# Solution of the Differential-Difference Equations by Optimal Homotopy Asymptotic Method

Authors: H. Ullah; S. Islam; M. Idrees; M. Fiza
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We applied a new analytic approximate technique, optimal homotopy asymptotic method (OHAM), for treatment of differential-difference equations (DDEs). To see the efficiency and reliability of the method, we consider Volterra equation in different form. It provides us with a convenient way to control the convergence of approximate solutions when it is compared with other methods of solution found in the literature. The obtained solutions show that OHAM is effective, simpler, easier, and explicit.


---

## Body

## 1. Introduction

In physical science the DDEs play a vital role in modeling of the complex physical phenomena. The DDEs models are used in vibration of particles in lattices, the flow of current in a network, and the pulses in biological chains. The models containing DDEs have been investigated by numerical techniques such as discretizations in solid state physics and quantum mechanics. In the last decades most of the research work has been done on DDEs. Levi and Yamilov [1, 2] attribute their work to classification of DDEs and connection of integrable partial differential equation (PDEs) and DDEs [3, 4]. In [5–8] the exact solutions of the DDEs have been studied. For the solution of the DDEs, Zou et al. extended the homotopy analysis method (HAM) [9, 10]. Wang et al. extended Adomian decomposition method (ADM) for solving nonlinear difference-differential equations (NDDEs) and got a good accuracy with the analytic solution [11]. Recently, Marinca et al. introduced optimal homotopy asymptotic method (OHAM) [12–16] for the solution of nonlinear problems. The validity of OHAM is independent of whether or not the nonlinear problems contain small parameters.The motivation of this paper is to extend OHAM for the solution of NDDEs. In [17–20] OHAM has been proved to be useful for obtaining an approximate solution of nonlinear boundary value problems of differential equations. In this work, we have proved that OHAM is also useful and reliable for the solution of NDDEs, hence showing its validity and great potential for the solution of NDDEs phenomenon in science and engineering.In the succeeding section, the basic idea of OHAM [12–16] is formulated for the solution of boundary value problems. In Section 3, the effectiveness of the modified formulation of OHAM for NDDEs has been studied.


## 2. Basic Mathematical Theory of OHAM for NDDE

Let us take OHAM to the following differential-difference equation:(1)
A
(
v
n
(
t
)
)
+
f
(
t
)
=
0
,
n
∈
Ω
,
with boundary conditions:
(2)
B
(
v
n
(
t
)
,
∂
v
n
(
t
)
∂
t
)
=
0
.
n
∈
Γ
,
where A is a differential operator, v
n
(
t
) is an unknown function, n and t denote spatial and temporal independent variables, respectively, Γ is the boundary of Ω, and f
n
(
t
) is a known analytic function. A can be divided into two parts L and N such that
(3)
A
=
L
+
N
,
where L is simpler part of the partial differential equation which is easier to solve and N contains the remaining part of A. So (1) can be written as
(4)
L
(
v
n
(
t
)
)
+
f
(
t
)
+
N
(
v
n
(
t
)
,
v
n
-
k
(
t
)
,
v
n
+
k
(
t
)
)
=
0
.
According to OHAM, one can construct an optimal homotopy ϕ
n
(
t
,
q
)
:
Ω
×
[
0,1
]
→
R satisfying
(5)
(
1
-
q
)
[
L
(
ϕ
n
(
t
,
q
)
)
+
f
(
t
)
]
=
H
(
q
)
[
L
(
ϕ
n
(
t
,
q
)
)
+
f
(
t
)
k
k
k
k
k
k
k
k
l
l
+
N
(
ϕ
n
(
t
,
q
)
,
ϕ
n
-
k
(
t
,
q
)
,
ϕ
n
+
k
(
t
,
q
)
)
]
,
(6)
B
(
ϕ
n
(
t
,
q
)
,
∂
ϕ
n
(
t
,
q
)
∂
t
)
=
0
,
where q
∈
[
0,1
] is an embedding parameter, ϕ
n
(
t
,
q
) is an unknown function, and H
(
q
) is a nonzero auxiliary function. The auxiliary function H
(
q
) is nonzero for q
≠
0 and H
(
0
)
=
0. Equation (5) is the structure of OHAM homotopy. Clearly we have
(7)
q
=
0
⟹
H
(
ϕ
n
(
t
;
0
)
,
0
)
k
=
L
(
ϕ
n
(
t
;
0
)
)
+
f
(
t
)
=
0
,
q
=
1
⟹
H
(
ϕ
n
(
t
;
1
)
,
1
)
k
=
H
(
1
)
{
A
(
ϕ
n
(
t
;
1
)
)
+
f
(
t
)
}
=
0
.
Obviously, when q
=
0 and q
=
1, we obtain
(8)
q
=
0
⟹
ϕ
n
(
t
,
0
)
=
v
n
0
(
t
)
,
q
=
1
⟹
ϕ
n
(
t
,
1
)
=
v
n
(
t
)
,
respectively. Thus, as q varies from 0 to 1, the solution ϕ
n
(
t
,
q
) varies from v
n
0
(
t
) to v
n
(
t
), where v
n
0
(
t
) is the zeroth order solution and can be obtained from (10).Expandingϕ
n
(
t
,
q
,
C
i
),   ϕ
n
-
k
(
t
,
q
,
C
i
),   ϕ
n
+
k
(
t
,
q
,
C
i
) by Taylors series and choosing H
(
q
) as given
(9)
k
k
k
k
H
(
q
)
=
q
C
1
+
q
2
C
2
+
q
3
C
3
+
⋯
+
q
m
C
m
,
ϕ
n
-
k
(
t
,
q
,
C
i
)
=
v
(
n
-
k
)
0
(
t
)
+
∑
m
=
1
∞
v
(
n
-
k
)
m
(
t
,
C
i
)
q
m
,
ϕ
n
(
t
,
q
,
C
i
)
=
v
(
n
)
0
(
t
)
+
∑
m
=
1
∞
v
(
n
)
m
(
t
,
C
i
)
q
m
,
ϕ
n
+
k
(
t
,
q
,
C
i
)
=
v
(
n
+
k
)
0
(
t
)
+
∑
m
=
1
∞
v
(
n
+
k
)
m
(
t
,
C
i
)
q
m
,
where C
i, i
=
1,2
,
3
,
… are constants and to be determined and m
∈
N.Now substituting (9) into (5) and equating the coefficient of like powers of q, we obtain
(10)
L
(
v
n
0
(
t
)
)
+
f
(
t
)
=
0
,
B
(
v
n
0
(
t
)
,
∂
v
n
0
(
t
)
∂
t
)
=
0
,
(11)
L
(
v
n
1
(
t
)
)
-
L
(
v
n
0
(
t
)
)
=
C
1
[
L
(
v
n
0
(
t
)
)
+
N
(
v
(
n
-
k
)
0
(
t
)
,
v
n
0
(
t
)
,
v
(
n
+
k
)
0
(
t
)
)
]
,
B
(
v
n
1
(
t
)
,
∂
v
n
1
(
t
)
∂
t
)
=
0
,
(12)
L
(
v
n
2
(
t
)
)
-
L
(
v
n
1
(
t
)
)
=
C
1
[
L
(
v
n
1
(
t
)
)
+
N
(
v
(
n
-
k
)
1
(
t
)
,
v
n
1
(
t
)
,
v
(
n
+
k
)
1
(
t
)
)
]
+
C
2
[
L
(
v
n
0
(
t
)
)
+
N
(
v
(
n
-
k
)
0
(
t
)
,
v
n
0
(
t
)
,
v
(
n
+
k
)
0
(
t
)
)
]
,
B
(
u
n
2
(
n
,
t
)
,
∂
u
n
2
(
n
,
t
)
∂
t
)
=
0
,
(13)
L
(
v
n
3
(
t
)
)
-
L
(
v
n
2
(
t
)
)
=
C
1
[
L
(
v
n
2
(
t
)
)
+
N
(
v
(
n
-
k
)
2
(
t
)
,
v
n
2
(
t
)
,
v
(
n
+
k
)
2
(
t
)
)
]
+
C
2
[
L
(
v
n
1
(
t
)
)
+
N
(
v
(
n
-
k
)
1
(
t
)
,
v
n
1
(
t
)
,
v
(
n
+
k
)
1
(
t
)
)
]
+
C
3
[
L
(
v
n
0
(
t
)
)
+
N
(
v
(
n
-
k
)
0
(
t
)
,
v
n
0
(
t
)
,
v
(
n
+
k
)
0
(
t
)
)
]
,
B
(
v
n
3
(
t
)
,
∂
v
n
3
(
t
)
∂
t
)
=
0
,
kkkk
⋮
(14)
L
(
v
n
j
(
t
)
)
-
L
(
v
n
(
j
-
1
)
(
t
)
)
=
∑
i
=
1
m
∑
j
=
m
-
1
0
C
i
[
L
(
v
n
j
(
t
)
)
k
k
k
k
k
k
k
k
k
k
k
k
k
l
+
N
(
v
n
j
(
t
)
,
v
(
n
-
k
)
j
(
t
)
,
v
(
n
+
k
)
j
(
t
)
)
]
,
B
(
v
n
j
(
t
)
,
∂
v
n
j
(
t
)
∂
t
)
=
0
.
We obtained the zeroth order solution and first, second, third, and the general order solutions by solving (10)–(14). In the same way the remaining solutions can be determined. It has been observed that the convergence of the series (9) depends on the auxiliary constants C
i. If it is convergent at  q
=
1, one has
(15)
v
n
(
t
;
C
i
)
=
v
n
0
(
x
,
t
)
+
∑
k
≥
1
v
n
k
(
t
;
C
i
)
.
Substituting (15) into (1), it results in the following expression for residual:
(16)
R
(
t
,
C
i
)
=
L
(
v
~
n
(
t
,
C
i
)
)
+
f
(
t
)
+
N
(
v
~
n
(
t
,
C
i
)
)
.
If R
(
t
,
C
i
)
=
0, then v
~
n
(
t
,
C
i
) is the exact solution of the problem. Generally it does not happen, especially in nonlinear problems.For the determinations of auxiliary constants,C
i, i
=
1,2
,
3
,
…, there are different methods like Galerkin’s method, Ritz method, least squares method, and collocation method. One can apply the method of least squares as follows:
(17)
J
(
C
i
)
=
∫
a
b
∫
Ω
R
2
(
t
,
C
i
)
d
n
d
t
,
where a and b are two values, depending on the nature of the given problem.The auxiliary constantsC
i can be optimally calculated as
(18)
∂
J
∂
C
1
=
∂
J
∂
C
2
=
∂
J
∂
C
3
⋯
∂
J
∂
C
m
=
0
.
The m
th order approximate solution can be obtained by these obtained constants.The convergence of OHAM is directly proportional to the number of optimal constantsC
i.


## 3. Application of OHAM to Difference-Differential Equations

To show the validity and effectiveness of OHAM formulation to difference-differential equation, we apply it to Voltera equations in different form.Model 1.
We consider Voltera equation of the form(19)
∂
v
n
(
t
)
∂
t
=
v
n
(
t
)
(
v
n
-
1
(
t
)
-
v
n
+
1
(
t
)
)
,
n
≻
0
,
t
∈
[
0
,
1
]
,
with initial conditions
(20)
v
n
(
0
)
=
n
.
The exact solution of (3.1) is given by
(21)
v
n
(
t
)
=
n
1
+
n
t
,
and we take
(22)
k
l
v
n
(
t
)
=
v
n
0
(
t
)
+
p
v
n
1
(
t
)
+
p
2
v
n
2
(
t
)
+
p
3
v
n
3
(
t
)
,
v
n
+
1
(
t
)
=
v
(
n
+
1
)
0
(
t
)
+
p
v
(
n
+
1
)
1
(
t
)
kkkkkkkk
+
p
2
v
(
n
+
1
)
2
(
t
)
+
p
3
v
(
n
+
1
)
3
(
t
)
,
v
n
-
1
(
t
)
=
v
(
n
-
1
)
0
(
t
)
+
p
v
(
n
-
1
)
1
(
t
)
kkkkkkkk
+
p
2
v
(
n
-
1
)
2
(
t
)
+
p
3
v
(
n
-
1
)
3
(
t
)
.
According to (1), we have
(23)
L
=
∂
v
n
(
t
)
∂
t
,
f
(
t
)
=
0
,
N
=
-
v
n
(
t
)
(
v
n
-
1
(
t
)
-
v
n
+
1
(
t
)
)
.
Using (19), (20), and (22) into (5) and using the method discussed in Section 2 lead to the following.
Zeroth Order Problem. Consider
(24)
∂
v
n
0
(
t
)
∂
t
=
0
,
with
(25)
v
n
0
(
0
)
=
n
,
from which we obtain
(26)
v
n
0
(
t
)
=
n
.
First Order Problem. One has
(27)
∂
v
n
1
(
t
)
∂
t
=
(
1
+
C
1
)
∂
v
n
0
(
t
)
∂
t
+
C
1
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
0
(
t
)
,
with
(28)
v
n
1
(
0
)
=
0
.
Its solution is
(29)
v
n
1
(
t
)
=
2
C
1
n
t
.
Second Order Problem. Consider
(30)
∂
v
n
2
(
t
)
∂
t
=
(
1
+
C
1
)
∂
v
n
1
(
t
)
∂
t
+
C
2
∂
v
n
0
(
t
)
∂
t
+
C
1
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
1
(
t
)
+
C
1
(
v
(
n
+
1
)
1
(
t
)
-
v
(
n
-
1
)
1
(
t
)
)
v
n
0
(
t
)
+
C
2
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
0
(
t
)
,
with
(31)
v
n
2
(
0
)
=
0
,
(32)
v
n
2
(
t
)
=
2
(
C
1
n
t
+
C
1
2
n
t
+
C
2
n
t
+
2
C
1
2
n
t
2
)
.
Third Order Problem. One has
(33)
∂
v
n
3
(
t
)
∂
t
=
(
1
+
C
1
)
∂
v
n
2
(
t
)
∂
t
+
C
2
∂
v
n
1
(
t
)
∂
t
+
C
3
∂
v
n
0
(
t
)
∂
t
+
C
1
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
2
(
t
)
+
C
1
(
v
(
n
+
1
)
1
(
t
)
-
v
(
n
-
1
)
1
(
t
)
)
v
n
1
(
t
)
+
C
2
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
1
(
t
)
+
C
1
(
v
(
n
+
1
)
2
(
t
)
-
v
(
n
-
1
)
2
(
t
)
)
v
n
0
+
C
2
(
v
(
n
+
1
)
1
(
t
)
-
v
(
n
-
1
)
1
(
t
)
)
v
n
0
(
t
)
+
C
3
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
0
(
t
)
,
with
(34)
v
n
3
(
0
)
=
0
,
(35)
v
n
3
(
t
)
=
2
3
(
3
C
1
n
t
+
6
C
1
2
n
t
+
3
C
1
3
n
t
+
3
C
2
n
t
+
3
C
3
n
t
k
k
k
k
+
12
C
1
2
n
t
2
+
12
C
1
3
n
t
2
+
6
C
1
C
2
n
t
2
+
4
C
1
3
n
t
3
)
.
Adding (26), (29), (32), and (35), we obtain
(36)
v
n
(
t
)
=
v
n
0
(
t
)
+
v
n
1
(
t
,
C
1
)
+
v
n
2
(
t
,
C
1
,
C
2
)
+
v
n
3
(
t
,
C
1
,
C
2
,
C
3
)
,
(37)
v
n
(
t
)
=
n
+
2
C
1
n
t
+
2
(
C
1
n
t
+
C
1
2
n
t
+
C
2
n
t
+
2
C
1
2
n
t
2
)
+
2
3
(
3
C
1
n
t
+
6
C
1
2
n
t
+
3
C
1
3
n
t
+
3
C
2
n
t
+
3
C
3
n
t
k
k
k
k
k
k
+
12
C
1
2
n
t
2
+
12
C
1
3
n
t
2
+
6
C
1
C
2
n
t
2
+
4
C
1
3
n
t
3
)
.
For the computation of the constants C
1, C
2, and C
3 applying the method of least square mentioned in (16)–(18), we get
(38)
C
1
=
-
0.23642908394894221
,
C
2
=
0.2344713680405294
,
C
3
=
-
0.4015040325725273
for
n
=
1
.
Putting these values in (37), we obtained the approximate solution of the form
(39)
v
n
(
t
)
=
n
-
0.472858
n
t
+
2
(
0.05
n
t
+
0.111797
n
t
2
)
+
2
3
(
-
0.914641
n
t
+
0.179576
n
t
2
-
0.0528643
n
t
3
)
.
The exact solution of the problem (3.1) is [11]
(40)
v
n
(
t
)
=
n
1
+
n
t
.
The Adomian solution with Padé approximation of (3.1) is given by [11]
(41)
v
n
(
t
)
=
n
1
+
n
t
.Model 2.
Now consider another form of Volterra equation(42)
∂
v
n
(
t
)
∂
t
=
v
n
(
t
)
(
v
n
-
1
(
t
)
-
v
n
+
1
(
t
)
)
+
v
n
(
t
)
(
v
n
-
2
(
t
)
-
v
n
+
2
(
t
)
)
,
n
≻
0
,
t
∈
[
0
,
1
]
,
with initial conditions
(43)
v
n
(
0
)
=
n
.
The exact solution of (42) is given by
(44)
v
n
(
t
)
=
n
1
+
6
t
.
We take
(45)
v
n
(
t
)
=
v
n
0
(
t
)
+
p
v
n
1
(
t
)
+
p
2
v
n
2
(
t
)
+
p
3
v
n
3
(
t
)
,
v
n
+
1
(
t
)
=
v
(
n
+
1
)
0
(
t
)
+
p
v
(
n
+
1
)
1
(
t
)
+
p
2
v
(
n
+
1
)
2
(
t
)
+
p
3
v
(
n
+
1
)
3
(
t
)
,
v
n
-
1
(
t
)
=
v
(
n
-
1
)
0
(
t
)
+
p
v
(
n
-
1
)
1
(
t
)
+
p
2
v
(
n
-
1
)
2
(
t
)
+
p
3
v
(
n
-
1
)
3
(
t
)
,
v
n
+
2
(
t
)
=
v
(
n
+
2
)
0
(
t
)
+
p
v
(
n
+
2
)
1
(
t
)
+
p
2
v
(
n
+
2
)
2
(
t
)
+
p
3
v
(
n
+
2
)
3
(
t
)
,
v
n
-
2
(
t
)
=
v
(
n
-
2
)
0
(
t
)
+
p
v
(
n
-
2
)
1
(
t
)
+
p
2
v
(
n
-
2
)
2
(
t
)
+
p
3
v
(
n
-
2
)
3
(
t
)
.
According to (1), we have
(46)
L
=
∂
v
n
(
t
)
∂
t
,
f
(
t
)
=
0
,
N
=
-
v
n
(
t
)
(
v
n
-
1
(
t
)
-
v
n
+
1
(
t
)
)
kk
-
v
n
(
t
)
(
v
n
-
2
(
t
)
-
v
n
+
2
(
t
)
)
.
Using (42), (43), and (45) into (5) and using the method discussed in Section 2 lead to the following.
Zeroth Order Problem. Consider
(47)
∂
v
n
0
(
t
)
∂
t
=
0
,
with
(48)
v
n
0
(
0
)
=
n
,
from which we obtain
(49)
v
n
0
(
t
)
=
n
.
First Order Problem. One has the following:
(50)
∂
v
n
1
(
t
)
∂
t
=
(
1
+
C
1
)
∂
v
n
0
(
t
)
∂
t
+
C
1
(
v
(
n
+
2
)
0
(
t
)
-
C
1
v
(
n
-
2
)
0
(
t
)
)
v
n
0
(
t
)
+
C
1
(
v
(
n
+
1
)
0
(
t
)
-
C
1
v
(
n
-
1
)
0
(
t
)
)
v
n
0
(
t
)
,
with
(51)
v
n
1
(
0
)
=
0
.
Its solution is
(52)
v
n
1
(
t
)
=
6
C
1
n
t
.
Second Order Problem. Consider(53)
∂
v
n
2
(
t
)
∂
t
=
(
1
+
C
1
)
∂
v
n
1
(
t
)
∂
t
+
C
2
∂
v
n
0
(
t
)
∂
t
+
C
1
(
v
(
n
+
2
)
1
(
t
)
-
v
(
n
-
2
)
1
(
t
)
)
v
n
0
(
t
)
+
C
2
(
v
(
n
+
2
)
0
(
t
)
-
v
(
n
-
2
)
0
(
t
)
)
v
n
0
+
C
1
(
v
(
n
+
2
)
0
(
t
)
-
v
(
n
-
2
)
0
(
t
)
)
v
n
1
(
t
)
+
C
1
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
1
(
t
)
+
C
2
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
0
(
t
)
+
C
1
(
v
(
n
+
1
)
1
(
t
)
-
v
(
n
-
1
)
1
(
t
)
)
v
n
0
(
t
)
,
with
(54)
v
n
2
(
0
)
=
0
,
(55)
v
n
2
(
t
)
=
6
n
t
(
C
1
+
C
2
+
C
1
2
(
1
+
6
t
)
)
.
Third Order Problem. One has
(56)
∂
v
n
3
(
t
)
∂
t
=
(
1
+
C
1
)
∂
v
n
2
(
t
)
∂
t
+
C
2
∂
v
n
1
(
t
)
∂
t
+
C
3
∂
v
n
0
(
t
)
∂
t
+
C
1
(
v
(
n
+
2
)
2
(
t
)
-
v
(
n
-
2
)
2
(
t
)
)
v
n
0
(
t
)
+
C
2
(
v
(
n
+
2
)
1
(
t
)
-
v
(
n
-
2
)
1
(
t
)
)
v
n
0
(
t
)
+
C
3
(
v
(
n
+
2
)
0
(
t
)
-
v
(
n
-
2
)
0
(
t
)
)
v
n
0
(
t
)
+
C
1
(
v
(
n
+
2
)
1
(
t
)
-
v
(
n
-
2
)
1
(
t
)
)
v
n
1
(
t
)
+
C
2
(
v
(
n
+
2
)
0
(
t
)
-
v
(
n
-
2
)
0
(
t
)
)
v
n
1
(
t
)
+
C
1
(
v
(
n
+
2
)
0
(
t
)
-
v
(
n
-
2
)
0
(
t
)
)
v
n
2
(
t
)
+
C
1
(
v
(
n
+
1
)
2
(
t
)
-
v
(
n
-
1
)
2
(
t
)
)
v
n
0
(
t
)
+
C
1
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
2
(
t
)
+
C
1
(
v
(
n
+
1
)
1
(
t
)
-
v
(
n
-
1
)
1
(
t
)
)
v
n
1
(
t
)
+
C
2
(
v
(
n
+
1
)
0
(
t
)
-
v
(
n
-
1
)
0
(
t
)
)
v
n
1
(
t
)
+
C
1
(
v
(
n
+
1
)
2
(
t
)
-
v
(
n
-
1
)
2
(
t
)
)
v
n
0
(
t
)
+
C
2
(
v
(
n
+
1
)
1
-
v
(
n
-
1
)
1
)
v
n
0
(
n
,
t
)
+
C
3
(
v
(
n
+
1
)
0
(
n
,
t
)
-
v
(
n
-
1
)
0
(
n
,
t
)
)
v
n
0
(
t
)
,
with
(57)
v
n
3
(
0
)
=
0
,
(58)
v
n
3
(
t
)
=
6
n
t
(
C
2
+
C
3
+
2
C
1
2
(
1
+
6
t
)
k
k
k
k
k
+
C
1
3
(
1
+
6
t
)
2
+
C
1
(
1
+
2
C
2
(
1
+
6
t
)
)
)
.
Adding (49), (52), (55), and (58), we obtain
(59)
v
n
(
t
)
=
v
n
0
(
t
)
+
v
n
1
(
t
,
C
1
)
+
v
n
2
(
t
,
C
1
,
C
2
)
+
v
n
3
(
t
,
C
1
,
C
2
,
C
3
)
,
v
n
3
(
t
)
=
n
+
6
C
1
n
t
+
6
n
t
(
C
1
+
C
2
+
C
1
2
(
1
+
6
t
)
)
+
6
n
t
(
C
1
3
C
2
+
C
3
+
2
C
1
2
(
1
+
6
t
)
+
C
1
3
(
1
+
6
t
)
2
kkkkkkk
+
C
1
3
(
1
+
6
t
)
2
+
C
1
(
1
+
2
C
2
(
1
+
6
t
)
)
)
.
For the computation of the constants C
1, C
2, and C
3 applying the method of least square mentioned in (16)–(18), we get
(60)
C
1
=
-
0.23640805995815892
,
C
2
=
-
0.039496242739004025
,
C
3
=
-
0.012135699742215487
for
n
=
1
.
The exact solution of problem (42) is [9]
(61)
v
n
(
t
)
=
n
1
+
6
t
.
The HAM solution of (42) is given by [9]
(62)
v
n
(
t
)
=
(
n
+
t
)
+
h
(
n
+
1
)
t
+
3
h
t
2
+
t
h
(
1
+
h
)
(
1
+
6
n
)
+
(
36
h
2
n
+
6
h
2
+
3
h
)
t
2
+
18
h
2
t
3
.


## 4. Results and Discussions

The formulation presented in Section2 provides highly accurate solutions for the problems demonstrated in Section 3. We have used Mathematica 7 for most of our computational work. Table 1 and Figure 1 give the comparisons of OHAM with ADM and exact solutions for Model 1. Also the absolute errors at different values of n at t
=
1 are given (Figure 4). Figure 3 shows the comparisons of OHAM results with HAM, exact at different values of n at t
=
1 for Model 2. The convergence of approximate orders to exact solutions is given in Table 2 for Model 1 and Figure 5 for Model 2. The residuals have been plotted for Model 1 in Figure 2 and for Model 2 in Figure 5. We have concluded that the results obtained by OHAM are identical to the results obtained by ADM, HAM, and exact. OHAM converges rapidly with increasing of the order of approximation.Table 1
Comparisons of OHAM, exact, and ADM results at different values ofn at  t
=
1.
n
OHAM
Exact
ADM
Abs error
1
0.333333
0.333333
0.333333
7.17103
×
10
-
7
2
0.666665
0.666665
0.666665
1.43421
×
10
-
6
3
0.999998
0.999998
0.999998
2.15131
×
10
-
6
4
1.33333
1.33333
1.33333
2.86841
×
10
-
6
5
1.66666
1.66666
1.66666
3.58551
×
10
-
6
6
2
2
2
4.30262
×
10
-
6
7
2.33333
2.33333
2.33333
5.01972
×
10
-
6
8
2.66666
2.66666
2.66666
5.73682
×
10
-
6
9
2.99999
2.99999
2.99999
6.45392
×
10
-
6
10
3.33333
3.33333
3.33333
7.17103
×
10
-
6Table 2
Comparisons of first, second, and third order and exact solutions at different values ofn  at t
=
1.
n
First order
Second order
Third order
Exact sol.
1
0.527142
0.858619
0.333333
0.333333
2
1.05428
1.71724
0.666665
0.666665
3
1.58143
2.57586
0.999998
0.999998
4
2.10857
3.43447
1.33333
1.33333
5
2.63571
4.29309
1.66666
1.66666
6
3.16285
5.15171
2
2
7
3.68999
6.01033
2.33333
2.33333
8
4.21713
6.86895
2.66666
2.66666
9
4.74428
7.72757
2.99999
2.99999
10
5.27142
8.58619
3.33333
3.33333Figure 1
Comparison of OHAM, exact, and ADM solutions att
=
1.Figure 2
Residual of (19) at t
=
1.Figure 3
Comparisons of OHAM, HAM and exact solutions att
=
1, h
=
-
1.Figure 4
Comparison of first, second, third of OHAM, HAM and exact solutions att
=
1, h
=
-
1.Figure 5
Residual of (42) at t
=
1.


## 5. Conclusion

In this work, we have seen the effectiveness of OHAM [12–16] to differential-difference equations. By applying the basic idea of OHAM to differential-difference equations, we found it simpler in applicability, more convenient to control convergence, and involved less computational overhead. Therefore, OHAM shows its validity and great potential for the differential-difference equations arising in science and engineering.



---

*Source: 520467-2014-05-29.xml*