The upper and lower solutions method is used to study the p-Laplacian fractional boundary value problem D0+γ(ϕp(D0+αu(t)))=f(t,u(t)), 0<t<1, u(0)=0, u(1)=au(ξ), D0+αu(0)=0, and D0+αu(1)=bD0+αu(η), 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 p-Laplacian fractional boundary value problems.

In this paper, we deal with the following p-Laplacian fractional boundary value problem: D0+γ(ϕp(D0+αu(t)))=f(t,u(t)),0<t<1,u(0)=0,u(1)=au(ξ),D0+αu(0)=0,D0+αu(1)=bD0+αu(η),
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-2s, 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 [<xref ref-type="bibr" rid="B4">9</xref>]).

The integral
I0+αy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds,
where α>0, is called the Riemann-Liouville fractional integral of order α.

Definition 2.2 (see [<xref ref-type="bibr" rid="B5">10</xref>]).

For a function y(t) given in the interval [0,∞), the expression
D0+αy(t)=1Γ(n-α)(ddt)n∫0t(t-s)n-α-1y(s)ds,
where n=[α]+1, and [α] denotes the integer part of number α, is called the Riemann-Liouville fractional derivative of order α.

Remark 2.3.

From the definition of the Riemann-Liouville fractional derivative, we quote for μ>-1, then
D0+αtμ=Γ(1+μ)Γ(1+μ-α)tμ-α.

In particular, D0+αtα-m=0,m=1,2,…,N, where N is the smallest integer greater than or equal to α.

Lemma 2.4 (see [<xref ref-type="bibr" rid="B6">4</xref>]).

Assume that u∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L(0,1). Then
I0+αD0+αu(t)=u(t)+C1tα-1+C2tα-2+⋯+CNtα-N,
for some Ci∈R,i=1,2,…N, where N is the smallest integer greater than or equal to α.

Lemma 2.5 (see [<xref ref-type="bibr" rid="B8">6</xref>]).

Let y∈C[0,1] and 1<α⩽2, the unique solution of
D0+αu(t)+y(t)=0,0<t<1,u(0)=0,u(1)=au(ξ),
is
u(t)=∫01G(t,s)y(s)ds,
where
G(t,s)={[t(1-s)]α-1-atα-1(ξ-s)α-1-(t-s)α-1(1-aξα-1)(1-aξα-1)Γ(α),0⩽s⩽t⩽1,s⩽ξ,[t(1-s)]α-1-(t-s)α-1(1-aξα-1)(1-aξα-1)Γ(α),0<ξ⩽s⩽t⩽1,[t(1-s)]α-1-atα-1(ξ-s)α-1(1-aξα-1)Γ(α),0⩽t⩽s⩽ξ⩽1,[t(1-s)]α-1(1-aξα-1)Γ(α),0⩽t⩽s⩽1,ξ⩽s.

Lemma 2.6.

Let y∈C[0,1], 1<α,γ⩽2, 0<ξ,η<1, 0⩽a,andb⩽1. The fractional boundary value problem
D0+γ(ϕp(D0+αu(t)))=y(t),0<t<1,u(0)=0,u(1)=au(ξ),D0+αu(0)=0,D0+αu(1)=bD0+αu(η),
has a unique solution
u(t)=∫01G(t,s)ϕq(∫01H(s,τ)y(τ)dτ)ds,
where
H(t,s)={[t(1-s)]γ-1-b1tγ-1(η-s)γ-1-(t-s)γ-1(1-b1ηγ-1)(1-b1ηγ-1)Γ(γ),0⩽s⩽t⩽1,s⩽η,[t(1-s)]γ-1-(t-s)γ-1(1-b1ηγ-1)(1-b1ηγ-1)Γ(γ),0<η⩽s⩽t⩽1,[t(1-s)]γ-1-b1tγ-1(η-s)γ-1(1-b1ηγ-1)Γ(γ),0⩽t⩽s⩽η⩽1,[t(1-s)]γ-1(1-b1ηγ-1)Γ(γ),0⩽t⩽s⩽1,η⩽s,b1=bp-1 and G(t,s) is defined by (2.7).

Proof.

At first, by Lemma 2.4, the (2.8) is equivalent to the integral equation
ϕp(D0+αu(t))=I0+γy(t)+C1tγ-1+C2tγ-2,C1,C1∈R,
that is,
ϕp(D0+αu(t))=∫0t(t-τ)γ-1Γ(γ)y(τ)dτ+C1tγ-1+C2tγ-2.
By the boundary conditions D0+αu(0)=0,D0+αu(1)=bD0+αu(η), we have
C2=0,C1=-∫01(1-τ)γ-1Γ(γ)(1-b1ηγ-1)y(τ)dτ+∫0ηb1(η-τ)γ-1Γ(γ)(1-b1ηγ-1)y(τ)dτ.
Therefore, the solution u(t) of boundary value problem (2.8) satisfies
ϕp(D0+αu(t))=∫0t(t-τ)γ-1Γ(γ)y(τ)dτ-∫01tγ-1(1-τ)γ-1Γ(γ)(1-b1ηγ-1)y(τ)dτ+∫0ηb1tγ-1(η-τ)γ-1Γ(γ)(1-b1ηγ-1)y(τ)dτ=-∫01H(t,τ)y(τ)dτ.
Consequently, D0+αu(t)+ϕq(∫01H(t,τ)y(τ)dτ)=0. So, fractional boundary value problem (2.8) is equivalent to the following problem:
D0+αu(t)+ϕq(∫01H(t,τ)y(τ)dτ)=0,0<t<1,u(0)=0,u(1)=au(ξ).
Lemma 2.5 implies that fractional boundary value problem (2.8) has a unique solution
u(t)=∫01G(t,s)ϕq(∫01H(s,τ)y(τ)dτ)ds.
The proof is completed.

Lemma 2.7.

Let 0⩽a,b⩽1, 0<ξ,η<1, 1<α,andγ⩽2. The functions G(t,s) and H(t,s) defined by (2.7) and (2.10), respectively, are continuous on [0,1]×[0,1] and satisfy

G(t,s)⩾0,H(t,s)⩾0,fort,s∈[0,1];

G(t,s)⩽G(s,s),H(t,s)⩽H(s,s)fort,s∈[0,1];

G(t,s)/G(1,s)⩾tα-1,fort,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
G(t,s)G(1,s)=[t(1-s)]α-1-atα-1(ξ-s)α-1-(t-s)α-1(1-aξα-1)aξα-1(1-s)α-1-a(ξ-s)α-1=tα-1·(1-s)α-1-a(ξ-s)α-1-(1-(s/t))α-1(1-aξα-1)aξα-1(1-s)α-1-a(ξ-s)α-1⩾tα-1.

If 0<ξ⩽s⩽t<1, then
G(t,s)G(1,s)=[t(1-s)]α-1-(t-s)α-1(1-aξα-1)aξα-1(1-s)α-1=tα-1·(1-s)α-1-(1-(s/t))α-1(1-aξα-1)aξα-1(1-s)α-1⩾tα-1.

If 0<t⩽s⩽ξ<1, then
G(t,s)G(1,s)=[t(1-s)]α-1-atα-1(ξ-s)α-1aξα-1(1-s)α-1-a(ξ-s)α-1=tα-1·(1-s)α-1-a(ξ-s)α-1aξα-1(1-s)α-1-a(ξ-s)α-1⩾tα-1.

If 0<t⩽s<1,ξ⩽s, then
G(t,s)G(1,s)=[t(1-s)]α-1aξα-1(1-s)α-1=tα-1·(1-s)α-1aξα-1(1-s)α-1⩾tα-1.
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 y(t)∈C[0,1], and y(t)⩾0, then fractional boundary value problem (2.5) has a unique solution u(t)⩾0,t∈[0,1].

Let E={u:u,ϕp(D0+αu)∈C2[0,1]}. Now one introduces the following definitions about the upper and lower solutions of fractional boundary value problem (1.1).

Definition 2.9.

A function α(t) is called a lower solution of fractional boundary value problem (1.1), if α(t)∈E and α(t) satisfies
D0+γ(ϕp(D0+αα(t)))⩽f(t,α(t)),0<t<1,1<α,γ⩽2,α(0)⩽0,α(1)⩽aα(ξ),D0+αα(0)⩾0,D0+αα(1)⩾bD0+αα(η).

Definition 2.10.

A function β(t) is called an upper solution of fractional boundary value problem (1.1), if β(t)∈E and β(t) satisfies
D0+γ(ϕp(D0+αβ(t)))⩾f(t,β(t)),0<t<1,1<α,γ⩽2,β(0)⩾0,β(1)⩾aβ(ξ),D0+αβ(0)⩽0,D0+αβ(1)⩽bD0+αβ(η).

3. Main Result

In this section, our objective is to give and prove our main results.

For the sake of simplicity, we assume that

f(t,u)∈C[(0,1)×(0,+∞),[0,+∞)] and f(t,u) is nonincreasing relative to u;

For any constant μ>0, 0<∫01H(s,s)f(s,μsα-1)ds<+∞;

There exist a continuous function m(t) and a positive number λ such that m(t)⩾λtα-1,t∈[0,1], and

We define P=:{u∈C[0,1]:thereexistsanumberλu>0suchthatu(t)⩾λutα-1,t∈[0,1]}. And define an operator T by Tu(t)=∫01G(t,s)ϕq(∫01H(s,τ)f(τ,u(τ))dτ)ds,∀u∈P.

Theorem 3.1.

Suppose that conditions (H1)~(H3) are satisfied, then the boundary value problem (1.1) has at least one positive solution ρ(t), which satisfies ρ(t)⩾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
∫01H(s,τ)f(τ,u(τ))dτ⩽∫01H(τ,τ)f(τ,λuτα-1)dτ<+∞.
Thus,
Tu(t)=∫01G(t,s)ϕq(∫01H(s,τ)f(τ,u(τ))dτ)ds⩽∫01G(s,s)ds·ϕq(∫01H(τ,τ)f(τ,λuτα-1)dτ)<+∞.

Secondly, it follows from Lemma 2.7 that
Tu(t)=∫01G(t,s)G(1,s)G(1,s)ϕq(∫01H(s,τ)f(τ,u(τ))dτ)ds⩾tα-1∫01G(1,s)ϕq(∫01H(s,τ)f(τ,u(τ))dτ)ds=λTutα-1,∀t∈[0,1].
Consequently, T is well defined and T(P)⊆P.

In the meanwhile, by direct computations, we can obtain
D0+γ(ϕp(D0+α(Tu)(t)))=f(t,u(t)),0<t<1,1<α,γ⩽2,(Tu)(0)=0,(Tu)(1)=a(Tu)(ξ),D0+α(Tu)(0)=0,D0+α(Tu)(1)=bD0+α(Tu)(η).

Step 2.

We will prove that the functions α(t)=Tn(t), β(t)=Tm(t) 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
m(t)⩽n(t)=Tm(t),Tn(t)⩽n(t)=Tm(t),t∈[0,1].
Therefore, α(t)⩽β(t).

By Step 1, we know α(t),β(t)∈P. And by (3.6)~(3.8), we obtain
D0+γ(ϕp(D0+α(α)(t)))-f(t,α(t))⩽D0+γ(ϕp(D0+α(Tn)(t)))-f(t,n(t))=0,α(0)=0,α(1)=aα(ξ),D0+αα(0)=0,D0+αα(1)=bD0+αα(η),D0+γ(ϕp(D0+α(β)(t)))-f(t,β(t))⩾D0+γ(ϕp(D0+α(Tm)(t)))-f(t,m(t))=0,β(0)=0,β(1)=aβ(ξ),D0+αβ(0)=0,D0+αβ(1)=bD0+αβ(η),
that is, α(t) and β(t) are lower and upper solutions of fractional boundary value problem (1.1), respectively.

Step 3.

We will show that the fractional boundary value problem
D0+γ(ϕp(D0+αu(t)))=g(t,u(t)),0<t<1,1<α,γ⩽2,u(0)=0,u(1)=au(ξ),D0+αu(0)=0,D0+αu(1)=bD0+αu(η),
has a positive solution, where
g(t,u(t))={f(t,α(t))ifu(t)<α(t),f(t,u(t))ifα(t)⩽u(t)⩽β(t),f(t,β(t))ifu(t)>β(t).
Thus, we consider the operator A:C[0,1]→C[0,1] defined as follows:
Au(t)=∫01G(t,s)ϕq(∫01H(s,τ)g(τ,u(τ))dτ)ds,
where G(t,s) is defined as (2.7), H(t,s) is defined as (2.10). It is clear that Au⩾0,forallu∈P, and a fixed point of the operator A is a solution of the fractional boundary value problem (3.10).

Since α(t)∈P, there exists a positive number λα such that α(t)⩾λαtα-1,t∈[0,1]. It follows from (H2) that
∫01H(τ,τ)g(τ,α(τ))dτ⩽∫01H(τ,τ)f(τ,α(τ))dτ⩽∫01H(τ,τ)f(τ,λατα-1)dτ<+∞.
Consequently, for all u(t)∈C[0,1], we have
Au(t)=∫01G(t,s)ϕq(∫01H(s,τ)g(τ,u(τ))dτ)ds⩽∫01G(s,s)ϕq(∫01H(τ,τ)g(τ,u(τ))dτ)ds⩽∫01G(s,s)ds·ϕq(∫01H(τ,τ)f(τ,α(τ))dτ)<+∞,
which implies that the operator A is uniformly bounded.

On the other hand, since G(t,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 ɛ>0, there exists a constant δ>0, such that any t1,t2∈[0,1] and |t1-t2|<δ,|G(t1,s)-G(t2,s)|<εϕ(∫01(τ,τ)f(τ,λατα-1)dτ).
Then, for all u(t)∈C[0,1],
|Au(t2)-Au(t1)|⩽∫01|G(t2,s)-G(t1,s)|ϕq(∫01H(τ,τ)g(τ,u(τ))dτ)ds⩽∫01|G(t2,s)-G(t1,s)|ϕq(∫01H(τ,τ)f(τ,α(τ))dτ)ds⩽∫01|G(t2,s)-G(t1,s)|ds·ϕq(∫01H(τ,τ)f(τ,λατα-1)dτ)<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.

Step 4.

We will prove that fractional boundary value problem (1.1) has at least one positive solution.

Suppose that ρ(t) is a solution of (3.10), we only need to prove that α(t)⩽ρ(t)⩽β(t),t∈[0,1].

Let ρ(t) be a solution of (3.10). We have
ρ(0)=0,ρ(1)=aρ(ξ),D0+αρ(0)=0,D0+αρ(1)=bD0+αρ(η).
In addition, the function f(t,u) is nonincreasing in u, we know that
f(t,β(t))⩽g(t,ρ(t))⩽f(t,α(t)),t∈[0,1].
It follows from (3.8) and (H3) that
f(t,n(t))⩽g(t,ρ(t))≤f(t,m(t)),t∈[0,1].
By (3.6) and m(t)∈P, we obtain
D0+γ(ϕp(D0+αβ(t)))=D0+γ(ϕp(D0+α(Tm)(t)))=f(t,m(t)),t∈[0,1].
Together with (3.7), (3.17)–(3.20), we obtain
D0+γ(ϕp(D0+αβ(t)))-D0+γ(ϕp(D0+αρ(t)))=f(t,m(t))-g(t,ρ(t))⩾0,t∈[0,1],(β-ρ)(0)=0,(β-ρ)(1)=a(β-ρ)(ξ),D0+α(β-ρ)(0)=0,D0+α(β-ρ)(1)=bD0+α(β-ρ)(η).
Let z(t)=ϕp(D0+αβ(t))-ϕp(D0+αρ(t)), we obtain
D0+γ(ϕp(D0+αβ(t)))-D0+γ(ϕp(D0+αρ(t)))⩾f(t,β(t))-g(t,ρ(t))⩾0,t∈[0,1]
and z(0)=0,z(1)-ϕp(b)z(η)=0.

By Lemma 2.8, we know z(t)⩽0,t∈[0,1], which implies that
ϕp(D0+αβ(t))⩽ϕp(D0+αρ(t)),t∈[0,1].
Since ϕp is monotone increasing, we have D0+αβ(t)⩽D0+αρ(t), that is, D0+α(β-ρ)(t)⩽0. By Lemma 2.8 and (3.21), we have (β-ρ)(t)⩾0. Therefore, β(t)⩾ρ(t),t∈[0,1].

In the similar way, we can prove that α(t)⩽ρ(t),t∈[0,1]. Consequently, ρ(t) is a positive solution of fractional boundary value problem (1.1). This completes the proof.

Theorem 3.2.

If f(t,u)∈C([0,1]×[0,+∞),[0,+∞)) is nonincreasing relative to u and f(t,u) does not vanish identically for any u>0, then the fractional boundary value problem (1.1) has at least one positive solution ρ(t), which satisfies ρ(t)⩾μtα-1 for some positive number μ.

The proof is similar to Theorem 3.1, we omit it here.

4. ExampleExample 4.1.

As an example, we consider the fractional boundary value problem
D0+4/3(ϕp(D0+3/2u(t)))=t2+1u,0<t<1,u(0)=0,u(1)=13u(12),D0+3/2u(0)=0,D0+3/2u(1)=12D0+3/2u(23).

Proof.

It is clear that (H1) holds. For any λ>0,
∫01H(τ,τ)f(τ,λτ((3/2)-1))dτ=∫01H(τ,τ)(τ2+1λτ1/4)dτ<+∞,
which implies that (H2) holds.

For 0<r<1, f(t,u)=t2+(1/√u)⩽t2+(1/√ru)⩽r-1/2f(t,u). Let a(t)=t1/2, by (3.5), we have
b(t)=:Ta(t)=∫01G(t,s)ϕq(∫01H(s,τ)f(τ,a(τ))dτ)ds∈P,
and Tb(t)=T2a(t)∈P, that is, there exist positive numbers λ1, λ2, such that Ta(t)⩾λ1a(t), T2a(t)⩾λ2a(t).

Choose a positive number λ0⩽{1,λ1,λ24/3} and combining the monotonicity of T, we have
T(λ0a(t))⩾Ta(t)⩾λ1a(t)⩾λ0a(t),T2(λ0a(t))⩾λ01/4T2a(t)⩾λ01/4λ2a(t)⩾λ0a(t).
Taking m(t)=λ0t1/2, then,
n(t)=Tm(t)=∫01G(t,s)ϕq(∫01H(s,τ)f(τ,λ0τ1/2)dτ)ds⩾λ0t1/2=m(t),Tn(t)=T2m(t)=∫01G(t,s)ϕq(∫01H(s,τ)f(τ,T(λ0τ1/2))dτ)ds⩾λ0t1/2=m(t),
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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