Consider an Initial Value Problem (IVP) for a nonlinear Ordinary Differential Equation (ODE) in the form,

L u + R u + N u = g , (2)

where L is the linear operator to be inverted, which usually is the highest order differential operator, N represents the nonlinear term, R is the linear remainder operator and g is the source term. Applying the inverse operator, L − 1 on both sides of Equation (2) gives,

L − 1 L u = L − 1 g − L − 1 [ R u + N u ] . (3)

Solving for u in Equation (3) we get,

u = φ ( 0 ) + L − 1 g − L − 1 [ R u + N u ] , (4)

where,

φ ( 0 ) = ( u ( 0 ) if   L = d d x , u ( 0 ) + x u ′ ( 0 ) if   L = d 2 d x 2 , u ( 0 ) + x u ′ ( 0 ) + x 2 2 ! u ″ ( 0 ) if   L = d 3 d x 3 . ⋮ u ( 0 ) + x u ′ ( 0 ) + x 2 2 ! u ″ ( 0 ) + ⋯ + x n n ! u n ( 0 ) if   L = d n + 1 d x n + 1 .

The ADM decomposes the solution in the form of Equation (1), and the nonlinear term Nu is decomposed into a series,

N u = ∑ n = 0 ∞     A n . (5)

The A_ns are obtained for the nonlinearity N u = f ( u ) by the formular in [10] which is given as,

A n = 1 n ! d n d ξ n [ F ( ∑ k = 0 ∞     u k ξ k ) ] ξ = 0 ,   n = 0 , 1 , 2 , ⋯ . (6)

Upon substituting Equations (1) and (5) into Equation (4) we obtain the following equation:

∑ n = 0 ∞     u n = φ ( 0 ) + L − 1 g − L − 1 [ R ∑ n = 0 ∞     u n + ∑ n = 0 ∞     A n ] . (7)

The solution components u n ( x ) are determined by the recursive scheme,

u 0 = φ ( 0 ) + L − 1 g

u n + 1 = − L − 1 [ R u n + A n ] , n ≥ 0 .

The n-term approximation of the solution is given by,

ϕ n ( x ) = ∑ k = 0 n − 1     u k ( x ) . (8)

Since its introduction, the ADM has seen several modifications with the view to improve the accuracy, reduce computational efficiency or improve on the range of application of the original method [11] [12] [13] [14] .

In this subsection, we present the modified Adomian decomposition method. The method is a modification of the Adomian Decomposition given in [15] [16] [17] and requires that we introduce the terms L − 1 [ ∑ n = 0 ∞     a n x n ] − p L − 1 [ ∑ n = 0 ∞     a n x n ] in the ADM calculations. Here p is an artificial parameter and ∀ n ∈ ( N ∪ { 0 } ) , a_ns are unknown coefficients to be determined. Thus we suggest that Equation (7) be rewritten as:

∑ n = 0 ∞     u n = φ ( 0 ) + L − 1 [ ∑ n = 0 ∞     a n x n ] − p L − 1 [ ∑ n = 0 ∞     a n x n ]   + L − 1 g − L − 1 [ R ( ∑ n = 0 ∞     u n ) + ∑ n = 0 ∞     A n ] . (9)

From Equation (9), the recursive relationship for MADM is expressed as follows:

u 0 = φ ( 0 ) + L − 1 [ ∑ n = 0 ∞     a n x n ] , u 1 = L − 1 g − p L − 1 [ ∑ n = 0 ∞     a n x n ] − L − 1 [ R ( u 0 ) + A 0 ] , ⋮ , u n + 1 = − L − 1 [ R ( u n ) + A n ] ,     for   n ≥ 1.

To avoid calculation of A n , n = 0,1,2, ⋯ , we determine the coefficients a n , for n = 0 , 1 , 2 , ⋯ . By setting u 1 = 0 we immediately verify that u n = 0 , ∀ n ≥ 1 . Letting p = 1 we write the solution of Equation (2) as

u ( x ) = φ ( 0 ) + L − 1 [ ∑ n = 0 ∞     a n x n ] .

We can observe that the new algorithm reduces the number of iterations because it uses u 0 and u 1 only hence the size of calculations is minimized compared to the standard ADM. It also reduces the number of Adomian polynomials to be constructed because it involves A 0 only. Thus MADM introduces an efficient algorithm that improves the performance of ADM.

We first present the inverse operator theorem, without proof, that is used in the new method. The proof of the theorem is found in reference [24] .

The Inverse Linear Operator Theorem [24]

Theorem 1. If u ′ ( a ) = α and u ′ ( b ) = β are Neumann boundary conditions of a second-order ordinary differential equation then,

L x x − 1 u ″ ( x ) = u ( x ) − ( x − Ω ) u ′ ( a ) − Ω 2 u ′ ( b ) − 1 Ω ∫ 0 Ω     u ( x ) d x d x ,   a ≤ x ≤ b

where,

L x x − 1 [ ⋅ ] = ∫ Ω x     d x ′ ∫ a x ′ [ ⋅ ] d x ′ ′ + 1 Ω ∫ 0 Ω     d x ′ ( x ′ ∫ b x ′ [ ⋅ ] d x ″ ) ,

where Ω is an arbitrary finite element.

To illustrate the NMADM let us consider the following equation,

u ″ ( x ) + m ( x ) u ′ ( x ) + n ( x ) h ( u ( x ) ) = g ( x ) ,   for     x ∈ [ a , b ] (28)

with Nuemann boundary conditions

u ′ ( a ) = β 1 ,   u ′ ( b ) = β 2 . (29)

We rewrite Equation (28) as

u ″ ( x ) = g ( x ) − m ( x ) u ′ ( x ) − n ( x ) h ( u ( x ) ) , (30)

where m ( x ) u ′ ( x ) is a linear term, n ( x ) h ( u ( x ) ) is a nonlinear term and g ( x ) is a source term. We combine the inverse linear operator theorem and MADM recursive relation to come up with the new solution equation for Equation (28) and condition Equation (29) as,

u 1 (31)

From Equation (31), the recursive scheme for NMADM is given as follows:

u 0 = ( x − Ω ) u ′ ( a ) + Ω 2 u ′ ( b ) + L x x − 1 [ ∑ n = 0 ∞     a n x n ] u 1 = 1 Ω ∫ 0 Ω     u 0 ( x ) d x − p L x x − 1 [ ∑ n = 0 ∞     a n x n ] + L x x − 1 g ( x ) − m ( x ) L x x − 1 [ u ′ 0 ( x ) ] − L x x − 1 [ A 0 ] u n + 1 = − m ( x ) L x x − 1 [ u ′ n ( x ) ] − L x x − 1 [ A n ] ,   n ≥ 1. } (32)

It should be noted that in the evaluation of u 0 and u 1 , Ω → 0 and consequently Ω 2 u ′ ( b ) = 0 and after evaluation, 1 Ω ∫ 0 Ω     u 0 ( x ) d x = 0 . We compute the coefficients a n , n ≥ 0 , by putting u 1 = 0 and setting p = 1 . This yields the solution of Equation (28) and the boundary condition Equation (29) in the form:

u ( x ) = x β 1 + L x x − 1 [ ∑ n = 0 ∞     a n x n ] . (33)

We now test the efficiency of the proposed method on different boundary value problems with Neumann conditions. Numerical simulations are done to compare the solutions from NMADM and the exact solutions.

Example 4.1 Let us consider the nonlinear Boundary Value Problem (BVP) which is taken from reference [24]

y ″ − ( y ′ ) 2 = 0,   0 ≤ x ≤ 1, (34)

with conditions,

y ′ ( 0 ) = − 1 ,   y ′ ( 1 ) = − 1 2 . (35)

Exact solution is given by y ( x ) = − log ( x + 1 ) . The NMADM solution scheme for equations (34) and (35) is given by

y 0 = ( x − Ω ) ( − 1 ) + Ω 2 ( − 1 2 ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) y 1 = 1 Ω ∫ 0 Ω     y 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) − L x x − 1 ( A 0 ) ⋮ } (36)

We can easily show that a 0 = 1 , a 1 = − 2 , a 2 = 3 , a 3 = − 4 , ⋯ so that the solution of Equations (34)-(35) is given by:

y ( x ) = − x + 1 2 x 2 − 1 3 x 3 + 1 4 x 4 − 1 5 x 5 + ⋯ .

The numerical results shown in Figure 10 implies the effectiveness of NMADM.

Example 4.2. In this example we look at the nonlinear Burger equation which is in reference [24]

y ″ + y y ′ + y = 1 2 sin ( 2 x ) ,   0 ≤ x ≤ π 2 , (37)

with conditions,

y ′ ( 0 ) = 1 ,   y ′ ( π 2 ) = 0. (38)

The exact solution is given by y ( x ) = sin ( x ) . The NMADM solution scheme for Equations (37) and (38) is given by

y 0 = ( x − Ω ) ( 1 ) + Ω 2 ( 0 ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) y 1 = 1 Ω ∫ 0 Ω     y 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) + L x x − 1 ( sin ( 2 x ) 2 ) − L x x − 1 ( y 0 ) − L x x − 1 ( A 0 ) ⋮ } (39)

It can be shown that if u 1 = 0 and p = 1 , then a 0 = 0 , a 1 = − 1 , a 2 = 0 , a 3 = 1 6 , ⋯ so that the solution of Equations ((37), (38)) is written as

y ( x ) = x − 1 3 ! x 3 + 1 5 ! x 5 − 1 7 ! x 7 + 1 9 ! x 9 + ⋯ .

Figure 11 shows that the result by NMADM is accurate.

Example 4.3. Consider the following linear partial BVP for the heat equation in [25] ,

∂ y ∂ t = ∂ 2 y ∂ x 2 ,   0 ≤ x ≤ 1 ,   t ≥ 0 , (40)

with specified conditions,

y x ( 0 , t ) = e t ,   y x ( 1 , t ) = e t cosh ( 1 ) , (41)

and the exact solution is given by y ( x , t ) = e t sinh ( x ) . We apply NMADM and write the solution scheme for Equations (40) and (41) as,

y 0 = ( x − Ω ) e t + Ω 2 ( e t cosh ( 1 ) ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) y 1 = 1 Ω ∫ 0 Ω     y 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) + L x x − 1 ( L t y 0 ) ⋮ } (42)

It can be shown that a 0 = 0 , a 1 = e t , a 2 = 0 , a 3 = e t 6 , a 4 = 0 , a 5 = e t 120 , ⋯ so that the solution of Equations ((40), (41)) is given by

y ( x , t ) = e t ( x + 1 3 ! x 3 + 1 5 ! x 5 + 1 7 ! x 7 + 1 9 ! x 9 + ⋯ ) .

From Figure 12, it is easily seen that the NMADM and the exact solutions are

the same.

Example 4.4. Let us consider the nonlinear Burger equation in [24]

y t + y y x − y x x = 0 ,   0 ≤ x ≤ π 2 ,   t ≥ 0 , (43)

with conditions,

y x ( 0 , t ) = 1 t − π 2 2 t 2 ,   y x ( 2 , t ) = 1 t − π 2 2 t 2 sech 2 ( π t ) . (44)

The exact solution is given as y ( x , t ) = x t − x 2 tanh ( π x 2 t ) . Using NMADM, the solution scheme for Equation (43) and conditions Equation (44) is given by

y 0 = ( x − Ω ) ( 1 t − π 2 2 t 2 ) + Ω 2 ( 1 t − π 2 2 t 2 sech 2 ( π t ) ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) y 1 = 1 Ω ∫ 0 Ω     y 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) + L x x − 1 ( L t y 0 ) + L x x − 1 ( A 0 ) ⋮ } (45)

where A 0 is the Adomian polynomial of the nonlinear term y y x . A 0 is given by y 0 y 0 x . By setting u 1 = 0 and p − 1 , we solve for the coefficients a n for n = 0 , 1 , 2 , 3 , ⋯ and obtain a 0 = 0 , a 1 = π 4 4 t 4 , a 2 = 0 , a 3 = π 4 6 t 5 − π 6 12 t 6 , a 4 = 0 , a 5 = π 4 20 t 6 − π 2 20 t 7 + 17 π 8 960 t 8 , ⋯ . Thus the solution of Equations ((43), (44)) is,

y ( x , t ) = x t − π t [ ( π x 2 t ) − 1 3 ( π x 2 t ) 3 + 2 15 ( π x 2 t ) 5 − 17 315 ( π x 2 t ) 7 + ⋯ ] + ξ 7 ,

where ξ 7 is a constant. It is noted from Figure 13 that NMADM is effective.

Example 4.5. We consider the following nonlinear oscillator equation which is found in reference [24]

y ″ + ω 2 y = λ y m ,   0 ≤ x ≤ 1 ,   t ≥ 0 , (46)

with conditions,

y ′ ( 0 ) = 1 ,   y ′ ( 1 ) = c n ( 1 | 1 4 ) d n ( 1 | 1 4 ) , (47)

where m is a positive integer. The problem has the exact solution y = s n ( x | 1 4 ) where m = 3 (Duffing oscillator), λ = 1 2 and ω 2 = 5 4 and s n , c n , d n are Jacobi elliptic functions. Using NMADM solution scheme the recursive relation for Equation (46) and conditions Equation (47) can be easily given as,

y 0 = ( x − Ω ) ( 1 ) + Ω 2 ( c n ( 1 | 1 4 ) d n ( 1 | 1 4 ) ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) y 1 = 1 Ω ∫ 0 Ω     y 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) − 5 4 L x x − 1 ( y 0 ) + 1 2 L x x − 1 ( A 0 ) ⋮ } (48)

where A 0 is the Adomian polynomial of the nonlinear term y 3 and it is given by y 0 3 . It can also be shown that a 0 = 0 , a 1 = − 5 4 , a 2 = 0 , a 3 = 73 96 , a 4 = 0 , a 5 = − 553 1536 , ⋯ so that the solution of Equations ((46), (47)) is given by,

y ( x ) = x − 5 24 x 3 + 73 1920 x 5 − 79 9216 x 7 + ⋯ .

In Figure 14 the comparison between NMADM and the exact solution is shown. It is clear that NMADM gives an accurate solution.

Example 4.6. Consider the following linear ordinary boundary problem [24] ,

u ″ ( x ) + u ( x ) + x = 0 ,   0 ≤ x ≤ 1 (49)

with boundary conditions,

u ′ ( 0 ) = − 1 + csc ( 1 ) ,   u ′ ( 1 ) = − 1 + cot ( 1 ) (50)

The NMADM solution scheme for Equation (49) and conditions Equation (50) is given by

u 0 = ( x − Ω ) ( − 1 + csc ( 1 ) ) + Ω 2 ( − 1 + cot ( 1 ) ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) u 1 = 1 Ω ∫ 0 Ω     u 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) − L x x − 1 ( x ) − L x x − 1 ( u 0 ) ⋮ } (51)

For u 1 = 0 and p = 1 , it can be shown that: a 0 = 0 , a 1 = − csc ( 1 ) , a 2 = 0 , a 3 = 1 6 csc ( 1 ) , a 4 = 0 , a 5 = − 1 120 csc ( 1 ) , ⋯ so that the solution of Equations ((49), (50)) is calculated as:

u ( x ) = − x + x csc ( 1 ) − x 3 120 csc ( 1 ) + x 7 5040 csc ( 1 ) + ⋯ = − x + csc ( 1 ) sin ( x ) ,

as obtained in reference [24] .

Example 4.7. Consider the following linear second-order two-point BVP [25] ,

y ″ + x y = ( 3 − x − x 2 + x 3 ) sin ( x ) + 4 x cos ( x ) ,   0 ≤ x ≤ 1 (52)

with boundary conditions

y ′ ( 0 ) = − 1 ,   y ′ ( 1 ) = 2 sin ( 1 ) . (53)

Exact solution is y ( x ) = ( x 2 − 1 ) sin x . The NMADM solution scheme for Equation (52) and conditions Equation (53) is given by

y 0 = ( x − Ω ) ( − 1 ) + Ω 2 ( 2 sin ( 1 ) ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) y 1 = 1 Ω ∫ 0 Ω     y 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) + L x x − 1 ( ( 3 − x − x 2 + x 3 ) sin x + 4 x cos x )               − L x x − 1 ( x y 0 ) ⋮ } (54)

For y 1 = 0 and p = 1 , it can be shown that, a 0 = 0 , a 1 = 7 , a 2 = 0 , a 3 = − 2 7 , a 4 = 0 , a 5 = 43 120 , ⋯ . Thus the solution to equations (52) and (53) is given by

y ( x ) = − x + 7 3 ! x 3 − 21 5 ! x 5 + 43 7 ! x 7 + ⋯ .

The numerical results shown in Figure 15 implies the effectiveness of the new method discussed in this section.

Example 4.8. Consider the following nonlinear second order two-point BVP [25] :

y ″ + e − 2 y = 0,   0 ≤ x ≤ 1 (55)

with boundary conditions

y ′ ( 0 ) = 1 ,   y ′ ( 1 ) = 1 2 . (56)

The exact solution is y ( x ) = ln ( 1 + x ) . The NMADM solution scheme for Equation (55) and conditions Equation (56) is given by

y 0 = ( x − Ω ) ( 1 ) + Ω 2 ( 1 2 ) + L x x − 1 ( ∑ n = 0 ∞     a n x n ) y 1 = 1 Ω ∫ 0 Ω     y 0 ( x , t ) d x − p L x x − 1 ( ∑ n = 0 ∞     a n x n ) − L x x − 1 ( A 0 ) ⋮ } (57)

where A 0 = e − 2 y 0 is the Adomian polynomial of the nonlinear term e − 2 y . For y 1 = 0 and p = 1 , it can be shown that a 0 = − 1 , a 1 = 2 , a 2 = − 3 , a 3 = 4 , a 4 = − 5 , a 5 = 6 , ⋯

Therefore the solution of Equations (55) and (56) is given as

y ( x ) = x − 1 2 x 2 + 1 3 x 3 − 1 4 x 4 + 1 5 x 5 − 1 6 x 6 + 1 7 x 7 + ⋯ .

In Figure 16, the NMADM solution is plotted against the exact solution. It is seen that NMADM is effective.