# Positive Solutions for the Eigenvalue Problem of Semipositone Fractional Order Differential Equation with Multipoint Boundary Conditions

Authors: Ge Dong
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study the existence of positive solution for the eigenvalue problem of semipositone
fractional order differential equation with multipoint boundary conditions by using known Krasnosel'skii's
fixed point theorem. Some sufficient conditions that guarantee the existence of at least one positive solution
for eigenvaluesλ>0 sufficiently small and λ>0 sufficiently large are established.


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## Body

## 1. Introduction

In this paper, we study the existence of positive solutions to the following eigenvalue problem of semipositone fractional order differential equation with multipoint boundary conditions:(1)-Dtαx(t)=λf(t,x(t),Dtγx(t)),t∈(0,1),Dtγx(0)=0,Dtγ+1x(0)=0,Dtγx(1)=∑j=1m-2‍ajDtγx(ξj),
where 3<α≤4, 0<γ≤α-2, 0<ξ1<ξ2<⋯<ξm-2<1, aj∈[0,+∞) with 0<∑j=1m-2‍ajξjα-γ-1<1, λ is a positive parameter, and Dtα,Dtγ are the standard Rieman-Liouville derivative. Throughout the paper, we assume that f is semipositone; that is, f:[0,1]×[0,+∞)→ℝ is continuous and there exists M>0, such that f(t,x)≥-M, for any (t,x)∈[0,1]×[0,∞).The multipoint boundary value problems (BVPs for short) for ordinary differential equations arise in a variety of different applied mathematics and physics. Recently, Feng and Bai [1] investigated the existence of positive solutions for a semipositone second-order multipoint boundary value problem:
(2)x′′(t)+λf(t,x(t))=0,t∈(0,1),x′(0)=∑j=1m-2‍ajx(ξj),x(1)=∑j=1m-2‍bjx(ξj).
By using Krasnosel'skii's fixed point theorem, some sufficient conditions that guarantee the existence of at least one positive solution are obtained. In [2], a (n-1,1)-type conjugate boundary value problem for the nonlinear fractional differential equation,
(3)Dtαx(t)+λf(t,x(t))=0,t∈(0,1),xj(0)=0,0≤j≤n-2,x(1)=0,
is considered. Based on the nonlinear alternative of Leray-Schauder type and Krasnosel'skii's fixed-point theorems, the existence of positive solution of the semipositone boundary value problems (3) for a sufficiently small λ>0 was given. In recent paper [3], Zhang et al. established the existence of multiple positive solutions for a general higher order fractional differential equation with derivatives and a negatively Carathèodory perturbed term:
(4)-Dαx(t)=p(t)f(t,x(t),Dμ1x(t),Dμ2x(t),…,Dμn-1x(t))-g(t,x(t),Dμ1x(t),Dμ2x(t),…,Dμn-1x(t)),Dμix(0)=0,1≤i≤n-1,Dμn-1+1x(0)=0,Dμn-1x(1)=∑j=1m-2‍ajDμn-1x(ξj).
Some local and nonlocal growth conditions were adopted to guarantee the existence of at least two positive solutions for the higher order fractional differential equation (4). For the recent work in application, the reader is referred to [4–20].Inspired by the above work, in this paper we study the existence of positive solutions to the semipositone BVP (1). Here we also emphasize that the main results of this paper contain not only the cases for λ>0 sufficiently small, but also for λ>0 sufficiently large, which is different from [2, 3].


## 2. Preliminaries and Lemmas

Definition 1 (see [21–24]).
The fractional integral of orderα>0 of a function x:(a,+∞)→R is given by
(5)Iαx(t)=1Γ(α)∫at‍(t-s)α-1x(s)ds,
provided that the right-hand side is pointwisely on (a,+∞).Definition 2 (see [21–24]).
The Riemann-Liouville fractional derivative of orderα>0 of a function x: (a,+∞)→R is given by
(6)Dtαx(t)=1Γ(n-α)(ddt)n∫at‍(t-s)n-α-1x(s)ds,
where n=[α]+1, [α] denotes the integer part of the number α, and t>a, provided that the right-hand side is defined on (a,+∞).Lemma 3 (see [21–24]).
Assuming thatx∈L1[0,1] with a fractional derivative of order α>0, then
(7)IαDtαx(t)=x(t)+c1tα-1+c2tα-2+⋯+cntα-n,
where ci∈R,i=1,2,…,n,n=[α]+1.Lemma 4 (see [3]).
Suppose thath∈L1[0,1]. Then the following boundary value problem
(8)Dtα-γx(t)+h(t)=0,t∈(0,1),x(0)=x′(0)=0,x(1)=∑j=1m-2‍ajx(ξj)
has a unique solution
(9)x(t)=∫01‍G(t,s)h(s)ds,
where
(10)G(t,s)=g(t,s)+tα-γ-11-∑j=1m-2ajξjα-γ-1∑j=1m-2‍ajg(ξj,s)
is the Green function of the boundary value problem (8) and
(11)g(t,s)={(t(1-s))α-γ-1-(t-s)α-γ-1Γ(α-γ),0≤s≤t≤1,(t(1-s))α-γ-1Γ(α-γ),0≤t≤s≤1.Lemma 5 (see [2]).
The functiong(t,s) in Lemma 4 has the following properties:(R1)
g(t,s)=g(1-s,1-t), for t,s∈[0,1];(R2)
Γ(α-γ)k(t)q(s)≤g(t,s)≤(α-γ-1)q(s), for t,s∈[0,1];(R3)
Γ(α-γ)k(t)q(s)≤g(t,s)≤(α-γ-1)k(t), for t,s∈[0,1], where
(12)k(t)=tα-γ-1(1-t)Γ(α-γ),q(t)=s(1-s)α-γ-1Γ(α-γ).Lemma 6.
The following boundary value problem(13)Dtα-γx(t)+λM=0,t∈(0,1),x(0)=x′(0)=0,x(1)=∑j=1m-2‍ajx(ξj)
has a unique solution w, which satisfies
(14)w(t)≤4λMσ(t)1-∑j=1m-2ajξjα-γ-1,σ(t)=(1-∑j=1m-2‍ajξjα-γ-1)k(t)+∑j=1m-2‍ajk(ξj)tα-γ-1≤tα-γ-1.Proof.
By Lemma4, the unique solution of (13) is
(15)w(t)=λM∫01‍G(t,s)ds.
So
(16)w(t)=λM∫01‍G(t,s)ds≤λM(α-γ-1)×∫01‍(k(t)+∑j=1m-2ajk(ξj)1-∑j=1m-2ajξjα-γ-1tα-γ-1)ds≤4λMσ(t)1-∑j=1m-2ajξjα-γ-1,
and by α-γ≥2, we have Γ(α-γ)≥1, so
(17)σ(t)=(1-∑j=1m-2‍ajξjα-γ-1)k(t)+∑j=1m-2‍ajk(ξj)tα-γ-1=1Γ(α-γ)(1-∑j=1m-2‍ajξjα-γ-1)tα-γ-1(1-t)+1Γ(α-γ)∑j=1m-2‍ajξjα-γ-1(1-ξj)tα-γ-1≤(1-∑j=1m-2‍ajξjα-γ-1)tα-γ-1+∑j=1m-2‍ajξjα-γ-1tα-γ-1=tα-γ-1.The basic space used in this paper isE=C([0,1];ℝ), where ℝ is the set of real numbers. Obviously, the space E is a Banach space if it is endowed with the norm as follows:
(18)∥y∥=max⁡t∈[0,1]|y(t)|,
for any y∈E. Let
(19)P={y∈E:y(t)≥18σ(t)∥y∥},
and then P is a cone of E.Now letv(t)=Dtγx(t); then the boundary value problem (1) is equivalent to the following boundary value problem:
(20)-Dtα-γv(t)=λf(t,Iγv(t),v(t)),t∈(0,1),v(0)=v′(0)=0,v(1)=∑j=1m-2‍ajv(ξj).
Define a modified function [·]* for any φ∈C[0,1] by
(21)[φ(t)]*={φ(t),φ(t)≥0,0,φ(t)<0,
and consider
(22)-Dtα-γy(t)=λ[f(t,Iγ[y(t)-w(t)]*,[y(t)-w(t)]*)+M],t∈(0,1),y(0)=y′(0)=0,y(1)=∑j=1m-2‍ajy(ξj).Lemma 7.
The BVP (1) and the BVP (22) are equivalent. Moreover, if y is a positive solution of the problem (22) and satisfies y(t)≥w(t), t∈[0,1], then Iγ[y(t)-w(t)] is a positive solution of the boundary value problem (1).Proof.
Sincey is a positive solution of the BVP (22) such that y(t)≥w(t) for any t∈[0,1], we have
(23)-Dtα-γy(t)=λ[f(t,Iγ[y(t)-w(t)],[y(t)-w(t)])+M],t∈(0,1),y(0)=y′(0)=0,y(1)=∑j=1m-2‍ajy(ξj).
Let v=y-w, and then we have
(24)Dtα-γv(t)=Dtα-γy(t)-Dtα-γw(t),w(0)=w′(0)=0,w(1)=∑j=1m-2‍ajw(ξj).
Substitute (24) into (23), that is (20), which implies that Iγ[y(t)-w(t)] is a positive solution of the BVP (1).It follows from Lemma4 that the BVP (22) is equivalent to the integral equation
(25)y(t)=λ∫01‍G(t,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)mi+MIγ[y(s)-w(s)]*].
Thus it is sufficient to find fixed points y(t)≥w(t),t∈[0,1] for the mapping T defined by
(26)(Ty)(t)=λ∫01‍G(t,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+MIγ[y(s)-w(s)]*].Lemma 8.
T
:
P
→
P is a completely continuous operator.Proof.
For any fixedy∈P, there exists a constant L>0 such that ∥y∥≤L, and
(27)0≤[y(s)-w(s)]*≤y(s)≤∥y∥≤L,0≤Iγ[y(s)-w(s)]*=∫0t‍(t-s)γ-1[y(s)-w(s)]*Γ(γ)ds≤LΓ(γ).
Take
(28)N=max⁡[0,1]×[0,L/Γ(γ)]×[0,L]f(t,u,v),
then
(29)(Ty)(t)=λ∫01‍G(t,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]≤λ(M+N)∫01‍G(t,s)ds<+∞.
This implies that the operator T:P→E is bounded.
Next for anyy∈P, by Lemma 5, we have
(30)∥Ty∥=max⁡0≤t≤1{λ∫01‍G(t,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]∫01‍}≤λ(α-γ-1)∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]+λ1-∑j=1m-2ajξjα-γ-1×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*Iγ)+M]ds≤4λ{∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]+11-∑j=1m-2ajξjα-γ-1×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*Iγ)+M]ds∫01‍}≤4λ1-∑j=1m-2ajξjα-γ-1×{∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+MIγ[y(s)-w(s)]*]+∑j=1m-2‍aj∫01‍g(ξj,s)×[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+MIγ[y(s)-w(s)]*]ds∫01‍}.
On the other hand, it follows from Lemma 5, Γ(α-γ)≥1, and σ(t)≤tα-γ-1 that
(31)(Ty)(t)≥λΓ(α-γ)k(t)×∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds+λtα-γ-11-∑j=1m-2ajξjα-γ-1×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds≥12λk(t)∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds+λtα-γ-11-∑j=1m-2ajξjα-γ-1×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds=12{λk(t)∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds+λtα-γ-11-∑j=1m-2ajξjα-γ-1×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds∫01‍}+λtα-γ-12(1-∑j=1m-2ajξjα-γ-1)×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds≥12λ{k(t)∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds+tα-γ-1∑j=1m-2ajk(ξj)1-∑j=1m-2ajξjα-γ-1×∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds∫01‍}+λtα-γ-12(1-∑j=1m-2ajξjα-γ-1)×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds=λσ(t)2(1-∑j=1m-2ajξjα-γ-1)×∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds+λσ(t)2(1-∑j=1m-2ajξjα-γ-1)×∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds=λσ(t)2(1-∑j=1m-2ajξjα-γ-1)×(∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds+∑j=1m-2‍aj∫01‍g(ξj,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds∫01‍q(s)).
So, by (30) and (31), we have
(32)(Ty)(t)≥18∥Ty∥σ(t),t∈[0,1],
which yields that T(P)⊂P.
At the end, using standard arguments, according to the Ascoli-Arzela Theorem, one can show thatT:P→P is completely continuous. Thus T:P→P is a completely continuous operator.Lemma 9 (see [25]).
LetE be a real Banach space, and let P⊂E be a cone. Assume that Ω1,Ω2 are two bounded open subsets of E with θ∈Ω1,Ω¯1⊂Ω2, and let T:P∩(Ω¯2∖Ω1)→P be a completely continuous operator such that either(1)
∥Tx∥≤∥x∥,x∈P∩∂Ω1 and ∥Tx∥≥∥x∥,x∈P∩∂Ω2, or(2)
∥Tx∥≥∥x∥,x∈P∩∂Ω1 and ∥Tx∥≤∥x∥,x∈P∩∂Ω2.Then T has a fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Result

Define(33)f∞=limsup⁡|x|+|y|→∞max⁡t∈[0,1]f(t,x,y)|x|+|y|,f∞=liminf⁡|x|+|y|→∞min⁡t∈[1/4,3/4]f(t,x,y)|x|+|y|.Theorem 10.
Suppose that(34)f∞=∞.
Then there exists a constant Λ>0 such that, for any λ∈(0,Λ], the BVP (1) has at least one positive solution.Proof.
Choosingy∈P with ∥y∥=1, then
(35)0≤[y(s)-w(s)]*≤y(s)≤∥y∥≤1,0≤Iγ[y(s)-w(s)]*=∫0t‍(t-s)γ-1[y(s)-w(s)]*Γ(γ)ds≤1Γ(γ).
Let
(36)N=max⁡(t,x,y)∈[0,1]×[0,1/Γ(γ)]×[0,1]f(t,x,y),Λ=min⁡{1-∑j=1m-2ajξjα-γ-134M,1-∑j=1m-2ajξjα-γ-14∑j=1m-2aj(M+N)}.
For any y∈∂B1, B1={y∈P:∥y∥≤1}, and λ>0 sufficiently small such that λ∈(0,Λ], we have
(37)∥Ty∥=max⁡0≤t≤1{λ∫01‍G(t,s)[f(s,Iγ[y(s)-w(s)]*,m[y(s)-w(s)]*)+M]λ∫01‍}≤λ(α-γ-1)×∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+MIγ[y(s)-w(s)]*]+λ(α-γ-1)1-∑j=1m-2ajξjα-γ-1×∑j=1m-2‍aj∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds≤4λ∑j=1m-2aj(M+N)1-∑j=1m-2ajξjα-γ-1≤1=∥y∥.
Therefore,
(38)∥Ty∥≤∥y∥,y∈∂B1.
On the other hand, take(39)ϵ=(1-∑j=1m-2‍ajξjα-γ)(14)α-γ,
and choose a large enough L>0 such that
(40)λLϵ64(14)α-γ∫1/43/4‍q(s)ds>1.
By (33), we know that f is an unbounded continuous function. Therefore, for any t∈[1/4,3/4], there exists a constant K>0 such that
(41)f(t,x,y)≥L(|x|+|y|),if|x|+|y|>K.
Choosing
(42)R>max⁡{64λM1-∑j=1m-2ajξjα-γ-1,1+K,16Kϵ},
then R>K>1. Let BR={y∈P:∥y∥≤R}. Then for any y∈∂BR and for any t∈[1/4,3/4], we have
(43)y(t)-w(t)≥y(t)-4λMσ(t)1-∑j=1m-2ajξjα-γ-1≥y(t)-32λMy(t)(1-∑j=1m-2ajξjα-γ-1)R≥12y(t)≥116σ(t)R≥116ϵR≥K>0.
Consequently, for s∈[1/4,3/4], it follows from (43) that
(44)|Iγ[y(s)-w(s)]*|+|[y(s)-w(s)]*|≥|[y(s)-w(s)]*|>K,
and then by (41) and (44), for s∈[1/4,3/4], we get
(45)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)≥L(|Iγ[y(s)-w(s)]*|+|[y(s)-w(s)]*|)≥L|[y(s)-w(s)]|≥164LϵR.
So for anyy∈∂BR and t∈[0,1], by (45), we have
(46)∥Ty∥≥λ∫01‍G(14,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*Iγ[y(s)-w(s)]*)+M]ds≥λ∫01‍G(14,s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)ds≥λ∫01‍g(14,s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)ds≥λΓ(α-γ)k(14)×∫01‍q(s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)ds≥λ(14)α-γ∫1/43/4‍q(s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*Iγ[y(s)-w(s)]*)ds≥λ(14)α-γ∫1/43/4‍q(s)164LϵRds≥R=∥y∥.
Thus, we have
(47)∥Ty∥≥∥y∥,y∈∂BR.
By Lemma 9, T has a fixed point y such that 1≤∥y∥≤R.
From(48)λ≤Λ≤1-∑j=1m-2ajξjα-γ-132M,
we have
(49)32λM1-∑j=1m-2ajξjα-γ-1≤1.
Thus
(50)y(t)≥18σ(t)∥y∥≥18σ(t)≥4λMσ(t)1-∑j=1m-2ajξjα-γ-1≥w(t).
By Lemma 7 and (50), the boundary value problem (1) has at least one positive solution. The proof of Theorem 10 is completed.Theorem 11.
Suppose that(51)f∞=0,
and there exist constants κ≥0 and θ>0 such that
(52)f(t,x,y)≥κ(|x|+|y|),(t,x,y)∈[14,34]×[θ4γΓ(γ+1),∞)×[θ,∞).
Then there exists a constant Λ>0 such that, for any λ∈[Λ,+∞), the BVP (1) has at least one positive solution.Proof.
Choosing(53)Λ1=[κϵ4α-γ+2(1+14γΓ(γ+1))∫1/43/4‍q(s)ds]-1,R1=max⁡{64λM1-∑j=1m-2ajξjα-γ-1,16θϵ},
and let BR1={y∈P:∥y∥≤R1}. Then for any ∈[Λ1,∞),y∈∂BR1, and t∈[1/4,3/4], we have
(54)y(t)-w(t)≥y(t)-4λMσ(t)1-∑j=1m-2ajξjα-γ-1≥y(t)-32λMy(t)(1-∑j=1m-2ajξjα-γ-1)R1≥12y(t)≥116σ(t)R1≥116ϵR1≥θ>0,(55)Iγ[y(t)-w(t)]≥Iγ(116ϵR1)≥ϵ16Γ(γ)∫0t‍(t-s)γ-1dsR1≥ϵR116×4γΓ(γ+1)≥θ4γΓ(γ+1)>0,
so for any y∈∂BR1 and t∈[0,1], by (52)–(55), we have
(56)∥Ty∥≥λ∫01‍G(14,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds≥λ∫01‍G(14,s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)ds≥λ∫01‍g(14,s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)ds≥λΓ(α-γ)k(14)∫01‍q(s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)ds≥λ(14)α-γ∫1/43/4‍q(s)f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)ds≥λκϵ4α-γ+2(1+14γΓ(γ+1))∫1/43/4‍q(s)dsR1≥R1=∥y∥.
Thus, we have
(57)∥Ty∥≥∥y∥,y∈∂BR1.
According to (51), it is clear that
(58)f∞=limsup⁡|x|+|y|→∞max⁡t∈[0,1]f(t,x,y)|x|+|y|=limsup⁡|x|+|y|→∞max⁡t∈[0,1]f(t,x,y)+M|x|+|y|=0.
Let us choose ε>0 such that
(59)4λ(Γ(γ+1)+1)εΓ(γ+1)<1.
Then there exists a large enough K>R1 such that
(60)f(t,x,y)+M≤ε(|x|+|y|),foranyt∈[0,1],|x|+|y|>K.
Thus, by (60), if
(61)|Iγ[y(s)-w(s)]*|+|[y(s)-w(s)]*|>K,
then
(62)f(t,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M≤ε(|Iγ[y(s)-w(s)]*|+|[y(s)-w(s)]*|)≤ε(1Γ(γ)∫0t‍(t-s)γ-1ds+1)∥y∥≤Γ(γ+1)+1Γ(γ+1)ε∥y∥,foranyt∈[0,1],|x|+|y|>K.
Now denote that(63)D=max⁡t∈[0,1],|x|+|y|≤Kf(t,x,y),
and choose
(64)R2=4λ(D+M)1-4λ(Γ(γ+1)+1)ε/Γ(γ+1)+K.
Then R2>K>R1.
Next letBR2={y∈P:||y||≤R2}. Then for any y∈∂BR1 and for any t∈[0,1], we have
(65)∥Ty∥=max⁡t∈[0,1](Ty)(t)=λmax⁡t∈[0,1]∫01‍G(t,s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds≤λ(α-γ-1)∫01‍q(s)[f(s,Iγ[y(s)-w(s)]*,[y(s)-w(s)]*)+M]ds≤λ(α-γ-1)×(max⁡t∈[0,1],|x|+|y|≤Kf(t,x,y)+M)∫01‍q(s)ds+λ(α-γ-1)∫01‍q(s)Γ(γ+1)+1Γ(γ+1)ε∥y∥ds≤4λ(D+M)∫01‍q(s)ds+4λΓ(γ+1)+1Γ(γ+1)ε∫01‍q(s)dsR2≤4λ(D+M)Γ(α-γ)+4λ(Γ(γ+1)+1)εΓ(α-γ)Γ(γ+1)R2≤4λ(D+M)+4λ(Γ(γ+1)+1)εΓ(γ+1)R2≤R2=∥u∥,
which implies that
(66)∥Ty∥≤∥y∥,y∈∂BR2.
By Lemma 9, T has at least a fixed points y∈(P∩BR2¯)∖BR1 such that R1≤||y||≤R2.
It follows fromR1≥64λM/(1-∑j=1m-2‍ajξjα-γ-1) that
(67)y(t)-w(t)≥y(t)-4λMσ(t)1-∑j=1m-2ajξjα-γ-1≥y(t)-32λMy(t)(1-∑j=1m-2ajξjα-γ-1)R1≥12y(t)≥116σ(t)R1≥0.
By Lemma 7 and (67), the boundary value problem (1) has at least one positive solution. The proof of Theorem 11 is completed.



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*Source: 925010-2014-04-15.xml*