# The Existence of Positive Solutions for Boundary Value Problem of the Fractional Sturm-Liouville Functional Differential Equation

Authors: Yanan Li; Shurong Sun; Zhenlai Han; Hongling Lu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study boundary value problems for the following nonlinear fractional Sturm-Liouville functional differential equations involving the Caputo fractional derivative:CDβ(p(t)CDαu(t)) + f(t,u(t-τ),u(t+θ))=0, t∈(0,1), CDαu(0)=CDαu(1)=(CDαu(0))′′=0, au(t)-bu′(t)=η(t), t∈[-τ,0], cu(t)+du′(t)=ξ(t), t∈[1,1+θ], where CDα, CDβ denote the Caputo fractional derivatives, f is a nonnegative continuous functional defined on C([-τ,1+θ],ℝ), 1<α≤2, 2<β≤3, 0<τ, θ<1/4 are suitably small, a,b,c,d>0, and η∈C([-τ,0],[0,∞)), ξ∈C([1,1+θ],[0,∞)). By means of the Guo-Krasnoselskii fixed point theorem and the fixed point index theorem, some positive solutions are obtained, respectively. As an application, an example is presented to illustrate our main results.


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## Body

## 1. Introduction

Fractional calculus is a branch of mathematics, it is an emerging field in the area of the applied mathematics that deals with derivatives and integrals of arbitrary orders as well as with their applications. The origins can be traced back to the end of the seventeenth century. During the history of fractional calculus, it was reported that the pure mathematical formulations of the investigated problems started to be addressed with more applications in various fields. With the help of fractional calculus, we can describe natural phenomena and mathematical models more accurately. Therefore, fractional differential equations have received much attention, and the theory and its applications have been greatly developed; see [1–6].Recently, there have been many papers focused on boundary value problems of fractional ordinary differential equations [7–19] and initial value problems of fractional functional differential equations [10, 20–27]. But the results dealing with the boundary value problems of fractional functional differential equations with delay are relatively scarce [28–32]. It is well known that in practical problems, the behavior of systems not only depends on the status just at the present but also on the status in the past [33]. Thus, in many cases, we must consider fractional functional differential equations with delay in order to solve practical problems. Consequently, our aim in this paper is to study the existence of solutions for boundary value problems of fractional functional differential equations.In 2005, by means of the fixed point index theorem, Bai and Ma [34] established some criteria for the existence of solutions for the boundary value problem expressed by second order differential equations with delay:
(1)-(p(t)y′(t))′+q(t)y(t)2003=f(t,y(t-τ),y(t+a)),t∈(0,1),αy(t)-βy′(t)=η(t),t∈[-τ,0],γy(t)+δy′(t)=ξ(t),t∈[1,1+φ],
where 0<τ, φ<1/4 are suitably small and p∈C1([0,1],(0,∞)), q∈C([0,1],[0,∞)), q≤0.In 2011, Li et al. [21] investigated the existence of positive solutions for the nonlinear Caputo fractional functional differential equation:
(2)D0+αu(t)+a(t)f(ut)=0,0<t<1,1<α≤2,
where D0+α is the Caputo fractional order derivative, subject to the following boundary conditions:
(3)-au(t)+bu′(t)=ξ(t),t∈[-τ,0],cu(t)+du′(t)=η(t),t∈[1,1+β].
They obtained the existence results of positive solutions by using some fixed point theorems.In 2012, Zhao et al. [22] studied the existence of positive solutions for the nonlinear Caputo fractional functional differential equation:
(4)D0+qy(t)+r(t)f(yt)=0,∀t∈(0,1),q∈(n-1,n],y(i)(0)=0,0≤i≤n-3,αy(n-2)(t)-βy(n-1)(t)=η(t),t∈[-τ,0],γy(n-2)(t)+δy(n-1)(t)=ξ(t),t∈[1,1+a].
By constructing a special cone and using the Guo-Krasnoselskii fixed point theorem, they obtained the existing results.Motivated by the works above, in this paper, we study the existence of positive solutions of boundary value problems for nonlinear fractional functional differential equation:(5)CDβ(p(t)CDαu(t))+f(t,u(t-τ),u(t+θ))=0,t∈(0,1),CDαu(0)=CDαu(1)=(CDαu(0))′′=0,au(t)-bu′(t)=η(t),t∈[-τ,0],cu(t)+du′(t)=ξ(t),t∈[1,1+θ],
where CDα,CDβ denote the Caputo fractional derivatives, f is a nonnegative continuous functional defined on C([-τ,1+θ],ℝ), 1<α≤2, 2<β≤3, 0<τ, θ<1/4 are suitably small, a,b,c,d>0 with ρ:=ac+ad+bc>0 and η∈C([-τ,0],[0,∞)), ξ∈C([1,1+θ],[0,∞)) with η(0)=ξ(1)=0, and p is a positive measurable continuous function defined on [0,1] and satisfies the following condition:
(6)0<∫EG(s,s)p(s)ds≤∫01‍G(s,s)p(s)ds<+∞,
where E={t∈[0,1]∣0≤t+a≤1,-τ≤a≤θ}, p(0)=p(1)=l, and l is a constant.Whenα=β=1, problem (5) is reduced to the problem of second order differential equations with delay and has been studied by Bai and Ma [34]. To the best of our knowledge, no one has studied the existence of positive solutions for boundary value problem (5). Key tools in finding our main results are fixed point index theorem and the Guo-Krasnoselskii fixed point theorem, and our main results of this paper are to extend and supplement some results in [21, 22, 34].The paper is organized as follows. In Section2, we will introduce some definitions and lemmas to prove our main results. In Section 3, we investigate the existence of positive solution for boundary value problems (5) by the Guo-Krasnoselskii fixed point theorem and the fixed point index theorem. As an application, an example is presented to illustrate our main results.


## 2. Preliminaries

In the following section, we introduce the definitions and lemmas which are used throughout the paper.Definition 1 (see [4]).
The fractional integral of orderα (α>0) of a function f:(t0,+∞)→ℝ is given by
(7)Iαf(t)=1Γ(α)∫t0t‍f(s)(t-s)1-αds,t>t0,
where Γ(·) is the gamma function, provided that the right side is pointwise defined on (t0,+∞).Definition 2 (see [4]).
The Caputo fractional derivative of orderα (n-1<α<n) of a function f:(t0,+∞)→ℝ is given by
(8)CDαf(t)=1Γ(n-α)∫t0t‍f(n)(s)(t-s)α+1-nds,t>t0,
where Γ(·) is the gamma function, provided that the right side is pointwise defined on (t0,+∞).
Obviously, the Caputo derivative for every constant function is equal to zero.From the definition of the Caputo derivative, we can acquire the following statements.Lemma 3 (see [5]).
Letf(t)∈L1[t0,∞). Then,
(9)CDα(Iαf(t))=f(t),t>t0,0<α<1.Lemma 4 (see [5]).
Letα>0. Then,
(10)IαCDαf(t)=f(t)-c1-c2t-⋯-cntn-1,
for some ci∈ℝ, i=1,2,…,n, where n is the smallest integer greater than or equal to α.Assume thatx0 is the solution of (5) with f≡0; then it can be expressed as
(11)x0(t)={x0(-τ;t)t∈[-τ,0]0t∈[0,1]x0(θ;t)t∈[1,1+θ],
where
(12)x0(-τ;t)=1beat/b∫t0‍e-as/bη(s)ds,x0(θ;t)=1de-ct/d∫1t‍ecs/dξ(s)ds.Next, we introduce the Green function of boundary value problems for fractional functional differential equations.Lemma 5.
Let1<α≤2 and f:[0,1]×(0,∞)×(0,∞)→ℝ be continuous. Then, the boundary value problem for fractional functional differential equation
(13)CDαu(t)+f(t,u(t-τ),u(t+θ))=0,t∈(0,1),(14)au(t)-bu′(t)=η(t),t∈[-τ,0],cu(t)+du′(t)=ξ(t),t∈[1,1+θ]
has a unique solution:
(15)u(t)={u(-τ;t)t∈[-τ,0]∫01‍G(t,s)f(s,u(s-τ),u(s+θ))dst∈[0,1]u(θ;t)t∈[1,1+θ],
where
(16)u(-τ;t)=e(a/b)t(1b∫t0‍e-(a/b)sη(s)ds+u(0)),u(θ;t)=e-(c/d)t(1d∫1t‍e(c/d)sξ(s)ds+u(1)),(17)G(t,s)=1Γ(α){g1(t,s)0≤s≤t≤1g2(t,s)0≤t≤s≤1,(18)g1(t,s)=-(t-s)α-1+b+atbc+ac+ad×(c(1-s)α-1+d(α-1)(1-s)α-2),g2(t,s)=b+atbc+ac+ad(c(1-s)α-1+d(α-1)(1-s)α-2).Proof.
It is easy to know thatu(t)=u(-τ;t) for t∈[-τ,0] and u(t)=u(θ;t) for t∈[1,1+θ]. From (13), we know that
(19)IαCDαu(t)=-Iαf(t,u(t-τ),u(t+θ)),1<α≤2.
From Lemma 4, we have
(20)u(t)=-1Γ(α)∫0t‍(t-s)α-1f(s,u(s-τ),u(s+θ))ds+c1+c2t.
Then, we get that
(21)u′(t)=-1Γ(α-1)∫0t‍(t-s)α-2f(s,u(s-τ),u(s+θ))ds+c2.
According to (14) and conditions η(0)=ξ(1)=0, we imply that
(22)au(0)-bu′(0)=0,cu(1)+du′(1)=0.
Then, we obtain
(23)c2=aac+bc+ad×(cΓ(α)∫01‍(1-s)α-1f(s,u(s-τ),u(s+θ))ds+dΓ(α-1)×∫01(1-s)α-2f(s,u(s-τ),u(s+θ))ds),c1=bac+bc+ad×(cΓ(α)∫01‍(1-s)α-1f(s,u(s-τ),u(s+θ))ds+dΓ(α-1)×∫01‍(1-s)α-2f(s,u(s-τ),u(s+θ))ds).
Therefore,
(24)u(t)=-1Γ(α)∫0t‍(t-s)α-1f(s,u(s-τ),u(s+θ))ds+at+bac+bc+ad×(cΓ(α)∫01‍(1-s)α-1f(s,u(s-τ),u(s+θ))ds+dΓ(α-1)×∫01‍(1-s)α-2f(s,u(s-τ),u(s+θ))ds)=∫01‍G(t,s)f(s,u(s-τ),u(s+θ))ds,
where G(t,s) is defined by (17). The proof is completed.Lemma 6.
Let1<α≤2, 2<β≤3 and f:[0,1]×(0,∞)×(0,∞)→ℝ be continuous. Then, the boundary value problem for fractional functional differential equation
(25)CDβ(p(t)CDαu(t))=f(t,u(t-τ),u(t+θ)),t∈(0,1),CDαu(0)=CDαu(1)=(CDαu(0))′′=0,αu(t)-bu′(t)=η(t),t∈[-τ,0],cu(t)+du′(t)=ξ(t),t∈[1,1+θ]
has a unique solution:
(26)u(t)={u(-τ;t)t∈[-τ,0]∫01‍G(t,s)×(1p(s)∫01‍H(s,v)×f(v,u(v-τ),u(v+θ))dv1p(s)∫01‍)dst∈[0,1]u(θ;t)t∈[1,1+θ],
where G(t,s) is defined by (17), and
(27)H(s,v)=1Γ(β){s(1-v)β-1-(s-v)β-10≤v≤s≤1s(1-v)β-10≤s≤v≤1.Proof.
From (25), we know that
(28)IβDCβ(p(t)DCαu(t))=Iβf(t,u(t-τ),u(t+θ)).
From Lemma 4, we have
(29)CDαu(t)=1p(t)Γ(β)×∫0t‍(t-v)β-1f(v,u(v-τ),u(v+θ))dv+c1+c2t+c3t2p(t).
According to the conditions CDαu(0)=CDαu(1)=(CDαu(0))′′=0 and p(0)=p(1)=l, we derive that
(30)c1=0,c3=0,c2=-1Γ(β)∫01‍(1-v)β-1f(v,u(v-τ),u(v+θ))dv.
Therefore,
(31)p(t)CDαu(t)=1Γ(β)∫0t‍(t-v)β-1f(v,u(v-τ),u(v+θ))dv2p-tΓ(β)∫01‍(1-v)β-1f(v,u(v-τ),u(v+θ))dv=1Γ(β)∫0t‍(t-v)β-1f(v,u(v-τ),u(v+θ))dv2p-tΓ(β)∫0t‍(1-v)β-1f(v,u(v-τ),u(v+θ))dv2p-tΓ(β)∫t1‍(1-v)β-1f(v,u(v-τ),u(v+θ))dv=-∫01‍H(t,v)f(v,u(v-τ),u(v+θ))dv.
Thus,
(32)p(t)CDαu(t)+∫01‍H(t,v)f(v,u(v-τ),u(v+θ))dv=0.
Then boundary value problem (25) is equivalent to the following problem:
(33)CDαu(t)+1p(t)×∫01‍H(t,v)f(v,u(v-τ),u(v+θ))dv=0,t∈[0,1],au(t)-bu′(t)=η(t),t∈[-τ,0],cu(t)+du′(t)=ξ(t),t∈[1,1+θ].
Lemma 5 implies that boundary value problem (33) has a unique solution:
(34)u(t)={u(-τ;t)t∈[-τ,0]∫01‍G(t,s)×(1p(s)∫01‍H(s,v)×f(v,u(v-τ),u(v+θ))dv1p(s)∫01‍)dst∈[0,1]u(θ;t)t∈[1,1+θ],
where G(t,s) and H(s,v) are defined as (17) and (27), respectively. The proof is completed.The following properties of the Green function play important roles in this paper.Lemma 7 (see [21]).
The functionG(t,s) defined by (17) satisfies the following conditions: (1)
G(t,s) is continuous on [0,1]×[0,1];(2)
for b>((2-α)/(α-1))a, we have G(t,s)>0 for t,s∈(0,1);(3)
G(t,s)≤G(s,s) for t,s∈(0,1);(4)
there exist positive numbers z,z* such that
(35)zY(s)≤G(t,s)≤z*Y(s),
where Y(s)=(b+as)((c/Γ(α))(1-s)α-1+(d/Γ(α-1))(1-s)α-2).Remark 8.
From the definition ofG(t,s) and Y(s), we know that G(s,s)=Y(s)/(ac+ad+bc). Furthermore, its easy to get from the proof of Lemma 3.3 in [21] that there exists a positive number γ1<1 such that
(36)G(t,s)≥γ1G(s,s)∀t,s∈(0,1),
where γ1=z(ac+ad+bc).Lemma 9 (see [18]).
The functionH(s,v) defined by (27) satisfies the following conditions: (1)
H(s,v)>0 for s,v∈(0,1);(2)
there exists a positive functionγ2(v)∈C(0,1) for v∈(0,1) such that
(37)min1/4≤s≤3/4H(s,v)≥γ2(v)M(v),max0≤s≤1H(s,v)≤M(v),
where
(38)M(v)=H(s0,v)=(β-2)(β-1)2/(2-β)(1-v)(β-1)2/(β-2)+v(1-v)β-1Γ(β),s0=(β-1)-1/(β-2)(1-v)(β-1)/(β-2)+v,s0∈[v,1].The following lemmas are fundamental in proof of our main results.Lemma 10 (see [35]).
LetE be a Banach space, and let K⊂E be a cone. Assume that Ω1, Ω2 are open and bounded subsets of E with 0∈Ω1, Ω-1⊂Ω2, and let T:K∩(Ω-2∖Ω1)→K be a completely continuous operator such that(i)
∥Tu∥≤∥u∥, u∈K∩∂Ω1, and ∥Tu∥≥∥u∥, u∈K∩∂Ω2;(ii)
∥Tu∥≥∥u∥, u∈K∩∂Ω1, and ∥Tu∥≤∥u∥, u∈K∩∂Ω2.
Then,T has a fixed point in K∩(Ω-2∖Ω1).Lemma 11 (see [36]).
Assume thatX is a Banach space and K⊂X is a cone. Let Kr={u∈K;∥u∥<r}. Furthermore, assume that T:K→K is compact and Tu≠u for u∈∂Kr={u∈K;∥u∥=r}. Thus, one has the following conclusions: (i)
if∥u∥≤∥Tu∥ for u∈∂Kr, then i(T,Kr,K)=0;(ii)
if∥u∥≥∥Tu∥ for u∈∂Kr, then i(T,Kr,K)=1.


## 3. Main Results

In this section, we discuss the existence of positive solutions for boundary value problem (5).LetX=C[-τ,1+θ] be a Banach space with the maximum norm ∥u∥[-τ,1+θ]=max-τ≤t≤1+θ|u(t)| for u∈X. Let K be a cone in X defined by
(39)K={u(t)≥γ1∥u∥foreacht∈[0,1]u∈X;u(t)≥0fort∈[-τ,1+θ];u(t)≥γ1∥u∥foreacht∈[0,1]}.Letx(t)=u(t)-x0(t). Noting that x(t)≡u(t) for t∈[0,1], we have
(40)x(t)={eat/bx(0)t∈[-τ,0]∫01‍G(t,s)×(1p(s)∫01‍H(s,v)×f((x+x0)(v+θ)v,(x+x0)(v-τ),(x+x0)(v+θ))dv∫1p(s))dst∈[0,1]e-ct/dx(1)t∈[1,1+θ],
where
(41)(x+x0)(v-τ):=x(v-τ)+x0(v-τ),(x+x0)(v+θ):=x(v+θ)+x0(v+θ).
Define an operator T on K as follows:
(42)(Tx)(t)={eat/b∫01‍G(0,s)×(1p(s)×∫01‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)×(v+θ))dv1p(s)∫01‍H(s,v))dst∈[-τ,0]∫01‍G(t,s)×(1p(s)×∫01‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)×(v+θ))dv1p(s)∫01‍H(s,v))dst∈[0,1]e-ct/d∫01‍G(1,s)×(1p(s)×∫01‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)×(v+θ))dv1p(s)∫01‍)dst∈[1,1+θ].Lemma 12.
The operatorT:K→K is completely continuous.Proof.
Forx∈K, we have (Tx)(t)≥0, t∈[-τ,1+θ] from properties of G(t,s) and H(s,v). It follows from (55) that we have for t∈[-τ,0],
(43)(Tx)(t)≤(Tx)(0),
and for t∈[1,1+θ],
(44)(Tx)(t)≤(Tx)(1).
Hence, we get that
(45)∥Tx∥[-τ,1+θ]=∥Tx∥[0,1].
By Lemma 7 and Remark 8, we have
(46)∥(Tx)(t)∥[0,1]≤∫01‍G(s,s)×(((v-τ))1p(s)∫01‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫01‍H(s,v))ds,
and
(47)(Tx)(t)≥∫01‍γ1G(s,s)×(((v-τ))1p(s)∫01‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫01‍)ds≥γ1∥(Tx)(t)∥[0,1]=γ1∥(Tx)(t)∥.
Thus, T(K)⊆K.
LetΩ be a bounded subset in K, that is, there exists a positive constant ζ>0 such that ∥x∥≤ζ, for all x∈Ω. Let Q:=max0≤v≤1,0≤x≤ζ|f(v,(x+x0)(v-τ),(x+x0)(v+θ))|+1. Then, for x∈Ω, in view of Lemma 9, we have
(48)(Tx)(t)=∫01‍G(t,s)×(1p(s)∫01‍H(s,v)×f(v,u(v-τ),u(v+θ))dv∫01‍)ds≤Q∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds.
Hence, T(Ω) is bounded in K.
Now, we divide three cases to prove thatT(Ω) is equicontinuous.
Case  1. If u∈Ω, 0<t<1.
Then,(49)|(Tx)′(t)|≤|-QΓ(α-1)×∫01‍(t-s)α-2(1p(s)∫01‍M(v)dv)ds+aQΓ(α)(bc+ac+ad)×∫01(c(1-s)α-1+d(α-1)(1-s)α-2)‍×(1p(s)∫01‍M(v)dv)ds|:=Q1.
Case  2. If u∈Ω, -τ≤t≤0.
Then,(50)|(Tx)′(t)|=|abeat/b∫01‍G(0,s)×(1p(s)∫01‍H(s,v)×f(v,u(v-τ),(v+θ))dv1p(s)∫01‍)dsabeat/b∫01‍|≤aQb∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds:=Q2.
Case  3. If u∈Ω, 1≤t≤1+θ.
Then,(51)|(Tx)′(t)|=|f(v,u,(v-τ),u(v+θ))dv1p(s)∫01‍)-cde-ct/d×∫01‍G(1,s)×(1p(s)∫01‍H(s,v)×f(v,u,(v-τ),u(v+θ))dv1p(s)∫01‍)ds|≤cQd∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds:=Q3.
Let δ1=ϵ1/max{Q1,Q2,Q3}. Then, for t1,t2∈[-τ,1+θ], |t2-t1|<δ1, we obtain
(52)|(Tx)(t1)-(Tx)(t2)|≤∫t1t2‍|(Tx)′(t)|dt≤max{Q1,Q2,Q3}|t2-t1|<ϵ1.
Thus, T(Ω) is equicontinuous.
Next, we show thatT:K→K is continuous. For any xn, x∈K, n=1,2… with ∥xn-x∥[-τ,1+θ]→0 as n→∞. Then, for t∈[0,1], we have
(53)|(Txn)(t)-(Tx)(t)|⩽supv∈[0,1]|f(v,(xn+x0)(v-τ),(xn+x0)(v+θ))-f(v,(x+x0)(v-τ),(x+x0)(v+θ))|×∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds.
This implies that ∥Txn-Tx∥[-τ,1+θ]→0 as n→∞. Hence, T is continuous. According to the Ascoli-Arzelà Theorem, T is completely continuous. The proof is completed.For convenience, we give some conditions, which will play roles in this paper forf(t,μ,ν) as follows(H1)
liminfμ+ν→0+mint∈[0,1]f(t,μ,ν)/(μ+ν)=∞.(H2)
liminfμ+ν→+∞mint∈[0,1](f(t,μ,ν))/(μ+ν)=∞.(H3)
limsupμ+ν→+∞mint∈[0,1](f(t,μ,ν))/(μ+ν)=0.Lemma 13.
Let(H1) and (H2) hold. Then, there exist positive numbers 0<r0<R0<∞ such that
(54)i(T,Kr,K)=0for0<r≤r0,i(T,KR,K)=0forR0≤R.Proof.
ChooseL>0 such that
(55)γ1L∫1/43/4‍G(12,s)(1p(s)∫1/43/4‍γ2(v)M(v)dv)ds>12.
By (H1), there is an r0>0 such that 0<r≤r0 implies that
(56)f(v,μ,ν)≥L(μ+ν),μ,ν≥0,μ+ν≤2r,v∈[0,1].
Thus, for x∈∂Kr, we have
(57)r=∥x∥≥x(v-τ)≥γ1∥x∥≥γ1r,v∈[14,34],r=∥x∥≥x(v+φ)≥γ1∥x∥≥γ1r,v∈[14,34].
In view of (55) and (56), we get that
(58)(Tx)(12)≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫1/43/4‍)ds≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×f(v,x(v-τ),x(v+θ))dv1p(s)∫1/43/4‍)ds≥L∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×(x(v-τ)+x(v+θ))dv1p(s)∫1/43/4‍)ds≥2γ1rL∫1/43/4‍G(12,s)(1p(s)∫1/43/4‍γ2(v)M(v)dv)ds>r=∥x∥.
Hence, ∥Tx∥>∥x∥. It is obvious that Tx≠x for x∈∂Kr. Therefore, by Lemma 11, we conclude that i(T,Kr,K)=0.
In the same way, for the sameL>0 satisfying (55), (H2) implies that there is R1>0 such that
(59)f(v,μ,ν)≥L(μ+ν),μ,ν≥0,μ+ν≥R1,v∈[0,1].
Choose
(60)R0>max{r,R12γ1}.
Thus, for x∈∂KR, R≥R0, we have
(61)R=∥x∥≥x(v-τ)≥γ1∥x∥≥γ1R≥γ1R0>R12,v∈[14,34],R=∥x∥≥x(v+θ)≥γ1∥x∥≥γ1R≥γ1R0>R12,v∈[14,34].
In view of (55) and (59), we get that
(62)(Tx)(12)≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫1/43/4‍)ds≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×f(v,x(v-τ),x(v+θ))dv1p(s)∫1/43/4‍)ds≥L∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×(x(v-τ)+x(v+θ))dv1p(s)∫1/43/4‍)ds≥2γ1LR∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)dv)ds>R=∥x∥.
Therefore, ∥Tx∥>∥x∥. By Lemma 11, we conclude that i(T,KR,K)=0. The proof is completed.Lemma 14.
Let(H3) hold. Then, there exists a 0<R0<∞ such that
(63)i(T,KR,K)=1forR0≤R.Proof.
By(H3), for any
(64)0<ϵ2<12(∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds)-1,
there exists R′>0 such that
(65)f(v,μ,ν)≤ϵ2(μ+ν),μ,ν≥0,μ+ν≥R′,v∈[0,1].
Setting
(66)C:=maxv∈[0,1]max0≤μ,ν,μ+ν≤R′|f(v,μ,ν)-ϵ2(μ+ν)|+1,
we get that
(67)f(v,μ,ν)≤ϵ2(μ+ν)+C,μ,ν≥0,v∈[0,1].
Choose
(68)R0≥(C+2ϵ2∥x0∥)∫01‍G(s,s)((1/p(s))∫01‍M(v)dv)ds(1-2ϵ2∫01‍G(s,s)((1/p(s))∫01‍M(v)dv)ds).
By (67), for x∈∂KR, R≥R0, we have
(69)(Tx)(t)=∫01‍G(t,s)×(1p(s)∫01‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫01‍H)ds≤∫01‍G(s,s)×(1p(s)∫01‍M(v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫01‍)ds≤∫01‍G(s,s)×(1p(s)∫01‍M(v)×(ϵ2((x+x0)(v-τ)+(x+x0)(v+θ))+C)dv1p(s)∫01‍)ds≤∫01‍G(s,s)×(1p(s)∫01‍M(v)(2ϵ2(∥x∥+∥x0∥)+C)dv)ds≤2ϵ2R∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds+(C+2ϵ2∥x0∥)×∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds<R=∥x∥.
Hence, ∥Tx∥<∥x∥. By Lemma 11, i(T,KR,K)=1. The proof is completed.Now, we prove the existence of solutions for boundary value problem (5) by using the Guo-Krasnoselskii fixed point theorem.Theorem 15.
Letb>(2-α)/(α-1), and let (H1) and (H3) hold. Then, boundary value problem (5) has at least a positive solution.Proof.
By(H3), there exists a Z1>0 such that
(70)f(v,μ,ν)≤ε1(μ+ν),μ,ν≥0,μ+ν≥Z1,v∈[0,1],
where
(71)0<ε1<12(∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds)-1.
Taking(72)C:=maxv∈[0,1]max0≤μ,ν,μ+ν≤Z1|f(v,μ,ν)-ε1(μ+ν)|+1,
we get that
(73)f(v,μ,ν)≤ε1(μ+ν)+C,μ,ν≥0,v∈[0,1].
Assume that
(74)Z1≥(C+2ε1∥x0∥)∫01‍G(s,s)((1/p(s))∫01‍M(v)dv)ds(1-2ε1∫01‍G(s,s)((1/p(s))∫01‍M(v)dv)ds).
By (73), for any x∈K satisfied ∥x∥=Z1 and t∈[0,1], we have
(75)|(Tx)(t)|=∫01‍G(t,s)×(1p(s)∫01‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫01‍)ds≤∫01‍G(s,s)×(1p(s)∫01‍M(v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫01‍)ds≤∫01‍G(s,s)×(1p(s)∫01‍M(v)×(ε1((x+x0)(v-τ)+(x+x0)(v+θ))+C)dv1p(s)∫01‍)ds≤∫01‍G(s,s)×(1p(s)∫01‍M(v)(2ε1(∥x∥+∥x0∥)+C)dv)ds≤2ε1Z1∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds+(C+2ε1∥x0∥)∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds<Z1=∥x∥.
Now, if we let
(76)Ω1={x∈X:∥x∥<Z1},
then (76) shows that
(77)∥Tx∥[-τ,1+θ]≤∥x∥[-τ,1+θ]forx∈K∩∂Ω1.
On the other hand, by(H1), there is an r0>0 such that 0<Z2≤r0 implies
(78)f(v,μ,ν)≥L(μ+ν),μ,ν≥0,μ+ν≤2Z2,v∈[0,1],
where L>0 satisfies
(79)γ1L∫1/43/4‍G(12,s)(1p(s)∫1/43/4‍γ2(v)M(v)dv)ds≥12.
Thus for x∈K and ∥x∥=Z2, t∈[0,1], we have
(80)Z2=∥x∥≥x(v-τ)≥γ1∥x∥≥γ1Z2,v∈[14,34],Z2=∥x∥≥x(v+φ)≥γ1∥x∥≥γ1Z2,v∈[14,34].
In view of (78) and (79), we get that
(81)|(Tx)(12)|≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫1/43/4‍)ds≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×f(v,x(v-τ),x(v+θ))dv1p(s)∫1/43/4‍)ds≥L∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×(x(v-τ)+x(v+θ))dv1p(s)∫1/43/4‍)ds≥2γ1Z2L×∫1/43/4‍G(12,s)(1p(s)∫1/43/4‍γ2(v)M(v)dv∫1/43/4‍)ds≥Z2=∥x∥.
Now, if we let
(82)Ω2={x∈X:∥x∥<Z2},
then (81) shows that
(83)∥Tx∥[-τ,1+θ]≥∥x∥[-τ,1+θ]forx∈K∩∂Ω2.
Thus, by the first part of Lemma10, T has a fixed point x∈K∩(Ω-2∖Ω1) with Z1≤∥x∥[-τ,1+θ]≤Z2, and accordingly, x is a positive solution of problem (5). The proof is completed.Theorem 16.
Letb>(2-α)/(α-1), and let (H2) and (H3) hold. Then, boundary value problem (5) has at least a positive solution.Proof.
By(H2) for any L>0 satisfying (126), there is a R0>0 such that
(84)f(v,μ,ν)≥L(μ+ν),μ,ν≥0,μ+ν≥R0,v∈[0,1].
Choose
(85)Z3>max{Z1,R02γ1}.
Thus for x∈K, ∥x∥=Z3 and t∈[0,1], we have
(86)Z3=∥x∥≥x(v-τ)≥γ1∥x∥≥γ1Z3>R02,v∈[14,34],Z3=∥x∥≥x(v+θ)≥γ1∥x∥≥γ1Z3>R02,v∈[14,34].
In view of (55) and (59), we get that
(87)|(Tx)(12)|≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫1/43/4‍)ds≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×f(v,x(v-τ),x(v+θ))dv1p(s)∫1/43/4‍)ds≥L∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×(x(v+θ)x(v-τ)+x(v+θ))dv1p(s)∫1/43/4‍)ds≥2γ1Z3L∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)dv)ds≥Z3=∥x∥.
Now, if we let(88)Ω1={x∈X:∥x∥<Z3},
then (144) shows that
(89)∥Tx∥[-τ,1+θ]≥∥x∥[-τ,1+θ]forx∈K∩∂Ω1.
On the other hand, from Theorem15, by (H3), there exists a Z1>0 such that
(90)f(v,μ,ν)≤ε2(μ+ν),μ,ν≥0,μ+ν≥Z1,v∈[0,1],
where
(91)0<ε2<12(∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds)-1.
Now, if we let
(92)Ω2={x∈X:∥x∥<Z1},
then (119) shows that
(93)∥Tx∥[-τ,1+θ]≤∥x∥[-τ,1+θ]forx∈K∩∂Ω2.
Thus, by the second part of Lemma10, T has a fixed point x∈K∩(Ω-2∖Ω1) with Z3≤∥x∥[-τ,1+θ]≤Z1, and accordingly, x is a positive solution of problem (5). The proof is completed.Next, we study the existence of solutions for boundary value problem (5) by the fixed point index theorem. For this purpose, we first give the following lemma.Lemma 17.
LetL0:=(λ∫01‍G(s,s)((1/p(s))∫01‍M(v)dv)ds)-1. Assume that
(94)limu→∞maxt∈[0,1]f(t,u)u=L~∈[0,L0).
If there exists a function g:[0,1]→ℝ such that
(95)f(t,u)≥g(t)fort∈[14,34],
Then boundary value problem for the following fractional functional differential equation:
(96)CDβ(p(t)CDαu(t))+λf(t,u)=0,t∈(0,1),CDαu(0)=CDαu(1)=(CDαu(0))′′=0,au(t)-bu′(t)=η(t),t∈[-τ,0],cu(t)+du′(t)=ξ(t),t∈[1,1+θ]
has at least a positive solution for λ>0.Proof.
Letλ>0. We define the operator Tλ on K by
(97)(Tλu)(t)=λ∫01‍G(t,s)(1p(s)∫01‍H(s,v)f(v,u)dv)ds.
In view of Lemma 12, it can be verified that Tλ is completely continuous. Then, we prove that Tλ has a fixed point in K.
Setϵ0=L0-L~. From the assumption, for any given ϵ0>0, there exists u0>0 such that
(98)f(t,u)≤(L~+ϵ0)u∀t∈[0,1],u≥u0.
For u∈K and ∥u∥=r1:=max{λ,γ1-1u0}, we have
(99)(Tλu)(t)≤λ∫01‍G(s,s)(1p(s)∫01‍M(v)f(v,u)dv)ds≤λL0r1∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds≤r1=∥u∥.
Now, if we let Ω1={u∈K:∥u∥<r1}, then it shows that
(100)∥(Tλu)(t)∥≤∥u∥,foru∈K∩∂Ω1.
Denote(101)M0:=∫1/43/4‍G(s,s)(1p(s)∫1/43/4‍γ2(v)M(v)g(v)dv)ds.
For u∈K and ∥u∥=λγ1M0, we have
(102)(Tλu)(t)≥λ∫1/43/4γ1G(s,s)×(1p(s)∫1/43/4‍γ2(v)M(v)g(v)dv)ds≥λγ1M0=∥u∥.
Now, if we let Ω2={u∈K:∥u∥<λγ1M0}, then it shows that
(103)∥(Tλu)(t)∥≥∥u∥,foru∈K∩∂Ω2.
Thus, by Lemma10, Tλ has a fixed point u~∈K∩(Ω-2∖Ω1) between r1 and λγ1M0. Accordingly, u~ is a positive solution to (96). The proof is completed.From Lemma17, we assume that λ1>0 is an eigenvalue of
(104)-CDβ(p(t)CDαu)=λ1u
subject to the following conditions:
(105)au(0)-bu′(0)=0,cu(1)+du′(1)=0,
and the corresponding eigenfunction φ1(t)>0 for t∈(0,1) such that
(106)∥φ1∥[0,1]=maxt∈[0,1]|φ1(t)|=1.For convenience, we give some conditions as follows.(H4)
liminfμ+ν→0+mint∈[0,1]f(t,μ,ν)/(μ+ν)>(1/2)λ1(1+M1), where
(107)M1=(τ+θ)(1-τ-θ)∥φ1′∥[0,1]+2(τ+θ)∫01-τ-θγ1φ1(t+τ)dt+∫τ+θ1γ1φ1(t-θ)dt.(H5)
liminfμ+ν→+∞mint∈[0,1]f(t,μ,ν)/(μ+ν)>(1/2)λ1(1+M2), where
(108)M2=(τ(1-τ)+θ(1-θ))∥φ1′∥[0,1]+τ+θ∫01-τγ1φ1(t+τ)dt+∫θ1γ1φ1(t-θ)dt.(H6)
limsupμ+ν→+∞maxt∈[0,1]f(t,μ,ν)/(μ+ν)>(1/2)λ1(1-m), where
(109)m=(∫(τ(1-τ)+θ(1-θ))∥φ1′∥[0,1]+τ+θ-∫1-τ1‍γ1φ1(t)dt-∫0θ‍γ1φ1(t)dt)×(∫01-τ‍γ1φ1(t+τ)dt+∫θ1‍γ1φ1(t-θ)dt+τ+θ)-1<1.(H7)
There is ah1>0 such that 0≤μ≤h1+∥x0∥[-τ,0], 0≤ν≤h1+∥x0∥[1,1+θ], and t∈[0,1] imply that
(110)f(t,μ,ν)<kh1,
where k=(∫01‍G(s,s)((1/p(s))∫01‍M(v)dv)ds)-1.(H8)
There is ah2>0 such that γ1h2≤μ≤h2, γ1h2≤ν≤h2, and t∈[0,1] imply that
(111)f(t,μ,ν)>ϑh2,
where ϑ=(∫1/43/4G((1/2),s)((1/p(s))∫1/43/4γ2(v)·M(v)dv)ds)-1.Lemma 18.
Let(H7) hold. Then, i(T,Kh1,K)=1.Proof.
Letx∈∂Kh1. Then,
(112)0≤(x+x0)(v-τ)≤∥x∥+∥x0∥[-τ,1-τ]=h1+∥x0∥[-τ,0],v∈[0,1],0≤(x+x0)(v+θ)≤∥x∥+∥x0∥[θ,1+θ]=h1+∥x0∥[1,1+θ],v∈[0,1].
Thus, for v∈[0,1], we have by (H7) that
(113)(Tx)(t)≤∫01‍G(s,s)×(1p(s)∫01‍M(v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫01‍)ds<kh1∫01‍G(s,s)(1p(s)∫01‍M(v)dv)ds=h1=∥x∥.
Hence, ∥Tx∥<∥x∥. By Lemma 11, we get i(T,Kh1,K)=1. The proof is complete.Lemma 19.
Let(H8) hold. Then, i(T,Kh2,K)=0.Proof.
Letx∈∂Kh2. Then,
(114)h2=∥x∥≥x(v-τ)≥γ1∥x∥≥γ1h2,v∈[14,34],h2=∥x∥≥x(v+θ)≥γ1∥x∥≥γ1h2,v∈[14,34].
Thus, for v∈[0,1], we have by (H8) that
(115)(Tx)(t)≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍H(s,v)×f(v,(x+x0)(v-τ),(x+x0)(v+θ))dv1p(s)∫1/43/4‍)ds≥∫1/43/4‍G(12,s)×(1p(s)∫1/43/4‍γ2(v)M(v)×f(v,x(v-τ),x(v+θ))dv1p(s)∫1/43/4‍)ds>ϑh2∫1/43/4‍G(12,s)(1p(s)∫1/43/4‍γ2(v)M(v)dv)ds=h2=∥x∥.
Hence, ∥Tx∥>∥x∥. By Lemma 11, we get i(T,Kh2,K)=0. The proof is complete.Theorem 20.
Let(H4), (H5), and (H7) hold. Then, boundary value problem (5) has at least two positive solutions u1 and u2 such that
(116)0<∥u1∥[0,1]<h1<∥u2∥[0,1].Proof.
According to Lemma18, we get that i(T,Kh1,K)=1. Let 0<m<1<n and f1(v,μ,ν)=(μ+ν)m+(μ+ν)n, for μ,ν≥0. Then, f1(v,μ,ν) satisfy (H1) and (H2).
DefineT1:K→K by
(117)(T1x)(t)={eat/b×∫01‍G(0,s)×(∫01‍1p(s)×∫01‍H(s,v)×f1(v,(x+x0)(v-τ),(x+x0)×(v+θ))dv1p(s)∫01‍)ds,t∈[-τ,0],∫01‍G(t,s)×(∫01‍1p(s)×∫01‍H(s,v)×f1(v,(x+x0)(v-τ),(x+x0)×(v+θ))dv1p(s)∫01‍)ds,t∈[0,1],e-ct/d∫01‍G(1,s)×(∫01‍1p(s)×∫01‍H(s,v)×f1(v,(x+x0)(v-τ),(x+x0)×(v+θ))dv1p(s)∫01‍)ds,t∈[1,1+θ].
It is easy to know that T1 is a completely continuous operator.
Also, according to Lemma13, we have that there exist 0<r0<R0<∞, such that
(118)i(T1,Kr,K)=0,0<r≤r0,(119)i(T1,KR,K)=0,R≥R0.
DefineA:[0,1]×K→K by A(s,x)=(1-s)Tx+sT1x. For any s1,s2∈[0,1], ω>0, we have
(120)|A(s1,x)-A(s2,x)|≤|(1-s1)Tx+s1T1x-(1-s2)Tx-s2T1x|≤|s1-s2|(∥T1x∥+∥Tx∥).
Note that ∥T1x∥+∥Tx∥ is uniformly bounded in Kω. Thus, A(s,x) is continuous on x∈Kω uniformly for s∈[0,1]. Then, we conclude that A is a completely continuous operator on [0,1]×Kω.
By(H4) and definition of f1, there exists a ϵ3>0 and 0<r1≤r0 such that
(121)f(t,μ,ν)≥12(λ1(1+M1)+ϵ3)(μ+ν),μ,ν≥0,μ+ν≤2r1,t∈[0,1],f1(t,μ,ν)≥12(λ1(1+M1)+ϵ3)(μ+ν),μ,ν≥0,μ+ν≤2r1,t∈[0,1].
Next, we prove thatA(s,x)≠x for all s∈[0,1] and x∈∂Kr1.
In fact, if there exists1∈[0,1] and x1∈∂Kr1 such that A(s1,x1)=x1, then x1(t) satisfies the following:
(122)-CDβ(p(t)CDαx1(t))=(1-s1)f(t,(x1+x0)(t-τ),(x1+x0)(t+θ))+s1f1(t,(x1+x0)(t-τ),(x1+x0)(t+θ))
and the following boundary conditions:
(123)ax1(t)-bx1′(t)=0,t∈[-τ,0],cx1(t)+dx1′(t)=0,t∈[1,1+θ].
Multiplying both sides of (122) by φ1(t), then integrating from 0 to 1 and using Green’s formula, we obtain
(124)λ1∫01‍x1(t)φ1(t)dt=∫01‍(-CDβ(p(t)CDαx1(t)))φ1(t)dt=(1-s1)×∫01‍f(t,(x1+x0)(t-τ),(x1+x0)(t+θ))φ1(t)dt+s1∫01‍f1(t,(x1+x0)(t-τ),(x1+x0)(t+θ))φ1(t)dt.
Thus, we know that
(125)λ1∫01‍x1(t)φ1(t)dt≥(1-s1)×∫τ1-θ‍f(t,(x1+x0)(t-τ),(x1+x0)(t+θ))×φ1(t)dt+s1∫τ1-θ‍f1(t,(x1+x0)(t-τ),(x1+x0)(t+θ))×φ1(t)dt=(1-s1)∫τ1-θ‍f(t,x1(t-τ),x1(t+θ))φ1(t)dt+s1∫τ1-θ‍f1(t,x1(t-τ),x1(t+θ))φ1(t)dt≥12(1-s1)(λ1(1+M1)+ϵ3)×∫τ1-θ‍(x1(t-τ)+x1(t+θ))φ1(t)dt+12s1(λ1(1+M1)+ϵ3)×∫τ1-θ‍(x1(t-τ)+x1(t+θ))φ1(t)dt=12(λ1(1+M1)+ϵ3)×(∫τ1-θ‍x1(t-τ)φ1(t)dt+∫τ1-θ‍x1(t+θ)φ1(t)dt)=12(λ1(1+M1)+ϵ3)×(∫01-θ-τ‍x1(t)φ1(t+τ)dt+∫τ+θ1‍x1(t)φ1(t-θ)dt).
Then, we have
(126)(λ1M1+ϵ3)×(∫01-θ-τ‍x1(t)φ1(t+τ)dt+∫τ+θ1‍x1(t)φ1(t-θ)dt)≤2λ1∫01‍x1(t)φ1(t)dt-λ1(∫01-θ-τ‍x1(t)φ1(t+τ)dt+∫τ+θ1‍x1(t)φ1(t-θ)dt)=λ1∫01-θ-τ‍x1(t)(φ1(t)-φ1(t+τ))dt+λ1∫1-τ-θ1‍x1(t)φ1(t)dt+λ1∫τ+θ1‍x1(t)(φ1(t)-φ1(t-θ))dt+λ1∫0τ+θ‍x1(t)φ1(t)dt≤λ1τ(1-θ-τ)∥φ1′∥[0,1]∥x1∥+λ1(θ+τ)∥φ1∥[0,1]∥x1∥+λ1θ(1-θ-τ)∥φ1′∥[0,1]∥x1∥+λ1(θ+τ)∥φ1∥[0,1]∥x1∥=λ1((θ+τ)(1-θ-τ)∥φ1′∥[0,1]+2(θ+τ))∥x1∥.
By the definition of K, we have
(127)∫01-θ-τ‍x1(t)φ1(t+τ)dt+∫τ+θ1‍x1(t)φ1(t-θ)dt≥γ1∥x1∥(∫01-θ-τ‍φ1(t+τ)dt+∫θ+τ1‍φ1(t-θ)dt).
Combining with (126), we get that
(128)(λ1M1+ϵ3)γ1(∫01-θ-τ‍φ1(t+τ)dt+∫θ+τ1‍φ1(t-θ)dt)≤λ1((θ+τ)(1-θ-τ)∥φ1′∥[0,1]+2(θ+τ)),
which is a contradiction for definition of M1 in (H4). Thus, A(s,x)≠x for x∈∂Kr and s∈[0,1]. In view of (118) and homotopy invariance of the fixed point index, we obtain
(129)i(T,Kr1,K)=i(H(0,·),Kr1,K)=i(H(1,·),Kr1,K)=i(T1,Kr1,K)=0.
On the other hand, by (H5) and definition of f1, there are ϵ4>0 and R1>h1 such that
(130)f(t,μ,ν)≥12(λ1(1+M2)+ϵ4)(μ+ν),μ,ν≥0,μ+ν>R′,t∈[0,1],f1(t,μ,ν)≥12(λ1(1+M2)+ϵ4)(μ+ν),μ,ν≥0,μ+ν>R′,t∈[0,1].
Setting
(131)C:=maxt∈[0,1]max0≤μ,ν,μ+ν≤R′|f(t,μ,ν)-(λ1(1+M2)+ϵ4)(μ+ν)|+maxt∈[0,1]max0≤μ,ν,μ+ν≤R′|f1(t,μ,ν)-(λ1(1+M2)+ϵ4)(μ+ν)|+1,
and it follows that
(132)f(t,μ,ν)≥12(λ1(1+M2)+ϵ4)(μ+ν)+C,μ,ν≥0,t∈[0,1],f1(t,μ,ν)≥12(λ1(1+M2)+ϵ4)(μ+ν)+C,μ,ν≥0,t∈[0,1].
Then, we prove that there exists a R1≥R′ such that A(s,x)≠x for all s∈[0,1] and x∈K, ∥x∥≥R1. In fact, if there exist s1∈[0,1] and x1∈K such that A(s1,x1)=x1, we get that
(133)λ1∫01‍x1(t)φ1(t)dt≥12(1-s1)(λ1(1+M2)+ϵ4)×∫01‍((x1+x0)(t-τ)+(x1+x0)(t+θ))×φ1(t)dt-(1-s1)C∫01‍φ1(t)dt+12s1(λ1(1+M2)+ϵ4)×∫01‍((x1+x0)(t-τ)+(x1+x0)(t+θ))×φ1(t)dt-s1C∫01‍φ1(t)dt≥12(λ1(1+M2)+ϵ4)×∫01‍(x1(t-τ)+x1(t+θ))φ1(t)dt-C∫01‍φ1(t)dt≥12(λ1(1+M2)+ϵ4)×(∫τ1‍x1(t-τ)φ1(t)dt+∫01-θ‍x1(t+θ)φ1(t)dt)+C∫01‍φ1(t)dt=12(λ1(1+M2)+ϵ4)×(∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt)+C∫01‍φ1(t)dt.
Then, we have
(134)(λ1M2+ϵ4)×(∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt)≤λ1∫01-τ‍x1(t)(φ1(t)-φ1(t+τ))dt+λ1∫1-τ1‍x1(t)φ1(t)dt+λ1∫θ1‍x1(t)(φ1(t)-φ1(t-θ))dt+λ1∫0θ‍x1(t)φ1(t)dt+2C∫01‍φ1(t)dt≤λ1τ(1-τ)∥φ1′∥[0,1]∥x1∥+λ1τ∥φ1∥[0,1]∥x1∥+λ1θ(1-θ)∥φ1′∥[0,1]∥x1∥+λ1θ∥φ1∥[0,1]∥x1∥+2C∥φ1∥[0,1]=λ1(τ(1-τ)+θ(1-θ)∥φ1′∥[0,1]+θ+τ)×∥x1∥+2C.
By the definition of K, we get that
(135)∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt≥γ1∥x1∥(∫01-τ‍φ1(t+τ)dt+∫θ1‍φ1(t-θ)dt).
Combining with (134), we conclude that
(136)(λ1M2+ϵ4)γ1∥x1∥(∫01-τ‍φ1(t+τ)dt+∫θ1‍φ1(t-θ)dt)≤λ1((τ(1-τ)+θ(1-θ))∥φ1′∥[0,1]+θ+τ)∥x1∥+2C,
that is,
(137)∥x1∥≤2Cϵ4γ1(∫01-τ‍φ1(t+τ)dt+∫θ1‍φ1(t-θ)dt)=R-.
Let R1=1+max{R′,R-}. Then, we obtain A(s,x)≠x for x∈∂Kr and s∈[0,1]. In view of (118) and homotopy invariance of the fixed point index, we obtain
(138)i(T,KR1,K)=i(A(0,·),KR1,K)=i(A(1,·),KR1,K)=i(T1,KR1,K)=0.
By using additivity, we get that
(139)i(T,KR1∖K-h1,K)=-1,i(T,Kh1∖K-r1,K)=1.
Since
(140)∥ui∥[0,1]=∥xi+x0∥[0,1]=∥xi∥[0,1]=∥Txi∥[0,1]=∥Txi∥=∥xi∥,
it follows that u1(t) and u2(t) satisfy
(141)0<∥u1∥[0,1]=∥x1∥<h1<∥x2∥=∥u2∥[0,1].
The proof is completed.Theorem 21.
Let(H6), (H7), and (H8) hold. Assume that 0<h1<h2; then boundary value problem (5) has at least two positive solutions u1 and u2 such that
(142)h1<∥u1∥[0,1]<h2<∥u2∥[0,1].Proof.
According to Lemmas18 and 19, we obtain
(143)i(T,Kh1,K)=1,(144)i(T,Kh2,K)=0.
By (H6) and definition of f1, there are ϵ5>0 and R2>h2 such that
(145)f(t,μ,ν)≤12(λ1(1-M3)-ϵ5)(μ+ν),μ,ν≥0,μ+ν>R′,t∈[0,1].
Let
(146)C:=maxt∈[0,1]max0≤μ,ν,μ+ν≤R′|f(t,μ,ν)-(λ1(1-M3)-ϵ5)(μ+ν)|+1,
and it follows that
(147)f(t,μ,ν)≤12(λ1(1-M3)-ϵ5)(μ+ν)+C,μ,ν≥0,t∈[0,1].
DefineA:[0,1]×K→K by A(s,x)=sTx. For any s1,s2∈[0,1], ω>0, we have
(148)|A(s1,x)-A(s2,x)|≤|s1-s2|∥Tx∥.
Thus, A(s,x) is continuous on x∈Kω uniformly for s∈[0,1]. Then, we conclude that A is also a completely continuous operator on [0,1]×Kω.
Next, we prove that there exists aR1≥R′ such that A(s,x)≠x for all s∈[0,1] and x∈K, ∥x∥≥R1.
In fact, if there exists2∈[0,1] and x2∈K such that A(s2,x2)=x2, then x2(t) satisfies the following:
(149)-CDβ(p(t)CDαx2(t))=s2f1(t,(x2+x0)(t-τ),(x2+x0)(t+θ)),t∈(0,1).
Multiplying both sides of (149) by φ1(t) then integrating from 0 to 1 and using Green’s formula, we get that
(150)λ1∫01‍x2(t)φ1(t)dt=∫01‍(-CDβ(p(t)CDαx2(t)))φ1(t)dt=s2∫01‍f1(t,(x2+x0)(t-τ),(x2+x0)(t+θ))φ1(t)dt.
Thus, we get that
(151)λ1∫01‍x1(t)φ1(t)dt=s1∫01‍f(t,(x1+x0)(t-τ),(x1+x0)(t+θ))φ1(t)dt≤12(λ1(1-M3)-ϵ5)×∫01‍((x1+x0)(t-τ)+(x1+x0)(t+θ))φ1(t)dt+C∫01‍φ1(t)dt=12(λ1(1-M3)-ϵ5)×(∫0τ‍(x1+x0)(t-τ)φ1(t)dt+∫τ1‍(x1+x0)(t-τ)φ1(t)dt+∫01-θ‍(x1+x0)(t+θ)φ1(t)dt+∫1-θ1‍(x1+x0)(t+θ)φ1(t)dt)+C∫01‍φ1(t)dt=12(λ1(1-M3)-ϵ5)×(∫0τ‍(x1+x0)(t-τ)φ1(t)dt+∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt+∫1-θ1‍(x1+x0)(t+θ)φ1(t)dt)+C∫01‍φ1(t)dt≤12(λ1(1-M3)-ϵ5)×(∫θ1‍τ∥x1+x0∥∥φ1∥[0,1]+∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt+θ∥x1+x0∥∥φ1∥[0,1])+C∥φ1∥[0,1]≤12(λ1(1-M3)-ϵ5)×((τ+θ)(∥x1∥+∥x0∥)+∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt+∥φ1∥[0,1])+C.
Then, we have
(152)(λ1M3+ϵ5)×(∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt+(τ+θ)(∥x1∥+∥x0∥))≤λ1∫01-τ‍x1(t)(φ1(t+τ)-φ1(t))dt+λ1∫θ1‍x1(t)(φ1(t-θ)-φ1(t))dt+λ1(θ+τ)(∥x1∥+∥x0∥)-λ1∫1-τ1‍x1(t)φ1(t)dt-λ1∫0θ‍x1(t)φ1(t)dt+2C≤λ1τ(1-τ)∥φ1′∥[0,1]∥x1∥+λ1θ(1-θ)∥φ1′∥[0,1]∥x1∥-λ1γ1∫1-τ1‍φ1(t)dt∥x1∥+λ1(θ+τ)(∥x1∥+∥x0∥)-λ1γ1∫0θ‍φ1(t)dt∥x1∥+2C.
By the definition of K, we get that
(153)∫01-τ‍x1(t)φ1(t+τ)dt+∫θ1‍x1(t)φ1(t-θ)dt≥γ1∥x1∥(∫01-τ‍φ1(t+τ)dt+∫θ1‍φ1(t-θ)dt).
This combined with (126) gives,
(154)(λ1M3+ϵ5)×γ1(∫01-τ‍φ1(t+τ)dt+∫θ1‍φ1(t-θ)dt+τ+θ)∥x1∥≤λ1(∫1θ‍τ(1-τ)∥φ1′∥[0,1]+θ(1-θ)∥φ1′∥[0,1]+θ+τ-∫1-τ1‍φ1(t)dt-∫1θ‍φ1(t)dt)∥x1∥+2C,
that is,
(155)∥x1∥≤2C+λ1(θ+τ)∥x0∥ϵ5γ1(∫01-τ‍φ1(t+τ)dt+∫θ1‍φ1(t-θ)dt+τ+θ)=R-.
Let R1=1+max{R′,R-}. Then, we obtain A(s,x)≠x for x∈∂Kr and s∈[0,1]. In view of (143), (144), and homotopy invariance of the fixed point index, we obtain
(156)i(T,KR1,K)=i(A(1,·),KR1,K)=i(A(0,·),KR1,K)=i(T1,KR1,K)=1.
By using additivity, we get that
(157)i(T,KR1∖K-h2,K)=1,i(T,Kh2∖K-h1,K)=-1.
Thus T has fixed points x1 and x2 in KR1∖K-h2 and Kh2∖K-h1, which means that u1(t)=x1(t)+x0(t) and u2(t)=x2(t)+x0(t) are positive solutions of boundary value problem (5) and
(158)h1<∥u1∥[0,1]=∥x1∥<h2<∥x2∥=∥u2∥[0,1].
The proof is completed.Corollary 22.
Let(H4), (H7), and (H8) hold. Assume that 0<h1<h2. Then, boundary value problem (5) has at least two positive solutions u1 and u2 such that
(159)0<∥u1∥[0,1]<h1<∥u2∥[0,1].Corollary 23.
Let(H5), (H7) and (H8) hold. Assume 0<h2<h1. Then boundary value problem (5) has at least two positive solutions u1 and u2 such that
(160)h2<∥u1∥[0,1]<h1<∥u2∥[0,1].


## 4. Example

In this section, we will present an example to illustrate our main results.Consider the following fractional functional differential equations with delay:(161)CD5/2((t-12)2CD3/2u(t))+(t+1)2+{u(t-15)+u(t+15)}1/3=0,t∈[0,1],CD3/2u(0)=CD3/2u(1)=(CD3/2u(0))′′=0,u(t)-2u′(t)=sint,t∈[-13,0]3u(t)+5u′(t)=cost,t∈[1,43].By a simple computation, we can get thatb>((2-α)/(α-1))a and
(162)mint∈[0,1]f(t,μ,ν)μ+ν=mint∈[0,1](t+1)2+(u+v)1/3u+v=1u+v+1(u+v)2/3⟶∞,
as u+v→0+.Similarly, we can obtainmint∈[0,1]f(t,μ,ν)/(μ+ν)→0 as u+v→+∞.Then, conditions(H1) and (H3) are satisfied. Then, by Theorem 15, boundary value problem (161) has at least a positive solution.



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*Source: 301560-2013-09-22.xml*