# The Existence of Positive Solutions for Fractional Differential Equations with Integral and Disturbance Parameter in Boundary Conditions

Authors: Mei Jia; Xiping Liu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study the existence and nonexistence of the positive solutions for the integral boundary value problem of the fractional differential equations with the disturbance parametera in the boundary conditions and the impact of the disturbance parameter a on the existence of positive solutions. By using the upper and lower solutions method, fixed point index theory and the Schauder fixed point theorem, we obtain sufficient conditions for that the problem has at least one positive solution, two positive solutions and no solutions. Under certain conditions, we also obtain the demarcation point which divides the disturbance parameters into two subintervals such that the boundary value problem has positive solutions for the disturbance parameter in one subinterval while no positive solutions in the other.


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## Body

## 1. Introduction

In this paper, we are concerned with the existence and nonexistence of positive solutions for the boundary value problem of the fractional differential equations(1)-CDδu(t)=f(t,u(t)),t∈J,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds+a,
where J=[0,1],  1<δ≤2,  f∈C(J×ℝ+,ℝ+),  mi≥0,  ni≥0,  mi2+ni2>0,  i=1,2,  g∈C(J,ℝ+), disturbance parameter a∈ℝ+, and CDδ is the Caputo fractional derivative of order δ.In the recent decades, since fractional differential equations have been applied widely and successfully in the description of complex dynamics, they have been regarded as a valuable tool being used in the fields to handle viscoelastic, physics, chemistry, electrical engineering, biology aspects, and so forth; see [1–12] and references therein. And besides, the boundary value problems for the differential equations appear in many applications; see [13–16]. As a result, the boundary value problems for the fractional differential equations are one of the most active fields in the researches of nonlinear differential equations theories and plenty of meaningful achievements have been gained in the related fields; see [1–8, 17–29] and the references therein. Due to that the boundary value problems with the integral boundary conditions include two-, three-, and multipoint boundary value problems as special cases and they can better describe the actual phenomenon; more and more emphases have been put on the researches of them; see [19, 20, 30–34] and the references therein.At the same time, while using the methods of the differential equations to solve the actual problems, it is inevitable that there always exists disturbance which will have great influence on the existence of the solutions. In paper [35, 36], the authors studied nonlinear nonlocal boundary value problem with nonhomogeneous boundary conditions
(2)u′′(t)+f(t,u(t),u′(t))=0,t∈(0,1),u(0)-∑i=1m‍aiu(ti)=λ1,u(1)-∑i=1m‍biu(ti)=λ2,
where they discussed the impact of disturbance parameters λ1,λ2 on the existence of the solution and obtained some meaningful conclusions. And then, the authors further studied the 2nth order nonlinear nonlocal boundary value problem with nonhomogeneous boundary conditions; see [37].The purpose of this paper is to study the impact of the disturbance parametera on the existence of positive solutions and obtain sufficient conditions for the boundary value problem (1) to have at least one positive solution, at least two solutions, and no solutions. Under certain conditions, we obtain that there exists a constant a*>0, which separates ℝ+ into two disjoint subintervals M=[0,a*] and N=(a*,+∞) such that the boundary value problem (1) has at least two positive solutions for each a∈(0,a*), at least one positive solution for a=0 and a=a*, and no positive solutions for a∈N. The main tools we applied are the upper and lower solutions method, fixed point index theory, and the Schauder fixed point theorem.This paper is organized as follows. In Section2, we introduce the basic definitions and the basic properties of the integral kernel. In Section 3, we study the comparison principles and the basic lemmas. In Section 4, we consider the existence and nonexistence of the positive solutions of the boundary value problem (1), and we study the impact of the disturbance parameter a on the existence of positive solutions.


## 2. Preliminaries

In this section, we give some basic definitions and lemmas which play an important role in our research.Definition 1 (See [2, 3]).
Letδ>0, for a function u:(0,+∞)→ℝ. The Riemann-Liouville fractional integral operator of order δ of u is defined by
(3)Iδu(t)=1Γ(δ)∫0t‍(t-s)δ-1u(s)ds,
provided the integral exists.
The Caputo derivative of orderδ for a function u:(0,+∞)→ℝ is given by
(4)CDδu(t)=1Γ(n-α)∫0t‍u(n)(s)(t-s)δ+1-nds,
provided the right side is pointwise defined on (0,+∞), where n=[δ]+1 and n-1<δ<n.
Ifδ=n, then CDδu(t)=u(n)(t).Definition 2.
LetAC[0,1] be the space of functions u which are absolutely continuous on [0,1]. We denote ACn[0,1] by the set of functions u which have continuous derivatives up to order n-1 on [0,1] such that u(n-1)∈AC[0,1]. In particular, AC1[0,1]=AC[0,1].Lemma 3 (See [2]).
Ifu∈ACn[0,1], then the Caputo fractional derivative CDδu(t) exists almost everywhere on [0,1], where n is the smallest integer greater than or equal to δ.Lemma 4 (See [2]).
Supposeδ>0 and u∈ACn(J). Then
(5)IδCDδu(t)=u(t)+c0+c1t+c2t2+⋯+cn-1tn-1,ck=u(k)(0)k!,k=0,1,2,…,n-1,
where n is the smallest integer greater than or equal to δ.Definition 5.
One saysu is a solution of the boundary value problem (1) if u∈AC2(J),  CDδu(t)∈C(J) and satisfies (1). One says u is a positive solution of the boundary value problem (1) if u is a solution of the boundary value problem (1) and u(t)≥0 and u(t)≢0 for t∈J.Throughout this paper, we assume the following conditions hold.(H1)
ρ>0,  g->0 and n2((δ-1)n1-(2-δ)m1)>0, where
(6)ρ=m1m2+m1n2+m2n1,g-=1ρ-∫01‍(n1+m1s)g(s)ds.(H2)
There exists a functionc∈AC2(J,(-∞,0]) and CDδc∈C(J) such that m1c(0)-n1c′(0)=0,  m2c(1)+n2c′(1)≥-1 and ∫01‍g(s)c(s)ds≤-2.(H3)
f(·,u) is monotonically increasing with respect to u and f≢0.Fory∈C(J), firstly we consider the boundary value problem
(7)-CDδu(t)=y(t),t∈J,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds.Lemma 6.
Suppose(H1) holds. Then the boundary value problem (7) has the unique solution
(8)u(t)=∫01‍H(t,s)y(s)ds,
where the function H is given by
(9)H(t,s)=G(t,s)+h(t,s),
where
(10)G(t,s)=1ρΓ(δ)×{(n1+m1t)(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2)-ρ(t-s)δ-1,0≤s<t≤1,(n1+m1t)(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2),0≤t≤s<1,(11)h(t,s)=g-(n1+m1t)∫01‍g(r)G(r,s)dr.Proof.
According to Lemma4, -CDδu(t)=y(t) is equivalent to the following equation:
(12)u(t)=-Iδy(t)+c0+c1t,
where c0=u(0),c1=u′(0).
By the boundary conditionm1u(0)-n1u′(0)=0 and m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds, we can show
(13)m1c0-n1c1=0,m2c0+(m2+n2)c1=∫01‍g(s)u(s)ds+m2Iδy(1)+n2Iδ-1y(1).
Hence, we can obtain(14)c0=n1ρ(∫01‍g(s)u(s)ds+m2Iδy(1)+n2Iδ-1y(1)),c1=m1ρ(∫01‍g(s)u(s)ds+m2Iδy(1)+n2Iδ-1y(1)).
It follows the definition Riemann-Liouville fractional integral that(15)u(t)=-1Γ(δ)∫0t‍(t-s)δ-1y(s)ds+n1+m1tρΓ(δ)×(∫01‍m2(1-s)δ-1y(s)ds+∫01‍n2(δ-1)(1-s)δ-2y(s)ds)+n1+m1tρ∫01‍g(s)u(s)ds,
that is,
(16)u(t)=∫01‍G(t,s)y(s)ds+n1+m1tρ∫01‍g(s)u(s)ds.
We multiply by the functiong on both sides of (16), integrate from 0 to 1,
(17)∫01‍g(s)u(s)ds=∫01‍g(s)(∫01‍G(s,r)y(r)dr)ds+∫01‍g(s)u(s)dsρ∫01‍(n1+m1s)g(s)ds=∫01‍(∫01‍g(r)G(r,s)dr)y(s)ds+∫01‍g(s)u(s)dsρ∫01‍(n1+m1s)g(s)ds,
and we can easily get
(18)∫01‍g(s)u(s)ds=ρρ-∫01‍(n1+m1s)g(s)ds×∫01‍(∫01‍g(r)G(r,s)dr)y(s)ds=ρg-∫01‍(∫01‍g(r)G(r,s)dr)y(s)ds.
So,
(19)u(t)=∫01‍G(t,s)y(s)ds+n1+m1tρ·ρg-∫01‍(∫01‍g(r)G(r,s)dr)y(s)ds=∫01‍G(t,s)y(s)ds+∫01‍h(t,s)y(s)ds,
where h(t,s)=g-(n1+m1t)∫01‍g(r)G(r,s)dr.
Hence,(20)u(t)=∫01‍H(t,s)y(s)ds.
It follows from (12) that
(21)u′(t)=-Iδ-1y(t)+m1ρ(∫01‍g(s)u(s)ds+m2Iδy(1)+n2Iδ-1y(1)).
We can obtainu∈AC2(J) and CDδu(t)∈C(J).Similarly, we can obtain the following lemma.Lemma 7.
Suppose(H1) holds and a∈ℝ. Then the boundary value problem
(22)-CDδu(t)=0,t∈J,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds+a
has the unique solution
(23)u*(t)=g-a(n1+m1t).Lemma 8.
Suppose(H1) holds. Then the solution u∈AC2(J) and CDδu(t)∈C(J) of the boundary value problem (1) is equivalent to the solution u∈C(J) of the integral equation
(24)u(t)=∫01‍H(t,s)f(s,u(s))ds+g-a(n1+m1t),t∈J.Proof.
Ifu∈AC2(J) and CDδu(t)∈C(J) is the solution of the boundary value problem (1), by Lemma 4, -CDδu(t)=f(t,u(t)) is equivalent to the equation
(25)u(t)=-Iδf(t,u(t))+c0+c1t,
where c0=u(0),c1=u′(0).
For convenience, we denotey(t)=f(t,u(t)).
Similar to the proof of Lemma6, we can obtain
(26)c0=n1ρ(∫01‍g(s)u(s)ds+m2Iδy(1)+n2Iδ-1y(1)+a),c1=m1ρ(∫01‍g(s)u(s)ds+m2Iδy(1)+n2Iδ-1y(1)+a).
Substitutingc0 and c1 into (25), we can obtain
(27)u(t)=∫01‍G(t,s)y(s)ds+n1+m1tρ∫01‍g(s)u(s)ds+(n1+m1t)aρ.
We multiply by the functiong on both sides of (27), integrate from 0 to 1, and get
(28)∫01‍g(s)u(s)ds=g-ρ∫01‍(∫01‍g(r)G(r,s)dr)y(s)ds+g-a∫01‍(n1+m1s)g(s)ds.
Since g-=1/(ρ-∫01‍(n1+m1s)g(s)ds), we have
(29)u(t)=∫01‍H(t,s)y(s)ds+g-a(n1+m1t).
That is,
(30)u(t)=∫01‍H(t,s)f(s,u(s))ds+g-a(n1+m1t).
On the other hand, ifu∈C(J) is the solution of the integral equation (24), we have
(31)u′(t)=-1Γ(δ-1)∫0t‍(t-s)δ-2f(s,u(s))ds+m1ρΓ(δ)∫01‍(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2)×f(s,u(s))ds+g-m1∫01‍(∫01‍g(r)G(r,s)dr)ds+g-am1.
It is easy to see that u′∈AC(J).
Hence,u∈AC2(J) and CDδu∈C(J).
We can easily verify thatu satisfies the boundary value problem (1).
Therefore,u∈AC2(J) and CDδu(t)∈C(J) is the solution of the boundary value problem (1).Lemma 9.
Suppose(H1) holds. Then the functions G,  h, and H have the following properties; (1)
G∈C([0,1]×[0,1)), 0≤G(t,s)≤(((m1+n1)(m2+n2(δ-1)))/Γ(δ)ρ)(1-s)δ-2 for (t,s)∈[0,1]×[0,1) and there exists a constant 0<γ1<1 such that for s∈[0,1)(32)min⁡t∈JG(t,s)≥γ1max⁡t∈JG(t,s);(2)
h(t,s)≥0 for (t,s)∈[0,1]×[0,1) and there exists a constant 0<γ2<1 such that for s∈[0,1)(33)min⁡t∈Jh(t,s)=γ2max⁡t∈Jh(t,s);(3)
H(t,s)≥0 for (t,s)∈[0,1]×[0,1) and for s∈[0,1)(34)min⁡t∈JH(t,s)≥γmax⁡t∈JH(t,s),
where γ=min⁡{γ1,γ2}.Proof.
(
1
) By the expression of G(t,s), it is easy to see G∈C([0,1]×[0,1)).
Since (H1) holds, we can show that for(t,s)∈[0,1]×[0,1)(35)G(t,s)≤(n1+m1)(m2+n2(δ-1))ρΓ(δ)(1-s)δ-2.
For 0≤t≤s<1, it follows
(36)G(t,s)≥n1n2(δ-1)(1-s)δ-2ρΓ(δ)≥n2(1-s)δ-2ρΓ(δ)(n1(δ-1)-(2-δ)m1).
For 0≤s<t≤1, by (10), we have
(37)∂G(t,s)∂t=1ρΓ(δ)(m1(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2)-ρ(δ-1)(t-s)δ-2),∂2G(t,s)∂t2=1ρΓ(δ)(-ρ(δ-1)(δ-2)(t-s)δ-3)>0.
It is easy to see that(38)max⁡t∈[s,1]∂G(t,s)∂t=∂G(t,s)∂t|t=1=1ρΓ(δ)(m1(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2)22222222222-ρ(δ-1)(1-s)δ-2(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2))=m2(1-s)δ-2ρΓ(δ)(m1(1-s)-(m1+n1)(δ-1)).
(H1) implies that (m1+n1)(δ-1)>m1. By (38), we have max⁡t∈[s,1](∂G(t,s)/∂t)≤0. Hence,
(39)min⁡t∈[s,1]G(t,s)=G(1,s)=n2(1-s)δ-2ρΓ(δ)((n1+m1)(δ-1)-m1(1-s))≥n2(1-s)δ-2ρΓ(δ)((n1+m1)(δ-1)-m1).
For(t,s)∈[0,1]×[0,1), we can show that
(40)G(t,s)≥n2(1-s)δ-2ρΓ(δ)((n1+m1)(δ-1)-m1)>0
from (36) and (39).
We denoteγ1=n2((n1+m1)(δ-1)-m1)/(n1+m1)(m2+n2(δ-1)). It is obvious that 0<γ1<1.
By (35) and (39), for (t,s)∈[0,1]×[0,1), we have
(41)min⁡t∈JG(t,s)≥n2((n1+m1)(δ-1)-m1)ρΓ(δ)(1-s)δ-2≥n2((n1+m1)(δ-1)-m1)ρΓ(δ)·ρΓ(δ)(n1+m1)(m2+n2(δ-1))·G(t,s)=γ1G(t,s).
Hence,min⁡t∈J⁡G(t,s)≥γ1max⁡t∈J⁡G(t,s).
(
2
) By the expression of h(t,s), see (11), it is easy to see h(t,s)≥0 for (t,s)∈[0,1]×[0,1) and
(42)min⁡t∈Jh(t,s)=γ2max⁡t∈Jh(t,s)fors∈[0,1),
where γ2=n1/(n1+m1).
(
3
) By (1) and (2), we have H(t,s)≥0 for (t,s)∈[0,1]×[0,1) and
(43)min⁡t∈JH(t,s)=min⁡t∈J(G(t,s)+h(t,s))≥min⁡t∈JG(t,s)+min⁡t∈Jh(t,s)≥γ1max⁡t∈JG(t,s)+γ2max⁡t∈Jh(t,s)≥γ(max⁡t∈JG(t,s)+max⁡t∈Jh(t,s))≥γmax⁡t∈JH(t,s).For the sake of the reader, we state the fixed point index theorem and Schauder’s fixed point theorem which will be used later.Lemma 10 (See [38]).
LetP be a cone of a real Banach space E,  Ω be a bounded open set in E, and θ∈Ω. Suppose A:P∩Ω¯→P is a completely continuous operator. If Au≠ρu for any u∈∂Ω∩P and ρ≥1, then
(44)i(A,Ω∩P,P)=1.Lemma 11 (Schauder’s fixed point theorem, see [39]).
LetE be a real Banach space, and let Ω⊂E be nonempty closed bounded and convex; A:Ω→Ω compact. Then A has a fixed point.


## 3. Comparison Principle and the Existence of Solutions

Definition 12.
Letα∈AC2(J) and CDδα∈C(J). One says that α is a lower solution of the boundary value problem (1), if
(45)-CDδα(t)≤f(t,α(t))+pα(t),t∈J,m1α(0)-n1α′(0)=0,
Where
(46)pα(t)={0,m2α(1)+n2α′(1)≤∫01g(s)α(s)ds+a,(CDδc(t))(∫01‍∫01g(s)α(s)ds+a-m2α(1)-n2α′(1)∫01‍)m2α(1)+n2α′(1)>∫01g(s)α(s)ds+a.
Letβ∈AC2(J) and CDδβ∈C(J). We say that β is an upper solution of the boundary value problem (1), if
(47)-CDδβ(t)≥f(t,β(t))+qβ(t),t∈J,m1β(0)-n1β′(0)=0,
Where
(48)qβ(t)={0,m2β(1)+n2β′(1)≥∫01‍g(s)β(s)ds+a,(CDδc(t))(∫01‍∫01‍g(s)β(s)ds+a-m2β(1)-n2β′(1)∫01‍),m2β(1)+n2β′(1)<∫01‍g(s)β(s)ds+a,
where c(t) is defined in (H2).The following comparison principle will play a very important role in our main results analysis.Lemma 13.
Let(H1) hold. Suppose that u∈AC2(J),  CDδu∈C(J), and satisfies
(49)-CDδu(t)≥0,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)≥∫01‍g(s)u(s)ds.
Then u(t)≥0 for t∈J.Proof.
Denote(50)-CDδu(t)=y(t),
then y(t)≥0 for t∈J. Let a≥0 such that
(51)m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds+a.
By Lemma8, we can get that the boundary value problem
(52)-CDδu(t)=y(t),t∈J,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds+a
has unique solution
(53)u(t)=∫01‍H(t,s)y(s)ds+u*(t),
where u*(t)=g-a(n1+m1t) and t∈J.
It follows thatu(t)≥0 for t∈J from Lemma 9.Lemma 14.
Let(H1) and(H2) hold. Suppose that u∈AC2(J), CDδu∈C(J), and satisfies
(54)-CDδu(t)-(CDδc(t))×(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))≥0,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)<∫01‍g(s)u(s)ds.
Then u(t)≥0 for t∈J.Proof.
Let(55)v(t)=u(t)+c(t)(∫01‍g(s)u(s)ds222222222222222-m2u(1)-n2u′(1)∫01‍).
Since (H2) holds, we havec(t)≤0 for t∈J, m1c(0)-n1c′(0)=0, m2c(1)+n2c′(1)≥-1, and ∫01‍g(s)c(s)ds≤-2.
Hence, by (54), we have
(56)u(t)≥v(t),fort∈J,-CDδv(t)=-CDδu(t)-(CDδc(t))×(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))≥0,m1v(0)-n1v′(0)=m1u(0)-n1u′(0)+(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))×(m1c(0)-n1c′(0))=m1u(0)-n1u′(0)=0,m2v(1)+n2v′(1)=m2u(1)+n2u′(1)+(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))×(m2c(1)+n2c′(1))≥m2u(1)+n2u′(1)-(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))=(-2)(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))+∫01‍g(s)u(s)ds≥∫01‍g(s)c(s)ds×(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))+∫01‍g(s)u(s)ds=∫01‍g(s)v(s)ds.
In view of Lemma 13, v(t)≥0 for t∈J, which implies that u(t)≥0.We can easily obtain the following lemma from the definition ofu*, where u* is defined by Lemma 7.Lemma 15.
Suppose(H1) holds. u is a positive solution of the boundary value problem (1) if and only if w=u-u* is a positive solution of the boundary value problem
(57)-CDδw(t)=f(t,w(t)+u*(t)),t∈J,m1w(0)-n1w′(0)=0,m2w(1)+n2w′(1)=∫01‍g(s)w(s)ds.LetE=C(J) be the Banach space with the norm ∥u∥=max⁡t∈J⁡|u(t)|,
(58)P={u∈E:u(t)≥0fort∈J},
and let
(59)P0={u∈E:u(t)≥0,u(t)≥γ∥u∥fort∈J}
be cones in E, and P0⊂P, where γ is defined by Lemma 9.Lemma 16.
Suppose(H1) holds. If w is a positive solution of the boundary value problem (57), then
(60)min⁡t∈Jw(t)≥γ∥w∥.Proof.
By Lemma6, we can show that the solution w of the boundary value problem (57) satisfies
(61)w(t)=∫01‍H(t,s)f(s,w(s)+u*(s))ds.
In view of Lemma9(3),
(62)min⁡t∈Jw(t)=min⁡t∈J∫01‍H(t,s)f(s,w(s)+u*(s))ds≥∫01‍γmax⁡t∈JH(t,s)f(s,w(s)+u*(s))ds≥γmax⁡t∈J∫01‍H(t,s)f(s,w(s)+u*(s))ds=γmax⁡t∈Jw(t)=γ∥w∥.We defineT:P→E by
(63)Tw(t)=∫01‍H(t,s)f(s,w(s)+u*(s))ds.By Lemma16, we have T:P→P0. Since f and H are nonnegative, then u is a positive solution of (1) if and only if w∈P is a fixed point of the operator T.Theorem 17.
Suppose (H1), (H2), and (H3) hold and there exist a nonnegative lower solutionα and an upper solution β of the boundary value problem (1) such that α≤β. Then the boundary value problem (1) has at least one solution u such that α≤u≤β.Proof.
Let(64)F(t,u)={f(t,β(t)),u>β(t),f(t,u),α(t)≤u≤β(t),f(t,α(t)),u<α(t).
We consider the boundary value problem
(65)-CDδu(t)=F(t,u(t)),t∈J,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds+a.
By Lemma15, u is a positive solution of the boundary value problem (65) if and only if w=u-u* is a positive solution of the following boundary value problem:
(66)-CDδw(t)=F(t,w(t)+u*(t)),t∈J,m1w(0)-n1w′(0)=0,m2w(1)+n2w′(1)=∫01‍g(s)w(s)ds.
We defineT^:P→E by
(67)T^w(t)=∫01‍H(t,s)F(s,w(s)+u*(s))ds.
Thenw is a solution of (66) if and only if w is a fixed point of the operator T^. It is easy to see T^:P→P0.
Next we can prove thatT^ is completely continuous.
LetD⊂P be a bounded set, and there exists a constant M0>0 such that ∥w∥≤M0 for w∈D. Because F is continuous, there exists a constant M1>0 such that max⁡s∈J,w∈D⁡|F(s,w(s)+u*(s))|≤M1. We have
(68)|T^w(t)|≤M1∫01‍H(t,s)ds≤M1max⁡t∈J∫01‍H(t,s)ds.
So T^(D) is uniformly bounded.
We denote(69)G1(t,s)=1ρΓ(δ)(n1+m1t)×(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2),G2(t,s)=1Γ(δ){-(t-s)δ-1,0≤s≤t≤1,0,0≤t<s≤1,z(s)=1ρΓ(δ)×(m2(1-s)δ-1+n2(δ-1)(1-s)δ-2)+g-∫01‍g(r)G(r,s)dr.
Hence,
(70)H(t,s)=G(t,s)+h(t,s)=G1(t,s)+G2(t,s)+h(t,s)=(n1+m1t)z(s)+G2(t,s),
where h(t,s) is defined by (11).
SinceG2(t,s) is continuous on J×J, we have G2 which is uniformly continuous on J×J. It implies that for any ε>0, there exists δ1>0, when t1,t2∈J; whenever |t2-t1|<δ1 and s∈J, we can obtain
(71)|G2(t2,s)-G2(t1,s)|<ε2M1.
By (H1), it is easy to see∫01‍z(s)ds≠0. We take 0<δ0≤min⁡{δ1,ε/(2M1(m1+1)∫01‍z(s)ds)}.
Therefore, ast1,t2∈J, whenever |t2-t1|<δ0 and u∈D, we can show that
(72)|T^w(t2)-T^w(t1)|=|∫01‍(H(t2,s)-H(t1,s))F(s,w(s)+u*(s))ds|≤M1∫01‍|H(t2,s)-H(t1,s)|ds≤M1∫01‍m1|t2-t1|z(s)ds+M1∫01‍|G2(t2,s)-G2(t1,s)|ds<ε.
Thus, we have provedT^ is equicontinuous.
By Arzela-Ascoli theorem, we know thatT^(D) is relatively compact.
We can easily show thatT^ is continuous since F is continuous. Hence, T^ is completely continuous.
SinceF is bounded and T^ is completely continuous, we can get T^ has at least one fixed point w by Schauder fixed point theorem, that is, there exists a solution w of the boundary value problem (66).
Thenu=w+u* is a solution of the boundary value problem (65).
Finally, we proveα(t)≤u(t)≤β(t), for t∈J.
We can prove that if each solutionu of the boundary value problem (65) satisfies α(t)≤u(t)≤β(t) for t∈J, then u is a solution of the boundary value problem (1).
Letv(t)=u(t)-α(t), for t∈J.
Ifm2α(1)+n2α′(1)≤∫01‍g(s)α(s)ds+a, since f(t,u) is monotonically increasing with respect to u and α is a lower solution of (1), we have
(73)-CDδv(t)=-CDδu(t)+CDδα(t)-CDδv(t)≥F(t,u(t))-f(t,α(t))≥0,m1v(0)-n1v′(0)=m1u(0)-n1u′(0)-(m1α(0)-n1α′(0))=0,m2v(1)+n2v′(1)=(m2u(1)+n2u′(1))-(m2α(1)+n2α′(1)),≥(∫01‍g(s)u(s)ds+a)-(∫01‍g(s)α(s)ds+a)=∫01‍g(s)v(s)ds.
It follows thatv(t)≥0 on J from Lemma 13.
Ifm2α(1)+n2α′(1)>∫01‍g(s)α(s)ds+a, we have
(74)-CDδv(t)-(CDδc(t))×(∫01‍g(s)v(s)ds-m2v(1)-n2v′(1))=-CDδu(t)-(CDδc(t))×(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))-(∫01‍-CDδα(t)-(CDδc(t))×(∫01‍g(s)α(s)ds-m2α(1)-n2α′(1)))≥F(t,u(t))-(CDδc(t))×(∫01‍g(s)u(s)ds-m2u(1)-n2u′(1))-(∫01‍f(t,α(t))+(CDδc(t))×(∫01‍g(s)α(s)ds+a-m2α(1)-n2α′(1))-(CDδc(t))(∫01‍g(s)α(s)ds2222222222222222222-m2α(1)-n2α′(1)∫01‍)∫01‍)=F(t,u(t))-f(t,α(t)).
By (H3), we can getF(t,u(t))-f(t,α(t))≥0, that is,
(75)-CDδv(t)-(CDδc(t))×(∫01‍g(s)v(s)ds-m2v(1)-n2v′(1))≥0.
So,
(76)m1v(0)-n1v′(0)=(m1u(0)-n1u′(0))-(m1α(0)-n1α′(0))=0,m2v(1)+n2v′(1)=m2u(1)+n2u′(1)-(m2α(1)+n2α′(1))<∫01‍g(s)u(s)ds+a-(∫01‍g(s)α(s)ds+a)=∫01‍g(s)v(s)ds.
It follows thatv(t)≥0 on J from Lemma 14.
Hence, we showα(t)≤u(t) on J.
Similarly, we can getu(t)≤β(t) on J.
Therefore, each solutionu of the boundary value problem (65) satisfies α(t)≤u(t)≤β(t) for t∈J. That is, F(t,u(t))=f(t,u(t)), and u is a solution of the boundary value problem (1).Theorem 18.
Suppose (H1), (H2), and (H3) hold:
(
1
) if there exists a constant a->0 such that the boundary value problem
(77)-CDδu(t)=f(t,u(t)),t∈J,m1u(0)-n1u′(0)=0,m2u(1)+n2u′(1)=∫01‍g(s)u(s)ds+a-
has a positive solution u-(t), then for each a with 0≤a≤a-, the boundary value problem (1) has a positive solution u and u*≤u≤u-, where u*(t)=g-a(n1+m1t) for t∈J;
(
2
) if there exists a constant a->0 such that the boundary value problem (77) does not have positive solutions, then for each a>a-, the boundary value problem (1) does not have positive solutions.Proof.
(
1
) Let u-*(t)=g-a-(n1+m1t) for t∈J. By Lemma 15, w=u--u-* is a positive solution of the boundary value problem
(78)-CDδw(t)=f(t,w(t)+u-*(t)),t∈J,m1w(0)-n1w′(0)=0,m2w(1)+n2w′(1)=∫01‍g(s)w(s)ds,
which implies u-*≤u-.
Sinceu-*(t)=g-a-(n1+m1t)>g-a(n1+m1t)=u*(t), we take α=u* and β=u-.
We can easily verify thatα and β are a lower solution and an upper solution of the boundary value problem (1) and α≤β.
By Theorem17, we have that the boundary value problem (1) has a positive solution u and u*≤u≤u-.
(
2
) If there exists a constant a0>a- such that the boundary value problem (77) has a positive solution, by (1), we can show that for each a with 0≤a≤a0, the boundary value problem (1) has a positive solution. So, the boundary value problem (1) has a positive solution, for a=a-, which is a contradiction.
Therefore, if there exists a constanta->0 such that the boundary value problem (77) does not have positive solutions, then for each a>a-, the boundary value problem (1) does not have positive solutions.


## 4. Impact of Disturbance Parametera on the Existence of Solutions

For convenience, we give the following notations:(79)f∞=limsup⁡u→+∞max⁡t∈Jf(t,u)u,f∞=liminf⁡u→+∞min⁡t∈Jf(t,u)u,f0=limsup⁡u→0max⁡t∈Jf(t,u)u,f0=liminf⁡u→0min⁡t∈Jf(t,u)u.We can see that∫01‍max⁡t∈JH(t,s)ds≠0 from (H1) and we denote η=1/(∫01‍max⁡t∈JH(t,s)ds).Lemma 19.
Suppose that(H1) holds, f∞>η/γ2, and Ω⊆[0,+∞) is a bounded set. Then for each a∈Ω, there exists a constant τ such that the solution u of the boundary value problem (1) satisfies ∥u∥<τ.Proof.
SinceΩ⊆[0,+∞) is a bounded set, there exists a constant σ such that each a∈Ω, and we have 0≤a≤σ.
Sincef∞>η/γ2, there exists a constant r>0 such that
(80)f(t,u)>ηγ2u
for any t∈J, and u≥r.
By Lemma15, u is a positive solution of the boundary value problem (1) if and only if w=u-u* is a positive solution of the boundary value problem (57).
In view of Lemma16, min⁡t∈J⁡w(t)≥γ∥w∥. We can get ∥w∥≤r/γ. Otherwise, if ∥w∥>r/γ, we have
(81)|w(t)+u*(t)|≥|w(t)|≥γ∥w∥≥r
for t∈J. Hence,
(82)w(t)=∫01‍H(t,s)f(s,w(s)+u*(s))ds>ηγ2∫01‍H(t,s)|w(s)+u*(s)|ds≥ηγ2·γ2∥w∥∫01‍max⁡t∈JH(t,s)ds=∥w∥,
which is a contradiction.
We takeτ=(r/γ)+g-σ(n1+m1)+1, then
(83)∥u∥≤∥w∥+∥u*∥≤rγ+g-σ(n1+m1)<τ.Theorem 20.
Suppose (H1) holds. If one of the following conditions is satisfied, then the boundary value problem (1) does not have positive solutions:(1)
f0>η/γ2,  a≥0 and a is small enough;(2)
f∞>η/γ2,  a>0 and a is large enough;(3)
there exist constantsK1≥0 and K2≥η/γ2 such that
(84)f(t,u)>K1+K2ufor (t,u)∈J×[0,+∞), and a≥0.Proof.
(
1
) If there exists a≥0 and a is small enough such that the boundary value problem (1) has a positive solution u, we have that w=u-u* is a positive solution of the boundary value problem (57) by Lemma 15, and min⁡t∈J⁡w(t)≥γ∥w∥ by Lemma 16.
Becauseliminf⁡u→0⁡min⁡t∈J⁡(f(t,u)/u)=f0, there exists a constant r1>0 such that
(85)f(t,u)>ηuγ2
for any t∈J, and 0<u≤r1.
We takea≥0 and a is small enough such that
(86)0<|w(t)+u*(t)|=w(t)+g-a(n1+m1t)≤r1
for t∈J. Hence,
(87)w(t)=∫01‍H(t,s)f(s,w(s)+u*(s))ds>ηγ2·γ∫01‍max⁡t∈JH(t,s)(w(s)+u*(s))ds≥ηγ2·γ2∥w∥∫01‍max⁡t∈JH(t,s)ds=∥w∥,
which is a contradiction.
(
2
) If there exists a constant a^>0 and a^ is large enough such that the boundary value problem (1) has a positive solution u^, we have that w^=u^-u^* is a positive solution by Lemma 15.
Byliminf⁡u→+∞⁡min⁡t∈J⁡(f(t,u)/u)=f∞, there exists a constant r2>0 such that
(88)f(t,u)>ηuγ2
for any t∈J, and u≥r2. We take a^>r2/g-n1, then
(89)|w^(t)+u^*(t)|≥|u^*(t)|=g-a^(n1+m1t)≥r2
for t∈J. Hence,
(90)w^(t)=∫01‍H(t,s)f(s,w^(s)+u^*(s))ds>ηγ2·γ∫01‍max⁡t∈JH(t,s)(w^(s)+u^*(s))ds≥ηγ2·γ2∥w^∥∫01‍max⁡t∈JH(t,s)ds=∥w^∥,
which is a contradiction.
(
3
) Since there exist constants K1≥0 and K2≥η/γ2 such that
(91)f(t,u)>K1+K2u
for (t,u)∈J×[0,+∞), we have
(92)w(t)=∫01‍H(t,s)f(s,w(s)+u*(s))ds>γ∫01‍max⁡t∈JH(t,s)(K1+K2(w(s)+u*(s)))ds≥K2γ2∥w∥∫01‍max⁡t∈JH(t,s)ds≥∥w∥,
which is a contradiction.Theorem 21.
Suppose (H1) holds(1)
If0≤f0<η, then the boundary value problem (1) has at least one positive solution when a≥0 and a is small enough.(2)
If0≤f∞<η, then the boundary value problem (1) has at least one positive solution when a≥0.Proof.
(
1
) Because limsup⁡u→0⁡max⁡t∈J⁡(f(t,u)/u)=f0, there exists a constant r1>0 such that
(93)f(t,u)<ηu
for any t∈J, and 0<u≤r1.
LetD={w∈P:∥w+u*∥≤r1}, where u*(t)=g-a(n1+m1t) for t∈J. So g-an1≤u*(t)≤g-a(n1+m1) for t∈J.
Let0≤a<r1/g-a(n1+m1), then D≠∅.
For anyw∈D, we have
(94)0≤Tw(t)=∫01‍H(t,s)f(s,w(s)+u*(s))ds≤η∫01‍max⁡t∈JH(t,s)|w(s)+u*(s)|ds≤∥w+u*∥η·1η≤r1.
That is T(D)⊂D. By the Schauder fixed point theorem, we can get that T has at least one fixed point on D. In view of Lemma 15, u=w+u* is a positive solution of the boundary value problem (1).
(
2
) Since limsup⁡u→+∞⁡max⁡t∈J⁡(f(t,u)/u)=f∞, there exist constants r-2>0 and ϵ>0 such that
(95)f(t,u)<(f∞+ϵ)u
for any t∈J and u≥r-2, where ϵ satisfies f∞+ϵ<η.
LetM=max⁡(t,u)∈J×[0,r-2]⁡f(t,u). We have
(96)f(t,u)≤M+(f∞+ϵ)ufor(t,u)∈J×[0,+∞).
We chooser2>max⁡{r-2,M/(η-(f∞+ϵ))} and D={w∈P:∥w+u*∥≤r2}, where u*(t)=g-a(n1+m1t) for t∈J. Then, for any w∈D,
(97)0≤Tw(t)=∫01‍H(t,s)f(s,w(s)+u*(s))ds≤∫01‍max⁡t∈JH(t,s)(M+(f∞+ϵ)|w(s)+u*(s)|)ds≤(M+∥w+u*∥(f∞+ϵ))·1η≤((η-(f∞+ϵ))r2+r2(f∞+ϵ))·1η=r2.
That is T(D)⊂D. By the Schauder fixed point theorem, we can get that T has at least one fixed point on D. In view of Lemma 15, u=w+u* is a positive solution of the boundary value problem (1).Theorem 22.
Suppose that (H1), (H2), and (H3) hold. If0≤f0<η and f∞>η/γ2, then there exists a constant a*∈(0,+∞) with the following properties.
(
1
)  a* separates [0,+∞) into two disjoint subintervals M=[0,a*] and N=(a*,+∞).
(
2
) The boundary value problem (1) has at least two positive solutions for each a∈(0,a*), has at least one positive solution for a=0 and a=a*, and does not have positive solutions for each a∈N.Proof.
We have the following four steps to prove the conclusions of Theorem22.
Step  1. Let
(98)Λ={a∈[0,+∞):Theboundaryvalueproblemhasatleastonepositivesolution}.
By Theorem 21(1), we have Λ≠∅.
For eacha~∈Λ, denote
(99)A(a~)={a:0≤a≤a~}.
Then, in view of Theorem18,   A(a~)⊆Λ if and only if a~∈Λ.
It follows thatΛ is a bounded set from f∞>η/γ2 and Theorem 20(2).
LetM=⋃a~∈ΛA(a~), then M is a bounded set. Hence, the set M has the supremum; we denote a*=sup⁡M and we have a*>0 from Theorem 21.
Step  2. We prove that the boundary value problem (1) has at least one positive solution when a=a*.
Sincea*=sup⁡M, there exist {ak}⊂M and ak<a* such that ak→a* as k→+∞.
Letuk be a solution of the boundary value problem (1) with a replaced by ak. By Lemma 8, the boundary value problem (1) with a=ak is equivalent to
(100)uk(t)=∫01‍H(t,s)f(s,uk(s))ds+g-ak(n1+m1t),k=1,2,….
In view of Lemma9, {uk} is uniformly bounded and equicontinuous. By Arzela-Ascoli, {uk} has a convergent subsequence; we also denote {ukm}, and {ukm} converge to u.
By Lebesgue dominated convergence theorem, we can get(101)u(t)=∫01‍H(t,s)f(s,u(s))ds+g-a*(n1+m1t).
Hence,u is a positive solution of the boundary value problem (1) with a=a*.
Step  3. By Theorem 18, when 0≤a≤a*, the boundary value problem (1) has at least one positive solution. That is, M=[0,a*].
LetN=(a*,+∞). Then a* separates [0,+∞) into two disjoint subintervals M=[0,a*] and N=(a*,+∞); we have that the boundary value problem (1) has at least one positive solution for each a∈M and does not have positive solutions for each a∈N.
Step  4. We prove the boundary value problem (1) has at least two positive solutions when a∈(0,a*).
For eacha∈(0,a*), there exist a_, a-∈M, such that 0<a_<a<a-.
Letu- be a solution of the boundary value problem (1) with a replaced by a-. By Theorem 18, the boundary value problem (1) has a positive solution u1 with u1≤u-.
Similarly, letu_ be a solution of the boundary value problem (1) with a replaced by a_ and u_≤u1. Hence, u_≤u1≤u-.
In fact, we can easily show thatu_<u1<u- from a_<a<a- and (H1).
Letα=u_ and β=u-. We can easily verify α and β are a lower solution and an upper solution of the boundary value problem (1), respectively, and α<β.
We choosea^∈N, then a<a*<a^.
DefineK:[a,a^]×P→E by
(102)K(κ,u)=g-κ(n1+m1t)+∫01‍H(t,s)f(s,u(s))ds.
Let
(103)F(t,u)={f(t,β(t)),u>β(t),f(t,u),α(t)≤u≤β(t),f(t,α(t)),u<α(t),
and define K^:[a,a^]×P→E by
(104)K^(κ,u)=g-κ(n1+m1t)+∫01‍H(t,s)F(s,u(s))ds.
Similar to the proof of Theorem17, we can prove K and K^ are completely continuous for each κ∈[a,a^]. By Lemma 8, u is a positive solution of the boundary value problem (1) if and only if u=K(a,u).
By Lemma19, there exists a constant τ such that the fixed point u of K satisfies ∥u∥<τ for each κ∈[a,a^].
Let(105)Ω={u∈P:∥u∥<τ,α(t)<u<β(t),t∈J}.
Clearly, Ω is a nonempty open-bounded subset of P.
SinceF is bounded, there exists a constant R>τ>0 such that |K^(κ,u)|<R for any (κ,u)∈[a,a^]×P. Let B(θ,R)={u∈E:∥u∥<R}; it is obvious that Ω⊂P∩B(θ,R) and u≠μK^u for u∈P∩∂B(θ,R), and μ∈[0,1]. Hence, by Lemma 10, for each κ∈[a,a^], we have
(106)i(K^(κ,u),P∩B(θ,R),P)=1.
BecauseK^ does not have a fixed point on P∩(B(θ,R)¯∖Ω), then for each κ∈[a,a^],
(107)i(K^(κ,u),P∩(B(θ,R)∖Ω¯),P)=0.
We notice that K^|Ω=K, by the excision property of the fixed point index, (106) and (107), for each κ∈[a,a^], we have
(108)i(K(κ,u),P∩Ω,P)=i(K^(κ,u),P∩B(θ,R),P)-i(K^(κ,u),P∩(B(θ,R)∖Ω¯),P)=1.
Sincea^∈N, we have K(a^,u)≠u for any u∈P. So,
(109)i(K(a^,u),P∩B(θ,R),P)=0.
We defineH:[0,1]×P∩B(θ,R)¯→E by
(110)H(t,u)=K((1-t)a+ta^,u).
It is obvious thatH is completely continuous.
We can proveH(t,u)≠u for (t,u)∈[0,1]×P∩∂B(θ,R).
Otherwise, if there exists(t0,u0)∈[0,1]×P∩∂B(θ,R) such that H(t0,u0)=u0, that is
(111)K((1-t0)a+t0a^,u0)=u0,∥u0∥=R.
Hence, u0 is a solution of the boundary value problem (1) with a replaced by κ=(1-t0)a+t0a^. We can get ∥u0∥<τ, which is a contradiction.
According to the homotopy invariance of the fixed point index and (109), it follows that
(112)i(K(a,u),P∩B(θ,R),P)=i(H(0,u),P∩B(θ,R),P)=i(H(1,u),P∩B(θ,R),P)=i(K(a^,u),P∩B(θ,R),P)=0.
By the additivity property of the fixed point index, (108) and (112), we obtain
(113)i(K(a,u),P∩B(θ,R)∖Ω¯,P)=-1.
Therefore, the boundary value problem (1) has a solution u2 in P∩(B(θ,R)∖Ω¯) and u1≠u2 from u1∈Ω. That is, the boundary value problem (1) has at least two positive solutions.



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*Source: 131548-2014-03-24.xml*