Hindawi Publishing Corporation
Boundary Value Problems
Volume 2011, Article ID 516481, 17 pages
doi:10.1155/2011/516481
Research Article
Existence of Solutions to Nonlinear
Langevin Equation Involving Two Fractional Orders
with Boundary Value Conditions
Anping Chen
1, 2
and Yi Chen
2
1
Department of Mathematics, Xiangnan University, Chenzhou, Hunan 423000, China
2
School of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan 411005, China
Correspondence should be addressed to Anping Chen, [email protected]
Received 30 September 2010; Revised 21 January 2011; Accepted 26 February 2011
Academic Editor: Kanishka Perera
Copyright q 2011 A. Chen and Y. Chen. 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 study a boundary value problem to Langevin equation involving two fractional orders. The
Banach fixed point theorem and Krasnoselskii’s fixed point theorem are applied to establish the
existence results.
1. Introduction
Recently, the subject of fractional differential equations has emerged as an important area of
investigation. Indeed, we can find numerous applications in viscoelasticity, electrochemistry,
control, electromagnetic, porous media, and so forth. In consequence, the subject of
fractional differential equations is gaining much importance and attention. For some recent
developments on the subject, see 1–8 and the references therein.
,
u
0
−u
T
,u
0
u
T
0,
1.1
2 Boundary Value Problems
where T is a positive constant, 1 <α≤ 2, 0 <β≤ 1,
C
D
α
,and
C
D
n−α−1
f
n
s
ds, n − 1 <α<n,n
α
1, 2.1
where α denotes the integer part of the real number α.
Definition 2.2. The Riemann-Liouville fractional integral of order α>0ofafunctionft,
t>0, is defined as
I
α
f
t
1
Γ
α
t
0
t
0
t −s
n−α−1
f
s
ds, 2.3
where n α1andα denotes the integer part of real number α, provided that the right
side is pointwise defined on 0, ∞.
Lemma 2.4 see 8. Let α>0, then the fractional differential equation
C
D
α
ut0 has solution
u
t
c
0
c
1
t c
2
t
2
··· c
n−1
t
n−1
, 2.5
for some c
i
∈ R, i 0, 1, 2, ,n− 1, n α1.
Boundary Value Problems 3
Lemma 2.6. The unique solution of the following boundary value problem
C
D
β
C
D
α
λ
u
t
y
t
,t∈
0,T
0
t − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
y
τ
dτ − λu
s
ds
−
1
ds
T
α
− 2t
α
2αT
α−1
T
0
T − s
α−2
Γ
α − 1
s
0
s − τ
β−1
Γ
2.8
can be written as
u
t
t
0
t −s
α−1
Γ
α
s
0
s −τ
β−1
Γ
β
0
Γ
α 1
1
αT
α−1
T
0
T − s
α−2
Γ
α −1
s
0
s − τ
β−1
Γ
s
0
s − τ
β−1
Γ
β
y
τ
dτ − λu
s
ds
−
T
2α
T
0
T − s
u
t
t
0
t − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
y
τ
β
y
τ
dτ − λu
s
ds
T
α
− 2t
α
2αT
α−1
T
0
T − s
α−2
Γ
α − 1
older inequality. Let p>1, 1/p1/q1, f ∈ L
p
a, b, g ∈ L
q
a, b,then
the following inequality holds:
b
a
f
x
g
x
dx ≤
b
a
f
(H1) there exists a real-valued function μt ∈ L
1/γ
0,T,R
for some γ ∈ 0, 1 such that
f
t, u
− f
t, v
≤ μ
t
|
u −v
|
, for almost all t ∈
0,T
,u,v∈ R. 3.1
If
Λ
α
Γ
α 1
< 1, 3.2
where γ ∈ 0, 1, β
/
γ,1 <α≤ 2, 0 <β≤ 1, μ
∗
T
0
μτ
1/γ
dτ
γ
,thenproblem1.1 has a
unique solution.
Boundary Value Problems 5
Proof. Define an operator F : Ω → Ω by
Fu
t
t
ds
−
1
2
T
0
T − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
f
τ, u
0
s − τ
β−1
Γ
β
f
τ, u
τ
dτ −λu
s
ds.
3.3
Let M sup
t∈0,T
|ft, 0| and choose
1
1 −δ
4α β
t
0
t − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
f
τ, u
τ
β
f
τ, u
τ
dτ − λu
s
ds
T
α
− 2t
α
2αT
α−1
T
0
T − s
α−2
Γ
α − 1
t
0
t − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
f
τ, u
T
0
T − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
f
τ, u
T
0
T − s
α−2
Γ
α −1
s
0
s − τ
β−1
Γ
β
f
τ, u
0
t − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
μ
τ
|
u
τ
|
s
0
s − τ
β−1
Γ
β
μ
τ
|
u
τ
|
f
τ, 0
β−1
Γ
β
μ
τ
|
u
τ
|
f
τ, 0
dτ
|
λu
s
μ
τ
dτ
ds M
t
0
t − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
dτ ds
α
s
0
s − τ
β−1
Γ
β
μ
τ
dτ
ds
M
2
T
0
T − s
α
ds
T
u
2α
T
0
T − s
α−2
Γ
α − 1
s
0
s − τ
β−1
Γ
Γ
β
dτ ds
T
|
λ
|
u
2α
T
0
T − s
α−2
Γ
α − 1
ds
≤
u
Γ
s
0
μ
τ
1/γ
dτ
γ
ds
M
Γ
α
Γ
β 1
t
0
t −s
0
T − s
α−1
s
0
s − τ
β−1
1/1−γ
dτ
1−γ
s
0
μ
τ
α
u
2Γ
α 1
T
u
2αΓ
α −1
Γ
β
T
0
T − s
α−2
2αΓ
α −1
Γ
β 1
T
0
T − s
α−2
s
β
ds
|
λ
|
T
α
u
2Γ
α 1
Γ
α
Γ
β 1
t
0
t − s
α−1
s
β
ds
μ
∗
u
2Γ
α
Γ
T − s
α−1
s
β
ds
Tμ
∗
u
2αΓ
α −1
Γ
β
1 −γ
β − γ
1−γ
T
0
T − s
α
u
Γ
α 1
Boundary Value Problems 7
μ
∗
u
t
αβ−γ
Γ
α
Γ
β
1 −γ
β − γ
1−γ
μ
∗
u
T
αβ−γ
2Γ
α
Γ
β
1 −γ
β −γ
1−γ
1
0
1 −η
α−1
η
β−γ
α −1
Γ
β
1 −γ
β − γ
1−γ
1
0
1 −η
α−2
η
β−γ
dη
MT
αβ
2αΓ
α −1
Γ
αβ−γ
Γ
α
Γ
β
1 −γ
β − γ
1−γ
1
0
1 −ξ
α−1
ξ
β−γ
dξ
MT
αβ
Γ
α
1−γ
1
0
1 −η
α−1
η
β−γ
dη
MT
αβ
2Γ
α
Γ
β 1
1
0
1 −η
α−1
η
MT
αβ
2αΓ
α −1
Γ
β 1
1
0
1 −η
α−2
η
β
dη
2
|
λ
|
T
α
r
Γ
Γ
α
Γ
β − γ 1
Γ
α β −γ 1
,
B
β 1,α
1
0
1 −ξ
α−1
ξ
β
dξ
1
α−2
η
β−γ
dη
Γ
α −1
Γ
β − γ 1
Γ
α β −γ
,
B
β 1,α− 1
1
0
1 −η
α−2
η
αβ−γ
Γ
β
Γ
α β −γ 1
1 −γ
β − γ
1−γ
MT
αβ
Γ
α β 1
rμ
∗
Γ
β − γ 1
T
αβ−γ
2αΓ
β
Γ
α β −γ
1 −γ
β − γ
1−γ
MT
αβ
2αΓ
α β
2
|
λ
|
T
α
r
Γ
α 1
λ
|
T
α
Γ
α 1
r
4α β
MT
αβ
2αΓ
α β 1
≤
Λ1 −δ
r
≤ r.
3.7
Therefore, Fut≤r.
For u, v ∈ Ω and for each t ∈ 0,T,basedonH
¨
β−1
Γ
β
f
τ, u
τ
− f
τ, v
τ
dτ
ds
|
λ
|
Γ
α
s
0
s − τ
β−1
Γ
β
f
τ, u
τ
− f
τ, v
τ
ds
T
2α
T
0
T − s
α−2
Γ
α −1
s
0
s − τ
β−1
Γ
β
f
α −1
|
u
s
− v
s
|
ds
≤
u − v
Γ
α
Γ
β
t
0
t − s
u −v
2Γ
α
Γ
β
T
0
T − s
α−1
s
0
s −τ
β−1
μ
τ
0
T − s
α−2
s
0
s − τ
β−1
μ
τ
dτ
ds
|
λ
|
T
α
2Γ
α 1
β−1
1/1−γ
dτ
1−γ
s
0
μ
τ
1/γ
dτ
γ
ds
u −v
2Γ
α
0
μ
τ
1/γ
dτ
γ
ds
T
u − v
2αΓ
α − 1
Γ
β
T
0
γ
ds
2
|
λ
|
T
α
Γ
α 1
u − v
≤
μ
∗
u −v
Γ
α
Γ
1 −γ
β − γ
1−γ
T
0
T − s
α−1
s
β−γ
ds
μ
∗
T
u −v
2αΓ
α − 1
Γ
β
μ
∗
u −v
t
αβ−γ
Γ
α
Γ
β
1 −γ
β −γ
1−γ
1
0
1 − ξ
α−1
ξ
β−γ
dξ
μ
μ
∗
u − v
T
αβ−γ
2αΓ
α −1
Γ
β
1 −γ
β − γ
1−γ
1
0
1 −η
α−2
η
β−γ
dη
Γ
α β −γ 1
1 −γ
β −γ
1−γ
2
|
λ
|
T
α
Γ
α 1
u − v
Λu − v.
3.8
Since Λ < 1, consequently F is a contraction. As a consequence of Banach fixed point theorem,
we deduce that F has a fixed point which is a solution of problem 1.1.
Corollary 3.2. Assume that
(H1)
2αΓ
α β 1
2
|
λ
|
T
α
Γ
α 1
< 1, 3.10
then problem 1.1 has a unique solution.
Theorem 3.3. Suppose that (H1) and the following condition hold:
(H2) There exists a constant l ∈ 0, 1 and a real-valued function mt ∈ L
1/l
0,T,R
such
that
f
t, u
1 −γ
β − γ
1−γ
|
λ
|
T
α
Γ
α 1
< 1. 3.12
Proof. Let us fix
4α β − l
Γ
β − l 1
m
∗
T
αβ−l
2αΓ
T
0
mτ
1/l
dτ
l
;considerB
r
{u ∈ Ω :
u
≤ r},thenB
r
is a closed, bounded,
and convex subset of Banach space Ω.WedefinetheoperatorsS and T on B
r
as
Su
t
t
0
t −s
Tu
t
−
1
2
T
0
T − s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
s
0
s − τ
β−1
Γ
β
f
τ, u
τ
dτ − λu
s
ds.
3.14
For u, v ∈ B
r
,basedonH
¨
older inequality, we find that
|
τ, u
τ
dτ
|
λu
s
|
ds
1
2
T
0
T − s
α−1
Γ
α
T
2α
T
0
T − s
α−2
Γ
α −1
s
0
s − τ
β−1
Γ
β
f
τ, v
s
0
s − τ
β−1
m
τ
dτ
ds
|
λ
|
u
t
0
t − s
α−1
Γ
α
dτ
ds
|
λ
|
u
2
T
0
T − s
α−1
Γ
α
ds
T
2αΓ
α − 1
Γ
2α
T
0
T − s
α−2
Γ
α −1
ds
≤
1
Γ
α
Γ
β
t
0
t − s
ds
1
2Γ
α
Γ
β
T
0
T − s
α−1
s
0
s − τ
β−1
1/1−l
0
T − s
α−2
s
0
s − τ
β−1
1/1−l
dτ
1−l
s
0
m
τ
1/l
T
0
T − s
α−1
Γ
α
ds
|
λ
|
T
u
2α
T
0
T − s
α−2
Γ
α −1
2Γ
α
Γ
β
1 −l
β − l
1−l
T
0
T − s
α−1
s
β−l
ds
m
∗
T
2αΓ
α − 1
≤
m
∗
T
αβ−l
Γ
α
Γ
β
1 − l
β − l
1−l
1
0
1 − ξ
α−1
ξ
β−l
dξ
T
αβ−l
2αΓ
α − 1
Γ
β
1 −l
β − l
1−l
1
0
1 −η
α−2
η
β−l
dη
2
|
λ
|
T
1−l
2
|
λ
|
T
α
r
Γ
α 1
≤ r.
3.15
Thus, Su Tv≤r,soSu Tv ∈ B
r
.
12 Boundary Value Problems
For u, v ∈ Ω and for each t ∈ 0,T, by the analogous argument to the p roof of
Theorem 3.1,weobtain
|
Tu
t
−
Tv
f
τ, u
τ
− f
τ, v
τ
dτ
ds
|
λ
|
2
T
0
T − s
α−1
Γ
0
s − τ
β−1
Γ
β
f
τ, u
τ
− f
τ, v
τ
dτ
ds
|
α
Γ
β
T
0
T − s
α−1
s
0
s − τ
β−1
μ
τ
dτ
ds
|
α−2
s
0
s − τ
β−1
μ
τ
dτ
ds
|
λ
|
T
α
2Γ
α 1
u − v
≤
s
0
μ
τ
1/γ
dτ
γ
ds
T
u − v
2αΓ
α −1
Γ
β
T
0
γ
ds
|
λ
|
T
α
Γ
α 1
u − v
≤
μ
∗
u −v
2Γ
α
Γ
β
1 −γ
β − γ
1−γ
T
0
T − s
α−2
s
β−γ
ds
|
λ
|
T
α
Γ
α 1
u − v
μ
μ
∗
u −v
T
αβ−γ
2αΓ
α −1
Γ
β
1 −γ
β − γ
1−γ
1
0
1 −η
α−2
η
β−γ
dη
α β −γ 1
1 −γ
β −γ
1−γ
|
λ
|
T
α
Γ
α 1
u −v
.
3.16
Boundary Value Problems 13
From the assumption
2α β − γ
Γ
< 1, 3.17
it follows that T is a contraction mapping.
The continuity of f implies that the operator S is continuous. Also, S is uniformly
bounded on B
r
as
Su
≤
Γ
β − l 1
m
∗
T
αβ−l
Γ
β
Γ
α β −l 1
1 −l
β − l
1/αβ
,
1
2
εΓ
α
2
|
λ
|
r
1/α
⎫
⎬
⎭
. 3.19
For each u ∈ B
r
, we will prove that if t
1
,t
2
∈ 0,T and 0 <t
2
− t
1
t
2
0
t
2
− s
α−1
Γ
α
s
0
s − τ
β−1
Γ
β
0
s −τ
β−1
Γ
β
f
τ, u
τ
dτ − λu
s
ds
τ
dτ − λu
s
ds
t
2
t
1
t
2
− s
α−1
Γ
α
s
0
s − τ
α
s
0
s − τ
β−1
Γ
β
f
τ, u
τ
dτ − λu
s
ds
0
s −τ
β−1
Γ
β
f
τ, u
τ
dτ − λu
s
ds
t
2
t
1
t
2
− s
14 Boundary Value Problems
≤
t
1
0
t
2
− s
α−1
−
t
1
− s
α−1
Γ
α
s
1
0
t
2
− s
α−1
−
t
1
− s
α−1
Γ
α
ds
t
2
t
1
t
2
− s
ds
|
λ
|
u
t
2
t
1
t
2
− s
α−1
Γ
α
ds
≤
N
Γ
α β 1
t
<T,wehave
|
Su
t
2
−
Su
t
1
|
≤
N
Γ
α β 1
t
αβ
2
− t
αβ
1
|
1
|
λ
|
r
Γ
α 1
ασ
α−1
t
2
− t
1
<
N
Γ
α β
σ
αβ
|
λ
|
|
Su
t
2
−
Su
t
1
|
≤
N
Γ
α β 1
t
αβ
2
− t
αβ
1
|
λ
t
α
2
<
N
Γ
α β 1
2σ
αβ
|
λ
|
r
Γ
α 1
2σ
α
<
ε
2
Γ
α 1
< 1. 3.24
Further assume that
(H2)
there exists a constant K>0 such that
f
t, u
≤ K, ∀t ∈
0,T
,u∈ R, 3.25
then problem 1.1 has at least one solution on 0,T.
4. Example
Let α 2, β 1, λ 1/8, T π/2. We consider the following boundary value problem
C
D
1
C
0
u
π
2
0,
4.1
where
f
t, u
1
t 2
2
u
1 u
,
t, u
∈
√
π/4
√
2. Further,
4α β −γ
Γ
β − γ 1
μ
∗
T
αβ−γ
2αΓ
β
Γ
α β −γ 1
1 −γ
β −γ
1−γ
2
2
|
λ
|
T
2
Γ
3
17π
3
15 × 64
π
2
32
≈ 0.86 < 1.
4.3
Then BVP 4.1 has a unique solution on 0,π/2 according to Theorem 3.1.
16 Boundary Value Problems
On the other hand, we find that
2α β −γ
Γ
β − γ 1
Γ
3/2
μ
∗
T
5/2
4Γ
7/2
|
λ
|
T
2
Γ
3
9π
3
64 × 15
π
2
18 pages, 2009.
8 S. Zhang, “Positive solutions for boundary-value problems of nonlinear fractional differential
equations,” Electronic Journal of Differential Equations, vol. 2006, pp. 1–12, 2006.
9 B. Ahmad and J. J. Nieto, “Solvability of nonlinear Langevin equation involving two fractional orders
with Dirichlet boundary conditions,” International Journal of Differential Equations, vol. 2010, Article ID
649486, 10 pages, 2010.
10
B. Ahmad and P. Eloe, “A nonlocal boundary value problem for a nonlinear fractional differential
equation with two indices,” Communications on Applied Nonlinear Analysis, vol. 17, no. 3, pp. 69–80,
2010.
Boundary Value Problems 17
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