Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2010, Article ID 971824, 12 pages
doi:10.1155/2010/971824
Research Article
Upper and Lower Solutions Method for a Class of
Singular Fractional Boundary Value Problems with
p-Laplacian Operator
Jinhua Wang and Hongjun Xiang
Department of Mathematics, Xiangnan University, Chenzhou 423000, China
Correspondence should be addressed to Hongjun Xiang, hunxhjxhj67@126.com
Received 24 May 2010; Accepted 25 August 2010
Academic Editor: Paul Eloe
Copyright q 2010 J. Wang and H. Xiang. 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.
The upper and lower solutions method is used to study the p-Laplacian fractional boundary value
γ
α
α
ut ft, ut, 0 < t < 1, u0 0, u1 auξ, D0
u0 0, and
problem D0 φp D0
α
α
D0 u1 bD0 uη, where 1 < α, γ 2, 0 a, b 1, 0 < ξ, η < 1. Some new results on
the existence of at least one positive solution are obtained. It is valuable to point out that the
nonlinearity f can be singular at t 0, 1 or u 0.
1. Introduction
It is well know that the upper and lower solutions method is a powerful tool for proving
the existence results for boundary value problem. It has been used to deal with many
multipoint boundary value problem of integer ordinary differential equations see, e.g., 1–3
and references therein.
Recently, boundary value problems of nonlinear fractional differential equations have
aroused considerable attention. Many people pay attention to the existence and multiplicity
of solutions or positive solutions for boundary value problems of nonlinear fractional
differential equations by means of some fixed point theorems, such as the Krasnosel’skii
fixed-point theorem, the Leggett-Williams fixed-point theorem, and the Schauder fixed-point
theorem see 4–8. To the best of our knowledge, the upper and lower solutions method
is seldom considered in the literatures, and there are few papers devoted to investigate pLaplacian fractional boundary value problems.
In this paper, we deal with the following p-Laplacian fractional boundary value
problem:
α
γ ut ft, ut, 0 < t < 1,
D0 φp D0
1.1
α
α
α
u0 0,
u1 auξ,
D0
u0 0,
D0
u1 bD0
u η ,
2
Abstract and Applied Analysis
where 1 < α, γ 2, 0 a, b 1, 0 < ξ, η < 1, and Dα is the standard Riemann-Liouville
fractional differential operator of order α, φp s |s|p−2 s, p > 1, φp −1 φq , 1/p 1/q 1. By using upper and lower solutions method, the existence results of at least one positive
solution for the above fractional boundary value problem are established, and an example is
given to show the effectiveness of our results. It is valuable to point out that the nonlinearity
f can be singular at t 0, 1 or u 0.
The remaining part of the paper is organized as follows. In Section 2, we will present
some definitions and lemmas. In Section 3, some results are given. In Section 4, we present an
example to demonstrate our results.
2. Basic Definitions and Preliminaries
In this section, we present some necessary definitions and lemmas.
Definition 2.1 see 9. The integral
α
I0
yt 1
Γα
t
t − sα−1 ysds,
2.1
0
where α > 0, is called the Riemann-Liouville fractional integral of order α.
Definition 2.2 see 10. For a function yt given in the interval 0, ∞, the expression
α
D0
yt
n t
d
1
t − sn−α−1 ysds,
Γn − α dt
0
2.2
where n α 1, and α denotes the integer part of number α, is called the RiemannLiouville fractional derivative of order α.
Remark 2.3. From the definition of the Riemann-Liouville fractional derivative, we quote for
μ > −1, then
Γ 1μ
α μ
t D0
tμ−α .
Γ 1μ−α
2.3
α α−m
In particular, D0
t
0, m 1, 2, . . . , N, where N is the smallest integer greater than
or equal to α.
Lemma 2.4 see 4. Assume that u ∈ C0, 1 ∩ L0, 1 with a fractional derivative of order α > 0
that belongs to C0, 1 ∩ L0, 1. Then
α
α
I0
D0
ut ut C1 tα−1 C2 tα−2 · · · CN tα−N ,
for some Ci ∈ R, i 1, 2, . . . N, where N is the smallest integer greater than or equal to α.
2.4
Abstract and Applied Analysis
3
Lemma 2.5 see 6. Let y ∈ C0, 1 and 1 < α 2, the unique solution of
α
D0
ut yt 0,
u0 0,
0 < t < 1,
2.5
u1 auξ,
is
ut 1
2.6
Gt, sysds,
0
where
Gt, s ⎧
t1 − sα−1 − atα−1 ξ − sα−1 − t − sα−1 1 − aξα−1
⎪
⎪
⎪
,
⎪
⎪
⎪
1 − aξα−1 Γα
⎪
⎪
⎪
⎪
⎪
⎪
⎪ t1 − sα−1 − t − sα−1 1 − aξα−1 ⎪
⎪
⎪
⎪
,
⎪
⎪
1 − aξα−1 Γα
⎪
⎪
⎪
⎪
⎨
⎪
t1 − sα−1 − atα−1 ξ − sα−1
⎪
⎪
⎪
,
⎪
⎪
1 − aξα−1 Γα
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
t1 − sα−1
⎪
⎪
⎪
⎪
⎪ 1 − aξα−1 Γα ,
⎪
⎩
0 s t 1, s ξ,
0 < ξ s t 1,
2.7
0 t s ξ 1,
0 t s 1, ξ s.
Lemma 2.6. Let y ∈ C0, 1, 1 < α, γ 2, 0 < ξ, η < 1, 0 a, and b 1. The fractional boundary
value problem
α
γ ut yt,
D0 φp D0
u0 0,
u1 auξ,
α
u0
D0
0,
0 < t < 1,
α
D0
u1
η ,
2.8
α
bD0
u
has a unique solution
ut 1
1
Gt, sφq
0
Hs, τyτdτ ds,
0
2.9
4
Abstract and Applied Analysis
where
⎧
γ−1
− t − sγ−1 1 − b1 ηγ−1
t1 − sγ−1 − b1 tγ−1 η − s
⎪
⎪
⎪
,
⎪
⎪
⎪
1 − b1 ηγ−1 Γ γ
⎪
⎪
⎪
⎪
⎪
⎪
γ−1
γ−1 ⎪
1 − b1 ηγ−1
⎪ t1 − s − t − s
⎪
⎪
,
⎪
⎪
⎪
1 − b1 ηγ−1 Γ γ
⎪
⎪
⎨
Ht, s γ−1
⎪
t1 − sγ−1 − b1 tγ−1 η − s
⎪
⎪
⎪
,
⎪
⎪
1 − b1 ηγ−1 Γ γ
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
t1 − sγ−1
⎪
⎪
,
⎪
⎪
γ−1 Γ γ
⎪
⎩ 1 − b1 η
0 s t 1, s η,
0 < η s t 1,
0 t s η 1,
0 t s 1, η s,
2.10
b1 bp−1 and Gt, s is defined by 2.7.
Proof. At first, by Lemma 2.4, the 2.8 is equivalent to the integral equation
α
γ
ut I0 yt C1 tγ−1 C2 tγ−2 ,
φp D0
C1 , C1 ∈ R,
2.11
t − τγ−1
yτdτ C1 tγ−1 C2 tγ−2 .
Γ γ
2.12
that is,
α
φp D0
ut t
0
α
α
α
By the boundary conditions D0
u0 0, D0
u1 bD0
uη, we have
C2 0,
C1 −
1
0
1 − τγ−1
yτdτ Γ γ 1 − b1 ηγ−1
η
0
γ−1
b1 η − τ
yτdτ.
Γ γ 1 − b1 ηγ−1
2.13
Therefore, the solution ut of boundary value problem 2.8 satisfies
1 γ−1
t 1 − τγ−1
t − τγ−1
yτdτ −
yτdτ
Γ γ
1 − b1 ηγ−1
0
0 Γ γ
γ−1
η
b1 tγ−1 η − τ
yτdτ
1 − b1 ηγ−1
0 Γ γ
1
− Ht, τyτdτ.
α
φp D0
ut t
0
2.14
Abstract and Applied Analysis
5
1
α
Consequently, D0
utφq 0 Ht, τyτdτ 0. So, fractional boundary value problem 2.8
is equivalent to the following problem:
α
ut
D0
φq
1
Ht, τyτdτ
0,
0 < t < 1,
2.15
0
u0 0,
u1 auξ.
Lemma 2.5 implies that fractional boundary value problem 2.8 has a unique solution
ut 1
1
Gt, sφq
2.16
Hs, τyτdτ ds.
0
0
The proof is completed.
Lemma 2.7. Let 0 a, b 1, 0 < ξ, η < 1, 1 < α, and γ 2. The functions Gt, s and Ht, s
defined by 2.7 and 2.10, respectively, are continuous on 0, 1 × 0, 1 and satisfy
1 Gt, s 0, Ht, s 0, for t, s ∈ 0, 1;
2 Gt, s Gs, s, Ht, s Hs, s for t, s ∈ 0, 1;
3 Gt, s/G1, s tα−1 , for t, s ∈ 0, 1.
Proof. The proofs of part 1 and part 2 can be obtained in 6. Here, we only prove part 3.
If 0 < s t < 1, s ξ, then
α−1
α−1
α−1 1 − aξα−1
Gt, s t1 − s − atα−1 ξ − s − t − s
G1, s
aξα−1 1 − sα−1 − aξ − sα−1
tα−1 ·
1 − sα−1 − aξ − sα−1 − 1 − s/tα−1 1 − aξα−1
aξα−1 1 − sα−1 − aξ − sα−1
2.17
tα−1 .
If 0 < ξ s t < 1, then
α−1
α−1 1 − aξα−1
Gt, s t1 − s − t − s
G1, s
aξα−1 1 − sα−1
tα−1 ·
1 − sα−1 − 1 − s/tα−1 1 − aξα−1
aξα−1 1 − sα−1
2.18
tα−1 .
If 0 < t s ξ < 1, then
Gt, s t1 − sα−1 − atα−1 ξ − sα−1
G1, s
aξα−1 1 − sα−1 − aξ − sα−1
t
α−1
·
1 − sα−1 − aξ − sα−1
aξα−1 1 − sα−1 − aξ − sα−1
2.19
t
α−1
.
6
Abstract and Applied Analysis
If 0 < t s < 1, ξ s, then
Gt, s
t1 − sα−1
1 − sα−1
tα−1 ·
tα−1 .
α−1
α−1
α−1
G1, s aξα−1 1 − s
aξ 1 − s
2.20
The above inequalities imply that part 3 of Lemma 2.7 holds.
From Lemmas 2.5 and 2.7, it is easy to obtain the following lemma.
Lemma 2.8. Let 0 a 1, 0 < ξ < 1, 1 < α 2. If yt ∈ C0, 1, and yt 0, then fractional
boundary value problem 2.5 has a unique solution ut 0, t ∈ 0, 1.
α
u ∈ C2 0, 1}. Now one introduces the following definitions about the
Let E {u : u, φp D0
upper and lower solutions of fractional boundary value problem 1.1.
Definition 2.9. A function αt is called a lower solution of fractional boundary value problem
1.1, if αt ∈ E and αt satisfies
α
γ D0 φp D0
αt ft, αt,
0 < t < 1, 1 < α, γ 2,
α0 0, α1 aαξ,
α
α
α
α0 0, D0
α1 bD0
α η .
D0
2.21
Definition 2.10. A function βt is called an upper solution of fractional boundary value
problem 1.1, if βt ∈ E and βt satisfies
α
γ D0 φp D0
βt f t, βt ,
β0 0,
α
D0
β0 0,
0 < t < 1, 1 < α, γ 2,
β1 aβξ,
α
α
D0
β1 bD0
β η .
2.22
3. Main Result
In this section, our objective is to give and prove our main results.
For the sake of simplicity, we assume that
H1 ft, u ∈ C0, 1 × 0, ∞, 0, ∞ and ft, u is nonincreasing relative to u;
1
H2 For any constant μ > 0, 0 < 0 Hs, sfs, μsα−1 ds < ∞;
H3 There exist a continuous function mt and a positive number λ such that mt λtα−1 , t ∈ 0, 1, and
1
1
Gt, sφq
0
Hs, τfτ, mτdτ ds nt mt,
0
1
1
Gt, sφq
0
0
Hs, τfτ, nτdτ ds mt.
3.1
Abstract and Applied Analysis
7
We define P : {u ∈ C0, 1 : there exists a number λu > 0 such that ut λu tα−1 , t ∈
0, 1}. And define an operator T by
T ut 1
1
Gt, sφq
∀u ∈ P.
Hs, τfτ, uτdτ ds,
0
3.2
0
Theorem 3.1. Suppose that conditions H1 ∼ H3 are satisfied, then the boundary value problem
1.1 has at least one positive solution ρt, which satisfies ρt μtα−1 for some positive number μ.
Proof. We will divide our proof into four steps.
Step 1. We prove that T P ⊆ P .
Firstly, from Lemma 2.7 and conditions H1 ∼H2 , for any u ∈ P , there exists λu > 0,
such that
1
Hs, τfτ, uτdτ 1
0
Hτ, τf τ, λu τ α−1 dτ < ∞.
3.3
0
Thus,
T ut 1
1
Gt, sφq
0
1
Hs, τfτ, uτdτ ds
0
Gs, sds · φq
1
Hτ, τf τ, λu τ
0
α−1
dτ
3.4
0
< ∞.
Secondly, it follows from Lemma 2.7 that
T ut 1
0
1
Gt, s
G1, sφq
Hs, τfτ, uτdτ ds
G1, s
0
1
1
tα−1
G1, sφq
0
λT u tα−1 ,
Hs, τfτ, uτdτ ds
3.5
0
∀t ∈ 0, 1.
Consequently, T is well defined and T P ⊆ P.
In the meanwhile, by direct computations, we can obtain
α
γ D0 φp D0
T ut ft, ut,
T u0 0,
α
D0
T u0 0,
0 < t < 1, 1 < α, γ 2,
T u1 aT uξ,
α
α
D0
T u1 bD0
T u η .
3.6
3.7
8
Abstract and Applied Analysis
Step 2. We will prove that the functions αt T nt, βt T mt are lower and upper
solutions of fractional boundary value problem 1.1, respectively.
From H1 , we know that T is nonincreasing relative to u. Combining H3 , we have
mt nt T mt,
T nt nt T mt,
t ∈ 0, 1.
3.8
Therefore, αt βt.
By Step 1, we know αt, βt ∈ P. And by 3.6∼3.8, we obtain
α
α
γ γ D0 φp D0
αt − ft, αt D0 φp D0
T nt − ft, nt 0,
α0 0,
α
α0 0,
D0
α1 aαξ,
α
α
D0
α1 bD0
α η ,
α α
γ γ β t − f t, βt D0 φp D0
D0 φp D0
T mt − ft, mt 0,
β0 0,
β1 aβξ,
α
D0
β0 0,
3.9
α
α
D0
β1 bD0
β η ,
that is, αt and βt are lower and upper solutions of fractional boundary value problem
1.1, respectively.
Step 3. We will show that the fractional boundary value problem
α
γ D0 φp D0
ut gt, ut,
u0 0,
u1 auξ,
0 < t < 1, 1 < α, γ 2,
α
u0 0,
D0
α
α
D0
u1 bD0
u η ,
3.10
has a positive solution, where
⎧
⎪
ft, αt if ut < αt,
⎪
⎪
⎨
gt, ut ft, ut if αt ut βt,
⎪
⎪
⎪
⎩ f t, βt if ut > βt.
3.11
Thus, we consider the operator A : C0, 1 → C0, 1 defined as follows:
Aut 1
1
Gt, sφq
0
Hs, τgτ, uτdτ ds,
3.12
0
where Gt, s is defined as 2.7, Ht, s is defined as 2.10. It is clear that Au 0, for all u ∈
P , and a fixed point of the operator A is a solution of the fractional boundary value problem
3.10.
Abstract and Applied Analysis
9
Since αt ∈ P , there exists a positive number λα such that αt λα tα−1 , t ∈ 0, 1. It
follows from H2 that
1
Hτ, τgτ, ατdτ 0
1
Hτ, τfτ, ατdτ
0
1
3.13
Hτ, τf τ, λα τ α−1 dτ < ∞.
0
Consequently, for all ut ∈ C0, 1, we have
Aut 1
1
Gt, sφq
0
Hs, τgτ, uτdτ ds
0
1
1
Gs, sφq
0
1
3.14
Hτ, τgτ, uτdτ ds
0
Gs, sds · φq
1
Hτ, τfτ, ατdτ
0
< ∞,
0
which implies that the operator A is uniformly bounded.
On the other hand, since Gt, s is continuous on 0, 1 × 0, 1, it is uniformly
continuous on 0, 1 × 0, 1. So, for fixed s ∈ 0, 1 and for any ε > 0, there exists a constant
δ > 0, such that any t1 , t2 ∈ 0, 1 and |t1 − t2 | < δ,
|Gt1 , s − Gt2 , s| <
φ
ε
1
τf τ, λα τ α−1 dτ
0 τ,
3.15
.
Then, for all ut ∈ C0, 1,
|Aut2 − Aut1 | 1
0
1
1
1
0
|Gt2 , s − Gt1 , s|φq
Hτ, τgτ, uτdτ ds
0
0
|Gt2 , s − Gt1 , s|φq
1
Hτ, τfτ, ατdτ ds
0
|Gt2 , s − Gt1 , s|ds · φq
1
Hτ, τf τ, λα τ α−1 dτ
3.16
0
< ε,
that is to say, A is equicontinuous. Thus, from the Arzela-Ascoli Theorem, we know that A is
a compact operator. By the Schauder’s fixed-point theorem, the operator A has a fixed point;
that is, the fractional boundary value problem 3.10 has a positive solution.
10
Abstract and Applied Analysis
Step 4. We will prove that fractional boundary value problem 1.1 has at least one positive
solution.
Suppose that ρt is a solution of 3.10, we only need to prove that αt ρt βt, t ∈ 0, 1.
Let ρt be a solution of 3.10. We have
ρ0 0,
ρ1 aρξ,
α
α
D0
ρ1 bD0
ρ η .
α
ρ0 0,
D0
3.17
In addition, the function ft, u is nonincreasing in u, we know that
f t, βt g t, ρt ft, αt,
t ∈ 0, 1.
3.18
t ∈ 0, 1.
3.19
It follows from 3.8 and H3 that
ft, nt g t, ρt ≤ ft, mt,
By 3.6 and mt ∈ P, we obtain
α
α
γ γ D0 φp D0
βt D0 φp D0
T mt ft, mt,
t ∈ 0, 1.
3.20
Together with 3.7, 3.17−3.20, we obtain
α
α
γ γ βt − D0 φp D0
ρt ft, mt − g t, ρt 0,
D0 φp D0
β − ρ 0 0,
β − ρ 1 a β − ρ ξ,
α
α
α
β − ρ 0 0,
D0
β − ρ 1 bD0
β−ρ η .
D0
t ∈ 0, 1,
3.21
α
α
Let zt φp D0
βt − φp D0
ρt, we obtain
α
α
γ γ D0 φp D0
βt − D0 φp D0
ρt f t, βt − g t, ρt 0,
t ∈ 0, 1
3.22
and z0 0, z1 − φp bzη 0.
By Lemma 2.8, we know zt 0, t ∈ 0, 1, which implies that
α
α
βt φp D0
ρt ,
φp D0
t ∈ 0, 1.
3.23
α
α
α
Since φp is monotone increasing, we have D0
βt D0
ρt, that is, D0
β − ρt 0. By
Lemma 2.8 and 3.21, we have β − ρt 0. Therefore, βt ρt, t ∈ 0, 1.
Abstract and Applied Analysis
11
In the similar way, we can prove that αt ρt, t ∈ 0, 1. Consequently, ρt is a
positive solution of fractional boundary value problem 1.1. This completes the proof.
Theorem 3.2. If ft, u ∈ C0, 1 × 0, ∞, 0, ∞ is nonincreasing relative to u and ft, u
does not vanish identically for any u > 0, then the fractional boundary value problem 1.1 has at least
one positive solution ρt, which satisfies ρt μtα−1 for some positive number μ.
The proof is similar to Theorem 3.1, we omit it here.
4. Example
Example 4.1. As an example, we consider the fractional boundary value problem
1
4/3
3/2
φp D0
D0
ut t2 √ ,
u
1
1
3/2
,
D0
u1 u
u0 0,
3
2
u0 0,
0 < t < 1,
3/2
D0
u1
1 3/2
2
.
D0
u
2
3
4.1
Proof. It is clear that H1 holds. For any λ > 0,
1
Hτ, τf τ, λτ
3/2−1
dτ 0
1
0
Hτ, τ τ √
2
1
λτ 1/4
dτ < ∞,
4.2
which implies that H2 holds.
√
√
For 0 < r < 1, ft, u t2 1/ u t2 1/ ru r −1/2 ft, u. Let at t1/2 , by
3.5, we have
bt : T at 1
1
Gt, sφq
0
Hs, τfτ, aτdτ ds ∈ P,
4.3
0
and T bt T 2 at ∈ P, that is, there exist positive numbers λ1 , λ2 , such that T at λ1 at,
T 2 at λ2 at.
Choose a positive number λ0 {1, λ1 , λ4/3
2 } and combining the monotonicity of T , we
have
T λ0 at T at λ1 at λ0 at,
1/4
2
T 2 λ0 at λ1/4
0 T at λ0 λ2 at λ0 at.
4.4
12
Abstract and Applied Analysis
Taking mt λ0 t1/2 , then,
nt T mt 1
1
Gt, sφq
0
Hs, τf τ, λ0 τ
1/2
dτ ds
0
λ0 t1/2 mt,
1
1
T nt T 2 mt Gt, sφq
0
Hs, τf τ, T λ0 τ 1/2
4.5
dτ ds
0
λ0 t1/2 mt,
that is to say, the condition H3 holds. Theorem 3.1 implies that the fractional boundary
value problem 4.1 has at least one positive solution.
Acknowledgments
This work was jointly supported by Science Foundation of Hunan Provincial under Grants
2009JT3042, 2010GK3008 and 10JJ6007, the Construct Program of the Key Discipline in Hunan
Province, Aid Program for Science and Technology Innovative Research Team in Higher
Educational Institutions of Hunan Province.
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