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 Chen1, 2 and Yi Chen2
1
2
Department of Mathematics, Xiangnan University, Chenzhou, Hunan 423000, China
School of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan 411005, China
Correspondence should be addressed to Anping Chen, chenap@263.net
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.
Langevin equation is widely used to describe the evolution of physical phenomena
in fluctuating environments. However, for systems in complex media, ordinary Langevin
equation does not provide the correct description of the dynamics. One of the possible generalizations of Langevin equation is to replace the ordinary derivative by a fractional derivative
in it. This gives rise to fractional Langevin equation, see for instance 9–12 and the references
therein.
In this paper, we consider the following boundary value problem of Langevin equation
with two different fractional orders:
C β C α
D
D λ ut ft, ut t ∈ 0, T,
1.1
u0 −uT, u 0 u T 0,
2
Boundary Value Problems
where T is a positive constant, 1 < α ≤ 2, 0 < β ≤ 1, C Dα , and C Dβ are the Caputo fractional
derivatives, f : 0, T × R → R is continuous, and λ is a real number.
The organization of this paper is as follows. In Section 2, we recall some definitions of
fractional integral and derivative and preliminary results which will be used in this paper. In
Section 3, we will consider the existence results for problem 1.1. In Section 4, we will give
an example to ensure our main results.
2. Preliminaries
In this section, we present some basic notations, definitions, and preliminary results which
will be used throughout this paper.
Definition 2.1. The Caputo fractional derivative of order α of a function f : 0, ∞ → R, is
defined as
t
1
C α
D ft 2.1
t − sn−α−1 f n sds, n − 1 < α < n, n α 1,
Γn − α 0
where α denotes the integer part of the real number α.
Definition 2.2. The Riemann-Liouville fractional integral of order α > 0 of a function ft,
t > 0, is defined as
t
1
2.2
I α ft t − sα−1 fsds,
Γα 0
provided that the right side is pointwise defined on 0, ∞.
Definition 2.3. The Riemann-Liouville fractional derivative of order α > 0 of a continuous
function f : 0, ∞ → R is given by
n t
d
1
2.3
Dα ft t − sn−α−1 fsds,
Γn − α dt
0
where n α 1 and α 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
Dα ut 0 has solution
ut c0 c1 t c2 t2 · · · cn−1 tn−1 ,
2.4
where ci ∈ R, i 0, 1, 2, . . . , n − 1, n α 1.
Lemma 2.5 see 8. Let α > 0, then
I αC Dα ut ut c0 c1 t c2 t2 · · · cn−1 tn−1 ,
for some ci ∈ R, i 0, 1, 2, . . . , n − 1, n α 1.
2.5
Boundary Value Problems
3
Lemma 2.6. The unique solution of the following boundary value problem
C
Dβ
C
Dα λ ut yt,
t ∈ 0, T, 1 < α ≤ 2, 0 < β ≤ 1,
2.6
u0 −uT,
u 0 u T 0,
is given by
ut t
0
t − sα−1
Γα
1
−
2
T
0
s
0
T − sα−1
Γα
T α − 2tα
2αT α−1
T
0
s − τβ−1
yτdτ − λus ds
Γ β
s
0
s − τβ−1
yτdτ − λus ds
Γ β
T − sα−2
Γα − 1
s
0
2.7
s − τβ−1
yτdτ − λus ds.
Γ β
Proof. Similar to the discussion of 9, equation 1.5, the general solution of
C
Dβ
C
Dα λ ut yt
2.8
can be written as
ut t
0
t − sα−1
Γα
s
0
c0
s − τβ−1
tα − c1 t − c2 .
yτdτ − λus ds −
Γα 1
Γ β
2.9
By the boundary conditions u0 uT 0 and u 0 u T 0, we obtain
1
c0
Γα 1 αT α−1
T
0
T − sα−2
Γα − 1
s
0
s − τβ−1
yτdτ − λus ds,
Γ β
c1 0,
1
c2 2
T
0
T
−
2α
T − sα−1
Γα
T
0
s
T − sα−2
Γα − 1
0
s − τβ−1
yτdτ − λus ds
Γ β
s
0
s − τβ−1
yτdτ − λus ds.
Γ β
2.10
4
Boundary Value Problems
Hence,
ut t
0
t − sα−1
Γα
1
−
2
T
0
s − τβ−1
yτdτ − λus ds
Γ β
s
0
T − sα−1
Γα
T α − 2tα
2αT α−1
T
0
s
0
s − τβ−1
yτdτ − λus ds
Γ β
T − sα−2
Γα − 1
s
0
2.11
s − τβ−1
yτdτ − λus ds.
Γ β
Lemma 2.7 Krasnoselskii’s fixed point theorem. Let E be a bounded closed convex subset of a
Banach space X, and let S, T be the operators such that
i Su Tv ∈ E whenever u, v ∈ E,
ii S is completely continuous,
iii T is a contraction mapping.
Then there exists z ∈ E such that z Sz Tz.
Lemma 2.8 Hölder inequality. Let p > 1, 1/p 1/q 1, f ∈ Lp a, b, g ∈ Lq a, b, then
the following inequality holds:
b
a
fxgxdx ≤
b
fxp dx
a
1/p b
1/q
|gx| dx
q
.
2.12
a
3. Main Result
In this section, our aim is to discuss the existence and uniqueness of solutions to the problem
1.1.
Let Ω be a Banach space of all continuous functions from 0, T → R with the norm
u supt∈0,T {|ut|}.
Theorem 3.1. Assume that
(H1) there exists a real-valued function μt ∈ L1/γ 0, T, R for some γ ∈ 0, 1 such that
ft, u − ft, v ≤ μt|u − v|,
for almost all t ∈ 0, T, u, v ∈ R.
3.1
4α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
2|λ|T α
< 1,
β−γ
Γα 1
2αΓ β Γ α β − γ 1
3.2
If
Λ
T
γ
where γ ∈ 0, 1, β /
γ, 1 < α ≤ 2, 0 < β ≤ 1, μ∗ 0 μτ1/γ dτ , then problem 1.1 has a
unique solution.
Boundary Value Problems
5
Proof. Define an operator F : Ω → Ω by
Fut t
0
t − sα−1
Γα
1
−
2
T
s
0
T − sα−1
Γα
0
T α − 2tα
2αT α−1
T
0
s − τβ−1
fτ, uτdτ − λus ds
Γ β
s
0
s − τβ−1
fτ, uτdτ − λus ds
Γ β
T − sα−2
Γα − 1
s
0
3.3
s − τβ−1
fτ, uτdτ − λus ds.
Γ β
Let M supt∈0,T |ft, 0| and choose
1
1−δ
4α β MT αβ
≤ r,
2αΓ α β 1
3.4
where δ is such that Λ ≤ δ < 1.
Now we show that FBr ⊂ Br , where Br {u ∈ Ω : u ≤ r}. For u ∈ Br , by Hölder
inequality, we have
|Fut|
t t − sα−1 s s − τβ−1
fτ, uτdτ − λus ds
0 Γα
Γ β
0
1
−
2
T
0
T − sα−1
Γα
T α − 2tα
2αT α−1
≤
t
0
T
0
t − sα−1
Γα
1
2
T
0
s
0
s − τβ−1
fτ, uτdτ − λus ds
Γ β
T − sα−2
Γα − 1
s
0
s
0
s − τβ−1
fτ, uτdτ − λus ds
Γ β
s − τβ−1 fτ, uτ − fτ, 0 fτ, 0 dτ |λus| ds
Γ β
T − sα−1
Γα
s
0
s − τβ−1 fτ, uτ − fτ, 0 fτ, 0 dτ |λus| ds
Γ β
s
T T T − sα−2
s − τβ−1 fτ, uτ − fτ, 0 fτ, 0 dτ |λus| ds
2α 0 Γα − 1
Γ β
0
≤
t
0
t − sα−1
Γα
1
2
T
0
s
0
T − sα−1
Γα
s − τβ−1 μτ|uτ| fτ, 0 dτ |λus| ds
Γ β
s
0
s − τβ−1 μτ|uτ| fτ, 0 dτ |λus| ds
Γ β
6
T
2α
T
0
T − sα−2
Γα − 1
Boundary Value Problems
s
s − τβ−1 μτ|uτ| fτ, 0 dτ |λus| ds
Γ β
0
s
t
t
t − sα−1
s − τβ−1
t − sα−1 s s − τβ−1
≤ u
μτdτ
ds
M
dτ ds
Γα
Γα
Γ β
Γ β
0
0
0
0
|λ|u
t
0
T
0
T − sα−1
Γα
s
0
s − τβ−1
μτdτ
Γ β
ds
s
|λ|u T T − sα−1
s − τβ−1
ds
dτ ds 2
Γα
Γ β
0
0
0
s
Tu T T − sα−2
s − τβ−1
μτdτ ds
2α 0 Γα − 1
Γ β
0
M
2
T
u
t − sα−1
ds Γα
2
TM
2α
T − sα−1
Γα
T
0
u
≤
ΓαΓ β
t
t − s
t
s − τ
1/1−γ 0
1−γ s
dτ
T
t − sα−1 sβ ds α−1
T − s
s
0
β−1
s − τ
1/1−γ T
0
α−2
dτ
T − s
β−1
s − τ
1/1−γ T − sα−2 sβ ds α−1 β−γ
t − s
γ ds
dτ
s
γ 1/γ
μτ
dτ
0
0
0
1/γ
μτ
dτ
1−γ s
0
1−γ t
|λ|T α u
2Γα 1
s
0
T
γ ds
0
T − sα−1 sβ ds T
1/γ
μτ
dτ
1−γ s
0
TM
2αΓα − 1Γ β 1
1−γ
β−γ
|λ|T α u
Γα 1
|λ|T α u
2Γα 1
M
ds ΓαΓ β 1
t
t − sα−1 sβ ds
0
T
μ∗ u
1 − γ 1−γ T
M
T − sα−1 sβ−γ ds T − sα−1 sβ ds
2ΓαΓ β β − γ
2ΓαΓ
β
1
0
0
Tμ∗ u
1 − γ 1−γ T
T − sα−2 sβ−γ ds
2αΓα − 1Γ β β − γ
0
T − sα−2
ds
Γα − 1
0
0
Tu
2αΓα − 1Γ β
β−1
T
0
M
2ΓαΓ β 1
μ∗ u
≤
ΓαΓ β
s
0
u
2ΓαΓ β
T|λ|u
s − τβ−1
dτ ds 2α
Γ β
0
α−1
M
ΓαΓ β 1
s
T − sα−2
Γα − 1
TM
2αΓα − 1Γ β 1
T
0
T − sα−2 sβ ds 2|λ|T α u
Γα 1
ds
Boundary Value Problems
μ∗ utαβ−γ
ΓαΓ β
1−γ
β−γ
μ∗ uT αβ−γ
2ΓαΓ β
7
1−γ 1
1 − ξ
α−1 β−γ
ξ
0
1−γ
β−γ
1−γ 1
Mtαβ
dξ ΓαΓ β 1
α−1 β−γ
1−η
η dη 0
1
1 − ξα−1 ξβ dξ
0
MT αβ
2ΓαΓ β 1
1
1−η
α−1
ηβ dη
0
α−2 β−γ
μ∗ uT αβ−γ
1 − γ 1−γ 1 1−η
η dη
β
−
γ
2αΓα − 1Γ β
0
MT αβ
2αΓα − 1Γ β 1
rμ∗ T αβ−γ
≤
ΓαΓ β
1−γ
β−γ
1
0
1−γ 1
α−2 β
2|λ|T α u
1−η
η dη Γα 1
1 − ξ
α−1 β−γ
ξ
0
MT αβ
dξ ΓαΓ β 1
1
1 − ξα−1 ξβ dξ
0
1
α−1 β−γ
α−1 β
rμ∗ T αβ−γ 1 − γ 1−γ 1 MT αβ
η dη η dη
1−η
1−η
2ΓαΓ β β − γ
2ΓαΓ β 1 0
0
α−2 β−γ
rμ∗ T αβ−γ
1 − γ 1−γ 1 1−η
η dη
β
−
γ
2αΓα − 1Γ β
0
MT αβ
2αΓα − 1Γ β 1
1
0
α−2 β
2|λ|T α r
1−η
.
η dη Γα 1
3.5
Take notice of Beta functions:
B β − γ 1, α 1
1 − ξ
α−1 β−γ
ξ
dξ 1
0
B β 1, α 1
0
0
1 − ξ
α−1 β
ξ dξ 1
0
α−1 β−γ
ΓαΓ β − γ 1
1−η
η dη ,
Γ αβ−γ 1
1−η
α−1
ΓαΓ β 1
η dη ,
Γ αβ1
β
Γα − 1Γ β − γ 1
α−2 β−γ
1−η
,
η dη B β − γ 1, α − 1 Γ αβ−γ
0
1
α−2 β
Γα − 1Γ β 1
1−η
B β 1, α − 1 η dη .
Γ αβ
0
1
We can get
rμ∗ Γ β − γ 1 T αβ−γ 1 − γ 1−γ
MT αβ
|Fut| ≤ β−γ
Γ β Γ αβ−γ 1
Γ αβ1
rμ∗ Γ β − γ 1 T αβ−γ 1 − γ 1−γ
MT αβ
2Γ β Γ α β − γ 1 β − γ
2Γ α β 1
3.6
8
Boundary Value Problems
rμ∗ Γ β − γ 1 T αβ−γ 1 − γ 1−γ
MT αβ
2|λ|T α r
β−γ
Γα 1
2αΓ β Γ α β − γ
2αΓ α β
4α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
2|λ|T α
r
β−γ
Γα 1
2αΓ β Γ α β − γ 1
4α β MT αβ
2αΓ α β 1
≤ Λ 1 − δr
≤ r.
3.7
Therefore, Fut ≤ r.
For u, v ∈ Ω and for each t ∈ 0, T, based on Hölder inequality, we obtain
|Fut − Fvt|
≤
t
0
t − sα−1
Γα
|λ|
t
T
|λ|
2
T
2α
≤
0
T − sα−1
Γα
T
0
T
|λ|T
2α
ds
0
s
0
s − τβ−1 fτ, uτ − fτ, vτdτ
Γ β
ds
T − sα−1
|us − vs|ds
Γα
T − sα−2
Γα − 1
T
0
u − v
ΓαΓ β
s
s − τβ−1 fτ, uτ − fτ, vτdτ
Γ β
0
ds
T − sα−2
|us − vs|ds
Γα − 1
t
t − sα−1
s
0
u − v
2ΓαΓ β
0
s − τβ−1 fτ, uτ − fτ, vτdτ
Γ β
t − sα−1
|us − vs|ds
Γα
0
1
2
s
s − τβ−1 μτdτ ds 0
T
α−1
T − s
0
Tu − v
2αΓα − 1Γ β
s
β−1
s − τ
0
T
0
T − sα−2
s
0
|λ|T α
u − v
Γα 1
μτdτ ds |λ|T α
u − v
2Γα 1
s − τβ−1 μτdτ ds |λ|T α
u − v
2Γα 1
Boundary Value Problems
u − v
≤
ΓαΓ β
t
α−1
t − s
0
T
α−1
T − s
T
s − τ
1/1−γ 1−γ s
dτ
α−2
T − s
1/γ
μτ
dτ
γ ds
0
s
0
β−1
s − τ
1/1−γ 0
1−γ t
1−γ s
dτ
1/γ
dτ
μτ
γ ds
0
1−γ
β−γ
μ∗ Tu − v
2αΓα − 1Γ β
μ∗ u − vtαβ−γ
ΓαΓ β
1−γ T
T − sα−1 sβ−γ ds
0
1−γ
β−γ
1−γ
β−γ
μ∗ u − vT αβ−γ
2αΓα − 1Γ β
t − sα−1 sβ−γ ds
0
μ∗ u − vT αβ−γ
2ΓαΓ β
≤
β−1
0
1−γ
β−γ
μ∗ u − v
2ΓαΓ β
s
2|λ|T α
u − v
Γα 1
μ∗ u − v
≤
ΓαΓ β
0
0
Tu − v
2αΓα − 1Γ β
s
γ 1/1−γ 1−γ s 1/γ
β−1
μτ
ds
dτ
dτ
s − τ
0
u − v
2ΓαΓ β
9
1−γ T
0
1−γ 1
1−γ
β−γ
1−γ
β−γ
T − sα−2 sβ−γ ds 2|λ|T α
u − v
Γα 1
1 − ξα−1 ξβ−γ dξ
0
1−γ 1
α−1 β−γ
η dη
1−η
0
1−γ 1
0
α−2 β−γ
2|λ|T α
η dη 1−η
u − v
Γα 1
4α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
2|λ|T α
u − v
β−γ
Γα 1
2αΓ β Γ α β − γ 1
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) There exists a constant L > 0 such that
ft, u − ft, v ≤ L|u − v|,
∀t ∈ 0, T, u, v ∈ R.
3.9
10
Boundary Value Problems
If
4α β LT αβ
2|λ|T α
< 1,
Γα 1
2αΓ α β 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 ∈ L1/l 0, T, R such
that
ft, u ≤ mt,
for almost every t ∈ 0, T, u ∈ R.
3.11
Then the problem 1.1 has at least one solution on 0, T if
2α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
|λ|T α
< 1.
β−γ
Γα 1
2αΓ β Γ α β − γ 1
3.12
4α β − l Γ β − l 1 m∗ T αβ−l
1 − l 1−l
≤ r;
2αΓ β Γ α β − l 1 1 − 2|λ|T α /Γα 1 β − l
3.13
Proof. Let us fix
T
l
here, m∗ 0 mτ1/l dτ ; consider Br {u ∈ Ω : u ≤ r}, then Br is a closed, bounded,
and convex subset of Banach space Ω. We define the operators S and T on Br as
Sut t
0
1
Tut −
2
t − sα−1
Γα
T
0
s
0
T − sα−1
Γα
T α − 2tα
2αT α−1
T
0
s − τβ−1
fτ, uτdτ − λus ds,
Γ β
s
0
s − τβ−1
fτ, uτdτ − λus ds
Γ β
T − sα−2
Γα − 1
s
0
s − τβ−1
fτ, uτdτ − λus ds.
Γ β
For u, v ∈ Br , based on Hölder inequality, we find that
|Su Tv|
≤
t
0
t − sα−1
Γα
1
2
T
0
s
0
T − sα−1
Γα
s − τβ−1 fτ, uτdτ |λus| ds
Γ β
s
0
s − τβ−1 fτ, vτ dτ |λvs| ds
Γ β
3.14
Boundary Value Problems
s
T T T − sα−2
s − τβ−1 fτ, vτ dτ |λvs| ds
2α 0 Γα − 1
Γ β
0
1
≤
ΓαΓ β
t
α−1
t − s
0
1
2ΓαΓ β
t
α−1
s − τ
mτdτ ds |λ|u
T − s
β−1
s − τ
0
T
α−2
T − s
0
α−1
t − s
s
s
β−1
s − τ
0
s − τβ−1
α−1
T − s
|λ|Tu T T − sα−2
ds
mτdτ ds 2α
0 Γα − 1
1−l s
l 1/1−l
mτ1/l dτ
dτ
0
T
2αΓα − 1Γ β
0
l s
1/1−l 1−l s
1/l
α−2
β−1
dτ
mτ dτ
ds
s − τ
T − s
0
0
t
0
T
|λ|u
|λ|Tu
t − sα−1
T − sα−1
ds ds Γα
2
Γα
2α
0
0
m∗
1 − l 1−l t
≤
t − sα−1 sβ−l ds
ΓαΓ β β − l
0
|λ|u
m∗
2ΓαΓ β
m∗ T
2αΓα − 1Γ β
≤
1−l
β−l
m∗ T αβ−l 1 − l
ΓαΓ β β − l
m∗ T αβ−l
2ΓαΓ β
ds
0
s
l 1/1−l 1−l s
β−1
1/l
ds
dτ
mτ dτ
s − τ
0
T
t − sα−1
ds
Γα
|λ|u T T − sα−1
ds
mτdτ ds 2
Γα
0
0
T
t
0
s
0
0
1
2ΓαΓ β
β−1
0
T
T
2αΓα − 1Γ β
1
≤
ΓαΓ β
s
11
1−l T
T
0
T − sα−2
ds
Γα − 1
T − sα−1 sβ−l ds
0
1−l
β−l
1−l 1
1−l T
T − sα−2 sβ−l ds 0
2|λ|T α r
Γα 1
1 − ξα−1 ξβ−l dξ
0
1−l
β−l
1−l 1
α−1 β−l
1−η
η dη
0
1−l 1
α−2 β−l
m∗ T αβ−l
1−l
2|λ|T α r
1−η
η dη β−l
Γα 1
2αΓα − 1Γ β
0
4α β − l Γ β − l 1 m∗ T αβ−l 1 − l 1−l 2|λ|T α r
β−l
Γα 1
2αΓ β Γ α β − l 1
≤ r.
3.15
Thus, Su Tv ≤ r, so Su Tv ∈ Br .
12
Boundary Value Problems
For u, v ∈ Ω and for each t ∈ 0, T, by the analogous argument to the proof of
Theorem 3.1, we obtain
|Tut − Tvt|
s
1 T T − sα−1
s − τβ−1 fτ, uτ − fτ, vτ dτ ds
≤
2 0
Γα
Γ β
0
T
T − sα−1
|us − vs|ds
Γα
0
s
T T T − sα−2
s − τβ−1 fτ, uτ − fτ, vτ dτ ds
2α 0 Γα − 1
Γ β
0
|λ|
2
T
T − sα−2
|us − vs|ds
0 Γα − 1
T
s
u − v
|λ|T α
≤
u − v
T − sα−1
s − τβ−1 μτdτ ds 2Γα 1
2ΓαΓ β 0
0
|λ|T
2α
Tu − v
2αΓα − 1Γ β
T
s
μτdτ ds |λ|T α
u − v
2Γα 1
0
0
s
T
1−γ s
γ u − v
β−1 1/1−γ 1/γ
α−1
≤
ds
s − τ dτ
μτ dτ
T − s
2ΓαΓ β 0
0
0
s
T
γ 1/1−γ 1−γ s
Tu − v
α−2
β−1
1/γ
ds
dτ
μτ dτ
T − s
s − τ
2αΓα − 1Γ β 0
0
0
≤
α−2
T − s
β−1
s − τ
|λ|T α
u − v
Γα 1
μ∗ u − v 1 − γ 1−γ T
T − sα−1 sβ−γ ds
2ΓαΓ β β − γ
0
μ∗ Tu − v
1 − γ 1−γ T
|λ|T α
u − v
T − sα−2 sβ−γ ds Γα 1
2αΓα − 1Γ β β − γ
0
μ∗ u − vT αβ−γ
2ΓαΓ β
1−γ
β−γ
1−γ 1
α−1 β−γ
1−η
η dη
0
α−2 β−γ
μ∗ u − vT αβ−γ 1 − γ 1−γ 1 |λ|T α
1−η
η dη u − v
β−γ
Γα 1
2αΓα − 1Γ β
0
2α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
|λ|T α
≤
u − v.
β−γ
Γα 1
2αΓ β Γ α β − γ 1
3.16
Boundary Value Problems
13
From the assumption
2α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
|λ|T α
< 1,
β−γ
Γα 1
2αΓ β Γ α β − 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 Br as
Γ β − l 1 m∗ T αβ−l 1 − l 1−l
|λ|T α r
.
Su ≤ β−l
Γα 1
Γ β Γ αβ−l1
3.18
On the other hand, let N maxt,u∈0,T ×Br |ft, ut| 1, for all ε > 0, setting
σ min
⎧ ⎨ 1 εΓ α β 1/αβ
⎩2
2N
⎫
1 εΓα 1/α ⎬
.
,
⎭
2 2|λ|r
3.19
For each u ∈ Br , we will prove that if t1 , t2 ∈ 0, T and 0 < t2 − t1 < σ, then
|Sut2 − Sut1 | < ε.
In fact, we have
|Sut2 − Sut1 |
t2 t − sα−1 s s − τβ−1
2
fτ, uτdτ − λus ds
0
Γα
Γ β
0
s
t1
t1 − sα−1
s − τβ−1
−
fτ, uτdτ − λus ds
Γα
Γ
β
0
0
t1 t − sα−1 s s − τβ−1
2
fτ, uτdτ − λus ds
0
Γα
Γ β
0
s
t2
t2 − sα−1
s − τβ−1
fτ, uτdτ − λus ds
Γα
Γ β
0
t1
s
t1
t1 − sα−1
s − τβ−1
−
fτ, uτdτ − λus ds
Γα
Γ
β
0
0
t1 t − sα−1 − t − sα−1 s s − τβ−1
2
1
fτ, uτdτ − λus ds
0
Γα
Γ β
0
s
t2
t2 − sα−1
s − τβ−1
fτ, uτdτ − λus ds
Γα
Γ β
0
t1
3.20
14
≤
t1
0
t2 − sα−1 − t1 − sα−1
Γα
s
0
s − τβ−1 fτ, uτdτ
Γ β
Boundary Value Problems
ds
t1
t2 − sα−1 − t1 − sα−1
ds
Γα
0
s
t2
t2
t2 − sα−1
s − τβ−1 t2 − sα−1
fτ, uτ dτ ds |λ|u
ds
Γα
Γα
Γ β
0
t1
t1
|λ|u
αβ αβ N
|λ|r α α ≤ t −t .
t2 − t1
Γα 1 2 1
Γ αβ1
3.21
In the following, the proof is divided into two cases.
Case 1. For σ ≤ t1 < t2 < T, we have
αβ αβ N
|λ|r α α t −t
t2 − t1
|Sut2 − Sut1 | ≤ Γα 1 2 1
Γ αβ1
N
|λ|r
≤ ασ α−1 t2 − t1 α β σ αβ−1 t2 − t1 Γα 1
Γ αβ1
N
|λ|r α
< σ
σ αβ Γα
Γ αβ
<
α
αβ
1 ε
ε
1
2
2
2 2
< ε.
3.22
Case 2. for 0 ≤ t1 < σ, t2 < 2σ, we have.
αβ αβ N
|λ|r α α t −t
|Sut2 − Sut1 | ≤ t2 − t1
Γα 1 2 1
Γ αβ1
N
|λ|r α
αβ
≤ t
t Γα 1 2
Γ αβ1 2
N
|λ|r
< 2σαβ 2σα
Γα 1
Γ αβ1
<
3.23
ε ε
2 2
ε.
Therefore, S is equicontinuous and the Arzela-Ascoli theorem implies that S is compact on
Br , so the operator S is completely continuous.
Boundary Value Problems
15
Thus, all the assumptions of Lemma 2.7 are satisfied and the conclusion of Lemma 2.7
implies that the boundary value problem 1.1 has at least one solution on 0, T.
Corollary 3.4. Suppose that the condition (H1) hold and, assume that
2α β LT αβ
|λ|T α
< 1.
Γα 1
2αΓ α β 1
3.24
Further assume that
(H2) there exists a constant K > 0 such that
ft, 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
D1
C
D2 u0 u
1
π
ut ft, ut, 0 ≤ t ≤ ,
8
2
π 2
0, u 0 u
π 2
4.1
0,
where
ft, u 1
u
, t, u ∈ 0, T × 0, ∞.
t 2 1 u
2
4.2
Because of |ft, u − ft, v| ≤ 1/4|u − v|, let μt ≡ 1/4, then μt ∈ L2 0, π/2, we have
√
T
π/2
1/2
γ
√
γ 1/2 and μ∗ 0 μτ1/γ dτ 0 1/42 dτ π/4 2. Further,
4α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
2|λ|T α
β−γ
Γα 1
2αΓ β Γ α β − γ 1
17/2Γ3/2μ∗ T 5/2 2|λ|T 2
4Γ1Γ7/2
Γ3
π2
17π 3
15 × 64 32
≈ 0.86 < 1.
Then BVP 4.1 has a unique solution on 0, π/2 according to Theorem 3.1.
4.3
16
Boundary Value Problems
On the other hand, we find that
2α β − γ Γ β − γ 1 μ∗ T αβ−γ 1 − γ 1−γ
|λ|T α
β−γ
Γα 1
2αΓ β Γ α β − γ 1
9/2Γ3/2μ∗ T 5/2 |λ|T 2
4Γ7/2
Γ3
9π 3
π2
64 × 15 64
4.4
≈ 0.44 < 1.
Then BVP 4.1 has at least one solution on 0, π/2 according to Theorem 3.3.
Acknowledgments
This work was supported by the Natural Science Foundation of China 10971173, the
Natural Science Foundation of Hunan Province 10JJ3096, the Aid Program for Science and
Technology Innovative Research Team in Higher Educational Institutions of Hunan Province,
and the Construct Program of the Key Discipline in Hunan Province.
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