# Optimal Homotopy Asymptotic Method to Nonlinear Damped Generalized Regularized Long-Wave Equation

Authors: R. Nawaz; S. Islam; I. A. Shah; M. Idrees; H. Ullah
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

A new semianalytical technique optimal homology asymptotic method (OHAM) is introduced for deriving approximate solution of the homogeneous and nonhomogeneous nonlinear Damped Generalized Regularized Long-Wave (DGRLW) equation. We tested numerical examples designed to confine the features of the proposed scheme. We drew 3D and 2D images of the DGRLW equations and the results are compared with that of variational iteration method (VIM). Results reveal that OHAM is operative and very easy to use.


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## Body

## 1. Introduction

Partial differential equations used in modeling different problems in physics, biology, chemical reactions, and engineering sciences problems are frequently too difficult to be solved exactly, and even if an exact solution is possible, the required calculations may be too difficult.The DGRLW equation is a partial differential equation that describes the amplitude of long-wave, which takes the following form:(1)ut-(φ(x,t)uxt)x-αuxx+ux+upux.
Here α>0, p≥1 is an integer, φ(x,t) is known function, and u(x,t) is the amplitude of the long-wave at the position x and time t. For α≠0, (1) features a balance between nonlinear and dispersive effects but also takes into account mechanisms of dissipation. In the physical sense, (1) with the dissipative term αuxx is suggested if the good predictive power is preferred; such type of problem arises in the bore propagation as well as for water waves [1].The Equal Width (EW) Wave, Regularized Long-Wave (RLW), and Generalized Regularized Long-Wave (GRLW) equations are special cases of the DGRLW equation [2–4].The EW equation corresponds toα=0, φ(x,t)=1, whereas the RLW or Benjamin-Bona-Mahony equation corresponds to α=0, φ(x,t)=1, and p=1. On the other hand, the GRLW equation corresponds to α=0, φ(x,t)=1, and p≥1.Different methods have been used for numerical solutions of DGRLW equations. VIM was presented by Demir et al. for numerical solutions for the DGRLW equation [5]. Yousefi et al. used Bernstein Ritz-Galerkin Method for solving the DGRLW equation [6]. Achouri et al. worked on an article called “A fully Galerkin method for the damped generalized regularized long-wave (DGRLW) equation,”namely, in [7]. For the mathematical theory and physical significance of DGRLW equation see [8–16] and references therein.Marinca and Herişanu proposed semianalytical technique OHAM for deriving approximate solution of nonlinear problems of thin film flow of a fourth grade fluid down a vertical cylinder [17–19]. The method has been used by many researchers for obtaining numerical approximations of linear and nonlinear differential equations [20–24]. The author has successfully applied the OHAM for deriving approximate solution of Equal Width Wave equation, Burger equations, and tenth order boundary value problems [25, 26]. The convergence criterion of proposed method is similar to that of homotopy analysis method (HAM) and homotopy perturbation method (HPM), but this method is more efficient and flexible. To improve the efficiency and accuracy of OHAM, Herişanu and Marinca introduced more generalized and new advances in OHAM which shows that the auxiliary function includes the functions of physical parameters in addition to the convergence control parameter [27, 28].Here, we investigate the approximate solution of the DGRLW equation with a variable coefficient using OHAM. The whole paper is divided into 3 sections. Section2 is devoted to the analysis of the proposed method. In Section 3, solution of homogeneous and non-homogenous (DGRLW) equations is presented by OHAM, and absolute errors are also compared with VIM. The 3D and 2D images of the approximate solution and exact solution are also drawn. In all cases, the proposed method yields very encouraging results.


## 2. Fundamental Theory of OHAM

Consider the partial differential equation of the following form:(2)L(u(x,t))+N(u(x,t))+g(x,t)=0,x∈Ω,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, x and t denote spatial and time variables, respectively, Ω is the problem domain, and g(x,t) is a known function.Using the basic idea of OHAM, the optimal homotopyψ(x,t;q):Ω×[0,1]→R is constructed which satisfies the following condition:
(3)(1-q){L(ψ(x,t;q))+g(x,t)}11=H(q){L(ψ(x,t;q))+N(ψ(x,t;q))+g(x,t)},
where q∈[0,1] is an embedding parameter and H(q) is a nonzero auxiliary function for q≠0, H(0)=0. In such a case, (3) is called optimal homotopy equation. Clearly, we have
(4)q=0⟹H(ψ(x,t;0),0)=L(ψ(x,t;0))+g(x,t)=0,(5)q=1⟹H(ψ(x,t;1),1)=H(1){L(ψ(x,t;q))11111111+N(ψ(x,t;q))+g(x,t)}=0,
when q=0 and q=1, then ψ(x,t;0)=u0(x,t) and ψ(x,t;1)=u(x,t) hold. Thus, as q varies from 0to1, 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:
(6)L(u0(x,t))+g(x,t)=0,B(u0,∂u0∂t)=0.
Next, we choose auxiliary function H(q) of the following general form:
(7)H(q)=qc1+q2c2+⋯.Herec1,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,
(8)ψ(x,t;q,ci)=u0(x,t)+∑k=1∞uk(x,t;ci)qk,i=1,2,….Substituting (8) into (3) 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 (9) and (10), respectively, and the general governing equations for uk(x,t) are given by (11) as follows:
(9)L(u1(x,t))=c1N0(u0(x,t)),B(u1,∂u1∂t)=0,(10)L(u2(x,t))-L(u1(x,t))=c2N0(u0(x,t))11+c1[L(u1(x,t))+N1(u0(x,t),u1(x,t))],11B(u2,∂u2∂t)=0(11)L(uk(x,t))-L(uk-1(x,t))1=ckN0(u0(x,t))11+∑i=1k-1ci[L(uk-i(x,t))11111+Nk-i(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:
(12)N(ψ(x,t;q,ci))=N0(u0(x,t))+∑k≥1Nk(u0,u1,u2,…,uk)qk.Hereuk for k≥0 are a set of linear equations with the linear boundary conditions, which can be easily solved.The convergence of the series in (8) depends upon the auxiliary constants c1,c2,… . If it is convergent at q=1, then
(13)u~(x,t;ci)=u0(x,t)+∑k≥1uk(x,t;ci).Substituting (13) into (1), it results with the following expression for residual:
(14)R(x,t;ci)=L(u~(x,t;ci))+g(x,t)+N(u~(x,t;ci)).IfR(x,t;Ci)=0, then u~(x,t;ci) will be the exact solution.For computing the optimal values of auxiliary constants,ci,i=1,2,…,m, there are many methods available like Galerkin’s, Ritz, Least Squares, and Collocation method. One can apply the method of Least Squares as under the following:
(15)J(Ci)=∫0t∫ΩR2(x,t,ci)dxdt,
where R is the residual,
(16)R(x,t;Ci)=L(u~(x,t;Ci))+g(x,t)+N(u~(x,t;Ci)),(17)∂J∂c1=∂J∂c2=⋯=∂J∂c3=0.The constantsci can also be determined by another method as under:
(18)R(h1;ci)=R(h2;ci)=⋯=R(hm;ci)=0,i=1,2,…,m
at any time t, where hi∈Ω. The convergence depends upon constants c1,c2,…, which can be optimally identified and minimized by (18).Example 1.
Consider (1) with α=1, p=2, and φ(x,t)=-(1/6)e-(2x+4t)f(x,t)=0 which in the simplest form is given as
(19)ut+16((e-(2x+4t))uxt)x-uxx+ux+u2ux=0.
The initial condition is u(x,0)=ⅇ-x and exact solution given by
(20)u(x,t)=ⅇ(-x+2t).Zeroth Order Problem. Consider the following:
(21)(u0)t=0,u0(x,0)=ⅇ-x.
Its solution is given as d under
(22)u0(x,t)=ⅇ-x.First Order Problem. Consider the following:
(23)-(u0)t-c(u0)t+(u1)t-c1(u0)x-c1u02(u0)x+13c1ⅇ4t-2x(u0)xt+c1(u0)xx+c1ⅇ4t-2x(u0)xxt=0.
Its solution is as follows:
(24)u1(x,t,c1)=-c1ⅇ-3x(1+2ⅇ2x)t.Second Order Problem. Consider the following:
(25)-c2(u0)t-(u1)t-c1(u1)t+(u2)t-c2(u0)x-c2(u0)2(u0)x-2c1u0u1(u0)x-c1(u1)x-c1(u0)2(u1)x+13c2ⅇ4t-2x(u0)x,t+13c1ⅇ4t-2x(u1)x,t=0.
Its solution is under
(26)u2(x,t,c1,c2)=18ⅇ-5x(5c12-5c12ⅇ4t+2c12ⅇ2x11111111-2c12ⅇ4t+2x-8c1ⅇ2xt-8c12ⅇ2xt11111111-8c2ⅇ2xt-16c1ⅇ4xt-16c12ⅇ4xt11111111-16c2ⅇ4xt+20c12t211111111+72c12ⅇ2xt2+16c12ⅇ4xt2).Third Order Problem. Consider the following:
(27)-c3(u0)t-c2(u1)t-(u2)t-c1(u2)t+(u3)t-c3(u0)x-c3(u0)2(u0)x-2c2u0u1(u0)x-c1(u1)2(u0)x-2c1u0u2(u0)x-c2(u1)x-c2(u0)2(u1)x-2c1u0u1(u1)x-c1(u2)x-c1(u0)2(u2)x+13c3ⅇ4t-2x(u0)x,t+13c2ⅇ4t-2x(u1)x,t+13c1ⅇ4t-2x(u2)x,t+c3(u2)x,t+c2(u1)x,x+c1(u2)x,x-16c3ⅇ4t-2x(u0)x,x,t-16c2ⅇ4t-2x(u1)x,x,t-16c1ⅇ4t-2x(u2)x,x,t=0.Its solution is given as follows:(28)u3(x,t,c1,c2,c3)=196ⅇ-7x(245c13-70c13ⅇ4t-175c13ⅇ8t1111111111+120c12ⅇ2x-60c13ⅇ2x+120c1c2ⅇ2x1111111111+48c12ⅇ4x-12c13ⅇ4x+48c1c2ⅇ4x1111111111-120c12ⅇ4t+2x+90c13ⅇ4t+2x-120c1c2ⅇ4t+2x1111111111-30c13ⅇ8t+2x-48c12ⅇ4t+4x+12c13ⅇ4t+4x1111111111-48c1c2ⅇ4t+4x-420c13t+700c13ⅇ4tt1111111111-1920c13ⅇ2xt-96c1ⅇ4xt-192c12ⅇ4xt1111111111-384c13ⅇ4xt-96c2ⅇ4xt-192c1c2ⅇ4xt1111111111-96c3ⅇ4xt-192c1ⅇ6xt-384c12ⅇ6xt1111111111-192c13ⅇ6xt-192c2ⅇ6xt-384c1c2ⅇ6xt1111111111-192c3ⅇ6xt+1080c13ⅇ4t+2xt+48c13ⅇ4t+4xt1111111111+480c12ⅇ2xt2+480c13ⅇ2xt2+480c1c2ⅇ2xt21111111111+1728c12ⅇ4xt2+1728c13ⅇ4xt21111111111+1728c1c2ⅇ4xt2+384c12ⅇ6xt2+384c13ⅇ6xt21111111111+384c1c2ⅇ6xt2-784c13t3-4480c13ⅇ2xt3111111111-4032c13ⅇ4xt3-128c13ⅇ6xt3).
The third order approximate solution is given by the following equation:
(29)u(x,t,c1,c2,c3)1=u0(x,t)+u1(x,t,c1)+u2(x,t,c1,c2)11+u3(x,t,c1,c2,c3),u=ⅇ-x-c1ⅇ-3x(1+2ⅇ2x)t111+18ⅇ-5x(5c12-5c12ⅇ4t+2c12ⅇ2x11111111111-2c12ⅇ4t+2x-8c1ⅇ2xt-8c12ⅇ2xt11111111111-8c2ⅇ2xt-16c1ⅇ4xt-16c12ⅇ4xt11111111111-16c2ⅇ4xt+20c12t211111111111+72c12ⅇ2xt2+16c12ⅇ4xt2)111+196ⅇ-7x(245c13-70c13ⅇ4t111111111111-175c13ⅇ8t+120c12ⅇ2x-60c13ⅇ2x111111111111+120c1c2ⅇ2x+48c12ⅇ4x-12c13ⅇ4x111111111111+48c1c2ⅇ4x-120c12ⅇ4t+2x+90c13ⅇ4t+2x111111111111-120c1c2ⅇ4t+2x-30c13ⅇ8t+2x111111111111-48c12ⅇ4t+4x+12c13ⅇ4t+4x111111111111-48c1c2ⅇ4t+4x-420c13t111111111111+700c13ⅇ4tt-1920c13ⅇ2xt111111111111-96c1ⅇ4xt-192c12ⅇ4xt-384c13ⅇ4xt111111111111-96c2ⅇ4xt-192c1c2ⅇ4xt-96c3ⅇ4xt111111111111-192c1ⅇ6xt-384c12ⅇ6xt-192c13ⅇ6xt111111111111-192c2ⅇ6xt-384c1c2ⅇ6xt-192c3ⅇ6xt111111111111+1080c13ⅇ4t+2xt+48c13ⅇ4t+4xt111111111111+480c12ⅇ2xt2+480c13ⅇ2xt2111111111111+480c1c2ⅇ2xt2+1728c12ⅇ4xt2111111111111+1728c13ⅇ4xt2+1728c1c2ⅇ4xt2111111111111+384c12ⅇ6xt2+384c13ⅇ6xt2111111111111+384c1c2ⅇ6xt2-784c13t3-4480c13ⅇ2xt3111111111111-4032c13ⅇ4xt3-128c13ⅇ6xt3).The constantsc1, c2, and c3 are calculated using the Least Squares, we have their optimal values as follows:
(30)c1=-0.8410486261349961,c2=-0.022653709109618163,c3=0.0031930834039587894.The 3rd order OHAM solution yields very encouraging results after comparing with 3rd order VIM solution [5]. Tables 1(a–c), and Figures 1, 2, 3, and 4 show the effectiveness of OHAM for x=1.2, x=1.4and x=1.6.Table 1
(a) Comparison of absolute errors of 3rd  order OHAM solution and 3rd  order VIM solution for Example1 at x=1.2 and various values oft. (b) Comparison of absolute errors of 3rd  order OHAM solution and 3rd  order VIM solution for Example 1 at x=1.4 and various values oft.(c) Comparison of absolute errors of 3rd  order OHAM solution and 3rd  order VIM solution for Example 1 at x=1.6 and various values oft.
(a)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.313486
0.313486
1.92728
×
10
-
5
3.146
×
10
-
7
0.04
0.32628
0.32627
3.39912
×
10
-
5
1.01891
×
10
-
5
0.06
0.339596
0.339571
1.301192
×
10
-
4
2.46667
×
10
-
5
0.08
0.353455
0.353416
2.401071
×
10
-
4
3.87876
×
10
-
5
0.10
0.367879
0.367829
3.356920
×
10
-
4
5.08783
×
10
-
5
(b)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.256661
0.256661
1.331
×
10
-
7
3.29357
×
10
-
8
0.04
0.267135
0.267135
2.11230
×
10
-
5
5.45591
×
10
-
7
0.06
0.278037
0.278037
4.57127
×
10
-
5
1.19768
×
10
-
7
0.08
0.289384
0.289387
5.71818
×
10
-
5
2.34162
×
10
-
6
0.10
0.301194
0.301199
3.93977
×
10
-
5
4.79793
×
10
-
6
(c)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.210136
0.210136
1.4946
×
10
-
6
4.27298
×
10
-
7
0.04
0.218712
0.218712
7.2701
×
10
-
6
1.00227
×
10
-
7
0.06
0.227638
0.227635
7.6219
×
10
-
6
2.32488
×
10
-
6
0.08
0.236928
0.23692
6.6969
×
10
-
6
7.99082
×
10
-
6
0.10
0.246597
0.246577
4.44429
×
10
-
5
1.99024
×
10
-
5Figure 1
Plot of of 3rd order approximate solution for the homogeneous DGRLW equation (2).Figure 2
Variation profile of exact solution for the homogeneous DGRLW equation (2).Figure 3
Approximate solution plot fort=0.04.Figure 4
Exact solution plot fort=0.04.Example 2.
Consider (1) with α=1, p=2 and φ(x,t)=-(1/6)e-(2x+4t) which in the simplest form is given as
(31)ut+16((e-(2x+4t))uxt)x-uxx+ux+u2ux=0.
The initial condition isu(x,0)=sech[x/4]2 and exact solution given by
(32)u(x,t)=sech[x4-t3]2.Zeroth Order Problem. Consider the following:
(33)(u0)t=0,u0(x,0)=sech[x4]2.
Its solution is given by
(34)u0(x,t)=sech[x4]2.First Order Problem. Consider the following:
(35)-(u0)t-c1(u0)t+(u1)t-c1(u0)x-c1u02(u0)x+13c1ⅇ4t-2x(u0)xt+c1(u0)xx+c1ⅇ4t-2x(u0)xxt=0.
Its solution is
(36)u1(x,t,c1)=-18t(-c1sech[x4]4+4c1sech[x4]2tanh[x4]1111111+4c1sech[x4]6tanh[x4]1111111+2c1sech[x4]2tanh[x4]2).Second Order Problem
(37)
-
c
2
(
u
0
)
t
-
(
u
1
)
t
-
c
1
(
u
1
)
t
+
(
u
2
)
t
-
c
2
(
u
0
)
x
-
c
2
u
0
2
(
u
0
)
x
-
2
c
1
u
0
u
1
(
u
0
)
x
-
c
1
(
u
1
)
x
-
c
1
u
0
2
(
u
1
)
x
+
1
3
c
2
ⅇ
4
t
-
2
x
(
u
0
)
x
,
t
+
1
3
c
1
ⅇ
4
t
-
2
x
(
u
1
)
x
,
t
+
c
2
(
u
0
)
x
,
x
+
c
1
(
u
1
)
x
,
x
-
1
6
c
2
ⅇ
4
t
-
2
x
(
u
0
)
x
,
x
,
t
-
1
6
c
1
ⅇ
4
t
-
2
x
(
u
1
)
x
,
x
,
t
=
0
.Its approximate solutionu2(x,t,c1,c2) is obtained in similar manner. The second order approximate solution is given by
(38)u(x,t,c1,c2)=u0(x,t)+u1(x,t,c1)+u2(x,t,c1,c2)=sech[x4]2-18t(-c1sech[x4]41111111111111111+4c1sech[x4]2tanh[x4]1111111111111111+4c1sech[x4]6tanh[x4]111111111111111+2c1sech[x4]2tanh[x4]2)11+u2(x,t,c1,c2).Using method of Least Squares, the optimal values of constantsc1andc2 are computed and are given as under
(39)c1=-0.8122876966282708,c2=-0.011502183336221636.Table2(a) shows the effectiveness of OHAM for x=15, x=20, and x=25, while Table 2(b), and Figures 5, 6, 7, and 8 shows the effectiveness of OHAM for various values of x and t.Table 2
(a) Absolute error of the solution of Example2 by optimal homotopy asymptotic method (OHAM) at  x=15, x=20, and x=25 and various values of t. (b) Absolute error of the solution of Example 2 by optimal homotopy asymptotic method (OHAM) at  various values of x and t.
(a)
t
x
=
15
x
=
20
x
=
25
0.01
2.7905
×
10
-
6
2.30504
×
10
-
7
1.89307
×
10
-
8
0.02
2.83577
×
10
-
6
2.35074
×
10
-
7
1.93117
×
10
-
8
0.03
8.32571
×
10
-
6
6.87803
×
10
-
7
5.6488
×
10
-
8
0.04
5.60061
×
10
-
6
4.64385
×
10
-
7
3.81506
×
10
-
8
0.05
1.37955
×
10
-
5
1.13979
×
10
-
6
9.36098
×
10
-
8
(b)
x
t
=
0.1
t
=
0.2
t
=
0.3
t
=
0.4
t
=
0.5
15
2.24995
×
10
-
5
4.36594
×
10
-
5
6.27619
×
10
-
5
9.29866
×
10
-
5
9.16758
×
10
-
5
16
1.36916
×
10
-
5
26576
×
10
-
5
3.82161
×
10
-
5
5.66063
×
10
-
5
5.58636
×
10
-
5
17
8.32101
×
10
-
6
1.61544
×
10
-
5
2.32343
×
10
-
5
3.441
×
10
-
5
3.39789
×
10
-
5
18
5.05308
×
10
-
6
9.81109
×
10
-
6
1.41127
×
10
-
5
2.0899
×
10
-
5
2.06447
×
10
-
5
19
3.0671
×
10
-
6
5.95551
×
10
-
6
8.56725
×
10
-
6
1.26863
×
10
-
5
1.25347
×
10
-
5
20
1.86112
×
10
-
6
3.61396
×
10
-
6
5.19906
×
10
-
6
7.69844
×
10
-
6
7.60747
×
10
-
6
21
1.12913
×
10
-
6
2.19263
×
10
-
6
3.1544
×
10
-
6
4.67075
×
10
-
6
4.61593
×
10
-
6
22
6.84967
×
10
-
7
1.33013
×
10
-
6
1.91361
×
10
-
6
2.83347
×
10
-
6
2.80036
×
10
-
6
23
4.15495
×
10
-
7
8.06854
×
10
-
7
1.1608
×
10
-
6
1.71878
×
10
-
6
1.69874
×
10
-
6
24
2.52026
×
10
-
7
4.89414
×
10
-
7
7.04113
×
10
-
7
1.04256
×
10
-
6
1.03043
×
10
-
6
25
1.52867
×
10
-
7
2.96856
×
10
-
7
4.27085
×
10
-
7
6.32372
×
10
-
7
6.25018
×
10
-
7Figure 5
Surface 2nd order approximate solution for (4.12).Figure 6
Exact solution for (4.12).Figure 7
Approximate solution plot fort=0.05.Figure 8
Exact solution plot fort=0.05.Example 3.
Consider (1) with α=1, p=1, and φ(x,t)=1-(3cosh[x/4-t/3]sinh[x/4 − t/3]/(-3+2cosh[t/3-x/4]2)) which in the simplest form is given as
(40)ut+(1-3cosh[x/4-t/3]sinh[x/4-t/3]-3+2cosh[t/3-x/4]2uxt)x-uxx+ux+uux=0.
The initial condition is u(x,0)=sech[x/4]2 and exact solution given by
(41)u(x,t)=sech[x4-t3]2.Zeroth Order Problem. Consider the following:
(42)(u0)t=0,u0(x,0)=sech[x4]2.
Its solution is as follows:
(43)u0(x,t)=sech[x4]2.First Order Problem. Consider the following:
(44)-(u0)t-c1(u0)t+(u1)t-c1(u0)x-c1u0(u0)x-c1(3cosh[t/3-x/4]24(-3+2cosh[t/3-x/4]2)11111-3cosh[t/3-x/4]2sinh[t/3-x/4]2(-3+2cosh[t/3-x/4]2)211111+3sinh[t/3-x/4]24(-3+2cosh[t/3-x/4]2))(u0)x,t+c1(u0)x,x-c1(-1-3cosh[t/3-x/4]sinh[t/3-x/4]-3+2cosh[t/3-x/4]2)1111×(u0)x,x,t=0.Its solution is(45)u1(x,t,c1)=-18t-c1sech[x4]4+4c1sech[x4]2tanh[x4]11+4c1sech[x4]4tanh[x4]+2c1sech[x4]2tanh[x4]2.The first order approximate solution is given by(46)u(x,t,c1)=u0(x,t)+u1(x,t,c1)(47)u(x,t,c1)=sech[x4]2-18t(-c1sech[x4]41111111111111111+4c1sech[x4]2tanh[x4]1111111111111111+4c1sech[x4]4tanh[x4]111111111111111+2c1sech[x4]2tanh[x4]2).
The constants c1 is calculated using the Least Squares that we have its optimal values as follows:
(48)c1=-0.802767563787412.
The first order optimum solution using OHAM is as follows:
(49)u(x,t)=sech[x4]211-18t(0.802767563787412sech[x4]411111111-3.211070255149648sech[x4]2tanh[x4]11111111-3.211070255149648sech[x4]4tanh[x4]11111111-1.605535127574824sech[x4]2tanh[x4]2).The first order OHAM solution yields very encouraging results after comparing with 2nd order VIM solution [5]. Tables 3(a–d), and Figures 9, 10, 11, and 12 show the effectiveness of OHAM for x=-2.5, x=0, x=2.5 and x=5.Table 3
(a) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example3 at x=-2.5 and various values oft.(b) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 3 at x=0 and various values oft.(c) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 3 at x=-2.5 and various values oft.(d) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 3 at x=-2.5 and various values oft.
(a)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.687297
0.687094
1.510466
×
10
-
3
2.02246
×
10
-
4
0.04
0.682171
0.68177
3.017151
×
10
-
3
4.0071
×
10
-
4
0.06
0.677041
0.676445
4.521014
×
10
-
3
5.96353
×
10
-
4
0.08
0.671911
0.67112
6.023003
×
10
-
3
7.90121
×
10
-
4
0.10
0.666779
0.665796
7.524051
×
10
-
3
9.82948
×
10
-
4
(b)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.999956
0.997993
2.455556
×
10
-
3
1.96248
×
10
-
3
0.04
0.999822
0.995986
4.822243
×
10
-
3
3.83608
×
10
-
3
0.06
0.9996
0.993979
7.100106
×
10
-
3
5.62086
×
10
-
3
0.08
0.999289
0.991972
9.289225
×
10
-
3
7.3169
×
10
-
3
0.10
0.99889
0.989965
1.138971
×
10
-
3
8.92431
×
10
-
3
(c)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.697537
0.697529
1.247750
×
10
-
3
7.71606
×
10
-
6
0.04
0.702649
0.702639
2.501243
×
10
-
3
9.6889
×
10
-
6
0.06
0.707754
0.707749
3.761480
×
10
-
3
4.91862
×
10
-
6
0.08
0.712851
0.712859
5.029472
×
10
-
3
7.60727
×
10
-
6
0.10
0.71794
0.717969
6.306245
×
10
-
3
2.89136
×
10
-
5
(d)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.283601
0.283512
6.719964
×
10
-
4
889401
×
10
-
5
0.04
0.286816
0.286609
1.315023
×
10
-
3
20685
×
10
-
4
0.06
0.29006
0.289706
1.928998
×
10
-
3
353812
×
10
-
4
0.08
0.293333
0.292803
2.513842
×
10
-
3
529904
×
10
-
4
0.10
0.296636
0.295901
3.069485
×
10
-
3
735198
×
10
-
4Figure 9
1st order approximate solution for (4.12).Figure 10
Exact solution plot for (4.12).Figure 11
Approximate solution plot fort=0.1.Figure 12
Exact solution plot fort=0.1.Example 4.
Let us consider the inhomogeneous DGRLW equation:(50)ut-(φ(x,t)uxt)x-uxx+ux+uux=f(x,t),
where φ(x,t)=xt and f(x,t)=(-tcos[x]+xtsin[x]+cos[x]+sin[x]cos[x]ⅇ-t)ⅇ-t.
The initial condition is(51)u(x,0)=sin[x]
and exact solution given by
(52)u(x,t)=sin[x]ⅇ-t.
According to OHAM scheme presented in Section2.Zeroth Order Problem. Consider the following:
(53)(u0)t=0,u0(x,0)=sin[x].
Its solution is
(54)u0(x,t)=sin[x].First Order Problem. Consider the following:
(55)c1ⅇ-t(ⅇ-tcos[x]-tcos[x]+txsin[x]11+ⅇ-tcos[x]sin[x])111-(u0)t-c1(u0)t+(u1)t-c1(u0)x-c1u0(u0)x111+c1t(u0)x,t+c1(u0)x,x+c1tx(u0)x,x,t=0.
Its solution is as follows:
(56)u1(x,t,c1)=-12c1ⅇ-2t(2ⅇttcos[x]-2ⅇ2ttcos[x]11111111111-2ⅇ2ttsin[x]-2ⅇtxsin[x]11111111111+2ⅇ2txsin[x]-2ⅇttxsin[x]11111111111-cos[x]sin[x]+ⅇ2tcos[x]sin[x]11111111111-2ⅇ2ttcos[x]sin[x]).Second Order Problem. Consider the following:
(57)c2ⅇ-t(ⅇ-tcos[x]-tcos[x]1+txsin[x]+ⅇ-tcos[x]sin[x])1-c2(u0)t-(u1)t-c1(u1)t+(u2)t1-c2(u0)x-c2u0(u0)x-c1u1(u0)x1-c1(u1)x-c1u0(u1)x+c2t(u0)x,t1+c1t(u1)x,x+c2(u0)x,x+c1(u1)x,x1+c2tx(u0)x,x,t+c1tx(u0)x,x,t=0.Its approximate solution is under(58)u2(x,t,c1,c2)=116ⅇ-2t(-16c12ⅇt+16c12ⅇ2t-8c12ⅇtt1111111111-8c12ⅇ2tt+80c12ⅇtcos[x]1111111111-80c12ⅇ2tcos[x]-16c1ⅇttcos[x]1111111111+32c12ⅇttcos[x]-16c2ⅇttcos[x]1111111111+16c1ⅇ2ttcos[x]+48c12ⅇ2ttcos[x]1111111111+16c2ⅇ2ttcos[x]+8c12ⅇ2tt2cos[x]1111111111-144c12ⅇtxcos[x]+144c12ⅇ2txcos[x]1111111111-128c12ⅇttxcos[x]-16c12ⅇ2ttxcos[x]1111111111-64c12ⅇtt2xcos[x]+8c12ⅇ2tt2xcos[x]1111111111-8c12cos[2x]+32c12ⅇtcos[2x]1111111111-24c12ⅇ2tcos[2x]-8c12tcos[2x]1111111111+24c12ⅇttcos[2x]+8c12ⅇ2tt2cos[2x]1111111111+c12sin[x]-96c12ⅇtsin[x]1111111111+95c12ⅇ2tsin[x]-80c12ⅇttsin[x]1111111111+16c1ⅇ2ttsin[x]+2c12ⅇ2ttsin[x]1111111111+16c2ⅇ2ttsin[x]-32c12ⅇtt2sin[x]1111111111+6c12ⅇ2tt2sin[x]+16c1ⅇtxsin[x]1111111111-16c12ⅇtxsin[x]+16c2ⅇtxsin[x]1111111111-16c1ⅇ2txsin[x]+16c12ⅇ2txsin[x]1111111111-16c2ⅇ2txsin[x]+16c1ⅇttxsin[x]1111111111+16c2ⅇttxsin[x]-16c12ⅇ2ttxsin[x]1111111111+8c12ⅇ2tt2xsin[x]+32c12ⅇtx2sin[x]1111111111-32c12ⅇ2tx2sin[x]+32c12ⅇttx2sin[x]1111111111+16c12ⅇtt2x2sin[x]+4c1sin[2x]1111111111-4c12sin[2x]+4c2sin[2x]1111111111-4c1ⅇ2tsin[2x]+4c12ⅇ2tsin[2x]1111111111-4c2ⅇ2tsin[2x]+8c1ⅇ2ttsin[2x]1111111111-8c12ⅇ2ttsin[2x]+8c2ⅇ2ttsin[2x]1111111111+24c12ⅇ2tt2sin[2x]+8c12xsin[2x]1111111111-32c12ⅇtxsin[2x]+24c12ⅇ2txsin[2x]1111111111+16c12txsin[2x]-16c12ⅇttxsin[2x]1111111111-16c12ⅇ2ttxsin[2x]+16c12ⅇ2tt2xsin[2x]1111111111-3c12sin[3x]+3c12ⅇ2tsin[3x]111111111-6c12ⅇ2ttsin[3x]+6c12ⅇ2tt2sin[3x]).
The second order approximate solution is given by the following equation,
(59)u(x,t,c1,c2)=u0(x,t)+u1(x,t,c1)+u2(x,t,c1,c2).
Using the method of Least Squares the optimum values of c1 and c2 are computed which are as follows:
(60)c1=-1.0433989069917953,c2=-0.0018859250375958814.The 2nd order OHAM solution yields very encouraging results after comparing with 2nd order VIM solution [5]. Tables 4(a–d), and Figures 13, 14, 15, and 16 show the effectiveness of OHAM for x=0.2, x=0.4, x=0.6, and x=1.Table 4
(a) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example4 at x=0.2 and various values oft. (b) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 4 at x=0.4 and various values oft. (c) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 4 at x=0.6 and various values oft. (d) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 4 at x=1 and various values oft.
(a)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.194735
0.194388
3.596760
×
10
-
4
3.46922
×
10
-
4
0.04
0.190879
0.189563
1.374338
×
10
-
3
1.31649
×
10
-
3
0.06
0.1871
0.184295
2.950802
×
10
-
3
2.8045
×
10
-
3
0.08
0.183395
0.178685
5.000174
×
10
-
3
4.71013
×
10
-
3
0.10
0.179763
0.172828
7.437789
×
10
-
3
6.93584
×
10
-
3
(b)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.381707
0.381449
2.832281
×
10
-
4
2.58242
×
10
-
4
0.04
0.374149
0.373216
1.043581
×
10
-
3
9.32938
×
10
-
4
0.06
0.36674
0.364862
2.151953
×
10
-
3
1.87854
×
10
-
3
0.08
0.359478
0.356523
3.485485
×
10
-
3
2.95526
×
10
-
3
0.10
0.35236
0.348331
4.927456
×
10
-
3
4.02886
×
10
-
3
(c)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.553462
0.55333
1.676249
×
10
-
4
1.31821
×
10
-
4
0.04
0.542503
0.542094
5.651230
×
10
-
4
4.08832
×
10
-
4
0.06
0.53176
0.531101
1.040306
×
10
-
3
6.59364
×
10
-
4
0.08
0.521231
0.520511
1.4483802
×
10
-
3
7.194
×
10
-
4
0.10
0.51091
0.510477
1.651669
×
10
-
3
4.32267
×
10
-
4
(d)
t
Exact solution
OHAM solution
Absolute error VIM
Absolute error OHAM
0.02
0.703151
0.703174
1.439370
×
10
-
4
2.23769
×
10
-
4
0.04
0.689228
0.689438
6.672725
×
10
-
4
2.10203
×
10
-
4
0.06
0.675581
0.676318
1.702709
×
10
-
3
7.37797
×
10
-
4
0.08
0.662203
0.663974
3.376802
×
10
-
3
1.77096
×
10
-
3
0.10
0.649091
0.652558
5.810408
×
10
-
3
3.46734
×
10
-
3Figure 13
2nd order approximate solution for (4.12).Figure 14
The surface shows Exact solution for (4.12).Figure 15
Approximate solution plot fort=0.1 for (4.12).Figure 16
Exact solution plot fort=0.1 for (4.12).


## 3. Conclusion

In this paper, the OHAM has been successfully implemented for the approximate solution of solutions of the Nonlinear Damped Generalized Regularized Long-Wave (DGRLW) equations. The results obtained by OHAM are very consistent in comparison with VIM.



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*Source: 503137-2013-10-28.xml*