# Positive Solutions to a Fractional-Order Two-Point Boundary Value Problem withp-Laplacian Operator

Authors: Xiangshan Kong; Haitao Li
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

This paper systematically investigates positive solutions to a kind of two-point boundary value
problem (BVP) for nonlinear fractional differential equations withp-Laplacian operator and presents
a number of new results. First, the considered BVP is converted to an operator equation by using the
property of the Caputo derivative. Second, based on the operator equation and some fixed point theorems,
several sufficient conditions are presented for the nonexistence, the uniqueness, and the multiplicity
of positive solutions. Finally, several illustrative examples are given to support the obtained new
results. The study of illustrative examples shows that the obtained results are effective.


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## Body

## 1. Introduction

Fractional differential equation has recently attracted many scholars’ interest due to its wide applications [1–3] in engineering, technology, biology, chemical process, and so on. The first issue for the theory of fractional differential equations is the existence of solutions to kinds of boundary value problems (BVPs), which has been studied recently by many scholars, and lots of excellent results have been obtained [4–17] by means of fixed point theorems, upper and lower solutions technique, and so forth.As an important branch of BVPs,p-Laplacian equation was firstly introduced in [18] to model the following turbulent flow in a porous medium:
(1)(φp(u′(t)))′=f(t,u(t),u′(t)),
where Φp(s)=|s|p-2s, p>1, Φp-1=Φq, and 1/p+1/q=1. Then, it was investigated in both integer-order BVPs [19, 20] and fractional BVPs [21–25]. In [21], Chen and Liu considered the antiperiodic boundary value problem of fractional differential equation with p-Laplacian operator and obtained the existence of one solution by using Schaefer’s fixed point theorem under certain nonlinear growth conditions. Han et al. [22] investigated a class of fractional boundary value problem with p-Laplacian operator and boundary parameter and presented several existence results for a positive solution in terms of the boundary parameter. It is noted that although there exist several results on the existence of one solution to fractional p-Laplacian BVPs, there are, to our best knowledge, relatively few results on the nonexistence, the uniqueness, and the multiplicity of positive solutions to fractional p-Laplacian BVPs.In this paper, we study the following two-point boundary value problem of nonlinear fractional differential equations withp-Laplacian operator:
(2)D0+βΦp(D0+αx(t))=f(t,x(t)),t∈[0,1],x(0)=γx(1),D0+αx(0)=ηD0+αx(1),
where Φp(s)=|s|p-2s, p>1, Φp-1=Φq, 1/p+1/q=1, 0<α, β≤1, 1<α+β≤2, 0<γ, η<1, f:[0,1]×[0,+∞)→[0,+∞), and D0+α is the Caputo derivative. We first convert BVP (2) into an equivalent operator equation by using the property of the Caputo derivative and then present some sufficient conditions for the nonexistence, the uniqueness, and the multiplicity of positive solutions to this problem. Finally, we give several illustrative examples to support our new results.The main contributions of this paper are as follows. (i) We systematically study the nonexistence, the uniqueness, and the multiplicity of positive solutions to BVP (2) and propose some techniques to deal with fractional p-Laplacian BVPs, which enriches this academic area. (ii) We present a sufficient condition for the existence of three positive solutions to BVP (2) by introducing a free constant to construct a proper concave functional. The main feature of this free constant is that one can weaken the conditions by regulating the free constant (see Remark 12).Throughout this paper, we consider BVP (2) in the real Banach space E=C[0,1] with the norm ∥x∥=maxt∈[0,1]|x(t)|. Let P={x∈E:x(t)≥0, for all t∈[0,1]}. Then, P is a normal solid cone of E with P∘={x∈E:x(t)>0, for all t∈[0,1]}. A solution x(t)∈E is said to be a positive solution to BVP (2), if x∈P and x(t)≢0. We will study the existence of positive solutions to BVP (2) in P.The rest of this paper is structured as follows. Section2 contains some preliminaries on the Caputo derivative. Section 3 investigates the nonexistence, the uniqueness, and the multiplicity of positive solutions to BVP (2) and presents the main results of this paper. In Section 4, three illustrative examples are worked out to support our obtained results.


## 2. Preliminaries

In this section, we give some necessary preliminaries on the Caputo derivative, which will be used in the sequel. For details, please refer to [1–3] and the references therein.Definition 1 (see [3]).
The Riemann-Liouville fractional integral of orderα>0 of a function y:(0,+∞)→ℝ is given by
(3)I0+αy(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds,
provided that the right side is pointwise defined on (0,+∞).Definition 2 (see [3]).
The Caputo fractional derivative of orderα>0 of a continuous function y:(0,+∞)→ℝ is given by
(4)D0+αy(t)=1Γ(n-α)∫0t‍y(n)(s)(t-s)α-n+1ds,
where n is the smallest integer greater than or equal to α, provided that the right side is pointwise defined on (0,+∞).One can easily obtain the following property from the definition of the Caputo derivative.Proposition 3 (see [3]).
Letα>0. Assume that y, D0+αy∈L(0,1). Then, the following equality holds:
(5)I0+αD0+αy(t)=y(t)+C0+C1t+⋯+Cn-1tn-1,
for some Ci∈ℝ, i=0,1,…,n-1, where n is the smallest integer greater than or equal to α.


## 3. Main Results

In this section, we first convert BVP (2) into an equivalent operator equation and then present some new results on the nonexistence, the uniqueness, and the multiplicity of positive solutions to BVP (2).Firstly, we convert BVP (2) into an equivalent operator equation.Theorem 4.
Given thath∈C[0,1], the unique solution of
(6)D0+βΦp(D0+αx(t))=h(t),t∈[0,1],x(0)=γx(1),D0+αx(0)=ηD0+αx(1)
is
(7)x(t)=I0+αΦq(I0+βh(t)+Ah(t))+Bh(t)=1Γ(α)∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1h(τ)dτ+Ah(s))ds+Bh(t),
where
(8)Ah(t)=Φp(η)1-Φp(η)I0+βh(t)|t=1=Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-s)β-1h(s)ds,∀t∈[0,1],Bh(t)=γ1-γI0+αΦq(I0+βh(t)+Ah(t))|t=1=γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1h(τ)dτ+Ah(s))ds,∀t∈[0,1].Proof.
Assume thatx(t) satisfies (6). Then, from Proposition 3, we have
(9)Φp(D0+αx(t))=I0+βh(t)+c0,c0∈ℝ.
From the boundary value conditionD0+αx(0)=ηD0+αx(1), one can see that
(10)c0=Φp(η)1-Φp(η)I0+βh(t)|t=1=Ah(t).
Thus, we have
(11)x(t)=I0+αΦq(I0+βh(t)+Ah(t))+c1,c1∈ℝ,
which together with the boundary value condition x(0)=γx(1) yields that
(12)c1=γ1-γI0+αΦq(I0+βh(t)+Ah(t))|t=1=Bh(t).
The proof is completed.Lemma 5.
Consider thatx(t)∈E is a solution to BVP (2), if and only if x(t)=Tx(t), where
(13)Tx(t)=I0+αΦq(I0+βNx(t)+ANx(t))+BNx(t)=1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds+γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds,
and Nx(t)=f(t,x(t)).Proof.
The necessity is obvious. Next, we prove the sufficiency.
In fact, ifx(t)=Tx(t), a straightforward calculation shows that x(0)=γx(1).
Since(14)D0+αx(t)=Φq(I0+βNx(t)+ANx(t))=Φq(1Γ(β)∫0t‍(t-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β)),
we have D0+αx(0)=ηD0+αx(1), and
(15)D0+βΦp(D0+αx(t))=D0+βΦp(Φq(I0+βNx(t)+ANx(t)))=D0+β(I0+βf(t,x(t))+ANx(t))=f(t,x(t)).
Therefore,x(t) is a solution to BVP (2).In the following, we consider the nonexistence of positive solutions to BVP (2) and present the following result.Theorem 6.
Consider BVP (2). Assume that
(16)f(t,x)≤(m1x)p-1,∀(t,x)∈[0,1]×[0,+∞),
where m1<(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1 is a positive constant. Then, BVP (2) has no positive solution.Proof.
In fact, if BVP (2) has a positive solution x(t)∈P with x(t)≢0. Then, one can see from (16) that
(17)|x(t)|=|Tx(t)|=I0+αΦq(I0+βNx(t)+ANx(t))+BNx(t)=1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds+γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds≤m1∥x∥1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1dτ1Γ(β))ds+m1∥x∥γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1dτ1Γ(β))ds=m1∥x∥1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β+1)sβ+Φp(η)1-Φp(η)1Γ(β+1))ds+m1∥x∥γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β+1)sβ+Φp(η)1-Φp(η)1Γ(β+1))ds≤m1∥x∥1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))ds+m1∥x∥γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))ds=m1∥x∥1Γ(α)×[∫0t‍(t-s)α-1ds]Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))+m1∥x∥γ1-γ1Γ(α)×[∫01‍(1-s)α-1ds]Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))=m1∥x∥1Γ(α+1)tαΦq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))+m1∥x∥γ1-γ1Γ(α)Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))≤m1∥x∥×1Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1+m1∥x∥×γ(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1≤m1∥x∥×1(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1<∥x∥,∀t∈[0,1],
which is a contradiction. Hence, BVP (2) has no positive solution.Next, we study the existence of a unique positive solution to BVP (2). To this end, we need the following fixed point theorem [26].Definition 7.
LetP be a solid cone in a real Banach space E, T:P∘→P∘ an operator, and 0≤θ<1. Then, T is called a θ-concave operator if
(18)T(ku)≥kθTu,∀0<k<1,u∈P∘.Lemma 8.
Assume thatP is a normal solid cone in a real Banach space E, 0≤θ<1, and T:P∘→P∘ is a θ-concave increasing operator. Then T has only one fixed point in P∘.Based on Lemma8, we have the following result.Theorem 9.
Consider BVP (2). Assume that for all t∈[0,1], f:[0,1]×(0,+∞)→(0,+∞) is increasing in x, and there exists a constant 0≤θ<1 such that
(19)f(t,kx)≥k(p-1)θf(t,x),∀(t,x)∈[0,1]×(0,+∞),∀0<k<1.
Then, BVP (2) has a unique positive solution.Proof.
Sincef:[0,1]×(0,+∞)→(0,+∞), one can see that T:P∘→P∘.
For anyx1,x2∈P∘ with x1≤x2, from the monotonicity of f and xq-1, we have
(20)Tx2(t)-Tx1(t)≥0,∀t∈[0,1].
Thus, T is increasing in P∘.
Next, we prove thatT is a θ-concave operator.
In fact, from (19), for any 0<k<1, x∈P∘, it is easy to see that
(21)T(kx)(t)=1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,kx(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,kx(τ))dτ1Γ(β))ds+γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,kx(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,kx(τ))dτ1Γ(β))ds≥kθ1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds+kθγ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds=kθT(x)(t),
which implies that T is a θ-concave operator.
By Lemma8, BVP (2) has a unique positive solution.Finally, we investigate the multiplicity of positive solutions to BVP (2).We first recall the famous Leggett-Williams fixed point theorem [26].LetE=(E,∥·∥) be a Banach space and P⊂E be a cone on E. A continuous mapping ω:P→[0,+∞) is said to be a concave nonnegative continuous functional on P, if ω satisfies ω(λx+(1-λ)y)≥λω(x)+(1-λ)ω(y) for all x,y∈P and λ∈[0,1].Leta, b, d>0 be constants. Define Pd={x∈P:∥x∥<d}, Pd¯={x∈P:∥x∥≤d}, and P(ω,a,b)={x∈P:ω(x)≥a,∥x∥≤b}.Lemma 10.
LetE=(E,∥·∥) be a Banach space, P⊂E a cone of E, and c>0 a constant. Suppose that there exists a concave nonnegative continuous functional ω on P with ω(x)≤∥x∥ for all x∈Pc¯. Let T:Pc¯→Pc¯ be a completely continuous operator. Assume that there are numbers a, b, and d with 0<d<a<b≤c, such that(i)
{x∈P(ω,a,b):ω(x)>a}≠∅ and ω(Tx)>a for all x∈P(ω,a,b);(ii)
∥Tx∥<d for all x∈Pd¯;(iii)
ω(Tx)>a for all x∈P(ω,a,c) with ∥Tx∥>b.
Then,T has at least three fixed points x1, x2, and x3 in Pc¯. Furthermore, x1∈Pa; x2∈{x∈P(ω,a,c):ω(x)>a}; x3∈Pc¯∖(P(ω,b,c)∪Pa¯).Now, based on Lemma10, we consider the existence of three positive solutions to BVP (2).Letξ∈[0,1) be a constant. Define ω(x)=mint∈[ξ,1]x(t), for all x∈P. Then, one can easily see that ω is a concave non-negative continuous functional. We have the following result.Theorem 11.
Consider BVP (2). Suppose that f∈C([0,1]×[0,+∞)→[0,+∞)). Assume that there exist positive constants a and c with a<η(1-ξ)β(q-1)[(1-γ)ξα+γ(1-ξ)α]c, such that(H1)
there exists a constant0<m2≤(1-γ)Γ(α+1)(Γ(β+1))q-1(1-Φp(η))q-1, such that
(22)f(t,x)≤(m2c)p-1,∀(t,x)∈[0,1]×[0,c];(H2)
there exists a constant(1-γ)Γ(α+1) (Γ(β+1))q-1(1-Φp(η))q-1 / η(1-ξ)β(q-1) [(1-γ)ξα+γ(1-ξ)α]≤m3<m2c/a, such that
(23)f(t,x)>(m3a)p-1,∀(t,x)∈[ξ,1]×[a,c];(H3)
there exists a constant0<m4<(1-γ)p-1(Γ(α+1))p-1Γ(β+1)(1-Φp(η)), such that
(24)limx→0+¯supt∈[0,1]f(t,x)xp-1=m4.Then, BVP (2) has at least three positive solutions.Proof.
Let us divide the proof into 4 steps.
Step
1. By (H1) and Lemma 5, for any x∈Pc¯, we have
(25)|x(t)|=|Tx(t)|=I0+αΦq(I0+βNx(t)+ANx(t))+BNx(t)=1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds+γ1-γ1Γ(α)∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds≤m2c1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1dτ1Γ(β))ds+m2cγ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1dτ+Φp(η)1-Φp(η)1Γ(β)∫01‍(1-τ)β-1dτ)ds=m2c1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β+1)sβ+Φp(η)1-Φp(η)1Γ(β+1))ds+m2cγ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β+1)sβ+Φp(η)1-Φp(η)1Γ(β+1))ds≤m2c1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))ds+m2cγ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))ds=m2c1Γ(α)[∫0t‍(t-s)α-1ds]Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))+m2cγ1-γ1Γ(α)[∫01‍(1-s)α-1ds]Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))=m2c1Γ(α+1)tαΦq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))+m2cγ1-γ1Γ(α)Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))≤m2c×1Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1+m2c×γ(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1≤m2c×1(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1≤c,∀t∈[0,1],
which implies that∥Tx∥≤c. Thus, T:Pc¯→Pc¯.
Next, let us show thatT:Pc¯→Pc¯ is completely continuous.
By the continuity off and uq-1, we can get that T:Pc¯→Pc¯ is continuous and T(Pc¯) is bounded. Moreover, there exists a constant M>0 such that |I0+βNx(t)+ANx(t)|≤M, for all x∈Pc¯, t∈[0,1].
In addition, for anyt1,t2∈[0,1] and x∈Pc¯, we have
(26)|Tx(t2)-Tx(t1)|=|I0+αΦq(I0+βNx(t)+ANx(t))|t=t2-I0+αΦq(I0+βNx(t)+ANx(t))|t=t2|=1Γ(α)|∫0t2‍(t2-s)α-1Φq(I0+βNx(s)+ANx(s))ds-∫0t1‍(t1-s)α-1Φq(I0+βNx(s)+ANx(s))ds∫0t1‍|=1Γ(α)|∫0t1‍((t2-s)α-1-(t1-s)α-1)Φq(I0+βNx(s)+ANx(s))ds+∫t1t2‍(t2-s)α-1Φq(I0+βNx(s)+ANx(s))ds∫t1t2‍|≤Mq-1Γ(α){∫0t1‍[(t1-s)α-1-(t2-s)α-1]ds+∫t1t2‍(t2-s)α-1ds}=Mq-1Γ(α+1)[t1α-t2α+2(t2-t1)α]⟶0,|t2-t1|⟶0.
Thus, the Arzela-Ascoli theorem guarantees thatT:Pc¯→Pc¯ is compact. Therefore, T:Pc¯→Pc¯ is completely continuous.
Step
2. Choose a constant b=c. Let x0(t)=(a+b)/2, for all  t∈[0,1]. Then, ω(x0)=(a+b)/2>a and ∥x0∥=(a+b)/2<b. Thus, x0∈{x∈P(ω,a,b):ω(x)>a}≠∅.
Now, let us prove thatω(Tx)>a holds for all x∈P(ω,a,b). In fact, x∈P(ω,a,b) implies that a≤x(t)≤b, for all t∈[ξ,1]. For any t∈[ξ,1], one can obtain from (H2) that
(27)Tx(t)≥I0+αΦq(ANx(t))+BNx(t)≥1Γ(α)×∫0t‍(t-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)×∫ξ1‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds+γ1-γ1Γ(α)×∫ξ1‍(1-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)×∫ξ1‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds>1Γ(α)×∫0t‍(t-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)×∫ξ1‍(1-τ)β-1(m3a)p-1dτ1Γ(β))ds+γ1-γ1Γ(α)×∫ξ1‍(1-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)×∫ξ1‍(1-τ)β-1(m3a)p-1dτ1Γ(β))ds=m3aΓ(α+1)tα(Φp(η)1-Φp(η)1Γ(β+1)(1-ξ)β)q-1+γ1-γm3aΓ(α+1)(1-ξ)α×(Φp(η)1-Φp(η)1Γ(β+1)(1-ξ)β)q-1≥m3aΓ(α+1)ξα(Φp(η)1-Φp(η)1Γ(β+1)(1-ξ)β)q-1+γ1-γm3aΓ(α+1)(1-ξ)α×(Φp(η)1-Φp(η)1Γ(β+1)(1-ξ)β)q-1=(Φp(η))q-1(1-ξ)β(q-1)((1-γ)ξα+γ(1-ξ)α)(1-γ)Γ(α+1)(Γ(β+1))q-1(1-Φp(η))q-1×m3a≥a.
Hence, condition (i) of Lemma 10 holds.
Step
3. It is easy to see from (H3) that for all t∈[0,1], for all 0 < ɛ < (1-γ)p-1(Γ(α+1))p-1Γ(β+1)(1-Φp(η))-m4, there exists δ>0, such that
(28)f(t,x)<(m4+ɛ)xp-1,0≤x<δ.
Let 0<d<min{δ,a}. Now, we prove that ∥Tx∥<d for all x∈Pd¯={x∈P:∥x∥≤d}. As a matter of fact, for all x∈Pd¯, one can see that
(29)|x(t)|=|Tx(t)|=I0+αΦq(I0+βNx(t)+ANx(t))+BNx(t)=1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds+γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1f(τ,x(τ))dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1f(τ,x(τ))dτ1Γ(β))ds≤(m4+ɛ)q-1∥x∥1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1dτ1Γ(β))ds+(m4+ɛ)q-1∥x∥γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β)∫0s‍(s-τ)β-1dτ+Φp(η)1-Φp(η)1Γ(β)×∫01‍(1-τ)β-1dτ1Γ(β))ds=(m4+ɛ)q-1∥x∥1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β+1)sβ+Φp(η)1-Φp(η)1Γ(β+1))ds+(m4+ɛ)q-1∥x∥γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β+1)sβ+Φp(η)1-Φp(η)1Γ(β+1))ds≤(m4+ɛ)q-1∥x∥1Γ(α)×∫0t‍(t-s)α-1Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))ds+(m4+ɛ)q-1∥x∥γ1-γ1Γ(α)×∫01‍(1-s)α-1Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))ds=(m4+ɛ)q-1∥x∥1Γ(α+1)tαΦq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))+(m4+ɛ)q-1∥x∥γ1-γ1Γ(α)Φq(1Γ(β+1)+Φp(η)1-Φp(η)1Γ(β+1))≤(m4+ɛ)q-1∥x∥×1Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1+(m4+ɛ)q-1∥x∥×γ(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1≤(m4+ɛ)q-1×d1(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1<d,∀t∈[0,1],
which implies that ∥Tx∥<d. Thus, ∥Tx∥<d, for all x∈Pd¯.
Step
4. Let us prove that ω(Tx)>a holds for all x∈P(ω,a,c) with ∥Tx∥>b.
Forx∈P(ω,a,c) with ∥Tx∥>b, we have a≤x(t)≤c, for all t∈[ξ,1]. From (H2), one can see that
(30)Tx(t)≥I0+αΦq(ANx(t))+BNx(t)≥1Γ(α)×∫ξt‍(t-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)∫ξ1‍(1-τ)β-1f(τ,x(τ))dτ)ds+γ1-γ1Γ(α)×∫ξ1‍(1-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)∫ξ1‍(1-τ)β-1f(τ,x(τ))dτ)ds>1Γ(α)×∫ξt‍(t-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)∫ξ1‍(1-τ)β-1(m3a)p-1dτ)ds+γ1-γ1Γ(α)×∫ξ1‍(1-s)α-1Φq(Φp(η)1-Φp(η)1Γ(β)∫ξ1‍(1-τ)β-1(m3a)p-1dτ)ds=m3aΓ(α+1)ξα(Φp(η)1-Φp(η)1Γ(β+1)(1-ξ)β)q-1+γ1-γm3aΓ(α+1)(1-ξ)α×(Φp(η)1-Φp(η)1Γ(β+1)(1-ξ)β)q-1=(Φp(η))q-1(1-ξ)β(q-1)((1-γ)ξα+γ(1-ξ)α)(1-γ)Γ(α+1)(Γ(β+1))q-1(1-Φp(η))q-1×m3a≥a,∀t∈[ξ,1].
Therefore, condition (iii) of Lemma 10 is satisfied.
To sum up, all conditions of Lemma10 hold. By Lemma 10, BVP (2) has at least three positive solutions.Remark 12.
It is noted that (H2) is dependent of the selection onξ. Specifically, to obtain a weaker condition, one can choose ξ*∈[0,1) which maximizes
(31)k(ξ)=ξα(1-ξ)β(q-1)(1-γ)+γ(1-ξ)α+β(q-1).


## 4. Illustrative Examples

In this section, we give three illustrative examples to support our new results.Example 1.
Consider the following BVP:(32)D0+1/2Φ3/2(D0+2/3x(t))=(1+t2)x(t)20,t∈[0,1],x(0)=12x(1),D0+2/3x(0)=12D0+2/3x(1).
A simple calculation shows that(1-γ)Γ(α+1)(1-Φp(η))q-1(Γ(β+1))q-1≈0.0301.
Setm1=1/100<0.0301, then one can see that
(33)f(t,x)=(1+t2)x20≤x100,∀(t,x)∈[0,1]×[0,∞],
which implies that (16) holds. By Theorem 6, BVP (32) has no positive solution.Example 2.
Consider the following BVP:(34)D0+βΦ2(D0+αx(t))=(1+t2)x(t)4,t∈[0,1],x(0)=γx(1),D0+αx(0)=ηD0+αx(1),
where 0<α, β≤1, 1<α+β≤2, and 0<γ, η<1 are arbitrary.Setθ=1/4; then it is easy to see that
(35)f(t,kx)=(1+t2)kx4=k1/4f(t,x),∀(t,x)∈[0,1]×(0,+∞),∀0<k<1,
Thus, (19) is satisfied. By Theorem 9, BVP (34) has a unique positive solution.Letα=β=1 and γ=η=1/2; we can draw the graph of the unique positive solution to BVP (34) by using MATLAB, shown in Figure 1.Figure 1
The unique positive solution to BVP (34) with α=β=1 and γ=η=1/2.Example 3.
Consider the following BVP:(36)D0+2/3Φ3(D0+1/2x(t))=f(t,x(t)),t∈[0,1],x(0)=13x(1),D0+1/2x(0)=34D0+1/2x(1),
where
(37)f(t,x)=(1+t2){x240,(t,x)∈[0,1]×[0,12),159980x-79980,(t,x)∈[0,1]×[12,1),10,(t,x)∈[0,1]×[1,+∞).Seta=1, c=50, ξ=0, m2=1/5, m3=2, and m4=1/20. By a straightforward calculation, one can easily see that (H1)–(H3) hold. By Theorem 11, BVP (36) has at least three positive solutions.



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*Source: 898206-2013-11-10.xml*