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Adomian Decomposition and Modified Variational

Iteration Methods for Linear Fractional

Integro-Differential Equations

M. H.Saleh, D.Sh.Mohamed,M.H.Ahmed and M.K. Marjan*

Mathematics Department, Faculty of Science,Zagazig University, Zagazig, Egypt

*- E-mail: [email protected]

Abstract

In this paper, we present a comparative study between the modified variational iteration methodand Adomian decomposition method. The study outlines the significant features of the two methods,for solving linear fractional integro-differential equations. From the computational viewpoint, thevariational iteration method is more efficient, convenient and easy to use.

keywords: Fractional integro-differential equations, Adomian decomposition method (ADM),Modified variational iteration method (MVIM), Caputo fractional derivative.

1. Introduction

In recent years various analytical and numerical methods have been applied for the approximate so-lutions of fractional integro-differential equations. Fractional models have been shown by many scientiststo adequately describe the operation of variety of physical and biological processes and systems ([4-7]).Consequently, considerable attention has been given to the solution of fractional integral equations ofphysical interest. Since most fractional integro-differential equations do not have exact analytic solu-tions, approximation and numerical techniques, therefore, are used extensively. Recently, the Adomiandecomposition method ([2],[12]) and Variational iteration method ([10],[11],[13],[14]) have been used forsolving a wide range of problems.To construct an approximate solution to linear integro-differential equations of fractional order. Thevariational iteration method was first proposed by He ([10],[11]) and has been worked out over a numberof years by many authors. This method has been shown to effectively, easily and accurately solve a largeclass of linear problems. Generally, one or two iterations lead to high accurate solutions.In 1980, Adomian proposed decomposition method popularly known as Adomian decomposition method(ADM) has been intensively applied by researchers like Shawagheh (1999,2002), Wazawaz (2001,2005),Odiabat and Momani (2006) as it provides analytical approximate solutions for nonlinear equations with-out linearization, perturbation or discretization. Also Daftardar Gejji and Jafri (2005,2006), Momani andAl-Khaled (2005) provided solution of system of fractional differential equations using Adomian decom-position method [3].Recently Adomian decomposition method is extended for multiterm diffusion-wave equations of fractionalorder(Gejji and (2008))also Mittal et al(2008) solved fractional integro-differential equations by Adomiandecomposition method. The method works very well and solutions of nonlinear problems can be obtainedvery easily and accurately. The method is well reviewed in (Adomian (1988,1992,1994)). The methodinvolving splitting an equation into linear and nonlinear parts and assume that it has a infinite seriessolution. This series has to be truncated for practical purpose but by adding more terms it is possible to

M K Marjan et al, Int.J.Computer Technology & Applications,Vol 6 (2),177-187

IJCTA | Mar-Apr 2015 Available [email protected]

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get arbitrary close to the exact solution in a specific domain.In this paper, we are concerned with the numerical solution of following linear fractional integro-differentialequations:

DαΨ(x) = g(x) +

∫ x

0

k(x, t)Ψ(t)dt, 0 ≤ x, t ≤ 1, (1.1)

with the following supplementary conditions:

Ψ(i)(0) = δi, n− 1 < α ≤ n, n ∈ N , (1.2)

where DαΨ(x) indicates α caputo fractional derivative of Ψ(x), g(x) and k(x, t) are given funtions,x and t are real variables varying in the interval [0, 1], and Ψ(x) is the unknown function to bedetermined.

2. Basic definitions

In this section some basic definitions and properties of fractional calculus theory which are necessaryfor the formulation of the problem are given.

Definition 2.1: A real function f(x), x > 0, is said to be in the space Cµ, µ ∈ R, if there existsa real number p > µ such that f(x) = xp f1(x), where f1(x) ∈ C[0, 1).

Definition 2.2: A function f(x), x > 0, is said to be in the space Cmµ , m ∈ N ∪ {0}, if f (m) ∈ Cµ.

Definition 2.3: The left sided Riemann-Liouville fractional integral operator of order α ≥ 0 of afunction f ∈ Cµ, µ ≥ −1, is defined as [15].

Jαf(x) =1

Γ(α)

∫ x

0

f(t)

(x− t)1−α dt, α > 0, x > 0, (2.1)

J0f(x) = f(x).

Definition 2.4: Let f ∈ Cm−11, m ∈ N ∪ {0}. Then the Caputo fractional derivative of f(x) is de-fined as [4-7]

Dαf(x) =

Jm−αfm(x), m− 1 < α ≤ m, m ∈ N,

Dmf(x)Dxm , α = m.

(2.2)

Hence, we have the following properties:

1. Jα Jνf = Jα+νf, α, ν > 0 , f ∈ Cµ , µ > 0. (2.3)

2. Jα xγ =Γ(γ + 1)

Γ(α+ γ + 1)xα+γ , α > 0, γ > −1, x > 0. (2.4)

3. Jα Dαf(x) = f(x)−m−1∑k=0

f (k)(0+)xk

k!, x > 0, m− 1 < α ≤ m. (2.5)

4. Dα Jα f(x) = f(x), x > 0, m− 1 < α ≤ m. (2.6)

5. Dα C = 0, C is a constant. (2.7)



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6. Dαxβ =

0, β ∈ N0, β < [α],

Γ(β+1)Γ(β−α+1) x

β−α, β ∈ N0, β ≥ [α].

(2.8)

where [α] denoted the smallest integer greater than or equal to α and N0 = {0, 1, 2, . . .}.

3.Analysis of The Modified Variational Iteration Method

In this section, for the convenience of the reader, we first present a brief review of He’s variationaliteration method [16]. Then we will propose the reliable modification of the VIM [1] for solving fractionalintegro-differential equations. Here, we consider the following fractional functional equation

Lu+Ru+Nu = g(x), (3.1)

where L is the fractional order derivative, R is a linear differential operator, N represents the nonlinearterms, and g is the source term. By using (2.5) and applying the inverse operator L−1

x to both sides of(3.1), and using the given conditions, we obtain

u = f − L−1x [Ru]− L−1[Nu] (3.2)

where the function f represents the terms arising from integrating the source term g and from using thegiven conditions, all are assumed to be prescribed.The basic character of He’s method is the construction of a correction functional for (3.1), which reads

un+1(x) = un(x) +

∫ x

0

λ(s)[Lun(s) +Run(s) +Nun(s)− g(s)]ds, (3.3)

where λ is a Lagrange multiplier which can be identified optimally via variational theory [9], un is the nthapproximate solution, and un denotes a restricted variation, i.e., δun = 0.

To solve (3.1) by He’s VIM, we first determine the Lagrange multiplier λ that will be identifiedoptimally via integration by parts. Then the successive approximations un(x), n ≥ 0, of the solutionu(x) will be readily obtained upon using the obtained Lagrange multiplier and by using any selectivefunction u0. The approximation u0 may be selected by any function that just satisfies at least the initialand boundary conditions. With determined λ, then several approximations un(x), n ≥ 0, follow im-mediately. Consequently, the exact solution may be obtained by using

limn−→1

un(x) = u(x). (3.4)

In summary, we have the following variational iteration formula for (3.2){u0(x) is an arbitrary initial guess,

un+1(x) = un(x) +∫ x

0λ(s)[Lun(s) +Run(s) +Nun(s)− g(s)]ds,

(3.5)

or equivalently, for (3.2), according to [16]:{u0(x) is an arbitrary initial guess,

un+1(x) = f(x)− L−1x [Ru]− L−1

x [Nu],(3.6)

where the multiplier Lagrange λ, has been identified.It is important to note that He’s VIM suggests that the u0 usually defined by a suitable trial-function

with some unknown parameters or any other function that satisfies at least the initial and boundaryconditions. This assumption made by He ([8],[9]) and others will be slightly varied, as will be seen in thediscussion.

The MVIM, that was introduced by Ghorbani et al [1], can be established based on the assumption



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that the function f(x) of the iterative relation (3.6) can be divided into two parts, namely f0(x) andf1(x). Under this assumption, we set

f(x) = f0(x) + f1(x). (3.7)

According to the assumption, (3.7), and by the relationship (3.6), we construct the following variationaliteration formula

u0(x) = f0(x),

u1(x) = f(x)− L−1x [Rf0(x)]− L−1

x [Nf0(x)],

un+1(x) = f(x)− L−1x [Run(x)]− L−1

x [Nun(x)],

where the multiplier Lagrange λ, has been identified. Here, a proper selection was proposed for thecomponents u0(x) and u1(x). The suggestion was that only the part f0 be assigned to the zerothcomponent i.e u0. An important observation that can be made here is that the success of the proposedmethod depends mainly on the proper choice of the functions f0 and f1. As will be seen from theexamples below, this selection of u0 will result in a reduction of the computational work and acceleratethe convergence. Furthermore, this proper selection of the components u0 and u1 may provide the solutionby using one iteration only. To give a clear overview of the content of this study. We have chosen severalfractional integro-differential equations with the nonlocal boundary conditions.

4.Analysis of The Adomian Decomposition Method

Consider the equation (1.1) with initial condition (1.2) where Dα is the operator defined as (2.2).Operating with Jα on both sides of the equation (1.1) as follows:

Ψ(x) =

m−1∑k=0

Ψ(k)(0+)xk

k!+ Jαg(x) + Jα

∫ x

0

k(x, t)Ψ(t)dt, 0 ≤ x, t ≤ 1, (4.1)

Adomian decomposition method defines the solution by the series:

Ψ(x) =∞∑i=0

Ψi(x), (4.2)

where the components Ψ0(x),Ψ1(x),Ψ2(x), ...,Ψi(x), ... are determined recursively by

Ψ0(x) =

m−1∑k=0

Ψ(k)(0+)xk

k!+ Jαg(x),

Ψi+1(x) = Jαg(x) + Jα∫ x

0

k(x, t)Ψi(t)dt, (4.3)

decomposition method suggests that 0th component Ψ0(x) be defined by the initial conditions and thefunction g(x) as described above. The other components namely Ψ0(x),Ψ1(x), . . .etc. are derived recur-rently.

5. Numerical examples

In this section, some numerical examples of linear fractional integro-differential equations are pre-sented to illustrate the above results. All results are obtained by using Maple 16.

Example 5.1Consider the following fractional integro-differential equation:

D57 Ψ(x) =

2401

276

x237

Γ( 27 )− 1

5x5 sin(x)− x sin(x) +

∫ x

0

sin(x)Ψ(t)dt, (5.1)



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subject to Ψ(0) = 1 with the exact solution Ψ(x) = 1 + x4.

By assuming L = D57 and applying the inverse operator L−1

x to both sides of (5.1), we have

Ψ(x) = 1 + J57 [

2401

276

x237

Γ( 27 )

]− J 57 [

1

5x5 sin(x) + x sin(x)] + J

57

∫ x

0

sin(x)Ψ(t)dt. (5.2)

According to the original VIM (3.5), and corresponding the recursive scheme (3.6) and by using (3.7),we obtain

g(x) = g0(x) + g1(x) = 1 + J57 [ 2401

276x

237

Γ( 27 )

]− J 57 [ 1

5x5 sin(x) + x sin(x)],

= 1 + x4 − J 57 [ 1

5x5 sin(x) + x sin(x)],

by assuming

g0(x) = 1 + x4, g1(x) = −J 57 [

1

5x5 sin(x) + x sin(x)], (5.3)

and with starting of the initial approximation Ψ0(x) = g0(x) = 1 + x4, we have

Ψ0(x) = 1 + x4,

Ψ1(x) = 1 + x4 − J 57 [ 1

5x5 sin(x) + x sin(x)] + L−1

x [Ψ0(x)],

Ψ1(x) = 1 + x4 − J 57 [

1


57 [

1

5x5 sin(x) + x sin(x)] = 1 + x4, (5.4)

Ψn+1(x) = 1 + x4 − J 57 [ 1

5x5 sin(x) + x sin(x)] + L−1

x [Ψn(x)]⇒ Ψn+1(x) = 1 + x4, n ≥ 0.

In view of (5.4), we obtain the approximate solution Ψ(x) = 1 + x4, which is the same of the exactsolution.

Now applying Adomian decomposition method for equation (5.2) by using equation (4.2) we get:

∞∑i=0

Ψi(x) = 1 + J57 [

2401

276

x237

Γ( 27 )

]− J 57 [

1


57

∫ x

0

sin(x)∞∑i=0

Ψi(t)dt, (5.5)

by using equation (5.5) we obtain as the follows:

Ψ0(x) = 1 + J57 g(x)

Ψ0(x) = 1 + x4 − 0.07160086639x477 + 0.007058904187x

617 − 0.0003033879182x

757 + . . . ,

Ψ1(x) = J57λ∫ x

0k(x, t)Ψ0(t)dt,

Ψ1(x) = 0.05264769589x477 − 0.007058904186x

617 + 0.0003033879182x

757 + . . . ,

Ψ2(x) = J57λ∫ x

0k(x, t)Ψ1(t)dt,

Ψ2(x) = −5.773754204× 10−12x1697 + 1.089352272× 10−13x

1837 − 1.315207672× 10−15x

1977 + . . . ,

Ψ3(x) = J57λ∫ x

0k(x, t)Ψ2(t)dt,

Ψ3(x) = −2.129464558× 10−10x1607 + 1.152235978× 10−11x

1747 − 3.718717854× 10−24x

2727 + . . . ,

hance∑∞i=0 Ψi(x) = Ψ0 + Ψ1 + Ψ2 + Ψ3 + . . . .

The approximation solution for α = 5/7 is

Ψ(x) = 1+x4−2.129464558×10−10x1607 +1.152235978×10−11x

1747 −3.718717854×10−24x

2727 − . . . .



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Table 1 and figure 1 show the comparison between Modified variational method and Adomian Decompo-sition method, we find that Modified variational method is better than Adomian decomposition method.

Table 1x Exact Approx variational Approx adomian Errer variational Errer adomian0 1 1 1 0 0

0.1 1.0001 1.0001 1.000099996 0 3.659466400× 10−9

0.2 1.0016 1.0016 1.001599616 0 3.843611821× 10−7

0.3 1.0081 1.0081 1.008094147 0 0.0000058526144810.4 1.0256 1.0256 1.025559568 0 0.000040431724990.5 1.0625 1.0625 1.062318800 0 0.00018120002540.6 1.1296 1.1296 1.128982201 0 0.00061779906990.7 1.2401 1.2401 1.238355374 0 0.0017446263650.8 1.4096 1.4096 1.405307042 0 0.0042929571950.9 1.6561 1.6561 1.646589249 0 0.0095107504601.0 2.0000 2.0000 1.980601618 0 0.01939838302

Figure 1: Numerical results of Example 5.1.



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D23 Ψ(x) =

3

4

x13

√3 Γ( 2

3 )(3x− 2)

π− 1

4x4 exp(x)− 1

3x3 exp(x) +

∫ x

0

exp(x) tΨ(t)dt, (5.6)

subject to Ψ(0) = 0 with the exact solution Ψ(x) = x2 − x.By assuming L = D

23 and applying the inverse operator L−1


Ψ(x) = 0 + J23 [

3

4

x13

√3 Γ( 2

3 )(3x− 2)

π]− J 2

3 [1

4x4 exp(x) +

1

3x3 exp(x)] + J

23

∫ x

0

exp(x) t Ψ(t)dt, (5.7)


g(x) = g0(x) + g1(x) = 0 + J23 [ 3

4

x13√

3 Γ( 23 )(3x−2)

π ]− J 23 [ 1

4x4 exp(x) + 1

3x3 exp(x)],

= x2 − x− J 23 [ 1

4x4 exp(x) + 1

3x3 exp(x)],

by assuming

g0(x) = x2 − x, g1(x) = −J 23 [

1

4x4 exp(x) +

1

3x3 exp(x)], (5.8)

and with starting of the initial approximation Ψ0(x) = g0(x) = x2 − x, we have

Ψ0(x) = x2 − x,

Ψ1(x) = x2−x−J 23 [

1

4x4 exp(x)+

1

3x3 exp(x)]+L−1

x [Ψ0(x)], (5.9)

Ψ1(x) = x2 − x− J 34 [ 1

4x4 exp(x) + 1

3x3 exp(x)] + J

34 [ 1

4x4 exp(x) + 1

3x3 exp(x)] = x2 − x,

Ψn+1(x) = x2 − x− J 34 [ 1

4x4 exp(x) + 1

3x3 exp(x)] + L−1

x [Ψn(x)]⇒ Ψn+1(x) = x2 − x, n ≥ 0.

In view of (5.9), we obtain the approximate solution Ψ(x) = x2 − x, which is the same of the exactsolution.

Now applying Adomian decomposition method for equation (5.7) by using equation (4.2), we get:

∞∑i=0

Ψi(x) = 0 + J23 [

3

4

x13

√3 Γ( 2

3 )(3x− 2)

π]− J 2

3 [1

4x4 exp(x) +

1

3x3 exp(x)] + J

23

∫ x

0

exp(x)t∞∑i=0

Ψi(t)dt,

(5.10)by using equation (5.10) we obtain as the follows:

Ψ0(x) = J23 g(x),

Ψ0(x) = x2 − x+ 0.02913191738x14/3 − 0.02570463301x17/3 − 0.01927847474x20/3 − . . . ,

Ψ1(x) = J23λ∫ x

0k(x, t)Ψ0(t)dt,

Ψ1(x) = 0.02570463300x17/3 + 0.01927847474x20/3 + 0.01056125138x23/3 − 0.02913191740x14/3 + . . . ,

Ψ2(x) = J23λ∫ x

0k(x, t)Ψ1(t)dt,

Ψ2(x) = −0.006885695970x19/3−0.007399887556x22/3−0.003123602081x25/3+0.0005423582779x34/3+ . . . ,

Ψ3(x) = J23λ∫ x

0k(x, t)Ψ2(t)dt,

Ψ3(x) = 4.873687408×10−8x19+0.000007245575142x16+0.00001413636375x15−0.0003449563982x10+ . . . ,



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hance∑∞i=0 Ψi(x) = Ψ0 + Ψ1 + Ψ2 + Ψ3 + . . . .


Ψ(x) = x2−x−2×10−11x14/3−1×10−11x17/3−6×10−13x28/3−1×10−13x37/3−5×10−13x31/3 +. . . .



0.1 -0.09 -0.09 -0.09 0 9.999999987× 10−25

0.2 -0.16 -0.16 -0.16 0 2.047983077× 10−21

0.3 -0.21 -0.21 -0.2100000000 0 1.768962184× 10−19

0.4 -0.24 -0.24 -0.2399999998 0 2.000000040× 10−10

0.5 -0.25 −0.25 -0.2499999965 0 3.600000003× 10−9

0.6 -0.24 -0.24 -0.2399999619 0 3.799999695× 10−8

0.7 -0.21 -0.21 -0.2099997080 0 2.918998698× 10−7

0.8 -0.16 -0.16 -0.1599982462 0 0.0000017536968700.9 -0.09 -0.09 -0.08999127985 0 0.0000087201486691.0 0 0 0.00003727337387 0 0.00003727337387




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D35 Ψ(x) =

125

14

x125 sin( 2

5π) Γ( 35 )

π− 1

3x

132 +

∫ x

0

√x t2Ψ(t)dt, (5.11)

subject to Ψ(0) = 0 with the exact solution Ψ(x) = 2x3.

By assuming L = D35 and applying the inverse operator L−1


Ψ(x) = 0 + J35 [

125

14

x125 sin( 2

5π) Γ( 35 )

π]− J 3

5 [1

3x

132 ] + J

35

∫ x

0

√x t2Ψ(t)dt, (5.12)


g(x) = g0(x) + g1(x) = 0 + J35 [ 125

14

x125 sin( 2

5π) Γ( 35 )

π ]− J 35 [ 1

3x132 ]

= 2x3 − J 35 [ 1

3x132 ]

by assuming

g0(x) = 2x3, g1(x) = −J 35 [

1

3x

132 ], (5.13)

and with starting of the initial approximation Ψ0(x) = g0(x) = 2x3, we have

Ψ0(x) = 2x3,

Ψ1(x) = 2x3 − J 35 [

1

3x

132 ] + L−1

x [Ψ0(x)], (5.14)

Ψ1(x) = 2x3 − J 35 [ 1

3x132 ] + J

35 [ 1

3x132 ] = 2x3,

Ψn+1(x) = 2x3 − J 35 [ 1

3x132 ] + L−1

x [Ψn(x)]⇒ Ψn+1(x) = 2x3, n ≥ 0.

In view of (5.14), we obtain the approximate solution Ψ(x) = 2x3, which is the same of the exact solution.

Now applying Adomian decomposition method for equation (5.12) by using equation (4.2), we get:

∞∑i=0

Ψi(x) = 0 + J35 [

125

14

x125 sin( 2

5π) Γ( 35 )

π]− J 3

51

3x

132 + J

35

∫ x

0

√x t2

∞∑i=0

Ψi(t)dt, (5.15)

by using equation (5.15) we obtain as the follows:

Ψ0(x) = J35 g(x),

Ψ0(x) = 2x3 − 0.1011008935x71/10,

Ψ1(x) = J35λ∫ x

0k(x, t)Ψ0(t)dt,

Ψ1(x) = −0.002324006305x56/5 + 0.1011008935x71/10,

Ψ2(x) = J35λ∫ x

0k(x, t)Ψ1(t)dt,

Ψ2(x) = −0.00003160247042x153/10 + 0.002324006305x56/5,

Ψ3(x) = J35λ∫ x

0k(x, t)Ψ2(t)dt,

Ψ3(x) = −2.896664654× 10−7x97/5 + 0.00003160247042x153/10,

hance∑∞i=0 Ψi(x) = Ψ0 + Ψ1 + Ψ2 + Ψ3 + . . . .



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Ψ(x) = 2x3 − 2.896664654× 10−7x97/5.



0.1 0.002 0.002 002 0 1.153182970× 10−26

0.2 0.016 0.016 0.016 0 7.977744732× 10−21

0.3 0.054 0.054 0.054 0 2.079938824× 10−17

0.4 0.128 0.128 0.128 0 5.519021066× 10−15

0.5 0.250 0.250 0.250 0 4.187128641× 10−13

0.6 0.432 0.432 0.4320000000 0 1.438906178× 10−11

0.7 0.686 0.686 0.6859999997 0 2.862860564× 10−10

0.8 1.024 1.024 1.023999996 0 3.818070713× 10−9

0.9 1.458 1.458 1.457999962 0 3.751482313× 10−8

1.0 2.000 2.000 1.999999710 0 2.896664654× 10−7




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ConclusionIn this paper we study the numerical solution of three examples by using Modified variational iteration

method and Adomian decomposition method which derive a good approximation. The study shows thatthe techniques require less computational work than existing approaches while supplying quantitativelyreliable results.

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