Numerical solution of a system of fractional ordinary differential
equations by a modified variational iteration procedure
ABDALLAH AL-HABAHBEH
Department of Mathematics
Tafila Technical University
Tafila
JORDAN
Abstract: - In this paper, a robust modification of the variational iteration method that gives a numerical solution
for a system of linear/nonlinear differential equations of fractional order was proposed. This technique does not
need the perturbation theory or linearization. The conformable fractional derivative initiated by the authors Khalil
et al. is considered. The efficiency of the modified method is established via illustrative examples. For linear and
nonlinear systems, the approximate solutions are in excellent agreement with the exact solutions.
Key-Words: - Conformable fractional derivative; Variational iteration method; Lagrange multiplier.
Received: June 29, 2021. Revised: April 9, 2022. Accepted: May 12, 2022. Published: June 8, 2022.
1 Introduction
The differential equations of fractional order, as well
as their exact and approximate solutions, have fun-
damental significance in several branches of science
and engineering [12, 24, 23, 22, 7, 12, 28, 10, 21, 29].
The analytic and approximate solutions of linear and
nonlinear systems of ordinary differential equations
of fractional order have been discussed by several au-
thors, see [9, 11, 3, 20, 26, 27].
The authors Khalil et al. defined a new fractional
derivative in [17]. It is based on the definition of the
basic limit of the derivative. The new simple frac-
tional derivative is named the conformable fractional
derivative, which then had been the focus of many
studies [1, 2, 5, 6, 8, 19, 18, 25].
The variational iteration approach was developed for
the first time by He [14]. This technique and its
modifications [15, 16] have potentially been used to
solve nonlinear differential equations. In [31], a com-
parative study between the Adomian decomposition
method and the variational iteration method has been
presented. The method has been used in [13] to pro-
vide an approximate solution for fractional differen-
tial equations with modified Riemann–Liouville frac-
tional derivative.
The purpose of this paper is to extend the analysis of
the variational iteration method to solve the system of
fractional ordinary differential equations which is as
follows:
Dα1x1(t) = f1(t, x1, x2, . . . , xn),
Dα2x2(t) = f2(t, x1, x2, . . . , xn),
.
.
.
Dαnxn(t) = fn(t, x1, x2, . . . , xn),
(1)
where Dαi=dαi
dtαiis the conformable fractional
derivative of order αi∈(0,1], for i= 1,2, . . . , n.
The system is subject to the initial conditions
x1(0) = c1, x2(0) = c2, . . . , xn(0) = cn.
The article is organized as follows: In Section 2 we
discuss the basic definitions and properties of the con-
formable fractional derivative. The variational itera-
tion method is presented in Section 3. Section 4 pro-
vides a series of examples to demonstrate the effi-
ciency of the implemented method. Section 5 con-
cludes.
2 Preliminaries and Notations
In this section, we introduce the main concepts and
properties of the conformable fractional derivative.
Definition 2.1. The conformable fractional deriva-
tive of order α,0< α ≤1of f: (0,∞)→Ris
defined by
Dα(f)(x) := lim
ϵ→0
f(x+ϵx1−α)−f(x)
ϵ
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for all x > 0. If the limit exists, we say
that fis α−differentiable at x. Moreover, if f
is α−differentiable in some (0, a),a > 0, and
lim
x→0+Dα(f)(x)exists, then define
Dα(f)(0) := lim
x→0+Dα(f)(x).
Remark 1. Let α∈(0,1] and fbe differentiable and
α−differentiable for all x > 0. Then
Dα(f)(x) = x1−αdf
dx(x).
The fractional exponential function, denoted by
e1
αxα, is defined by
e1
αxα=
∞
∑
j=0
xαj
αjj!.
Remark 2. The conformable fractional derivative of
common functions are
•Dα(c) = 0,for any constant c.
•Dα(xr) = rxr−α, r ∈R.
•Dα(sin 1
αxα) = cos 1
αxα.
•Dα(cos 1
αxα) = −sin 1
αxα.
•Dα(e1
αxα) = e1
αxα.
Next, we introduce the fractional integral of order
α.
Definition 2.2. Let α∈(0,1] and x∈[a, ∞),a≥
0. The conformable fractional integral of order αis
given by
(Ia
αf)(x) := ∫x
a
f(t)dα(t, a) = ∫x
a
(t−a)α−1f(t)dt.
When a= 0, we use Iαand dα(t).
Theorem 2.1. Let f: [0,∞)→Rbe differentiable
and α∈(0,1]. Then for all t > 0we have
IαDα(f)(t) = f(t)−f(0).
Proof. From definition and since fis differentiable
we obtain
IαDα(f)(t) = ∫t
0
xα−1Dαf(x)dx
=∫t
0
xα−1x1−αf′(x)dx =f(t)−f(0).
3 Variational iteration method
In 1999, He [14] proposed an analytical approach for
a non-linear problem based on a general Lagrange
multiplier. The method is called the variational iter-
ation method, where no perturbation or linearization
are needed, has been used effectively to solve a vast
class of nonlinear problems.
To demonstrate the main concept of the method, we
examine the following general nonlinear system:
Lu +Nu =g(t),(2)
where Land Nare a linear and nonlinear operators,
respectively.
A correctional functional can be given by
un+1(t) = un(t)+∫t
o
λ[Lun(τ)+N˜un(τ)−g(τ)]dτ,
(3)
where λis a general Lagrange multiplier which can be
determined optimally through variational theory, and
˜unis considered such that δ˜un= 0.
Accordingly, the exact solution can be given by
x(t) = lim
k→∞
xk(t).
Now, consider the general fractional differential equa-
tion:
Dαx(t) = Lx(t) + N x(t) + g(t),(4)
where α∈(0,1],Dαis the conformable fractional
derivative of x(t)of order α.
The author has modified the above iteration method
into:
xk+1(t) = xk(t)+
∫t
0
λ(τ)[Dαxk(τ)−(Lxk(τ)+N˜xk(τ)+g(τ)]dα(τ),
which can be rewritten as
xk+1(t) = xk(t)+ (5)
∫t
0
λ(τ)τα−1[Dαxk(τ)−(Lxk(τ)+N˜xk(τ)+g(τ)]dτ,
where any selective function can be used for x0, we
usually use the initial condition x(0). To find the op-
timal value of Lagrange multiplier λ, we perform the
following, we take the variation of (5) with respect to
x(t)
δxk+1(t) = δxk(t)+
δ∫t
0
λ(τ)τα−1[Dαxk(τ)−(Lxk(τ)+N˜xk(τ)+g(τ))]dτ.
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This yields the stationary conditions
λ′(τ) = 0,
1 + λ(τ) = 0.
Hence, we obtain
λ(τ) = −1.(6)
The correction functionals for system (1) can be ex-
pressed as
xk+1
1(t) = xk
1(t)+
∫t
0
λ1(τ)(Dα1xk
1(τ)−f1(τ, ˜xk
1(τ), . . . , ˜xk
n(τ)))dα1(τ),
xk+1
2(t) = xk
2(t)+
∫t
0
λ2(τ)(Dα2xk
2(τ)−f2(τ, ˜xk
1(τ), . . . , ˜xk
n(τ)))dα2(τ),
.
.
.
xk+1
n(t) = xk
n(t)+
∫t
0
λn(τ)(Dαnxk
n(τ)−fn(τ, ˜xk
1(τ), . . . , ˜xk
n(τ)))dαn(τ),
(7)
Substituting (6) into (7) gives
xk+1
1(t) = xk
1(t)−
∫t
0
τα1−1(Dα1xk
1(τ)−f1(τ, ˜xk
1(τ), . . . , ˜xk
n(τ)))dτ,
xk+1
2(t) = xk
2(t)−
∫t
0
τα2−1(Dα2xk
2(τ)−f2(τ, ˜xk
1(τ), . . . , ˜xk
n(τ)))dτ,
.
.
.
xk+1
n(t) = xk
n(t)−
∫t
0
ταn−1(Dαnxk
n(τ)−fn(τ, ˜xk
1(τ), . . . , ˜xk
n(τ)))dτ.
(8)
4 Applications
In this section, we illustrate the efficiency of the our
modified version of VIM by presenting four exam-
ples. The first two examples are considered for linear
systems of fractional ordinary differential equations.
The accuracy of the proposed method is appraised by
comparison with the exact solutions. The third and
fourth examples are considered for nonlinear systems.
All computations are performed by Mathematica.
Example 4.1. Consider the linear system of fractional
differential equations
Dα1x(t) = x(t)−y(t),
Dα2y(t) = x(t) + y(t),(9)
where 0< α1, α2≤1and Dα=∂α
∂tαis the con-
formable fractional derivative, subject to the initial
conditions
x(0) = 1, y(0) = 0.(10)
The exact solution of system (9) with initial condi-
tions (10), when α1=α2=α, refer to [4] for more
details, is
x(t) = etα
αcos tα
α,
y(t) = etα
αsin tα
α.
(11)
According to the formulas (8), the variational iteration
formulas for system (9) are given by
xk+1(t) = xk(t)−∫t
0
τα1−1[Dα1xk(τ)−xk(τ) + yk(τ))]dτ,
yk+1(t) = yk(t)−∫t
0
τα2−1[Dα2yk(τ)−xk(τ)−yk(τ))]dτ,
(12)
where x0(t) = 1 and y0(t) = 0.
Consequently, we obtain the following approxima-
tions
x1(t) =1 + tα1
α1
,
y1(t) =tα2
α2
,
x2(t) =1 + tα1
α1
+t2α1
2α2
1
−tα1+α2
α2(α1+α2),
y2(t) =tα2
α2
+tα1+α2
α1(α1+α2)+t2α2
2α2
2
,
x3(t) =1 + tα1
α1
+t2α1
2α2
1
+t3α1
6α3
1
−tα1+α2
α2(α1+α2)−
t2α1+α2
α1α2(2α1+α2)−tα1+2α2
2α2
2(α1+ 2α2),
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y3(t) =tα2
α2
+t2α2
2α2
2
+t3α2
6α3
2
+tα1+α2
α1(α1+α2)+
t2α1+α2
2α2
1(2α1+α2)−tα1+2α2
α2(α1+α2)(α1+ 2α2)+
tα1+2α2
α1(α1+α2)(α1+ 2α2),
x4(t) =1 + tα1
α1
+t2α1
2α2
1
+t3α1
6α3
1
+t4α1
24α4
1
−
tα1+α2
α2(α1+α2)−t2α1+α2
α1α2(2α1+α2)−
tα1+2α2
2α2
2(α1+ 2α2)−tα1+3α2
6α3
2(α1+ 3α2)−
t3α1+α2
α1α2(2α1+α2)(3α1+α2)−
t2α1+2α2
4α2
2(α1+α2)(α1+ 2α2)−
t3α1+α2
2α2
1(2α1+α2)(3α1+α2),
y4(t) =tα2
α2
+t2α2
2α2
2
+t3α2
6α3
2
+t4α2
24α4
2
+
tα1+α2
α1(α1+α2)+t3α1+α2
6α3
1(3α1+α2)+
t2α1+α2
2α1(α1+α2)(2α1+α2)+
α2t2α1+α2
2α2
1(α1+α2)(2α1+α2)+
+t2α1+2α2
4α2
1(α1+α2)(2α1+α2)−
t2α1+2α2
2α1α2(α1+α2)(2α1+α2)−
tα1+3α2
2α2
2(α2
1+ 5α1α2+ 6α2
2),
.
.
. (13)
If α1=α2=α, then (13) become
x1(t) = 1 + tα
α,
y1(t) = tα
α,
x2(t) = 1 + tα
α,
y2(t) = tα
α+t2α
α2,
x3(t) = 1 + tα
α−t3α
3α3,
y3(t) = tα
α+t2α
α2+t3α
3α3,
x4(t) = 1 + tα
α−t3α
3α3−t4α
6α4,
y4(t) = tα
α+t2α
α2+t3α
3α3,
.
.
.
(14)
The solution of (9) in series form is given by
x(t) = 1 + tα
α−t3α
3α3−t4α
6α4+. . . ,
y(t) = tα
α+t2α
α2+t3α
3α3+. . . ,
(15)
which converges to the exact solution (11).
Figures 1 and 2 show the exact solution x(t)and y(t)
and the approximate solution x4(t)and y4(t)for sys-
tem (9) for different values of α. The figures show
that our approximate solutions are in good agreement
with the exact solutions.
Example 4.2. Consider the linear system of fractional
order differential equations
Dα1x(t) = x(t)−y(t)+4z(t),
Dα2y(t) = 3x(t)+2y(t)−z(t),
Dα3z(t) = 2x(t) + y(t)−z(t),
(16)
subject to the initial conditions
x(0) = −1, y(0) = 7, z(0) = 3.(17)
The exact solution of the system (16), when α1=
α2=α3=α, is
x(t) = −etα
α+e3tα
α−e−2tα
α,
y(t) = 4etα
α+ 2e3tα
α+e−2tα
α,
z(t) = etα
α+e3tα
α+e−2tα
α.
(18)
According to the formulas (8), the variational iteration
formulas for system (16) are given by
xk+1(t) = xk(t)−
∫t
0
τα1−1[Dα1xk(τ)−xk(τ) + yk(τ)−4zk(τ)]dτ,
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(a)
(b)
(c)
Figure 1: A Comparison for exact x(t)and the ap-
proximate solution x4(t)in Example (4.1) where
α1=α2= 1,0.9,0.5.
yk+1(t) = yk(t)−
∫t
0
τα2−1[Dα2yk(τ)−3xk(τ)−2yk(τ) + zk(τ)]dτ,
zk+1(t) = zk(t)−
∫t
0
τα3−1[Dα3zk(τ)−2xk(τ)−yk(τ) + zk(τ)]dτ.
Begin with x0(t) = −1, y0(t) = 7 and z0(t) = 3,
we obtain
x1(t) = −1+4tα1
α1
,
y1(t) = 7 + 8tα2
α2
,
z1(t) = 3 + 2tα3
α3
,
x2(t) = −1 + 4tα1
α1
+ 2t2α1
α2
1
−8tα1+α2
α2(α1+α2)+
8tα1+α3
α3(α1+α3),
(a)
(b)
(c)
Figure 2: A Comparison for exact y(t)and the ap-
proximate solution y4(t)in Example (4.1) where
α1=α2= 1,0.9,0.5.
y2(t) = 7 + 8tα2
α2
+8t2α2
α2
2
+12tα1+α2
α1(α1+α2)−2tα2+α3
α3(α2+α3),
z2(t) = 3 + 2tα3
α3
−t2α3
α2
3
+8tα1+α3
α1(α1+α3)+8tα2+α3
α2(α2+α3),
x3(t) = −1 + 4tα1
α1
+2t2α1
α2
1
+2t3α1
3α3
1
−8tα1+α2
α2(α1+α2)−
12t2α1+α2
α1(α1+α2)(2α1+α2)+8tα1+α3
α3(α1+α3)−
8t2α1+α2
α2(α1+α2)(2α1+α2)−8tα1+2α2
α2
2(α1+ 2α2)+
32t2α1+α3
α1(α1+α3)(2α1+α3)+8t2α1+α3
α3(α1+α3)(2α1+α3)+
32tα1+α2+α3
α2(α2+α3)(α1+α2+α3)+
2tα1+α2+α3
α3(α2+α3)(α1+α2+α3)−4tα1+2α2
α2
3(α1+ 2α3),
y3(t) = 7 + 8tα2
α2
+8t2α2
α2
2
+16
3
t3α2
3α3
2
+12tα1+α2
α1(α1+α2)+
6t2α1+α2
α2
1(2α1+α2)+24tα1+2α2
α1(α1+α2)(α1+ 2α2)−
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24tα1+2α2
α2(α1+α2)(α1+ 2α2)−2tα2+α3
α3(α2+α3)−
8tα1+α2+α3
α1(α1+α3)(α1+α2+α3)+
24tα1+α2+α3
α3(α1+α3)(α1+α2+α3)−
8t2α2+α3
α2(α2+α3)(2α2+α3)−
4t2α2+α3
α3(α2+α3)(2α2+α3)+tα2+2α3
α2
3(α2+ 2α3),
z3(t) = 3 + 2tα3
α3
−t2α3
α2
3
+t3α3
3α3
3
+8tα1+α3
α1(α1+α3)+
4t2α1+α3
α2
1(2α1+α3)+8tα2+α3
α2(α2+α3)−
16tα1+α2+α3
α2(α1+α2)(α1+α2+α3)+8t2α2+α3
α2
2(2α2+α3)−
8tα1+2α3
α1(α1+α3)(α1+ 2α3)+
16tα1+2α3
α3(α1+α3)(α1+ 2α3)−
8tα2+2α3
α2(α2+α3)(α2+ 2α3)−
2tα2+2α3
α3(α2+α3)(α2+ 2α3)+
12tα1+α2+α3
α1(α1+α2)(α1+α2+α3),
.
.
.
(19)
If α1=α2=α3=α, then the solution of (16) in
series form is given by
x(t) = −1+4tα
α+ 2t2α
α2+17t3α
3α3+. . . ,
y(t) =7 + 8tα
α+ 13t2α
α2+25t3α
3α3+. . . ,
z(t) =3 + 2tα
α+ 7t2α
α2+10t3α
3α3+. . . ,
(20)
which converges to the exact solution (18).
The graphics of the exact and approximate solutions
are given in Figure 3.
Example 4.3. Consider the nonlinear predator-prey
(a)
(b)
(c)
Figure 3: A Comparison for exact x(t), y(t), z(t)
and the approximate solution x3(t), y3(t), z3(t), re-
spectively in Example (4.2) where α1=α2=α3=
1.
system of fractional order differential equations
Dα1x(t) = x(t) + y2(t),
Dα2y(t) = y(t)
2,(21)
subject to the initial conditions
x(0) = 0, y(0) = 1.(22)
The exact solution of the system (21), when α1=
α2=α, is
x(t) = tα
αetα
α,
y(t) = etα
2α.
(23)
According to the formulas (8), the variational iteration
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formulas for system (21) are given by
xk+1(t) = xk(t)−
∫t
0
τα1−1[Dα1xk(τ)−xk(τ)−(yk(τ))2]dτ,
yk+1(t) = yk(t)−∫t
0
τα2−1[Dα2yk(τ)−yk(τ)
2]dτ.
(24)
Begin with x0(t) = 0 and y0(t) = 1, we obtain
x1(t) = tα1
α1
,
y1(t) = 1 + tα2
2α2
,
x2(t) = tα1
α1
+t2α1
2α2
1
+tα1+α2
α2(α1+α2)+tα1+2α2
4α2
2(α1+ 2α2),
y2(t) = 1 + tα2
2α2
+t2α2
8α2
2
,
x3(t) = tα1
α1
+t2α1
2α2
1
+t3α1
6α3
1
+tα1+α2
α2(α1+α2)+
t2α1+α2
α2(α1+α2)(2α1+α2)+α1tα1+2α2
4α2
2(α1+ 2α2)2+
3α1tα1+2α2
4α2(α1+α2)(α1+ 2α2)2+tα1+3α2
8α3
1(α1+ 3α2)+
tα1+2α2
(α1+α2)(α1+ 2α2)2+tα1+4α2
64α4
2(α1+ 4α2)+
α1tα1+2α2
4α2
2(α1+α2)(α1+ 2α2)+
t2α1+2α2
8α2
2(α1+α2)(α1+ 2α2),
y3(t) = 1 + tα2
2α2
+t2α2
8α2
2
+t3α2
48α3
2
,
.
.
.
(25)
If α1=α2=α, then
x3(t) = tα
α+t2α
α2+37
72
t3α
α3,
y3(t) = 1 + tα
2α+t2α
8α2+t3α
48α3,
(26)
Figures 4 and 5 show the approximate solutions x3(t)
and y3(t)and the exact solutions x(t)and y(t),
respectively for different values of α1=α2=
1,0.9,0.5.
(a)
(b)
(c)
Figure 4: A Comparison for exact x(t)and the ap-
proximate solution x3(t)in Example (4.3) where
α1=α2= 1,0.9,0.5.
(a)
(b)
(c)
Figure 5: A Comparison for exact y(t)and the ap-
proximate solution y3(t)in Example (4.3) where
α1=α2= 1,0.9,0.5.
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Example 4.4. Consider the nonlinear system of frac-
tional order differential equations
Dα1x(t) = t z(t),
Dα2y(t) = x(t)y(t),
Dα3z(t) = 2x2(t),
(27)
subject to the initial conditions
x(0) = 1, y(0) = 1, z(0) = 0.(28)
According to the formulas (8), the variational iteration
formulas for system (27) are given by
xk+1(t) = xk(t)−∫t
0
τα1−1[Dα1xk(τ)−τ zk(τ)]dτ,
yk+1(t) = yk(t)−
∫t
0
τα2−1[Dα2yk(τ)−xk(τ)yk(τ)]dτ,
zk+1(t) = zk(t)−∫t
0
τα3−1[Dα3zk(τ)−2(xk(τ))2]dτ.
(29)
Begin with x0(t)=1, y0(t)=1and z0(t)=0, we
obtain
x1(t) = 1,
y1(t) = 1 + tα2
α2
,
z1(t) = 2tα3
α3
,
x2(t) = 1 + 2tα1+α3+1
α3(α1+α3+ 1),
y2(t) = 1 + tα2
α2
+t2α2
2α2
2
,
z2(t) = 2tα3
α3
,
x3(t) = 1 + 2tα1+α3+1
α3(α1+α3+ 1),
y3(t) = 1 + tα2
α2
+t2α2
2α2
2
−t3α2
6α3
2
+
2tα1+α2+α3+1
α3(α1+α3+ 1)(α1+α2+α3+ 1)+
2tα1+2α2+α3+1
α2α3(α1+α3+ 1)(α1+ 2α2+α3+ 1)+
tα1+3α2+α3+1
α2
2α3(α1+α3+ 1)(α1+ 3α2+α3+ 1),
z3(t) = 2tα3
α3
+ 4 tα1+2α3+1
α3(α1+α3+ 1)(α1+ 2α3+ 1),
(30)
5 Conclusions
In this article, we presented a modified technique of
the variational iteration method that approximate the
solutions of linear and nonlinear systems of differ-
ential equations of fractional order. Several exam-
ples were examined to show the efficiency of our
new method. We have solved three systems: lin-
ear and nonlinear of fractional differential equations
by the proposed technique. The numerical solutions
are given in series form, that converge to the exact
solution obtained by conformable Laplace transform
method modified by the author. We observed that our
approach is effective in obtaining numerical solutions
for linear and nonlinear systems.
Future research might apply our procedure to obtain
numerical solution for nonlinear fractional differen-
tial equations arise in physics and engineering.
Conflict of Interests
The author declares no conflict of interest with re-
gards to the publication of this paper.
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