Gen. Math. Notes, Vol. 3, No. 2, April 2011, pp. 13-26
ISSN 2219-7184; Copyright © ICSRS Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
A Study on Series Solutions of Two Species
Lotka Volterra Equations by Adomian
Decomposition and Homotopy Perturbation
Methods
D. Venu Gopala Rao
Associate Professor, Dept. of Applied Mathematics, GITAM College of Science,
GITAM University, Visakhapatnam, AP, India - 530045
E-mail: dvgrgitam@gmail.com
(Received: 19-1-11/ Accepted: 26-3-11)
Abstract
In this work, series solutions are obtained for the two species Lotka –Volterra,
prey-predator interaction species which are governed by a system of non linear
differential equations. For this purpose, the Adomian decomposition method
(ADM) and Homotopy Perturbation method (HPM) are employed and it is shown
that the Homotopy Perturbation method is much easier and a more efficient
method.
Keywords: Adomian Decomposition, Homotopy Perturbation, Lotka Volterra, prey-predator.
1
Introduction
In the study of non linear system of differential equations such as the Lotka –
Volterra equations, analytical solutions are usually unknown. In order to analyze
14
D. Venu Gopala Rao
the behaviour of the system, one usually resorts to numerical integration
techniques such as perturbation techniques. Perturbation technique depends on the
existence of small or large parameters in the nonlinear problems. In this paper, the
Adomian decomposition method (ADM) and Homotopy Perturbation method
(HPM) will be employed to obtain series solutions to two species Lotka – Volterra,
prey – predator interaction species which are governed by a system of non linear
differential equations. The first method is Adomian decomposition method [3]
which was introduced by Adomian in the 1980s, an iterative method which
provides approximate analytical solutions in the form of an infinite series for non
linear equations. This method avoids linearization, discretization and scientifically
unrealistic assumptions. The second method is the Homotopy perturbation method
[4] which was proposed by ‘He’ in 1999. In this method, the solutions are
obtained as an infinite series, the summation of which is an analytical solution. In
this method, a homotopy is constructed with an embedding parameter p∈ (0, 1),
which is valid not only for small parameters but also for very large parameters.
We also show with the help of an example that although the solutions obtained by
both ADM and HPM are the same, HPM is convenient and efficient when
compared to ADM.
2
Adomian Decomposition Method
2.1
Analysis of the Method for Interacting Species
In this section we consider the system of equations of the form
dx1
= x1 (a1 − b1 x1 − c1 x 2 )
dt
dx 2
(2.1)
= x 2 (− a 2 + c 2 x1 )
dt
where a1, b1, c1, a2, c2 are positive constants.
x1 and x2 represent the prey and predator population at time, t. We non
dimensionalize the system of equations (2.1) by setting
c
c1c2
x2 (t ) ,
u(τ) = 2 x1 (t ) ,
ν(τ) =
τ = a1t
a2
a1c2 − a 2 b1
a
ba
α= 2 ,
β= 1 2
(2.2)
a1
c2 a1
The equations (2.1) are then transformed to
du
= u (τ ) (1 − β u (τ ) − (1 − β ) ν (τ ) )
dτ
dv
(2.3)
= αν (τ ) (− 1 + u (τ ) )
dτ
Considering f and g to be non linear functions of u and (u, v) respectively, the
equations (2.3) can now be written in the form
A Study on Series Solutions of Two Species…
15
du
= u (τ ) − βf (u ) − (1 − β ) g (u , v )
dτ
dv
(2.4)
= − αν (τ ) + αg (u , v )
dτ
with initial conditions u(0) = u0, u(0) = u0.
Now, to find the solutions of u and v that satisfy equation (2.4), the decomposition
method is used. It consists of approximating the solutions of (2.4) as an infinite
series in the following form.
∞
u=
∑U
n =0
∞
v=
n
∑V
n =0
(2.5)
n
Now, decomposing f and g as
∞
f(u) =
∑A
n =0
(2.6)
n
∞
g(u, v) =
∑B
n=0
(2.7)
n
where An, Bn are Adomian polynomials that can be generated for any form of non
linearity. Writing equation (2.4) as
Lu = u(τ) - βf (u(τ)) – (1– β) g (u(τ), v(τ))
Lv = – αv(τ) + αg(u(τ), v(τ))
(2.8)
∂
where L =
, the linear differential operator. Assume the inverse operator L-1
∂τ
τ
exists, which stands for integrations as defined by L-1 =
∫ (.)dτ . Applying L
-1
on
0
both sides of equation (2.8), we obtain
u(τ) = u(0) + L-1 u(τ) - β L-1 f (u(τ)) – (1-β) L-1 g (u(τ), v(τ))
v(τ) = – α L-1 v(τ) + α L-1g (u(τ), v(τ))
(2.9)
Using equations (2.5), (2.6) & (2.7), we get
∞
∑ un = u(0) + L−1
n =0
∞
∑v
n =0
n
∞
∑ un − β L−1
n=0
∞
−1
= v(0) − α L
∑v
n =0
n
+ α L−1
∞
∑ An − (1 − β ) L−1
n =0
∞
∞
∑B
n =0
n
∑B
n =0
n
(2.10)
Now, the iterates are determined in the following recursive way.
u(0) = u0
un+1 = L-1 un - β L-1An – (1 – β) L-1 Bn
v(0) = v0
(2.11)
16
D. Venu Gopala Rao
vn + 1 = – α L-1 vn + α L-1 Bn
(2.12)
We then define the solution (u, v) as
n

(u, v) =  nlim
→∞ ∑ uk ,
k =0

n
lim
n→∞

∑ vk 
(2.13)
k =0
2.2 Analysis of the Method for Interacting Species with
Negative Feedback
In this section, we consider the nonlinear system of the form
dx1
= x1 (a1 − b1 x1 − c1 x2 )
dt
dx 2
= x 2 (− a 2 + b2 x1 − c 2 x 2 )
dt
(2.14)
Here x1 and x2 are prey, predator populations at time t respectively where a1, b1, c1,
a2, b2, c2 are positive constants. Each of the terms – b1x12 and – c2x22 is called
negative feed back. The equilibrium of the system is positive provided the
inequality a1b2 > a2b1 holds.
We non dimensionalize the system by setting
u(τ) =
b1 c 2 + b2 c1
x1 (t ),
a1 c 2 + a 2 c1
Taking α =
a1c 2 + a 2 c1
,
b1c 2 + b2 c1
v(τ) =
β=
b1c 2 + b2 c1
x 2 (t ), and τ = a1t
a1b2 − a 2 b1
a1b2 − a 2 b1
, the equations (2.14) are then
b1c 2 + b2 c1
b
c
du
= u (τ ) − 1 α u 2 (τ ) − 1 β u (τ ) v(τ )
dτ
a1
a1
a
b
c
dv
= − 2 v(τ ) + 2 α u (τ ) v(τ ) − 2 β v 2 (τ )
dτ
a1
a1
a1
With k0 =
(2.15)
(2.16)
a2
c
b
c
b
, k1 = 1 α , k2 = 1 β , k3 = 2 α , k4 = 2 β the equation (2.16)
a1
a1
a1
a1
a1
is now
du
= u (τ ) − k 1 u 2 (τ ) − k 2 u (τ ) v (τ )
dT
A Study on Series Solutions of Two Species…
17
dv
(2.17)
= − k 0 v (τ ) + k 3u (τ ) v (τ ) − k 4 v 2 (τ )
dT
Considering f, g and h to be non linear functions of u, v and (u, v) respectively, the
equations (2.17) can now be written in the form
du
= u (τ ) − k1 f (u ) − k 2 g (u , v )
dτ
du
= − k 0 v (τ ) + k 3 g (u , v ) − k 4 h(v)
dτ
(2.18)
with initial values u(0) = u0, v(0) = v0. To find the solutions of u and v that satisfy
equation (2.18), the decomposition method is used. It consists of approximating
the solutions of (2.18) as an infinite series in the following form.
∞
u=
∞
∑ un
v=
n =0
∑v
n =0
(2.19)
n
Decomposing f, h and g as
∞
f(u) =
∞
∑ An
h(v) =
n =0
∞
∑ Cn
g(u, v) =
n =0
∑B
n =0
n
(2.20)
where An, Bn & Cn are Adomian polynomials that can be generated for any form
of non linearity. Applying the decomposition method, the system (2.18) can be
written as
Lu = u (τ ) − k1 f (u ) − k 2 g (u , v)
Lv = − k 0 v (τ ) + k 3 g (u , v ) − k 4 h(v )
where L =
(2.21)
∂
, the linear differential operator.. Applying the integration inverse
∂τ
τ
operator L−1 = ∫ (.) dτ to equation (2.21),
0
u(τ) = u(0) + L u(τ) – k1 L-1 f (u(τ)) – k2 L-1 g (u (τ), v(τ))
v(τ) = v0 – k0 L-1 v(τ) + k3 L-1g (u(τ), v(τ) – k4 L-1 h(v(τ))
-1
(2.22)
Using equations (2.19) and (2.20),
∞
∑u
n
= u0 + L
∑v
n
= v 0 − k 0 L−1
n =0
∞
n =0
−1
∞
∑u
n=0
∞
−1
n
∑v
n =0
n
− k1 L
+ k 3 L−1
∞
∑A
n =0
∞
n
∑B
n=0
−1
− k2 L
n
− k 4 L−1
∞
∑B
n =0
∞
n
∑C
n=0
n
(2.23)
18
D. Venu Gopala Rao
Now, the iterates are determined in the following recursive way.
u(0) = u0
un + 1 = L−1 u n − k 1 L−1 An − k 2 L−1 B n , n = 0, 1, 2.....
v(0) = v0
vn + 1 = − k 0 L−1 v n + k 3 L−1 B n − k 4 L−1 C n , n = 0, 1, 2.....
(2.24)
We then define the solution of the initial value problem (2.18) as
n

uk ,
(u, v) =  nlim
∑
→∞
k =0

3
n
lim
n→∞
∑v
k =0
k



(2.25)
Application of Adomian Decomposition Method
Example 3.1
We first consider (2.1) with initial values u(0) = u0, v(0) = v0 and proceeding as in
Section 2.1.
We take f(u) = u2, g(u, v) = uv
(3.1)
The Adomian polynomials can be derived as follows:
An =
1 dn   ∞ n  
 f  ∑ λ u n  
n dx n   n =0
  λ =0
f(u) = u 2 =
∞
∑A
n =0
n
n>0
= (u 0 + u1 + u 2 + ......)
(3.2)
2
= (u0)2 + (2u0u1) + (2u0u2 + u12) + (2u0u3 + 2u1u2) + (2u0u4 + 2u1u3 + u22)
+ (2u0u5 + 2u1u4 + 2u2u3) + ……………..
(3.3)
Therefore, we get the following Adomian polynomials
A0 = u02
A1 = 2u0u1
A2 = 2u0u2 + u12
A3 = 2u0u3 + 2u1u2
A4 = 2u0u4 + 2u1u3 + u22
A5 = 2u0u5 + 2u1u4 + 2u2u3 and so on.
∞
g(u, v) = uv =
∑B
n =0
n
 ∞
  ∞ 
=  ∑ u n   ∑ vn 
 n =0   n =0 
(3.4)
A Study on Series Solutions of Two Species…
∞
=
n =0
n
Therefore Bn =
∑u v
k =0
k n −k

n
∑  ∑ u
k =0
k

vn− k 

, n = 0, 1, 2,..........
19
(3.5)
(3.6)
We have Adomian Polynomials
B0 = u0v0
B1 = u0v1 + u1v0
B2 = u0v2 + u1v1 + u2v0
B3 = u0v3 + u1v2 + u2v1+ u3v0
B4 = u0v4 + u1v3 + u2v2+ u3v1+ u4v0 and so on.
For numerical purpose, we take a1 = a2 = b1 = 1, b2 = 2, c1 =
(3.7)
1
1
, c2 =
in (2.1)
4
2
and initial values u0 = 1, v0 = 2. Therefore,
u0 = 1
v0 = 2
u1 = L−1 u 0 − β L−1 A0 − (1 − β ) L−1 B0 = τ
v1 = α L−1 B0 − α L−1 v 0 = 0
u2 = L−1 u1 − β L−1 A1 − (1 − β ) L−1 B1 = − 0.5τ 2
v2 = α L−1 B1 − α L−1v1 = τ 2
u3 = L−1 u 2 − β L−1 A2 − (1 − β ) L−1 B2 = − 0.167τ 3
v3 = α L−1 B3 − α L−1 v 2 = − 0.334τ 3
(3.8)
u4 = L−1 u 3 − β L−1 A3 − (1 − β ) L−1 B3 = 0.708τ 4
v4 = α L−1 B3 − α L−1v3 = 0.167τ 4
u5 = L−1 u 4 − β L−1 A4 − (1 − β ) L−1 B4 = − 0.241τ 5
v5 = α L−1 B4 − α L−1v 4 = 0.116τ 5
The rest of the terms of the decomposition series have been calculated using
Mathcad 7. Substituting these terms into (2.5), we obtain
u (τ ) = u 0 + u1 (τ ) + u 2 (τ ) + u 3 (τ ) + u 4 (τ ) + u 5 (τ ) + .................
u (τ ) = 1 + τ − 0.5τ 2 − 0.167τ 3 + 0.708τ 4 − 0.241τ 5 + ............
v(τ ) = v0 (τ ) + v1 (τ ) + v 2 (τ ) + v3 (τ ) + v 4 (τ ) + v5 (τ ) + .................
(3.9)
v(τ ) = 2 + τ 2 − 0.334τ 3 + 0.167τ 4 + 0.116τ 5 + ....................
(3.10)
Example 3.2
We now consider (2.23) with initial values u(0) = u0, v(0) = v0 and proceeding as
in section 2.2.
20
D. Venu Gopala Rao
We take f(u) = u2, g(u, v)= uv, h(v) = v2
(3.11)
The Adomian polynomials for f(u) = u2, g(u, v) = uv are just same as obtained in
(3.4), (3.7) in example (3.1) and
h(v) = v2 =
∞
∑C
n=0
n
= (v0 + v1 + v2 + ......)
2
(3.12)
= v02 + (2v0v1) + (2v0v1) + (2v0v2 + v12) + (2v0v3 + 2v1v2)
+ (2v0v4+ 2v1v3 + v22) + (2v0v5 + 2v1v4 + 2v2v3) + ……….
Therefore, we get the following Adomian polynomials
C0 = v02
C1 = 2v0v1
C2 = 2v0v2 + v12
C3 = 2v0v3 + 2v1 v2
C4 = 2v0v4 + 2v1v3 + v22
C5 = 2v0v5 + 2v1v4 + 2v2v3 and so on
(3.13)
For numerical evolution, we take u(0) = 1, v(0) = 2 and
1
1
a1 = a2 = b1 = 1, b2 = 2, c1 = , c2 =
in (2.21) and proceeding as in section 2.2,
4
2
we obtain
u0 = 1
v0 = 2
u1 = L−1 u 0 − k 1 L−1 A0 − k 2 L−1 B0 = − 0.25τ
v1 = − k 0 L−1 v 0 + k 3 L−1 B 0 − k 4 L−1 C 0 = − τ
u2 = L−1 u1 − k 1 L−1 A1 − k 2 L−1 B1 = − 0.25τ 2
v2 = − k 0 L−1 v1 + k 3 L−1 B1 − k 4 L−1 C1 = 0.3750 τ 2
u3 = L−1 u 2 − k 1 L−1 A2 − k 2 L−1 B 2 = − 0.1510τ 3
v3 = − k 0 L−1 v 2 + k 3 L−1 B 2 − k 4 L−1 C 2 = 0.0208τ 3
u4 = L u 3 − k 1 L A3 − k 2 L B3 = 0.0814τ
−1
−1
−1
(3.14)
4
v4 = − k 0 L−1 v 3 + k 3 L−1C 3 − k 4 L−1 C 3 = − 0.1563τ 4
u5 = L−1 u 4 − k 1 L−1 A4 − k 2 L−1 B4 = − 0.0411τ 5
v5 = − k 0 L−1 v 4 + k 3 L−1 B 4 −k 4 L−1 C 4 = 0.1577τ 5 and so on.
The rest of the terms of the decomposition series have been calculated using
Mathcad 7. Substituting these terms into (2.17)
u (τ ) = u 0 + u1 (τ ) + u 2 (τ ) + u 3 (τ ) + u 4 (τ ) + u 5 (τ ) + .................
A Study on Series Solutions of Two Species…
u (τ ) = 1 − 0.25τ − 0.25τ 2 − 0.510τ 3 + 0.0814τ 4 − 0.0411τ 5 + ........
ν (τ ) =ν 0 (τ ) + ν 1 (τ ) + ν 2 (τ ) + ν 3 (τ ) + ν 4 (τ ) + ν 5 (τ ) + .......
= 2 − τ + 0.3750τ 2 + 0.0208τ 3 − 0.1563τ 4 + 0.1577τ 5 .......
4
21
(3.15)
(3.16)
Homotopy Perturbation Method
The combination of the peturbation method and homotpy method is called the
homotopy perturbation method which has eliminated the limitations of the
traditional perturbation method. Consider the nonlinear differential equation.
M(u) –g(s) = 0, s ∈ S
(4.1)
with boundary conditions
 ∂u 
N  u,  = 0, s ∈ Γ
 ∂n 
where M is a differential operator, N is a boundary operator, g(s) is a known
analytic function. Γ is the boundary of the domains. Divide the operator M into
linear operator R and nonlinear operator Q. Therefore (4.1) can be expressed as
R(u) + Q(u) = g(s)
(4.2)
By the homotopy technique, constructing a homotopy
ν(s,p): S x [0,1] R, which satisfies
L(ν, p) = (1–p) [R(u) –R(u0) + p (M (ν) – g(s)] = 0 p ∈ (0, 1), s ∈ S.
L(ν,p) = R(u) –R(u0) + pR(u0) + p [M(ν) –g(s)] =0
(or)
(4.3)
where p ε [0,1] is an embedding parameter, u0 is an initial approximation of (4.1)
which satisfies the boundary conditions.
L(ν,0) = R(u) –R(u0) = 0
H(v,1) = M (ν) –g(s)] = 0
(4.4)
(4.5)
Changing p from zero to unity is similar to that of changing ν[s,p] from u0(s) to
u(s). By the method of HPM, we use the embedding parameter p as a small
parameter and assume that the solutions of (4.3) and (4.4) can be written as a
power series of p.
ν = ν0 + p ν1 +p2ν2 + p3ν3 + ….
(4.6)
Setting p 1 results inn the approximating solutions of (4.1) as
u = lim ν =ν 0 + ν 1 + ν 2 + ....
p →1
(4.7)
22
D. Venu Gopala Rao
The series (4.7) is convergent for most cases. However, the convergent rate
depends on the nonlinear operator M(v).
5
Application of Homotopy Perturbation Method
Example 5.1
We now solve (2.3) using HPM with initial conditions u(0)=1, ν(0) =2 as chosen
in example 3.1. We rewrite (2.3) as
[
du
= − p u (τ ) − β u 2 (τ ) − (1 − β )u (τ ) ν (τ )
dτ
dν
= − p[α ν (τ ) − α u (τ ) ν (τ )]
dτ
]
(5.1)
where p ∈ (0,1) is an embedding parameter. The solutions of (5.1) when
expressed as a power series is in the following form
u = u 0 + pu1 + p 2u 2 + ......
ν =ν 0 + pν 1 + p 2ν 2 + ......
(5.2)
(5.3)
Substituting (5.2), (5.3) in (5.1) and equating the like coefficients of p, we obtain
the following differential equations
u 0′ = 0
p :  v 0′ = 0
u (0) = u 0 , ν (0) = ν 0 .
u ′ = u − βu 2 − (1 − β )u ν
0
0
0 0
 1
 ′
p 1 : ν 1 = −α ν 0 + α u 0ν 0
u1 (0) = 0, ν 1 (0) = 0

u ′ = u − (1 − β ) (u ν + u ν ) − 2β u u
1
0 1
1 0
0 1
 2
′
2
p : ν 2 = α (u1ν 0 + u 0ν 1 ) − α v1

u 2 (0) = 0, ν 2 (0) = 0

0
A Study on Series Solutions of Two Species…
23
u ′ = u − (1 − β ) (u ν + u ν + u ν ) − 2β u u − βu 2
2
0 2
1 1
2 0
0 2
1
 3
′
3
p : ν 3 = α (u 0ν 2 + u1ν 1 + u 2 v 0 ) − α ν 2

u 3 (0) = 0, ν 3 (0) = 0

u ′ = u − (1 − β ) (u ν + u ν + u ν + u ν ) − 2 β u u − 2β u u
3
0 3
1 2
2 1
3 0
0 3
1 2
 4
′
p 4 : ν 4 = α (u 0ν 3 + u 2ν 1 + u1ν 2 + u 3 v0 ) − α ν 3

u 4 (0) = 0, ν 4 (0) = 0

(5.4)
2
′
u = u − β (2u u + 2u u + u ) − (1 − β )( u ν + u ν + u ν + u ν + u ν )
4
0 4
1 3
2
o 4
1 3
2 2
3 1
4 0
 5
′
5
p : ν 4 = α (u 0ν 4 + u1ν 3 + u 2 v 2 + u 3 v1 + u 4ν 0 ) − α ν 4

u 5 (0) = 0, ν 5 (0) = 0

where the coefficients of p denote differentiation with respect to p. Solving the
above system of equations for ui, νi (i = 1,2,3,4,5)
u0
u1
u2
u3
u4
u5
=1
=τ
= − 0.5τ 2
= − 0.167τ 3
= 0.708τ 4
= − 0.241τ 5
ν0 = 2
ν1 = 0
(5.5)
ν2 =τ 2
3
ν 3 = −0.334τ
ν 4 = 0.167τ 4 Substituting these ui, νi (i = 1,2…..5) into
ν 5 = 0.116τ 5 (5.2) and (5.3) respectively, we have
u = 1 + pτ − 0.5 p 2τ 2 − 0.167 p 3τ 3 + 0.708 p 4τ 4 − 0.241 p 5τ 5 − ......
v = 2 + p 2τ 2 − 0.334 p 3τ 3 + 0.167 p 4τ 4 + 0.116 p 5τ 5 + .......
(5.6)
(5.7)
letting p 1
u = 1 + τ − 0.5τ 2 − 0.167τ 3 + 0.708τ 4 − 0.241τ 5 + ....
ν = 2 + τ 2 − 0.334τ 3 + 0.167τ 4 + 0.116τ 5 + .......
(5.8)
(5.9)
which are exactly the same solutions obtained in (3.9) and (3.10) respectively
Example 5.2
We now solve (2.24) using HPM with initial conditions u(0) =1, ν(0) = 2 as taken
in example 3.2. We rewrite (2.24) as
[
du
= p u (τ ) − k1u 2 (τ ) − k 2 u (τ )ν (τ )
dτ
]
24
D. Venu Gopala Rao
[
dv
= p k 0ν (τ ) + k 3 u (τ )ν (τ ) − k 4ν 2 (τ )
dτ
]
(5.10)
where p ∈ (0, 1) is an embedding parameter and the solutions of (5.10) are
expressed as a power series as in (5.2) & (5.3).
Substituting (5.2), (5.3) in (5.10) and collecting the like coefficients p, we obtain
the following differential equations
p0 :
p1 :
p2 :
p3 :
p4 :
p5 :
u ′ = 0
 0
ν ′ = 0
 0
u (0) =1, ν (0) = 2.

u ′ = u − k u 2 − k u ν
0
1 0
2 0 0
 1
′
ν = −k ν + k u ν − k v 2
0 0
3
0 0
4 0
 1
u1 (0) = 0, ν 1 (0) = 0

u ′ = u − 2k u u − k (u ν + u ν )
1
1 0 1
2
0 1
1 0
 2
ν ′ = −k ν + k ( u ν + u ν ) − 2k v ν
0
1
3
0 1
1 0
4 0 1
 2
u 2 (0) = 0, ν 2 (0) = 0

u ′ = u − k (u 2 + 2u u ) − k (u ν + u ν + u v )
2
1
1
0 2
2
0 2
1 1
2 0
 3
ν ′ = − k ν + k (u ν + u ν + u ν ) − k (2 v v + v 2 )
0 2
3
0 2
1 1
2 0
4
0 2
1
 3
u 3 (0) = 0, ν 3 (0) = 0

u ′ = u − k (2u u + 2u u ) − k (u ν + u ν + u ν + u ν )
3
1
0 3
1 2
2
0 3
1 2
2 1
3 0
 4
ν ′ = − k ν + k (u ν + u v + u v + u ν ) − k (2v v + 2v v )
(5.11)
0 3
3
0 3
1 2
2 1
3 0
4
0 3
1 2
 4
u 4 (0) = 0, ν 4 (0) = 0

u ′ = u − k (2u u + 2u u + u 2 ) − k ( u ν + u ν + u ν + u ν + u ν )
4
1
4 0
3 1
2
2
o 4
1 3
2 2
3 1
4 0
 5
ν ′ = − k ν + k (u ν + u ν + u v + u v + u ν ) − k (2ν ν + 2ν ν + v 2 )
0 4
3
0 4
1 3
2 2
3 1
4 0
4
0 4
1 3
2
 5
u 5 (0) = 0, ν 5 (0) = 0

Thus, solving the above system of equations yields
A Study on Series Solutions of Two Species…
u0 = 1
u1 = − 0.25τ
25
ν0 = 2
ν 1 = −τ
u 2 = − 0.25τ 2 ν 2 = 0.3750τ 2
u 3 = 0.1510τ 3 ν 3 = 0.0208τ 3
u 4 = 0.0814τ 4
ν 4 = −0.1563τ 4
u 5 = − 0.0411τ 5
ν 5 = 0.1577τ 5
(5.12)
Substituting these ui, νi (1 = 1,2…..5) into (5.2) and (5.3) respectively, we have
u (τ ) =1 − 0.25 pτ − 0.25 p 2τ 2 − 0.1510 p 3τ 3 + 0.0814 p 4τ 4 − 0.0411p 5τ 5 + .... (5.13)
u (τ ) = 2 − pτ + 0.3750 p 2τ 2 + 0.0208 p 3τ 3 − 0.1563 p 4τ 4 + 0.1577 p 5τ 5 + ....... (5.14)
letting p 1, we get
u (τ ) =1 − 0.25τ − 0.25τ 2 − 0.1510τ 3 + 0.0814τ 4 − 0.0411τ 5 + ....
u (τ ) = 2 − τ + 0.3750τ 2 + 0.0208τ 3 − 0.1563τ 4 + 0.1577τ 5 + .......
(5.15)
(5.16)
which are exactly the same solutions as those obtained in (3.15) and (3.16) in
example 3.2, section 3.
6
Conclusion
In each of the examples described in the paper for the purpose of validation of the
work, we have solved the same problem of finding the series solutions of the two
species Lotka–Volterra prey-predator interaction species using two different
methods i.e. the Adomian Decomposition method & the Homotopy Perturbation
method. Though we obtain the same solutions for both methods, the advantage of
the Homotopy Perturbation method is its straightforwardness and calculation
procedure is simple when compared to the Adomian Decomposition method,
where the calculation of Adomian polynomials is tedious. Hence, we conclude
that the Homotopy Perturbation method is much easier and a more efficient
method than the Adomian Decomposition method. These are the main results of
the paper.
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26
D. Venu Gopala Rao
[4]
J.H. He, Homotopy perturbation technique, Computer Methods in Applied
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