# Application of Optimal Homotopy Asymptotic Method to Burger Equations

Authors: R. Nawaz; H. Ullah; S. Islam; M. Idrees
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We apply optimal homotopy asymptotic method (OHAM) for finding approximate solutions of the Burger's-Huxley and Burger's-Fisher equations. The results obtained by proposed method are compared to those of Adomian decomposition method (ADM) (Ismail et al., (2004)). As a result it is concluded that the method is explicit, effective, and simple to use.


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## Body

## 1. Introduction

Nonlinear phenomena play a vital role in applied mathematics, physics, and engineering sciences. The Burger’s equation models efficiently certain problems of a fluid flow nature, in which either shocks or viscous dissipation is a significant factor. It can be used as a model for any nonlinear wave propagation problem subject to dissipation [1]. The first steady-state solutions of Burger equation were given by Young et al. [2] However, the equation gets its name from the extensive research of Burger’s [3]. The generalized Burger’s-Huxley introduced by Satsuma shows a prototype model for describing the communication among reaction mechanisms, convection effects, and diffusion transports [4]. Burger-Fisher equation has significant applications in various fields of applied mathematics and has physical applications such as gas dynamic, traffic flow, convection effect, and diffusion transport [5–12]. Marinca and Herişanu et al. introduced a new semianalytic method OHAM for approximate solution of nonlinear problems of thin film flow of a fourth-grade fluid down a vertical cylinder. In progression of papers Marinca and Herişanu et al. have applied this method for the solution of nonlinear equations arising in the steady state flow of a fourth-grade fluid past a porous plate and for the solution of nonlinear equations arising in heat transfer [13–15]. The method has been applied by a number of researchers for solution of ordinary and partial differential equations [16–21]. The motivation of this paper is to show the effectiveness of OHAM for the solution of Burger’s-Huxley and Burger’s-Fisher equations. We consider Burger’s-Huxley equation of the form
(1)∂u(x,t)∂t+αuδ(x,t)∂u(x,t)∂x-∂2u(x,t)∂x2-βu(x,t)(1-uδ(x,t))(uδ(x,t)-γ)=0,βu(x,t)55(1-uδ(x,t))∀0≤x≤1,t≥0
and Burger’s-Fisher equation of the form
(2)∂u(x,t)∂t+αuδ(x,t)∂u(x,t)∂x-∂2u(x,t)∂x2-βu(x,t)(1-uδ(x,t))=0,5555555-βu(x1)∀0≤x≤1,t≥0,
where α, β, γ, and δ are parameters and β≥0, δ≥0, γ∈(0,1).The present paper is divided into three sections. In Section2 fundamental mathematical theory of OHAM is presented. In Section 3 comparisons are made between the results of the proposed method and HAM for Burger’s-Huxley. In Section 4 solution of Burger’s-Fisher equation is presented, and absolute error of approximate solution of proposed method is compared with approximate solution of HAM. In all cases the proposed method yields better results than those of ADM.


## 2. Fundamental Theory of OHAM

Here we start by describing the basic idea of OHAM. Consider the partial differential equation of the form:(3)ℒ(u(x,t))+𝒩(u(x,t))+ℊ(x,t)=0,x∈Ω,(4)ℬ(u,∂u∂t)=0,
where ℒ is a linear operator and 𝒩 is nonlinear operator. ℬ is boundary operator, u(x,t) is an unknown function, and x and t denote spatial and time variables, respectively; Ω is the problem domain and g(x,t) is a known function.According to the basic idea of OHAM, one can construct the optimal homotopyψ(x,t;q):Ω×[0,1]→R which satisfies
(5)(1-q){ℒ(ψ(x,t;q))+g(x,t)}=ℋ(q){ℒ(ψ(x,t;q))+𝒩(ψ(x,t;q))+ℊ(x,t)},
where q∈[0,1] is an embedding parameter, H(q) is a nonzero auxiliary function for q≠0, H(0)=0. Equation (3) is called optimal homotopy equation. Clearly, we have
(6)q=0⟹ℋ(ψ(x,t;0),0)=ℒ(ψ(x,t;0))+ℊ(𝓍,t)=0,(7)q=1⟹ℋ(ψ(x,t;1),1)=ℋ(1){ℒ(ψ(x,t;q))+𝒩(ψ(x,t;q))+ℊ(x,t)}=0.
Clearly, when q=0 and q=1, it holds that ψ(x,t;0)=u0(x,t) and ψ(x,t;1)=u(x,t), respectively. Thus, as q varies from 0 to 1, the solution ψ(x,t;q) approaches from u0(x,t) to u(x,t), where u0(x,t) is obtained from (3) for q=0:
(8)ℒ(u0(x,t))+g(x,t)=0,ℬ(u0,∂u0∂t)=0.
Next, we choose auxiliary function H(q) in the form
(9)ℋ(q)=qC1+q2C2+⋯.
Here C1,C2,… are constants to be determined later.To get an approximate solution, we expandψ(x,t;q,Ci) in Taylor’s series about q in the following manner:
(10)ψ(x,t;q,Ci)=u0(x,t)+∑k=1∞uk(x,t;Ci)qk,i=1,2,….
Substituting (10) into (4) and equating the coefficient of like powers of q, we obtain Zeroth-order problem, given by (6), the first- and second-order problems are given by (11)-(12), respectively, and the general governing equations for uk(x,t) are given by (13):
(11)ℒ(u1(x,t))=C1𝒩0(u0(x,t)),ℬ(u1,∂u1∂t)=0,(12)ℒ(u2(x,t))-ℒ(u1(x,t))=C2𝒩0(u0(x,t))+C1[ℒ(u1(x,t))+𝒩1(u0(x,t),u1(x,t))],ℬ(u2,∂u2∂t)=0,(13)ℒ(uk(x,t))-ℒ(uk-1(x,t))=Ck𝒩0(u0(x,t))+∑i=1k-1Ci[ℒ(uk-i(x,t))+𝒩k-i+∑i=1k-1Ci×(u0(x,t),u1(x,t),…,uk-i(x,t))],ℬ(uk,∂uk∂t)=0,k=2,3,…,
where 𝒩k-i(u0(x,t),u1(x,t),…,uk-i(x,t)) is the coefficient of qk-i in the expansion of 𝒩(ψ(x,t;q)) about the embedding parameter q:
(14)𝒩(ψ(x,t;q,Ci))=𝒩0(u0(x,t))+∑k≥1𝒩k(u0,u1,u2,…,uk)qk.
Here uk for k≥0 are set of linear equations with the linear boundary conditions, which can be easily solved.The convergence of the series in (10) depends upon the auxiliary constants C1,C2,…. If it is convergent at q=1, one has:
(15)u~(x,t;Ci)=u0(x,t)+∑k≥1uk(x,t;Ci).
Substituting (15) into (1) results in the following expression for residual:
(16)ℛ(x,t;Ci)=ℒ(u~(x,t;Ci))+ℊ(x,t)+𝒩(u~(x,t;Ci)).
If R(x,t;Ci)=0, then u~(x,t;Ci) will be the exact solution.For computing the auxiliary constants,Ci, i=1,2,…,m, there are many methods like Galerkin’s Method, Ritz Method, Least Squares Method, and Collocation Method to find the optimal values of Ci, i=1,2,3,…, One can apply the Method of Least Squares as
(17)ℐ(Ci)=∫0t∫Ωℛ2(x,t,Ci)dxdt,
where R is the residual, R(x,t;Ci)=L(u~(x,t;Ci))+g(x,t)+N(u~(x,t;Ci)), and
(18)∂ℐ∂C1=∂ℐ∂C2=⋯=∂ℐ∂Cm=0.
The constants Ci can also be determined by another method as
(19)ℛ(h1;Ci)=ℛ(h2;Ci)=⋯=ℛ(hm;Ci)=0,ℛ(h1;Ci)=ℛ(h2;Ci)=⋯1.=i=1,2,…,m,
at any time t, where hi∈Ω. The convergence depends upon constants C1,C2,…, can be optimally identified and minimized by (18).


## 3. Application of OHAM

In this section we apply OHAM for the two problems: the first is the Burger’s-Huxley equation (1) and the second is the Burger’s-Fisher equation (2).


### 3.1. Application of OHAM for Burger’s-Huxley Equation

Let us consider Burger’s-Huxley equation of form (1):
(20)∂u(x,t)∂t+αuδ(x,t)∂u(x,t)∂x-∂2u(x,t)∂x2-βu(x,t)(1-uδ(x,t))(uδ(x,t)-γ)=0,-βu(x,t)5555(1-uδ(x,t))∀0≤x≤1,t≥0,
Subject to constant initial condition
(21)u(x,0)=(0.5γ+0.5γtanh(A1x))1/δ.
The exact solution of (26) with given condition is given by
(22)u(x,0)=(0.5γ+0.5γtanh(A1(x-A2t)))1/δ,
where
(23)A1=-αδ+δα2+4β(1+δ)4(1+δ)γ,A2=αγ(1+δ)-(1+δ-γ)(-α+α2+4β(1+δ))2(1+δ).
For computational work, we have taken α=1, β=1, δ=1, and γ=0.001 for various values of x and t.Following the basic idea of OHAM presented in preceding section we start withZeroth-Order Problem
(24)
∂
u
0
(
x
,
t
)
∂
t
=
0
,
u
0
(
x
,
0
)
=
(
0.0005
+
0.0005
tanh
(
0.00025
x
)
)
.
Its solution is
(25)u0(x,t)=(0.0005+0.0005tanh(0.00025x)).First-Order Problem
(26)
∂
u
1
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
0
(
x
,
t
)
∂
t
-
C
1
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
+
C
1
∂
2
u
0
(
x
,
t
)
∂
x
2
-
0.001
C
1
u
0
(
x
,
t
)
+
0.001
C
1
u
0
2
(
x
,
t
)
-
C
1
u
0
3
(
x
,
t
)
=
0
,
u
1
(
x
,
0
)
=
0
.
Its solution is
(27)u1(x,t)=-t(-2.4987500000000003×10-7C1=-t-6.25×10-11C1sech2(0.00025x)=-t+1.249999999999317=-t×10-10C1tanh(0.00025x)=-t-1.25×10-10C1sech2(0.00025x)=-t×tanh(0.00025x)=-t+2.49875×10-7C1tanh2(0.00025x)=-t-1.25×10-10C1tanh3(0.00025x)).Second-Order Problem
(28)
∂
u
2
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
1
(
x
,
t
)
∂
t
-
C
2
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
+
C
2
∂
2
u
0
(
x
,
t
)
∂
x
2
-
C
2
∂
u
0
(
x
,
t
)
∂
t
-
C
1
u
1
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
C
1
u
0
(
x
,
t
)
∂
u
1
(
x
,
t
)
∂
x
+
C
1
∂
2
u
1
(
x
,
t
)
∂
x
2
-
0.001
C
2
u
0
(
x
,
t
)
+
1.001
C
2
u
0
2
(
x
,
t
)
-
C
2
u
0
3
(
x
,
t
)
-
0.001
C
1
u
1
(
x
,
t
)
+
2.002
C
1
u
0
(
x
,
t
)
u
1
(
x
,
t
)
-
3
C
1
u
0
2
(
x
,
t
)
u
1
(
x
,
t
)
=
0
,
u
2
(
x
,
0
)
=
0
.
Its solution is
(29)u2(x,t,C1)=-t-2.4987500000000003×10-7C1-6.25×10-11C1sech2(0.00025x)+1.249999999999317×10-10C1×tanh(0.00025x)-1.25×10-10C1×sech2(0.00025x)tanh(0.00025x)+2.49875×10-7C1tanh2(0.00025x)-1.25×10-10C1tanh3(0.00025x).Third-Order Problem
(30)
∂
u
3
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
2
(
x
,
t
)
∂
t
-
C
3
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
+
C
3
∂
2
u
0
(
x
,
t
)
∂
x
2
-
C
3
∂
u
0
(
x
,
t
)
∂
t
-
C
2
u
1
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
C
2
u
0
(
x
,
t
)
∂
u
1
(
x
,
t
)
∂
x
+
C
2
∂
2
u
1
(
x
,
t
)
∂
x
2
-
C
2
∂
u
1
(
x
,
t
)
∂
t
-
C
1
u
2
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
C
1
u
1
(
x
,
t
)
∂
u
1
(
x
,
t
)
∂
x
+
C
1
∂
2
u
2
(
x
,
t
)
∂
x
2
-
C
1
u
0
(
x
,
t
)
∂
u
2
(
x
,
t
)
∂
x
-
0.001
C
3
u
0
(
x
,
t
)
+
1.001
C
3
u
0
2
(
x
,
t
)
-
C
3
u
0
3
(
x
,
t
)
-
0.001
C
2
u
1
(
x
,
t
)
-
3
C
2
u
0
2
(
x
,
t
)
u
1
(
x
,
t
)
+
1.001
C
1
u
1
2
(
x
,
t
)
-
3
C
1
u
0
(
x
,
t
)
u
1
2
(
x
,
t
)
-
0.001
C
1
u
2
(
x
,
t
)
+
2.002
C
1
u
0
(
x
,
t
)
u
2
(
x
,
t
)
-
3
C
1
u
0
2
(
x
,
t
)
u
2
(
x
,
t
)
+
2.002
C
2
u
0
(
x
,
t
)
u
1
(
x
,
t
)
=
0
,
u
3
(
x
,
0
)
=
0
.
Its solution is
(31)u3(x,t,C1,C2,C3)=(1(1+e0.5x)2tsech2(0.25x)=×(-0.0625C2-0.0625C3=×+C1(-0.0625+C1(-0.125-0.015625t)=×+C1+C2(-0.125-0.015625t)=×+C1-0.000651042C12(5.0718+t)=×+C1×(18.9282+t)C1)=×+(-24C2t+5C12(C12288C2+288C3+C15=×+×(288+576C2+C1(576-24t)5=×+×-24C2t-3C12(-6.58301+t)5=×+××(14.583+t)cosh(0.25x)5=×+×+(-96C2-96C3+C15=×+×+×(-96-192C2+C1(-192-24t)55=×+×+-24C2t+5C12(-7.396+t)5=×+×+×(2.596+t)192C2))sinh(0.25x))=×+-24C2t+5C12(-7.396+t)(2.596+t)))=××sinh(0.25x)=××(-0.000651042cosh(0.75x)=××-0.000651042sinh(0.75x)0.0625C2-0.0625C3))1(1+e0.5x)2).
Adding (25), (27), (29), and (31) we obtain
(32)u~(x,t,C1,C2)=u0(x,t)+u1(x,t,C1)+u2(x,t,C1,C2)+u3(x,t,C1,C2,C3).
For the calculations of the constants C1, C2, and C3 using the collocation method, we have computed thatC1=-1.0000010231545267,C2=-9.98159444155818×10-7,C3=-2.041939789528322×10-12.Putting the values of these constants into (32) the third order approximate solution using OHAM is
(33)u3(x,t)=0.5-5.2607845247854×10-6tsech2(0.25x)=+1(1+e0.5x)2=×(6.360090415451543×10-9=×+(1.3153067008227217×10-6=×++0.0006512058813062292t10-6)t=×+(1.2720180853076355×10-8=×+-0.002604823525224917t2)cosh(0.5x)=×+(6.36009042653818×10-9=×++(-1.3153067008227217×10-6=×+++0.0006512058813062292t10-6)t)cosh(x)=×+(1.2720180853076355×10-8=×+-0.002604823525224917t2)sinh(0.5x)=×+(6.36009042653818×10-9=×++(-1.3153067008227217×10-9=×+++0.0006512058813062292t10-8)t)sinh(x)=×-0.5tanh(0.25x)+0.007813813661895406t2=××sech2(0.25x)tanh(0.25x)+sech2(0.25x)=××(0.06250525442670962t10-8)).Table1 shows a comparison between OHAM solution and ADM solution for α=1 and γ=0.001. For α=0 (1) is reduced to the generalized Huxley equation which describes nerve pulse propagation in nerve fibers and wall motion in liquid crystals [22]. Tables 2 and 3 show a comparison between ADM solution and OHAM solution for α=0 and β=1 respectively. Table 4 shows absolute errors of OHAM solution for larger domain for α=0,1, β=1, and δ=1,2 respectively.Table 1
Comparison of absolute errors of OHAM and ADM [5] for   α=1, β=1, δ=1, and γ=0.001.
t
ADMx=0.1
OHAMx=0.1
ADMx=0.5
OHAMx=0.5
ADMx=0.9
OHAMx=0.9
0.05
1
.
93715
×
10
-
7
1.87406
×
10
-
8
1.9373
×
10
-
7
1.87406
×
10
-
8
1.93745
×
10
-
7
1.87406
×
10
-
8
0.1
3.87434
×
10
-
7
3.74812
×
10
-
8
3.87464
×
10
-
7
3.74812
×
10
-
8
3.87494
×
10
-
7
3.74812
×
10
-
8
1
3.87501
×
10
-
6
3.74812
×
10
-
7
3.87531
×
10
-
6
3.74812
×
10
-
7
3.87561
×
10
-
6
3.74812
×
10
-
7Table 2
Comparison of absolute errors obtained by OHAM and ADM [5] for α=0, β=1, δ=1, and γ=0.001.
t
ADMx=0.1
OHAMx=0.1
ADMx=0.5
OHAMfor x=0.5
ADMx=0.9
OHAMx=0.9
0.05
1
.
93715
×
10
-
7
2.49875
×
10
-
8
1.9373
×
10
-
7
2.49875
×
10
-
8
1.93745
×
10
-
7
2.49875
×
10
-
8
0.1
3.87434
×
10
-
7
4.9975
×
10
-
8
3.87464
×
10
-
7
4.9975
×
10
-
8
3.87494
×
10
-
7
4.9975
×
10
-
8
1
3.87501
×
10
-
6
4.9975
×
10
-
7
3.87531
×
10
-
6
4.9975
×
10
-
7
3.87561
×
10
-
6
4.9975
×
10
-
7Table 3
Comparison of absolute errors obtained by OHAM and ADM [5] for                                                α=0, β=1, δ=2, and γ=0.001.
t
Error ADM
Error OHAM
0.05
5.58836
×
10
-
7
2.7938
×
10
-
7
0.1
1.11766
×
10
-
6
5.58771
×
10
-
7
1
1.00741
×
10
-
5
5.5896
×
10
-
6Table 4
Absolute errors of OHAM forα=0,1, β=1, δ=1,2, and x=2.
t
α
=
1, β=1, δ=1, and γ=0.001
α
=
0, β=1, δ=1, and γ=0.001
α
=
0, β=1, δ=2, and γ=0.001
0.1
3.74812
×
10
-
8
2.49875
×
10
-
8
2.23403
×
10
-
6
0.2
7.49625
×
10
-
8
4.9975
×
10
-
8
4.46806
×
10
-
6
0.3
1.12444
×
10
-
7
7.49625
×
10
-
8
6.70209
×
10
-
6
0.4
1.49925
×
10
-
7
9.995
×
10
-
8
8.93612
×
10
-
6
0.5
1.87406
×
10
-
7
1.24937
×
10
-
7
1.11702
×
10
-
5
0.6
2.24887
×
10
-
7
1.49925
×
10
-
7
1.34042
×
10
-
5
0.7
2.62369
×
10
-
7
1.74912
×
10
-
7
1.56382
×
10
-
5
0.8
2.9985
×
10
-
7
1.999
×
10
-
7
1.78722
×
10
-
5
0.9
3.37331
×
1
0
-
7
2.24887
×
10
-
7
2.01063
×
10
-
5
1.0
3.74812
×
10
-
7
2.49875
×
10
-
7
2.23403
×
10
-
5


### 3.2. Application of OHAM for Burger’s-Fisher Equation

Consider the Burger’s-Fisher equation of form (2):
(34)∂u(x,t)∂t+αuδ(x,t)∂u(x,t)∂x-∂2u(x,t)∂x2-βu(x,t)(1-uδ(x,t))=0,∂u(x,t)∂t+αuδ(x,t)∀0≤x≤1,t≥0,
subject to constant initial condition
(35)u(x,0)=(0.5+0.5tanh(-αδ2(δ+1)x))1/δ,
with exact solution given by
(36)u(x,t)=0.5+0.5tanh×((x-(α(δ+1)+β(δ+1)α)t)-αδ2(δ+1)××(x-(α(δ+1)+β(δ+1)α)t))1/δ.
For computational work, we have taken α=0.001, β=0.001, and δ=1 for various values of x and t.Zeroth-Order Problem
(37)
∂
u
0
(
x
,
t
)
∂
t
=
0
,
u
0
(
x
,
0
)
=
(
0.5
+
0.5
tanh
(
-
0.001
4
x
)
)
.
Its solution is
(38)u0(x,t)=(0.5-0.5tanh(0.0014x)).First-Order Problem
(39)
∂
u
1
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
0
(
x
,
t
)
∂
t
-
0.001
C
1
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
0.001
C
1
u
0
(
x
,
t
)
(
1
-
u
0
(
x
,
t
)
)
+
C
1
∂
2
u
0
(
x
,
t
)
∂
x
2
=
0
,
u
1
(
x
,
0
)
=
0
.
Its solution is
(40)u1(x,t,C1)=-t(0.00025C1+6.25×10-8C1=-t×sech2(0.00025x)=-t-0.00025C1tanh2(0.00025x)).Second-Order Problem
(41)
∂
u
2
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
1
(
x
,
t
)
∂
t
+
0.001
C
2
u
0
(
x
,
t
)
×
(
1
-
u
0
(
x
,
t
)
)
+
C
2
∂
2
u
0
(
x
,
t
)
∂
x
2
-
C
2
∂
u
0
(
x
,
t
)
∂
t
-
0.001
C
2
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
0.001
C
1
u
1
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
0.001
C
1
u
0
(
x
,
t
)
×
∂
u
1
(
x
,
t
)
∂
x
+
C
1
∂
2
u
1
(
x
,
t
)
∂
x
2
-
0.001
C
1
u
1
(
x
,
t
)
-
0.002
C
1
u
0
(
x
,
t
)
u
1
(
x
,
t
)
=
0
,
u
2
(
x
,
0
)
=
0
.
Its solution is
(42)u2(x,t,C1,C2)=sech5(0.00025x)×(+C12(-0.000187547t)-0.000187547C1t-0.000187547C2t×+C12(-0.000187547t))cosh(0.00025x)+-0.0000625156C1t-0.0000625156C2t+C12(-0.0000625156t)+2.71051×10-20C1tsinh(0.00025x)+2.71051×10-20C12tsinh(0.00025x)+2.71051×10-20C2tsinh(0.00025x)+3.12656×10-8C12t2sinh(0.00025x)+3.12656×10-8C12t2sinh(0.00075x).
The third order approximate solution using OHAM is given by
(43)u~(x,t,C1,C2)=u0(x,t)+u1(x,t,C1)+u2(x,t,C1,C2)+u3(x,t,C1,C2,C3),
where u3(x,t,C1,C2,C3) is obtained in same lines as for first problem.For the calculations of the constantsC1, C2, and  C3 using the collocation method we have computed that
(44)C1=-5.928318703338053×10-7,C2=-465.9630543691778,C3=1.8651679832921486.The third order OHAM solution yields very encouraging results after being compared with Fourth order approximate solution by ADM [5].Table5 shows a comparison between OHAM solution and ADM solution for α=0.001, β=0.001,  and  δ=1. Table 6 compares between OHAM solution and ADM solution for α=1, β=1, and δ=2. Table 7 shows the reliability of OHAM for larger domain.Table 5
Comparison of absolute errors obtained by OHAM and ADM [5] for α=0.001, β=0.001, and δ=1.
t
ADM forx=0.1
OHAMx=0.1
ADMx=0.5
OHAMx=0.5
ADMx=0.9
OHAMx=0.9
0.005
9.68763
×
10
-
6
1.12257
×
10
-
7
9.68691
×
10
-
6
2.28888
×
10
-
7
9.68619
×
10
-
6
2.28888
×
10
-
7
0.001
1.93753
×
10
-
6
2.24513
×
10
-
8
1.93738
×
10
-
6
4.57775
×
10
-
8
1.93724
×
10
-
6
4.57775
×
10
-
8
0.01
1.93752
×
10
-
5
2.24514
×
10
-
7
1.93738
×
10
-
5
4.57777
×
10
-
7
1.93724
×
10
-
5
4.57777
×
10
-
7Table 6
Comparison of absolute errors obtained by OHAM and ADM [5] for  α=1, β=1, and δ=2.
t
ADM forx=0.1
OHAMx=0.1
ADM  x=0.5
OHAM  x=0.5
ADMx=0.9
OHAM  x=0.9
0.0005
1.40177
×
10
-
3
5.87633
×
10
-
5
1.34526
×
10
-
3
1.06736
×
10
-
5
1.27699
×
10
-
3
4.64718
×
10
-
5
0.0001
2.80396
×
10
-
4
1.17539
×
10
-
5
2.69094
×
10
-
4
5.33686
×
10
-
5
2.55438
×
10
-
4
9.29303
×
10
-
6
0.001
2.80301
×
10
-
3
1.17512
×
10
-
4
2.69000
×
10
-
3
1.06739
×
10
-
4
2.55346
×
10
-
3
9.296
×
10
-
4Table 7
Absolute errors of OHAM forx=2 and t∈[0.1,1].
t
α
=
0.001, β=0.001, and δ=1
α
=
0.001, β=0.001, and δ=2
0.1
1.98526
×
10
-
9
1.09926
×
10
-
5
0.2
3.20807
×
10
-
8
2.19856
×
10
-
5
0.3
1.63084
×
10
-
7
2.9789
×
10
-
5
0.4
5.16881
×
10
-
7
4.39726
×
10
-
5
0.5
1.26475
×
10
-
6
5.49666
×
10
-
5
0.6
2.62763
×
10
-
6
6.5961
×
10
-
5
0.7
4.87621
×
10
-
6
7.69557
×
10
-
5
0.8
8.33106
×
10
-
6
8.79507
×
10
-
5
0.9
1.33626
×
10
-
5
9.89461
×
10
-
5


## 3.1. Application of OHAM for Burger’s-Huxley Equation

Let us consider Burger’s-Huxley equation of form (1):
(20)∂u(x,t)∂t+αuδ(x,t)∂u(x,t)∂x-∂2u(x,t)∂x2-βu(x,t)(1-uδ(x,t))(uδ(x,t)-γ)=0,-βu(x,t)5555(1-uδ(x,t))∀0≤x≤1,t≥0,
Subject to constant initial condition
(21)u(x,0)=(0.5γ+0.5γtanh(A1x))1/δ.
The exact solution of (26) with given condition is given by
(22)u(x,0)=(0.5γ+0.5γtanh(A1(x-A2t)))1/δ,
where
(23)A1=-αδ+δα2+4β(1+δ)4(1+δ)γ,A2=αγ(1+δ)-(1+δ-γ)(-α+α2+4β(1+δ))2(1+δ).
For computational work, we have taken α=1, β=1, δ=1, and γ=0.001 for various values of x and t.Following the basic idea of OHAM presented in preceding section we start withZeroth-Order Problem
(24)
∂
u
0
(
x
,
t
)
∂
t
=
0
,
u
0
(
x
,
0
)
=
(
0.0005
+
0.0005
tanh
(
0.00025
x
)
)
.
Its solution is
(25)u0(x,t)=(0.0005+0.0005tanh(0.00025x)).First-Order Problem
(26)
∂
u
1
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
0
(
x
,
t
)
∂
t
-
C
1
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
+
C
1
∂
2
u
0
(
x
,
t
)
∂
x
2
-
0.001
C
1
u
0
(
x
,
t
)
+
0.001
C
1
u
0
2
(
x
,
t
)
-
C
1
u
0
3
(
x
,
t
)
=
0
,
u
1
(
x
,
0
)
=
0
.
Its solution is
(27)u1(x,t)=-t(-2.4987500000000003×10-7C1=-t-6.25×10-11C1sech2(0.00025x)=-t+1.249999999999317=-t×10-10C1tanh(0.00025x)=-t-1.25×10-10C1sech2(0.00025x)=-t×tanh(0.00025x)=-t+2.49875×10-7C1tanh2(0.00025x)=-t-1.25×10-10C1tanh3(0.00025x)).Second-Order Problem
(28)
∂
u
2
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
1
(
x
,
t
)
∂
t
-
C
2
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
+
C
2
∂
2
u
0
(
x
,
t
)
∂
x
2
-
C
2
∂
u
0
(
x
,
t
)
∂
t
-
C
1
u
1
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
C
1
u
0
(
x
,
t
)
∂
u
1
(
x
,
t
)
∂
x
+
C
1
∂
2
u
1
(
x
,
t
)
∂
x
2
-
0.001
C
2
u
0
(
x
,
t
)
+
1.001
C
2
u
0
2
(
x
,
t
)
-
C
2
u
0
3
(
x
,
t
)
-
0.001
C
1
u
1
(
x
,
t
)
+
2.002
C
1
u
0
(
x
,
t
)
u
1
(
x
,
t
)
-
3
C
1
u
0
2
(
x
,
t
)
u
1
(
x
,
t
)
=
0
,
u
2
(
x
,
0
)
=
0
.
Its solution is
(29)u2(x,t,C1)=-t-2.4987500000000003×10-7C1-6.25×10-11C1sech2(0.00025x)+1.249999999999317×10-10C1×tanh(0.00025x)-1.25×10-10C1×sech2(0.00025x)tanh(0.00025x)+2.49875×10-7C1tanh2(0.00025x)-1.25×10-10C1tanh3(0.00025x).Third-Order Problem
(30)
∂
u
3
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
2
(
x
,
t
)
∂
t
-
C
3
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
+
C
3
∂
2
u
0
(
x
,
t
)
∂
x
2
-
C
3
∂
u
0
(
x
,
t
)
∂
t
-
C
2
u
1
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
C
2
u
0
(
x
,
t
)
∂
u
1
(
x
,
t
)
∂
x
+
C
2
∂
2
u
1
(
x
,
t
)
∂
x
2
-
C
2
∂
u
1
(
x
,
t
)
∂
t
-
C
1
u
2
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
C
1
u
1
(
x
,
t
)
∂
u
1
(
x
,
t
)
∂
x
+
C
1
∂
2
u
2
(
x
,
t
)
∂
x
2
-
C
1
u
0
(
x
,
t
)
∂
u
2
(
x
,
t
)
∂
x
-
0.001
C
3
u
0
(
x
,
t
)
+
1.001
C
3
u
0
2
(
x
,
t
)
-
C
3
u
0
3
(
x
,
t
)
-
0.001
C
2
u
1
(
x
,
t
)
-
3
C
2
u
0
2
(
x
,
t
)
u
1
(
x
,
t
)
+
1.001
C
1
u
1
2
(
x
,
t
)
-
3
C
1
u
0
(
x
,
t
)
u
1
2
(
x
,
t
)
-
0.001
C
1
u
2
(
x
,
t
)
+
2.002
C
1
u
0
(
x
,
t
)
u
2
(
x
,
t
)
-
3
C
1
u
0
2
(
x
,
t
)
u
2
(
x
,
t
)
+
2.002
C
2
u
0
(
x
,
t
)
u
1
(
x
,
t
)
=
0
,
u
3
(
x
,
0
)
=
0
.
Its solution is
(31)u3(x,t,C1,C2,C3)=(1(1+e0.5x)2tsech2(0.25x)=×(-0.0625C2-0.0625C3=×+C1(-0.0625+C1(-0.125-0.015625t)=×+C1+C2(-0.125-0.015625t)=×+C1-0.000651042C12(5.0718+t)=×+C1×(18.9282+t)C1)=×+(-24C2t+5C12(C12288C2+288C3+C15=×+×(288+576C2+C1(576-24t)5=×+×-24C2t-3C12(-6.58301+t)5=×+××(14.583+t)cosh(0.25x)5=×+×+(-96C2-96C3+C15=×+×+×(-96-192C2+C1(-192-24t)55=×+×+-24C2t+5C12(-7.396+t)5=×+×+×(2.596+t)192C2))sinh(0.25x))=×+-24C2t+5C12(-7.396+t)(2.596+t)))=××sinh(0.25x)=××(-0.000651042cosh(0.75x)=××-0.000651042sinh(0.75x)0.0625C2-0.0625C3))1(1+e0.5x)2).
Adding (25), (27), (29), and (31) we obtain
(32)u~(x,t,C1,C2)=u0(x,t)+u1(x,t,C1)+u2(x,t,C1,C2)+u3(x,t,C1,C2,C3).
For the calculations of the constants C1, C2, and C3 using the collocation method, we have computed thatC1=-1.0000010231545267,C2=-9.98159444155818×10-7,C3=-2.041939789528322×10-12.Putting the values of these constants into (32) the third order approximate solution using OHAM is
(33)u3(x,t)=0.5-5.2607845247854×10-6tsech2(0.25x)=+1(1+e0.5x)2=×(6.360090415451543×10-9=×+(1.3153067008227217×10-6=×++0.0006512058813062292t10-6)t=×+(1.2720180853076355×10-8=×+-0.002604823525224917t2)cosh(0.5x)=×+(6.36009042653818×10-9=×++(-1.3153067008227217×10-6=×+++0.0006512058813062292t10-6)t)cosh(x)=×+(1.2720180853076355×10-8=×+-0.002604823525224917t2)sinh(0.5x)=×+(6.36009042653818×10-9=×++(-1.3153067008227217×10-9=×+++0.0006512058813062292t10-8)t)sinh(x)=×-0.5tanh(0.25x)+0.007813813661895406t2=××sech2(0.25x)tanh(0.25x)+sech2(0.25x)=××(0.06250525442670962t10-8)).Table1 shows a comparison between OHAM solution and ADM solution for α=1 and γ=0.001. For α=0 (1) is reduced to the generalized Huxley equation which describes nerve pulse propagation in nerve fibers and wall motion in liquid crystals [22]. Tables 2 and 3 show a comparison between ADM solution and OHAM solution for α=0 and β=1 respectively. Table 4 shows absolute errors of OHAM solution for larger domain for α=0,1, β=1, and δ=1,2 respectively.Table 1
Comparison of absolute errors of OHAM and ADM [5] for   α=1, β=1, δ=1, and γ=0.001.
t
ADMx=0.1
OHAMx=0.1
ADMx=0.5
OHAMx=0.5
ADMx=0.9
OHAMx=0.9
0.05
1
.
93715
×
10
-
7
1.87406
×
10
-
8
1.9373
×
10
-
7
1.87406
×
10
-
8
1.93745
×
10
-
7
1.87406
×
10
-
8
0.1
3.87434
×
10
-
7
3.74812
×
10
-
8
3.87464
×
10
-
7
3.74812
×
10
-
8
3.87494
×
10
-
7
3.74812
×
10
-
8
1
3.87501
×
10
-
6
3.74812
×
10
-
7
3.87531
×
10
-
6
3.74812
×
10
-
7
3.87561
×
10
-
6
3.74812
×
10
-
7Table 2
Comparison of absolute errors obtained by OHAM and ADM [5] for α=0, β=1, δ=1, and γ=0.001.
t
ADMx=0.1
OHAMx=0.1
ADMx=0.5
OHAMfor x=0.5
ADMx=0.9
OHAMx=0.9
0.05
1
.
93715
×
10
-
7
2.49875
×
10
-
8
1.9373
×
10
-
7
2.49875
×
10
-
8
1.93745
×
10
-
7
2.49875
×
10
-
8
0.1
3.87434
×
10
-
7
4.9975
×
10
-
8
3.87464
×
10
-
7
4.9975
×
10
-
8
3.87494
×
10
-
7
4.9975
×
10
-
8
1
3.87501
×
10
-
6
4.9975
×
10
-
7
3.87531
×
10
-
6
4.9975
×
10
-
7
3.87561
×
10
-
6
4.9975
×
10
-
7Table 3
Comparison of absolute errors obtained by OHAM and ADM [5] for                                                α=0, β=1, δ=2, and γ=0.001.
t
Error ADM
Error OHAM
0.05
5.58836
×
10
-
7
2.7938
×
10
-
7
0.1
1.11766
×
10
-
6
5.58771
×
10
-
7
1
1.00741
×
10
-
5
5.5896
×
10
-
6Table 4
Absolute errors of OHAM forα=0,1, β=1, δ=1,2, and x=2.
t
α
=
1, β=1, δ=1, and γ=0.001
α
=
0, β=1, δ=1, and γ=0.001
α
=
0, β=1, δ=2, and γ=0.001
0.1
3.74812
×
10
-
8
2.49875
×
10
-
8
2.23403
×
10
-
6
0.2
7.49625
×
10
-
8
4.9975
×
10
-
8
4.46806
×
10
-
6
0.3
1.12444
×
10
-
7
7.49625
×
10
-
8
6.70209
×
10
-
6
0.4
1.49925
×
10
-
7
9.995
×
10
-
8
8.93612
×
10
-
6
0.5
1.87406
×
10
-
7
1.24937
×
10
-
7
1.11702
×
10
-
5
0.6
2.24887
×
10
-
7
1.49925
×
10
-
7
1.34042
×
10
-
5
0.7
2.62369
×
10
-
7
1.74912
×
10
-
7
1.56382
×
10
-
5
0.8
2.9985
×
10
-
7
1.999
×
10
-
7
1.78722
×
10
-
5
0.9
3.37331
×
1
0
-
7
2.24887
×
10
-
7
2.01063
×
10
-
5
1.0
3.74812
×
10
-
7
2.49875
×
10
-
7
2.23403
×
10
-
5


## 3.2. Application of OHAM for Burger’s-Fisher Equation

Consider the Burger’s-Fisher equation of form (2):
(34)∂u(x,t)∂t+αuδ(x,t)∂u(x,t)∂x-∂2u(x,t)∂x2-βu(x,t)(1-uδ(x,t))=0,∂u(x,t)∂t+αuδ(x,t)∀0≤x≤1,t≥0,
subject to constant initial condition
(35)u(x,0)=(0.5+0.5tanh(-αδ2(δ+1)x))1/δ,
with exact solution given by
(36)u(x,t)=0.5+0.5tanh×((x-(α(δ+1)+β(δ+1)α)t)-αδ2(δ+1)××(x-(α(δ+1)+β(δ+1)α)t))1/δ.
For computational work, we have taken α=0.001, β=0.001, and δ=1 for various values of x and t.Zeroth-Order Problem
(37)
∂
u
0
(
x
,
t
)
∂
t
=
0
,
u
0
(
x
,
0
)
=
(
0.5
+
0.5
tanh
(
-
0.001
4
x
)
)
.
Its solution is
(38)u0(x,t)=(0.5-0.5tanh(0.0014x)).First-Order Problem
(39)
∂
u
1
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
0
(
x
,
t
)
∂
t
-
0.001
C
1
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
0.001
C
1
u
0
(
x
,
t
)
(
1
-
u
0
(
x
,
t
)
)
+
C
1
∂
2
u
0
(
x
,
t
)
∂
x
2
=
0
,
u
1
(
x
,
0
)
=
0
.
Its solution is
(40)u1(x,t,C1)=-t(0.00025C1+6.25×10-8C1=-t×sech2(0.00025x)=-t-0.00025C1tanh2(0.00025x)).Second-Order Problem
(41)
∂
u
2
(
x
,
t
)
∂
t
-
(
1
+
C
1
)
∂
u
1
(
x
,
t
)
∂
t
+
0.001
C
2
u
0
(
x
,
t
)
×
(
1
-
u
0
(
x
,
t
)
)
+
C
2
∂
2
u
0
(
x
,
t
)
∂
x
2
-
C
2
∂
u
0
(
x
,
t
)
∂
t
-
0.001
C
2
u
0
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
0.001
C
1
u
1
(
x
,
t
)
∂
u
0
(
x
,
t
)
∂
x
-
0.001
C
1
u
0
(
x
,
t
)
×
∂
u
1
(
x
,
t
)
∂
x
+
C
1
∂
2
u
1
(
x
,
t
)
∂
x
2
-
0.001
C
1
u
1
(
x
,
t
)
-
0.002
C
1
u
0
(
x
,
t
)
u
1
(
x
,
t
)
=
0
,
u
2
(
x
,
0
)
=
0
.
Its solution is
(42)u2(x,t,C1,C2)=sech5(0.00025x)×(+C12(-0.000187547t)-0.000187547C1t-0.000187547C2t×+C12(-0.000187547t))cosh(0.00025x)+-0.0000625156C1t-0.0000625156C2t+C12(-0.0000625156t)+2.71051×10-20C1tsinh(0.00025x)+2.71051×10-20C12tsinh(0.00025x)+2.71051×10-20C2tsinh(0.00025x)+3.12656×10-8C12t2sinh(0.00025x)+3.12656×10-8C12t2sinh(0.00075x).
The third order approximate solution using OHAM is given by
(43)u~(x,t,C1,C2)=u0(x,t)+u1(x,t,C1)+u2(x,t,C1,C2)+u3(x,t,C1,C2,C3),
where u3(x,t,C1,C2,C3) is obtained in same lines as for first problem.For the calculations of the constantsC1, C2, and  C3 using the collocation method we have computed that
(44)C1=-5.928318703338053×10-7,C2=-465.9630543691778,C3=1.8651679832921486.The third order OHAM solution yields very encouraging results after being compared with Fourth order approximate solution by ADM [5].Table5 shows a comparison between OHAM solution and ADM solution for α=0.001, β=0.001,  and  δ=1. Table 6 compares between OHAM solution and ADM solution for α=1, β=1, and δ=2. Table 7 shows the reliability of OHAM for larger domain.Table 5
Comparison of absolute errors obtained by OHAM and ADM [5] for α=0.001, β=0.001, and δ=1.
t
ADM forx=0.1
OHAMx=0.1
ADMx=0.5
OHAMx=0.5
ADMx=0.9
OHAMx=0.9
0.005
9.68763
×
10
-
6
1.12257
×
10
-
7
9.68691
×
10
-
6
2.28888
×
10
-
7
9.68619
×
10
-
6
2.28888
×
10
-
7
0.001
1.93753
×
10
-
6
2.24513
×
10
-
8
1.93738
×
10
-
6
4.57775
×
10
-
8
1.93724
×
10
-
6
4.57775
×
10
-
8
0.01
1.93752
×
10
-
5
2.24514
×
10
-
7
1.93738
×
10
-
5
4.57777
×
10
-
7
1.93724
×
10
-
5
4.57777
×
10
-
7Table 6
Comparison of absolute errors obtained by OHAM and ADM [5] for  α=1, β=1, and δ=2.
t
ADM forx=0.1
OHAMx=0.1
ADM  x=0.5
OHAM  x=0.5
ADMx=0.9
OHAM  x=0.9
0.0005
1.40177
×
10
-
3
5.87633
×
10
-
5
1.34526
×
10
-
3
1.06736
×
10
-
5
1.27699
×
10
-
3
4.64718
×
10
-
5
0.0001
2.80396
×
10
-
4
1.17539
×
10
-
5
2.69094
×
10
-
4
5.33686
×
10
-
5
2.55438
×
10
-
4
9.29303
×
10
-
6
0.001
2.80301
×
10
-
3
1.17512
×
10
-
4
2.69000
×
10
-
3
1.06739
×
10
-
4
2.55346
×
10
-
3
9.296
×
10
-
4Table 7
Absolute errors of OHAM forx=2 and t∈[0.1,1].
t
α
=
0.001, β=0.001, and δ=1
α
=
0.001, β=0.001, and δ=2
0.1
1.98526
×
10
-
9
1.09926
×
10
-
5
0.2
3.20807
×
10
-
8
2.19856
×
10
-
5
0.3
1.63084
×
10
-
7
2.9789
×
10
-
5
0.4
5.16881
×
10
-
7
4.39726
×
10
-
5
0.5
1.26475
×
10
-
6
5.49666
×
10
-
5
0.6
2.62763
×
10
-
6
6.5961
×
10
-
5
0.7
4.87621
×
10
-
6
7.69557
×
10
-
5
0.8
8.33106
×
10
-
6
8.79507
×
10
-
5
0.9
1.33626
×
10
-
5
9.89461
×
10
-
5


## 4. Conclusion

We successfully applied OHAM for solution of Burger’s-Huxley and Burger’s-Fisher equations. The method is simple in applicability and is fast converging to the exact solution. The results obtained by OHAM are very consistent in comparison with ADM.



---

*Source: 387478-2013-07-16.xml*