# Existence of Positive Solutions for a Kind of Fractional Boundary Value Problems

Authors: Hongjie Liu; Xiao Fu; Liangping Qi
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We are concerned with the following nonlinear three-point fractional boundary value problem:D0+αut+λatft,ut=0, 0<t<1, u0=0, and u1=βuη, where 1<α≤2, 0<β<1, 0<η<1, D0+α is the standard Riemann-Liouville fractional derivative, at>0 is continuous for 0≤t≤1, and f≥0 is continuous on 0,1×0,∞. By using Krasnoesel'skii's fixed-point theorem and the corresponding Green function, we obtain some results for the existence of positive solutions. At the end of this paper, we give an example to illustrate our main results.


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. With the development of nonlinear science, the researchers found that nonlinear fractional differential equations could describe something's changing rules more accurately. Therefore, it is significant to study nonlinear fractional differential equations to solve the nonlinear problems. Recently, many researchers paid attention to existence and multiplicity of solution of the boundary value problem for fractional differential equations with different boundary conditions, such as [1–10]. Bai and Lü [1] investigated the existence and multiplicity of positive solutions for nonlinear fractional boundary value problem:
(1)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u(1)=0,
where 1<α≤2 is a real number, D0+α is the standard Riemann-Liouville fractional derivative, and f:[0,1]×[0,∞)→[0,∞) is continuous. By means of some fixed-point theorems on cone, some existence and multiplicity results of positive solutions are obtained. Agarwal et al. [2] investigated the existence of positive solutions for the singular fractional boundary value problem:
(2)D0+αu(t)+f(t,u(t),Dμu(t))=0,0<t<1,u(0)=u(1)=0,
where 1<α≤2,μ>0 are real numbers, f is positive, and D0+α is the standard Riemann-Liouville fractional derivative. By means of a fixed point theorem on cone, the existence of positive solutions is obtained. Delbosco and Rodino [3] investigated the nonlinear Dirichlet-type problem:
(3)xα-1(D0+αy)(x)=f(y(x)),1<α<2,0≤x≤1,y(0)=y(1)=0.
They proved that if f(y) is a Lipschitzian function, then the problem has at least one solution y(x) in a certain subspace of C[0,1]. In this paper, we study the following three-point fractional boundary value problem. Consider
(4)D0+αu(t)+λa(t)f(t,u(t))=0,0<t<1,u(0)=0,βu(η)=u(1),
where 1<α≤2,0<β<1,0<η<1,D0+α is the standard Riemann-Liouville fractional derivative, a(t)>0 is continuous, and f≥0 is continuous on [0,1]×[0,∞). By using Krasnoesel'skii's fixed-point theorem, we get the existence of at least one positive solution.


## 2. Background Materials and Preliminaries

For convenience of the readers, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature [1–5].Definition 1.
The fractional integral of orderα>0 of a function y:(0,∞)→R is given by
(5)I0+αy(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds,
provided that the right side is pointwise defined on (0,∞).Definition 2.
The fractional derivative of orderα>0 of a function y:(0,∞)→R is given by
(6)D0+αy(t)=1Γ(n-α)(ddt)n∫0t‍y(s)(t-s)α-n+1ds,n=[α]+1,
where n=[α]+1, provided that the right side is pointwise defined on (0,∞).Lemma 3.
Letα>0, if we assume u∈C(0,1)∩L(0,1), then the fractional differential equation D0+αu(t)=0 has
(7)u(t)=C1tα-1+C2tα-2+⋯CNtα-N,iiiiiiiiiiCi∈R,i=1,2,…,N
as unique solutions.Lemma 4.
Assume thatu∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L(0,1). Then
(8)I0+αD0+αu(t)=u(t)+C1tα-1+C2tα-2+⋯CNtα-N
for some Ci∈R,i=1,2,…,N.Definition 5.
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>0  implies  σx∈K;(2)
x∈K,-x∈K  implies  x=0.Lemma 6.
LetE be a Banach space and let K⊂E be a cone in E. Assume that Ω1 and Ω2 are open subsets of E with 0∈Ω1,Ω¯1⊂Ω2. Let T:K∩(Ω¯2∖Ω1)→K be a completely continuous operator. In addition, suppose that either (H1)
∥Tu∥≤∥u∥,∀u∈K∩∂Ω1 and ∥Tu∥≥∥u∥,∀u∈K∩∂Ω2 or(H2)
∥Tu∥≥∥u∥,∀u∈K∩∂Ω1 and ∥Tu∥≤∥u∥,∀u∈K∩∂Ω2
holds. ThenT has a fixed point in K∩(Ω¯2∖Ω1).In the following, we present the Green function of fractional differential equation boundary value problem.Lemma 7 (see [4]).
Lety∈C[0,1], then the boundary value problem
(9)D0+αu(t)+y(t)=0,0<t<1,u(0)=0,βu(η)=u(1)
has a unique solution
(10)u(t)=∫01‍G(t,s)y(s)ds,
where
(11)G(t,s)={([t(1-s)]α-1-βtα-1(η-s)α-1ii-(t-s)α-1(1-βηα-1))×((1-βηα-1)Γ(α))-1,0≤s≤min⁡{t,η}≤1;[t(1-s)]α-1-(t-s)α-1(1-βηα-1)(1-βηα-1)Γ(α),0<η≤s≤t≤1;[t(1-s)]α-1-βtα-1(η-s)α-1(1-βηα-1)Γ(α),0≤t≤s≤η<1;[t(1-s)]α-1(1-βηα-1)Γ(α),0≤max⁡{t,η}≤s≤1.Lemma 8.
The functionG(t,s) defined by (11) satisfies G(t,s)>0  for  t,s∈(0,1).Proof.
(1) For0≤s≤min⁡{t,η}≤1,
(12)G(t,s)=[t(1-s)]α-1-βtα-1(η-s)α-1-(t-s)α-1(1-βηα-1)(1-βηα-1)Γ(α).
Let F(t,s,η)=[t(1-s)]α-1-βtα-1(η-s)α-1-(t-s)α-1(1-βηα-1). There is
(13)∂F(t,s,η)∂η=-βtα-1(α-1)(η-s)α-2+(α-1)(t-s)α-1βηα-2=β(α-1)[(t-s)α-1ηα-2-tα-1(η-s)α-2]≤β(α-1)[tα-1ηα-2-tα-1(η-s)α-2]=β(α-1)tα-1[ηα-2-(η-s)α-2].
With 0<β<1,1<α≤2, there is ∂F/∂η<0. Hence, function F(t,s,η) is monotonically decreasing in η. With 0<η<1, F(t,s,η)>F(t,s,η)min⁡=F(t,s,1). Let η=1, there is
(14)F(t,s,1)=[t(1-s)]α-1-βtα-1(1-s)α-1-(t-s)α-1(1-β)=(1-β){[t(1-s)]α-1-(t-s)α-1}=(1-β)[(t-ts)α-1-(t-s)α-1].
With 0<s≤t<1,1<α≤2, we have (t-ts)α-1-(t-s)α-1≥0. With 1-β>0, there is F(t,s,1)≥0, then F(t,s,η)>F(t,s,1)≥0, there is G(t,s)>0.
(2) For0<η≤s≤t<1,
(15)G(t,s)=[t(1-s)]α-1-(t-s)α-1(1-βηα-1)(1-βηα-1)Γ(α).
Let F(t,s,η)=[t(1-s)]α-1-(t-s)α-1(1-βηα-1); it is obvious that function F(t,s,η) is monotonically increasing in η, when η=0,
(16)F(t,s,η)min⁡=F(t,s,0)=[t(1-s)]α-1-(t-s)α-1=(t-ts)α-1-(t-s)α-1.
With 0≤s≤t<1,0<α-1≤1, there is (t-ts)α-1≥(t-s)α-1; we have F(t,s,η)>F(t,s,0)≥0, then G(t,s)>0.
(3) For0<t≤s≤η<1,
(17)G(t,s)=[t(1-s)]α-1-βtα-1(η-s)α-1(1-βηα-1)Γ(α).
Let F(t,s,η)=[t(1-s)]α-1-βtα-1(η-s)α-1, then F(t,s,η) is monotonically decreasing in η,
(18)F(t,s,η)min⁡=F(t,s,1)=[t(1-s)]α-1-βtα-1(1-s)α-1=(1-β)[t(1-s)]α-1.
With 0<β<1,0<t≤s<1, there is F(t,s,η)>0, then F(t,s,η)>F(t,s,1)≥0, hence G(t,s)>0.
(4) For0≤max⁡{t,η}≤s≤1,
(19)G(t,s)=[t(1-s)]α-1(1-βηα-1)Γ(α)>0.
Therefore, G(t,s)>0 for t,s∈(0,1). The proof is complete.Lemma 9.
G
(
t
,
s
)
≤
G
(
s
,
s
) for t,s∈[0,1].Proof.
To getG(t,s)≤G(s,s), firstly, we prove the following inequality:
(20)ta-sa(t-s)a≤1-sa(1-s)a,
where a=α-1, 0≤s<t≤1. Let f(t)=(ta-sa)/(t-s)a, then
(21)f′(t)=ata-1(t-s)a-a(ta-sa)(t-s)a-1(t-s)2a=a(t-s)a-1(t-s)2a·[ta-1(t-s)-(ta-sa)]=a(t-s)a-1(t-s)2a·s(sa-1-ta-1).
With 1<α≤2, there is a-1≤0; with s<t, there is sa-1>ta-1; then f′(t)>0. It means that f(t) is monotonically increasing in t, then f(t)≤f(1); we have
(22)ta-sa(t-s)a≤1-sa(1-s)a.
Here we prove G(t,s)≤G(s,s).
(1) For0≤s≤min⁡{t,η}≤1,
(23)G(t,s)=[t(1-s)]α-1-βtα-1(η-s)α-1-(t-s)α-1(1-βηα-1)(1-βηα-1)Γ(α),G(s,s)=[s(1-s)]α-1-βsα-1(η-s)α-1(1-βηα-1)Γ(α).
With (1-βηα-1)Γ(α)>0,  G(t,s)≤G(s,s) is equivalent to
(24)ta-sa(t-s)a[(1-s)a-β(η-s)a]≤1-βηa
or
(25)(ta-sa)[(1-s)a-β(η-s)a]≤(1-βηa)(t-s)a.(a)
Fors=t,(ta-sa)[(1-s)a-β(η-s)a]=(1-βηa)(t-s)a=0;(b)
fors=t=η=1, (ta-sa)[(1-s)a-β(η-s)a]=(1-βηa)(t-s)a=0;(c)
fors=η≤t,
(26)(ta-sa)[(1-s)a-β(η-s)a]=(ta-ηa)(1-ηa),(1-βηa)(t-s)a=(1-βηa)(t-η)a,ift=η, (ta-sa)[(1-s)a-β(η-s)a]=(1-βηa)(t-s)a=0; if t>η, from (22), we have (ta-ηa)/(t-η)a≤(1-ηa)/(1-η)a≤(1-βηa)/(1-η)a;(d)
for0≤s<t≤1, 0≤s<η≤1, if (1-s)a-β(η-s)a<0; it is obvious that (ta-sa)[(1-s)a-β(η-s)a]≤(1-βηa)(t-s)a. If (1-s)a-β(η-s)a>0, by (22), there is (ηa-sa)/(η-s)a≤(1-sa)/(1-s)a; then
(27)(η-s)a≥1-sa(1-s)a(ηa-sa).
Consider(28)ta-sa(t-s)a[(1-s)a-β(η-s)a]≤1-sa(1-s)a[(1-s)a-β(η-s)a]=1-sa-β1-sa(1-s)a(η-s)a≤1-sa-β1-sa(1-s)a·(1-s)a1-sa·(ηa-sa)=1-βηa-(1-β)sa≤1-βηa.
It means that G(t,s)≤G(s,s) holds.
(2) For0<η≤s≤t≤1,
(29)G(t,s)=[t(1-s)]α-1-(t-s)α-1(1-βηα-1)(1-βηα-1)Γ(α),G(s,s)=[s(1-s)]α-1(1-βηα-1)Γ(α).
With (1-βηα-1)Γ(α)>0,   G(t,s)≤G(s,s) is equivalent to
(30)[t(1-s)]a-(t-s)a(1-βηa)≤[s(1-s)]a
or
(31)(ta-sa)(1-s)a≤(t-s)a(1-βηa).
From (22), we have ((ta-sa)/(t-s)a)(1-s)a≤1-sa; with η≤s,0<β<1, we have 1-sa≤1-ηa≤1-βηa; then ((ta-sa)/(t-s)a)(1-s)a≤1-βηa; with s≤t, there is
(32)(ta-sa)(1-s)a≤(t-s)a(1-βηa).
It means that G(t,s)≤G(s,s) holds.
(3) For0≤t≤s≤η<1,
(33)G(t,s)=[t(1-s)]α-1-βtα-1(η-s)α-1(1-βηα-1)Γ(α),G(s,s)=[s(1-s)]α-1-βsα-1(η-s)α-1(1-βηα-1)Γ(α).
With (1-βηα-1)Γ(α)>0, G(t,s)≤G(s,s) is equivalent to
(34)[t(1-s)]a-βta(η-s)a≤[s(1-s)]a-βsa(η-s)a
or
(35)(ta-sa)(1-s)a≤β(ta-sa)(η-s)a.
With 0<β<1, s≤η, there is β(η-s)a≤(η-s)a<(1-s)a; with t≤s, there is
(36)(ta-sa)(1-s)a≤β(ta-sa)(η-s)a.
It means that G(t,s)≤G(s,s) holds.
(4) For0<max⁡{t,η}≤s≤1,
(37)G(t,s)=[t(1-s)]α-1(1-βηα-1)Γ(α),G(s,s)=[s(1-s)]α-1(1-βηα-1)Γ(α).
It is obvious that G(t,s)≤G(s,s). The proof is complete.Lemma 10.
G
(
t
,
s
)
≥
q
(
k
)
G
(
s
,
s
) for t,s∈[1/k,1-(1/k)], where k>2 is an integer and
(38)q(k)=min⁡{(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1,iiiiiiiiiii(1-β)(1/k)2(α-1)(1/4)α-1,iiiiiiiiiii(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1(k-1)α-1/k2(α-1)(1/4)α-1(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1}.Proof.
From (11), we have
(39)G(s,s)={[s(1-s)]α-1-βsα-1(η-s)α-1(1-βηα-1)Γ(α),s≤η,[s(1-s)]α-1(1-βηα-1)Γ(α),s>η.
It is obvious that [s(1-s)]α-1/(1-βηα-1)Γ(α)≥([s(1-s)]α-1-βsα-1(η-s)α-1)/(1-βηα-1)Γ(α), for t,s∈[1/k,1-(1/k)],
(40)max⁡G(s,s)=[s(1-s)]α-1(1-βηα-1)Γ(α)|s=1/2=(1/4)α-1(1-βηα-1)Γ(α).
(1) For1/k≤s≤min⁡{t,η}≤1-(1/k), let
(41)F(t,s,η)=[t(1-s)]α-1-βtα-1(η-s)α-1-(t-s)α-1(1-βηα-1).
By Lemma 8, for η=1,
(42)F¯(t,s)=F(t,s,η)min⁡=F(t,s,1)=[t(1-s)]α-1-βtα-1(1-s)α-1-(t-s)α-1(1-β)=(1-β){[t(1-s)]α-1-(t-s)α-1}=(1-β)[(t-ts)α-1-(t-s)α-1],∂F¯∂s=(1-β)(α-1)[(t-s)α-2-tα-1(1-s)α-2].
With 1<α≤2,0≤s≤t≤1, then (t-s)α-2-tα-1(1-s)α-2>0; with 0<β<1,∂F¯/∂s>0. It means that F¯(t,s) is monotonically increasing in s. Similarly,
(43)∂F¯∂t=(1-β)(α-1)[(t-ts)α-2(1-s)-(t-s)α-2]<0.
It means that F¯(t,s) is monotonically decreasing in t. Therefore
(44)F¯(t,s)≥min⁡F¯(t,s)=F¯(t,s)|t=1-(1/k),s=1/k=(1-β)[(1-1k)2(α-1)-(1-2k)α-1],min⁡G(t,s)=(1-β)[((1/k)-(1/k2))α-1-(1-(2/k))α-1](1-βηα-1)Γ(α),min⁡G(t,s)max⁡G(s,s)=(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1.
(2) For1/k<η≤s≤t≤1-(1/k), let
(45)F(t,s,η)=[t(1-s)]α-1-(t-s)α-1(1-βηα-1).
By Lemma 8, for η=0,
(46)F¯(t,s)=F(t,s,η)min⁡=F(t,s,0)=[t(1-s)]α-1-(t-s)α-1=(t-ts)α-1-(t-s)α-1,∂F¯∂s=(α-1)[(t-s)α-2-tα-1(1-s)α-2]>0.
It means that F¯(t,s) is monotonically increasing in s(47)∂F¯∂t=(α-1)[(1-s)α-1tα-2-(t-s)α-2]<0.
It means that F¯(t,s) is monotonically decreasing in t(48)F¯(t,s)≥min⁡F¯(t,s)=F¯(t,s)|t=1-(1/k),s=(1/k)=(1-1k)2(α-1)-(1-2k)α-1,min⁡G(t,s)=(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1-βηα-1)Γ(α),min⁡G(t,s)max⁡G(s,s)=(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1.
(3) For1/k≤t≤s≤η<1-(1/k), let
(49)F(t,s,η)=[t(1-s)]α-1-βtα-1(η-s)α-1.
By Lemma 8, for η=1,
(50)F¯(t,s)=F(t,s,η)min⁡=F(t,s,1)=[t(1-s)]α-1-βtα-1(1-s)α-1=(1-β)[t(1-s)]α-1,∂F¯∂s=(α-1)tα-1(β-1)(1-s)α-2<0;
then F¯(t,s) is monotonically decreasing in s. Consider
(51)∂F¯∂t=(α-1)(1-β)tα-2(1-s)α-1>0;
then F¯(t,s) is monotonically increasing in t. Therefore,
(52)F¯(t,s)≥min⁡F¯(t,s)=F¯(t,s)|t=1/k,s=1-(1/k)=1-βk2(α-1),min⁡G(t,s)=(1-β)(1/k2(α-1))(1-βηα-1)Γ(α),min⁡G(t,s)max⁡G(s,s)=(1-β)(1/k)2(α-1)(1/4)α-1.
(4) For1/k≤max⁡{t,η}≤s≤1-(1/k), let
(53)F(t,s)=[t(1-s)]α-1.
It is obvious that F(t,s) is monotonically increasing in t and monotonically decreasing in s; then
(54)F(t,s)≥min⁡F(t,s)=F(t,s)|t=1/k,s=1-(1/k)=(1k-1k2)α-1;
we have
(55)min⁡G(t,s)=((1/k)-(1/k2))α-1(1-βηα-1)Γ(α);
there is
(56)min⁡G(t,s)max⁡G(s,s)=(k-1)α-1/k2(α-1)(1/4)α-1.
Let
(57)q(k)=min⁡{(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1,iiiiiiiiii(1-β)(1/k)2(α-1)(1/4)α-1,iiiiiiiiii(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1(k-1)α-1/k2(α-1)(1/4)α-1}.
Therefore, G(t,s)/G(s,s)≥min⁡G(t,s)/max⁡G(s,s)=q(k)≠0.
The proof is complete.We consider the Banach spaceX=C[0,1] equipped with standard norm ∥u∥=max⁡0≤t≤1|u(t)|,u∈X. We define a cone P by
(58)P={min⁡t∈[1/k,1-1/k]u∈X:u(t)≥0,t∈[0,1],iiimin⁡t∈[1/k,1-1/k]⁡u(t)≥q(k)∥u∥}.
Define an integral operator T:P→X by
(59)(Tu)(t)=λ∫01‍G(t,s)a(s)f(s,u(s))ds,iiiiiiiiiiiiiiiiii0≤t≤1,u∈P.Lemma 11.
It holds the following.(1)
T:P→P is completely continuous.(2)
u(t) is a positive solution of the fractional boundary value problem (4) if and only if u(t) is a fixed point of the operator T in cone P.Proof.
Foru∈P,t∈[1/k,1-(1/k)], by Lemma 10, G(t,s)≥q(k)G(s,s); for t,s∈[1/k,1-(1/k)],
(60)Tu(t)=λ∫01‍G(t,s)a(s)f(s,u(s))ds≥λq(k)∫01‍G(s,s)a(s)f(s,u(s))ds=λq(k)max⁡1/k≤t≤1-1/k⁡∫01‍G(t,s)a(s)f(s,u(s))ds=q(k)∥Tu(t)∥.
It is obvious that min⁡1/k≤t≤1-1/k(Tu)(t)≥q(k)∥Tu∥,TP⊆P. Thus TP⊂P. In addition, standard arguments show that T is completely continuous.
(2) It is obvious thatu(t) is the positive solution of BVP (4) if and only if
(61)u(t)=λ∫01‍G(t,s)a(s)f(s,u(s))ds,0≤t≤1.
It can be proved by the definition of integral operator T.


## 3. Main Results

We denote some important constants as follows:(62)A=∫01‍G(s,s)a(s)q(k)ds,B=∫01‍G(s,s)a(s)ds,F0=lim⁡u→0+⁡sup⁡max⁡0≤t≤1⁡f(t,u(t))u(t),f0=lim⁡u→+∞⁡inf⁡min⁡0≤t≤1⁡f(t,u(t))u(t),F∞=lim⁡u→0+⁡sup⁡max⁡0≤t≤1⁡f(t,u(t))u(t),f∞=lim⁡u→+∞⁡inf⁡min⁡0≤t≤1⁡f(t,u(t))u(t).
Here we assume that 1/Af∞=0 if f∞→∞, 1/BF0=∞ if F0→0, 1/Af0=0 if f0→∞, and 1/BF∞=∞ if F∞→0.Theorem 12.
Suppose thatAf∞>BF0, then for each λ∈(1/Af∞,1/BF0), BVP (4) has at least one positive solution.Proof.
We chooseε>0 sufficiently small such that (F0+ε)λB≤1. By the definition of F0, we can see that there exists l1>0, such that f(t,u(t))≤(F0+ε)u(t) for 0≤u≤l1. For u∈P with ∥u∥=l1, we have
(63)∥Tu(t)∥=λ∫01‍G(s,s)a(s)f(s,u(s))ds≤λ∫01‍G(s,s)a(s)(F0+ε)u(s)ds≤λ(F0+ε)∥u∥∫01‍G(s,s)a(s)ds=λB(F0+ε)∥u∥≤∥u∥.
Then we have ∥Tu∥≤∥u∥. Thus if we let Ω1={u∈X:∥u∥<l1}, then ∥Tu∥≤∥u∥  for  u∈P∩∂Ω1. We choose δ>0 and c∈(0,1/4) such that λ(f∞-δ)A≥1. There exists l2>l1>0, such that f(t,u)≥(f∞-δ)u for u>l2. Therefore, for each u∈P with ∥u∥=l2, we have
(64)∥Tu(t)∥=λ∫01‍G(s,s)a(s)f(s,u(s))ds≥λ∫c1‍G(s,s)a(s)(f∞-δ)u(s)ds≥λ(f∞-δ)∥u∥∫c1‍G(s,s)a(s)q(k)ds≥λ(f∞-δ)A·∥u∥≥∥u∥.
Thus if we let Ω2={u∈E:∥u∥<l2}, then Ω¯1⊂Ω2 and ∥Tu∥≥∥u∥ for u∈P∩∂Ω2. Condition (H1) of Krasnoesel'skii's fixed point theorem is satisfied. So there exists a fixed point of T in P. This completes the proof.Theorem 13.
Suppose thatAf0>BF∞, then for each λ∈(1/Af0,1/BF∞), BVP (4) has at least one positive solution.Proof.
Chooseε>0 sufficiently small such that (f0-ε)λA≥1. From the definition of f0, we see that there exists l1>0, such that f(t,u)≥(f0-ε)u for 0<u≤l1. If u∈P with ∥u∥=l1, we have
(65)∥Tu(t)∥=λ∫01‍G(s,s)a(s)f(s,u(s))ds≥λ(f0-ε)∥u∥A≥∥u∥.
Let Ω1={u∈X:∥u∥<l1}; then we have ∥Tu∥≥∥u∥ for u∈P∩∂Ω1. Choose δ>0,c∈(0,1/4); then we have λ(F∞+δ)A≤1. There exists l2>l1>0, such that f(t,u)≤(F∞+δ)u for u>l2. For each u∈Pwith∥u∥=l2, we have
(66)∥Tu(t)∥=λ∫01G(s,s)a(s)f(s,u(s))ds≤λ∫c1G(s,s)a(s)(F∞+δ)u(s)ds≤λ(F∞+δ)∥u∥∫c1G(s,s)a(s)q(k)ds≤λ(F∞+δ)A·∥u∥≤∥u∥.
Let Ω2={u∈E:∥u∥<l2}; then Ω¯1⊂Ω2, and we have ∥Tu∥≤∥u∥ for u∈P∩∂Ω2. Condition (H2) of Krasnoesel'skii's fixed-point theorem is satisfied. So there exists a fixed point of T in P. This completes the proof.


## 4. An Example

Example 1.
Consider the following three-point fractional boundary value problem:(67)D0+3/2u(t)+λf(t,u(t))=0,0<t<1,u(0)=0,u(1)=12u(13),
where f(t,u(t))=((t+1)(2014u3+u)/(u2+1))+u(sinu+1),a(t)=1.
By calculations,(68)G(s,s)={[s(1-s)]α-1-βsα-1(η-s)α-1(1-βηα-1)Γ(α),0≤s=t≤η≤1,[s(1-s)]α-1(1-βηα-1)Γ(α),0≤η≤t=s≤1,∫01G(s,s)‍ds=∫0η[s(1-s)]α-1-βsα-1(η-s)α-1(1-βηα-1)Γ(α)‍ds+∫η1[s(1-s)]α-1(1-βηα-1)Γ(α)‍ds=∫01[s(1-s)]α-1(1-βηα-1)Γ(α)‍ds-∫0ηβsα-1(η-s)α-1(1-βηα-1)Γ(α)‍ds=11-(1/2)(1/3)1/2Γ(3/2)×{∫01‍[s(1-s)]1/2dsii-∫01/3‍12s1/2(η-s)1/2ds}=0.58832.
Let k=10, then
(69)q(k)min⁡{(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1,iiiiiiiiiii(1-β)(1/k)2(α-1)(1/4)α-1,iiiiiiiiiii(k-1)α-1/k2(α-1)(1/4)α-1(1-β)[(1-(1/k))2(α-1)-(1-(2/k))α-1](1/4)α-1}.=0.00557,f∞=lim⁡u→+∞⁡inf⁡min⁡0≤t≤1⁡[(t+1)(2014u2+1)u2+1+sinu+1]=2014,F0=lim⁡u→0+sup⁡max⁡0≤t≤1[(t+1)(2014u2+1)u2+1+sinu+1]=4.
Consider
(70)A=∫01‍G(s,s)a(s)q(k)ds=0.00328,(71)B=∫01‍G(s,s)a(s)ds=0.58832.
The condition Af∞>BF0 is obtained. From Theorem 12, we see that if λ∈(0.15138,0.42494), the problem (67) has a positive solution.



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*Source: 602604-2014-05-20.xml*