# Fractional Variational Iteration Method for Fractional Cauchy Problems

Authors: Bao Si-yuan
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

The fractional variational iteration method is used to solve the fractional Cauchy problem. Some examples are given to elucidate the solution procedure and reliability of the obtained results. The variational iteration algorithm leads to exact solutions in the present study.


---

## Body

## 1. Introduction

Fractional problems have attracted many scholars’ attention as the actual problems can be much better simulated by using the fractional derivatives than by using traditional integral derivatives [1–6].In [7], the author gave a very lucid as well as elementary discussion of the variational iteration method. The variational iteration algorithm [7–14] is very simple, with results of high precision (sometimes exact solutions can be obtained), and is easy to understand and has been widely applied in various nonlinear problems. In addition, many authors have made a great effort to give sophisticated theoretical verification of the variational iteration method; for example, Odibat [15], Salkuyeh [16], and Tatari and Dehghan [17] proved that the variational iteration algorithm leads to convergent results. Due to its flexibility and ability to solve nonlinear equations accurately and conveniently, the method has been modified or improved to solve nonlinear problems more efficiently [18–21]; furthermore the method has been extended to handle fractional nonlinear models and the fractional variational iteration method (FVIM) has been presented [22].Many scholars have applied FVIM to solve linear or nonlinear fractional order differential equations [23–28]. For a relatively comprehensive survey on the concepts, theory, and applications of the method, readers are referred to review articles [10, 29].In [22] the improved VIM is called fractional variational iteration method and the new Lagrange multiplier is determined by using the Laplace transformation. In this paper, we use the fractional variational iteration method to discuss the fractional-order partial differential equation in the form
(1)
u
t
(
α
)
(
x
,
t
)
+
a
(
x
,
t
)
u
x
(
x
,
t
)
=
ϕ
(
x
)
,
x
∈
R
,
t
>
0
(2)
u
(
x
,
0
)
=
ψ
(
x
)
,
x
∈
R
.Whena
(
x
,
t
)
=
a is a constant and ϕ
(
x
)
=
0, (1) is a linear equation called the fractional transport equation which can describe many interesting phenomena such as the spread of AIDS and the moving of wind. When a
(
x
,
t
)
=
u
(
x
,
t
), the equation is called the nefractional inviscid Burgers’ equation arising in a one-dimensional stream of particles or fluid having zero viscosity. When α
=
1, (1) is the equation for traditional Cauchy problem [30, 31].


## 2. The Fractional Variational Iteration Method

To discuss the fractional problems, two definitions are introduced.Definition 1.
The Caputo derivative is given as(3)
D
0
C
t
α
f
(
t
)
=
1
Γ
(
m
-
α
)
∫
a
t
(
t
-
τ
)
m
-
α
-
1
f
(
m
)
(
τ
)
d
τ
,
t
>
a
,
m
-
1
<
α
<
m
∈
Z
+
.Definition 2.
The Riemann-Liouville (R-L) integration off
(
t
) is defined as
(4)
I
a
,
t
α
f
(
t
)
=
1
Γ
(
α
)
∫
a
t
(
t
-
τ
)
α
-
1
f
(
τ
)
d
τ
,
α
>
0
.To illustrate the basic concepts of the fractional variational iteration method, one considers the following general nonlinear fractional system:(5)
0
C
D
t
α
u
+
R
[
u
]
+
N
[
u
]
=
f
(
τ
)
,
where R
[
u
] is a linear term and N
[
u
] is a nonlinear one and f
(
t
) is a known analytic function.According to the fractional variational iteration method, a correction functional can be constructed as follows:(6)
u
n
+
1
=
u
n
+
∫
0
t
λ
(
t
,
τ
)
[
D
0
C
t
α
u
n
+
R
[
u
n
]
+
N
[
u
n
]
-
f
(
τ
)
]
d
τ
,
where λ
(
t
,
τ
) is a general Lagrange multiplier [7–12], which can be identified optimally via the variational theory, the subscript n denotes the nth approximation, and R
[
u
n
] and N
[
u
n
] are considered as a restricted variation [7]; that is δ
R
[
u
n
]
=
0, δ
N
[
u
n
]
=
0.Taking Laplace transform on the correction functional equation established via the R-L integration, Wu and Baleanu [22] present a new way to identify the Lagrange multiplier. The Lagrange multiplier can be identified as
(7)
λ
(
t
,
τ
)
=
(
-
1
)
α
(
τ
-
t
)
α
-
1
Γ
(
α
)
.The variational iteration formula (6) can be improved as
(8)
u
n
+
1
=
u
n
+
∫
0
t
(
-
1
)
α
(
τ
-
t
)
α
-
1
Γ
(
α
)
[
D
0
C
t
α
u
n
+
R
[
u
n
]
u
n
+
∫
0
t
(
-
1
)
α
(
τ
-
t
)
α
-
1
Γ
(
α
)
+
N
[
u
n
]
-
f
(
τ
)
D
0
C
t
α
]
d
τ
.The initial guessu
0 can be freely chosen with possible unknown constants; it can also be solved from its corresponding linear homogeneous equation.The above iteration formula (8) is also valid for differential equations when α is an arbitrary positive integer.The fractional variational iteration method can solve effectively, easily, and accurately a large class of nonlinear fractional problems with approximations converging rapidly to the accurate solution.


## 3. Applications

Since our focus is on the ideas and basic principles, we will consider only the simplest possible equations to clearly illustrate the solution procedure. In particular, we will focus on pure Cauchy problems. These problems are initial value problems.According to (8), we can construct a correction functional to (1) which reads
(9)
u
n
+
1
(
x
,
t
)
=
u
n
(
x
,
t
)
+
∫
0
t
(
-
1
)
α
(
τ
-
t
)
α
-
1
Γ
(
α
)
[
D
0
C
t
α
u
n
(
x
,
τ
)
+
∫
0
t
(
-
1
)
α
(
τ
-
t
)
α
-
1
Γ
(
α
)
+
a
(
x
,
τ
)
u
x
(
x
,
τ
)
+
∫
0
t
(
-
1
)
α
(
τ
-
t
)
α
-
1
Γ
(
α
)
-
ϕ
(
x
)
0
C
D
t
α
]
d
τ
.Example 3.
Consider the fractional transport equation:(10)
u
t
(
α
)
(
x
,
t
)
+
a
u
x
(
x
,
t
)
=
0
,
x
∈
R
,
t
>
0
u
(
x
,
0
)
=
x
2
,
x
∈
R
.
According to the iteration formula (8), we can obtain
(11)
u
n
+
1
=
u
n
-
∫
0
t
(
τ
-
t
)
α
Γ
(
α
)
[
u
n
τ
(
α
)
(
x
,
τ
)
+
a
u
n
x
(
x
,
t
)
]
d
τ
,
u
n
τ
(
α
)
(
x
,
τ
)
+
a
u
n
x
(
x
,
t
)
(
x
,
t
)
iiiiiii
x
∈
R
,
t
>
0
.The initial iterative value is selected asu
0
(
x
,
0
)
=
u
(
x
,
0
)
=
x
2 from the given initial condition. Using (10), we obtain the following successive approximations:
(12)
u
1
(
x
,
t
)
=
x
2
-
2
a
x
t
α
Γ
(
α
+
1
)
u
2
(
x
,
t
)
=
x
2
-
2
a
x
t
α
Γ
(
α
+
1
)
+
2
a
2
t
2
α
Γ
(
2
α
+
1
)
u
3
(
x
,
t
)
=
x
2
-
2
a
x
t
α
Γ
(
α
+
1
)
+
2
a
2
t
2
α
Γ
(
2
α
+
1
)
⋮
u
n
(
x
,
t
)
=
x
2
-
2
a
x
t
α
Γ
(
α
+
1
)
+
2
a
2
t
2
α
Γ
(
2
α
+
1
)
.So the exact analytical solution is yielded as(13)
u
(
x
,
t
)
=
x
2
-
2
a
x
t
α
Γ
(
α
+
1
)
+
2
a
2
t
2
α
Γ
(
2
α
+
1
)
.Example 4.
Consider the fractional nonlinear Cauchy problem(14)
u
t
(
α
)
(
x
,
t
)
+
x
u
x
(
x
,
t
)
=
0
,
x
∈
R
,
t
>
0
u
0
(
x
,
0
)
=
u
(
x
,
0
)
=
x
2
,
x
∈
R
.
According to the iteration formula (8), we can obtain
(15)
u
n
+
1
=
u
n
-
∫
0
t
(
τ
-
t
)
α
Γ
(
α
)
[
u
t
(
α
)
(
x
,
t
)
+
x
u
x
(
x
,
t
)
]
d
ξ
,
x
∈
R
,
t
>
0
.The initial iterative value is selected asu
0
(
x
,
0
)
=
u
(
x
,
0
)
=
x
2 from the given initial condition. Using (15), we obtain the following successive approximations:
(16)
u
0
(
x
,
t
)
=
x
2
u
1
(
x
,
t
)
=
x
2
-
2
x
2
t
α
Γ
(
α
+
1
)
u
2
(
x
,
t
)
=
x
2
-
2
x
2
t
α
Γ
(
α
+
1
)
+
x
2
(
2
t
α
)
2
Γ
(
2
α
+
1
)
u
3
(
x
,
t
)
=
x
2
-
2
x
2
t
α
Γ
(
α
+
1
)
+
x
2
(
2
t
α
)
2
Γ
(
2
α
+
1
)
-
x
2
(
2
t
α
)
3
Γ
(
3
α
+
1
)
⋮
u
n
(
x
,
t
)
=
x
2
[
1
-
2
t
α
Γ
(
α
+
1
)
+
(
2
t
α
)
2
Γ
(
2
α
+
1
)
-
(
2
t
α
)
3
Γ
(
3
α
+
1
)
+
(
2
t
α
)
4
Γ
(
4
α
+
1
)
-
(
2
t
α
)
5
Γ
(
5
α
+
1
)
+
⋯
]
.
The VIM admits the use of u
=
li
m
n
→
∞
u
n, which gives the exact solution
(17)
u
(
x
,
t
)
=
x
2
E
α
(
-
2
t
α
)
,
where E
α is Mittag-Leffler function
(18)
E
α
(
z
)
=
∑
n
=
0
∞
z
n
Γ
(
α
n
+
1
)
.Example 5.
Consider the following nonhomogeneous fractional Cauchy problem:(19)
u
t
(
α
)
(
x
,
t
)
+
x
u
x
(
x
,
t
)
=
x
,
x
∈
R
,
t
>
0
u
(
x
,
0
)
=
e
x
,
x
∈
R
.According to the iteration formula (9), we can obtain
(20)
u
n
+
1
=
u
n
-
∫
0
t
(
τ
-
t
)
α
Γ
(
α
)
[
u
t
(
α
)
(
x
,
t
)
+
x
u
x
(
x
,
t
)
-
x
]
d
ξ
.
The following successive approximations are obtained by using (20) with the selected initial value u
0
=
e
x:
(21)
u
1
(
x
,
t
)
=
e
x
+
(
x
-
e
x
)
t
α
Γ
(
α
+
1
)
u
2
(
x
,
t
)
=
x
2
+
(
x
-
e
x
)
t
α
Γ
(
α
+
1
)
+
(
e
x
-
1
)
t
2
α
Γ
(
2
α
+
1
)
u
3
(
x
,
t
)
=
x
2
+
(
x
-
e
x
)
t
α
Γ
(
α
+
1
)
+
(
e
x
-
1
)
t
2
α
Γ
(
2
α
+
1
)
-
e
x
t
3
α
Γ
(
3
α
+
1
)
⋮
u
n
(
x
,
t
)
=
[
x
-
Γ
(
α
+
1
)
t
α
Γ
(
2
α
+
1
)
]
t
α
Γ
(
α
+
1
)
+
e
x
[
1
-
t
α
Γ
(
α
+
1
)
+
t
2
α
Γ
(
2
α
+
1
)
-
t
3
α
Γ
(
3
α
+
1
)
+
t
4
α
Γ
(
4
α
+
1
)
-
t
5
α
Γ
(
5
α
+
1
)
+
⋯
]
.
By using u
=
li
m
n
→
∞
u
n, the exact solution of the equation is obtained as
(22)
u
(
x
,
t
)
=
[
x
-
Γ
(
α
+
1
)
t
α
Γ
(
2
α
+
1
)
]
t
α
Γ
(
α
+
1
)
+
e
x
E
α
(
-
t
α
)
.
Particularly when α
=
1, the result in (22) is u
(
x
,
t
)
=
(
x
-
(
t
/
2
)
)
t
+
e
x
-
t, which is the same as that in [30].Example 6.
Consider the fractional inviscid Burgers’ equation(23)
u
t
(
α
)
(
x
,
t
)
+
u
(
x
,
t
)
u
x
(
x
,
t
)
=
0
,
x
∈
R
,
t
>
0
u
(
x
,
0
)
=
x
,
x
∈
R
.
According to the iteration formula (8), we can obtain
(24)
u
n
+
1
=
u
n
-
∫
0
t
(
τ
-
t
)
α
Γ
(
α
)
[
u
t
(
α
)
(
x
,
τ
)
+
u
(
x
,
τ
)
u
x
(
x
,
τ
)
]
d
τ
.
Starting with initial approximation u
0
(
x
,
0
)
=
u
(
x
,
0
)
=
x and proceeding in a similar way illustrated above, we obtain the following successive approximations:
(25)
u
0
(
x
,
t
)
=
1
-
t
α
Γ
(
α
+
1
)
u
1
(
x
,
t
)
=
1

-

t
α
Γ
(
α
+
1
)

+

2
t
2
α
Γ
(
2
α
+
1
)

-

Γ
(
2
α
+
1
)
t
3
α
Γ
(
3
α
+
1
)
Γ
2
(
α
+
1
)
u
2
(
x
,
t
)
=
1
-
t
α
Γ
(
α
+
1
)
+
2
t
2
α
Γ
(
2
α
+
1
)
+
a
3
t
3
α
Γ
(
3
α
+
1
)
+
a
4
t
4
α
Γ
(
4
α
+
1
)
+
a
5
t
5
α
Γ
(
5
α
+
1
)
+
a
6
t
6
α
Γ
(
6
α
+
1
)
+
a
7
t
7
α
Γ
(
7
α
+
1
)
⋮
,
where
(26)
a
3
=
-
4
Γ
(
1
+
3
α
)
-
Γ
(
1
+
2
α
)
Γ
2
(
1
+
α
)
Γ
(
1
+
3
α
)
a
4
=
2
α
Γ
(
4
α
)
+
Γ
(
2
α
)
α
2
Γ
2
(
α
)
Γ
(
4
α
)
+
3
Γ
(
3
α
)
2
α
2
Γ
(
α
)
Γ
(
2
α
)
Γ
(
4
α
)
a
5
=
-
4
Γ
(
1
+
4
α
)
Γ
2
(
1
+
2
α
)
-
2
Γ
(
1
+
2
α
)
Γ
(
1
+
4
α
)
Γ
3
(
1
+
α
)
Γ
(
1
+
3
α
)
a
6
=
4
Γ
(
1
+
5
α
)
Γ
2
(
1
+
α
)
Γ
(
1
+
3
α
)
(27)
a
7
=
-
Γ
2
(
1
+
2
α
)
Γ
(
1
+
6
α
)
Γ
4
(
1
+
α
)
Γ
2
(
1
+
3
α
)
.
Particularly for the case α
=
1, the result for u
3 in (23) will be
(28)
u
3
=
x
(
1
-
t
+
t
2
-
t
3
+
2
3
t
4
-
2
3
t
5
+
1
9
t
6
-
1
63
t
7
)
,
which is the same as that in [30].In order to show the convergence of the iteration solutions, we introduce a functionv
i
(
x
,
t
):
(29)
v
i
(
x
,
t
)
=
u
i
(
x
,
t
)
x
.
It can be seen that v
i is a function of t. When α
=
1, the accurate solution of (23) is u
=
x
/
(
1
+
t
); then v
=
1
/
(
1
+
t
).Figures1, 2, and 3 show the curves of v
i (where i means the iteration times) changing with time t when α
=
0.5, 1, and 1.5. When α
=
1, the fifth iteration result v
5 is almost the same as the exact solution at t
∈
[
0,1
]. When α
=
1.5, the iteration results converge fast and the fourth iteration result v
4 is almost the same as v
5 at t
∈
[
0,3
].Figure 1
v
i
(
t
)’s curve changing with time t when α
=
0.5.Figure 2
v
i
(
t
)’s curve changing with time t when α
=
1.Figure 3
v
i
(
t
)’s curve changing with time t when α
=
1.5.


## 4. Conclusions

The variational iteration method has been proved by many authors to be a powerful mathematical tool for various kinds of nonlinear problems.In this paper, we extend the use of fractional variational iteration method to fractional Cauchy problems and give the numerical examples. Compared with the classical VIM, the modified version method is powerful for solving differential equations with fractional derivatives. The higher order approximate solutions of the Cauchy equation illustrate the method’s efficiency and high accuracy.



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*Source: 925016-2014-02-23.xml*