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Mathematical Problems in Engineering
Volume 2011 (2011), Article ID 724927, 19 pages
http://dx.doi.org/10.1155/2011/724927
Research Article

On the Solutions of Nonlinear Higher-Order Boundary Value Problems by Using Differential Transformation Method and Adomian Decomposition Method

Department of Mathematics, University Putra Malaysia, 43400 UPM Serdang, Selangor, Malaysia

Received 8 November 2010; Accepted 7 February 2011

Academic Editor: Alexei Mailybaev

Copyright © 2011 Che Haziqah Che Hussin and Adem Kiliçman. 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 study higher-order boundary value problems (HOBVP) for higher-order nonlinear differential equation. We make comparison among differential transformation method (DTM), Adomian decomposition method (ADM), and exact solutions. We provide several examples in order to compare our results. We extend and prove a theorem for nonlinear differential equations by using the DTM. The numerical examples show that the DTM is a good method compared to the ADM since it is effective, uses less time in computation, easy to implement and achieve high accuracy. In addition, DTM has many advantages compared to ADM since the calculation of Adomian polynomial is tedious. From the numerical results, DTM is suitable to apply for nonlinear problems.

1. Introduction

Recently, many researchers use ADM to approximate numerical solutions. In [1], Wazwaz proposed a modification of ADM method in series solution to accelerate its rapid convergence, and, in [2], Wazwaz also presented several numerical examples of higher-order boundary value problems for first-order linear equation and second-order nonlinear equation by applying modified decomposition method.

In addition, Wazwaz [3, 4] provided first-order linear and second-order nonlinear problems to solve fifth-order and sixth-order boundary value problems by the modified decomposition method. Later, Meštrović [5] solved eight-order boundary value problems for first-order linear and second-order nonlinear boundary value problems. Similarly, in [6], Hosseini and Jafari used Adomian decomposition method to solve high-order and system of nonlinear differential equations.

The Adomian decomposition method is widely used in applied science to compute the series solution accurately because it provides rapid convergent series to solve the problem. The big application of ADM in research area is stochastic and deterministic problems involving differential, integrodifferential, integral, differential delay, and systems of such equations; for example, see [3, 7]. For example, we use the same method for solving several different problems, such as, in calculus of variations, see [8], for eikonal partial differential equation, see [9], for the Fitzhugh-Nagumo equation which models the transmission of nerve impulses, see [10], for linear and nonlinear systems of Volterra functional equations using Adomian-Pade technique, see [11], for coupled Burgers equations by using Adomian-Pade technique see [12], for solution of a nonlinear time-delay model in biology by using semi-analytical approaches, see [13], for solving the pantograph equation of order 𝑚, see [14], and for nonclassic problem for one-dimensional hyperbolic equation by using the decomposition procedure, see [15].

Further, Ray and Bera in [16] used ADM to solve analytical solution of a fractional diffusion equation. They performed the explicit solution of the equation in the closed form by using initial value problem.

In [17], Ayaz investigated initial value problem of partial differential equation (PDE) to solve two-dimensional differential transformation method, and we compare the results with Adomian decomposition method. The results show that the solutions of the present method are exactly the same as the decomposition method, but the calculation of DTM is simple and reduces the difficulty of calculations.

Ayaz in [18] performed two- and three-dimensional differential transformation methods to find exact solutions of linear and nonlinear partial differential equations. Results are compared to decomposition method, and DTM has less computational effort. After that, Ertürk and Momani in [19] presented numerical solution by comparing the differential transformation method (DTM) and Adomian decomposition method (ADM) for solving linear and nonlinear fourth-order boundary value problems and proved that DTM is very accurate and efficient in numerical solution.

Recently, Arikoglu and Ozkol solved fractional differential equations by using differential transform method. They applied fractional differential equations to various types of problems such as the Bagley-Torvik, Ricatti, and composite fractional oscillation equations; see [20].

In this study, we make comparison among differential transformation method, Adomian decomposition method and exact solutions. We prove that DTM is more powerful technique than ADM and can be applied to nonlinear problems easily.

2. Differential Transformation Method

Suppose that the function 𝑦(𝑥) is continuously differentiable in the interval (𝑥0𝑟,𝑥0+𝑟) for 𝑟>0, then we have the following definition.

Definition 2.1. The differential transform of the function 𝑦(𝑥) for the 𝑘th derivative is defined as follows: 1𝑌(𝑘)=𝑑𝑘!𝑘𝑦(𝑥)𝑑𝑥𝑘𝑥=𝑥0,(2.1) where 𝑦(𝑥) is the original function and 𝑌(𝑘) is the transformed function. The inverse differential transform of 𝑌(𝑘) is defined as 𝑦(𝑥)=𝑘=0𝑥𝑥0𝑘𝑌(𝑘).(2.2) Note that, the substitution of (2.1) into (2.2) yields the following equation: 𝑦(𝑥)=𝑘=0𝑥𝑥0𝑘1𝑑𝑘!𝑘𝑦(𝑥)𝑑𝑥𝑘𝑥=𝑥0(2.3) which is the Taylor's series for 𝑦(𝑥) at 𝑥=𝑥0.

The following theorems are operations of differential transforms.

Theorem 2.2. If 𝑡(𝑥)=𝑟(𝑥)±𝑝(𝑥), then 𝑇(𝑘)=𝑅(𝑘)±𝑃(𝑘).

Theorem 2.3. If 𝑡(𝑥)=𝛼𝑟(𝑥), then 𝑇(𝑘)=𝛼𝑅(𝑘).

Theorem 2.4. If 𝑡(𝑥)=𝑑𝑟(𝑥)/𝑑𝑥, then 𝑇(𝑘)=(𝑘+1)𝑅(𝑘+1).

Theorem 2.5. If 𝑡(𝑥)=𝑑2𝑟(𝑥)/𝑑𝑥2, then 𝑇(𝑘)=(𝑘+1)(𝑘+2)𝑅(𝑘+2).

Theorem 2.6. If 𝑡(𝑥)=(𝑑𝑏𝑟(𝑥))/𝑑𝑥𝑏, then 𝑇(𝑘)=(𝑘+1)(𝑘+2)(𝑘+𝑏)𝑅(𝑘+𝑏).

Theorem 2.7. If 𝑡(𝑥)=𝑟(𝑥)𝑝(𝑥), then 𝑇(𝑘)=𝑘𝑙=0𝑃(𝑙)𝑅(𝑘𝑙).

Theorem 2.8. If 𝑡(𝑥)=𝑥𝑏, then 𝑇(𝑘)=𝛿(𝑘𝑏), where 𝛿(𝑘𝑏)=1,if𝑘=𝑏,0,if𝑘𝑏.(2.4)

Theorem 2.9. If 𝑡(𝑥)=exp(𝜆𝑥), then 𝑇(𝑘)=𝜆𝑘/𝑘!.

Theorem 2.10. If 𝑡(𝑥)=(1+𝑥)𝑏, then 𝑇(𝑘)=𝑏(𝑏1)(𝑏𝑘+1)/𝑘!.

Theorem 2.11. If 𝑡(𝑥)=sin(𝑗𝑥+𝛼), then 𝑇(𝑘)=(𝑗𝑘/𝑘!)sin((𝜋𝑘/2)+𝛼).

Theorem 2.12. If 𝑡(𝑥)=cos(𝑗𝑥+𝛼), then 𝑇(𝑘)=(𝑗𝑘/𝑘!)cos((𝜋𝑘/2)+𝛼).

See the details in [21].

3. General Differential Transformation

Now we prove the following generalized theorem.

Theorem 3.1. The general differential transformation for nonlinear 𝑛th-order BVPs, 𝑦(𝑛)(𝑥)=𝑒𝑥𝑦𝑚(𝑥), is given by 𝑌(𝑛+𝑘)=𝑘!(𝑛+𝑘)!𝑘𝑘𝑚=0𝑘𝑚𝑘𝑚1=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑚𝑖=2𝑌𝑘𝑖𝑘𝑖1𝑌𝑘𝑘𝑚.(3.1)

Proof. We prove this theorem by induction method. First of all, we prove the generalization of the differential equation 𝑒𝑥𝑦𝑚(𝑥). Let 𝑚=2, then 𝑦(𝑛)(𝑥)=𝑒𝑥𝑦𝑚(𝑥).(3.2) For 𝑘1, we have 𝑦(𝑛+𝑘)(𝑥)=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑒𝑥𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘𝑘2)𝑦(𝑥),(𝑛+𝑘)(𝑥)𝑥=0=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘𝑘2)(𝑥).(3.3) By definition, we have (𝑛+𝑘)!𝑌(𝑛+𝑘)=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑘2𝑘1𝑘!𝑌2𝑘1𝑘𝑘2!𝑌𝑘𝑘2,(𝑛+𝑘)!𝑌(𝑛+𝑘)=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2,𝑌(𝑛+𝑘)=𝑘!(𝑛+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2.(3.4) From (3.1), for 𝑚=2, we have 𝑌(𝑛+𝑘)=𝑘!(𝑛+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2.(3.5) This implies that the theorem holds for 𝑚=2. Now assume that for 𝑚=𝑝, the DT is as follows: 𝑌(𝑛+𝑘)=𝑘!(𝑛+𝑘)!𝑘𝑘𝑝=0𝑘𝑝𝑘𝑝1=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑝𝑖=2𝑌𝑘𝑖𝑘𝑖1𝑌𝑘𝑘𝑝,𝑌(𝑛+𝑘)=𝑘!(×𝑛+𝑘)!𝑘𝑘𝑝=0𝑘𝑝𝑘𝑝1=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘3𝑘2𝑘𝑝𝑘𝑝1𝑌𝑘𝑘𝑝.(3.6) Note that 𝑚=𝑝 we have 𝑦(𝑛)(𝑥)=𝑒𝑥𝑦𝑝(𝑥).(3.7) For 𝑘1, we have 𝑦(𝑛+𝑘)(𝑥)𝑥=0=𝑘𝑘𝑝=0𝑘𝑝𝑘𝑝1=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘3𝑘2)(𝑥)𝑦(𝑘𝑝𝑘𝑝1)(𝑥)𝑦(𝑘𝑘𝑝)(𝑥)𝑘1!𝑘2𝑘1!𝑘3𝑘2𝑘!𝑝𝑘𝑝1!𝑘𝑘𝑝!.(3.8) Thus, for 𝑚=𝑝+1, we have 𝑦(𝑛)(𝑥)=𝑒𝑥𝑦𝑝(𝑥)𝑦(𝑥).(3.9) For 𝑘1, we have 𝑦(𝑛+𝑘)(𝑥)𝑥=0=𝑘𝑘𝑝+1=0𝑘𝑝+1𝑘𝑝=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘3𝑘2)(𝑥)𝑦(𝑘𝑝+1𝑘𝑝)(𝑥)𝑦(𝑘𝑘𝑝+1)(𝑥)𝑘1!𝑘2𝑘1!𝑘3𝑘2𝑘!𝑝+1!𝑘𝑘𝑘𝑝+1!.(3.10) By definition, we have (𝑛+𝑘)!𝑌(𝑛+𝑘)=𝑘𝑘𝑝+1=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘2𝑘1𝑘!𝑌2𝑘1𝑘𝑘𝑝+1!𝑌𝑘𝑘𝑝+1𝑘1!𝑘2𝑘1!𝑘𝑘𝑝+1!,𝑌(𝑛+𝑘)=𝑘!(𝑛+𝑘)!𝑘𝑘𝑝+1=0𝑘𝑝+1𝑘𝑝=0𝑘2𝑘1=0(1)𝑘1𝑘1𝑌𝑘2𝑘1𝑌𝑘3𝑘2𝑘𝑝+1𝑘𝑝𝑌𝑘𝑘𝑝+1.(3.11) From (3.1), for 𝑚=𝑝+1, we have 𝑌(𝑛+𝑘)=𝑘!(𝑛+𝑘)!𝑘𝑘𝑝+1=0𝑘𝑝+1𝑘𝑝=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑝+1𝑖=2𝑌𝑘𝑖𝑘𝑖1𝑌𝑘𝑘𝑝+1,𝑌(𝑛+𝑘)=𝑘!(𝑛+𝑘)!𝑘𝑘𝑝+1=0𝑘𝑝+1𝑘𝑝=0𝑘2𝑘1=0(1)𝑘1𝑘1𝑌𝑘2𝑘1𝑌𝑘3𝑘2𝑘𝑝+1𝑘𝑝𝑌𝑘𝑘𝑝+1.(3.12) This implies that the theorem holds for 𝑚=𝑝+1. Now, we prove the generalization of the BVPs, 𝑦(𝑛)(𝑥)=𝑒𝑥𝑦𝑚(𝑥).(3.13) For that purpose, we fixed 𝑚=2.
For 𝑛=1 and 𝑘1, we have 𝑦(1+𝑘)(𝑥)𝑥=0=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘𝑘2)(𝑥).(3.14) By definition, we have (𝑛+𝑘)!𝑌(𝑛+𝑘)=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑘2𝑘1𝑘!𝑌2𝑘1𝑘𝑘2!𝑌𝑘𝑘2,𝑌(1+𝑘)=𝑘!(1+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2.(3.15) From (3.1), for 𝑛=1, we have 𝑌(1+𝑘)=𝑘!(1+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!2𝑖=2𝑌𝑘𝑖1𝑘𝑖2𝑌𝑘𝑘2,𝑌(1+𝑘)=𝑘!(1+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2.(3.16) Now that is the required result; thus, the theorem holds for 𝑛=1.
Now assume that for 𝑛=𝑞, the DT is as follows: 𝑌(𝑞+𝑘)=𝑘!(𝑞+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2.(3.17) For 𝑘1, we have 𝑦(𝑞+𝑘)(𝑥)𝑥=0=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘𝑘2)(𝑥).(3.18) Thus, for 𝑛=𝑞+1, we have 𝑦(𝑞+1)(𝑥)=(𝑦(𝑥))2𝑒(𝑥).(3.19) For 𝑘1, we have 𝑦(𝑞+1+𝑘)(𝑥)𝑥=0=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘𝑘2)(𝑥),(𝑞+1+𝑘)!𝑌(𝑞+1+𝑘)=𝑘𝑘2=0𝑘2𝑘1=0𝑘!(1)𝑘1𝑘1!𝑘2𝑘1!𝑘𝑘2!𝑦(𝑘2𝑘1)(𝑥)𝑦(𝑘𝑘2)𝑌(𝑥),(𝑞+1+𝑘)=𝑘!(𝑞+1+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2.(3.20) From (3.1), for 𝑛=𝑞+1, we have 𝑌(𝑞+1+𝑘)=𝑘!(𝑞+1+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!2𝑖=2𝑌𝑘𝑖1𝑘𝑖2𝑌𝑘𝑘2,𝑌(𝑞+1+𝑘)=𝑘!(𝑞+1+𝑘)!𝑘𝑘2=0𝑘2𝑘1=0(1)𝑘1𝑘1!𝑌𝑘2𝑘1𝑌𝑘𝑘2.(3.21)

Note that the theorem holds for 𝑛=𝑞+1.

Example 3.2. By using DTM, we solve the following nonlinear equation of order three for fifth-order BVP: 𝑦(5)(𝑥)=𝑒𝑥(𝑦(𝑥))3,0<𝑥<1(3.22) subject to the boundary conditions 𝑦(0)=1,𝑦1(0)=2,𝑦1(0)=4,𝑦(1)=𝑒1/2,𝑦1(1)=2𝑒1/2.(3.23) By using Theorem 3.1, (3.22) is transformed to the following equation: 𝑌(𝑟+5)=𝑟!(𝑟+5)!𝑟𝑙𝑙𝑠𝑠𝑝(1)𝑝𝑝!𝑌(𝑠𝑝)𝑌(𝑙𝑠)𝑌(𝑟𝑙).(3.24) By using (2.1) at 𝑥=0, we obtain the following transformed boundary conditions: 1𝑌(0)=0,𝑌(1)=21,𝑌(2)=8,(3.25) where 𝑡=𝑦(0)/3!=𝑌(3) and 𝑤=𝑦(4)(0)/4!=𝑌(4).
By using the transformed equation (3.24) together with the transformed boundary conditions (3.25), we can easily solve 𝑌(𝑟), for 𝑟5. The constants 𝑡 and 𝑤 can be evaluated by using boundary conditions (3.23) at 𝑥=1: 21𝑟=0𝑌(𝑟)=𝑒1/2,21𝑟=01𝑟𝑌(𝑟)=2𝑒1/2.(3.26) These equations give 𝑡=0.03041272159 and 𝑤=0.01577358145.
Finally, the following series solution can be formed by applying the inverse transformation in (2.2) up to 𝑁=14: 𝑦(𝑥)=1.0+0.5𝑥+0.125𝑥2+0.304127215×101𝑥30.1577358145×101𝑥4+0.833333333×102𝑥5+0.694444444×103𝑥6+0.4960317460×104𝑥7+0.737671102×105𝑥80.3474145349×105𝑥9+8.094962522×107𝑥10+4.176090542×108𝑥111.102928451×108𝑥12+1.334963335×108𝑥137.232116854×109𝑥14.(3.27)

Example 3.3. Next, we solve the following fourth-order nonlinear for fifth-order BVP by using DTM: 𝑦(5)(𝑥)=𝑒𝑥(𝑦(𝑥))4,0<𝑥<1(3.28) subject to the boundary conditions 𝑦(0)=1,𝑦1(0)=3,𝑦1(0)=9,𝑦(1)=𝑒1/3,𝑦1(1)=3𝑒1/3.(3.29) By using Theorem 3.1, (3.28) is transformed to the following equation: 𝑌(𝑟+5)=𝑟!(𝑟+5)!𝑟𝑙𝑙𝑠𝑠𝑝𝑝𝑗(1)𝑗𝑗!𝑌(𝑝𝑗)𝑌(𝑠𝑝)𝑌(𝑙𝑠)𝑌(𝑟𝑙).(3.30) By using (2.1) at 𝑥=0, we obtain the following transformed boundary conditions: 𝑌(0)=1,𝑌(1)=131,𝑌(2)=,18(3.31) where 𝑡=𝑦(0)/3!=𝑌(3) and 𝑤=𝑦(4)(0)/4!=𝑌(4).
By using the transformed equation (3.30) together with the transformed boundary conditions (3.31), we can easily solve 𝑌(𝑟), for 𝑟5. The constants 𝑡 and 𝑤 can be evaluated by using boundary conditions (3.29) at 𝑥=1: 20𝑘=0𝑌(𝑟)=𝑒1/3,20𝑘=01𝑟𝑌(𝑟)=3𝑒1/3.(3.32) These equations give 𝑡=0.001331809417 and 𝑤=0.01491571112.
Finally, the following series solution can be formed by applying the inverse transformation in (2.2) up to 𝑁=20: 𝑦(𝑥)=1.00.3333333333𝑥+0.05555555556𝑥2+0.001331809417𝑥30.1491571112×101𝑥4+0.008333333333𝑥50.462962963×103𝑥6+0.220458553×104𝑥7+0.354847562×105𝑥80.404801946×105𝑥9+0.110569490×105𝑥102.741069465×108𝑥111.250500836×108𝑥121.134323528×108𝑥133.968854379×109𝑥14+2.378413332×1010𝑥15+1.215436605×1010𝑥162.331970237×1011𝑥17+1.269686222×1011𝑥18.(3.33)

Example 3.4. We perform the following third order of nonlinear equation for sixth-order BVP by using DTM: 𝑦(6)(𝑥)=𝑒𝑥(𝑦(𝑥))3,0<𝑥<1(3.34) subject to the boundary conditions 𝑦(0)=1,𝑦1(0)=2,𝑦1(0)=4,𝑦(1)=𝑒1/2,𝑦1(1)=2𝑒1/2,𝑦1(1)=4𝑒1/2.(3.35) By using Theorem 3.1, (3.34) is transformed to the following equation: 𝑌(𝑟+6)=𝑟!(𝑟+6)!𝑟𝑙𝑙𝑠𝑠𝑝(1)𝑝𝑝!𝑌(𝑠𝑝)𝑌(𝑙𝑠)𝑌(𝑟𝑙).(3.36) By using (2.1) at 𝑥=0, we obtain the following transformed boundary conditions: 1𝑌(0)=1,𝑌(1)=21,𝑌(2)=8,(3.37) where 𝑡=𝑦(0)/3!=𝑌(3), 𝑤=𝑦(4)(0)/4!=𝑌(4), and 𝑧=𝑦(5)(0)/5!=𝑌(5).
By using the transformed equation (3.36) together with the transformed boundary conditions (3.37), we can easily solve 𝑌(𝑟), for 𝑟6. The constants 𝑡 and 𝑤 can be evaluated by using boundary conditions (3.35) at 𝑥=1: 20𝑟=0𝑌(𝑟)=𝑒1/2,20𝑟=01𝑟𝑌(𝑟)=2𝑒1/2,20𝑘=01𝑟(𝑟1)𝑌(𝑟)=4𝑒1/2.(3.38) These equations give 𝑡=0.02083333265, 𝑤=0.2604165679𝐸2, and 𝑧=0.2604149398𝐸3.
Consequently, the following series solution can be formed by applying the inverse transformation equation in (2.2) up to 𝑁=20: 𝑦(𝑥)=1.00.5𝑥+0.125𝑥20.2194097232×101𝑥3+0.600832059×102𝑥40.003832229731𝑥5+0.001388888889𝑥60.00009920634921𝑥7+0.6200396825×105𝑥83.994089004×107𝑥9+8.476606127×108𝑥103.299619968×108𝑥11+6.203319625×109𝑥122.551978071×1010𝑥13+3.05519710×1011𝑥141.574229877×1011𝑥15+5.71359200×1012𝑥168.01789691×1013𝑥177.09484742×1014𝑥18.(3.39)

4. Adomian Decomposition Method

Consider the differential equation𝐿𝑢+𝑅𝑢+𝑁𝑢=𝑔,(4.1) where 𝐿 is invertible and is the highest-order derivative, 𝑅 is a linear differential operator, where the order of 𝐿 must be greater than 𝑅, 𝑁 is a nonlinear operator, and 𝑔 is the source term. By using the given conditions and applying the inverse operator 𝐿1 to both sides of (4.1), we get the following equation:𝑢=𝑓𝐿1(𝑅𝑢)𝐿1(𝑁𝑢),(4.2) where the function 𝑓 is arising from integrating the source term 𝑔 and from applying the given conditions which are prescribed.

The series solution of 𝑢(𝑥) by the standard Adomian method is given as follows:𝑢(𝑥)=𝑛=0𝑢𝑛(𝑥).(4.3) On the other hand, the nonlinear function 𝑢(𝑥) by an infinite series of polynomial is given as follows:𝑢(𝑥)=𝑛=0𝐴𝑛.(4.4) The components 𝑢𝑛 are determined by using the following relation:𝑢0𝑢=𝑓,(4.5)𝑘+1=𝐿1𝑅𝑢𝑘𝐿1𝑁𝑢𝑘(4.6) for 𝑘0.

From the above equations, we observe that the component 𝑢0 is identified by the function 𝑓. We can apply modification by assuming that the function 𝑓 is divisible into two parts, 𝑓0 and 𝑓1. Let the function be as follows:𝑓=𝑓0+𝑓1.(4.7) Under this assumption, we have a slight variation for components 𝑢0 and 𝑢1, where 𝑓0 assigned to 𝑢0 and 𝑓1 is combined with the other terms in (4.5) to assign 𝑢1. The modified recursive algorithm is as follows:𝑢0=𝑓0,𝑢1=𝑓1𝐿1𝑅𝑢0𝐿1𝑁𝑢0,𝑢𝑘+2=𝐿1𝑅𝑢𝑘+1𝐿1𝑁𝑢𝑘+1,(4.8) for 𝑘0. There are several rules that are needed to follow for Adomian polynomials of nonlinear operator 𝐹(𝑢):𝐴0𝑢=𝐹0,𝐴1=𝑢1(𝑥)𝐹𝑢0,𝐴2=𝑢2𝐹𝑢0+𝑢21𝐹2!𝑢0,𝐴3=𝑢3𝐹𝑢0+𝑢1𝑢2𝐹𝑢0+𝑢31𝐹3!𝑢0,(4.9) and so on; see [3].

For comparison purpose, we solve the boundary value problems in Examples 3.2, 3.3, and 3.4 by using the Adomian decomposition method.

Example 4.1. By using ADM, we solve the following nonlinear equation of order three for fifth-order BVP: 𝑢(5)(𝑥)=𝑒𝑥(𝑢(𝑥))3,0<𝑥<1(4.10) subject to the boundary conditions 𝑢(0)=1,𝑢1(0)=2,𝑢1(0)=4,𝑢(1)=𝑒1/2,𝑢1(1)=2𝑒1/2.(4.11) Equation (4.10) can be rewritten in operator form as follows: 𝐿𝑢=𝑒𝑥(𝑢(𝑥))3,0<𝑥<1.(4.12) Operating with fivefold integral operator 𝐿1 on (4.12) and using the boundary conditions at 𝑥=0, we obtain the following equation: 1𝑢(𝑥)=1+21𝑥+8𝑥2+16𝐴𝑥3+124𝐵𝑥4+𝐿1𝑒𝑥𝑢3.(𝑥)(4.13) Then, determine the constants 𝑢(0)=𝐴, 𝑢(4)(0)=𝐵. Substituting the decomposition series (4.3) for 𝑢(𝑥) and the series of polynomials (4.4) into (4.13) yields 𝑛=0𝑢𝑛1(𝑥)=1+21𝑥+8𝑥2+16𝐴𝑥3+124𝐵𝑥4+𝐿1𝑒𝑥𝑛=0𝐴𝑛.(4.14) Then, we split the terms into two parts which are assigned to 𝑢0(𝑥) and 𝑢1(𝑥) that are not included under 𝐿1 in (4.14). We can obtain the following recursive relation: 𝑢0𝑢(𝑥)=1,11(𝑥)=21𝑥+8𝑥2+16𝐴𝑥3+124𝐵𝑥4+𝐿1𝐴0,𝑢𝑘+1(𝑥)=𝐿1𝐴𝑘,𝑘1.(4.15) To determine the constants 𝐴 and 𝐵, we use the boundary conditions in (4.11) at 𝑥=1 on the four-term approximant 𝜙4, where 𝜙4=𝑘=3𝑘=0𝑦𝑘.(4.16) Then, solving the above equation yields 𝐴=0.376764,𝐵=2.154638929.(4.17) Then we get the series solution as follows: 𝑦(𝑥)=1.0+0.5𝑥+0.125𝑥2+0.6279×101𝑥30.8977662204×101𝑥4+0.583333333×101𝑥50.6944444444×102𝑥60.0005456349206𝑥70.0001800740079𝑥8+0.673748941×104𝑥90.3295218362×104𝑥10+0.741814372×105𝑥110.00000119022282𝑥127.604880242×108𝑥131.306718703×1010𝑥14.(4.18)

Example 4.2. Now, we solve the following fourth-order nonlinear for fifth-order boundary value problems by using ADM: 𝑢(5)(𝑥)=𝑒𝑥(𝑢(𝑥))4,0<𝑥<1(4.19) subject to the boundary conditions 𝑢(0)=1,𝑢1(0)=3,𝑢1(0)=9,𝑢(1)=𝑒1/3,𝑢1(1)=3𝑒1/3.(4.20) Equation (4.19) can be rewritten in operator form as follows: 𝐿𝑢=𝑒𝑥(𝑢(𝑥))4,0<𝑥<1.(4.21) Operating with fivefold integral operator 𝐿1 on (4.21) and using the boundary conditions at 𝑥=0, we obtain the following equation: 1𝑢(𝑥)=131𝑥+𝑥182+16𝐴𝑥3+124𝐵𝑥4+𝐿1𝑒𝑥𝑢4.(𝑥)(4.22) Then, determine the constants 𝑢(0)=𝐴, 𝑢(4)(0)=𝐵. Substituting the decomposition series (4.3) for 𝑢(𝑥) and the series of polynomials (4.4) into (4.22) yields 𝑛=0𝑢𝑛1(𝑥)=131𝑥+𝑥182+16𝐴𝑥3+124𝐵𝑥4+𝐿1𝑒𝑥𝑛=0𝐴𝑛.(4.23) Then, we split the terms into two parts which are assigned to 𝑢0(𝑥) and 𝑢1(𝑥) that are not included under 𝐿1 in (4.23). We can obtain the following recursive relation: 𝑢0𝑢(𝑥)=1,11(𝑥)=31𝑥+𝑥182+16𝐴𝑥3+124𝐵𝑥4+𝐿1𝐴0,𝑢𝑘+1(𝑥)=𝐿1𝐴𝑘,𝑘1.(4.24) To determine the constants 𝐴 and 𝐵, we use the boundary conditions in (4.20) at 𝑥=1 on the four-term approximant 𝜙4, where 𝜙5=𝑘=4𝑘=0𝑦𝑘.(4.25) Then, solving the above equation yields 𝐴=1.178489258,𝐵=0.1205488584.(4.26) Finally, we get the series solution as follows: 𝑦(𝑥)=10.3333333333𝑥+0.05555555556𝑥20.4453076220𝑥30.4040993754×101𝑥4+0.84845560𝑥50.298115585𝑥60.680956807×101𝑥70.2456515229×102𝑥8+0.6206348264×103𝑥90.1181956121×103𝑥100.0001615659149𝑥110.7466165099×104𝑥120.2012942457×104𝑥130.6094555007×105𝑥140.1080643342×105𝑥158.736244687×108𝑥161.417232124×109𝑥17+2.178184252×1011𝑥18.(4.27)

Example 4.3. Finally, we perform third order of nonlinear function for sixth-order BVP for ADM as follows: 𝑢(6)(𝑥)=𝑒𝑥(𝑢(𝑥))3,0<𝑥<1(4.28) subject to the boundary conditions 𝑢(0)=1,𝑢1(0)=2,𝑢1(0)=4,𝑢(1)=𝑒1/2,𝑢1(1)=2𝑒1/2,𝑢1(1)=4𝑒1/2.(4.29) Equation (4.28) can be rewritten in operator form as follows: 𝐿𝑢=𝑒𝑥(𝑢(𝑥))3,0<𝑥<1.(4.30) Operating with sixfold integral operator 𝐿1 on (4.30) and using the boundary conditions at 𝑥=0, we obtain the following equation: 1𝑢(𝑥)=121𝑥+8𝑥2+16𝐴𝑥3+124𝐵𝑥4+1120𝐶𝑥5+𝐿1𝑒𝑥𝑢4.(𝑥)(4.31) Then, determine the constants 𝑢(0)=𝐴, 𝑢(4)(0)=𝐵, and 𝑢(5)(0)=𝐶. Substituting the decomposition series (4.3) for 𝑢(𝑥) and the series of polynomials (4.4) into (4.31) yields 𝑛=0𝑢𝑛1(𝑥)=121𝑥+8𝑥2+16𝐴𝑥3+124𝐵𝑥4+1120𝐶𝑥5+𝐿1𝑒𝑥𝑛=0𝐴𝑛.(4.32) Then similarly, we split the terms into two parts which are assigned to 𝑢0(𝑥) and 𝑢1(𝑥) that are not included under 𝐿1 in (4.32). We can obtain the following recursive relation: 𝑢0𝑢(𝑥)=1,11(𝑥)=21𝑥+8𝑥2+16𝐴𝑥3+124𝐵𝑥4+1120𝐶𝑥5+𝐿1𝐴0,𝑢𝑘+1(𝑥)=𝐿1𝐴𝑘,𝑘1.(4.33) To determine the constants 𝐴, 𝐵, and 𝐶, we use the boundary conditions in (4.29) at 𝑥=1 on the four-term approximant 𝜙4, where 𝜙4=𝑘=3𝑘=0𝑦𝑘.(4.34) Then, solving the above equation yields 𝐴=14.69918030,𝐵=102.1542319,𝐶=219.2146156.(4.35) Finally, we get the series solution as follows: 𝑦(𝑥)=10.5𝑥+0.125𝑥22.449863383𝑥3+4.256426329𝑥41.826788463𝑥5+0.138888888×102𝑥6+0.1984126984×103𝑥7+0.0000248015873𝑥8+0.2292633427×103𝑥90.343791062×104𝑥100.680212248×104𝑥11+0.406878271×104𝑥120.00005031070540𝑥13+0.00003795993196𝑥140.128793671×104𝑥15+0.000001752215748𝑥165.518698119×1012𝑥17.(4.36)

5. Results

We provide the results of the given examples in Tables 1, 2, and 3.

tab1
Table 1: Numerical result for Example 3.2 and Example 4.1.
tab2
Table 2: Numerical result for Example 3.3 and Example 4.2.
tab3
Table 3: Numerical result for Example 3.4 and Example 4.3.

6. Conclusion

From the results, the proposed method, as well as the Differential Transformation Method, is more accurate than the Adomian Decomposition Method. The errors between the solutions of Differential Transformation Method and the exact solutions are smaller compared to the errors between the solutions of Adomian Decomposition Method and the exact solutions. In addition, Differential Transformation Method also shows less computational effort because it needs less time in calculation. Besides that, it is hard to calculate Adomian polynomials. From the results we obtained, it can reinforce conclusion made by many researchers that Differential Transformation Method is more efficient and accurate than Adomian Decomposition Method. Therefore, we can conclude that Differential Transformation Method is applicable for such problems in the bounded domains. The computations in all examples were performed by using Maple 13.

Acknowledgments

The authors gratefully acknowledge that this research was partially supported by the University Putra Malaysia under the Research University Grant scheme 05-01-09-0720RU and Fundamental Research Grant scheme 01-11-09-723FR. The authors also thank the referee(s) for very constructive comments and suggestions that improved the paper.

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