Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 708192, 21 pages
doi:10.1155/2012/708192
Research Article
Positive Solutions for Nonlinear Fractional
Differential Equations with Boundary Conditions
Involving Riemann-Stieltjes Integrals
Jiqiang Jiang,1 Lishan Liu,1 and Yonghong Wu2
1
2
School of Mathematical Sciences, Qufu Normal University, Shandong, Qufu 273165, China
Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia
Correspondence should be addressed to Jiqiang Jiang, qfjjq@163.com
and Lishan Liu, lls@mail.qfnu.edu.cn
Received 10 March 2012; Accepted 23 July 2012
Academic Editor: Dumitru Baleanu
Copyright q 2012 Jiqiang Jiang et al. 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 consider the existence of positive solutions for a class of nonlinear integral boundary value
problems for fractional differential equations. By using some fixed point theorems, the existence
and multiplicity results of positive solutions are obtained. The results obtained in this paper
improve and generalize some well-known results.
1. Introduction
This paper is concerned with the existence of positive solutions to the following boundary
value problem BVP for fractional differential equation:
α
ut f t, ut, u t, . . . , un−2 t 0,
D0
u0 u 0 · · · un−2 0 0,
un−2 1 β un−2 ,
0 < t < 1,
1.1
1
where βv 0 vtdAt is a linear functional on C0, 1 given by a Riemann-Stieltjes
α
is the Riemannintegral with A representing a suitable function of bounded variation, D0
n−1
→ R satisfies the
Liouville fractional derivative of order n − 1 < α ≤ n, n ≥ 2, f : 0, 1 × R
Carathéodory type conditions, R −∞, ∞ and R 0, ∞.
2
Abstract and Applied Analysis
Fractional differential equations arise in the modeling and control of many realworld systems and processes particularly in the fields of physics, chemistry, aerodynamics,
electrodynamics of complex media, and polymer rheology. Fractional differential equations
also serve as an excellent tool for the description of hereditary properties of various materials
and processes. Hence, intensive research has been carried out worldwide to study the
existence of solutions of nonlinear fractional differential equations see 1–25. For example,
by means of a mixed monotone method, Zhang 11 studied a unique positive solution for
the singular boundary value problem
α
ut qtf t, ut, u t, . . . , un−2 t 0,
D0
0 < t < 1,
u0 u 0 · · · un−2 0 un−2 1 0,
1.2
α
where α ∈ n − 1, n, n ≥ 2, D0
is the standard Riemann-Liouville derivative, f g h is
nonlinear, and g and h have different monotone properties.
Recently, nonlocal boundary value problems for fractional differential equations were
investigated intensively 13–23. In 14, Bai concerned the existence and uniqueness of a
positive solution for the following nonlocal problem:
α
ut ft, ut 0, 0 < t < 1,
D0
u0 0,
βu η u1,
1.3
α
is the standard Riemann-Liouville
where 1 < α ≤ 2, 0 < βηα−1 < 1, 0 < η < 1, D0
differentiation. The function f is continuous on 0, 1 × R .
In 20, El-Shahed and Nieto investigated the existence of nontrivial solutions for the
following nonlinear m-point boundary value problem of fractional type:
α
R D0 ut
ft, ut 0,
t ∈ 0, 1, α ∈ n − 1, n, n ∈ N,
u0 u 0 · · · un−2 0 0,
u1 m−2
ai u ηi ,
1.4
i1
where n ≥ 2, ai > 0 i 1, 2, . . . , m − 2, 0 < η1 < η2 < · · · < ηm−2 < 1, f ∈ C0, 1 × R, R. Also
the authors considered the analogous problem using the Caputo fractional derivative:
α
C D0 ut
ft, ut 0,
t ∈ 0, 1, α ∈ n − 1, n, n ∈ N,
u0 u 0 · · · un−2 0 0,
u1 m−2
ai u ηi .
1.5
i1
Under certain growth conditions on the nonlinearity, several sufficient conditions for
the existence of nontrivial solution are obtained by using the Leray-Schauder nonlinear
alternative.
Abstract and Applied Analysis
3
Inspired by the work of the above papers, the aim of this paper is to establish the
existence and multiplicity of positive solutions of the BVP 1.1. We discuss the boundary
value problem with the Riemann-Stieltjes integral boundary conditions, that is, the BVP 1.1,
which includes fractional order two-point, three-point, multipoint, and nonlocal boundary
value problems as special cases. Moreover, the β· in 1.1 is a linear function on C0, 1
denoting the Riemann-Stieltjes integral; the A in the Riemann-Stieltjes integral is of bounded
variation, namely, dA can be a signed measure. By using the Krasnosel’skii fixed point
theorem, the Leray-Schauder nonlinear alternative and the Leggett-Williams fixed point
theorem, some existence and multiplicity results of positive solutions are obtained.
The rest of this paper is organized as follows. In Section 2, we present some lemmas
that are used to prove our main results. In Section 3, the existence and multiplicity of positive
solutions of the BVP 1.1 are established by using some fixed point theorems. In Section 4,
we give four examples to demonstrate the application of our theoretical results.
2. Basic Definitions and Preliminaries
We begin this section with some preliminaries of fractional calculus. Let α > 0 and n α 1,
where α is the largest integer smaller than or equal to α. For a function f : 0, ∞ → R, we
define the fractional integral of order α of f as
α
ft I0
1
Γα
t
t − sα−1 fs ds,
2.1
0
provided the integral exists. The fractional derivative of order α > 0 of a continuous function
f is defined by
α
D0
ft
n t
d
1
t − sn−α−1 fs ds,
Γn − α dt
0
2.2
provided the right-hand side is pointwise defined on 0, ∞. We recall the following
properties 26, 27 which are useful for the sequel. For α > 0, β > 0, we have
β
αβ
α
I0
I0 ft I0 ft,
α α
D0
I0 ft ft.
2.3
α
As an example, we can choose a function f such that f, D0
f ∈ C0, ∞ ∩ L1loc 0, ∞.
α
ut 0 with
For α > 0, the general solution of the fractional differential equation D0
u ∈ C0, 1 ∩ L0, 1 is given by
ut c1 tα−1 c2 tα−2 · · · cn tα−n ,
2.4
where ci ∈ R i 1, 2, . . . , n. Hence for u ∈ C0, 1 ∩ L0, 1, we have
α
α
I0
D0
ut ut c1 tα−1 c2 tα−2 · · · cn tα−n .
2.5
4
Abstract and Applied Analysis
Set
⎧
⎨t1 − sα−n1 − t − sα−n1 ,
1
G0 t, s Γα − n 2 ⎩t1 − sα−n1 ,
0 ≤ s ≤ t ≤ 1,
2.6
0 ≤ t ≤ s ≤ 1.
Lemma 2.1 see 11. Let y ∈ Cr 0, 1Cr 0, 1 {y ∈ C0, 1, tr y ∈ C0, 1, 0 ≤ r < 1}. Then
the boundary value problem,
α−n2
vt yt,
D0
0 < t < 1, n − 1 < α ≤ n, n ≥ 2,
v0 0,
2.7
v1 0,
has a unique solution
vt 1
2.8
G0 t, sys ds.
0
Lemma 2.2 see 11. The function G0 t, s defined by 2.6 satisfies the following properties:
i G0 t, s ≥ 0, G0 t, s ≤ G0 s, s for all t, s ∈ 0, 1;
ii there exist a positive function ρ ∈ C0, 1 and 0 < ξ < η < 1 such that
min G0 t, s ≥ ρsG0 s, s,
t∈ξ,η
s ∈ 0, 1,
2.9
where
⎧
α−n1
α−n1 ⎪
− η−s
η1 − s
⎪
⎪
,
⎪
⎨
s1 − sα−n1
ρs α−n1
⎪
⎪
ξ
⎪
⎪
⎩
,
s
s ∈ 0, r,
2.10
s ∈ r, 1.
Abstract and Applied Analysis
5
By 2.4, the unique solution of the problem
α−n2
D0
vt 0,
0 < t < 1, n − 1 < α ≤ n, n ≥ 2,
v0 0,
2.11
v1 βv
is γt tα−n1 , with βv replaced by 1. As in 21, the Green’s function for boundary value
problem 2.11 is given by
Gt, s where Gs :
1
0
γt
Gs G0 t, s,
1−β γ
2.12
G0 t, sdAt.
Lemma 2.3. Let 0 ≤ βγ < 1 and Gs ≥ 0 for s ∈ 0, 1, the Green function Gt, s defined by
2.12 has the following properties:
i Gt, s ≥ 0, Gt, s ≤ 1 β1/1 − βγG0 s, s for all t, s ∈ 0, 1;
ii mint∈ξ,η Gt, s ≥ mint∈ξ,η G0 t, s ≥ ρsG0 s, s, s ∈ 0, 1.
Proof. By Lemma 2.2, it is easy to prove this lemma, so we omit it.
Let X C0, 1. It follows that X, · is a Banach space, where · is defined by the
supernorm x supt∈0,1 |xt|. P {x ∈ X : xt ≥ 0, t ∈ 0, 1}. Clearly P is a cone of X.
Now, in the following, we give the assumptions to be used throughout the rest of this paper.
H1 A is a function of bounded variation, Gs ≥ 0 for s ∈ 0, 1 and 0 ≤ βγ < 1.
H2 f : 0, 1 × Rn−1 → R satisfies the following conditions of Carathéodory type:
i f·, x is Lebesgue measurable for each fixed x ∈ Rn−1 ;
ii ft, · is continuous for a.e. t ∈ 0, 1.
In order to overcome the difficulty due to the dependance of f on derivatives, we
consider the following modified problem:
α−n2
n−2
n−3
1
D0
vt f t, I0
vt, I0
vt, . . . , I0
vt, vt 0,
v0 0,
where n − 1 < α ≤ n, n ≥ 2.
v1 βv,
0 < t < 1,
2.13
6
Abstract and Applied Analysis
Lemma 2.4. The nonlocal fractional order boundary value problem 1.1 has a positive solution if and
only if the nonlinear fractional integrodifferential equation 2.13 has a positive solution.
Proof. If u is a positive solution of the fractional order boundary value problem 1.1, let
n−2
ut. Then from the boundary value conditions of 1.1 and the definition of the
vt D0
Riemann-Liouville fractional integral and derivative, we have
n−2
ut un−2 t,
vt D0
1
vt
I0
1 n−2
I0
u
t
1
Γ1
2
2 n−2
vt I0
u
I0
t 1
Γ2
t
un−2 s ds un−3 t,
0
t
t − sun−2 s ds un−4 t,
0
2.14
..
.
t
1
t − sn−3 un−2 s ds ut,
Γn − 2 0
n t
d
1
α−n2
D0
vt t − s2n−α−3 un−2 s ds
Γ2n − α − 2 dt
0
n t
d
1
t − s2n−α−4 un−3 s ds
Γ2n − α − 3 dt
0
n−2
vt
I0
n−2 n−2
I0
u
t
2.15
···
n t
d
1
t − sn−α−1 us ds
Γn − α dt
0
α
D0
ut,
which imply that v0 un−2 0 0, v1 un−2 1 βun−2 βv. Thus vt is a
positive solution of the nonlinear fractional integrodifferential equation 2.13.
On the other hand, if v is a positive solution of the nonlinear fractional integrodiffern−2
vt, then by 2.3 and the definition of the Riemannential equation 2.13, let ut I0
Liouville fractional derivative, we have
1
1 n−2
1 1 n−3
n−3
ut D0
I0 vt D0
I0 I0 vt I0
vt,
u t D0
2
2 n−2
2 2 n−4
n−4
u t D0
ut D0
I0 vt D0
I0 I0 vt I0
vt,
..
.
n−3
n−3 n−2
n−3 n−3 1
1
un−3 t D0
ut D0
I0 vt D0
I0 I0 vt I0
vt,
n−2
n−2 n−2
un−2 t D0
ut D0
I0 vt vt,
Abstract and Applied Analysis
7
dn n−α
dn n−α n−2
dn 2n−α−2
α−n2
I
ut
I
I
vt
I
vt D0
vt
dtn 0
dtn 0 0
dtn 0
n−2
n−3
1
−f t, I0
vt, I0
vt, . . . , I0
vt, vt
α
ut D0
−f t, ut, u t, . . . , un−3 t, un−2 t ,
0 < t < 1,
2.16
which imply that u0 u 0 · · · un−3 0 0, un−2 0 v0 0, un−2 1 n−2
that
v1 βv βun−2 . Moreover, it follows from the monotonicity and property of I0
n−2
n−2
I0 vt ∈ C0, 1, R . Consequently, ut I0 vt is a positive solution of the fractional
order boundary value problem 1.1.
By Lemma 2.4, we will concentrate our study on 2.13. We here define an operator
T : P → P by
T vt 1
0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds,
t ∈ 0, 1.
2.17
Clearly, v is a fixed point of T in P , and so v is a positive solution of BVP 2.13.
In order to prove our main results, we need the following lemmas.
Lemma 2.5 see 28. Let X be a real Banach space, Ω be a bounded open subset of X, where θ ∈ Ω,
T : Ω → X is a completely continuous operator. Then, either there exist x ∈ ∂Ω, μ ∈ 0, 1 such that
μT x x, or there exists a fixed point x∗ ∈ Ω.
Lemma 2.6 see 29. Let X be a real Banach space, P be a cone in X. Assume that Ω1 and Ω2 are
two bounded open sets of X with θ ∈ Ω1 and Ω1 ⊂ Ω2 . Let T : P ∩ Ω2 \ Ω1 → P be a completely
continuous operator such that either
i T x ≤ x, x ∈ P ∩ ∂Ω1 and T x ≥ x, x ∈ P ∩ ∂ Ω2 , or
ii T x ≥ x, x ∈ P ∩ ∂Ω1 and T x ≤ x, x ∈ P ∩ ∂ Ω2 .
Then T has a fixed point in P ∩ Ω2 \ Ω1 .
Lemma 2.7 see 30, 31. Let P be a cone in a real Banach space X, Pc {x ∈ P : x < c},
ϕ be a nonnegative continuous concave functional on P such that ϕx ≤ x for all x ∈ P c , and
P ϕ, b, d {x ∈ P : b ≤ ϕx, x ≤ d}. Suppose that T : P c → P c is completely continuous and
there exist positive constants 0 < a < b < d ≤ c such that
C1 {x ∈ P ϕ, b, d : ϕx > b} / φ and ϕT x > b for x ∈ P ϕ, b, d,
C2 T x < a for x ∈ P a ,
C3 ϕT x > b for x ∈ P ϕ, b, c with T x > d.
Then T has at least three fixed points x1 , x2 , and x3 satisfying
x1 < a,
b < ϕx2 ,
a < x3 with ϕx3 < b.
2.18
8
Abstract and Applied Analysis
Remark 2.8. If d c, then condition C1 of Lemma 2.7 implies condition C3 of Lemma 2.7.
For notational convenience, we introduce the following constants:
L1 η
β1
1
1−β γ
L2 ρsG0 s, s ds,
ξ
1
G0 s, s ds,
2.19
0
and a nonnegative continuous concave functional ϕ on the cone P defined by
ϕv min|vt|.
2.20
ξ≤t≤η
3. Main Results
In this section, we present and prove our main results.
Theorem 3.1. Assume that (H1 ) and (H2 ) hold and there exist nonnegative functions
h1 , h2 , . . . , hn−1 ∈ L0, 1 such that
n−1
ft, x1 , x2 , . . . , xn−1 − f t, y1 , y2 , . . . , yn−1 ≤
hi txi − yi ,
3.1
i1
for almost every t ∈ 0, 1 and all x1 , x2 , . . . , xn−1 , y1 , y2 , . . . , yn−1 ∈ Rn−1 .
If
1
0<
n−1
G0 s, s hi s ds <
0
i1
β1
1
1−β γ
−1
,
3.2
then BVP 1.1 has a unique positive solution.
Proof. We will show that T is a contraction mapping. For any v1 , v2 ∈ P and 1 ≤ i ≤ n − 2, by
the definition of fractional integral, we obtain
t
1
i
i−1
i
v2 t t − s v1 s − v2 s ds
I0 v1 t − I0
i − 1! 0
1
≤
i − 1!
1
≤
i − 1!
t
t − si−1 |v1 s − v2 s| ds
0
t
t − si−1 dsv1 − v2 0
1 i
t v1 − v2 ≤ v1 − v2 .
i!
3.3
Abstract and Applied Analysis
9
So, for any v1 , v2 ∈ P , by 3.3 and Lemma 2.3, we have
1
n−2
n−3
1
vs, I0
vs, . . . , I0
vs, vs
|T v1 t − T v2 t| Gt, s f s, I0
0
n−2
n−3
1
−f s, I0 vs, I0 vs, . . . , I0 vs, vs ds
≤
1
0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs
−f
n−2
n−3
1
s, I0
vs, I0
vs, . . . , I0
vs, vs
ds
3.4
n−1
β1
n−1−i
n−1−i
≤
1
v1 s − I0
v2 s ds
G0 s, s hi sI0
1−β γ
0
i1
1
n−1
β1
≤ 1
G0 s, s hi s dsv1 − v2 .
1−β γ
0
i1
1
This implies that
T v1 − T v2 ≤ κv1 − v2 ,
3.5
1
where κ 1 β1/1 − βγ 0 G0 s, s n−1
i1 hi s ds ∈ 0, 1. By the Banach contraction
mapping principle, we deduce that T has a unique fixed point v∗ . Thus, by Lemma 2.4, u∗ t n−2 ∗
v t is a unique positive solution of BVP 1.1.
I0
Lemma 3.2. Assume that (H1 ) and (H2 ) hold and the following conditions are satisfied.
H3 There exist nonnegative real-valued functions q, p1 , p2 , . . . , pn−1 ∈ L0, 1 such that
ft, x1 , x2 , . . . , xn−1 ≤ qt n−1
pi t|xi |,
3.6
i1
for almost every t ∈ 0, 1 and all x1 , x2 , . . . , xn−1 ∈ Rn−1 .
Then T : P → P is a completely continuous operator.
Proof. For any v ∈ P , as Gt, s ≥ 0 for all t, s ∈ 0, 1, we have T vt ≥ 0, so T P ⊂ P . Let
D ⊂ P be any bounded set. Then there exists a constant L > 0 such that v ≤ L for any v ∈ D.
Moreover for any v ∈ D, s ∈ 0, 1, vs ≤ v ≤ L. Proceeding as for 3.3, we obtain
n−1−i
n−1−i
vs ≤ v ≤ L,
I0 vs I0
i 1, 2, . . . , n − 1.
3.7
10
Abstract and Applied Analysis
Thus,
1
|T vt| 0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds
n−1
β1
n−1−i
≤
1
G0 s, s qs pi sI0 vs ds
1−β γ
0
i1
1
n−1
β1
≤ 1
G0 s, s qs pi sv ds
1−β γ
0
i1
1
n−1
β1
≤ 1
L 1 G0 s, s qs pi s ds
1−β γ
0
i1
1
3.8
< ∞.
Therefore, T D is uniformly bounded.
Now we show that T D is equicontinuous on 0, 1. Since Gt, s is continuous on
0, 1 × 0, 1, Gt, s is uniformly continuous on 0, 1 × 0, 1. Hence, for any ε > 0, there exists
a constant δ0 > 0 such that for any s ∈ 0, 1, t, t ∈ 0, 1, when |t − t | < δ0 , it holds
−1
1
n−1
Gt, s − G t , s < 1 L 1
qs pi s ds ε.
0
3.9
i1
Consequently, for any t, t ∈ 0, 1 and |t − t | < δ0 , we have
T vt − T v t ≤
1
0
≤
1
Gt, s − G t , s f s, I n−2 vs, I n−3 vs, . . . , I 1 vs, vs ds
0
0
0
n−1
Gt, s − G t , s qs pi sv ds
0
≤ L 1
i1
1
0
3.10
n−1
Gt, s − G t , s qs pi s ds
i1
< ε.
This implies that T D is equicontinuous. Thus according to the Ascoli-Arzela Theorem, T D
is a relatively compact set.
In the end, we show that T : P → P is continuous. Assume that vm , v0 ∈ P m 1, 2, . . ., vm → v0 m → ∞, then
|vm t − v0 t| ≤ vm − v0 −→ 0,
3.11
Abstract and Applied Analysis
11
and vm ≤ L m 0, 1, 2, . . ., where L is a positive constant. Keeping in mind that f satisfies
Carathéodory conditions on 0, 1 × Rn−1 , we have
n−2
n−3
1
lim f t, I0
vm t, I0
vm t, . . . , I0
vm t, vm t
m → ∞
f
n−2
n−3
1
t, I0
v0 t, I0
v0 t, . . . , I0
v0 t, v0 t
,
3.12
for a.e. t ∈ 0, 1.
Proceeding as for 3.3, for m ∈ N we obtain
n−1−i
n−1−i
vm s ≤ vm ≤ L
I0 vm s I0
i 1, 2, . . . , n − 1.
3.13
This together with 3.6,
n−1
n−2
n−3
1
vm t, I0
vm t, . . . , I0
vm t, vm t ≤ qt L pi t.
0 ≤ f t, I0
3.14
i1
The Lebesgue dominated convergence theorem gives
1 n−2
n−3
1
vm s, I0
vm s, . . . , I0
vm s, vm s
lim
f s, I0
m → ∞
0
−f
n−2
n−3
1
s, I0
v0 s, I0
v0 s, . . . , I0
v0 s, v0 s
ds 0.
3.15
Now we deduce from 3.15, Lemma 2.3
|T vm t − T v0 t|
1
n−2
n−3
1
Gt, s f s, I0
vm s, I0
vm s, . . . , I0
vm s, vm s
0
−f
≤
β1
1
1−β γ
n−2
n−3
1
s, I0
v0 s, I0
v0 s, . . . , I0
v0 s, v0 s
1
0
ds
n−2
n−3
1
G0 s, sf s, I0
vm s, I0
vm s, . . . , I0
vm s, vm s
n−2
n−3
1
−f s, I0
v0 s, I0
v0 s, . . . , I0
v0 s, v0 s ds
≤
β1
1
1−β γ
max G0 s, s
s∈0,1
1 n−2
n−3
1
vm s, I0
vm s, . . . , I0
vm s, vm s
f s, I0
0
n−2
n−3
1
−f s, I0
v0 s, I0
v0 s, . . . , I0
v0 s, v0 s ds
3.16
12
Abstract and Applied Analysis
that T vm − T v0 → 0, as m → ∞. So T : P → P is continuous. Therefore T : P → P is
completely continuous.
Remark 3.3. If f : 0, 1 × Rn−1 is continuous, by similar argument as above, we can show that
T is completely continuous.
Theorem 3.4. Assume that (H1 )–(H3 ) hold. If
1
G0 s, s
0
n−1
pi s ds <
1
i1
β1
1 − βγ
−1
,
3.17
then BVP 1.1 has at least one positive solution.
Proof. Let
1
G0 s, sqsds
1 β1/ 1 − β γ
0
where r ,
1
1 − 1 β1/ 1 − β γ
G0 s, s n−1
i1 pi sds
0
3.18
Ω {v ∈ P : v < r},
we have Ω ⊂ P . From Lemma 3.2, we know that T : Ω → P is completely continuous. If there
exists v ∈ ∂Ω, μ ∈ 0, 1 such that
v μT v,
3.19
then by H3 and 3.19, we have
vt μT vt μ
≤μ
1
0
1
0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds
β1
≤μ 1
1−β γ
1
0
G0 s, sqs ds 1
0
3.20
n−1
G0 s, s pi sdsv ,
i1
which implies that
1
1
n−1
β1
G0 s, sqs ds r G0 s, s pi sds
v ≤ μ 1 1−β γ
0
0
i1
1
1
n−1
β1
G0 s, sqsds r G0 s, s pi sds r.
< 1
1−β γ
0
0
i1
3.21
This means that v ∈
/ ∂Ω. By Lemma 2.5, T has a fixed point v ∈ Ω. By Lemma 2.4, BVP 1.1
n−2
v t.
has at least one positive solution u
t I0
Abstract and Applied Analysis
13
Theorem 3.5. Assume that (H1 )–(H3 ) hold. If there exist two positive constants r1 < r2 such that
i ft, x1 , x2 , . . . , xn−1 ≤ L−1
2 r2 , for t, x1 , x2 , . . . , xn−1 ∈ 0, 1 × 0, r2 × · · · × 0, r2 ,
ii ft, x1 , x2 , . . . , xn−1 ≥ L−1
1 r1 , for t, x1 , x2 , . . . , xn−1 ∈ 0, 1 × 0, r1 × · · · × 0, r1 ,
where L1 , L2 are defined by 2.19, then BVP 1.1 has at least one positive solution.
Proof. Let Ω2 {v ∈ P : v < r2 }. For any v ∈ ∂Ω2 , we have v r2 and 0 ≤ vt ≤ r2 for
every t ∈ 0, 1. Similar to 3.7, for 0 ≤ vs ≤ r2 , we have
n−1−i
n−1−i
0 ≤ I0
vs I0
vs ≤ v ≤ r2 ,
i 1, 2, . . . , n − 1.
3.22
It follows from condition i that
n−2
n−3
1
vs, I0
vs, . . . , I0
vs, vs ≤ L−1
f s, I0
2 r2 ,
for s, v ∈ 0, 1 × 0, r2 .
3.23
Thus, for any v ∈ ∂Ω2 , by 3.23 and Lemma 2.3, we have
|T vt| 1
0
≤
1
0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds
Gt, sL−1
2 r2
r2 v,
ds ≤ 1 β1/ 1 − β γ L−1
2 r2
1
G0 s, sds
3.24
0
t ∈ 0, 1,
which means that
T v ≤ v,
v ∈ ∂Ω2 .
3.25
On the other hand, let Ω1 {v ∈ P : v < r1 }. For any v ∈ ∂Ω1 , we have v r1 and
0 ≤ vt ≤ r1 for every t ∈ 0, 1. Similar to 3.23, from condition ii, we can get
n−2
n−3
1
vs, I0
vs, . . . , I0
vs, vs ≥ L−1
f s, I0
1 r1 ,
for s, v ∈ 0, 1 × 0, r1 .
3.26
Hence for any t ∈ ξ, η, v ∈ ∂Ω1 , by 3.26 and Lemma 2.3 we have
|T vt| 1
0
≥
1
0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds
−1
Gt, sL−1
1 r1 ds ≥ L1 r1
≥ L−1
1 r1
η
ρsG0 s, sds
ξ
r1 v.
η
Gt, sds
ξ
3.27
14
Abstract and Applied Analysis
Thus we get
T v ≥ v,
3.28
v ∈ ∂Ω1 .
By 3.25, 3.28, and Lemma 2.6, T has a fixed point v ∈ Ω2 \ Ω1 such that r1 ≤ v ≤ r2 . By
n−2
v t.
Lemma 2.4, BVP 1.1 has at least one positive solution u
t I0
Theorem 3.6. Assume that (H1 )–(H3 ) hold. If there exist constants 0 < a < b < c such that
I ft, x1 , x2 , . . . , xn−1 < L−1
2 a, for t, x1 , x2 , . . . , xn−1 ∈ 0, 1 × 0, a × · · · × 0, a,
II ft, x1 , x2 , . . . , xn−1 ≤ L−1
2 c, for t, x1 , x2 , . . . , xn−1 ∈ 0, 1 × 0, c × · · · × 0, c,
n−2
, c ×
III ft, x1 , x2 , . . . , xn−1 ≥ L−1
1 b, for t, x1 , x2 , . . . , xn−1 ∈ ξ, η × b/n − 2!ξ
n−3
b/n − 3!ξ , c × · · · × bξ, c × b, c,
where L1 , L2 are defined by 2.19, then BVP 1.1 has at least three positive solutions u1 , u2 , and u3
satisfying
n−2 n−2 n−2
b < ϕ D0
u2 < D0
u2 ≤ c,
D0 u1 < a,
n−2 n−2
ϕ D0
a < D0
u3 ,
u3 < b.
3.29
Proof. We will show that all conditions of Lemma 2.7 are satisfied.
First, if v ∈ P c , then v ≤ c. So we have 0 ≤ vt ≤ c, t ∈ 0, 1. Similar to 3.23, it
follows from condition II that
n−2
n−3
1
f s, I0
vs, I0
vs, . . . , I0
vs, vs ≤ L−1
2 c,
for s, v ∈ 0, 1 × 0, c.
3.30
Thus, for any v ∈ P c , by 3.30, we have
|T vt| 1
0
≤
1
0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds
Gt, sL−1
2 c ds
≤
β1
1
1−β γ
L−1
2 c
1
G0 s, sds
3.31
0
c,
which means that T v ≤ c, v ∈ P c . Therefore, T : P c → P c . By Lemma 3.2, we know that
T : P c → P c is completely continuous.
Next, similar to 3.30 and 3.31, it follows from condition I that if v ∈ P a then
T v < a. So the condition C2 of Lemma 2.7 holds.
Abstract and Applied Analysis
15
Now, we take vt bc/2, t ∈ 0, 1. It is easy to see that vt bc/2 ∈ P ϕ, b, c,
and so
ϕv min|vt| ξ≤t≤η
bc
> b,
2
3.32
where ϕv is defined by 2.20. This proves that {v ∈ P ϕ, b, c : ϕv > b} / φ.
On the other hand, if v ∈ P ϕ, b, c, then b ≤ vt ≤ c, t ∈ ξ, η. By the definition of
fractional integral, for any t ∈ ξ, η, 1 ≤ i ≤ n − 2, we obtain
b i
b
ξ ≤
i!
i − 1!
c
≤
i − 1!
≤
t
0
t
i
vt t − si−1 ds ≤ I0
1
i − 1!
t
t − si−1 vsds
0
3.33
t − si−1 ds
0
c i
η ≤ c.
i!
It follows from 3.33 and condition III that
n−2
n−3
1
f s, I0
vs, I0
vs, . . . , I0
vs, vs ≥ L−1
1 b,
for s ∈ ξ, η , v ∈ P ϕ, b, c .
3.34
Hence we have
ϕT v min|T vt|
ξ≤t≤η
min
ξ≤t≤η
≥
1
0
n−2
n−3
1
Gt, sf s, I0
vs, I0
vs, . . . , I0
vs, vs ds
L−1
1 bmin
ξ≤t≤η
1
Gt, sds >
0
L−1
1 b
η
3.35
ρsG0 s, sds b,
ξ
which implies that ϕT v > b, for v ∈ P ϕ, b, c. This shows that condition C1 of Lemma 2.7
is also satisfied.
By Lemma 2.7 and Remark 2.8, BVP 2.13 has at least three positive solutions v1 , v2 ,
and v3 such that v1 < a, b < ϕv2 < v2 ≤ c, and a < v3 , ϕv3 < b. By Lemma 2.4,
n−2
vi t, i 1, 2, 3. By 2.3, we have
BVP 1.1 has at least three positive solutions ui t I0
n−2
n−2 n−2
D0 ui t D0 I0 vi t vi , i 1, 2, 3. So u1 , u2 , u3 are three positive solutions of BVP 1.1
satisfying
n−2 n−2 n−2
b < ϕ D0
u2 < D0
u2 ≤ c,
D0 u1 < a,
n−2 n−2
ϕ D0
a < D0
u3 ,
u3 < b.
The proof of Theorem 3.6 is completed.
3.36
16
Abstract and Applied Analysis
4. Examples
Example 4.1. Consider the following problem:
1
1
1 − t3 et u
t2 sin2 u tu 0, 0 < t < 1,
4
1 et 1 u 2
u 1 β u .
u0 u 0 u 0 0,
7/2
D0
ut 4.1
Let βu 1/2 u 1/2. Then
t1 − s1/2 − t − s1/2 ,
1
G0 t, s Γ3/2 t1 − s1/2 ,
1
1
, s ≥ 0,
Gs G0
2
2
β1 1
0
1
dAt ,
2
0 ≤ s ≤ t ≤ 1,
0 ≤ t ≤ s ≤ 1.
β γ 4.2
√
1
t
1/2
dAt 0
2
< 1.
4
4.3
Let
f t, x, y, z 1
1
1 − t3 et x
t2 sin2 y tz,
4
1 et 1 x 2
1 − t3 et
h1 t ,
1 et
1
h2 t t2 ,
2
4.4
1
h3 t t.
4
Then f is a nonnegative continuous function on 0, 1 × R 3 and, for any t, x1 , y1 , z1 and
t, x2 , y2 , z2 ∈ 0, 1 × R 3 , satisfies
f t, x1 , y1 , z1 − f t, x2 , y2 , z2 ≤ h1 t|x1 − x2 | h2 ty1 − y2 h3 t|z1 − z2 |.
4.5
So we have
1
0
G0 s, s
3
hi s ds ≤
i1
1
Γ3/2
1
s1 − s1/2 1 − s3 s2 s ds
0
33 √
B3/2, 9/2 B7/2, 3/2 B5/2, 3/2
π ≈ 0.4569608
Γ3/2
128
−1
√
β1
4− 2
< 1
√ ≈ 0.5638901,
1−β γ
6− 2
4.6
where B·, · denotes a Beta function. So all conditions of Theorem 3.1 are satisfied. Thus, by
Theorem 3.1, BVP 4.1 has at least one positive solution.
Abstract and Applied Analysis
17
Example 4.2. Consider the following problem:
5/2
D0
ut where βu 1
0
1
1
t − t2 ln1 u t2 u t3 sin t 0,
2
2
u0 u 0 0,
u 1 β u ,
0 < t < 1,
4.7
u sdAs with
⎧
⎪
⎪
⎪0,
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
As 2,
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩1,
1
,
s ∈ 0,
4
1 9
,
,
s∈
4 16
s∈
4.8
9
,1 .
16
Set
1
1
f t, x, y t − t2 ln1 x t2 y t3 sin t,
2
2
1
p2 t t2 ,
2
p1 t 1
t − t2 ,
2
4.9
qt t 1.
3
Then f : 0, 1 × 0, ∞ × 0, ∞ → 0, ∞ is continuous and
f t, x, y ≤ p1 tx p2 ty qt.
As in 21, βγ 1
0
1
0
4.10
1
γtdAt 2 1/4 −1 × 9/16 1/4 < 1, Gs 0 G0 t, sdAt ≥ 0,
1
G0 s, s p1 s p2 s ds ≤
Γ3/2
1
s1 − s1/2
s − s2 s2 ds
0
B5/2, 5/2 B7/2, 3/2 1 √
π ≈ 0.22155673
Γ3/2
8
−1
β1
3
< 1
≈ 0.42857143,
7
1−β γ
4.11
where B·, · denotes a Beta function and G0 t, s is defined by 4.2. So all conditions of
Theorem 3.4 are satisfied. Thus, by Theorem 3.4, BVP 4.7 has at least one positive solution.
18
Abstract and Applied Analysis
Example 4.3. Consider the following problem:
5/2
ut D0
where βu 1
0
tet u
t2
sin t 1
ln1 u 0,
t
5
20
2
10 10e
u0 u 0 0,
u 1 β u ,
0 < t < 1,
4.12
u sdAs with As as given by 4.8. Set
t2
tet y
sin t 1
f t, x, y ln1 x ,
t
5
20
2
10 10e
tet
p2 t ,
10 10et
p1 t t2
,
5
sin t 1
.
qt 20
2
4.13
Then f : 0, 1 × 0, ∞ × 0, ∞ → 0, ∞ is continuous and
f t, x, y ≤ p1 tx p2 ty qt.
4.14
1
1
By Example 4.2, βγ 0 γtdAt 1/4 < 1, Gs 0 G0 t, sdAt ≥ 0, where G0 t, s is
defined by 4.2. As in 1, 3, we also take ξ 1/4, η 3/4, then
L−1
1
3/4
≈ 13.6649,
ρsG0 s, s ds
1/4
L−1
2 −1
β1
1
1−β γ
−1 1
−1
G0 s, s ds
0
4.15
12
√ ≈ 0.967182.
7 π
Choosing r1 1/30, r2 1, we have
f t, x, y ≤ 0.85 ≤ L−1
2 r2 ,
f t, x, y ≥ 0.5 ≥ L−1
1 r1 ,
for t, x, y ∈ 0, 1 × 0, 1 × 0, 1,
1
1
for t, x, y ∈ 0, 1 × 0,
× 0,
.
30
30
4.16
So all conditions of Theorem 3.5 are satisfied. Thus, by Theorem 3.5, BVP 4.12 has at least
one positive solution.
Example 4.4. Consider the following problem:
5/2
D0
ut f t, ut, u t 0,
u0 u 0 0,
0 < t < 1,
u 1 β u ,
4.17
Abstract and Applied Analysis
where βu 1
0
19
u sdAs with As as given by 4.8. Set
⎧ 2
t
t
⎪
2
⎪
⎨ 100 ln1 x 15y 1000 ,
f t, x, y ⎪
29 1
t
⎪ t2
⎩
ln1 x y
,
100
2
2
1000
t2
,
100
p1 t t, x, y ∈ 0, 1 × 0, 1 × 0, 1,
t, x, y ∈ 0, 1 × R 2 \ 0, 12 , 4.18
p2 t 15,
qt 29
t
.
1000
2
Then
f t, x, y ≤ p1 tx p2 ty qt.
1
By Example 4.3, βγ L−1
1
0
γtdAt 1/4 < 1, Gs 3/4
0
G0 t, sdAt ≥ 0,
−1
≈ 13.6649,
ρsG0 s, sds
1/4
L−1
2 1
1
4.19
β1
1−β γ
−1 1
−1
G0 s, sds
0
4.20
12
√ ≈ 0.967182.
7 π
Choosing a 1/20, b 1, c 100, we have
1
1
f t, x, y ≤ 0.044845 <
× 0,
,
≈ 0.048359, for t, x, y ∈ 0, 1 × 0,
20
20
1
1 3
−1
f t, x, y ≥ 14.5025 ≥ L1 b ≈ 13.6649, for t, x, y ∈
,
× , 100 × 1, 100,
4 4
4
f t, x, y ≤ 65.501 ≤ L−1
for t, x, y ∈ 0, 1 × 0, 100 × 0, 100.
2 c ≈ 96.7182,
L−1
2 a
4.21
So all conditions of Theorem 3.6 are satisfied. Thus, by Theorem 3.6, BVP 4.17 has at least
three positive solutions u1 , u2 , and u3 , satisfying
1
,
maxu1 t <
0≤t≤1
20
1<
min u2 t < maxu2 t ≤ 100,
1/4≤t≤3/4
1
≤ maxu3 t ≤ 100
20 0≤t≤1
with
0≤t≤1
min u3 t < 1.
4.22
1/4≤t≤3/4
Acknowledgments
The first and second authors were supported financially by the National Natural Science
Foundation of China 11071141, 11126231 and the Natural Science Foundation of Shandong
Province of China ZR2011AQ008, ZR2010AM017. The third author was supported
financially by the Australia Research Council through an ARC Discovery Project grant.
20
Abstract and Applied Analysis
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