2 2  C3 x 2  C 4 ,
and observe that
h(0)  C4  0.
h(k 2 )  C1 k 23  C 2 k 22  C3 k 2  C 4  0.
We note that h(0)  0 and h(k 2 )  0, showing the existence of x 2 in the interval
0  x2  k 2 . Now, the sufficient condition for x 2 to be unique positive real is
h( x2 )  0 , where
h( x2 )  3C1 x2 2  2C2 x2  C3 x2  0.
(8.2.16)
For x1 to be positive, we must have
x2 
r2   2 k 2
r2
.
(8.2.17)
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And for y  to be positive, we must have
x1 
r3
(8.2.18)
.

Hence the equilibrium E4 x1 x2 , y,0 exists under the above conditions.
Existence of E5 x1, x2 ,0, z 
Here x1, x2 and z  are the positive solutions of the system of algebraic equations
given below:
 x   xz 
r1 x11  1   1 1   1 x1   2 x2  0,
 k1  1  x1
(8.2.19)
 x 
r2 x2 1  2    1 x1   2 x2  0,
 k2 
(8.2.20)
 cz  
 2 x1
1  x1
 0.
(8.2.21)
From equations (8.2.21), we get
z  
 2 x1
c1  x1
.
Putting the value of z  in equation (8.2.19), we get
x 2 
 x  
x1 
 1 2 x1
 r1 1  1 .
 1 
2
2 
c1  x1
 k 1 
Putting the value of x 2 in equation (8.2.20), we get
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S1 x1 7  S 2 x1 6  S 3 x1 5  S 4 x1 4  S 5 x1 3  S 6 x1 2  S 7 x1  S 8  0 ,
(8.2.22)
where
S1  T10 , S 2  T1  T11  2T10 , S 3  T2  2T1  T10  T12  2T11  T16 ,
S 4  T1  T3  2T2  T6  T11  T13  2T12  T17  2T16 ,
S 5  T2  T4  2T3  T7  T12  T14  2T13  T16  T18  2T17 ,
S 6  T3  T5  2T4  T8  T13  T15  2T14  T17  r1  2T18 ,
S 7  T4  2T5  T9  T14  2T15  T18  2r1 , S 8  T5  T15  r1 , T1 
r1r2 1
,
k1k 2 22

2r 
r 
2 1

r1
2r 
 1  r1  1 , T3   2 1 1 
 1 2 
 1 ,
k1 
 2  k 2 2 k 2 2 c k 2 2 k1 k 2 2 

r 

r 
r
r 
r 
T4   2 1  2  1  1 , T5   2 1 , T6  2 1 2 2  1  1 ,
2 
k 2 2 k 2 2 
2
k1 
k 2 2 c 
T2 
r2 1
k 2 22
T7  
r2 1 2
 2c

2 1
r
2r1
1 
 1 
 k 2  2 k 2  2 k 1 k 2 2

,




r1
2r1 
 2  1  1 2 
,

k 2 2 k 2 2 c k1 k 2 2 k 2 2 

 r 
r 
r 2r
r 2r
rr
r 2r
2r 2 r
T9   1 2 2 1  1 , T10  2 1 2 2 , T11   1 2 2  1 2 12  1 2 2  2 1 2 2 ,
 2 c  k 2 2 
k1 k 2 2
k1k 2 2 k1k 2 2 k1k 2 2 k1 k 2 2
T8  
r2 1 2
 2c
r1 r2 1 r12 r2
2r12 r2
r1r2
2r1r2 1 r1r2 1 2
2r12 r2
r12 r2
T12  







,
k 2 22 k 2 22 k1k 2 22 k1 2 k1k 2 22 k1k 2 c 22 k1k 2 22 k12 k 2 22
2r1 r2 1 r1r2 1 2 2r12 r2
r12 r2
2r1 r2
r1 r2 1
r12 r2
T13 







,
2
k 2 22
k 2 c 22
k 2 22 k1 k 2 22 k1 2 k1k 2 22 k1k 2 22
r1 r2
T14 
2r1 r2
2
T17  r1 
r1 r2 1 r12 r2
rr
rr
r


 1 2 , T15  1 2 , T16  1 ,
2
2
2
k1
k 2 2 k 2 2 k1 2
2r1

r
, T18  1 2  1  2r1 .
k1
c
k1
Let us consider a function of x1,
F ( x1)  S1 x1 7  S 2 x1 6  S 3 x1 5  S 4 x1 4  S 5 x1 3  S 6 x1 2  S 7 x1  S 8 ,
and observe that
F (0)  S 8  0,
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F k1   S1 k17  S 2 k16  S 3 k15  S 4 k14  S 5 k13  S 6 k12  S 7 k1  S 8  0.
(8.2.23)
We note that F (0)  0 and F k1   0, showing the existence of x1 in the interval
0  x1  k1. Now, the sufficient condition for x1 to be unique is F ( x1)  0 at x1 ,
where
F x1  7 S1 x1 6  6S 2 x1 5  5S 3 x1 4  4S 4 x1 3  3S 5 x1 2  2S 6 x1  S 7 .
(8.2.24)
Hence the equilibrium E5 x1, x2 ,0, z  exists under the above conditions.
Existence of E6 ~
x1 , ~
x2 , ~
y, ~
z
Here ~
x1 , ~
x2 , ~
y and ~z are the positive solutions of the system of algebraic equations
given below:
 ~
x  ~
x1 ~
z
r1 ~
x1 1  1   1 ~
  1~
x1   2 ~
x2   ~
x1 ~
y~
x1 ~
y~
z  0,
k
1

x
1 
1

(8.2.25)
 ~
x 
r2 ~
x2 1  2    1 ~
x1   2 ~
x2  0,
k
2 

(8.2.26)
~

z
y ~
r3 1    1 ~   ~
x1  ~
x1 ~
z  0,
k
1

y
3 

(8.2.27)
 ~x
 ~y
 c~
z  2 ~1  2 ~  0.
1  x1 1  y
(8.2.28)
From equation (8.2.28), we get
x
 ~y 
1  ~
~
z   2 ~1  2 ~  .
c  1  x1 1  y 
(8.2.29)
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206
Putting the value of ~z in equation (8.2.25), we get
~
  2~
x1   ~
x1 
x1
 2 ~y 
1  1
~
~
~





 .
x2 
 1  y  
 y

 r1 1 
~
~
 2   k1 
c 1 ~
x1
1

x
1

y
1


(8.2.30)
Putting the value of ~z in equation (8.2.27) and the value of ~
x2 in equation (8.2.26),
we get
~

x1  2 ~
y ~
 
y
1  2~
~
~
~



  x1  1~  ,
f x1 , y   r3 1     x1   ~ 
~
c  1  x1 1  y  
1 y 
 k3 
r

r~
x
g ~
x1 , ~
y    2  1 J 1   1  2 12 J 12 ,
k 2 2
2

(8.2.31)
(8.2.32)
where
 ~
  ~
x 
x
 ~y 
1 
J 1   1  r1 1  1    ~
y   1~  ~
y   2 ~1  2 ~  .
c  1  x1
 k1 
  1  x1 1  y 
From (8.2.31), we note the following: when ~
y~
y a where
x1  0 then ~
 2 

r3 c ~ 3  2r3 c
rc
y a  
 r3 c  ~
y a   1  2  3  2r3 c  ~
y a  r3 c  0.
k3
k3
 k3



(8.2.33)
The equation (8.2.33) has positive solution if
k 3  2, 1  2 k 3  r3 c  2r3 ck 3 .

 2 1
f
  
2
~
x1
c 1  ~
y 1  ~
x1 

 2 ~
y 2  ~
x1  2 ~
x1


,
2
~
~
 1 y


c
1

x
1

(8.2.34)
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207
r
1
f
 3 
2
~
y
k 3 c 1  ~
y
 2~
x
 ~y 
2
 ~1  2 ~  
~2
 1  x1 1  y  c 1  y 
~
 
 x1  1~  .
1 y 

(8.2.35)
Now, from (8.2.34) and (8.2.35)
f
~
~
x1
dy


f
d~
x1
~
y

A1
.
B1

 2 1
Let A1   
2
c 1  ~
y 1  ~
x1 

r
1
B1   3 
2
k 3 c 1  ~
y
 2 ~
y 2  ~
x1  2 ~
x1

,

2
~
 1 ~
y
c 1  x1 

 2~
x
 ~y 
2
 ~1  2 ~  
~2
 1  x1 1  y  c 1  y 
~
 
 x1  1~  .
1 y 

d~
y
It is clear that ~  0 , if either
dx1
(i) A1  0 and B1  0 , or
(ii) A1  0 and B1  0
(8.2.36)
hold.
y~
yb where
From (8.2.32), we note the following: when ~
x1  0 then ~
r
2
  2  c   2 ~y b2  r2   2  c  cr1   1  2   r2 c 1 ~y b  cr2 1  r1 r2   2   0. (8.2.37)
The equation (8.2.37) has positive solution if
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208
 c   2 , r2   2 ,  c  1  2  cr1 , r2 1  r1 r2   2 
(8.2.38)
hold.
r2 J 12 2r2 ~
x1 J 1 J 2
g  r2   2 
 J 2  2 
 
,
~
x1   2 
 2 k2
 22 k 2
(8.2.39)
2r2 ~
x1 J 1 J 3
g  r2   2 



J

.
3
~


y   2 
 22 k 2
(8.2.40)
Where
J2 
r1
1

2
k1 c 1  ~
x1 
J3   
 2~
x
 ~y 
2
 ~1  2 ~  
~ 2
 1  x1 1  y  c 1  x1 
x
 ~y 
2
1  2~
 ~1  2 ~  
2
c  1  x1 1  y  c 1  ~
y
 1

 ~  ~
y  ,
 1  x1

 1

 ~  ~
y  .
 1  x1

Now, from (8.2.39) and (8.2.40)
g
~
~
x1
dy


g
d~
x1
~
y

A2
.
B2
 r  2 
r J 2 2r ~
xJJ
 J 2  22 1  2 21 1 2 ,
Let A2   2
 2 k2
 2 k2
 2 
 r2   2 
2r2 ~
x1 J 1 J 3
 J 3 
B2  
.
 22 k 2
 2 
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209
d~
y
It is clear that ~  0 , if either
dx1
(i) A1  0 and B1  0 , or
(ii) A1  0 and B1  0
(8.2.41)
hold.
From the above analysis we note that the two isoclines (8.2.31) and (8.2.32) interest at
a unique ~
x1 , ~
y  if in addition to conditions (8.2.36) and (8.2.41), the inequality
~
ya  ~
yb holds. Knowing the value of ~
x1 and ~y , the value of ~
x 2 and ~z can be
calculated from equations (8.2.30) and (8.2.29) respectively. This completes the
existence of E6 ~
x1 , ~
x2 , ~
y, ~
z .
8.3 LOCAL STABILITY
To determine the local stability of system (8.1.1), we compute the variational
matrix of system (8.1.1). The entries of general variational matrix are given by
differentiating the right side of system (8.1.1) with respect to x1 , x2 , y and z i.e.
 a11

a
V E    21
a
 31
a
 41
a12
a13
a 22
0
0
a33
0
a 43
a14 

0 
,
a34 

a 44 
where
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210
a11  r1 
2r1 x1
1 z

  1   y  y z , a12   2 , a13   x1  x1 z,
k1
(1  x1 )
a14  x1 y 
a33  r3 
a 43 
 1 x1
1  x1
, a21   1 , a 22  r2   2 
2r2 x 2
, a31   y  yz,
k2
2r3 y
2z
1 z
 y
,
  x1  x1 z 
, a34   1  x1 y, a 41 
2
1 y
k3
1  x1 2
1  y 
2 z
1  y 
2
, a44  2cz 
 2 x1
1  x1

2 y
1 y
.
The variational matrix V E 0  at equilibrium point E0
 r1   1

 1
V E 0   
0

 0

2
r2   2
0
0
r3
0
0
0
0

0
.
0

0 
Since one eigenvalue is zero and other are always positive. So equilibrium point E0 is
always an unstable point.
The variational matrix V E1  at equilibrium point E1
 r1   1   k 3

1


 k3
V E1   


0


2
r2   2
0
0
 r3
0
0
0


0 
k 
 1 3 .
1  k3 
 2 k3 
1  k 3 
0
The characteristic polynomial for the variational matrix V E1  is given by
4  B1* 3  B2* 2  B3*   B4*  0,
(8.3.1)
where
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211
B1*   k 3  r1  r2  r3 
k3  2
 1   2 ,
1  k3
B2*  r2   2 r1  r3   k 3   r3 r1   1    k 3 r3  r2 1 
k3  2
r1  r2  r3   k3   1   2  ,
1  k3
k3  2
r3 r2   2  r1   1   k 3    2 r1   k 3   r2  1   k 3  r1  r3  2 r1   k 3   r2  1   k 3  r1 ,
1 k3
 rk
B4*  2 3 3 r2  k 3  r1   1    2 r1   k 3 .
1  k3
B3* 
According to Routh-Hurwitz criterion, equilibrium point E1 is locally asymptotically
stable provided the following conditions are satisfied:
B1*  0, B2*  0, B3*  0, B4*  0, B1* B2*  B3*  0, B3* B1* B2*  B3*   B1*  B4*  0,
2
where Bi* , i  1,2,3,4 are the coefficient equation of V E1   t ij ; i, j  1,2,3,4.
The variational matrix V E2  at equilibrium point E2
 b11 b12

b
b
V E 2    21 22
0
0

 0
0

b13
0
b33
0
b14 

0 
,
0 

b44 
where
b11  r1 
b14  
2r1 xˆ1
  1 , b12   2 , b13   xˆ1 ,
k1
1 xˆ1
1  xˆ1
, b21   1 , b22  r2   2 
 xˆ
2r2 xˆ 2
, b33  r3   xˆ1 , b44  2 1 .
1  xˆ1
k2
The characteristic polynomial for the variational matrix V E2  is given by
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212
4  C1* 3  C 2* 2  C3*   C 4*  0 ,
(8.3.2)
where
C1*  b11  b22  b33  b44 ,
C2*  b12 b21  b11b22  b11b33  b22 b33  b11b44  b22 b44  b33b44 ,
C3*  b12 b21b33  b11b22 b33  b12 b21b44  b11b22 b44  b11b33 b44  b22 b33 b44 ,
C 4*  b11b22 b33 b44  b12 b21b33b44 .
According to Routh-Hurwitz criterion, equilibrium point E2 is locally asymptotically
stable provided the following conditions are satisfied:

  
2
C1*  0, C 2*  0, C3*  0, C 4*  0, C1*C 2*  C3*  0, C3* C1*C 2*  C3*  C1* C 4*  0 ,
 
where C i* , i  1,2,3,4 are the coefficient equation of V E 2   t ij ; i, j  1,2,3,4.
The variational matrix V E3  at equilibrium point E 3
 c11

c
V E3    21
c
 31
c
 41
c12
0
c 22
0
0
c33
0
c 43
0 

0 
,
c34 

c 44 
where
c11  r1  1 z   1   y  y z , c12   2 , c21   1 , c22  r2   2 , c31   y  y z ,
c33  r3 
 y
2r3 y
1 z
2 z
 y

, c34   1 , c41   2 z , c 43 
, c 44  2cz  2 .
2
2
1 y
1 y
k3
1  y 
1  y 
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213
The characteristic polynomial for the variational matrix V E3  is given by
4  D1* 3  D2* 2  D3*   D4*  0,
(8.3.3)
where
D1*  c11  c22  c33  c44 ,
D2*  c12 c21  c11c22  c11c33  c22 c33  c11c44  c22 c44  c33 c44  c34 c43 ,
D3*  c12 c 21 c33  c11 c 22 c33  c11 c34 c 43  c 22 c34 c 43  c11 c 22 c 44  c11 c33 c 44  c 22 c33 c 44  c12 c 21 c 44 ,
D4*  c11c22 c33 c44  c11c22 c34 c43  c12 c21c34 c43  c12 c21c33 c44 .
According to Routh-Hurwitz criterion, equilibrium point E 3 is locally asymptotically
stable provided the following conditions are satisfied:

  
2
D1*  0, D2*  0, D3*  0, D4*  0, D1* D2*  D3*  0, D3* D1* D2*  D3*  D1* D4*  0,
 
where Di* , i  1,2,3,4 are the coefficient equation of V E3   t ij ; i, j  1,2,3,4.
The variational matrix V E4  at equilibrium point E4
 d11

d
V E 4    21
d
 31
 0

d12
d13
d 22
0
0
d 33
0
0
d14 

0 
,
d 34 

d 44 
where
d11  r1 
2r1 x1
  1   y  , d12   2 , d13   x1 ,
k1
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d14  x1 y  
d 33  r3 
 1 x1
1  x1
, d 21   1 , d 22  r2   2 
2r2 x2
, d 31   y ,
k2
 x  y 
2r3 y 
 y
  x1 , d 34   1  x1 y , d 44  2 1  2 .
1  y
1  x1 1  y 
k3
The characteristic polynomial for the variational matrix V E4  is given by
4  E1*3  E2*2  E3*  E4*  0 ,
(8.3.4)
where
E1*  d11  d 22  d 33  d 44 ,
E 2*  d12 d 21  d11d 22  d11d 33  d 22 d 33  d11d 44  d 22 d 44  d 33 d 44  d13 d 31 ,
E3*  d13 d 22 d 31  d11d 22 d 33  d12 d 21d 33  d12 d 21d 44  d11d 22 d 44  d13 d 31d 44
 d11d 33 d 44  d 22 d 33 d 44 ,
E 4*  d11d 22 d 33 d 44  d13 d 22 d 31d 44  d12 d 21d 33 d 44 .
According to Routh-Hurwitz criterion, equilibrium point E4 is locally asymptotically
stable provided the following conditions are satisfied:

  
2
E1*  0, E 2*  0, E3*  0, E 4*  0, E1* E 2*  E3*  0, E3* E1* E 2*  E3*  E1* E 4*  0,
 
where Ei* , i  1,2,3,4 are the coefficient equation of V E 4   t ij ; i, j  1,2,3,4.
The Variational matrix V E5  at equilibrium point E 5 is given by
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215
 l11 l12

l
l
V E 5    21 22
0 0

l
 41 0
l13
0
l 33
l 43
l14 

0
,
0

l 44 
where
l11  r1 
l14  
l 41 
2r1 x1 1 z 

  1 , l12   2 , l13   x1  x1z ,
k1
(1  x1)
 1 x1
1  x1
, l 21   1 , l 22  r2   2 
 2 z 
1  x1
2
2r2 x2
, l33  r3   x1  x1z   1 z ,
k2
, l 43   2 z , l 44  2cz  
 2 x1
1  x1
.
The characteristic polynomial for the variational matrix V E5  is given by
4  F1* 3  F2* 2  F3*   F4*  0 ,
(8.3.5)
where
F1*  l11  l 22  l33  l 44 ,
F2*  l12l 21  l11l 22  l11l33  l 22l33  l14l 41  l11l 44  l 22l 44  l33l 44 ,
F3*  l12l 21l33  l11l 22l33  l14l 22l 41  l14l33l 41  l12l 21l 44  l11l 22l 44  l11l33l 44  l 22l33l 44 ,
F4*  l14l 22l33l 41  l12l 21l33l 44  l11l 22l33l 44 .
According to Routh-Hurwitz criterion, equilibrium point E 5 is locally asymptotically
stable provided the following conditions are satisfied:
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
  
2
F1*  0, F2*  0, F3*  0, F4*  0, F1* F2*  F3*  0, F3* F1* F2*  F3*  F1* F4*  0,
 
where Fi * , i  1,2,3,4 are the coefficient equation of V E5   t ij ; i, j  1,2,3,4.
The Variational matrix V E 6  at equilibrium point E6
 m11

m
V E 6    21
m
 31
m
 41
m12
m13
m22
0
0
m33
0
m 43
m14 

0 
,
m34 

m44 
where
m11  r1 
2r1 ~
x1
 ~z
x1  ~
x1 ~
z,
 1 ~ 1  ~
y~
y~
z , m12   2 , m13   ~
k1
(1  x1 )
m14  ~
x1 ~
y
m33  r3 
m43 
1 ~x1
2r2 ~
x2
,
, m31   ~
y~
y~
z,
,
m


m

r



21
1
22
2
2
~
1  x1
k2
 2 ~z
2r3 ~
y
1 ~z
 1 ~y ~ ~
m

,
m



x
y
,
 ~
x1  ~
x1 ~
z
,
34
1
41
1 ~
y
k3
1  ~y 2
1  ~x1 2
2~
x1  2 ~
y
~
,
m


2
c
z


.
44
2
~
~
1  x1 1  ~
y
1  y 
 2 ~z
The characteristic polynomial for the variational matrix V E 6  is given by
4  G1* 3  G2* 2  G3*   G4*  0,
(8.3.6)
where
G1*  m11  m22  m33  m44 ,
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217
G2*  m12 m21  m11m22  m13 m31  m11m33  m22 m33  m14 m41  m34 m43  m11m44
 m22 m44  m33 m44 ,
G3*  m13 m22 m31  m12 m21m33  m11m22 m33  m14 m22 m41  m14 m41m33
 m13 m34 m41  m14 m31m43  m11m34 m43  m22 m34 m43  m21m12 m44
 m11m22 m44  m44 m31m13  m11m33 m44  m22 m33 m44 ,
G4*  m14 m22 m33 m41  m13 m22 m34 m41  m14 m22 m31m43  m12 m21m34 m43
 m11m 22 m34 m43  m13 m22 m31m44  m12 m21m33 m44  m11m22 m33 m44 .
According to Routh-Hurwitz criterion, equilibrium point E6 is locally asymptotically
stable provided the following conditions are satisfied:

  
2
G1*  0, G2*  0, G3*  0, G4*  0, G1*G2*  G3*  0, G3* G1*G2*  G3*  G1* G4*  0,
where Gi* , i  1,2,3,4 are the coefficient equation of V E6   t ij ; i, j  1,2,3,4.
8.4 PERSISTENCE
Lemma: If
p  0, q  0 and
lim sup u (t ) 
q
p
t 
du
   ut  q  put , ut 0   0, then we have
dt

q
 lim inf u (t )   .
t 
p

Theorem: The system (8.1.1) persistence provided,
(i) 1 M 2*   1   M 1*  r1 ,
(ii)  2  r2 ,
(iii)  M *  1 M 2*  r3 .
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218
Proof: Let x1 , x2 , y, z  be unique positive solution of the system (8.1.1) with initial
conditions x1 0, x2 0, y0, z0 . Then with the help of the system (8.1.1) and lemma
we obtain
S
(i) lim sup x(t ) 

t 
 M *,
where xt   x1 t   x2 t  , S 
k1 r1   
k r   
 2 2
.
4r1
4r2
2
2
(ii) lim sup y(t )  k 3  M 1* ,
t 
(iii) lim sup z (t ) 
 2 M *   2 M 1*
c
t 
 M 2* .
Now, with the help of M * , M 1* , M 2* and the system (8.1.1) we get
(i) lim inf x1 (t ) 
k1 r1   1 M 2*   1   M 1* 
 m1* ,
r1
(ii) lim inf x2 (t ) 
k 2 r2   2 
 m2* ,
r2
t 
t 
k 3 r3  1 M 2*   M * 
(iii) lim inf y(t ) 
 m3* ,
t 
r3
(iv) lim inf z (t ) 
t 
 2 m1*

c 1 M
*

 2 m3*
 c1  M 
*
1
 m4* .
Since all the parameters r1 , k1 ,  1 ,  2 ,  1 ,  2 , r2 , k 2 , r3 , 1 ,  2 , c and  are positive.
Hence we have
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219
(i) lim inf x1 (t )  0 ,
t 
(ii) lim inf x 2 (t )  0 ,
t 
(iii) lim inf y (t )  0 ,
t 
(iv) lim inf z (t )  0 .
t 
provided the following holds
(i) 1 M 2*   1   M 1*  r1 ,
(ii)  2  r2 ,
(8.4.1)
(iii)  M *  1 M 2*  r3 .
Thus, model system (8.1.1) is persistent under the conditions given in (8.4.1).
8.5 NUMERICAL SIMULATION
In this section, we present numerical simulation to illustrate results obtained in
previous sections. The system (8.1.1) is integrated using fourth order Runge – Kutta
Method with the help of MATLAB software package under the following set of
parameters
r1  5, r2  8, r3  3, k1  20, c  0.4,  1  0.2,  2  0.1, k 2  30, k 3  120,  1  2,  2  4,
1  0.2,  2  0.1,   1.
(8.5.1)
We find that all the equilibrium points for the system (8.1.1) exist and given by
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E0 0,0,0,0, E1 0,0,120,0 , E2 26.2635,23.4131,0,0 , E3 0,0,119.984,0.247934 ,
E4 0.682477,15.3338,92.7009,0 , E5 26.2584,23.4119,0,0.240829 and
E6 1.15548,15.5571,91.3717,0.38131.
At the equilibrium point E6  the equation (8.3.6) becomes
4  60.877 3  326.093 2  382.729   45.931  0.
(8.5.2)
The roots of equation (8.5.2) are
 55.0828,4.18522,1.47378,0.135189 , this implies that E6 is stable equilibrium
point of the system.
100
90
Total Population
80
70
x1
60
x2
y
z
50
40
30
20
10
0
0
100
200
300
400
500
600
700
800
900
1000
Time(t)
Figure 1. Graph of x1 , x2 , y and z versus t and other parameters are same as (8.5.1).
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1
0.5
0
-0.5
Equilibrium Point
-1
10
12
14
0
Prey(x2) 16
5
18
10
20
15
22
Prey(x1)
20
Figure 2. x1 , x2 approach to their equilibrium values in finite time and other values of
parameters are same as (8.5.1).
100
Equilibrium Point
90
Prey(y)
80
70
60
50
40
0
2
4
6
8
10
12
14
16
18
Prey(x1)
Figure 3. x1 , y approach to their equilibrium values in finite time and other values of
parameters are same as (8.5.1).
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1
0.9
Predator (z)
0.8
0.7
0.6
0.5
0.4
Equilibrium Point
0
5
Prey (x1)
10
15
Figure 4. x1 , z approach to their equilibrium values in finite time and other values of
parameters are same as (8.5.1).
100
90
Equilibrium Point
Prey (y)
80
70
60
50
40
10
12
14
16
18
20
22
Prey (x2)
Figure 5. x2 , y approach to their equilibrium values in finite time and other values of
parameters are same as (8.5.1).
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1
0.9
Predator (z)
0.8
0.7
0.6
0.5
0.4
10
Equilibrium Point
11
12
13
14
15
16
17
18
19
20
Prey (x2)
Figure 6. x2 , z approach to their equilibrium values in finite time and other values of
parameters are same as (8.5.1).
1
Equilibrium Point
100
0.5
80
0
-0.5
60
Prey (y)
-1
40
0.4
0.5
0.6
Predator (z) 0.7
0.8
0.9
1
20
Figure 7. y, z approach to their equilibrium values in finite time and other values of
parameters are same as (8.5.1).
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1
Prey (x2)
0.5
0
-0.5
-1
0.2
0.3
0.4
Predator (Z)
26.2
26.22
0.5
26.24
0.6
26.26
0.7
26.28
26.3
Prey (x1)
Figure 8. Phase Portrait of the system (8.1.1) in absence of prey y and other values of
parameters are same as (8.5.1).
The results of numerical simulation are displayed graphically. In Figure (1) prey ( x1 ),
prey ( x2 ) , prey ( y ) and predator ( z ) population are plotted against time, from this
figure it is noted that for a given initial value
the population tend to their
corresponding value of equilibrium point E 6 and hence coexist in the form of stable
steady state, assuring the local stability of E 6 . Now in Figure (2)-(6) we can see the
relationship between prey ( x1 ) and prey ( x2 ) , prey ( x1 ) and prey ( y ) , prey ( x1 ) and
predator (z ) , prey ( x2 ) and prey ( y ) , prey ( x2 ) and predator (z ) , prey ( y ) and
predator (z ) . We study the effect of team approach on prey population in two preys
and one predator model. The above situation is described by means of a system with
four nonlinear differential equations. From comparing Figure (4) and Figure (8) we
can see effect of team approach on our model. In Figure (8) prey ( y ) is absent, the
system is showing unstable behavior, on other way in Figure (4) prey ( y ) is present,
the system become stable. A biological realization of our study is the herds of zebras
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225
and gazelles living side by side and attacked by one type of predator (lion). Hence our
mathematical model is biologically implemented in the case of two preys (gazelles,
zebras) and one predator (lion) model. When prey (zebras) is absent, the population of
predator (lion) increases out of limit. It is harmful for population of prey (gazelles).
Our system becomes unstable. In presence of prey (zebras), the population of predator
(lion) decreases. The population of prey (gazelles) is also decreases due to
competition with prey (zebras). In this situation our model remains stable. Our model
represents the nature of forest approximately. With the help of results of this model,
we can save the wild life. We may make reserve area for prey in man-made parks.
8.6 CONCLUSION
In this chapter, a nonlinear mathematical model is proposed and analyzed two
preys (gazelles, zebras) one predator (lion) model with one prey dispersal in a two
patch environment. We have obtained conditions for the existence of different
equilibria and discussed their stabilities in local manner by using stability theory of
differential equations.We have shown the effect of team approach in Section 8.5.
Many animals live in groups. Different groups share one habitat hence these groups
may cooperate, compete with each other or form prey- predator system. From our
analysis, we have found prey reserve and team approach are most important for
species from getting extinct. This model in applicable to many species of wild life.
For better management of wild life these fact should be taken into consideration. For
an example the bird reserve in managed forests, the individualistic nonlinear
responses of bird species to canopy retention levels emphasize the importance of
setting specific management objectives. The canopy retention level should be
sufficient to support and protect the required number of bird species.
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