# Variational Iteration Transform Method for Fractional Differential Equations with Local Fractional Derivative

Authors: Yong-Ju Yang; Liu-Qing Hua
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We propose the variational iteration transform method in the sense of local fractional derivative, which is derived from the coupling method of local fractional variational iteration method and differential transform method. The method reduces the integral calculation of the usual variational iteration computations to more easily handled differential operation. And the technique is more orderly and easier to analyze computing result as compared with the local fractional variational iteration method. Some examples are illustrated to show the feature of the presented technique.


---

## Body

## 1. Introduction

Fractional differential equation has been considered with great importance due to its demonstrated applications in various areas such as electrical networks, fluid flow, biology, and dynamical systems [1–8]. And many classical differential equations possess a fractional analogy. As a result, a substantial number of methods for solving the fractional differential equations are developed to solve fractional differential equations [9–25].Recently, local fractional derivative and calculus theory has been introduced by the researcher in [18, 19], which is set up on fractal geometry and which is the best method for describing the nondifferential function defined on Cantor sets. The physical explanation of the local fractional derivative can be seen in [26, 27]. A great deal of research work has been directed for the nondifferentiable phenomena in fractal domain concerning local fractional derivative, for example, [11, 12, 15, 18–25].Motivated by the ongoing research method of local fractional differential equation, we present variational iteration transform method, which is the coupling method of local fractional variational iteration method and differential transform method. It is worth mentioning that since the method makes the complex iteration calculation more orderly, it may be considered as an efficient modification of variational iteration method. The rest of the paper is organized as follows. In Section2, the basic mathematical fundamentals are presented briefly. In Section 3, the local fractional function differential transform method for solving the differential equations with local fractional derivative is investigated. In Section 4, several examples are illustrated. Finally, in Section 5, the conclusion is given.


## 2. Mathematical Fundamentals

In this section, we introduce the basic notions of local fractional derivative, local fractional integral, and local fractional differential transform method.


### 2.1. Local Fractional Derivatives and Integrals

The local fractional derivative ofu
(
x
) of order α
(
0
<
α
≤
1
) at x
0 is defined by [18, 19]
(1)
u
(
α
)
(
x
0
)
=
lim
⁡
x
→
x
0
Δ
α
(
u
(
x
)
-
u
(
x
0
)
)
(
x
-
x
0
)
α
=
D
x
(
α
)
u
(
x
0
)
=
d
α
u
(
x
)
d
x
α
|
x
=
x
0
,
where Δ
α
(
u
(
x
)
-
u
(
x
0
)
)
≅
Γ
(
1
+
α
)
(
u
(
x
)
-
u
(
x
0
)
).In the interval[
a
,
b
], local fractional integral of u
(
x
) of order α is given by [18, 19]
(2)
I
b
(
α
)
a
u
(
x
)
=
1
Γ
(
1
+
α
)
lim
⁡
Δ
t
→
0
∑
j
=
0
j
=
N
-
1
u
(
t
j
)
(
Δ
t
j
)
α
=
1
Γ
(
1
+
α
)
∫
a
b
u
(
t
)
(
d
t
)
α
,
where  Δ
t
j
=
t
j
+
1
-
t
j, Δ
t
=
max
⁡
{
Δ
t
1
,
Δ
t
2
,
Δ
t
j
,
…
} and [
t
j
,
t
j
+
1
], j
=
0
,
…
,
N
-
1, and t
0
=
a
,
t
N
=
b is a partition of the interval [
a
,
b
].


### 2.2. Local Fractional Differential Transform Method

Similarly in [28–34], the local fractional different transform U
(
x
,
t
) (or u
(
x
), resp.) of the function u
(
x
,
t
) (or u
(
x
), resp.) is defined by the following formula:
(3)
U
(
x
,
k
)
=
1
Γ
(
1
+
k
α
)
[
∂
k
α
∂
t
k
α
u
(
x
,
t
)
]
t
=
0
(
or
U
(
k
)
=
1
Γ
(
1
+
k
α
)
[
d
k
α
u
(
t
)
d
t
k
α
]
t
=
0
,
resp
.
)
.
Obviously
(4)
u
(
x
,
t
)
=
∑
k
=
0
∞
{
U
(
x
,
k
)
t
k
α
}
(
or
u
(
t
)
=
∑
k
=
0
∞
{
U
(
k
)
t
k
α
}
,
resp
.
)
.
If we let
(5)
U
(
x
,
k
)
=
D
T
k
{
u
(
x
,
t
)
}
(
or
U
(
k
)
=
D
T
k
{
u
(
t
)
}
,
resp
.
)
and let
(6)
D
T
k
-
1
{
U
(
x
,
k
)
}
=
u
(
k
α
)
(
x
,
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
(
or
D
T
k
-
1
{
U
(
k
)
}
=
u
(
k
α
)
(
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
,
resp
.
)
then we have
(7)
u
(
x
,
t
)
=
∑
k
=
0
∞
D
T
k
-
1
{
U
(
x
,
k
)
}
(
or
u
(
t
)
=
∑
k
=
0
∞
D
T
k
-
1
{
U
(
k
)
}
,
resp
.
)
.In Table1, we list a few operations concerning local fractional differential transform, where m is a nonnegative integer.Table 1
Original function
Transformed function
f
(
x
)
=
x
m
α
F
(
k
)
=
{
1
,
k
=
m
0
,
k
≠
m
u
(
x
)
v
(
x
)
∑
r
=
0
k
{
U
(
r
)
V
(
k
-
r
)
}
l
l
l
l
l
u
(
m
α
)
(
x
)
Γ
[
1
+
(
k
+
m
)
α
]
Γ
(
1
+
k
α
)
U
(
k
+
m
)
;
k
≥
-
m
+
1
l
l
l
l
l
l
f
(
x
)
=
0
I
x
α
⋯
0
I
x
α
︸
k
0
u
(
τ
)
F
(
k
)
=
Γ
(
1
+
(
k
-
k
0
)
α
)
Γ
(
1
+
k
α
)
U
(
k
-
k
0
)


## 2.1. Local Fractional Derivatives and Integrals

The local fractional derivative ofu
(
x
) of order α
(
0
<
α
≤
1
) at x
0 is defined by [18, 19]
(1)
u
(
α
)
(
x
0
)
=
lim
⁡
x
→
x
0
Δ
α
(
u
(
x
)
-
u
(
x
0
)
)
(
x
-
x
0
)
α
=
D
x
(
α
)
u
(
x
0
)
=
d
α
u
(
x
)
d
x
α
|
x
=
x
0
,
where Δ
α
(
u
(
x
)
-
u
(
x
0
)
)
≅
Γ
(
1
+
α
)
(
u
(
x
)
-
u
(
x
0
)
).In the interval[
a
,
b
], local fractional integral of u
(
x
) of order α is given by [18, 19]
(2)
I
b
(
α
)
a
u
(
x
)
=
1
Γ
(
1
+
α
)
lim
⁡
Δ
t
→
0
∑
j
=
0
j
=
N
-
1
u
(
t
j
)
(
Δ
t
j
)
α
=
1
Γ
(
1
+
α
)
∫
a
b
u
(
t
)
(
d
t
)
α
,
where  Δ
t
j
=
t
j
+
1
-
t
j, Δ
t
=
max
⁡
{
Δ
t
1
,
Δ
t
2
,
Δ
t
j
,
…
} and [
t
j
,
t
j
+
1
], j
=
0
,
…
,
N
-
1, and t
0
=
a
,
t
N
=
b is a partition of the interval [
a
,
b
].


## 2.2. Local Fractional Differential Transform Method

Similarly in [28–34], the local fractional different transform U
(
x
,
t
) (or u
(
x
), resp.) of the function u
(
x
,
t
) (or u
(
x
), resp.) is defined by the following formula:
(3)
U
(
x
,
k
)
=
1
Γ
(
1
+
k
α
)
[
∂
k
α
∂
t
k
α
u
(
x
,
t
)
]
t
=
0
(
or
U
(
k
)
=
1
Γ
(
1
+
k
α
)
[
d
k
α
u
(
t
)
d
t
k
α
]
t
=
0
,
resp
.
)
.
Obviously
(4)
u
(
x
,
t
)
=
∑
k
=
0
∞
{
U
(
x
,
k
)
t
k
α
}
(
or
u
(
t
)
=
∑
k
=
0
∞
{
U
(
k
)
t
k
α
}
,
resp
.
)
.
If we let
(5)
U
(
x
,
k
)
=
D
T
k
{
u
(
x
,
t
)
}
(
or
U
(
k
)
=
D
T
k
{
u
(
t
)
}
,
resp
.
)
and let
(6)
D
T
k
-
1
{
U
(
x
,
k
)
}
=
u
(
k
α
)
(
x
,
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
(
or
D
T
k
-
1
{
U
(
k
)
}
=
u
(
k
α
)
(
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
,
resp
.
)
then we have
(7)
u
(
x
,
t
)
=
∑
k
=
0
∞
D
T
k
-
1
{
U
(
x
,
k
)
}
(
or
u
(
t
)
=
∑
k
=
0
∞
D
T
k
-
1
{
U
(
k
)
}
,
resp
.
)
.In Table1, we list a few operations concerning local fractional differential transform, where m is a nonnegative integer.Table 1
Original function
Transformed function
f
(
x
)
=
x
m
α
F
(
k
)
=
{
1
,
k
=
m
0
,
k
≠
m
u
(
x
)
v
(
x
)
∑
r
=
0
k
{
U
(
r
)
V
(
k
-
r
)
}
l
l
l
l
l
u
(
m
α
)
(
x
)
Γ
[
1
+
(
k
+
m
)
α
]
Γ
(
1
+
k
α
)
U
(
k
+
m
)
;
k
≥
-
m
+
1
l
l
l
l
l
l
f
(
x
)
=
0
I
x
α
⋯
0
I
x
α
︸
k
0
u
(
τ
)
F
(
k
)
=
Γ
(
1
+
(
k
-
k
0
)
α
)
Γ
(
1
+
k
α
)
U
(
k
-
k
0
)


## 3. Local Fractional Variational Iteration Transform Method

To introduce the basic ideas of the variational iteration transform method, we consider the following nonlinear fractional differential equation in the sense of local fractional derivative:(8)
∂
k
o
α
∂
t
k
0
T
(
x
,
t
)
+
L
α
T
(
x
,
t
)
+
R
α
T
(
x
,
t
)
=
g
(
x
,
t
)
,
where L
α is the linear local fractional operator of order less than k
0
(
k
0
≥
2
), R
α is the nonlinear local fractional operator of order less than k
0, and g
(
x
,
t
) is a source term of the nondifferential function.According to the local fractional variational iteration method [20, 35], we can construct local fractional variational iteration algorithm:
(9)
T
n
+
1
(
x
,
t
)
=
T
n
(
x
,
t
)
+
I
t
0
(
(
k
o
-
1
)
α
)
×
{
λ
α
(
τ
)
Γ
(
1
+
α
)
(
∂
k
o
α
∂
t
k
0
T
n
(
x
,
τ
)
+
L
α
T
n
(
x
,
τ
)
h
h
h
h
h
h
h
h
h
h
h
h
h
+
R
α
T
~
n
(
x
,
τ
)
-
g
(
x
,
τ
)
∂
k
o
α
∂
t
k
0
)
λ
α
(
τ
)
Γ
(
1
+
α
)
}
,
where λ
α
(
τ
)
/
Γ
(
1
+
α
) is a fractal Lagrange multiplier.Supposeδ
α
T
~
n
(
x
,
τ
) is a restricted local fractional variation; that is, δ
α
T
~
n
(
x
,
τ
)
=
0 [20, 35]; we get [21]
(10)
λ
α
(
τ
)
Γ
(
1
+
α
)
=
(
τ
-
t
)
(
k
0
-
1
)
α
Γ
[
1
+
(
k
0
-
1
)
α
]
.
Substituting (10) into (9), a fractional iteration procedure is constructed as follows:
(11)
T
n
+
1
(
x
,
t
)
=
T
n
(
x
,
t
)
+
I
t
0
(
(
k
o
-
1
)
α
)
×
{
(
x
-
t
)
(
k
0
-
1
)
α
Γ
[
1
+
(
k
0
-
1
)
α
]
h
h
h
h
h
×
{
∂
k
o
α
∂
t
k
0
T
n
(
x
,
τ
)
h
h
h
h
h
h
h
h
+
L
α
T
n
(
x
,
τ
)

+
R
α
T
n
(
x
,
τ
)

-
g
(
x
,
τ
)
∂
k
o
α
∂
t
k
0
}
}
.Applying local fractional differential transform with respect tot on both sides of (11) and using local fractional derivative algorithm [18], we obtain
(12)
D
T
k
(
T
n
+
1
(
x
,
t
)
)
=
D
T
k
(
T
0
(
x
,
t
)
)
,
(
0
≤
k
≤
k
0
-
1
)
and also
(13)
D
T
k
{
T
n
+
1
(
x
,
t
)
}
=
-
D
T
k
{
I
t
0
(
k
α
)
(
L
α
T
n
(
x
,
τ
)
+
R
α
T
n
(
x
,
τ
)
-
g
(
x
,
τ
)
)
}
,
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
(
k
≥
k
0
)
,
where T
0
(
x
,
t
) is an initial value.By virtue of (5), (12), (13), and the operation displayed in Table 1, we can yield
(14)
T
n
+
1
(
x
,
t
)
=
-
∑
k
=
k
0
∞
D
T
k
-
1
[
Γ
(
1
+
(
k
-
k
0
)
α
)
Γ
(
1
+
k
α
)
W
n
(
x
,
k
-
k
0
)
]
+
∑
k
=
0
k
0
-
1
D
T
k
-
1
{
D
T
k
(
T
0
(
x
,
k
)
)
}
=
-
∑
k
=
k
0
∞
D
T
k
-
1
[
Γ
(
1
+
(
k
-
k
0
)
α
)
Γ
(
1
+
k
α
)
W
n
(
x
,
k
-
k
0
)
]
+
∑
k
=
0
k
0
-
1
T
0
(
k
α
)
(
x
,
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
,
where
(15)
W
n
(
x
,
k
)
=
D
T
k
-
1
(
w
n
(
x
,
t
)
)
and where
(16)
w
n
(
x
,
t
)
=
L
α
T
n
(
x
,
t
)
+
R
α
T
n
(
x
,
t
)
-
g
(
x
,
t
)
.So, we have the exact solution of (8) in the form:
(17)
T
(
x
,
t
)
=
lim
⁡
n
→
∞
T
n
+
1
(
x
,
t
)
=
lim
⁡
n
→
∞
{
-
∑
k
=
k
0
∞
D
T
k
-
1
h
h
h
h
h
h
h
×
[
Γ
(
1
+
(
k
-
k
0
)
α
)
Γ
(
1
+
k
α
)
W
n
(
x
,
k
-
k
0
)
]
h
h
h
h
h
h
h
+
∑
k
=
0
k
0
-
1
T
0
(
k
α
)
(
x
,
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
∑
k
=
k
0
∞
D
T
k
-
1
}
.Now we discuss the solution of the following quadratic linear fractional differential equations which is the simple example of (8):
(18)
∂
2
α
T
(
x
,
t
)
∂
t
2
α
+
p
1
∂
2
α
T
(
x
,
t
)
∂
x
2
α
+
p
2
∂
α
T
(
x
,
t
)
∂
t
α
+
p
3
∂
α
T
(
x
,
t
)
∂
x
α
+
p
4
T
(
x
,
t
)
=
0
,
where p
1, p
2, p
3, and p
4 are all constant numbers.According to (9), the correction iteration algorithm is given as follows:
(19)
T
n
+
1
(
x
,
t
)
=
T
n
(
x
,
t
)
+
I
t
α
0
×
{
(
τ
-
t
)
(
α
)
Γ
(
1
+
α
)
{
L
t
t
(
2
α
)
T
n
+
p
1
L
x
x
(
2
α
)
T
n
+
p
2
L
t
(
α
)
T
n
hhhhhhhhhhhhih
+
p
3
L
x
(
α
)
T
n
+
p
4
T
n
}
(
τ
-
t
)
(
α
)
Γ
(
1
+
α
)
}
.
Taking local fractional differential transform with respect to t on (19), we deduce
(20)
D
T
k
(
T
n
+
1
(
x
)
)
=
-
D
T
k
-
2
(
p
1
L
x
x
(
2
α
)
T
n
+
p
2
L
t
(
α
)
T
n
+
p
3
L
x
(
α
)
T
n
+
p
4
T
n
)
,
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhii
(
k
≥
2
)
D
T
k
(
T
n
+
1
(
x
)
)
=
D
T
k
(
T
0
(
x
)
)
,
(
k
=
0,1
)
,
where
(21)
T
n
(
x
,
t
)
=
u
n
(
t
)
f
n
(
x
)
and where
(22)
U
n
(
k
)
=
D
T
k
(
u
n
(
t
)
)
.
By using (20), (21), (22), and the results included in Table 1, we obtain
(23)
U
n
+
1
(
k
)
f
n
+
1
(
x
)
=
-
(
p
1
f
n
(
2
α
)
(
x
)
+
p
3
f
n
(
α
)
(
x
)
+
p
4
f
n
(
x
)
)
×
D
T
k
{
I
t
(
2
α
)
0
(
u
n
)
}
-
p
2
f
n
(
x
)
D
T
k
{
I
t
(
α
)
0
(
u
n
)
}
=
-
(
p
1
f
n
(
2
α
)
(
x
)
+
p
3
f
n
(
α
)
(
x
)
+
p
4
f
n
(
x
)
)
U
(
k
-
2
)
k
α
(
k
-
1
)
α
-
p
2
f
n
(
x
)
U
(
k
-
1
)
k
α
,
(
k
≥
2
)
and also
(24)
U
n
+
1
(
k
)
f
n
+
1
(
x
)
=
U
0
(
k
)
f
n
+
1
(
x
)
,
(
0
≤
k
≤
1
)
.
If we let
(25)
p
1
f
0
(
2
α
)
(
x
)
+
p
3
f
0
(
α
)
(
x
)
+
p
4
f
0
(
x
)
=
p
5
f
0
(
x
)
,
then
(26)
f
n
(
x
)
=
f
0
(
x
)
.
By (23), (24), and (26), we have
(27)
U
n
+
1
(
k
)
=
-
p
5
U
n
(
k
-
2
)
k
α
(
k
-
1
)
α
-
p
2
U
n
(
k
-
1
)
k
α
,
(
k
≥
2
)
U
n
+
1
(
k
)
=
U
0
(
k
)
,
(
k
=
0,1
)
.
According to (14) and (27), we get
(28)
u
n
+
1
(
t
)
=
∑
k
=
2
∞
D
T
k
-
1
[
-
p
5
U
n
(
k
-
2
)
k
α
(
k
-
1
)
α
-
p
2
U
n
(
k
-
1
)
k
α
]
+
∑
k
=
0
1
u
0
(
k
α
)
(
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
.
Taking the limit, we arrive at
(29)
u
(
t
)
=
lim
⁡
n
→
∞
u
n
+
1
(
t
)
=
lim
⁡
n
→
∞
{
∑
k
=
2
∞
D
T
k
-
1
[
-
p
5
U
n
(
k
-
2
)
k
α
(
k
-
1
)
α
-
p
2
U
n
(
k
-
1
)
k
α
]
hhhhhhh
+
∑
k
=
0
1
u
0
(
k
α
)
(
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
}
.Hence, under the condition equation (25), the exact solution of (18) reads as follows:
(30)
T
(
x
,
t
)
=
u
(
t
)
f
0
(
x
)
.


## 4. Illustrative Examples

In porous media, there are fractals [35]. In fractal media, there are lots of mathematical physical models by the local differential fractional equation, for example, [36, 37]. In this section, we will present the solution of several equations or system of equations to assess the validity of variational iteration transform method.Case 1.
Consider the following nonlinear local fractional wave-like differential equation with variable coefficients:(31)
∂
2
α
T
∂
t
2
α
-
x
2
α
Γ
(
1
+
2
α
)
∂
α
(
D
x
(
α
)
(
T
)
D
x
(
2
α
)
(
T
)
)
∂
x
α
+
x
2
α
Γ
(
1
+
2
α
)
(
∂
2
α
T
∂
x
2
α
)
2
+
T
=
0
,
hhhhhhhhhhhh
(
t
>
0
,
0
<
x
<
1
)
,
where
(32)
T
=
T
(
x
,
t
)
,
subject to the initial conditions
(33)
T
(
x
,
0
)
=
0
,
∂
α
∂
t
α
T
(
x
,
0
)
=
x
2
α
Γ
(
1
+
2
α
)
.From (33) we take the initial value, which reads
(34)
T
0
(
x
,
t
)
=
x
2
α
Γ
(
1
+
2
α
)
t
α
Γ
(
1
+
α
)
.
By using (31), we structure a local fractional iteration procedure as follows:
(35)
T
n
+
1
=
T
n
+
I
t
(
α
)
0
×
{
+
x
α
Γ
(
1
+
α
)
(
∂
2
α
T
n
∂
x
2
α
)
2
+
T
n
∂
α
(
D
x
(
α
)
(
T
n
)
D
x
(
2
α
)
(
T
n
)
)
∂
x
α
)
(
τ
-
t
)
α
Γ
(
1
+
α
)
(
∂
2
α
T
n
∂
t
2
α
-
x
2
α
Γ
(
1
+
2
α
)
h
h
h
h
h
h
h
h
h
h
h
h
h
h
×
∂
α
(
D
x
(
α
)
(
T
n
)
D
x
(
2
α
)
(
T
n
)
)
∂
x
α
h
h
h
h
h
h
h
h
h
h
h
h
h
h
+
x
α
Γ
(
1
+
α
)
(
∂
2
α
T
n
∂
x
2
α
)
2
+
T
n
∂
α
(
D
x
(
α
)
(
T
n
)
D
x
(
2
α
)
(
T
n
)
)
∂
x
α
)
}
,
where
(36)
T
n
=
T
n
(
x
,
t
)
.Let(37)
w
n
(
x
,
τ
)
=
-
x
2
α
Γ
(
1
+
2
α
)
∂
α
(
D
x
(
α
)
(
T
n
)
D
x
(
2
α
)
(
T
n
)
)
∂
x
α
+
x
α
Γ
(
1
+
α
)
(
∂
2
α
T
n
∂
x
2
α
)
2
+
T
n
.According to (14), we can derive
(38)
T
1
(
x
,
t
)
=
∑
k
=
0
1
T
0
(
k
α
)
(
x
,
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
-
∑
k
=
2
∞
D
T
k
-
1
[
1
k
α
·
(
k
-
1
)
α
W
0
(
x
,
k
-
2
)
]
=
T
0
(
x
,
t
)
-
∑
k
=
2
∞
D
T
k
-
1
{
1
k
α
·
(
k
-
1
)
α
W
0
(
x
,
k
-
2
)
}
=
T
0
(
x
,
t
)
-
D
T
2
-
1
(
W
0
(
x
,
1
)
Γ
(
1
+
3
α
)
)
=
x
2
α
Γ
(
1
+
2
α
)
(
t
α
Γ
(
1
+
α
)
-
t
3
α
Γ
(
1
+
3
α
)
)
;
T
2
(
x
,
t
)
=
∑
k
=
0
1
T
0
(
k
α
)
(
x
,
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
-
∑
k
=
2
∞
D
T
k
-
1
[
1
k
α
·
(
k
-
1
)
α
W
1
(
x
,
k
-
2
)
]
=
T
1
(
x
,
t
)
-
D
T
4
-
1
{
1
5
α
·
4
α
W
1
(
x
,
3
)
}
=
x
2
α
Γ
(
1
+
2
α
)
(
t
α
Γ
(
1
+
α
)

-
t
3
α
Γ
(
1
+
3
α
)

+
t
5
α
Γ
(
1
+
5
α
)
)
;
⋮
Proceeding in this manner, we get
(39)
T
n
(
x
,
t
)
=
x
2
α
Γ
(
1
+
2
α
)
(
t
α
Γ
(
1
+
α
)
-
t
3
α
Γ
(
1
+
3
α
)
+
t
5
α
Γ
(
1
+
5
α
)
hhhhhhhhhhhi
-
⋯
+
(
-
1
)
n
+
1
t
(
2
n
-
1
)
α
Γ
(
1
+
(
2
n
-
1
)
α
)
)
.Thus, the exact solution of (31) is given in the form:
(40)
T
(
x
,
t
)
=
lim
⁡
n
→
∞
T
n
(
x
,
t
)
=
x
2
α
Γ
(
1
+
2
α
)
×
(
t
α
Γ
(
1
+
α
)
-
t
3
α
Γ
(
1
+
3
α
)
+
t
5
α
Γ
(
1
+
5
α
)
hhhhi
-
⋯
+
(
-
1
)
n
+
1
t
(
2
n
-
1
)
α
Γ
(
1
+
(
2
n
-
1
)
α
)
+
⋯
)
.
=
x
2
α
Γ
(
1
+
2
α
)
si
n
α
(
t
α
)
.Case 2.
Consider the following local fractional equation:(41)
∂
2
α
T
(
x
,
t
)
∂
t
2
α
+
2
α
∂
2
α
T
(
x
,
t
)
∂
x
2
α
+
9
α
T
(
x
,
t
)
=
0
subject to the initial conditions
(42)
T
(
x
,
0
)
=
si
n
α
(
2
x
α
)
,
∂
α
∂
t
α
T
(
x
,
0
)
=
0
.According to (41), we structure the following iteration formula:
(43)
T
n
+
1
(
x
,
t
)
=
T
n
(
x
,
t
)
+
I
t
(
α
)
0
×
{
(
τ
-
t
)
α
Γ
(
1
+
α
)
(
∂
2
α
T
(
x
,
τ
)
∂
t
2
α
+
2
α
∂
2
α
T
(
x
,
τ
)
∂
x
2
α
h
h
h
h
h
h
h
h
h
h
h
h
h
h
+
9
α
T
(
x
,
τ
)
∂
2
α
T
(
x
,
τ
)
∂
t
2
α
)
}
.
By using (25) and (42), we take an initial value as
(44)
T
0
(
x
,
t
)
=
si
n
α
(
2
x
α
)
.
According to (23) and (43), we can get
(45)
U
n
+
1
(
k
)
=
-
U
n
(
k
-
2
)
k
α
·
(
k
-
1
)
α
,
(
k
≥
2
)
and moreover
(46)
T
n
+
1
(
x
,
t
)
=
si
n
α
(
2
x
α
)
∑
k
=
0
1
u
0
(
k
α
)
(
x
,
0
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
+
si
n
α
(
2
x
α
)
∑
k
=
2
∞
D
T
k
-
1
(
-
U
n
(
k
-
2
)
(
k
-
1
)
α
·
k
α
)
=
T
0
(
x
,
t
)
+
si
n
α
(
2
x
α
)
∑
k
=
2
∞
D
T
k
-
1
(
-
U
n
(
k
-
2
)
(
k
-
1
)
α
·
k
α
)
.
Thus we have
(47)
T
1
(
x
,
t
)
=
T
0
(
x
,
t
)
+
si
n
α
(
2
x
α
)
∑
k
=
2
∞
D
T
k
-
1
(
-
U
0
(
k
-
2
)
(
k
-
1
)
α
·
k
α
)
=
si
n
α
(
2
x
α
)
-
si
n
α
(
2
x
α
)
D
T
2
-
1
(
U
0
(
0
)
α
·
2
α
)
=
si
n
α
(
2
x
α
)
(
1
α
-
t
2
α
Γ
(
1
+
2
α
)
)
;
T
2
(
x
,
t
)
=
T
0
(
x
,
t
)
+
si
n
α
(
2
x
α
)
∑
k
=
2
∞
D
T
k
-
1
(
-
U
1
(
k
-
2
)
(
k
-
1
)
α
·
k
α
)
=
si
n
α
(
2
x
α
)
(
1
α
-
t
2
α
Γ
(
1
+
2
α
)
)
-
D
T
4
-
1
(
E
α
(
x
α
)
U
1
(
2
)
4
α
·
3
α
)
=
si
n
α
(
2
x
α
)
(
1
α
-
t
2
α
Γ
(
1
+
2
α
)
+
t
4
α
Γ
(
1
+
4
α
)
)
;
T
3
(
x
,
t
)
=
si
n
α
(
2
x
α
)
×
(
1
α
-
t
2
α
Γ
(
1
+
2
α
)
+
t
4
α
Γ
(
1
+
4
α
)
-
t
6
α
Γ
(
1
+
6
α
)
)
;
⋮
Continuing in this manner, we obtain
(48)
T
n
(
x
,
t
)
=
si
n
α
(
2
x
α
)
(
∑
k
=
0
∞
(
-
1
)
k
t
2
k
α
Γ
(
1
+
2
k
α
)
)
.
Thus, the expression of the exact solution of (41) is given by
(49)
T
(
x
,
t
)
=
lim
⁡
n
→
∞
T
n
(
x
,
t
)
=
si
n
α
(
2
x
α
)
(
∑
k
=
0
∞
(
-
1
)
k
t
2
k
α
Γ
(
1
+
2
k
α
)
)
=
si
n
α
(
2
x
α
)
cos
⁡
α
(
-
t
α
)
.According to the ideas of local fractional variational iteration transform method, we can also deal with fractional differential equation system with a similar method.Case 3.
Consider the system of local fractional coupled Helmholtz equations(50)
∂
2
α
u
1
(
x
,
t
)
∂
t
2
α
+
∂
2
α
u
2
(
x
,
t
)
∂
x
2
α
-
u
1
(
x
,
t
)
=
0
,
∂
2
α
u
2
(
x
,
t
)
∂
t
2
α
+
∂
2
α
u
1
(
x
,
t
)
∂
x
2
α
-
u
2
(
x
,
t
)
=
0
with the initial conditions(51)
∂
α
∂
t
α
u
1
(
x
,
0
)
=
E
α
(
x
α
)
,
u
1
(
x
,
0
)
=
0
.Now we can structure the similar local fractional iteration procedure.Using (51), we take an initial value as
(52)
u
1,0
(
x
,
t
)
=
t
α
E
α
(
x
α
)
Γ
(
1
+
α
)
.According to variational iteration method [38], we can similarly construct local fractional variational iteration formula
(53)
u
1
,
n
+
1
(
x
,
t
)
=
u
1
,
n
(
x
,
t
)
+
I
t
(
α
)
0
{
(
τ
-
t
)
α
Γ
(
1
+
α
)
(
∂
2
α
u
1
,
n
(
x
,
τ
)
∂
τ
2
α
hhhhhhhhhhhhhhhhi
+
∂
2
α
u
2
,
n
(
x
,
τ
)
∂
x
2
α
-
u
1
,
n
)
}
,
u
2
,
n
+
1
(
x
,
t
)
=
u
2
,
n
(
x
,
t
)
+
I
t
(
α
)
0
{
(
τ
-
t
)
α
Γ
(
1
+
α
)
(
∂
2
α
u
2
,
n
(
x
,
τ
)
∂
τ
2
α
hhhhhhhhhhhhhhhhi
+

∂
2
α
u
1
,
n
(
x
,
τ
)
∂
x
2
α


-
u
2
,
n
)
}
.Applying local fractional differential transform and integral transformation on (53), we can easily get
(54)
D
T
k
{
u
1
,
n
+
1
(
x
,
t
)
}
=
-
D
T
k
{
I
t
(
α
)
0
I
t
(
α
)
0
{
∂
2
α
u
2
,
n
(
x
,
τ
)
∂
x
2
α
-
u
1
,
n
(
x
,
τ
)
}
}
,
D
T
k
{
u
2
,
n
+
1
(
x
,
t
)
}
=
-
D
T
k
{
I
t
(
α
)
0
I
t
(
α
)
0
{


∂
2
α
u
1
,
n
(
x
,
τ
)
∂
x
2
α
-
u
2
,
n
(
x
,
τ
)
}
}
,
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
(
k
≥
2
)
.
Analyzing (52) and (54), we can deduce
(55)
D
T
k
{
u
1
,
n
+
1
(
x
,
t
)
}
=
-
D
T
k
{
u
2
,
n
+
1
(
x
,
t
)
}
,
D
T
k
{
u
2
,
n
+
1
(
x
,
t
)
}
=
2
D
T
k
{
I
t
(
α
)
0
I
t
(
α
)
0
(
u
2
,
n
(
x
,
t
)
)
}
.
According to (54) and (55), we have
(56)
U
n
+
1
(
x
,
k
)
=
2
U
n
(
x
,
k
-
2
)
k
α
·
(
k
-
1
)
α
,
(
k
≥
2
)
.
From (52), we can deduce
(57)
U
n
+
1
(
x
,
k
)
=
2
U
n
(
x
,
k
-
2
)
k
α
·
(
k
-
1
)
α
(
k
≥
2
,
k
=
2
k
′
-
1
,
k
′
∈
N
)
,
U
n
+
1
(
x
,
k
)
=
0
,
(
k
≥
2
,
k
=
2
k
′
,
k
′
∈
N
)
.
Then we get
(58)
u
1
,
n
+
1
(
x
,
t
)
=
∑
k
=
2
∞
D
T
k
-
1
(
2
U
1
,
n
(
x
,
k
-
2
)
k
α
(
k
-
1
)
α
)
+
∑
k
=
0
1
u
1,0
(
k
α
)
(
x
,
k
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
.
By virtue of (58), we have
(59)
u
1,1
(
x
,
t
)
=
∑
k
=
0
1
u
1,0
(
k
α
)
(
x
,
k
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
+
∑
k
=
2
∞
D
T
k
-
1
(
2
U
1,0
(
x
,
k
-
2
)
k
α
·
(
k
-
1
)
α
)
=
t
α
E
α
(
x
α
)
Γ
(
1
+
α
)
+
D
T
3
-
1
(
2
U
1,0
(
x
,
1
)
3
α
·
2
α
)
=
E
α
(
x
α
)
(
t
α
Γ
(
1
+
α
)
+
2
t
3
α
Γ
(
1
+
3
α
)
)
;
u
1,2
(
x
,
t
)
=
∑
k
=
0
1
u
1,0
(
k
α
)
(
x
,
k
)
Γ
(
1
+
k
α
)
t
k
α
Γ
(
1
+
k
α
)
+
∑
k
=
2
∞
D
T
k
-
1
(
2
U
1,1
(
x
,
k
-
2
)
k
α
(
k
-
1
)
α
)
=
E
α
(
x
α
)
(
t
α
Γ
(
1
+
α
)
+
2
t
3
α
Γ
(
1
+
3
α
)
)
+
D
T
5
-
1
(
2
U
1,1
(
x
,
3
)
4
α
·
5
α
)
=
E
α
(
x
α
)
(
t
α
Γ
(
1
+
α
)
+
2
t
3
α
Γ
(
1
+
3
α
)
+
4
t
5
α
Γ
(
1
+
5
α
)
)
;
u
1,3
(
x
,
t
)
=
E
α
(
x
α
)
(
t
α
Γ
(
1
+
α
)
+
2
t
3
α
Γ
(
1
+
3
α
)
h
h
h
h
h
h
h
h
h
h
i
+
4
t
5
α
Γ
(
1
+
5
α
)
+
8
t
7
α
Γ
(
1
+
7
α
)
)
;
⋮
Then we derive the following result:
(60)
u
1
,
n
(
x
,
t
)
=
E
α
(
x
α
)
(
∑
k
=
0
n
2
k
t
(
2
k
+
1
)
α
Γ
(
1
+
(
2
k
+
1
)
α
)
)
.
Obviously
(61)
u
2
,
n
(
x
,
t
)
=
-
u
1
,
n
(
x
,
t
)
=
-
E
α
(
x
α
)
(
∑
k
=
0
n
2
k
t
(
2
k
+
1
)
α
Γ
(
1
+
(
2
k
+
1
)
α
)
)
.
Thus, the expression of the final solution of (50) reads as follows:
(62)
u
1
(
x
,
t
)
=
lim
⁡
n
→
∞
u
1
,
n
(
x
,
t
)
=
E
α
(
x
α
)
sinh
⁡
α
(
2
t
)
α
2
,
u
2
(
x
,
t
)
=
lim
⁡
n
→
∞
u
2
,
n
(
x
,
t
)
=
-
E
α
(
x
α
)
sinh
⁡
α
(
2
t
)
α
2
.Case 4.
Consider the system of local fractional coupled Burger’s equations(63)
∂
α
u
1
(
x
,
t
)
∂
t
α
+
∂
2
α
u
1
(
x
,
t
)
∂
x
2
α
-
2
∂
α
u
1
(
x
,
t
)
∂
x
α
u
1
(
x
,
t
)
+
∂
α
[
u
1
(
x
,
t
)
u
2
(
x
,
t
)
]
∂
x
α
=
0
,
∂
α
u
2
(
x
,
t
)
∂
t
α
+
∂
2
α
u
2
(
x
,
t
)
∂
x
2
α
-
2
∂
α
u
2
(
x
,
t
)
∂
x
α
u
2
(
x
,
t
)
+
∂
α
[
u
1
(
x
,
t
)
u
2
(
x
,
t
)
]
∂
x
α
=
0
with the initial conditions
(64)
u
1
(
x
,
0
)
=
u
2
(
x
,
0
)
=
E
α
(
x
α
)
,
u
1
(
x
,
0
)
=
u
1
(
x
,
0
)
=
E
α
(
x
α
)
.Now we can structure the similar local fractional iteration procedure.Using (64), we take an initial value as
(65)
u
1,0
(
x
,
t
)
=
u
2,0
(
x
,
t
)
=
E
α
(
t
α
)
.According to variational iteration method [38], we can similarly construct local fractional variational iteration formula
(66)
u
1
,
n
+
1
(
x
,
t
)
=
u
1
,
n
(
x
,
t
)
+
I
x
(
α
)
0
{
(
τ
-
x
)
α
Γ
(
1
+
α
)
(
∂
α
u
1
,
n
∂
t
α
+
∂
2
α
u
1
,
n
∂
x
2
α
hhhhhhhhhhhhhhhhhh
-
2
∂
α
u
1
,
n
∂
x
α
u
1
,
n
+
∂
α
(
u
1
,
n
u
2
,
n
)
∂
x
α
)
}
,
u
2
,
n
+
1
(
x
,
t
)
=
u
2
,
n
(
x
,
t
)
+
I
x
(
α
)
0
{
(
τ
-
x
)
α
Γ
(
1
+
α
)
(
∂
α
u
2
,
n
∂
t
α
+
∂
2
α
u
2
,
n
∂
x
2
α
hhhhhhhhhhhhhhhhhhi
-
2
∂
α
u
2
,
n
∂
x
α
u
2
,
n
+
∂
α
(
u
1
,
n
u
2
,
n
)
∂
x
α
)
}
.Applying local fractional differential transform and integral transformation on (66), we can easily get
(67)
D
T
k
{
u
1
,
n
+
1
(
x
,
t
)
}
=
-
D
T
k
{
{
∂
α
u
1
,
n
∂
t
α
-
2
∂
α
u
1
,
n
∂
x
α
u
1
,
n
+
∂
α
(
u
1
,
n
u
2
,
n
)
∂
x
α
}
I
x
(
α
)
0
I
x
(
α
)
0
hhhhhhhhh
×
{
∂
α
u
1
,
n
∂
t
α
-
2
∂
α
u
1
,
n
∂
x
α
u
1
,
n
+
∂
α
(
u
1
,
n
u
2
,
n
)
∂
x
α
}
}
,
D
T
k
{
u
2
,
n
+
1
(
x
,
t
)
}
=
-
D
T
k
{
{
∂
α
u
2
,
n
∂
t
α
-
2
∂
α
u
2
,
n
∂
x
α
u
2
,
n
+
∂
α
(
u
1
,
n
u
2
,
n
)
∂
x
α
}
I
x
(
α
)
0
I
x
(
α
)
0
hhhhhhhhh
×
{
∂
α
u
2
,
n
∂
t
α
-
2
∂
α
u
2
,
n
∂
x
α
u
2
,
n
+
∂
α
(
u
1
,
n
u
2
,
n
)
∂
x
α
}
}
,
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
(
k
≥
2
)
.
Analyzing (64) and (67), we can deduce
(68)
D
T
k
{
u
1
,
n
+
1
(
x
,
t
)
}
=
D
T
k
{
u
2
,
n
+
1
(
x
,
t
)
}
,
D
T
k
{
u
2
,
n
+
1
(
x
,
t
)
}
=
-
D
T
k
{
I
x
(
α
)
0
I
x
(
α
)
0
(
u
2
,
n
(
x
,
t
)
)
}
.
According to (67) and (68), we have
(69)
U
1
,
n
+
1
(
k
,
t
)
=
-
U
1
,
n
(
k
-
2
,
t
)
k
α
·
(
k
-
1
)
α
,
(
k
≥
2
)
.
From (64), we can deduce
(70)
U
1
,
n
+
1
(
k
,
t
)
=
-
U
1
,
n
(
k
-
2
,
t
)
k
α
·
(
k
-
1
)
α
(
k
≥
2
,
k
=
2
k
′
,
k
′
∈
N
)
,
U
n
+
1
(
k
,
t
)
=
0
,
(
k
≥
2
,
k
=
2
k
′
-
1
,
k
′
∈
N
)
.
Then we get
(71)
u
1
,
n
+
1
(
x
,
t
)
=
∑
k
=
2
∞
D
T
k
-
1
(
-
U
1
,
n
(
k
-
2
,
t
)
k
α
(
k
-
1
)
α
)
+
∑
k
=
0
1
u
1,0
(
k
α
)
(
k
,
t
)
Γ
(
1
+
k
α
)
x
k
α
Γ
(
1
+
k
α
)
.
By virtue of (71), we have
(72)
u
1,1
(
x
,
t
)
=
∑
k
=
0
1
u
1,0
(
k
α
)
(
k
,
t
)
Γ
(
1
+
k
α
)
x
k
α
Γ
(
1
+
k
α
)
+
∑
k
=
2
∞
D
T
k
-
1
(
-
U
1,0
(
k
-
2
,
t
)
k
α
·
(
k
-
1
)
α
)
=
E
α
(
t
α
)
+
D
T
2
-
1
(
-
U
1,0
(
0
,
t
)
2
α
·
α
)
=
E
α
(
t
α
)
(
1
α
-
x
2
α
Γ
(
1
+
2
α
)
)
;
u
1,2
(
x
,
t
)
=
∑
k
=
0
1
u
1,0
(
k
α
)
(
k
,
t
)
Γ
(
1
+
k
α
)
x
k
α
Γ
(
1
+
k
α
)
+
∑
k
=
2
∞
D
T
k
-
1
(
-
U
1,1
(
k
-
2
,
t
)
k
α
(
k
-
1
)
α
)
=
E
α
(
t
α
)
(
1
α
-
x
2
α
Γ
(
1
+
2
α
)
)
+
D
T
4
-
1
(
-
U
1,1
(
2
,
t
)
4
α
·
3
α
)
=
E
α
(
t
α
)
(
1
α
-
x
2
α
Γ
(
1
+
2
α
)
+
x
4
α
Γ
(
1
+
4
α
)
)
;
u
1,3
(
x
,
t
)
=
E
α
(
t
α
)
(
1
α
-
x
2
α
Γ
(
1
+
2
α
)
+
x
4
α
Γ
(
1
+
4
α
)
-
x
6
α
Γ
(
1
+
6
α
)
)
;
⋮
Then we derive the following result:
(73)
u
1
,
n
(
x
,
t
)
=
E
α
(
t
α
)
(
∑
k
=
0
n
(
-
1
)
k
x
2
k
α
Γ
(
1
+
2
k
α
)
)
.
Obviously
(74)
u
2
,
n
(
x
,
t
)
=
u
1
,
n
(
x
,
t
)
=
E
α
(
t
α
)
(
∑
k
=
0
n
(
-
1
)
k
x
2
k
α
Γ
(
1
+
2
k
α
)
)
.
Thus, the expression of the final solution of (63) reads as follows:
(75)
u
1
(
x
,
t
)
=
lim
⁡
n
→
∞
u
1
,
n
(
x
,
t
)
=
E
α
(
t
α
)
cos
⁡
α
(
x
α
)
,
u
2
(
x
,
t
)
=
lim
⁡
n
→
∞
u
2
,
n
(
x
,
t
)
=
E
α
(
t
α
)
cos
⁡
α
(
x
α
)
.


## 5. Conclusions

In this work, we proposed the local fractional iteration transform method. The applications of the methods for solving fractional differential equations with local fractional derivative are discussed in several cases. The method provides the variational iteration formula in the form of polynomial or Maclaurin’s series, which is more easily to deal with convergence and approximate problem. Of course, the new problems are beyond the scope of the present work.



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*Source: 760957-2014-06-18.xml*