Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2011, Article ID 587068, 14 pages
doi:10.1155/2011/587068
Research Article
Solution to the Linear Fractional Differential
Equation Using Adomian Decomposition Method
Jin-Fa Cheng1 and Yu-Ming Chu2
1
2
Department of Mathematics, Xiamen University, Xiamen 361005, China
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 25 March 2010; Accepted 1 June 2010
Academic Editor: Geraldo Silva
Copyright q 2011 J.-F. Cheng and Y.-M. Chu. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
We obtain the analytical general solution of the linear fractional differential equations with
constant coefficients by Adomian decomposition method under nonhomogeneous initial value
condition, which is in the sense of the Caputo fractional derivative.
1. Introduction
Fractional differential equations are hot topics both in mathematics and physics. Recently, the
fractional differential equations have been the subject of intensive research. There are several
methods to obtain the solution, such as the Laplace transform method, power series method,
and Green function method. Many remarkable results for the fractional differential equations
can be found in the literature 1–11. In particular, the Adomian decomposition method has
attracted the attention of many mathematicians 12–15.
For a better understanding of the fractional derivatives and for a physical understanding of the fractional equations, the readers can refer to the recent publications in 16, 17.
Ebaid 18 suggested a modification of the Adomian method, and a few iterations lead
to exact solution. Das 19 compared the variational iteration method with the Adomian
method for fractional equations and found that the variational iteration method is much more
effective. For other methods of the fractional differential equations, especially the homotopy
perturbation method, variational iteration method and differential transform method were
presented in 20, 21.
2
Mathematical Problems in Engineering
Consider the following n-term fractional differential equation with constant coeffi-
cients:
an
C
D
βn
yt an−1
C
C
β1
Dβn−1 yt · · · a1 C D yt a0 Dβ0 yt ft,
1.1
where n 1 > β ≥ n > βn−1 > · · · > β1 > β0 and ai i 0, 1, . . . , n is a real constant. In 12,
the authors obtain the particular solution of 1.1 of the homogeneous initial value problem
of the form
yi 0 0,
1.2
i 0, 1, . . . , n.
However, it seems also more meaningful and more complicated for solving general
solution of 1.1 under nonhomogeneous initial value condition. Therefore, in this paper, we
will remove the restriction of the homogeneous initial value, consider the nonhomogeneous
initial value problems of the form
yji 0 ciji ,
i 0, 1, . . . , n, ji 1, 2, . . . , li , li − 1 βi < li ,
1.3
and obtain the analytical general solution of 1.1, which generalizes the result in 12.
We organize the paper as follows. In Section 2, we give some basic definitions and
properties. In Section 3, we obtain the analytical general solution of the linear fractional
differential equations by Adomian decomposition method. Some explicit examples are given
in Section 4.
2. Basic Definitions and Notations
Definition 2.1 see 1. The Riemann-Liouville integral of order p is defined by
1
Γ p
−p
a D t ft
t
t − τp−1 fτdτ,
p > 0.
2.1
a
From Definition 2.1 , we clearly see that
t − aν Γ1 ν
t − aνα ,
Γ1 ν α
2.2
t − aν Γ1 ν
t − aν−α ,
Γ1 ν − α
2.3
−α
a Dt
α
a Dt
where α > 0 and ν is a real number.
Mathematical Problems in Engineering
3
Definition 2.2 see 1. For ft ∈ Cm m ∈ N, the Caputo fractional derivative of ft is
defined by
⎧
⎪
Dm−p f m t,
⎪
⎪
⎪
⎪
⎨ m
d
C p
ft
ft,
D
a t
⎪
dtm
⎪
⎪
⎪
⎪
⎩ Dp ft,
a
t
m − 1 < p < m,
2.4
p m,
p 0.
Therefore,
c −α
t
a Dt
− aν a D−α
t − aν t
Γ1 ν
t − aνα .
Γ1 ν α
2.5
Lemma 2.3 see 1. If ft is continuous, then
C −n n
n
t a D−n
t
a Dt f
t f
ft −
n−1 j
f at − aj
,
Γ j 1
j0
n ∈ Z.
2.6
Lemma 2.4 see 1. If ft is continuous, then
C −p C −q
a Dt
a Dt
−p
ft a Dt
−q
a Dt
−p−q
ft a Dt
−p−q
ft C
a Dt
−q C −p
a Dt
ft C
a Dt
ft.
2.7
Lemma 2.5. If ft is continuous, then
C −p C q
a Dt
a Dt
q−p
ft a Dt ft −
m−1
t − aj−qp
f j a ,
Γ j −qp1
j0
2.8
where m − 1 < q < m and q p.
Proof. From Definition 2.2 and Lemmas 2.3–2.4, we get
C −p C q
a Dt
a Dt
−p
ft a Dt
q−m
a Dt
−pq
f m t a Dt
−pq
−m
m
t
a Dt
a Dt tf
q−p
a Dt ft −
m−1
j0
−q
a Dt
q−m
a Dt
⎡
−pq
⎣ft
a Dt
f j at − aj−qp
.
Γ j −qp1
−
m−1
j0
f m t
⎤
f j at − aj
⎦
Γ j 1
2.9
4
Mathematical Problems in Engineering
It is easy to see that
C −p C p
a Dt
a Dt
ft ft −
m−1
t − aj
f j a .
Γ j 1
j0
2.10
Proposition 2.6 see 12. One has
∞
···
k1 0
∞
ak1 ,k2 ···kn−1 kn kn 0
∞
ak1 ,k2 ···kn−1 kn .
2.11
m0 k1 ,...,kn−1 ,kn 0
k1 ···kn−1 kn m
Proposition 2.7 see 12. More over, one has
∞
ak1 ,k2 ···kn−1 kn m0 k1 ,...,kn−1 ,kn 0
k1 ···kn−1 kn m
∞
∞
ak1 ,k2 ···kn−1 kn .
2.12
s0 k1 ,...,kn−1 0 kn 0
k1 ···kn−1 s
3. The Analytical Solution of the Linear Constant Coefficient
Fractional Differential Equation
p
−p
C
C p
C −p
For simplicity, if a 0, then we denote C
a Dt or a Dt by D or D .
In this section, we use Adomian decomposition method to discuss the general form
of the linear fractional differential equations with constant coefficients, and apply and some
basic transformation and integration to obtain the solution of the equations.
Let us consider the following n-term linear fractional differential equations with
constant coefficients:
an
C
C
C
C
Dβn yt an−1 Dβn−1 yt · · · a1 Dβ1 yt a0 Dβ0 yt ft,
yji 0 ciji ,
3.1
i 0, 1, . . . , n, ji 1, 2, . . . , li , li − 1 βi < li ,
p
where n 1 > βn n > βn−1 · · · > β1 > β0 , ai and ciji are real constants, C Dp C
0 D t denotes
Caputo fractional derivative of order α.
Applying C D−βn to both sides of 1.1 and utilizing Lemma 2.5, we get
yt an−1 C βn−1 −βn
a0 C β0 −βn
yt · · · yt
D
D
an
an
li −1
n
1 C −βn
tβn −βi ji
1
ai yji 0 .
D ft an
an i0 ji 0
Γ 1 βn − βi ji
3.2
Mathematical Problems in Engineering
5
By the Adomian decomposition method, we obtain the recursive relationship as
follows:
y0 t li −1
n
tβn −βi ji
1
1 C −βn
ai yji 0 ,
D ft an
an i0 ji 0
Γ 1 βn − βi ji
a0 C β0 −βn
an−1 C βn−1 −βn
··· y0 t,
D
D
an
an
a0 C β0 −βn 2
2 an−1 C βn−1 −βn
··· y0 t,
y2 t −1
D
D
an
an
y1 t −
3.3
..
.
s
ys t −1
a0 C β0 −βn
an−1 C βn−1 −βn
··· D
D
an
an
s
y0 t,
..
.
By Adomian decomposition method, adding all terms of the recursion, we obtain the
solution of 1.1 as
yt ∞
ys t
s0
∞
a0 C β0 −βn s
an−1 C βn−1 −βn
−1s
··· y0 t
D
D
an
an
s0
s
∞
1
a0
an−1 βn−1 −βn
−1s
D
· · · Dβ0 −βn D−βn ft
an s0
an
an
3.4
s li −1
n
∞
tβn −βi ji
1
a0
an−1 βn−1 −βn
D
· · · Dβ0 −βn
ai yji 0 .
−1s
an s0
an
an
Γ 1 βn − βi ji
i0 ji 0
Let
I1 s
∞
1
a0
an−1 βn−1 −βn
D
· · · Dβ0 −βn D−βn ft
−1s
an s0
an
an
n li −1
∞
tβn −βi ji
1
a0 β0 −βn s s an−1 βn−1 −βn
−1
D
··· D
ai yji 0 I2 .
an s0
an
an
Γ 1 βn − βi ji
i0 ji 0
3.5
Then,
yt I1 I2 .
Next, we estimate I1 and I2 , respectively.
3.6
6
Mathematical Problems in Engineering
For I1 , by 12 we obtain
s
∞
1
a0
an−1 βn−1 −βn
D
· · · Dβ0 −βn D−βn ft,
−1s
an s0
an
an
t
0
∞
1 −1m
an m0 m!
×
n−2 ap
p0 a
n
m; k0 , k1 , . . . , kn−2 k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−2 m
kp
3.7
t − τβn −βn−1 mβn m
βn −βn−1 ,βn n−2
j0 βn−1 −βj kj
×E
−
n−2
j0 βn−1 −βj kj −1
an−1 βn −βn−1
D
fτdτ.
an
For I2 , by the initial conditions 2.10 we get
s li −1
n
∞
1
tβn −βi ji
an−1 βn−1 −βn
a0
−1s
D
· · · Dβ0 −βn
ai yji 0 an s0
an
an
Γ 1 βn − βi ji
i0 ji 0
∞
1
−1s
an s0
k0 ,k1 ,...,kn−1 0
k0 k1 ···kn−1 s
k0
an−1 kn−1 an−2 kn−2
s!
a0
···
k0 !k1 ! · · · kn−1 ! an
an
an
× Dkn−1 βn−1 −βn kn−2 βn−2 −βn ···k0 β0 −βn li −1
n
tβn −βi ji
ai ciji .
Γ 1 βn − βi ji
i0 ji 0
Using formulas 2.2 and 2.3, the above expression can be written as
∞
1
−1s
an s0
×
k0 ,k1 ,...,kn−1 0
k0 k1 ···kn−1 s
li −1
n
ai ciji
i0
ji 0
ΓAtβn −βi ji kn−1 βn −βn−1kn−2 βn −βn−2···k0 βn −β0 ΓA · Γ A kn−1 βn − βn−1 kn−2 βn − βn−2 · · · k0 βn − β0
∞
1
−1s
an s0
×
k0
s!
a0
an−1 kn−1 an−2 kn−2
···
k0 !k1 ! · · · kn−1 ! an
an
an
k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−1 s
k0
s!
an−1 kn−1 an−2 kn−2
a0
···
k0 !k1 ! · · · kn−1 ! an
an
an
li −1
n
tβn −βi ji kn−1 βn −βn−1 kn−2 βn −βn−2 ···k0 βn −β0 ai ciji Γ A kn−1 βn − βn−1 kn−2 βn − βn−2 · · · k0 βn − β0
i0 ji 0
3.8
Mathematical Problems in Engineering
⎡
li −1
n
∞
⎢1
s
ai ciji Dβi −ji −1 ⎢
⎣ an −1
s0
i0 ji 0
×
an−1
an
7
k0 ,k1 ,...,kn−1 0
k0 k1 ···kn−1 s
kn−1 s!
k0 !k1 ! · · · kn−1 !
kn−2
an−2
an
···
a0
an
k0
⎤
⎥
⎥
tβn −1kn−1 βn −βn−1 kn−2 βn −βn−2 ···k0 βn −β0 ⎥
× ⎥,
Γ βn kn−1 βn − βn−1 kn−2 βn − βn−2 · · · k0 βn − β0 ⎥
⎦
3.9
where A denotes 1 βn − βi ji .
Using Propositions 2.6 and 2.7, the above solution is equivalent to the following form:
⎡
li −1
∞
n
∞ ⎢1 ai ciji Dβi −ji −1 ⎢
⎣ an
i0 ji 0
k 0 m0
×
−1kn−1 m
k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−2 m
n−1
×
an−1
an
kn−1 an−2
an
kn−2
···
a0
an
kn−1 m!
k0 !k1 ! · · · kn−1 !
k0
⎤
⎥
tβn −1kn−1 βn −βn−1 kn−2 βn −βn−2 ···k0 βn −β0 ⎥
Γ βn kn−1 βn − βn−1 kn−2 βn − βn−2 · · · k0 βn − β0 ⎦
⎡
li −1
n
∞
⎢1 −1m
ai ciji Dβi −ji −1 ⎢
⎣ an
m!
m0
i0 ji 0
×
×
k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−2 m
n−2 ar kr
r0
∞
an
−1kn−1
kn−1 0
tβn −βn−1 m
an−1
an
kn−1
m!
k0 !k1 ! · · · kn−2 !
n−2
ξ0 βn−1 −βξ kξ βn −1
kn−1 m!
kn−1 !
⎤
×
⎥
t
⎥
n−2 ⎦
Γ kn−1 βn − βn−1 βn − βn−1 m δ0 βn−1 − βδ kδ βn
kn−l βn −βn−1 8
Mathematical Problems in Engineering
⎡
n
ai
i0
li −1
∞
⎢1 −1m
ciji Dβi −ji −1 ⎢
⎣ an
m!
m0
ji 0
×
m; k0 , k1 , . . . , kn−2 k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−2 m
n−2 ar kr
r0
an
tβn −βn−1 m
m
βn −βn−1 , n−2
δ0 βn−1 −βl kδ βn
×E
n−2
ξ0 βn−1 −βξ kξ βn −1
−
⎤
an−1 βn −βn−1 ⎥
⎥.
t
⎦
an
3.10
Therefore,
yt I1 I2
t
∞
1 −1m
0 an m0 m!
×
n−2 ap kp
p0
an
t − τβn −βn−1 mβn m
βn −βn−1 ,βn n−2
j0 βn−1 −βj kj
×E
⎡
n
i0
ai
m; k0 , k1 , . . . , kn−2 k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−2 m
−
n−2
j0 βn−1 −βj kj −1
an−1 βn −βn−1
D
fτdτ
an
ji 0
∞
⎢1 −1m
ciji Dβi −ji −1 ⎢
⎣ an
m!
li −1
m0
×
3.11
m; k0 , k1 , . . . , kn−2 k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−2 m
n−2 ar kr
r0
an
tβn −βn−1 m
n−2
ξ0 βn−1 −βξ kξ βn −1
⎤
an−1 βn −βn−1 ⎥
m
⎥,
×E
t
−
n−2
⎦
an
βn −βn−1 , βn−1 −βl kδ βn
δ0
where
m; k0 , k1 , . . . , kn−2 m!
,
k0 !k1 ! · · · kn−2 !
3.12
i
and Eλ,μ y is the Mittag-Leffler function
i Eλ,μ y
∞
i j !yj
di
Eλ,μ y .
dyi
j0 j!Γ λj λi μ
3.13
Mathematical Problems in Engineering
9
Substituting the Green function
Gn t ∞
−1m
1 an m0 m!
×
n−2 a kp
p
an
p0
m; k0 , k1 , . . . , kn−2 k0 ,k1 ,...,kn−2 0
k0 k1 ···kn−2 m
tβn −βn−1 mβn m
βn −βn−1 ,βn n−2
j0 βn−1 −βj kj
n−2
j0 βn−1 −βj kj −1
×E
−
an−1 βn −βn−1
D
an
3.14
into the above expression, we know that
yt t
Gn t − τfτdτ 0
li −1
n
β −j −1
ai ciji Gn i i t
i0
ji 0
3.15
is the analytical general solution of 1.1.
4. Illustrative Examples
In order to verify our conclusions, we give some examples.
1. Consider an initial value problem for the relaxation-oscillation equation see 1
C
α
D yt Ayt ft,
yj 0 bj ,
j 1, 2, . . . , m − 1,
t > 0,
4.1
m − 1 < α < m,
where bj are real constants.
Utilizing Lemma 2.5 and applying C D−α to both sides of 4.1, we obtain
yt AD−α yt D−α ft m−1
tj
yj 0 .
Γ j 1
j0
4.2
10
Mathematical Problems in Engineering
According to the above procedure of solving the linear fractional differential equations
with constant coefficients and using the Adomian decomposition method, let
y0 t D−α ft m−1
tj
yj 0 ,
Γ j 1
j0
y1 t −AD−α y0 t,
y2 t −AD−α y1 t −A2 D−2α y0 t,
4.3
..
.
ys t −AD−α ys−1 t −As D−sα y0 t,
..
.
Adding all of the above terms, we obtain the solution of the equation by Adomian
decomposition method as follows:
yt ∞
ys t
s0
∞
−As D−sα y0 t
s0
∞
−As D−s−1α ft s0
∞
−As
s0
t ∞
1
Γs 1α
t − τ
α−1
0 s0
t
t
∞
m−1
tj
−As D−sα yj 0 Γ j 1
s0
j0
t − τs1α−1 fτdτ 0
t
bj
j0
∞
tjsα
−As Γ j sα 1
s0
4.4
s
m−1
∞
α s
−At − τα α−j−1
α−1 −At fτdτ ·
bj D
t
Γsα α
Γsα α
s0
j0
m−1
bj Dα−j−1 tα−1 Eα,α −Atα t − τα−1 Eα,α −At − τα fτdτ 0
m−1
j0
G2 t − τfτdτ 0
m−1
bj Dα−j−1 G2 t,
j0
where G2 t tα−1 Eα,α −Atα .
It is easy to see that
yt t
0
G2 t − τfτdτ b0 Dα−1 G2 t,
0 < α < 1.
4.5
Mathematical Problems in Engineering
11
2. Consider an initial value problem for the nonhomogeneous Bagley-Torvik
equation see 5
Ay t BC D3/2 yt Cyt ft,
t > 0,
4.6
i
y 0 ai , i 0, 1,
where ai are real constants.
Utilizing Lemma 2.5 and applying C D−2 to both sides of 4.6, we obtain
yt 1
BD−1/2 yt CD−2 yt
A
1
1
ti
t1/2i
1 −2
B
D ft yi 0
yi 0
A
Γi 1 A i0
Γ3/2 i
i0
4.7
1
ti
B
t1/2i
1 −2
i
·
D ft y 0
A
Γi 1 A Γ3/2 i
i0
According to the above procedure of solving the linear fractional differential equation
with constant coefficients and using the Adomian decomposition method, let
1
ti
B
t1/2i
1 −2
i
·
,
y0 t D ft y 0
A
Γi 1 A Γ3/2 i
i0
y1 t −
C B
1
BD−1/2 y0 t CD−2 y0 t −
I D−3/2 D−1/2 y0 t,
A
A C
y2 t −
2
C B
C 2 B
I D−3/2 D−1/2 y1 t −
I D−3/2 D−2/2 y0 t,
A C
A
C
..
.
n
C B
C n B
−3/2
−1/2
−3/2
I D
I D
D
yn−1 t −
D−n/2 y0 t,
yn t −
A C
A
C
..
.
4.8
12
Mathematical Problems in Engineering
Adding all of the above terms, we obtain the solution of the equation by Adomian
decomposition method as follows:
yt ∞
yn t
n0
n
∞ C n B
I D−3/2 D−n/2 y0 t
−
A
C
n0
n
∞ C n B
1 −2
I D−3/2 D−n/2
D ft
−
A
C
A
n0
n
∞ 1
ti
B
t1/2i
C n B
−3/2
−n/2
i
I D
·
D
y 0
−
A
C
Γi 1 A Γ3/2 i
n0
i0
n
n−k
∞ 1
C n
B
I
Cnn−k
D−3k/2 D−n/2 D−2 ft
−
A n0
A k0
C
n
n−k
1
∞ ti
B
t1/2i
C n
n−k B
−3k/2 −n/2
I
·
C
D
D
ai
−
A k0 n
C
Γi 1 A Γ3/2 i
i0 n0
n
n−k
∞ n!
B
1
C n
I
D−3k/2n/22 ft
−
A n0
A k0 n − k!k! C
n
n−k
1
∞ ti
B
t1/2i
n!
B
C n
−3k/2n/2
I
·
ai
D
−
A k0 n − k!k! C
Γi 1 A Γ3/2 i
i0 n0
1
A
t ∞ 0 n0
−
C
A
n−k
n!
1
B
I
t − τ3k/2n/22−1 fτdτ
C
Γ3k/2
n/2
2
−
k!k!
n
k0
n n
n
n−k 1
∞ ti3k/2n/2
n!
C n
B
I
ai
−
A k0 n − k!k! C
Γi 3k/2 n/2 1
i0 n0
1
A
t ∞ ∞ 0 k0 m0
C
−
A
km
t1/2i3k/2n/2
B
·
A Γi 3k/2 n/2 3/2
1
k m! B m
I
m!k!
C
Γ3/2k k m/2 2
× t − τ3k/2km/21 fτdτ
1
ai D
ai D
1/2−i
i0
1
i0
∞ ∞ t3k/2km/21
C km k m! B m
I
−
A
m!k!
C
Γ3/2k k m/2 2
k0 m0
1−i
∞ ∞ t3k/2km/21
C km k m! B m
I
−
A
m!k!
C
Γ3/2k k m/2 2
k0 m0
Mathematical Problems in Engineering
13
t ∞
m
∞
−B/At − τ1/2 −C/Ak
k m!
2k1
t − τ
·
fτdτ
k!
m!
Γk/2 m/2 3k/2 2
0 k0
m0
⎡
m ⎤
1/2
k
∞
1
∞
−B/At
−
τ
−C/A
k m!
⎥
⎢1
A ai D1−i ⎣
·
t − τ2k1
⎦
A
k!
m!
Γk/2
m/2
3k/2
2
m0
i0
k0
1
A
A
1
ai D
1/2−i
i0
∞
1
−C/Ak
t − τ2k1
A k0
k!
⎤
m
1/2
∞
−B/At
−
τ
k m!
⎥
·
×
⎦
m!
Γk/2
2
m/2
3k/2
m0
t ∞
B
−C/Ak
k
t − τ2k1 E1/2,3k/22 − t − τ1/2 fτdτ
k!
A
0 k0
k
1
∞
B 1/2
−C/A 2k1 k
1−i 1
t
E1/2,3k/22 − t
A ai D
A k0
k!
A
i0
1
∞
B 1/2
−C/Ak 2k1 k
1/2−i 1
A ai D
t
E1/2,3k/22 − t
A k0
k!
A
i0
t
1
G3 t − τfτdτ A ai D1−i G3 t D1/2−i G3 t ,
1
A
0
i0
4.9
where G3 t 1/A
∞
k
2k1 k
E1/2,3k/22 −B/At1/2 .
k0 −C/A /k!t
Acknowledgment
The authors wish to thank the anonymous referees for their very careful reading of
the manuscript and fruitful comments and suggestions. This work was supported by
the National Natural Science Foundation of China No. 60850005, the Natural Science
Foundation of Zhejiang Province Nos. D7080080 and Y607128 and the Innovation Team
Foundation of the Department of Education of Zhejiang Province No. T200924.
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