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Int Jr. of Mathematics Sciences & Applications
Vol. 2, No. 2, May 2012
Copyright  Mind Reader Publications
ISSN No: 2230-9888
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STABILITY ANALYSIS OF TWO COMPETITIVE INTERACTING
SPECIES WITH LINEARLY VARYING RESERVE FOR ONE OF THE
SPECIES
R.Archana Reddy1 & N.Ch.Pattabhi Ramacharyulu2
1
Asso.Professor in Mathematics, Department of H & Sc, SREC, Warangal.
2
Former Faculty, National Institute of Technology, Warangal.
ABSTRACT
The present work deals with a study on Mathematical models of ecological competition
between two interacting species with a variable reserve for one of the species. The basic
equations for each model have been formulated as a coupled system of first order non-linear
ordinary differential equations. The equilibrium states are identified and the criteria for their
stability have been established. The numerical solutions for the growth rate equations are
computed using Runge Kutta fourth order method.
Key Words: Carrying Capacities Competition, Equilibrium States, Stability Criteria, Variable
Reserve.
1. Introduction: Mathematical modeling in mathematical biosciences is an attempt to identify and describe
some instances of day-to-day life in the language of mathematics. The mathematical model, structured on a
biological base, would enable in the prediction/quantitative estimation of population strength at subsequent
instants of time. The diverse techniques adopted in mathematical modeling have been illustrated exhaustively by
Kapur [5] in his treatise on mathematical modeling. Detailed research in theoretical ecology was initiated in
1925 by Lotka [7] and in 1931 by Volterra [11]. Exhaustive information on ecological species is given in the
treatises of authors such as Cushing [2], Freedman [3] and Paul Colinvaux [8]. The mathematical model
equations for interactions such as Prey-Predation, Competition, Commensalism, Neutralism etc. have been
presented in the treatises of authors such as Kapur [4], Svirezhev and Logofet [10] and Freedman [3]. Phani
Kumar and Pattabhiramacharyulu [6] investigated a Prey-Predator ecological model with partial cover for prey
and alternate food for the predator. A Stability analysis of Competetive species was carried out by Bhaskara
Rama Sharma and Pattabhiramacharyulu [1]. Mutualism between two species was examined by Ravindra Reddy
[9]. This investigation is on a two species competing model with a variable reserve for one of the species. The
mathematical model is characterized by a couple of first order non-linear ordinary differential equations. The
equilibrium points of the model are identified and stability criteria are discussed. Solutions for the linearized
perturbed equations are found and results are illustrated. The numerical solutions for the growth rate equations
are computed using Runge Kutta fourth order method.
2. Basic Equations: Notation:
The basic equations for a two species competing model are given by the following system of non-linear ordinary
differential equations employing the following notation.
N1 , N 2 are the population of species S1 & S 2 ,
a1 , a 2 are the rate of natural growth of S1 & S 2 ,
a1 / k1 , a 2 / k 2 are the rate of decrease of S1 & S 2 due to limited natural resources,
a12 , a 21 are the rate of decrease of either species due to competitive inhibition by the other species,
k1 , k 2 are the carrying capacities of the first and second species,
a + bN1 : is the first species population provided with a linearly varying reserve of safety with constant values
of a and b, from the inhibition/attacks of the second species.
521
R.Archana Reddy & N.Ch.Pattabhi Ramacharyulu
dN 1
N
= a1 N1 (1 − 1 ) − a12 {N1 − ( a + bN 1 )}N 2
dt
k1
dN 2
N
= a2 N 2 (1 − 2 ) − a21{N1 − (a + bN1 )}N 2
dt
k2
(1)
(2)
The growth rate equations are coupled and non-linear in N 1 , N 2 . As such they are not amenable to yield exact
solutions, satisfying properly posed positive initial conditions.
N 1 (0) = N 10 , N 2 (0) = N 20 .
3. Equilibrium States ( N 1 , N 2 ) :
The system under investigation has five equilibrium states. They are
I. The fully washed out state.
N1 = 0 ; N 2 = 0
(3)
a1
= k1 ; N 2 = 0
a11
In this state, the species S1 only survives and S 2 is washed out.
k [2a2 qr + a21 (b − 1) p]
p
III. N 1 =
; N2 = 2
2qa2
2q
II.
N1 =
(4)
(5)
p 2 = 4qaa12 k 2 r .
k [a pr − aa12 a21rk2 (1 − b)]
N2 = 2 2
a2 p
In this state both the species co-exist only when
IV.
N1 =
aa12 k 2 r
;
p
V.
N1 =
p 2 − aa12 k 2 rq
;
pq
N2 =
(6)
k 2 [a 2 pqr + a 21 (1 − b){a12 aqrk 2 − p 2 }]
a 2 pq
In the states IV & V both the species co-exist and this would happen only when
p=
r=
1
[a1a 2 − a 2 a12 k 2 (1 − b)(a 2 + 2aa 21 )]
a2
q=
(7)
p 2 > 4qaa12 k 2 r where
1
[−a1a 2 + a12 a 21k1 (k 2 + b 2 − 2bk 2 )]
a 2 k1
1
(a 2 + aa 21 )
a2
4. The Stability of Equilibrium States: Let
N = ( N1 , N 2 ) = N + U
where
U = (u1 , u 2 ) is the
perturbation over the equilibrium state N = ( N 1, N 2 ) . The basic equations (1) and (2) are linearized to obtain
the equations for the perturbed state.
dU
= AU where
dt
2a1 N1

a1 − k − a12 (1 − b) N 2
1
A=

a21 (b − 1) N 2

(8)


u 
 & U =  1  (9)
2a N
u 2 
a2 − 2 2 − a21{N1 − (a + bN1 )}

k2
The characteristic equation for the system is det[ A − λI ] = 0
(10)
− a12 {N1 − (a + bN1 )}
The equilibrium state is stable if both the roots of the equation (10) are negative in case they are real or have
negative real parts in case they are complex.
4. I. Equilibrium state I:
4. I.A Stability of the Equilibrium state I:
The linearized equations for the perturbations ( u1 , u 2 ) are
522
STABILITY ANALYSIS OF TWO COMPETITIVE INTERACTING…
a
d
[u ] =  1
dt
0
aa12   u1 
a2 + aa21  u 2 
(11)
and the corresponding characteristic equation is
[λ − a1 ][λ − {a2 + aa21}] = 0
Clearly both the roots are positive. Hence this state is unstable. The trajectories in this case are
u1 =
1
[u 20 aa12 e λ2t + {u10 (λ 2 − a1 ) − u 20 aa12 }e a1t ]
λ 2 − a1
where u10 , u 20 are the initial values of
u 2 = u 20 e λ2t ;
(12)
u1 & u 2 . The solution curves are illustrated in Fig. 1 - 3.
Case1: if u10 < u 20
u1
u2
u 20
u10
0
t*
t→
Fig.1
In this case initially the second species outnumbers the first species and continues till a time instant t = t
given by (13) after which the first species dominates the second species. Also both the species are going away
from the equilibrium point.
t = t* =
*
u (λ − a1 ) − u 20 aa12
1
ln[ 10 2
]
λ2 − a1
u 20 {λ2 − a1 − aa12 }
(13)
Case 2: if u10 > u 20
u1
u2
u10
u 20
0
Fig.2
t→
In this case initially the first species outnumbers the second species and continues throughout its growth. Also
we notice that both the species are diverging from the equilibrium point.
4.II.B. Trajectories of Perturbed Species:
↑
u2
u 20
p1 < q1
p1 = q1
p1 > q1
0
Fig.3
u1
→
u10
523
R.Archana Reddy & N.Ch.Pattabhi Ramacharyulu
The
where
trajectories
q1 =
in
the
u1 − u 2
plane
are
given
by
(q1 − 1)u1 = cu 2q1 − aa12 u 2
a1
, p1 = aa12 , & c = arbitrary constant. The solution curves are illustrated in Fig.3
a 2 + aa 21
4.II Equilibrium State II:
a1
= k1 ; N 2 = 0
a11
The linearized equations for the perturbations ( u1 , u 2 ) are
4.II.A Stability of Equilibrium State II:
−a
d
[u ] =  1
dt
 0
N1 =
a12 {a + (b − 1)k1 }   u1 
a 2 + a 21{a + (b − 1)k1} u 2 
(14)
and the corresponding characteristic equation is
[λ + a1 ][λ − {a 2 + a 21 (a + (b − 1)k1 )}] = 0
One root of the equation is negative and the other root is
Case A: when
u1 =
(15)
λ2 = a 2 + {a + (b − 1)k1 }a 21
λ 2 > 0 , the equilibrium state is unstable. The trajectories in this case are
1
[u 20 a12 {a + (b − 1)k1 }e λ2t +
λ 2 + a1
{u10 (λ 2 + a1 ) − u 20 a12 (a + (b − 1)k1 )}e − a1t
u 2 = u 20 e λ2t
(17)
u1 & u 2 .
where u10 ,u 20 are the initial values of
Case 1: when u10 < u 20 and
(16)
λ 2 > a1
u2
u1
u 20
u10
t→
0
Fig.4
In this case initially the second species outnumbers the first species and continues throughout its growth. Also
both the species are diverging from the equilibrium point. This is illustrated in Fig.4.
Case 2: when u10 > u 20 and
λ 2 > a1
u2
u1
u10
u 20
0
t*
t→
Fig.5
In this case initially the first species dominates over the second species till the time instant t = t given in (18)
after which the second species outnumbers the first species. Also both the species are diverging from the
equilibrium point. This is illustrated in Fig.5.
*
524
STABILITY ANALYSIS OF TWO COMPETITIVE INTERACTING…
t = t* =
u (λ + a1 ) − u 20 {a + (b − 1)k1 }
1
ln[ 10 2
]
λ2 + a1
u 20 {a + (b − 1)k1 − (λ2 + a1 )}
Case 3: when u10 < u 20 and
(18)
λ 2 < a1
u2
u 20
u10
u1
0
Fig.6
t1*
In this case initially the second species dominates over the first species and continues throughout its growth.
Further the second species is noted to be going away from the equilibrium point while the first species would
*
become extinct at the instant (t1 ) of time given by (19) as seen in Fig.6.
u {a + (b − 1)k1} − u10 (d 2 + a1 )
1
ln[ 20
]
d 2 + a1
u 20 {a + (b − 1)k1
Case 4: when u10 > u 20 and λ 2 < a1
t = t1* =
(19)
u2
u10
u 20
0
u1
t*
Fig.7
t1*
*
In this case initially the first species outnumbers the second species till the time instant ( t ) given by (18) after
which the second species outnumbers the first species. Also the second species is noted to be going away from
*
the equilibrium point while the first species would become extinct at the instant (t1 ) of time given by (19) as
seen in Fig.7.
Case B: when λ 2 < 0 , the equilibrium state is stable. The trajectories in this case are same as in Case A. The
solution curves are illustrated in Fig.8 & 9.
Case 1: when u10 < u 20
u 20
u10
u2
u1
t1* t →
0
Fig.8
In this case initially the second species dominates over the first species and continues throughout its growth.
Further both the species converge asymptotically to the equilibrium point and the first species would become
*
extinct at the instant (t1 ) of time given by (19) as illustrated in Fig.8.
525
R.Archana Reddy & N.Ch.Pattabhi Ramacharyulu
Case 2: when u10 > u 20
u10
u 20
u2
0
u1
t
t1*
Fig.9
*
*
Initially the first species dominates the second species till the time instant ( t ) given by (18) after which the
dominance is reversed. However the second species converges asymptotically to the equilibrium point while the
*
first species would become extinct at the instant (t1 ) of time given by (19) as illustrated in Fig.9.
Case C: when λ 2 = 0 , one root of the equation is λ1 = − a1 , which is negative and the other root is zero.
Hence the equilibrium state is neutrally stable. The trajectories in this case are
u1 =
&
1
[u 20 a12 {a + (b − 1)k1 } + {u10 a1 − u 20 a12 {a + (b − 1)k1 }e − a1t ]
a1
u 2 = u 20
(20)
Case 1: when u10 < u 20
u 20
u2
u10
u1
u1*
0
Fig.10
t→
In this case the second species always outnumbers the first species. In course of time
which is evident from the equation (21). This is illustrated in Fig.10.
u1* =
u 20 {a + (b − 1)k1 }
a1
u1 is asymptotic to u1*
(21)
Case 2: when u10 > u 20 and u 20 > u1
*
u10
u 20
u2
u1
t→
0 t*
Fig.11
Initially the first species outnumbers the second species and it continues up to the time ( t = t ) given by the
*
equation (22) after which the second species outnumbers the first species. In course of time
u1 is asymptotic to
*
1
u which is evident from the equation (21). This is illustrated in Fig.11.
u a + u 20 a12 k1
1
]
t = t * = ln[ 10 1
a1
u 20 (a1 + a12 k1 )
526
(22)
STABILITY ANALYSIS OF TWO COMPETITIVE INTERACTING…
Case 3: when u10 > u 20 and u 20 < u1
*
u10
u1
u1*
u 20
0
u2
t
t→
*
Fig.12
Initially the first species outnumbers the second species and it continues to do so. In course of time
*
1
asymptotic to u which is evident from the equation (21). This is illustrated in Fig.12.
4.II.B. Trajectories of Perturbed Species:
↑
u2
u 20
p2 < q2
p2 = q2
p2 > q2
0
Fig.13
u1
→
u10
u1 − u 2 plane are given by (q 2 − 1)u1 = cu 2q2 − p 2 u 2 (23)
a + (b − 1)k1
− a1
where p 2 =
; q2 =
a 2 + {a + (b − 1) k1 }a 21
a 2 + {a + (b − 1)k1 }a 21
c = arbitrary constant. The solution curves are illustrated in Fig.13.
The trajectories in the
5.4. III. Equilibrium State III i.e., the normal steady state:
5.4. III.A. Stability of the equilibrium state:
k 2 {2a 2 qr + a 21 (b − 1) p}
2qa 2
The linearized equations for the perturbations (u1 , u 2 ) are
2a


a1 − 1 N 1 − a 21 (1 − b) N 2 aa12 − a12 (1 − b) N 1 

u 
d
k1
 1 
[u ] = 
a
dt

 u 2 
− a 21 (1 − b) N 2
− 2 N2


k2
2
The characteristic equation is λ − Aλ + B = 0
(25)
2a N 1
a N2
where
A = a1 − 1
− a12 (1 − b) N 2 − 2
k1
k2
N1 =
B=−
p
2q
;
N2 =
(24)
a2 N 2
2a N 1
{a1 − 1
− a12 (1 − b) N 2 } + a12 a 21 (1 − b) N 2 {a − (1 − b) N 1 }
k2
k1
527
u1 is
R.Archana Reddy & N.Ch.Pattabhi Ramacharyulu
Case A : when A > 0 and
trajectories in this case are
B > 0 , both the roots are positive. Hence the equilibrium state is unstable. The
2a1
N 1 + a 21 (1 − b) N 2 } − u 20 {a12 (a + b − 1)}
k1
u1 = [
]e λ1t
λ 2 − λ1
2a
u10 (λ1 − {a1 − 1 N 1 + a 21 (1 − b) N 2 } − u 20 {a12 (a + b − 1)}
k1
+[
]e λ2t
λ1 − λ 2
2a
u 20 (λ1 − {a1 − 1 N 1 + a 21 (1 − b) N 2 }) − u10 a 21 (1 − b)}
k1
u2 = [
]e λ1t
λ1 − λ 2
2a
u 20 (λ 2 − {a1 − 1 N 1 + a 21 (1 − b) N 2 } − u10 a 21 (1 − b)
k1
+[
]e λ2t (27)
λ1 − λ 2
u10 (λ 2 − {a1 −
(26)
Case 1: when u10 < u 20
u1
u2
u 20
u10
0
t*
Fig.14
Initially the second species dominates over the first species till the time instant t = t given in (28) after which
the first species outnumbers the second species. Also both the species are going away from the equilibrium
point. This is illustrated in Fig.14.
*
u (a + b2 ) − u 20 (a 4 + b1 )
1
ln[ 10 3
]
λ1 − λ2 u10 (a 4 − b2 ) − u 20 (b1 − a3 )
2a1
N 1 + a12 (1 − b) N 2 } ; b1 = a12 ( a + b − 1)
where a 3 = λ1 − {a1 −
k1
2a
a 4 = λ2 − {a1 − 1 N 1 + a12 (1 − b) N 2 } ; b2 = a 21 (1 − b )
k1
Case 2: when u10 > u 20
u1
u2
t = t* =
u10
u 20
0
Fig.15
528
t→
(28)
STABILITY ANALYSIS OF TWO COMPETITIVE INTERACTING…
In this case initially the first species outnumbers the second species and continues throughout its growth. Also
both the species are going away from the equilibrium point as illustrated in Fig.15.
Case B: when B < 0 , one root of the equation is negative while the other root is positive. The trajectories in
this case are same as in CaseA.
Case 1: when u10 < u 20
u1
u2
u 20
u10
0
t*
Fig.16
Initially the second species outnumbers the first species and this continues up to the time instant t = t given in
(28) after which the first species outnumbers the second species. Also the second species decreases in the
beginning and gradually increases. Both the species are diverging from the equilibrium point as illustrated in
Fig.16.
*
Case 2: when u10 > u 20
u1
u2
u10
u 20
0
Fig.17
t→
In this case initially the first species outnumbers the second species and continues throughout. The second
species decreases initially and gradually increases. Also we notice that both the species are going away from the
equilibrium point. Hence the equilibrium state is unstable. This is illustrated in Fig.17.
Case C: when A < 0 and B > 0 , both the roots are negative. Hence the equilibrium state is stable. The
trajectories in this case are same as in Case A.
Case 1: when u10 < u 20
u 20
u10
u1
0
u2
t
t→
Fig.18
*
529
R.Archana Reddy & N.Ch.Pattabhi Ramacharyulu
Initially the second species outnumbers the first species and this continues up to the time instant t = t given in
*
(28) after which the first species outnumbers the second species. As t → ∞ both
equilibrium point. Hence the equilibrium point is stable. This is illustrated in Fig.18.
u1 and u 2 approach the
Case 2: when u10 > u 20
u10
u 20
u2
u1
t→
0
Fig.19
In this case the first species always outnumbers the second species and both are converging asymptotically to
the equilibrium point. Hence the equilibrium point is stable as illustrated in Fig.19.
When A < 4 B , the roots are complex conjugate with negative real part. Hence the equilibrium state is
stable. The solution curves are illustrated in Fig.20.
2
u 20
u2
u10
u1
t→
Fig.20
4. IV.B. Trajectories of perturbed species:
The trajectories in the u1 − u 2 plane are given by
cu
( v1 − v2 )
2
=
(u1 − v1u 2 ) p −bv
3
1
(u1 − v 2 u 2 ) p3 −bv2
2
where, v1 & v 2 are the roots of bv + ( p 3 − q 3 )v − a = 0 ,
p3 = −
a2
N2;
k2
b = − a 21 (1 − b) N 2
a = − a12 {a − (1 − b) N 1 }
q3 = a1 −
↑
u2
u1 →
Fig.21
530
(29)
2a1
N 1 − a12 N 2 (1 − b) ;
k1
STABILITY ANALYSIS OF TWO COMPETITIVE INTERACTING…
5. Numerical Solution of the Growth Rate Equations:
The numerical solution of the growth rate equations (1) and (2) are computed employing the fourth order
Runge-Kutta method for specific values of the various parameters that characterize the model and the initial
conditions as given in table 1.
Table 1
Case
a1
a11
a12
a2
a 22
a 21
N10
N 20
a
b
1
0.2
0.5
0.5
1
0.04
0.2
2
1
1.3
0.6
2
1.2
0.01
1
0.004
0.02
1
3
1.3
1.1
3
2.2
0.5
0.5
1
0.04
0.2
2
1
1.3
0.6
4
1
0.1
0.01
2
0.4
0.02
1
2.8
0.03
0.06
5
2
0.05
0.05
1
0.004
0.02
1
2
1.3
0.6
6
2
0.05
0.05
1
0.004
0.002
3
1
0.9
0.6
0.01
Case1: Initially the first species is dominant over the second species up to a time instant t = 0.4 after which
the second species dominates over the first species. In this case we observe that the second species initially has a
steady increase and in course of time both the species coexist with a steady variation and no appreciable growth.
This is seen in Fig.22.
*
Plot of
N and N
1
2
6
N1
and N2
3
N
4
1
5
N2
2
1
0
2
4
6
8
Fig.22
time (t)
10
Case2: In this case initially the second species dominates over the first species up to a time instant t = 2.3
after which the first species being a strong competitor suppresses the second species. The rise in the first species
is steep whereas there is a slight variation in the growth of the second species. This is illustrated in Fig.23.
*
531
R.Archana Reddy & N.Ch.Pattabhi Ramacharyulu
Plot of
N and N
1
2
150
N2
100
1 and
N2
N1
N
50
0
0
2
4
6
8
10
time (t)
Fig.23
Case3: In this case the first species outnumbers the second species initially up to an instant t = 2.8 after
which the second species dominates the first species. Further we notice that both the species survive hand in
hand, the second species suppressing the first species as illustrated in Fig.24.
*
Plot of
N and N
1
2
5
3
N
1 and
N2
4
N1
2
N2
1
0
2
4
6
Fig.24
8
10
time (t)
Case4: In this case initially the second species has a steep rise and then suffers a fall. The second species
dominates the first species up to an instant t = 4.1 after which the first species dominates the second species.
Further we notice that the first species exists with a steady growth rate forcing the second species to extinction
*
t1* of time. This is seen in Fig.25.
Plot of
N and N
1
2
20
N1
15
N2
N2
1 and
at
10
N
5
0
0
2
4
6
8
10
time (t)
Fig. 25
532
STABILITY ANALYSIS OF TWO COMPETITIVE INTERACTING…
Case 5: In this case the second species initially dominates the first species and from a time instant
t * = 0.7 to 3.8 the first species dominates the second species. Further the second species again suppresses
the first species and both the species coexist. This is illustrated in Fig.26.
Plot of
N and N
1
2
50
and N2
20
N
30
1
40
N1
10
N2
0
0
2
4
6
8
10
time (t)
Fig.26
Case 6: In this case the first species is dominant over the second species up to t = 3.5 after which the second
species dominates the first species suppressing the first species and forcing it to extinct in course of time. This is
*
Plot of
N1 and N 2
500
N1
N
1 and
N2
400
N2
300
200
100
0
0
illustrated in Fig.27.
2
4
6
Fig.27
8
10
time (t)
References:
[1] Bhaskara Rama Sharma, Pattabhiramacharyulu.N.Ch.; “Stability Analysis of two species ecosystem”. International
Journal of logic based intelligent systems, Vol.2 No 1 Jan-June(2008).
[2] Cushing.J.M.: Integro- Differential equations and Delay Models in Population Dynamics, Lect. Notes in
Biomathematics, Vol 20.- Springer –Verlag,Heidelberg.1977.
[3] Freedman, H.I.: Deterministic Mathematical models in Population Ecology, Marcel – Decker, New York, 1980.
[4] Kapur.J.N. : Mathematical Models in Biology and Medicine, Affiliated East-West, 1985.
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