# Existence of Positive Solutions for m-Point Boundary Value Problem for Nonlinear Fractional Differential Equation

Authors: Moustafa El-Shahed; Wafa M. Shammakh
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We investigate an m-point boundary value problem for nonlinear fractional differential equations.  The associated Green function for the boundary value problem is given at first, and some useful properties of the Green function are obtained. By using the fixed point theorems of cone expansion and compression of norm type and Leggett-Williams fixed point theorem, the existence of multiple positive solutions is obtained.


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## Body

## 1. Introduction

In recent years, the existence of positive solutions multipoint boundary value problems of fractional order differential equations has been studied by many authors using various methods (see [1–7]).The study of multipoint boundary value problems for linear second-order ordinary differential equations was initiated by II'in and Moiseev [8, 9].Since then, nonlinear multipoint boundary value problems have been studied by several authors (see [10–14]). Recently, in [15], the authors have studied the existence of at least one positive solution for the following nth-order three-point boundary value problem:(1.1)u(n)(t)+h(t)f(t,u(t))=0,t∈[a,b],u(a)=αu(η),u′(a)=u′′(a)=⋯=u(n-2)(a)=0,u(b)=βu(η),
where a<η<b,0≤α<1,f∈C([a,b]×[0,∞),[0,∞)) and h∈C([a,b]×[0,∞)) may be singular at t=a and t=b.Goodrich [16] considered the BVP for thehigher-dimensional fractional differential equation as follows:(1.2)-D0+νy(t)=f(t,y(t)),0<t<1,n-1<ν≤n,y(i)(0)=0,0≤i≤n-2,[D0+αy(t)]t=1=0,1≤α≤n-2,
and  a Harnack-like inequality associated with the Green function related to the above problem is obtained improving the results in [17].Motivated by the aforementioned results and techniques in coping with those boundary value problems of fractional differential equations, we then turn to investigate the existence and multiplicity of positive solutions for the following BVP:(1.3)CDa+αu(t)+f(t,u(t))=0,a≤t≤b,n-1≤α<n,n>2,(1.4)u′(a)=∑i=1m-2βiu′(ηi),u′′(a)=u′′′(a)=⋯=u(n-1)(a)=0,u(b)=∑i=1m-2γiu(ηi),
where a<η1<η2<⋯<ηm-2<b,∑i=1m-2βi<1,∑i=1m-2γi<1 and CDa+α are the Caputo fractional derivative.In this paper, we study the existence of at least one positive solution, existence of two positive solutions associated with the BVP (1.3)-(1.4) by applying the fixed point theorems of cone expansion and compression of norm type, and the existence of at least three positive solutions for BVP (1.3)-(1.4) by using Leggett-Williams fixed point theorem.The rest of the paper is organized as follows. In Section2, we introduce some basic definitions and preliminaries used later. In Section 3, the existence of multipoint boundary value problem (1.3)-(1.4) will be discussed.


## 2. Preliminaries

In this section, we introduce definitions and preliminary facts which are used throughout this paper.Definition 2.1 (see [18]).
For a functiony:(a,∞)→R, the Caputo derivative of fractional order α>0 is defined as
(2.1)DCa+αy(t)=1Γ(n-α)∫at(t-s)n-α-1y(n)(s)ds,n-1<α≤n.Definition 2.2 (see [18]).
The standard Riemann-Liouville fractional derivative of orderα>0 of  a continuous function  y:(a,∞)→R  is given by
(2.2)Da+αy(t)=1Γ(n-α)(ddt)n∫at(t-s)n-α-1y(s)ds,
where n=[α]+1, provided that the integral on the right-hand side converges.Definition 2.3 (see [18]).
The Riemann-Liouville fractional integral of orderα>0 of a function y:(a,∞)→R is given by
(2.3)Ia+αy(t)=1Γ(α)∫at(t-s)α-1y(s)ds
provided that the integral on the right-hand side converges.Definition 2.4 (see [19]).
LetE be a real Banach space. A nonempty closed convex set K⊂E is called  cone of E if it satisfies the following conditions:(1)
x∈K,σ≥0impliesσx∈K;(2)
x∈K,-x∈Kimpliesx=0.Definition 2.5.
An operator is called completely continuous if it is continuous and maps bounded sets into precompact  sets.Theorem 2.6 (see [20]).
LetE be a Banach space and K⊂E is a cone in E. Assume that Ω1 and Ω2 are open subsets of E with 0∈Ω1 and  Ω̅1⊂Ω2. Let T:K∩(Ω̅2∖Ω1)→K be completely continuous operator. In addition, suppose either(i)
∥Tu∥≤∥u∥,forallu∈K∩∂Ω1,and∥Tu∥≥∥u∥,forallu∈K∩∂Ω2or(ii)
∥Tu∥≤∥u∥,forallu∈K∩∂Ω2,and∥Tu∥≥∥u∥,forallu∈K∩∂Ω1holds. Then, T has a fixed point in K∩(Ω̅2∖Ω1).Lemma 2.7.
Forα>0, the general solution of the fractional differential equation DCa+αu(t)=0 is given by
(2.4)u(t)=c0+c1(t-a)+c2(t-a)2+⋯+cn-1(t-a)n-1,
where  ci∈R,i=0,1,2,…,n-1.Remark 2.8 (see [18]).
In view of Lemma2.7, it follows that
(2.5)Ia+αDCa+αu(t)=u(t)+c0+c1(t-a)+c2(t-a)2+⋯+cn-1(t-a)n-1,
for some ci∈R,i=0,1,2,…,n-1.Definition 2.9.
The mapθ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E provided that θ:P→[0,∞) is continuous and
(2.6)θ(λx+(1-λ)y)≥λθ(x)+(1-λ)θ(y),
for  all x,y∈P,0≤λ≤1.Lemma 2.10 (see [21]).
LetP be a cone in a real Banach space E,Pc={x∈P:∥x∥<c},θ is a nonnegative continuous concave functional on P such that  θ(x)≤∥x∥, for all x∈Pc, and P(θ,b,d)={x∈P:b≤θ(x),x≤d}. Suppose that T:Pc→Pc is completely continuous and there exist positive constants 0<a<b<d≤c such that (C1)
{x∈P(θ,b,d):θ(x)>b}≠ϕandθ(x)>bforx∈P(θ,b,d),(C2)
∥Tx∥<aforx∈Pa,(C3)
θ(Tx)>bforx∈P(θ,b,d)with∥Tx∥>d,then T has at least three fixed points x1,x2, and x3 with
(2.7)‖x1‖<a,b<θ(x2),a<‖x3‖withθ(x3)<b.Lemma 2.11.
For a giveny(t)∈C[a,b] and n-1≤α<n, the unique solution of the boundary value problem
(2.8)DCa+αu(t)+y(t)=0,a≤t≤b,n-1≤α<n,n>2,n∈N,(2.9)u′(a)=∑i=1m-2βiu′(ηi),u′′(a)=u′′′(a)=⋯=u(n-1)(a)=0,u(b)=∑i=1m-2γiu(ηi),
is given by
(2.10)u(t)=∫abG(t,s)y(s)ds+∫abH(t,s;η1,…,ηm-2)y(s)ds,
where
(2.11)G(t,s)=1Γ(α){(b-s)α-1-(t-s)α-1,a≤s≤t≤b,(b-s)α-1,a≤t≤s≤b,H(t,s;η1,…,ηm-2)={∑i=1m-2γi[(b-s)α-1-(ηi-s)α-1]δ2Γ(α)+μ(t)∑i=1m-2βi(ηi-s)α-2δ1δ2Γ(α-1),a≤s≤ηi,i=1,2,…,m-2,∑i=1m-2γi(b-s)α-1δ2Γ(α),ηi≤s≤b,i=1,2,…,m-2,δ1=1-∑i=1m-2βi,δ2=1-∑i=1m-2γi,μ(t)=(b-∑i=1m-2γiηi)-δ2t.Proof.
Using Remark2.8, for arbitrary constants ci∈R,i=0,1,2,…,n-1, we have
(2.12)u(t)=-1Γ(α)∫at(t-s)α-1y(s)ds+c0+c1(t-a)+c2(t-a)2+⋯+cn-1(t-a)n-1=-Ia+αy(t)+c0+c1(t-a)+c2(t-a)2+⋯+cn-1(t-a)n-1.
In view of the relations CDa+αIa+αu(t)=u(t) and Ia+αIa+βu(t)=Ia+α+βu(t) for α,β>0, we obtain
(2.13)u′(t)=-Ia+α-1y(t)+c1+2c2(t-a)+⋯+(n-1)cn-1(t-a)n-2,u′′(t)=-Ia+α-2y(t)+2c2+⋯+(n-1)(n-2)cn-1(t-a)n-3,⋮u(n-1)(t)=-Ia+α-n+1y(t)+(n-1)!cn-1.
Applying the boundary conditions (2.9), we find that
(2.14)c2=c3=⋯=cn-1=0,(1-∑i=1m-2βi)c1=-∑i=1m-2βiIa+α-1y(ηi),
then c1=-∑i=1m-2βiIa+α-1y(ηi)/δ1, and
(2.15)c0=Ia+αy(b)(1-∑i=1m-2γi)-∑i=1m-2γiIa+αy(ηi)(1-∑i=1m-2γi)+[(b-a)-∑i=1m-2γi(ηi-a)]∑i=1m-2βiIa+α-1y(ηi)(1-∑i=1m-2βi)(1-∑i=1m-2γi)=Ia+αy(b)δ2-∑i=1m-2γiIa+αy(ηi)δ2+[(b-a)-∑i=1m-2γi(ηi-a)]∑i=1m-2βiIa+α-1y(ηi)δ1δ2.
Substituting the values of the constants ci,i=0,1,2,…,n-1, in (2.12), we obtain
(2.16)u(t)=-Ia+αy(t)+Ia+αy(b)δ2-∑i=1m-2γiIa+αy(ηi)δ2+[(b-a)-∑i=1m-2γi(ηi-a)]∑i=1m-2βiIa+α-1y(ηi)δ1δ2-∑i=1m-2βiIa+α-1y(ηi)δ1(t-a)=-Ia+αy(t)+Ia+αy(b)δ2-∑i=1m-2γiIa+αy(ηi)δ2+μ(t)δ1δ2∑i=1m-2βiIa+α-1y(ηi)=-Ia+αy(t)+Ia+αy(b)+(1-δ2)δ2Ia+αy(b)-∑i=1m-2γiIa+αy(ηi)δ2+μ(t)δ1δ2∑i=1m-2βiIa+α-1y(ηi)=-∫at(t-s)α-1Γ(α)y(s)ds+∫ab(b-s)α-1Γ(α)y(s)ds+∑i=1m-2γiδ2∫ab(b-s)α-1Γ(α)y(s)ds-∑i=1m-2γiδ2∫aηi(ηi-s)α-1Γ(α)y(s)ds+μ(t)δ1δ2∑i=1m-2βi∫aηi(ηi-s)α-2Γ(α-1)y(s)ds=-∫at(t-s)α-1Γ(α)y(s)ds+∫ab(b-s)α-1Γ(α)y(s)ds+∑i=1m-2γiδ2∫aηi[(b-s)α-1-(ηi-s)α-1Γ(α)]y(s)ds+∑i=1m-2γiδ2∫ηib(b-s)α-1Γ(α)y(s)ds+μ(t)δ1δ2∑i=1m-2βi∫aηi(ηi-s)α-2Γ(α-1)y(s)ds.Lemma 2.12.
μ(t)=(b-∑i=1m-2γiηi)-δ2t≥0, for t,ηi∈[a,b],i=1,2,…,m-2.Proof.
Wehave(2.17)μ=(b-∑i=1m-2γiηi)-δ2t≥(b-∑i=1m-2γiηi)-(1-∑i=1m-2γi)b≥∑i=1m-2γi(b-ηi)>0.Lemma 2.13.
The functionsG(t,s),H(t,s;η1,η2,…,ηm-2) defined by (2.11) satisfy(i)
G(t,s)≥0,H(t,s;η1,η2,…,ηm-2)≥0,forallt,s∈[a,b],(ii)
minτ1≤t≤τ2G(t,s)≥τ0maxa≤t≤bG(t,s)=τ0G(s,s),  for all t,s∈(a,b),a<τ1<τ2<b,τ0=minτ1≤t≤τ2φ(t)=(b-τ2)/(b-a),(iii)
N2q(s)≤H(t,s;η1,η2,…,ηm-2)≤N1q(s), where(2.18)q(s)=(b-s)α-2δ1δ2Γ(α),N1=(α-1)[δ1(b-a)∑i=1m-2γi+b∑i=1m-2βi],N2=min⁡{δ1∑i=1m-2γi(α-1),δ1∑i=1m-2γi(b-ηi)},(iv)
minτ1≤t≤τ2H(t,s;η1,η2,…,ηm-2)≥τ*maxa≤t≤bH(t,s;η1,η2,…,ηm-2),s∈(a,b), where
(2.19)τ*=[(b-∑i=1m-2γiηi)-δ2τ2][(b-∑i=1m-2γiηi)-δ2a]<1,a<τ1<τ2<b.Proof.
It is clear that (i) holds. So, we prove that (ii) is true.
(ii) Forα>1, in view of the expression for G(t,s), it follows that G(t,s)≤G(s,s) for all s,t∈[a,b], where G(s,s)=(b-s)α-1/Γ(α).
Ifa≤s≤t≤b, we have
(2.20)G(t,s)G(s,s)=[(b-s)α-1-(t-s)α-1](b-s)α-1=[(b-s)α-2(b-s)-(t-s)α-2(t-s)](b-s)α-1≥(b-s)α-2[(b-s)-(t-s)](b-s)α-1=(b-t)(b-a)∶=φ(t).
If  a≤t≤s≤b, then we have
(2.21)G(t,s)G(s,s)=1≥(b-t)(b-a)∶=φ(t).
Thus
(2.22)max⁡a≤t≤bG(t,s)=G(s,s),φ(t)G(s,s)≤G(t,s)≤G(s,s),∀t,s∈(a,b).
Therefore,
(2.23)min⁡τ1≤t≤τ2G(t,s)≥τ0max⁡a≤t≤bG(t,s)=τ0G(s,s),∀t,s∈(a,b),a<τ1<τ2<b.(iii) If a≤s≤ηi,i=1,2,…,m-2, then
(2.24)H(t,s;η1,η2,…,ηm-2)=∑i=1m-2γi[(b-s)α-1-(ηi-s)α-1]δ2Γ(α)+μ(t)∑i=1m-2βi(ηi-s)α-2δ1δ2Γ(α-1)≤∑i=1m-2γi(b-s)α-1δ2Γ(α)+(α-1)b∑i=1m-2βi(ηi-s)α-2δ1δ2Γ(α)=1δ1δ2Γ(α)[δ1∑i=1m-2γi(b-s)α-1+(α-1)b∑i=1m-2βi(ηi-s)α-2]≤(α-1)(b-s)α-2δ1δ2Γ(α)[δ1∑i=1m-2γi(b-s)+b∑i=1m-2βi]≤(α-1)(b-s)α-2δ1δ2Γ(α)[δ1(b-a)∑i=1m-2γi+b∑i=1m-2βi]:=N1q(s),H(t,s;η1,η2,…,ηm-2)=∑i=1m-2γi[(b-s)α-1-(ηi-s)α-1]δ2Γ(α)+μ(t)∑i=1m-2βi(ηi-s)α-2δ1δ2Γ(α-1)≥∑i=1m-2γi[(b-s)α-1-(ηi-s)α-1]δ2Γ(α)=∑i=1m-2γi[(b-s)α-2(b-s)-(ηi-s)α-2(ηi-s)]δ2Γ(α)=(b-s)α-2δ2Γ(α)∑i=1m-2γi(b-ηi)≥N2q(s).
If  ηi≤s≤b,i=1,2,…,m-2, then we have
(2.25)H(t,s;η1,η2,…,ηm-2)=∑i=1m-2γi(b-s)α-1δ2Γ(α)=δ1∑i=1m-2γi(b-s)(b-s)α-2δ1δ2Γ(α)≤δ1∑i=1m-2γi(b-a)q(s)<N1q(s),H(t,s;η1,η2,...,ηm-2)=∑i=1m-2γi(α-1)(b-s)α-1(α-1)δ2Γ(α)≥∑i=1m-2γi(b-s)α-2(α-1)δ2Γ(α)≥δ1∑i=1m-2γi(α-1)q(s)≥N2q(s).
(iv) Since∂H(t,s;η1,η2,…,ηm-2)/∂t=-(∑i=1m-2βi(ηi-s)α-2/δ1Γ(α-1))≤0, then H(t,s;η1,η2,…,ηm-2) is nonincreasing in t, so
(2.26)max⁡a≤t≤bH(t,s;η1,η2,…,ηm-2)=H(a,s;η1,η2,…,ηm-2)=∑i=1m-2γi[(b-s)α-1-(ηi-s)α-1]δ2Γ(α)+[(b-∑i=1m-2γiηi)-δ2a]∑i=1m-2βi(ηi-s)α-2δ1δ2Γ(α-1),min⁡τ1≤t≤τ2H(t,s;η1,η2,…,ηm-2)=∑i=1m-2γi[(b-s)α-1-(ηi-s)α-1]δ2Γ(α)+[(b-∑i=1m-2γiηi)-δ2τ2]∑i=1m-2βi(ηi-s)α-2δ1δ2Γ(α-1)=∑i=1m-2γi[(b-s)α-1-(ηi-s)α-1]δ2Γ(α)+τ*[(b-∑i=1m-2γiηi)-δ2a]∑i=1m-2βi(ηi-s)α-2δ1δ2Γ(α-1)>τ*max⁡a≤t≤bH(t;η1,η2,…,ηm-2,s).


## 3. Main Results

Let us denote byE=C[a,b] the Banach space of all continuous real functions on [a,b] endowed with the norm ∥u∥=maxa≤t≤b|u(t)| and P the cone (3.1)P={u∈E:u≥0,min⁡τ1≤t≤τ2u(t)≥τ‖u‖,t∈[a,b]},
where τ=min{τ0,τ*},  since τ0, τ* are constants do not depend on t.Let the nonnegative continuous concave functionalθ on the cone P be defined by θ(u)=minτ1≤t≤τ2u(t).SetT:P→E by(3.2)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds,a≤t≤b,
where  G(t,s),H(t,s;η1,η2,…,ηm-2) are defined as in Lemma 2.11.From (3.2) and Lemma 2.13, we have(3.3)min⁡τ1≤t≤τ2(Tu(t))≥∫ab[τ0G(s,s)+τ*max⁡a≤t≤bH(t,s;η1,…,ηm-2)]f(s,u(s))ds≥τ‖Tu‖.
Hence, we have T(P)⊂P.By standard argument, one can prove thatT:P→P is a completely continuous operator.The Existence of One Positive Solution
We introduce the following definitions:(3.4)f̅(u):=sup⁡t∈[a,b]f(t,u),f̲(u):=inf⁡t∈[a,b]f(t,u),f0=limsup⁡u→0+f̅(u)u,f0=liminf⁡u→0+f̲(u)u,f∞=limsup⁡u→∞f̅(u)u,f∞=liminf⁡u→∞f̲(u)u,M=(∫ab[G(s,s)+N1q(s)]ds)-1,N=(∫τ1τ2τ[G(s,s)+N2q(s)]ds)-1.
Theorem 3.1.
Letf(t,u) be continuous on [a,b]×[0,∞)→[0,∞). If there exist two positive constants r2>r1>0 such that (H1)
f(t,u)≤Mr2,for(t,u)∈[a,b]×[0,r2],(H2)
f(t,u)≥Nr1,for(t,u)∈[a,b]×[0,r1],then the BVP (1.3)-(1.4) has at least a positive solution.Proof.
We know that the operatorT:P→P defined by (3.2) is completely  continuous.
(a) LetΩ2={u∈E:∥u∥<r2}. For any u∈P∩∂Ω2, we have ∥u∥=r2 which implies that 0≤u(t)≤r2 for every t∈[a,b]:
(3.5)Tu(t)=∫ab[G(t,s)+H(t,s;η1,…,ηm-2)]f(s,u(s))ds≤Mr2∫ab[G(s,s)+N1q(s)]ds≤r2=‖u‖,
which implies that
(3.6)‖Tu‖≤‖u‖,u∈P∩∂Ω2.
(b) LetΩ1={u∈E:∥u∥<r1}. For any t∈[τ1,τ2],u∈P∩∂Ω1. We have
(3.7)Tu(t)=∫ab[G(t,s)+H(t,s;η1,…,ηm-2)]f(s,u(s))ds≥∫abφ(t)[G(s,s)+N2q(s)]f(s,u(s))ds≥Nr1∫τ1τ2τ[G(s,s)+N2q(s)]ds=r1=‖u‖,
which implies that
(3.8)‖Tu‖≥‖u‖,u∈P∩∂Ω1.
In view of Theorem 2.6, T has a fixed point u0∈P∩(Ω̅2∖Ω1) which is a solution of  the BVP (1.3)-(1.4).The Existence of Two Positive Solutions
Theorem 3.2.
Assume that all assumptions of Theorem3.1, hold. Moreover, one assumes that f(t,u) also satisfies (H3)
f∞=∞.Then, the BVP (1.3)-(1.4) has at least two positive solutions.Proof.
At first, it follows from condition(H1) that
(3.9)‖Tu‖≤‖u‖,u∈P∩∂Ω2.
Further, it follows from condition (H2) that
(3.10)‖Tu‖≥‖u‖,u∈P∩∂Ω1.
Finally, since f∞=∞, there exists ψ>(τ2∫τ1τ2[G(s,s)+N2q(s)]ds)-1 and r3>r2 such that
(3.11)f(t,u)≥ψu(t),t∈[a,b],u≥r3.
Let r*=max{2r2,τ-1r3} and set Ω3={u∈E:∥u∥<r*}, then u∈P∩∂Ω3impliesminτ1≤t≤τ2u(t)≥τ∥u∥≥τr*≥r3,
(3.12)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds≥∫abφ(t)[G(s,s)+N2q(s)]f(s,u(s))ds≥ψ∫abφ(t)[G(s,s)+N2q(s)]u(s)ds≥r*ψ∫τ1τ2τ2[G(s,s)+N2q(s)]ds>r*=‖u‖.
Therefore, we have
(3.13)‖Tu‖≥‖u‖,u∈P∩∂Ω3.
Thus, from (3.6), (3.8), (3.13), and Theorem 2.6, T has a fixed point  u1, in P∩(Ω̅2∖Ω1) and a fixed point u2, in P∩(Ω̅3∖Ω2). Both are positive solutions of BVP (1.3)-(1.4) and satisfy
(3.14)0<‖u1‖<r2<‖u2‖.Theorem 3.3.
Assume thatf(t,u) be continuous on [a,b]×[0,∞)→[0,∞). If the following assumptions hold:(H1)
f0>ψ,(H2)
f∞>ψ,(H3)
there exists a constantρ>0 such that f(t,u)≤ρM,(t,u)∈[a,b]×[0,ρ],then the BVP (1.3)-(1.4)  has at least two positive solutions u1 and u2 such that
(3.15)0<‖u1‖<ρ<‖u2‖.Proof.
At first, it follows from condition(H1) that we may choose ρ1∈(0,ρ) such that
(3.16)f(t,u)>ψu,0<u≤ρ1,
where ψ is defined as in Theorem 3.2. Set Ω1={u∈E:∥u∥<ρ1}, and u∈P∩∂Ω1; from (3.2) and Lemma 2.13, for a≤t≤b, we have
(3.17)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds≥∫abφ(t)[G(s,s)+N2q(s)]f(s,u(s))ds≥ψ∫abφ(t)[G(s,s)+N2q(s)]u(s)ds≥ρ1ψ∫τ1τ2τ2[G(s,s)+N2q(s)]ds>ρ1=‖u‖.
Therefore, we have
(3.18)‖Tu‖≥‖u‖,u∈P∩∂Ω1.
Further, it follows from condition (H2) that there exists ρ2>ρ such that
(3.19)f(t,u)>ψu(t),u≥ρ2.
Let ρ*=max{2ρ,τ-1ρ2}, set Ω2={u∈E:∥u∥<ρ*}, then u∈P∩∂Ω2impliesminτ1≤t≤τ2u(t)≥τ∥u∥≥τρ*≥ρ2,
(3.20)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds≥∫abφ(t)[G(s,s)+N2q(s)]f(s,u(s))ds≥ψ∫abφ(t)[G(s,s)+N2q(s)]u(s)ds≥ρ*ψ∫τ1τ2τ2[G(s,s)+N2q(s)]ds>ρ*=‖u‖.
Therefore, we have
(3.21)‖Tu‖≥‖u‖,u∈P∩∂Ω2.
Finally, let Ω3={u∈E:∥u∥<ρ} and u∈P∩∂Ω3. By condition (H3), we have
(3.22)Tu(t)≤∫ab[G(s,s)+N1q(s)]f(s,u(s))ds≤Mρ∫ab[G(s,s)+N1q(s)]ds=ρ=‖u‖,
which implies
(3.23)‖Tu‖≤‖u‖,u∈P∩∂Ω3.
Thus, from (3.18), (3.21), (3.23), and Theorem 2.6, T has a fixed point  u1 in P∩(Ω̅3∖Ω1) and a fixed point u2, in P∩(Ω̅2∖Ω3). Both are positive solutions of BVP (1.3)-(1.4)  and satisfy
(3.24)0<‖u1‖<ρ<‖u2‖.Theorem 3.4.
Assume thatf(t,u) be continuous on [a,b]×[0,∞)→[0,∞). If the following assumptions hold:(H1′)
f0=∞,(H2′)
f∞=∞,(H3′)
there exists a constantρ′>0 such that f(t,u)≤ρ′M,(t,u)∈[a,b]×[0,ρ′],then the BVP (1.3)-(1.4)  has at least two positive solutions u1 and u2 such that
(3.25)0<‖u1‖<ρ′<‖u2‖.
The proof of Theorem 3.4 is very similar to that of Theorem 3.3 and therefore is omitted.Theorem 3.5.
Assume thatf(t,u) be continuous on [a,b]×[0,∞)→[0,∞). If the following assumptions hold:(H1)
f0<M,(H2)
f∞<M,(H3)
there exists a constantl>0 such that f(t,u)≥Nl,(t,u)∈[a,b]×[τl,l],then the BVP (1.3)-(1.4) has at least two positive solutions u1 and u2 such that
(3.26)0<‖u1‖<l<‖u2‖.Proof.
It follows from condition(H1) that we may choose ρ3∈(0,l) such that
(3.27)f(t,u)<Mu,0<u≤ρ3.
Set Ω4={u∈E:∥u∥<ρ3}, and u∈P∩∂Ω4; from (3.2) and Lemma 2.13, for a≤t≤b, we have
(3.28)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds<M∫ab[G(s,s)+N1q(s)]ds⋅‖u‖=M⋅M-1‖u‖=‖u‖.
Therefore, we have
(3.29)‖Tu‖≤‖u‖,u∈P∩∂Ω4.
It follows from condition (H2) that there exists ρ4>l such that
(3.30)f(t,u)<Mu,u≥ρ4,
and we consider two cases.Case 1.
Suppose thatf is unbounded, there exists l*>ρ4 such that f(t,u)≤f(t,l*) for 0<u≤l*.Then, for u∈P and ∥u∥=l*, we have
(3.31)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds≤∫ab[G(s,s)+N1q(s)]f(s,l*)ds<Ml*∫ab[G(s,s)+N1q(s)]ds=l*=‖u‖.Case 2.
Iff is bounded, that is, f(t,u)≤k for all u∈[0,∞), taking l*≥max{2l,kM-1}, for u∈P and ∥u∥=l*, then we have
(3.32)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds≤k∫ab[G(s,s)+N1q(s)]ds=kM-1≤l*=‖u‖.
Hence, in either case, we always may set Ω5={u∈E:∥u∥<l*} such that
(3.33)‖Tu‖≤‖u‖,u∈P∩∂Ω5.
Finally, set Ω6={u∈E:∥u∥<l}, then u∈P∩∂Ω6 and
(3.34)min⁡τ1≤t≤τ2u(t)≥τ‖u‖=τl,
and by condition (H3) and (3.2), we have
(3.35)Tu(t)=∫ab[G(t,s)+H(t,s;η1,η2,…,ηm-2)]f(s,u(s))ds≥Nl∫abφ(t)[G(s,s)+N2q(s)]ds≥Nl∫τ1τ2τ[G(s,s)+N2q(s)]ds=l=‖u‖.
Hence,  we have
(3.36)‖Tu‖≥‖u‖,u∈P∩∂Ω6.
Thus, from (3.29), (3.33), (3.36) and Theorem 2.6, T has a fixed point  u1 in P∩(Ω̅6∖Ω4) and a fixed point u2 in P∩(Ω̅5∖Ω6).Both are positive solutions of BVP (1.3)-(1.4)  and satisfy
(3.37)0<‖u1‖<l<‖u2‖.Theorem 3.6.
Assume thatf(t,u) be continuous on [a,b]×[0,∞)→[0,∞). If the following assumptions hold:(H1′)
f0=0,(H2′)
f∞=0,(H3′)
there exists a constantρ′′>0 such that f(t,u)≥Nρ′′,(t,u)∈[a,b]×[τρ′′,ρ′′],then the BVP (1.3)-(1.4)  has at least two positive solutions u1 and u2 such that
(3.38)0<‖u1‖<ρ′′<‖u2‖.
The proof of Theorem 3.6 is very similar to that of Theorem 3.5 and therefore omitted.The Existence of Three Positive Solutions
Theorem 3.7.
Letf(t,u) be continuous on [a,b]×[0,∞)→[0,∞). If there exist constants 0<a1<a2≤a3 such that the following assumptions(i)
f(t,u)<Ma1,(t,u)∈[a,b]×[0,a1],(ii)
f(t,u)≤Ma3,(t,u)∈[a,b]×[0,a3],(iii)
f(t,u)≥Na2,(t,u)∈[τ1,τ2]×[a2,a2/τ],hold, then BVP (1.3)-(1.4)  has at least three positive solutions u1,u2, and u3 with
(3.39)‖u1‖<a1,a2<θ(u2)<‖u2‖≤a3,a1<‖u3‖,θ(u3)<a2.Proof.
We will show that all conditions of Lemma2.10, are satisfied.
First, ifu∈P̅a3, then ∥u∥≤a3. So, 0≤u(t)≤a3,t∈[a,b].
By condition (ii), we have(3.40)Tu(t)=∫ab[G(t,s)+H(t,s;η1,…,ηm-2)]f(s,u(s))ds≤∫ab[G(s,s)+N1q(s)]Ma3ds=Ma3∫ab[G(s,s)+N1q(s)]ds=a3,
which implies that ∥Tu∥≤a3,u∈P̅a3. Hence T:P̅a3→P̅a3.
Next, by using the analogous argument, it follows from condition (i) that ifu∈Pa1, then ∥Tu∥<a1.
Chooseu(t)=(a2+a2/τ)/2,t∈[a,b], it is easy to see that u(t)=(a2+a2/τ)/2∈P(θ,a2,a3),θ(u)=(a2+a2/τ)/2>a2.
Therefore,{u∈P(θ,a2,a2/τ)∣θ(u)>a2}≠ϕ. On the other hand, if u∈P(θ,a2,a2/τ), then a2≤u(t)≤a2/τ,t∈[τ1,τ2]. By condition (iii), we have f(t,u(t))≥Na2.
Hence,(3.41)θ(Tu(t))=min⁡τ1≤t≤τ2Tu(t)=min⁡τ1≤t≤τ2∫ab[G(t,s)+H(t,s;η1,…,ηm-2)]f(s,u(s))ds≥Na2∫abτ[G(s,s)+N2q(s)]ds>Na2∫τ1τ2τ[G(s,s)+N2q(s)]ds=a2,
which implies that θ(Tu)>a2, for u∈P(θ,a2,a2/τ).
Finally, ifu∈P(θ,a2,a3) and ∥Tu∥>a2/τ, then
(3.42)θ(u)=min⁡τ1≤t≤τ2Tu(t)≥τ‖Tu‖>a2.
Thus, all the conditions of the Leggett-Williams fixed point theorem are satisfied by taking d=a2/τ. Hence, the BVPs have at least three solutions in P, that is, three positive solutions ui(i=1,2,3) such that
(3.43)‖u1‖<a1,a2<θ(u2)<‖u2‖≤a3,a1<‖u3‖,θ(u3)<a2.Example 3.8.
Consider the problem(3.44)D0+(4.2)u(t)+13(1+ueu)=0,t∈(0,1),u′(0)=14u′(12),u′′(0)=u′′′(0)=u′′′′(0)=0,u(1)=34u(12),
where α=4.2, a=0, b=1, β=0.25, γ=0.75, η=0.5, τ1=0.25, τ2=0.75,N1=2.6,N2=0.1758, N=168.9596,  and M=1.6968,f(t,u)=(1/3)(1+ueu).
Sincef(t,u)=(1/3)(1+ueu) is a monotone increasing function on [0,∞), we take r1=0.001,r2=0.8. We can get
(3.45)f(t,u)≤f(0.8)=0.9268<Mr2,f(t,u)≥f(0)=0.3333>Nr1.
So, conditions (H1) and (H2) hold. By Theorem 3.1, the BVP (3.44)  has at least one positive solution.



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*Source: 986575-2011-06-12.xml*