Exact Solutions for a Class of Singular Two-Point Boundary Value Problems Using Adomian Decomposition Method

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1 Applied Mathematical Sciences, Vol 6, 212, no 122, Exact Solutions for a Class of Singular Two-Point Boundary Value Problems Using Adomian Decomposition Method Abdelhalim Ebaid 1 and Mona D Aljoufi 2 Department of Mathematics, Faculty of Science, Tabuk University PO Box 741, Tabuk 71491, Saudi Arabia halimgamil@yahoocom 1 aljufi-mona@hotmailcom 2 Abstract Some of the most common problems in applied sciences and engineering are usually formulated as singular two-point boundary value problems A well known fact is that the exact solutions in closed form of such problems were not obtained in many cases In this paper, the exact solutions for a wide class of singular two-point boundary value problems are obtained by using Adomian decomposition method Keywords: Adomian decomposition method; singular two-point boundary value problem; Exact solution 1 Introduction In this work, we consider a class of singular two-point boundary value problems: Subject to the boundary conditions (x α y = f(x, y, x 1 (1 y( = A, y(1 = B, (2 where α (, 2], A and B are finite constants Many problems in applied mathematics, such as gas dynamics, nuclear physics, chemical reaction, studies of atomic structures and atomic calculations, lead to singular boundary value problems given by Eq (1 subject to the boundary conditions (2 For example, when α = and f(x, y =q(xy σ, Eq (1 is known as the generalized Emden-Fowler equation with negative exponent and arises frequently in applied mathematics, (see [1,2] and the references cited therein Also when α = and f(x, y =x 1/2 y 3 2, Eq (1 is known as Thomas-Fermi equation [3], given by

2 698 A Ebaid and M D Aljoufi the singular equation: y = x 1/2 y 3/2, which arises in the study of the electrical potential in an atom When α = p {, 1, 2}, another example is given by the singular equation: (x p y = x p f(y, which results from an analysis of heat conduction through a solid with heat generation The function f(y represents the heat generation within the solid, y is the temperature and the constant p is equal to, 1 or 2 depending on whether the solid is a plate, a cylinder or a sphere [4] In recent years, the class of the singular two-point boundary value problems given by (1 and (2 was studied by many mathematicians and a number of numerical methods [4-14] and analytical methods [15,16] have been proposed Although these numerical methods have many advantages, a huge amount of computational work is required in obtaining accurate approximations The most important notice here is that all the previous attempts during the last two decades were devoted only to obtain approximate solutions wether numerical or analytical In this research, the exact solutions for a wide class of singular two-point boundary value problems will be obtained by implementing the Adomian decomposition method (ADM and its modification (MADM made by Wazwaz [17] 2 Analysis of Adomian decomposition method The Adomian decomposition method and its modification have been used during the past three decades to solve effectively and easily a large class of linear and nonlinear ordinary and partial differential equations In this section, the class of singular BVPs (1-(2, will be handled more easily, quickly and elegantly by implementing this method (ADM rather than by the traditional methods for the exact solutions without making massive computational work Let us begin our analysis by writing Eq (1 in an operator form: L(y =f(x, y, (3 where the linear differential operator L is defined by L[] = d ( x α d dx dx [] (4 The inverse operator L 1 is therefore defined by L 1 [] = (x α []dx dx (5 Operating with L 1 on Eq (3, it then follows y = y( (x α [f(x, y] dx dx (6 Notice that only y( = A, is sufficient to carry out the solution and the other condition y(1 = B, can be used to show that the obtained solution satisfies this

3 Exact solutions for a class of singular two-point BVP 699 given condition In this paper, we consider the linear case: f(x, y =g(xr(xy, so Eq (6 becomes y = y( (x α [g(x] dx dx (x α [r(xy] dx dx (7 Now we decompose the solution y as y = y n (8 n= It is important to note that the standard ADM suggests that the zeroth component y usually defined by the function h(x: h(x =y( (x α [g(x] dx dx, (9 that represents the terms arising from integrating the source term g(x and from using the given conditions and all are assumed to be prescribed Accordingly, the solution can be computed by using the recurrence relation y = h(x, y n1 = (x α [r(xy n ] dx dx, n (1 The modified decomposition method needs only a slight variation from the standard Adomian decomposition method and has been shown to be computationally efficient in several examples It is based on dividing the function h(x into two parts, namely h 1 (x and h 2 (x Under this assumption, the solution can be computed by using the recurrence scheme y = h 1 (x, y 1 = h 2 (x (x α [r(xy ] dx dx, (11 y n1 = (x α [r(xy n ] dx dx, n 1 It is also indicated in [17] that the power of the MADM depends mainly on the proper selection of the functions h 1 (x and h 2 (x In the next section, we show that algorithms (1 and (11 are very effective and powerful in obtaining the exact solutions for a wide class of linear singular two-point boundary value problems in the form given by (1 and (2 3 Applications and exact solutions 31 Example 31 Firstly, we consider a simple linear singular equation [1]: ( x 1/2 y =5x 5/ , (12

4 61 A Ebaid and M D Aljoufi subject to the boundary conditions y( =, y(1 = (13 Applying algorithm (1 for this problem, we get [ y = (x 1/2 5x 5/2 15 ] dx dx = 5 ( x 4 x 3/2, y n1 =, n (14 Using Eq (8, the solution is then given by which is the exact solution of Eqs ( Example 32 y = 5 ( x 4 x 3/2, (15 14 In this example, we also consider the simple singular equation [11]: subject to the boundary conditions Firstly, Eq (16 can be written as y 1 x y = cos(x 1 sin(x, (16 x y( = 1, y(1 = cos(1 (17 (xy = x cos(x sin(x (18 On using algorithm (1 for this singular BVP, we obtain y =1 (x 1 [ x cos(x sin(x] dx dx = cos(x, y n1 =, n (19 In view of (8, the exact solution: y = cos(x is immediately obtained 33 Example 33 Consider the linear singular equation [15,16]: y 1 x y y = 5 4 x2 16, (2

5 Exact solutions for a class of singular two-point BVP 611 subject to the boundary conditions y( = 1 y(1 = (21 Here, we can also rewrite Eq (2 in the form (xy = xy 5x 4 x3 16 (22 In this example, we show that algorithm (11 is effective in obtaining the exact solution as a result of the appearance of the self-cancelling noise-terms phenomenon Before doing so, we may discuss the way of choosing the functions h 1 (x and h 2 (x From Eq (9, we find that the function h(x is given by h(x =1 ( x x x 1 ( 5x 4 x3 16 dx dx =1 5x2 16 x4 256 (23 As shown in [17], the reduction in the number of terms of y will result in a reduction of the computational work So, in this example we choose h 1 (x =1 and h 2 (x = 5x2 x4, consequently, algorithm (11 becomes y =1, y 1 = 5x2 16 x4 (x [xy ] dx dx, = (x 1 [xy n ] dx dx, n 1 (24 y n1 From this recurrence relation, we can easily obtain y =1, y 1 = x2 16 x4 256, y 2 = x4 256 x6 9216, y 3 = x x , (25 y 4 = x x , x 1 y 5 = x , Considering the two components y 1 and y 2, we observe the appearance of the self-cancelling noise-terms x 4 /256 in y 1 and x 4 /256 in y 2 Also many other selfcancelling noise-terms appear in the components y 2, y 3, y 4, and y 5 By cancelling these noise-terms, we obtain the exact solution: y =1x 2 /16, in the limit

6 612 A Ebaid and M D Aljoufi 34 Example 34 We next consider the linear singular equation [1]: y 2 x y (1 x 2 y = x 4 2x 2 7, (26 subject to y( = 1, y(1 = (27 We also rewrite Eq (26 in the form ( x 2 y =(x 2 x 4 y x 5 2x 3 7x (28 Following the same analysis of the previous example, the solution of this singular BVP can be elegantly computed by using the recursive scheme: y =1, y 1 = 7x2 6 x4 1 x6 x x (x 42 2 y n1 = (x 2 [ (x 2 x 4 y n ] dx dx, n 1 From this recursive relation, we can easily obtain y =1, [ (x 2 x 4 y ] dx dx, (29 y 1 = x 2 x4 2 x6 42, y 2 = x4 2 x6 4 31x8 324 x1 462, y 3 = x6 84 x x x x , (3 y 4 = x8 648 x x x x x , By investigating the two components y 1 and y 2, we observe the appearance of the self-canceling noise-terms x 4 /2 in y 1 and x 4 /2 in y 2 Considering the components y 1, y 2 and y 3, we observe the appearance of the noise-terms x 6 /42, x 6 /4 and x 6 /84 which vanish when added By canceling these noise-terms beside the others that appear in the next components, we obtain the exact solution: y =1 x 2, in the limit

7 Exact solutions for a class of singular two-point BVP Example 35 Consider the linear singular equation [16]: with the boundary conditions y (x 1 x y (xy(x =4 9x x 2 x 3, (31 y( =, y(1 = (32 In this problem, Eq (31 has a singular point at x = and can be written as (xy = xy 4x 9x 2 x 3 x 4 (33 Again, we use algorithm (11 for this singular BVP to obtain y [ = (x 1 4x 9x 2 x 3 x 4] dx dx = x 2 x 3, y 1 = x4 16 x5 x x (x 25 [xy ] dx dx =, (34 y n1 =, n 1 Also in view of Eq (8, we obtain the exact solution: y = x 2 x 3 36 Example 36 Consider the Bessel equation of order zero [12,14]: with the boundary conditions The given Bessel equation can be written as On applying algorithm (1, we obtain y (x 1 x y (xy(x =, (35 y( = 1/J (1, y(1 = 1 (36 y =1/J (1, = (x 1 y n1 (xy = xy (37 [xy n ] dx dx, n (38

8 614 A Ebaid and M D Aljoufi This recursive scheme gives y = 1 J (1, y 1 = x2 4J (1 = x (1! 2 J (1, y 2 = x4 64J (1 = ( 12 x (2! 2 J (1, y 3 x 6 = 234J (1 = ( 13 x (3! 2 J (1, y 4 x 8 = J (1 = ( 14 x (4! 2 J (1, (39 x 1 y 5 = J (1 = ( 15 x (n! 2 J (1, y n = ( 1n x 2n 2 2n (n! 2 J (1, n The solution y is now given by y = y n = 1 n= J (1 which is also the exact solution 37 Example 37 n= We next consider the linear singular equation [13]: subject to ( 1 n x 2n 2 2n (n! 2 = J (x J (1, (4 y 2 x y 4y = 2, (41 y( = 1 csch(2, 2 11 y(1 = 2 (42 Here, we rewrite Eq (41 again in the form ( x 2 y =4x 2 y 2x 2 (43 Proceeding as above, the solution is determined by using the recursive scheme: y = γ, γ = 1 2 csch(2, y 1 = x2 x x (x 6 [ ] 2 4x 2 y dx dx, (44 [ ] y n1 = (x 2 4x 2 y n dx dx, n 1

9 Exact solutions for a class of singular two-point BVP 615 On using (44 we obtain y = γ, y 1 = 1 8x y 2 = 1 8x y 3 = 1 8x y 4 = 1 8x The Eq (8 gives the solution y as (4γ 2(2x3, 3! (4γ 2(2x5, 5! (4γ 2(2x7, 7! (4γ 2(2x9, (45 9! y n = 1 (2x2n1 (4γ 2 8x (2n 1!, n 1 y = y n = γ 1 n= 8x (4γ 2 (2x 2n1 n=1 (2n 1! = γ 1 (4γ 2 [sinh(2x 2x] 8x (46 Inserting the value of γ into Eq (46, the exact solution: y = sinh(2x x sinh(2,is easily obtained 38 Example 38 Finally, we consider the following class of singular two-point BVPs [1-12]: y (x α x y (x =βx β 2 [ α β 1βx β] y(x, β > (47 with the boundary conditions We can rewrite Eq (47 as y( = 1, y(1 = e (48 (x α y = βx αβ 2 [ α β 1βx β] y(x (49 This class has been studied extensively by many authors [1-12] at certain values of α and β by using several numerical techniques Although, the numerical solution obtained by these methods is qualitatively accurate, a massive computational work was required However, little attention was devoted to the analytical

10 616 A Ebaid and M D Aljoufi investigation of this class So, we aim here to obtain an analytical solution for this class and to show that the exact solution can be also obtained for any α and β> Proceeding as in the previous examples, we obtain the following recursive scheme for n 1: y =1, y n1 = β x (x α x From this recurrence relation, we can easily obtain [ (α β 1x αβ 2 βx α2β 2] y n dx dx (5 y =1, y 1 = x β βx 2β 2(α 2β 1, y 2 (α β 1x2β = 2(α 2β 1 β(3α 5β 3x 3β 6(α 2β 1(α 3β 1 β 2 x 4β 8(α 2β 1(α 4β 1, y 3 = (α β 1 2 x 3β 6(α 2β 1(α 3β 1 β(α β 1(3α 7β 3x 4β 12(α 2β 1(α 3β 1(α 4β 1 [3(α β 1(α 3β 1 4(3α 5β 3(α 4β 1]βx 5β 12(α 2β 1(α 3β 1(α 4β 1(α 5β 1 β 3 x 6β 48(α 2β 1(α 4β 1(α 6β 1 (51 To show the possibility of obtaining the exact solution of Eq (47, let us begin by calculating the approximate solutions Φ 1,Φ 2,Φ 3 and Φ 4 as Φ 1 = y =1, Φ 2 = y y 1 =1x β βx 2β 2(α 2β 1, Φ 3 = y y 1 y 2 =1x β (xβ 2 β(3α 5β 3x 3β 2! 6(α 2β 1(α 3β 1 β 2 x 4β 8(α 2β 1(α 4β 1, Φ 4 = y y 1 y 2 y 3 =1x β (xβ 2 (xβ 3 2! 3! 3β 2 (α 3β 1 2β(α β 1(3α 7β 3x 4β 24(α 2β 1(α 3β 1(α 4β 1 [3(α β 1(α 3β 1 4(3α 5β 3(α 4β 1]βx 5β 12(α 2β 1(α 3β 1(α 4β 1(α 5β 1 β 3 x 6β 48(α 2β 1(α 4β 1(α 6β 1 (52

11 Exact solutions for a class of singular two-point BVP 617 By investigating the approximate solutions Φ 2,Φ 3 and Φ 4, we find that they agree with the Maclaurin expansion of the function e xβ up to x β, x 2β and x 3β, respectively So, by evaluating more terms of the decomposition series, we obtain the exact solution: y = e xβ in the limit 4 Conclusions Based on Adomian s method, an efficient approach is proposed in this paper for solving the linear singular two-point BVPs The proposed approach is shown to be very effective and powerful in finding the exact solutions of a wide class of linear singular second-order boundary value problems Moreover, the approach not only used in a straightforward manner but also requires less computational works in comparison with the other methods References [1] J Janus, J Myjak, A generalized Emden-Fowler equation with a negative exponent, Nonlinear Anal, 23 (8 (1994, [2] PK Palamides, Monotone positive solutions for singular boundary value problems, J Comput Appl Math, 146 (22, [3] G Adomian, Solution of the Thomas-Fermi Equation, Appl Math Lett, 11 (3 (1998, [4] MK Kadalbajoo, VK Aggarwal, Numerical solution of singular boundary value problems via Chebyshev polynomial and B-spline, Appl Math Comput, 16 (25, [5] PM Lima, AM Oliveira, Numerical solution of a singular boundary value problem for a generalized Emden-Fowler equation, Appl Numer Math, 45 (23, [6] T Aziz, M Kumar, A fourth-order finite-difference method based on nonuniform mesh for a class of singular two-point boundary value problems, J Comput Appl Math, 136 (21, [7] M Kumar, A fourth-order finite difference method for a class of singular two-point boundary value problems, Appl Math Comput, 133 (22, [8] A Ebaid, Exact solutions for a class of nonlinear singular two-point boundary value problems: The decomposition method, Z Naturforsch A, 65 (21,

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