# The Optimal Homotopy Asymptotic Method for the Solution of  Higher-Order Boundary Value Problems in Finite Domains

Authors: Javed Ali; S. Islam; Hamid Khan; Syed Inayat Ali Shah
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We solve some higher-order boundary value problems by the optimal homotopy asymptotic method (OHAM). The proposed method is capable to handle a wide variety of linear and nonlinear problems effectively. The numerical results given by OHAM are compared with the exact solutions and the solutions obtained by Adomian decomposition (ADM), variational iteration (VIM), homotopy perturbation (HPM), and variational iteration decomposition method (VIDM). The results show that the proposed method is more effective and reliable.


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## Body

## 1. Introduction

In this paper, we consider a well-posednth-order problem of the form(1.1)u(n)=ψ(u,u′,…,u(n-1))+ϕ(r),a<r<b,
with boundary conditions:   u(k)(a)=αi and u(k)(b)=βi, where k(<n) is a nonnegative integer, αi and βi are real finite constants, and ϕ(r) is a continuous function on [a,b].Such types of problems have been investigated by many authors [1, 2] due to their mathematical importance and the potential for applications in hydrodynamic and hydromagnetic stability. Fifth-order boundary value problems arise in the mathematical modeling of viscoelastic flows. Sixth- and eighth-order differential equation govern physics of some hydrodynamic stability problems. When an infinite horizontal layer of fluid is heated from below and is subject to the action of rotation, instability sets in. When this instability is as ordinary convection, the ordinary differential equation is sixth order; when the instability sets in as overstability, it is modeled by an eighth-order ordinary differential equation. If an infinite horizontal layer of fluid is heated from below, with the supposition that a uniform magnetic field is also applied across the fluid in the same direction as gravity and the fluid is subject to the action of rotation, instability sets in. When instability sets in as ordinary convection, it is modeled by tenth-order boundary value problem.So for the solution of these problems is concerned, many methods appeared in literature. The recent analytic methods are Adomian decomposition method (ADM) [3–5], variational iteration method (VIM) [6], homotopy perturbation method (HPM) [7–9], homotopy analysis method (HAM) [10, 11], differential transform method (DTM) [12], and so forth. Classical perturbation methods are based on the assumptions of small or large parameters, and they cannot produce a general form of an approximate solution. The nonperturbation methods like ADM and DTM can deal strongly with nonlinear problems, but the convergence region of their series solution is generally small. The HPM, which is an elegant combination of homotopy and perturbation technique, overcomes the restrictions of small or large parameters in the problems. It deals with a wide variety of nonlinear problems effectively. Recently, Marinca et al. [13–17] introduced OHAM for approximate solution of nonlinear problems of thin film flow of a fourth-grade fluid down a vertical cylinder. In their work, they have used this method to understand the behavior of nonlinear mechanical vibration of electrical machine. They also used the same 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 equation arising in heat transfer. This method is straight forward, reliable, and explicitly defined. It provides a convenient way to control the convergence of the series solution and allows adjustment of convergence region where it is needed.Fifth- and sixth-order linear and nonlinear problems were solved by Wazwaz [18, 19], while using decomposition method. Noor et al. [20–25] investigated these type of problems using variational iteration method (VIM), homotopy perturbation method (HPM), and variational iteration decomposition method (VIDM). Modified variational iteration method (MVIM) and iterative method (ITM) were used by Mohyud-Din et al. [26, 27] for such type of problems. Kasi Viswanadham and Murali Krishna [28] used Quintic B-Spline Galerkin method for fifth-order boundary value problems. Siraj-ul-Islam et al. [29, 30] used numerical scheme for the solution of fifth- and sixth-order boundary value problems.Recently, Ali et al. [31, 32] used OHAM for the solution of multipoint boundary value problems and twelfth-order boundary value problems. We use OHAM to find the approximate analytic solution of some higher-order BVPs. The results of OHAM are compared with those of exact solution, and the errors are compared with the existing results. This paper is organized as follows: Section 2 is devoted to the analysis of the proposed method. Some numerical examples are presented in Section 3. In Section  4, we concluded by discussing results of the numerical simulation using Mathematica.


## 2. Method Analysis for Two-Point Boundary Value Problems

Consider the differential equation(2.1)Lu=Nu+ϕ,
along with boundary conditions:
(2.2)B(u,∂u∂r)=0,
where ℒ is linear, 𝒩 is a nonlinear, and ℬ is a boundary operator. ϕ is a known function which is continues for r:r∈Ω. According to OHAM, we can construct a homotopy defined by(2.3)(1-p)L(u(r;p))-h(p)[L(u(r;p))-N(u(r;p))-ϕ(r)]=0,
where p∈[0,1] is an embedding parameter, and  h(p) is a nonzero auxiliary function for p≠0 and h(0)=0. Equation (2.3) satisfies (2.4)Lu=0,forp=0,Lu=Nu+ϕ,forp=1.
The solution, u(r,0)=v0(r), of ℒu=0 traces the solution curve u(r) of (2.1), continuously as p approaches to 1, where v0 is the solution of the zeroth-order problem, that will come in the next few lines.The auxiliary functionh(p) is chosen in the form (it is a commonly used form) (2.5)h(p)=∑i=1mpiCi,
where Ci:i=1,2,…,m are the convergence controlling constants which are to be determined. We will use this function unless otherwise stated. The auxiliary function can be chosen in a variety of ways, as reported by Marinca et al. [13–17]. We will use some other forms of h(p) as well.To get an approximate solution, we expandu(r,p) in Taylor’s series about p in the following manner:(2.6)u(r;p)=v0(r)+∑m=1∞vm(r,C1,C2,…,Cm)pm.
Substituting (2.5) and (2.6) into (2.3) and equating the coefficient of like powers of p, we obtain the following linear equations which can be integrated directly.Zeroth-order problem:(2.7)Lv0=0,B(v0,∂v0∂n)=0.First-order problem:(2.8)Lv1=(1+C1)Lv0-C1(N0v0-ϕ),B(v1,∂v1∂n)=0.Second-order problem:(2.9)Lv2=(1+C1)Lv1-C1N1(v0,v1)+C2(Lv0-N0v0-ϕ),B(v2,∂v2∂n)=0.
Though we can construct higher-order problems easily, solutions upto the second-order problems are enough to produce excellent results.If the series (2.6) is convergent at p=1, then the approximate solution in our case is,(2.10)ũ(r)=v(r)=v0(r)+v1(r,C1)+v2(r,C1,C2).
By substituting (2.10) into (2.1), the resulting residual is(2.11)R(r,C1,C2)=L(ũ(r))-N(ũ(r))-ϕ(r).
If ℛ=0, ũ will be the exact solution. Otherwise, we minimize ℛ over domain of the problem. To find the optimal values of Ciwhich minimizes ℛ, many methods can be applied [13–17]. We follow two methods: the method of least squares and the Galerekin’s method. According to the method of least squares, we first construct the functional(2.12)J(C1,C2)=∫abR2dr,
and then minimizing it, we have(2.13)∂J∂C1=∂J∂C2=0.
According to the Galerekin’s method, we solve the following system for C1 andC2:(2.14)∫abR∂ũ∂C1dr=0,∫abR∂ũ∂C2dr=0.
Knowing C1 and C2, the approximate solution is well determined.


## 3. Some Numerical Examples

Example 3.1 (fifth-order linear).
Consider the following problem:(3.1)y(v)(x)=y-15ex-10xex,0<x<1,
with boundary conditions
(3.2)y(0)=0,y′(0)=1,y′′(0)=0,y(1)=0,y′(1)=-e.
The exact solution of this problem is y(x)=x(1-x)ex.
We choose the auxiliary function ash(p)=p(C1+C2x). Plugging in this value in (2.3) of Section 2, we obtain the following linear problems which can be integrated directly.
Zeroth-order problem:(3.3)y0(5)(x)=0,y0(0)=0,y0′(0)=0,y0′′(0)=0,y0(1)=0,y0′(1)=-e.
First-order problem:(3.4)y1(5)(x,C1,C2)=5ex(3+2x)(C1+C2x)-(C1+C2x)y0(x),y1(0)=0,y1′(0)=0,y1′′(0)=0,y1(1)=0,y1′(1)=0.
Second-order problem:(3.5)y2(5)(x,C1,C2)=(1+C1+C2x)y1(5)(x,C1,C2)-(C1+C2x)y1(x,C1,C2),y2(0)=0,y2′(0)=0,y2′′(0)=0,y2(1)=0,y2′(1)=0.
Adding up the solutions of these problems, the second-order approximate solution,
(3.6)ỹ(x)=y0(x)+y1(x,C1,C2)+y2(x,C1,C2)+O(x15),
is determined by knowing the optimal values of the auxiliary constants, C1 and C2. Using Galerkin’s method, we obtain C1=-1.000245451, C2=0.000124615.
By considering these values, (3.6) becomes
(3.7)ỹ(x)=x-0.5x3-0.333333x4-0.125x5-0.0333333x6-0.00694444x7-0.00119049x8-0.000173601x9-0.0000220495x10-2.48013×10-6x11-2.50501×10-7x12-2.27501×10-8x13-2.0326×10-9x14+O(x15).
Numerical results of the solution (3.7) are displayed in Table 1.Table 1
It shows comparison of the solutions obtained by OHAM (3.7), ADM [18], HPM [20], VIM [22], ITM [24], and VIHPM [21]. From the numerical results, it is clear that OHAM is more efficient and accurate.
xExact sol.OHAM sol.E* (3.7)E* (ADM)E* (HPM)E* (VIM)E* (ITM)E* (VIHPM)0.00.000000.00000.00000.00000.00000.00000.00000.00000.10.0994653830.099465383-1E-12-3E-11-3E-11-3E-11-3E-11-3E-110.20.1954244410.195424441-7E-12-2E-10-2E-10-2E-10-2E-10-2E-100.30.283470350.28347035-1E-11-4E-10-4E-10-4E-10-4E-10-4E-100.40.3580379270.358037927-1E-11-8E-10-8E-10-8E-10-8E-10-8E-100.50.4121803180.412180318-8E-12-1E-9-1E-9-1E-9-1E-9-1E-90.60.4373085120.4373085123E-14-2E-9-2E-9-2E-9-2E-9-2E-90.70.4228880680.4228880686E-12-2E-9-2E-9-2E-9-2E-9-2E-90.80.3560865480.3560865482E-12-2E-9-2E-9-2E-9-2E-9-2E-90.90.2213642800.221364280-3E-11-1E-9-1E-9-1E-9-1E-9-1E-91.00.00000.0000.0000.0000.0000.0000.0000.000E* = exact – approximate.Example 3.2 (another fifth-order linear).
Consider the following problem:(3.8)y(5)(x)+xy(x)=19xCos⁡(x)-2x3Cos⁡(x)-41Sin⁡(x)+2x2Sin⁡(x),
with boundary conditions
(3.9)y(-1)=y(1)=Cos⁡(1),y′(-1)=y′(1)=-4Cos⁡(1)+Sin⁡(1),y′′(-1)=3Cos⁡(1)-8Sin⁡(1).
Exact solution of this problem is y(x)=(2x2-1)Cos(x).
Considering the second-order solutionỹ(x)=y0(x)+y1(x,C1)+y2(x,C1,C2)+O(x13), we use the method of least squares to obtain C1=0.9940605306, C2=-3.9762851376.
Having these values, our solution in this case is(3.10)ỹ(x)=-0.999978+2.49992x2-1.04155x4+0.0846365x6-0.00276866x8+0.0000420239x10+3.38286×10-7x12+O(x13).
Numerical results of the solution (3.10) are displayed in Table 2.Table 2
The maximum absolute error as reported in [28] is 1.8775×10-5, while in our case, it is 8.5757×10-8.
xExact sol.OHAM sol. (3.10)E* (3.10)−1.00.5403023060.5403023013.90E-9−0.80.1950778790.1950778798.44E-9−0.6−0.231093972−0.2310940103.74E-8−0.4−0.626321476−0.6263215436.70E-8−0.2−0.901661252−0.9016613348.22E-80.0−1.000000000−1.0000000868.58E-80.2−0.901661252−0.9016613348.22E-80.4−0.626321476−0.6263215436.70E-80.6−0.231093972−0.2310940103.74E-80.80.1950778790.1950778798.44E-91.00.5403023060.5403023013.90E-9E* = exact − approximate.Example 3.3 ([33] fifth-order nonlinear).
Consider the following problem:(3.11)y(v)(x)=y3(x)e-x,0<x<1,
with boundary conditions
(3.12)y(0)=1,y′(0)=1/2,y′′(0)=1/4,y(1)=e1/2,y′(1)=12e1/2.
The exact solution for this problem is  y(x)=ex/2.
We consider the second-order solution,ỹ(x)=y0(x)+y1(x,C1)+y2(x,C1,C2)+O(x15).
Using Galerkin’s procedure in Section2, we obtain the following values:
(3.13)C1=0.010868466,C2=-0.029423113.
The second-order approximate solution is
(3.14)ỹ(x)=1+x2+x28+0.0205993x3+0.0030533x4+0.0000630671x5+5.2556×10-6x6+3.754×10-7x7+2.29401×10-8x8+1.74892×10-9x9-3.2692×10-11x10-1.48596×10-12x11-6.20243×10-14x12-2.58495×10-15x13+2.75287×10-17x14+O(x15).
Numerical results of the solution (3.14) are displayed in Table 3.Table 3xExact solutionOHAM sol. (3.14)E* (3.14)E* (DTM [33])0.00.000000.00000.00000.00000.11.1051709181.105170919-9.2E-101.0E-90.21.2214027581.221402763-5.0E-92.0E-90.31.3498588081.349858818-1.1E-81.0E-80.41.4918246981.491824713-1.5E-82.0E-80.51.6487212711.648721287-1.6E-83.1E-80.61.8221188001.822118814-1.4E-83.7E-80.72.0137527072.013752717-9.9E-94.1E-80.82.2255409282.225540934-5.6E-93.1E-80.92.4596031112.459603112-1.1E-91.4E-81.02.7182818282.7182818240.0000.000E* = exact – approximate.Example 3.4 (sixth-order nonlinear).
Consider the following problem:(3.15)y(vi)(x)=exy2(x),0<x<1,
with boundary conditions
(3.16)y(0)=1,y′(0)=1,y′′(0)=1,y(1)=e-1,y′(1)=-e-1,y′′(1)=e-1.
The exact solution is y(x)=e-x.
For this problem, we take the auxiliary functionh(p)=p(C1+C2e-x),
(3.17)ỹ(x)=y0(x)+y1(x,C1,C2)+y2(x,C1,C2)+O(x13).
Using Galerkin’s method, we obtain C1=0.41243798998, C2=0.0014069149.
OHAM solution in this case is(3.18)ỹ(x)=1-0.999999994x+x22-0.166666775x3+x424-0.008332465x5+0.001387441x6-0.000197784x7+0.000025071x8-3.002×10-6x9+2.918×10-7x10-6.7×10-9x11-2.257×10-9x12+1.806×10-10x13-1.974×10-11x14+O(x15).
Numerical results of the solution (3.18) are displayed in Table 4.Table 4
It shows comparison of the OHAM solution (3.18) with the exact solution and the errors obtained by decomposition method (ADM) [19], homotopy perturbation method (HPM) [24], and the variational iteration method [22]. It is clear from the results that the method we applied is more efficient and accurate than the other methods.
xExact solutionOHAM sol (3.18)E* (3.18)E* (ADM)E* (HPM)E* (VIM)0.01100000.10.9048374180.904837418-4.82E-10-1.2E-4-1.2E-4-1.2E-40.20.8187307530.818730753-4.92E-10-2.3E-4-2.3E-4-2.3E-40.30.7408182210.740818221-2.37E-11-3.2E-4-3.2E-4-3.2E-40.40.6703200460.6703200465.11E-10-3.8E-4-3.8E-4-3.8E-40.50.606530660.606530666.42E-10-4.0E-4-4.0E-4-4.0E-40.60.5488116360.5488116362.02E-10-3.9E-4-3.9E-4-3.9E-40.70.4965853040.496585304-5.37E-10-3.3E-4-3.3E-4-3.3E-40.80.4493289640.449328965-1.02E-09-2.4E-4-2.4E-4-2.4E-40.90.406569660.40656966-8.23E-10-1.2E-4-1.2E-4-1.2E-41.00.3678794410.367879441-2.05E-122.0E-92.0E-92.0E-9E* = exact − approximate.Example 3.5 (eighth-order nonlinear).
Consider the following problem:(3.19)y(viii)(x)=e-xy2(x),0<x<1,
with boundary conditions
(3.20)y′(0)=1,y′(0)=1,y′′(0)=1,y′′′(0)=1,y(1)=e,y′(1)=e,y′′(1)=e.
Considering the second-order solution ỹ(x)=y0(x)+y1(x,C1)+y2(x,C1,C2)+O(x13), the following values of the convergence controlling constants are obtained by using Galerkin’s method:
(3.21)C1=-1.451894673×10-11,C2=-0.000647581.
The approximate solution in this case is
(3.22)ỹ(x)=1+x+0.5x2+0.166667x3+0.0416275x4+0.00884857x5+0.00117027x6+0.00033161x7+1.6061×10-8x8+1.78456×10-9x9+1.78456×10-10x10+1.62233×10-11x11+1.3494×10-12x12+O(x13).
If the method of least squares is used to determine C’s, we have then
(3.23)C1=1.793×10-8,C2=-1.001347284.
The approximate solution in this case is
(3.24)ỹ(x)=1+x+0.5x2+0.166667x3+0.0416667x4+0.00833313x5+0.00138918x6+0.000198233x7+0.000024835x8+2.75944×10-6x9+2.75944×10-7x10+2.50859×10-8x11+2.08656×10-9x12+O(x13).
Let us use the auxiliary function h(p)=p(C1+C2e-x) and consider the second-order solution
(3.25)ỹ(x)=y0(x)+y1(x,C1,C2)+y2(x,C1,C2)+O(x13).
Using Galerkin’s method, we obtain C1=-0.9993171458, C2=-0.0012314995.
The OHAM solution in this case is(3.26)ỹ(x)=1+x+x22!+x33!+0.041666667x4+0.008333333x5+0.001388889x6+1.984×10-4x7+2.480×10-5x8+2.756×10-6x9+2.756×10-7x10+2.505×10-8x11+2.088×10-9x12+O(x13).
Numerical results of the solutions (3.22), (3.24), and (3.26) are displayed in Table 5.Table 5xExactOHAM sol.E*
(3.22)E*
(3.24)E*
(3.26)E* ([25])0.01100000.11.1051709181.1051709152.60E-9-3.5E-121.39E-161.27E-50.21.2214027581.2214027322.62E-8-3.5E-115.32E-162.43E-50.31.3498588081.3498587297.86E-8-1.1E-101.48E-153.35E-50.41.4918246981.4918245621.36E-7-1.8E-103.36E-153.94E-50.51.6487212711.6487211091.61E-7-2.2E-102.28E-144.16E-50.61.8221188001.8221186621.39E-7-1.9E-102.21E-133.96E-50.72.0137527072.0137526258.22E-8-1.1E-101.64E-123.38E-50.82.2255409282.2255409002.80E-8-2.8E-119.36E-122.45E-50.92.4596031112.4596031082.84E-94.1E-114.36E-111.29E-51.02.7182818282.7182818281.14E-131.8E-101.73E-101.00E-9E* = exact − approximate.Example 3.6 (nineth-order linear).
Consider the following problem:(3.27)y(9)(x)=y(x)-9ex,
with boundary conditions
(3.28)y(0)=1,y′(0)=0,y′′(0)=-1,y′′′(0)=-2,y′′′′(0)=-3,y(1)=0,y′(1)=-e,y′′(1)=-2e,y′′′(1)=-3e.
Exact solution is   y(x)=(1-x)ex.
For this linear problem, we takeh(p)=p(C1+C2x), and according to the rest of the procedure of OHAM, the second-order solution, ỹ(x)=y0(x)+y1(x,C1,C2)+y2(x,C1,C2)+O(x13), is determined by the values of Ci:i=1,2. Following the Galerkin’s method, we obtain C1=-1, C2=0, for a=0 and b=1.
The second-order approximate solution is(3.29)y(x)=1-x22-x33-x44-0.03333333333x5-0.0069444444x6-0.0011904762x7-0.0001736111x8-0.00002204586x9-2.48016×10-6x10-2.50521×10-7x11-2.29644×10-8x12+O(x13).
Numerical results of the solution (3.29) are displayed in Table 6.Table 6xExactOHAM sol.E* (3.29)E* ([27])0.0110.00000.00000.10.9946538260.9946538261.03E-16-2.0E-100.20.9771222060.9771222061.33E-16-2.0E-100.30.9449011650.944901165-2.12E-16-2.0E-100.40.8950948180.895094818-1.30E-14-2.0E-100.50.8243606350.824360635-2.44E-13-2.0E-100.60.7288475200.728847520-2.64E-12-6.0E-100.70.6041258120.604125812-1.97E-11-1.0E-090.80.4451081860.445108186-1.13E-10-2.0E-090.90.2459603110.245960311-5.26E-10-3.4E-091.00.00002.09E-09-2.09E-090.000E* = exact − approximate.Example 3.7 (tenth-order nonlinear).
Consider the following problem:(3.30)y(X)(x)=e-xy2(x),0<x<1,y(0)=1,y′(0)=1,y′′(0)=1,y′′′(0)=1,y(iv)(0)=1,y(1)=e,y′(1)=e,y′′(1)=e,y′′′(1)=e,y(iv)(1)=e.
We consider the second-order solution ỹ(x)=y0(x)+y1(x,C1)+y2(x,C1,C2)+O(x13).
To find the values ofCi, we apply the Galarkin’s method. So solving the system
(3.31)∫abR∂ṽ∂C1dr=0,∫abR∂ṽ∂C2dr=0,
we have C1=0, C2=-1.023966086.
In this case, the approximate solution is(3.32)ỹ(x)=1+x+x22+x36+x424+0.008333323x5+0.00138894x6+0.000198312x7+0.000024898x8+2.712×10-6x9+2.8218×10-7x10+2.5652×10-8x11+2.1377×10-9x12+O(x13).
Numerical results of the solution (3.32) are displayed in Table 7.Table 7xExactOHAM sol.E* (3.32)E* ([27])0.011000.11.1051709181.1051709188.12E-17-1.41E-60.21.2214027581.2214027582.18E-18-2.69E-60.31.3498588081.349858808-2.84E-18-3.70E-60.41.4918246981.4918246987.95E-16-4.35E-60.51.6487212711.6487212711.91E-14-4.58E-60.61.8221188001.8221188002.15E-13-4.36E-60.72.0137527072.0137527071.63E-12-3.71E-60.82.2255409282.2255409289.33E-12-2.69E-60.92.4596031112.4596031114.36E-11-1.42E-61.02.7182818282.7182818281.73E-102.00E-9E* = exact − approximate.


## 4. Conclusions

In this paper, we have used OHAM to find the approximate analytic solution to higher-order two-point boundary value problems in finite domain. It is observed that the method is explicit, effective, and reliable. It works well for higher-order problems and represents the fastest convergence as well as a remarkable low error. The OHAM also provides us with a very simple way to control and adjust the convergence of the series solution using the auxiliary constantsCi’s which are optimally determined. Furthermore, by using different forms of the auxiliary function, more accuracy can be obtained. It has been also observed that for determining the optimal values of C’s, the performance of both the least squares and the Galerkin’s method is problem dependent. One can select one of these two which best suits the problem solution.



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*Source: 401217-2011-12-25.xml*