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The purpose of this paper is to employ the Adomian Decomposition Method (ADM) and Restarted Adomian Decomposition Method (RADM) with new useful techniques to resolve Bratu’s boundary value problem by using a new integral operator. The solutions obtained in this way require the use of the boundary conditions directly. The obtained results indicate that the new techniques give more suitable and accurate solutions for the Bratu-type problem, compared with those for the ADM and its modification.

In 1914, the Bratu problem was first set up by Bratu, then the problem was named after him [

The classical nonlinear Bratu’s boundary value problem in one-dimensional planar coordinates is in the form [

u ″ + λ e u = 0 , 0 < x < 1 ， u ( 0 ) = u ( 1 ) = 0. (1)

The exact solution to Equation (1) is given in [

u ( x ) = − 2 ln [ cosh ( ( x − 1 2 ) θ 2 ) cosh ( θ 4 ) ] . (2)

where θ satisfies

θ = 2 λ cosh ( θ 4 ) .

There are three cases for λ :

1) If λ > λ c , then the Bratu problem has zero solution.

2) If λ = λ c , then the Bratu problem has one solution.

3) If λ < λ c , then the Bratu problem has two solutions.

where the critical value λ c satisfies the equation

1 = 1 4 2 λ c sinh ( θ c 4 ) ,

where

λ c = 3.513830719.

The Bratu problem was gained the attention of researchers for two important reasons:

First: it appears in a wide variety of application areas such as physical, chemical and engineering. Through these applications, we can mention the reaction of thermal, the theory of chemical reactor and nanotechnology. Specific models include Chandrasekhar model of the expansion of the universe [

Second: because of its simplicity, the equation is widely used as a benchmarking tool for various numerical techniques. Several numerical methods, such as the finite difference method [

In 1980, the Adomian decomposition method (ADM) was first introduced and developed by Adomian [

As far as our information, few of researches have used ADM and its modifications to obtain the approximate solution of the Bratu’s boundary value problem, such as ADM used in [

In present study, we apply the ADM and restarted Adomian decomposition method (RADM) [

This research consists of an introduction and three four sections as follows: In Section 2, the ADM and the RADM with new integral operator will be presented as it applies to Bratu problem. In Section 3, the Comparison of this method to two test problems is presented, together with the Adomian decomposition method and its modifications. Finally, the conclusion is as follows in Section 4.

In this section, the standard ADM and RADM by using new integral operator are introduced, and the scheme is implemented for the solution of Bratu’s boundary value problem.

We have two cases to apply the standard ADM with new integral operator for solving boundary value problem of the Bratu-type. For first case, we consider the nonlinear term as e u , and we use Taylor series of e u in second case.

Case 1. We consider the boundary value problem of Bratu-type with nonlinear term as e u

u ″ + λ e u = 0 , 0 < x < 1 , u ( 0 ) = u ( 1 ) = 0. (3)

The operator form of Equation (3) can be written as

L u = − λ e u , u ( 0 ) = u ( 1 ) = 0. (4)

where the differential operator L is L = d 2 d x 2 .

The inverse L − 1 is assumed a new two-fold integral operator [

L − 1 ( ⋅ ) = ∫ 0 x ∫ 0 x ( ⋅ ) d x d x − x ∫ 0 1 ∫ 0 x ( ⋅ ) d x d x .

By applying L − 1 on both sides of Equation (4) and using the boundary condition u ( 0 ) = u ( 1 ) = 0 we get

u ( x ) = − λ L − 1 ( e u ) . (5)

The decomposition method assumes u ( x ) and the nonlinear term

N ( u ) = e u as infinite series given by

u ( x ) = ∑ n = 0 ∞ u n , N ( u ) = ∑ n = 0 ∞ A n (6)

where A n are the so-called Adomian polynomials of u 0 , ⋯ , u n given by

A n = 1 n ! d n d λ n [ N ( ∑ i = 0 n λ i u i ) ] λ = 0 , n = 0 , 1 , 2 , ⋯

substituting Equation (6) into Equation (5) yields

∑ n = 0 ∞ u n = − λ L − 1 ( ∑ n = 0 ∞ A n ) .

Identifying the zeroth component u 0 by 0, the remaining components

u n , n ≥ 1 can be obtained recurrently by using the relation

u 0 = 0 , u n + 1 = − λ L − 1 ( A n ) , n ≥ 0 , (7)

where A n of the nonlinear term e u , we can obtain

A 0 = e u 0 , A 1 = u 1 e u 0 , A 2 = ( u 2 + 1 2 u 1 2 ) e u 0 , A 3 = ( u 3 + u 1 u 2 + 1 6 u 1 3 ) e u 0 , ⋮

The solution components u n + 1 from Equation (7) can be calculated as

u 1 = − 1 2 λ x 2 + 1 2 λ x , u 2 = 1 24 λ 2 x 4 − 1 12 λ 2 x 3 + ⋯ , u 3 = − 1 180 λ 3 x 6 + 1 60 λ 3 x 5 − ⋯ , u 4 = 17 20160 λ 4 x 8 − 17 5040 λ 4 x 7 + ⋯ , ⋮

The approximate solution u ( x ) is obtained in a series form

u ( x ) = ∑ n = 0 ∞ u n = u 0 + u 1 + u 2 + u 3 + u 4 + ⋯ .

Case 2. For accelerating the convergence of the ADM when used for nonlinear differential equations, we replaced nonlinear terms by their Taylor expansion.

Toward this end, we can consider e u as

e u ≃ 1 + u + u 2 2 ! + u 3 3 ! .

The Equation (3) can be written as

u ″ + λ ( 1 + u + u 2 2 ! + u 3 3 ! ) = 0 , 0 < x < 1 , u ( 0 ) = u ( 1 ) = 0. (8)

Applying the standard ADM with new integral operator in Equation (8), we get

∑ n = 0 ∞ u n = − λ ( L − 1 ( 1 ) + L − 1 ( ∑ n = 0 ∞ u n ) + 1 2 ! L − 1 ( ∑ n = 0 ∞ A n ) + 1 3 ! L − 1 ( ∑ n = 0 ∞ B n ) ) ,

where

A n = 1 n ! d n d λ n [ ( ∑ i = 0 n λ i u i ) 2 ] λ = 0 , B n = 1 n ! d n d λ n [ ( ∑ i = 0 n λ i u i ) 3 ] λ = 0 , n = 0 , 1 , 2 , ⋯

Identifying the zeroth component u 0 by – λ L − 1 ( 1 ) , the remaining components u n , n ≥ 1 can be obtained recurrently by using the relation

u 0 = − 1 2 λ x 2 + λ 1 2 x , u n + 1 = − λ ( L − 1 ( u n ) + 1 2 L − 1 ( A n ) + 1 6 L − 1 ( B n ) ) , n ≥ 0. (9)

The solution components u n + 1 from Equation (9) can be calculated as

u 1 = 1 2688 λ 4 x 8 − 1 672 λ 4 x 7 + ⋯ , u 2 = − 1 3913728 λ 7 x 14 + 1 559104 λ 7 x 13 − ⋯ , u 3 = 5 28554559488 λ 10 x 20 − 25 14277279744 λ 10 x 19 + ⋯ , u 4 = − 449 3741789475307520 λ 13 x 26 + 449 287829959639040 λ 13 x 25 − ⋯ , ⋮

The approximate solution u ( x ) is obtained in a series form

u ( x ) = ∑ n = 0 ∞ u n = u 0 + u 1 + u 2 + u 3 + u 4 + ⋯ .

Generally, the RADM has the same structure as the ADM, but the ADM is used more than once. In this section, we extend the RADM for solving the Bratu’s boundary value problem. Noting that we can apply the RADM with new integral operator for the translated equation

u ″ + λ ( 1 + u + u 2 2 ! + u 3 3 ! ) = 0 , 0 < x < 1 , u ( 0 ) = u ( 1 ) = 0. (10)

Applying the standard ADM with new integral operator in Equation (10), we get

∑ n = 0 ∞ u n = − λ ( L − 1 ( 1 ) + L − 1 ( ∑ n = 0 ∞ u n ) + 1 2 ! L − 1 ( ∑ n = 0 ∞ A n ) + 1 3 ! L − 1 ( ∑ n = 0 ∞ B n ) ) . (11)

This gives

u 0 = − 1 2 λ x 2 + λ 1 2 x , u n + 1 = − λ ( L − 1 ( u n ) + 1 2 L − 1 ( A n ) + 1 6 L − 1 ( B n ) ) , n ≥ 0. (12)

Adding and subtracting g ( x ) to right side of Equation (11) to obtain

∑ n = 0 ∞ u n = g ( x ) − λ ( L − 1 ( 1 ) + L − 1 ( ∑ n = 0 ∞ u n ) + 1 2 ! L − 1 ( ∑ n = 0 ∞ A n ) + 1 3 ! L − 1 ( ∑ n = 0 ∞ B n ) ) − g ( x ) .

By equating the terms we can get

u 0 = g ( x ) , u 1 = − λ ( L − 1 ( 1 ) + L − 1 ( u 0 ) + 1 2 L − 1 ( A 0 ) + 1 6 L − 1 ( B 0 ) ) − g ( x ) , u n + 2 = − λ ( L − 1 ( u n + 1 ) + 1 2 L − 1 ( A n + 1 ) + 1 6 L − 1 ( B n + 1 ) ) , n ≥ 0. (13)

Step 1: In this step, g ( x ) is calculated from Equation (12) as follows:

u 0 = − 1 2 λ x 2 + 1 2 λ x , u 1 = 1 2688 λ 4 x 8 − 1 672 λ 4 x 7 + ⋯ ,

So

g ( x ) = ϕ 1 ( x ) = u 0 + u 1 .

Step 2: Now, components of the RADM is computed from Equation (13) as follows:

u 0 = g ( x ) , u 1 = − 1 75744726220800 λ 13 x 26 + 1 5826517401600 λ 13 x 25 − ⋯ , u 2 = 1 2070913964648082230476800 λ 22 x 44 − 1 94132452938549192294400 λ 22 x 43 + ⋯ , u 3 = − 1921 110351113887194570160875411425198080000 λ 31 x 62 + ⋯ , u 4 = 18470161 29478690975630536257743382096194416238070333440000000 λ 40 x 80 − ⋯ ,

So

ϕ 2 ( x ) = u 0 + u 1 + u 2 + u 3 + u 4 .

The approximate solution u ( x ) is obtained in a series form

u ( x ) = ϕ 2 ( x ) = u 0 + u 1 + u 2 + u 3 + u 4 .

In this section, we present our numerical results for Bratu’s problem. In order to demonstrate the robustness of the schemes, we will consider of two parameters for the eigenvalue λ. The two methods will be illustrated through the following examples.

We can consider the Bratu’s problem

u ″ + e u = 0 , 0 < x < 1 , u ( 0 ) = u ( 1 ) = 0.

In

The approximation’s accuracy is reflected in

We can consider the Bratu’s problem

u ″ + e u = 0 , 0 < x < 1 , u ( 0 ) = u ( 1 ) = 0.

The approximation’s accuracy is reflected in

x | Exact | ADM | ADM with Taylor | RADM with Taylor | ADM [ | LADM [ |
---|---|---|---|---|---|---|

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 | 0 0.0498467900 0.0891899350 0.1176090956 0.1347902526 0.1405392142 0.1347902526 0.1176090956 0.0891899350 0.0498467900 0 | 0 2.9 × 10^{−}^{5} 5.6 × 10^{−}^{5} 7.9 × 10^{−}^{5} 9.4 × 10^{−}^{5} 9.9 × 10^{−}^{5} 9.4 × 10^{−}^{5} 7.9 × 10^{−}^{5} 5.6 × 10^{−}^{5} 2.9 × 10^{−}^{5} 2.5 × 10^{−}^{11} | 0 1.4 × 10^{−}^{6} 2.8 × 10^{−}^{6} 3.9 × 10^{−}^{6} 4.6 × 10^{−}^{6} 4.9 × 10^{−}^{6 } 4.6 × 10^{−}^{6} 3.9 × 10^{−}^{6} 2.8 × 10^{−}^{6} 1.4 × 10^{−}^{6} 0 | 0 4.8 × 10^{−}^{7} 9.4 × 10^{−}^{7} 1.3 × 10^{−}^{7} 1.6 × 10^{−}^{7} 1.7 × 10^{−}^{7} 1.6 × 10^{−}^{7} 1.3 × 10^{−}^{7 } 9.4 × 10^{−}^{7} 4.8 × 10^{−}^{7} 1 × 10^{−}^{11} | 2.7 × 10^{−}^{3} 2.0 × 10^{−}^{3} 1.5 × 10^{−}^{4} 2.2 × 10^{−}^{3} 3.0 × 10^{−}^{3} 2.2 × 10^{−}^{3} 1.5 × 10^{−}^{4} 2.0 × 10^{−}^{3} 2.7 × 10^{−}^{3} | 2.0 × 10^{−}^{6} 3.9 × 10^{−}^{6} 5.9 × 10^{−}^{6} 7.7 × 10^{−}^{6} 9.5 × 10^{−}^{6} 1.1 × 10^{−}^{5} 1.3 × 10^{−}^{5} 1.3 × 10^{−}^{5} 1.2 × 10^{−}^{5} |

x | Exact | ADM | ADM with Taylor | RADM with Taylor | ADM [ | LADM [ |
---|---|---|---|---|---|---|

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 | 0 0.1144107440 0.2064191156 0.2738793116 0.3150893646 0.3289524216 0.3150893646 0.2738793116 0.2064191156 0.1144107440 0 | 0 1.3 × 10^{−3} 2.5 × 10^{−3} 3.6 × 10^{−3} 4.2 × 10^{−3} 4.5 × 10^{−3} 4.2 × 10^{−3} 3.6 × 10^{−3} 2.5 × 10^{−3} 1.3 × 10^{−3} 9 × 10^{−}^{11} | 0 2.0 × 10^{−}^{4} 3.9 × 10^{−}^{4} 5.4 × 10^{−}^{4} 6.5 × 10^{−}^{4} 6.8 × 10^{−}^{4} 6.5 × 10^{−}^{4} 5.4 × 10^{−}^{4} 3.9 × 10^{−}^{4} 2.0 × 10^{−}^{4} 0 | 0 6.5 × 10^{−}^{5} 1.3 × 10^{−}^{4} 1.8 × 10^{−}^{4} 2.1 × 10^{−}^{4} 2.3 × 10^{−}^{4} 2.1 × 10^{−}^{4} 1.8 × 10^{−}^{4} 1.3 × 10^{−}^{4} 6.5 × 10^{−}^{5} 2.8 × 10^{−}^{10} | 1.5 × 10^{−}^{2} 1.5 × 10^{−}^{2} 5.9 × 10^{−}^{2} 3.3 × 10^{−}^{2} 7.0 × 10^{−}^{2} 3.3 × 10^{−}^{2} 5.9 × 10^{−}^{2} 1.5 × 10^{−}^{2} 1.5 × 10^{−}^{2} | 2.1 × 10^{−}^{2} 4.2 × 10^{−}^{2} 6.2 × 10^{−}^{2} 8.0 × 10^{−}^{2} 9.6 × 10^{−}^{2} 1.1 × 10^{−}^{2} 1.2 × 10^{−}^{2} 1.2 × 10^{−}^{2} 1.1 × 10^{−}^{2} |

perfect match of those solutions, bating the approximate solutions of ADM.

The ADM and RADM with new integral operator as presented in this paper have been shown to be more efficient for solving Bratu’s boundary value problem. The main advantage of the method is that it can use the boundary conditions directly. In addition, it is capable of greatly reducing the size of computational work while still maintaining high accuracy of the numerical solution. Furthermore, this paper shows the validity of the ADM and RADM with new operator for nonlinear problems in science and engineering.

Al-Mazmumy, M., Al-Mutairi, A. and Al-Zahrani, K. (2017) An Efficient Decomposition Method for Solving Bratu’s Boundary Value Problem. American Journal of Computational Mathematics, 7, 84-93. https://doi.org/10.4236/ajcm.2017.71007