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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 401217, 14 pages
http://dx.doi.org/10.1155/2012/401217
Research Article

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

1Islamia College Peshawar (Chartered University), Khyber Pakhtunkhawa, Peshawar 25120, Pakistan
2Department of Mathematics, CIIT, H-8, Islamabad 44000, Pakistan
3FAST NU, Peshawar 25100, Pakistan

Received 28 July 2011; Accepted 12 October 2011

Academic Editor: Muhammad Aslam Noor

Copyright © 2012 Javed Ali et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction

In this paper, we consider a well-posed 𝑛th-order problem of the form𝑢(𝑛)=𝜓𝑢,𝑢,,𝑢(𝑛1)+𝜙(𝑟),𝑎<𝑟<𝑏,(1.1) 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) [35], variational iteration method (VIM) [6], homotopy perturbation method (HPM) [79], 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. [1317] 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. [2025] 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 𝑢=𝒩𝑢+𝜙,(2.1) along with boundary conditions: 𝑢,𝜕𝑢𝜕𝑟=0,(2.2) where is linear, 𝒩 is a nonlinear, and is a boundary operator. 𝜙 is a known function which is continues for 𝑟𝑟Ω. According to OHAM, we can construct a homotopy defined by[](1𝑝)(𝑢(𝑟;𝑝))𝒽(𝑝)(𝑢(𝑟;𝑝))𝒩(𝑢(𝑟;𝑝))𝜙(𝑟)=0,(2.3) where 𝑝[0,1] is an embedding parameter, and 𝒽(𝑝) is a nonzero auxiliary function for 𝑝0 and (0)=0. Equation (2.3) satisfies 𝑢=0,for𝑝=0,𝑢=𝒩𝑢+𝜙,for𝑝=1.(2.4) The solution, 𝑢(𝑟,0)=𝑣0(𝑟), of 𝑢=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 function 𝒽(𝑝) is chosen in the form (it is a commonly used form) 𝒽(𝑝)=𝑚𝑖=1𝑝𝑖𝐶𝑖,(2.5) 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. [1317]. We will use some other forms of 𝒽(𝑝) as well.

To get an approximate solution, we expand 𝑢(𝑟,𝑝) in Taylor’s series about 𝑝 in the following manner:𝑢(𝑟;𝑝)=𝑣0(𝑟)+𝑚=1𝑣𝑚𝑟,𝐶1,𝐶2,,𝐶𝑚𝑝𝑚.(2.6) 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: 𝑣0𝑣=0,0,𝜕𝑣0𝜕𝑛=0.(2.7)

First-order problem: 𝑣1=1+𝐶1𝑣0𝐶1𝒩0𝑣0𝑣𝜙,1,𝜕𝑣1𝜕𝑛=0.(2.8)

Second-order problem: 𝑣2=1+𝐶1𝑣1𝐶1𝒩1𝑣0,𝑣1+𝐶2𝑣0𝒩0𝑣0𝑣𝜙,2,𝜕𝑣2𝜕𝑛=0.(2.9) 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,̃𝑢(𝑟)=𝑣(𝑟)=𝑣0(𝑟)+𝑣1𝑟,𝐶1+𝑣2𝑟,𝐶1,𝐶2.(2.10) By substituting (2.10) into (2.1), the resulting residual is𝑟,𝐶1,𝐶2=(̃𝑢(𝑟))𝒩(̃𝑢(𝑟))𝜙(𝑟).(2.11) If =0, ̃𝑢 will be the exact solution. Otherwise, we minimize over domain of the problem. To find the optimal values of 𝐶𝑖which minimizes , many methods can be applied [1317]. 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 functional𝒥𝐶1,𝐶2=𝑏𝑎2𝑑𝑟,(2.12) and then minimizing it, we have𝜕𝒥𝜕𝐶1=𝜕𝒥𝜕𝐶2=0.(2.13) According to the Galerekin’s method, we solve the following system for 𝐶1 and𝐶2:𝑏𝑎𝜕̃𝑢𝜕𝐶1𝑑𝑟=0,𝑏𝑎𝜕̃𝑢𝜕𝐶2𝑑𝑟=0.(2.14) Knowing 𝐶1 and 𝐶2, the approximate solution is well determined.

3. Some Numerical Examples

Example 3.1 (fifth-order linear). Consider the following problem: 𝑦(𝑣)(𝑥)=𝑦15𝑒𝑥10𝑥𝑒𝑥,0<𝑥<1,(3.1) with boundary conditions 𝑦(0)=0,𝑦(0)=1,𝑦(0)=0,𝑦(1)=0,𝑦(1)=𝑒.(3.2) The exact solution of this problem is 𝑦(𝑥)=𝑥(1𝑥)𝑒𝑥.
We choose the auxiliary function as (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: 𝑦0(5)𝑦(𝑥)=0,0(0)=0,𝑦0(0)=0,𝑦0(0)=0,𝑦0(1)=0,𝑦0(1)=𝑒.(3.3)
First-order problem: 𝑦1(5)𝑥,𝐶1,𝐶2=5𝑒𝑥𝐶(3+2𝑥)1+𝐶2𝑥𝐶1+𝐶2𝑥𝑦0𝑦(𝑥),1(0)=0,𝑦1(0)=0,𝑦1(0)=0,𝑦1(1)=0,𝑦1(1)=0.(3.4)
Second-order problem: 𝑦2(5)𝑥,𝐶1,𝐶2=1+𝐶1+𝐶2𝑥𝑦1(5)𝑥,𝐶1,𝐶2𝐶1+𝐶2𝑥𝑦1𝑥,𝐶1,𝐶2,𝑦2(0)=0,𝑦2(0)=0,𝑦2(0)=0,𝑦2(1)=0,𝑦2(1)=0.(3.5) Adding up the solutions of these problems, the second-order approximate solution, ̃𝑦(𝑥)=𝑦0(𝑥)+𝑦1𝑥,𝐶1,𝐶2+𝑦2𝑥,𝐶1,𝐶2𝑥+𝑂15,(3.6) 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 ̃𝑦(𝑥)=𝑥0.5𝑥30.333333𝑥40.125𝑥50.0333333𝑥60.00694444𝑥70.00119049𝑥80.000173601𝑥90.0000220495𝑥102.48013×106𝑥112.50501×107𝑥122.27501×108𝑥132.0326×109𝑥14𝑥+𝑂15.(3.7) Numerical results of the solution (3.7) are displayed in Table 1.

tab1
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.

Example 3.2 (another fifth-order linear). Consider the following problem: 𝑦(5)(𝑥)+𝑥𝑦(𝑥)=19𝑥Cos(𝑥)2𝑥3Cos(𝑥)41Sin(𝑥)+2𝑥2Sin(𝑥),(3.8) with boundary conditions 𝑦𝑦(1)=𝑦(1)=Cos(1),(1)=𝑦𝑦(1)=4Cos(1)+Sin(1),(1)=3Cos(1)8Sin(1).(3.9) Exact solution of this problem is 𝑦(𝑥)=(2𝑥21)Cos(𝑥).
Considering the second-order solution ̃𝑦(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 ̃𝑦(𝑥)=0.999978+2.49992𝑥21.04155𝑥4+0.0846365𝑥60.00276866𝑥8+0.0000420239𝑥10+3.38286×107𝑥12𝑥+𝑂13.(3.10) Numerical results of the solution (3.10) are displayed in Table 2.

tab2
Table 2: The maximum absolute error as reported in [28] is 1.8775×105, while in our case, it is 8.5757×108.

Example 3.3 ([33] fifth-order nonlinear). Consider the following problem: 𝑦(𝑣)(𝑥)=𝑦3(𝑥)𝑒𝑥,0<𝑥<1,(3.11) with boundary conditions 𝑦(0)=1,𝑦(0)=1/2,𝑦(0)=1/4,𝑦(1)=𝑒1/2,𝑦1(1)=2𝑒1/2.(3.12) The exact solution for this problem is 𝑦(𝑥)=𝑒𝑥/2.
We consider the second-order solution, ̃𝑦(𝑥)=𝑦0(𝑥)+𝑦1(𝑥,𝐶1)+𝑦2(𝑥,𝐶1,𝐶2)+𝑂(𝑥15).
Using Galerkin’s procedure in Section 2, we obtain the following values: 𝐶1=0.010868466,𝐶2=0.029423113.(3.13) The second-order approximate solution is 𝑥̃𝑦(𝑥)=1+2+𝑥28+0.0205993𝑥3+0.0030533𝑥4+0.0000630671𝑥5+5.2556×106𝑥6+3.754×107𝑥7+2.29401×108𝑥8+1.74892×109𝑥93.2692×1011𝑥101.48596×1012𝑥116.20243×1014𝑥122.58495×1015𝑥13+2.75287×1017𝑥14𝑥+𝑂15.(3.14) Numerical results of the solution (3.14) are displayed in Table 3.

tab3
Table 3

Example 3.4 (sixth-order nonlinear). Consider the following problem: 𝑦(𝑣𝑖)(𝑥)=𝑒𝑥𝑦2(𝑥),0<𝑥<1,(3.15) with boundary conditions 𝑦(0)=1,𝑦(0)=1,𝑦(0)=1,𝑦(1)=𝑒1,𝑦(1)=𝑒1,𝑦(1)=𝑒1.(3.16) The exact solution is 𝑦(𝑥)=𝑒𝑥.
For this problem, we take the auxiliary function (𝑝)=𝑝(𝐶1+𝐶2𝑒𝑥), ̃𝑦(𝑥)=𝑦0(𝑥)+𝑦1𝑥,𝐶1,𝐶2+𝑦2𝑥,𝐶1,𝐶2𝑥+𝑂13.(3.17) Using Galerkin’s method, we obtain 𝐶1=0.41243798998, 𝐶2=0.0014069149.
OHAM solution in this case is 𝑥̃𝑦(𝑥)=10.999999994𝑥+220.166666775𝑥3+𝑥4240.008332465𝑥5+0.001387441𝑥60.000197784𝑥7+0.000025071𝑥83.002×106𝑥9+2.918×107𝑥106.7×109𝑥112.257×109𝑥12+1.806×1010𝑥131.974×1011𝑥14𝑥+𝑂15.(3.18) Numerical results of the solution (3.18) are displayed in Table 4.

tab4
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.

Example 3.5 (eighth-order nonlinear). Consider the following problem: 𝑦(𝑣𝑖𝑖𝑖)(𝑥)=𝑒𝑥𝑦2(𝑥),0<𝑥<1,(3.19) with boundary conditions 𝑦(0)=1,𝑦(0)=1,𝑦(0)=1,𝑦(0)=1,𝑦(1)=𝑒,𝑦(1)=𝑒,𝑦(1)=𝑒.(3.20) Considering the second-order solution ̃𝑦(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: 𝐶1=1.451894673×1011,𝐶2=0.000647581.(3.21) The approximate solution in this case is ̃𝑦(𝑥)=1+𝑥+0.5𝑥2+0.166667𝑥3+0.0416275𝑥4+0.00884857𝑥5+0.00117027𝑥6+0.00033161𝑥7+1.6061×108𝑥8+1.78456×109𝑥9+1.78456×1010𝑥10+1.62233×1011𝑥11+1.3494×1012𝑥12𝑥+𝑂13.(3.22) If the method of least squares is used to determine 𝐶’s, we have then 𝐶1=1.793×108,𝐶2=1.001347284.(3.23) The approximate solution in this case is ̃𝑦(𝑥)=1+𝑥+0.5𝑥2+0.166667𝑥3+0.0416667𝑥4+0.00833313𝑥5+0.00138918𝑥6+0.000198233𝑥7+0.000024835𝑥8+2.75944×106𝑥9+2.75944×107𝑥10+2.50859×108𝑥11+2.08656×109𝑥12𝑥+𝑂13.(3.24) Let us use the auxiliary function (𝑝)=𝑝(𝐶1+𝐶2𝑒𝑥) and consider the second-order solution ̃𝑦(𝑥)=𝑦0(𝑥)+𝑦1𝑥,𝐶1,𝐶2+𝑦2𝑥,𝐶1,𝐶2𝑥+𝑂13.(3.25) Using Galerkin’s method, we obtain 𝐶1=0.9993171458, 𝐶2=0.0012314995.
The OHAM solution in this case is 𝑥̃𝑦(𝑥)=1+𝑥+2+𝑥2!33!+0.041666667𝑥4+0.008333333𝑥5+0.001388889𝑥6+1.984×104𝑥7+2.480×105𝑥8+2.756×106𝑥9+2.756×107𝑥10+2.505×108𝑥11+2.088×109𝑥12𝑥+𝑂13.(3.26) Numerical results of the solutions (3.22), (3.24), and (3.26) are displayed in Table 5.

tab5
Table 5

Example 3.6 (nineth-order linear). Consider the following problem: 𝑦(9)(𝑥)=𝑦(𝑥)9𝑒𝑥,(3.27) with boundary conditions 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.(3.28) Exact solution is 𝑦(𝑥)=(1𝑥)𝑒𝑥.
For this linear problem, we take (𝑝)=𝑝(𝐶1+𝐶2𝑥), and according to the rest of the procedure of OHAM, the second-order solution, ̃𝑦(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 𝑥𝑦(𝑥)=122𝑥33𝑥440.03333333333𝑥50.0069444444𝑥60.0011904762𝑥70.0001736111𝑥80.00002204586𝑥92.48016×106𝑥102.50521×107𝑥112.29644×108𝑥12𝑥+𝑂13.(3.29) Numerical results of the solution (3.29) are displayed in Table 6.

tab6
Table 6

Example 3.7 (tenth-order nonlinear). Consider the following problem:𝑦(𝑋)(𝑥)=𝑒𝑥𝑦2(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.(3.30) We consider the second-order solution ̃𝑦(𝑥)=𝑦0(𝑥)+𝑦1(𝑥,𝐶1)+𝑦2(𝑥,𝐶1,𝐶2)+𝑂(𝑥13).
To find the values of 𝐶𝑖, we apply the Galarkin’s method. So solving the system 𝑏𝑎𝑅𝜕̃𝑣𝜕𝐶1𝑑𝑟=0,𝑏𝑎𝑅𝜕̃𝑣𝜕𝐶2𝑑𝑟=0,(3.31) we have 𝐶1=0, 𝐶2=1.023966086.
In this case, the approximate solution is 𝑥̃𝑦(𝑥)=1+𝑥+22+𝑥36+𝑥424+0.008333323𝑥5+0.00138894𝑥6+0.000198312𝑥7+0.000024898𝑥8+2.712×106𝑥9+2.8218×107𝑥10+2.5652×108𝑥11+2.1377×109𝑥12𝑥+𝑂13.(3.32) Numerical results of the solution (3.32) are displayed in Table 7.

tab7
Table 7

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 constants 𝐶𝑖’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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