# Existence of Solutions for Fractional Integro-Differential Equation with Multipoint Boundary Value Problem in Banach Spaces

Authors: Yulin Zhao; Li Huang; Xuebin Wang; Xianyang Zhu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By means of the fixed-point theorem in the cone of strict-set-contraction operators, we consider the existence of a nonlinear multi-point boundary value problem of fractional integro-differential equation in a Banach space. In addition, an example to illustrate the main results is given.


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## Body

## 1. Introduction

The purpose of this paper is to establish the existence results of positive solution to nonlinear fractional boundary value problem(1.1)D0+qu(t)+f(t,u,u′,…,u(n-2),Tu,Su)=θ,0<t<1,n-1<q≤n,u(0)=u′(0)=⋯=u(n-2)(0)=θ,u(n-2)(1)=∑i=1m-2‍aiu(n-2)(ηi)
in a Banach space E, where θ is the zero element of E, and n≥2,0<η1<⋯<ηm-2<1,  ai>0(i=1,2,…,m-2),  D0+α is Riemann-Liouville fractional derivative, and
(1.2)Tu(t)=∫0t‍K(t,s)u(s)ds,Su(t)=∫01‍H(t,s)u(s)ds,
where K∈C[B,R+],B={(t,s)∈I×I:t≥s},H∈C[I×I,R+],  I=[0,1], and R+ denotes the set of all nonnegative numbers.Fractional differential equations have gained importance due to their numerous applications in many fields of science and engineering including fluid flow, rheology, diffusive transport akin to diffusion, electrical networks, and probability. For details see [1–3] and the references therein. In recent years, there are some papers dealing with the existence of the solutions of initial value problems or linear boundary value problems for fractional differential equations by means of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, lower and upper solutions method, and so forth), see for example, [4–23].In [8], by means of the fixed-point theorem for the mixed monotone operator, the authors considers unique existence of positive to singular boundary value problems for fractional differential equation
(1.3)D0+qu(t)+a(t)f(t,u,u′,…,u(n-2))=0,0<t<1,u(0)=u′(0)=⋯=u(n-2)(0)=u(n-2)(1)=0,
where D0+q is Riemann-Liouville fractional derivative of order n-1<q≤n,n≥2.In [11], El-Shahed and Nieto study the existence of nontrivial solutions for a multi-point boundary value problem for fractional differential equations
(1.4)D0+qu(t)+f(t,u(t))=0,0<t<1,n-1<q≤n,n∈N,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=∑i=1m-2‍aiu(ηi),
where n≥2,ηi∈(0,1),  ai>0(i=1,2,…,m-2), and  D0+q is Riemann-Liouville fractional derivative. Under certain growth conditions on the nonlinearity, several sufficient conditions for the existence of nontrivial solution are obtained by using Leray-Schauder nonlinear alternative. And then, Goodrich [24] was concerned with a partial extension of the problem (1.3)
(1.5)D0+qu(t)=f(t,u(t)),0<t<1,n-n<q≤n-2,u(i)(0)=0,0≤i≤n-2,D0+pu(1)=0,1≤p≤n-2,
and the authors derived the Green function for the problem (1.5) and showed that it satisfies certain properties.By the contraction mapping principle and the Krasnoselskii’s fixed-point theorem, Zhou and Chu [13] discussed the existence and uniqueness results for the following fractional differential equation with multi-point boundary conditions:
(1.6)CD0+qu(t)+f(t,u,Ku,Su)=0,0<t<1,1<q<2,a1u(0)-b1u′(0)=d1u(η1),a2u(1)-b2u′(1)=d2u(η2),
where CD0+q is the Caputo’s fractional derivative, a1,a2,b1,b2,d1, and d2 are real numbers, 0<η1, and η2<1.In [20], Stanĕk has discussed the existence of positive solutions for the singular fractional boundary value problem
(1.7)Dqu(t)+f(t,u,u′,Dpu)=0,2<q<3,0<p<1,u(0)=0,u′(0)=u′(1)=0.However, to the best of the author’s knowledge, a few papers can be found in the literature dealing with the existence of solutions to boundary value problems of fractional differential equations in Banach spaces. In [25], Salem investigated the existence of Pseudo solutions for the following nonlinear m-point boundary value problem of fractional type
(1.8)D0+qu(t)+a(t)f(t,u(t))=0,0<t<1,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=∑i=1m-2‍ζiu(ηi)
in a reflexive Banach space E, where D0+q is the Pseudo fractional differential operator of order n-1<q≤n,  n≥2.In [26], by the monotone iterative technique and mönch fixed-point theorem, Lv et al. investigated the existence of solution to the following Cauchy problems for differential equation with fractional order in a real Banach space E(1.9)CDqu(t)=f(t,u(t)),u(0)=u0,
where DCqu(t) is the Caputo’s derivative order, 0<q<1.By means of Darbo’s fixed-point theorem, Su [27] has established the existence result of solutions to the following boundary value problem of fractional differential equation on unbounded domain [0,+∞)(1.10)D0+qu(t)=f(t,u(t)),t∈[0,+∞),1<q≤2,u(0)=θ,D0+q-1u(∞)=u∞
in a Banach space E.  D0+q is the Riemann-Liouville fractional derivative.Motivated by the above mentioned papers [8, 13, 24, 25, 27, 28] but taking a quite different method from that in [26–29]. By using fixed-point theorem for strict-set-contraction operators and introducing a new cone Ω, we obtain the existence of at least two positive solutions for the BVP (1.1) under certain conditions on the nonlinear term in Banach spaces. Our results are different from those of [8, 13, 24, 25, 28, 30]. Note that the nonlinear term f depends on u and its derivatives u′,u′′,…,u(n-2).


## 2. Preliminaries and Lemmas

Let the real Banach spaceE with norm ∥·∥ be partially ordered by a cone P of E; that is, u≤v if and only if v-u∈P; and P is said to be normal if there exists a positive constant N such that θ≤u≤v implies ∥u∥≤N∥v∥, where the smallest N is called the normal constant of P. For details on cone theory, see [31].The basic space used in this paper isC[I,E]. For any u∈C[I,E], evidently, (C[I,E],∥·∥C) is a Banach space with norm ∥u∥C=supt∈I|u(t)|, and P={u∈C[I,E]:u(t)≥θ for t∈I} is a cone of the Banach space C[I,E].Definition 2.1 (see [31]).
LetV be a bounded set in a real Banach space E, and α(V)=inf{δ>0:V=∪i=1mVi, all the diameters of Vi≤δ}. Clearly, 0≤α(V)<∞.  α(V) is called the Kuratovski measure of noncompactness.
We useα,  αC to denote the Kuratowski noncompactness measure of bounded sets in the spaces E,C(I,E), respectively.Definition 2.2 (see [31]).
LetE1,E2 be real Banach spaces, S⊂E1.  T:S→E2 is a continuous and bounded operator. If there exists a constant k, such that α(T(S))≤kα(S), then T is called a k-set contraction operator. When k<1,  T is called a strict-set-contraction operator.Lemma 2.3 (see [31]).
IfD⊂C[I,E] is bounded and equicontinuous, then α(D(t)) is continuous on I and
(2.1)αC(D)=maxt∈Iα(D(t)),α({∫I‍u(t)dt:u∈D})≤∫I‍α(D(t))dt,
where D(t)={u(t):u∈D,t∈I}.Definition 2.4 (see [2, 3]).
The left-sided Riemann-Liouville fractional integral of orderq>0 of a function y:R+0→R is given by
(2.2)I0+qy(t)=1Γ(q)∫0t‍(t-s)q-1y(s)ds.Definition 2.5 (see [2, 3]).
The fractional derivative of orderq>0 of a function y:R+0→R is given by
(2.3)D0+qy(t)=1Γ(n-q)(ddt)n∫0t‍(t-s)n-q-1y(s)ds,
where n=[q]+1,[q] denotes the integer part of number q, provided that the right side is pointwise defined on R+0.Lemma 2.6 (see [2, 3]).
Letq>0. Then the fractional differential equation
(2.4)D0+qy(t)=0
has a unique solution y(t)=c1tq-1+c2tq-2+⋯+cntq-n,ci∈R,i=1,2,…,n; here n-1<q≤n.Lemma 2.7 (see [2, 3]).
Letq>0. Then the following equality holds for y∈L(0,1),D0+qy∈L(0,1),
(2.5)I0+qD0+qy(t)=y(t)+c1tq-1+c2tq-2+⋯+cNtq-N
for some ci∈R,  i=1,2,…,N; here N is the smallest integer greater than or equal to q.Lemma 2.8 (see [31]).
LetK be a cone in a Banach space E. Assume that Ω1,Ω2 are open subsets of E with 0∈Ω1,  Ω-1⊂Ω2. If T:K∩(Ω-2∖Ω1)→K is a strict-set-contraction operator such that either: (i)
∥Tx∥≤∥x∥,  x∈K∩∂Ω1, and ∥Tx∥≥∥x∥,  x∈K∩∂Ω2, or(ii)
∥Tx∥≥∥x∥,  x∈K∩∂Ω1, and ∥Tx∥≤∥x∥,  x∈K∩∂Ω2,then T has a fixed point in K∩(Ω-2∖Ω1).


## 3. Main Results

For convenience, we list some following assumptions.(H1)
There exista∈C[I,R+] and h∈C[R+n+1,R+] such that
(3.1)∥f(t,u1,…,un+1)∥≤a(t)h(∥u1∥,…,∥un+1∥),∀t∈I,uk∈P,k=1,…,n+1.(H2)
f:I×Prn+1→P, for any r>0,  f is uniformly continuous on I×Prn+1 and there exist nonnegative constants Lk,  k=1,…,n+1, with
(3.2)2Γ(q-n+2)η*(∑k=1n-2‍Lk(n-2-k)!+Ln-1+a*(n-3)!Ln+b*(n-3)!Ln+1)<1such that
(3.3)α(f(t,D1,D2,…,Dn+1))≤∑k=1n+1‍Lkα(Dk),∀t∈I,bounded setsDk∈Pr,
where a*=max{K(t,s):(t,s)∈B},b*=max{H(t,s):(t,s)∈I×I},  Pr={u∈P:∥u∥≤r},  η*=1-∑i=1m-2‍aiηiq-n+1.Lemma 3.1.
Giveny∈C[I,E] and 1-∑i=1m-2‍aiηiq-n+1≠0 hold. Then the unique solution of
(3.4)D0+q-n+2x(t)+y(t)=0,0<t<1,n-1<q≤n,n≥2,x(0)=0,x(1)=∑i=1m-2‍aix(ηi),
is
(3.5)x(t)=∫01‍G(t,s)y(s)ds,
where
(3.6)G(t,s)=g(t,s)+∑i=1m-2aig(ηi,s)η*tq-n+1,(3.7)g(t,s)={(t(1-s))q-n+1-(t-s)q-n+1Γ(q-n+2),s≤t,(t(1-s))q-n+1Γ(q-n+2),t≤s.Proof.
Deduced from Lemma2.7, we have
(3.8)x(t)=-I0+q-n+2y(t)+c1tq-n+1+c2tq-n
for some c1,c2∈R. Consequently, the general solution of (3.4) is
(3.9)x(t)=-∫0t‍(t-s)q-n+1Γ(q-n+2)y(s)ds+c1tq-n+1+c2tq-n.
By boundary value conditions x(0)=0,x(1)=∑i=1m-2‍aix(ηi), there is c2=0, and
(3.10)c1=11-∑i=1m-2aiηiq-n+1∫01‍(1-s)q-n+1Γ(q-n+2)y(s)ds-∑i=1m-2ai1-∑i=1m-2aiηiq-n+1∫0ηi‍(ηi-s)q-n+1Γ(q-n+2)y(s)ds.
Therefore, the solution of problem (3.4) is
(3.11)x(t)=-∫0t‍(t-s)q-n+1Γ(q-n+2)y(s)ds+tq-n+11-∑i=1m-2aiηiq-n+1∫01‍(1-s)q-n+1Γ(q-n+2)y(s)ds-∑i=1m-2aitq-n+11-∑i=1m-2aiηiq-n+1∫0ηi‍(ηi-s)q-n+1Γ(q-n+2)y(s)ds=∫01‍G(t,s)y(s)ds.
The proof is complete.Moreover, there is one paper [8] in which the following statement has been shown.Lemma 3.2.
The functiong(t,s) defined in (3.7) satisfying the following properties: (1)
g(t,s)≥0 is continuous on [0,1]×[0,1], and g(t,s)≤tq-n+1/Γ(q-n+2),g(t,s)≤g(s,s) for all 0≤t,s≤1;(2)
there exists a positive functionρ0∈C(0,1) such that minγ≤t≤δg(t,s)≥ρ0(s)g(t,s),s∈(0,1), where 0<γ<δ<1 and
(3.12)ρ0(s)={(δ(1-s))q-n+1-(δ-s)q-n+1(s(1-s))q-n+1,s∈(0,ξ],(γs)q-n+1,s∈[ξ,1),
where γ<ξ<δ is the solution of
(3.13)(δ(1-ξ))q-n+1-(δ-ξ)q-n+1=(γ(1-ξ))q-n+1.
For our purpose, one assumes that(H3)
η*=1-∑i=1m-2‍aiηiq-n+1>0 and 0<γ≤min{2δ-1,δ/2},  2/3≤δ<1, where γ,  δ are the constants in (2) of Lemma 3.2.Remark 3.3.
We note that if(H3) holds, then the function G(t,s) defined in (3.4) is satisfying the following properties: (i)
G(t,s)≥0   is continuous on [0,1]×[0,1], and G(t,s)≤Δtq-n+1,   for all 0≤t,  s≤1,   where Δ-1=η*Γ(q-n+2);(ii)
G(t,s)≤G(s) for all 0≤t,  s≤1, where
(3.14)G(s)=g(s,s)+∑i=1m-2aig(ηi,s)η*.Indeed, it is obvious from (1) of Lemma 3.2 and (3.6) that
(3.15)G(t,s)≤g(s,s)+1η*∑i=1m-2‍aig(ηi,s)=G(s)≤tq-n+1Γ(q-n+2)+∑i=1m-2aiηiq-n+1η*Γ(q-n+2)tq-n+1≤Δtq-n+1.Lemma 3.4.
Letu(t)=I0+n-2x(t),  x∈C[I,E]. Then the problem (1.1) can be transformed into the following modified problem:
(3.16)D0+q-n+2x(t)+f(s,I0+n-2x(s),…,I0+1x(s),x(s),T(I0+n-2x(s)),S(I0+n-2x(s)))=θ,x(0)=θ,x(1)=∑i=1m-2‍aix(ηi),
where 0<t<1,  n-1<q≤n,  n≥2. Moreover, if x∈C[I,E] is a solutions of problem (3.16), then the function u(t)=I0+n-2x(t) is a solution of (1.1).The proof follows by routine calculations.To obtain a positive solution, we construct a coneΩ by
(3.17)Ω={x(t)∈P:x(t)≥λ3x(s),t∈I*,s∈I},
where P={x∈C[I,E],x(t)≥θ,t∈I},  λ=min{minγ≤t≤δρ(t),γq-n+1},I*=[γ,δ].Let(3.18)(Ax)(t)=∫01‍G(t,s)f(s,I0+n-2x(s),…,x(s),T(I0+n-2x(s)),S(I0+n-2x(s)))ds,0≤t≤1.Lemma 3.5.
Assume that (H1)–(H3) hold. ThenA:Ω→Ω is a strict-set-contraction operator.Proof.
Letx∈Ω. Then, it follows from Remark 3.3. that
(3.19)(Ax)(t)≤∫01‍G(s)f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))ds=(∫0γ‍+∫γδ‍+∫δ1‍)G(s)f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))ds≤3∫γδ‍G(s)f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))ds,
here, by (H3), we know that γ<δ-γ and δ-γ>1-δ.
From (3.6) and (3.18), we obtain
(3.20)mint∈[γ,δ](Ax)(t)=mint∈[γ,δ]∫01‍G(t,s)f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))ds≥∫γδ‍(g(t,s)+tq-n+1η*∑i=1m-2‍aig(ηi,s))×f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))ds≥∫γδ‍(ρ0(s)g(s,s)+γq-n+1η*∑i=1m-2‍aig(ηi,s))×f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))ds≥λ∫γδ‍G(s)f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))ds≥λ3(Ax)(t′),t′∈I,
which implies that (Ax)(t)∈Ω;  that is,  A(Ω)⊂Ω.
Next, we prove thatA is continuous on Ω. Let {xj},{x}⊂Ω, and ∥xj-x∥Ω→0(j→∞). Hence {xj} is a bounded subset of Ω. Thus, there exists r>0 such that r=supj∥xj∥Ω<∞ and ∥x∥Ω≤r. It is clear that
(3.21)∥(Axj)(t)-(Ax)(t)∥=∫01‍G(t,s)∥f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))-f(s,I0+n-2x(s),…,x(s),T(I0+n-2x(s)),S(I0+n-2x(s)))∥ds≤Δtq-n+1∥f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))-f(s,I0+n-2x(s),…,x(s),T(I0+n-2x(s)),S(I0+n-2x(s)))∥ds.
According to the properties of f,  for all  ε>0, there exists J>0 such that
(3.22)∥f(s,I0+n-2xj(s),…,xj(s),T(I0+n-2xj(s)),S(I0+n-2xj(s)))-f(s,I0+n-2x(s),…,x(s),T(I0+n-2x(s)),S(I0+n-2x(s)))∥<εΔ,
for j≥J,  for all  t∈I.
Therefore,  for allε>0, for any t∈I and j≥J, we get
(3.23)∥(Axj)(t)-(Ax)(t)∥<tq-n+1ε≤ε.
This implies that A is continuous on Ω.
By the properties of continuous ofG(t,s), it is easy to see that A is equicontinuous on I.
Finally, we are going to show thatA is a strict-set-contraction operator. Let D⊂Ω be bounded. Then by condition (H1), Lemma 3.1 implies that αC(AD)=maxt∈Iα((AD)(t)). It follows from (3.18) that
(3.24)α((AD)(t))≤α(c-o{G(t,s)f(s,I0+n-2x(s),…,x(s),T(I0+n-2x(s)),S(I0+n-2x(s))):s∈[0,t],t∈I,x∈DI0+n-2}I0+n-2)≤Δ·α({f(s,I0+n-2x(s),…,x(s),T(I0+n-2x(s)),S(I0+n-2x(s))):s∈[0,t],t∈I,x∈DI0+n-2I0+n-2}I0+n-2I0+n-2)≤Δ·α(f(I×(I0+n-2D)(I)×⋯×D(I)×T(I0+n-2D)(I)×S(I0+n-2D)(I)))≤Δ·{∑k=1n-2‍Lkα((I0+n-1-kD)(I))+Ln-1α(D(I))+a*Lnα(T(I0+n-2D)(I))+b*Ln+1α(S(I0+n-2D)(I))∑k=1n-2‍},
which implies
(3.25)αC(AD)≤Δ·{∑k=1n-2‍Lkα((I0+n-1-kD)(I))+Ln-1α(D(I))+a*Lnα(T(I0+n-2D)(I))+b*Ln+1α(S(I0+n-2D)(I))∑k=1n-2‍}.
Obviously,
(3.26)α(I0+n-1-kD)(I)=α({∫0s‍(s-τ)n-2-k(n-2-k)!x(τ)dτ:τ∈[0,s],s∈I,k=1,…,n-2})≤1(n-2-k)!α(D(I)),(3.27)α(T(I0+n-2D))(I)=α({∫0t‍K(t,s)(∫0s‍(s-τ)n-2(n-3)!u(τ)dτ)ds:u∈D,t∈I})≤a*(n-3)!α({u(t):t∈I,u∈D})≤a*(n-3)!α(D(I)),(3.28)α(S(I0+n-2D))(I)=α({∫01‍H(t,s)(∫0s‍(s-τ)n-2(n-3)!u(τ)dτ)ds:u∈D,t∈I})≤b*(n-3)!α({u(t):t∈I,u∈D})≤b*(n-3)!α(D(I)).
Using a similar method as in the proof of Theorem  2.1.1 in [31], we have
(3.29)α(D(I))≤2αC(D).
Therefore, it follows from (3.26)–(3.29) that
(3.30)αC(AD)≤2Δ·(∑k=1n-2‍Lk(n-2-k)!+Ln-1+a*Ln(n-3)!+b*Ln+1(n-3)!)αC(D).
Noticing that (3.3), we obtain that T is a strict-set-contraction operator. The proof is complete.Theorem 3.6.
Let coneP be normal and conditions (H1)~(H3) hold. In addition, assume that the following conditions are satisfied. (H4)
There existu*∈P∖{θ},  c1∈C[I*,R+] and h1∈C[Pn+1,R+] such that
(3.31)f(t,u1,…,un+1)≥c1(t)h1(u1,…,un-1)u*,∀t∈I*,uk∈P,h1(u1,…,un-1)∑k=1n-1∥uk∥→∞,as∑k=1n-1‍∥uk∥→∞,uk∈P.(H5)
There existu*∈P∖{θ},  c2∈C[I*,R+], and h2∈C[Pn-1,R+] such that
(3.32)f(t,u1,…,un+1)≥c2(t)h2(u1,…,un-1)u*,∀t∈I*,uk∈P,h2(u1,…,un-1)∑k=1n-1∥ui∥→∞,as∑k=1n-1‍∥uk∥→0,uk∈P.(H6)
There exists aβ>0 such that
(3.33)NMβ∫01‍G(s)a(s)ds<β,where Mβ=maxuk∈Pβ{h(∥u1∥,…,∥un+1∥)}. Then problem (1.1) has at least two positive solutions.Proof.
Consider condition (H4), there exists anr1>0, such that
(3.34)h1(u1,…,un-1)≥3N2∑k=1n-1∥uk∥λ2∫γδ‍G(s)c1(s)ds·∥u*∥,∀uk∈P,∑k=1n-1‍∥uk∥≥r1.
Therefore,
(3.35)f(t,u1,·,un+1)≥3N2∑k=1n-1∥uk∥λ2∫γδ‍G(s)c1(s)ds·∥u*∥·c1(t)u*,∀uk∈P,∑k=1n-1‍∥uk∥≥r1.
Take
(3.36)r0>max{3Nλ-1r1,β}.
Then for t∈[γ,δ],  ∥x∥Ω=r0, we have, by (3.18),
(3.37)∥x(t)∥≥λ3N∥x∥Ω≥λ3Nr0>r1.
Hence,
(3.38)(Ax)(t)≥∫γδ‍G(t,s)f(s,I0+n-2x(s),…,x(s),T(I0+n-2x)(s),S(I0+n-2x)(s))ds≥3N2λ∫γδ‍G(s)c1(s)ds·∥u*∥∫γδ‍G(s)(∑k=1n-2‍∥I0+n-1-kx(s)∥+∥x(s)∥)c1(s)ds·u*≥3N2λ∫γδ‍G(s)c1(s)ds·∥u*∥∫γδ‍G(s)c1(s)∥x(s)∥ds·u*≥N∫γδ‍G(s)c1(s)ds·∥u*∥∥x∥Ω(∫γδ‍G(s)c1(s)ds)·u*=1∫γδ‍G(s)c1(s)ds·∥u*∥(∫γδ‍G(s)c1(s)ds∥u*∥)·N∥x∥Ω∥u*∥u*≥N∥x∥Ω∥u*∥·u*,
and consequently,
(3.39)∥Ax∥Ω≥∥x∥Ω,∀x∈Ω,∥x∥Ω=r0.
Similarly, by condition (H5), there exists r2>0, such that
(3.40)h2(u1,…,un-1)≥3N2∑k=1n-1∥uk∥λ∫γδ‍G(ξ,s)c2(s)ds·∥u*∥,∀uk∈P,0<∑k=1n-1‍∥uk∥≤r2,
where ξ is given in (2) of Lemma 3.2. Therefore,
(3.41)f(t,u1,…,un+1)≥3N2∑k=1n-1∥uk∥λ∫γδ‍G(ξ,s)c2(s)ds·∥u*∥·c2(t)u*,∀uk∈P,0<∑k=1n-1‍∥uk∥≤r2.
Choose
(3.42)0<r<min{(∑k=0n-2‍1k!)-1r2,β}.
Then for t∈[γ,δ],  x∈Ω,  ∥x∥Ω=r, we have
(3.43)(Ax)(ξ)=∫01‍G(ξ,s)f(s,I0+n-2x(s),…,x(s),T(I0+n-2x)(s),S(I0+n-2x)(s))ds≥∫γδ‍G(ξ,s)f(s,I0+n-2x(s),…,x(s),T(I0+n-2x)(s),S(I0+n-2x)(s))ds≥3N2λ∫γδ‍G(ξ,s)c2(s)ds·∥u*∥∫γδ‍G(ξ,s)(∑k=1n-2‍∥I0+n-1-kx(s)∥+∥x(s)∥)c2(s)ds·u*≥3N2λ∫γδ‍G(ξ,s)c2(s)ds·∥u*∥∫γδ‍G(ξ,s)∥x(s)∥c2(s)ds·u*≥N∫γδ‍G(ξ,s)c2(s)ds·∥u*∥∥x(s)∥Ω∫γδ‍G(ξ,s)c2(s)ds·u*=1∫γδ‍G(ξ,s)c2(s)ds·∥u*∥∥x(s)∥Ω(∫γδ‍G(ξ,s)c2(s)ds∥u*∥)·N∥x(s)∥Ω∥u*∥u*≥N∥x(s)∥Ω∥u*∥·u*,
which implies
(3.44)∥(Ax)(ξ)∥Ω≥∥x(s)∥Ω,∀x∈Ω,∥x∥Ω=r,
that is,
(3.45)∥Ax∥Ω≥∥x(s)∥Ω,∀x∈Ω,∥x∥Ω=r.
On the other hand, according to (ii) of Remark3.3 and (3.18), we get
(3.46)(Ax)(t)≤∫01‍G(s)f(s,I0+n-2x(s),…,x(s),T(I0+n-2x)(s),S(I0+n-2x)(s))ds.
By condition (H1), for t∈I,  x∈Ω,  ∥x∥Ω=β, we have
(3.47)∥f(t,u1,…,un+1)∥≤a(t)h(∥u1∥,…,∥un+1∥)≤Mβa(t).
Therefore,
(3.48)∥(Ax)(t)∥Ω≤NMβ·∫01‍G(s)a(s)ds<β=∥x∥Ω.
Applying Lemma2.7 to (3.39), (3.45), and (3.48) yields that T has a fixed-point x*∈Ω-r,β,r≤∥x*∥≤β, and a fixed-point x**∈Ω-β,r0,β≤∥x**∥≤r0. Noticing (3.48), we get ∥x*∥≠β and ∥x**∥≠β. This and Lemma 3.4 complete the proof.Theorem 3.7.
Let coneP be normal and conditions (H1)~(H4) hold. In addition, assume that the following condition is satisfied: (H7)
(3.49)h(∥u1∥,…,∥un+1∥)∑k=1n+1∥uk∥→0,asuk∈P,∑k=1n+1‍∥uk∥→0+.Then problem (1.1) has at least one positive solution.Proof.
By (H4), we can chooser0>3Nλ-1r1. As in the proof of Theorem 3.6, it is easy to see that (3.39) holds. On the other hand, considering (3.49), there exists r3>0 such that
(3.50)h(∥u1∥,…,∥un+1∥)≤ε0∑k=1n+1‍∥uk∥,fort∈I,uk∈P,0<∑k=1n+1‍∥uk∥≤r3,
where ε0>0 satisfies
(3.51)ε0=(N{∑k=1n-1‍1k!+a*+b*(n-3)!}∫01‍G(s)a(s)ds)-1.
Choose 0<r*<min{(∑k=1n-1‍(1/k!)+(a*+b*)/(n-3)!)-1r3,r0}. Then for t∈I,  x∈Ω,  ∥x∥Ω=r*, it follows from (3.46) that
(3.52)∥(Ax)(t)∥≤N∫01‍G(s)∥f(s,I0+n-2x(s),…,x(s),T(I0+n-2x)(s),S(I0+n-2x)(s))∥ds≤N∫01‍G(s)a(s)h(∥I0+n-2x(s)∥,…,∥x(s)∥,∥T(I0+n-2x)(s)∥,∥S(I0+n-2x)(s)∥)ds≤Nε0∫01‍G(s)a(s)(∑k=1n-1‍∥I0+n-1-kx(s)∥+∥T(I0+n-2x)(s)∥+∥S(I0+n-2x)(s)∥)ds≤Nε0(∑k=1n-1‍1k!+a*+b*(n-3)!)r*∫01‍G(s)a(s)ds=r*
and consequently,
(3.53)∥(Ax)(t)∥Ω≤∥x∥Ω,∀x∈Ω,∥x∥Ω≤r*.
Since 0<r*<r0, applying Lemma 2.7 to (3.39) and (3.53) yield that T has a fixed-point x*∈Ω-r*,r0,r*≤∥x*∥≤r0. This and Lemma 3.4 complete the proof.


## 4. An Example

Consider the following system of scalar differential equations of fractional order(4.1)-D5/2uk(t)=(1+t)3960k3{[u2k(t)+u3k′(t)+∑j=1∞‍u2j(t)+∑j=1∞‍uj′(t)]3+(3uk(t)+3uk+1′(t)+∑j=1∞‍uj(t)+∑j=1∞‍u2j′(t))1/2}+1+t336k5(∫0t‍e-(1+t)suk(s)ds)2/3+1+t224k4(∫01‍e-ssin2(t-s)πu2k(s)ds),t∈I,uk(0)=uk′(0)=0,uk′(1)=18uk′(14)+12uk′(49),k=1,2,3,….Conclusion.
The problem (4.1) has at least two positive solutions.Proof.
LetE=l1={u=(u1,u2,…,uk,…):∑k=1∞‍|uk|<∞} with the norm ∥u∥=∑k=1∞‍|uk|, and P={(u1,…,uk,…):uk≥0,k=1,2,3,…}. Then P is a normal cone in E with normal constant N=1, and system (4.1) can be regarded as a boundary value problem of the form (1.1). In this situation, q=5/2,n=3,a1=1/8,a2=1/2,η1=1/4,η2=4/9,η*=29/48,K(t,s)=e-(1+t)s,H(t,s)=e-ssin2(t-s)π,u=(u1,…,uk,…),f=(f1,f2,…,fk,…), in which
(4.2)fk(t,u,v,x,y)=(1+t)3960k3{(u2k+v3k+∑j=1∞‍u2j+∑j=1∞‍vj)3(1+t)3960k3+(3uk+3vk+1+∑j=1∞‍uj+∑j=1∞‍v2j)1/2}+1+t336k5xk1/5+1+t224k4y2k.
Observing the inequality ∑k=1∞‍(1/k3)<3/2, we get, by (4.2),
(4.3)∥f(t,u,v,x,y)∥=∑k=1∞‍|fk(t,u,v,x,y)|≤(1+t)32(140(∥u∥+∥v∥)3+1160(∥u∥+∥v∥)1/2+112∥x∥1/5+18∥y∥).
Hence (H1) is satisfied for a(t)=(1+t)3/2 and
(4.4)h(u,v,x,y)=140(u+v)3+1160(u+v)1/2+112x1/5+18y.
Now, we check condition (H2). Obviously,f:I×Pr4→P, for any r>0, and f is uniformly continuous on I×Pr4. Let f=f(1)+f(2), where f(1)=(f1(1),…,fk(1),…) and f(2)=(f1(2),…,fk(2),…), in which
(4.5)fk(1)(t,u,v,x,y)=(1+t)3960k3{(u2k+v3k+∑j=1∞‍u2j+∑j=1∞‍vj)3(1+t)3960k3+(3uk+3vk+1+∑j=1∞‍uj+∑j=1∞‍v2j)1/2}+1+t336k5xk1/5,(k=1,2,3,…),fk(2)(t,u,v,x,y)=1+t224k4y2k,(k=1,2,3,…).
For any t∈I and bounded subsets Di⊂E,i=1,2,3,4, from (4.5) and by the diagonal method, we have
(4.6)α(f(1)(t,D1,D2,D3,D4))=0,∀t∈I,bounded  setsDi⊂E,i=1,2,3,4,α(f(2)(I,D1,D2,D3,D4))≤112α(D4),∀t∈I,Di⊂E,i=1,2,3,4.
It follows from (4.6) that
(4.7)α(f(I,D1,D2,D3,D4))≤112α(D4),∀t∈I,Di⊂E,i=1,2,3,4,2Γ(q-n+2)η*(∑k=1n-2‍Lk(n-2-k)!+Ln-1+a*(n-3)!Ln+b*(n-3)!Ln+1)≈0.1565<1.
that is, condition (H2) holds for L1=L2=L3=0,L4=1/12.
On the other hand, takeγ=1/4,δ=3/4. Then 1/4=γ≤min{δ/2,2δ-1}=3/8,2/3<δ, which implies that condition (H3) holds. By (4.2), we have
(4.8)fk(t,u,v,x,y)≥(1+t)3960k3(∥u∥+∥v∥)3,∀t∈I*,u,v,x,y∈P,(k=1,2,3,…),fk(t,u,v,x,y)≥(1+t)3960k3∥u∥+∥v∥,∀t∈I*,u,v,x,y∈P,(k=1,2,3,…).
Hence condition (H4) is satisfied for
(4.9)c1(t)=(1+t)3960,h1,k(u,v)=(∥u∥+∥v∥)3,u*=(1,…,1k3,…),
in this situation,
(4.10)h1,k=lim∥u∥+∥v∥→∞(∥u∥+∥v∥)3∥u∥+∥v∥=∞.
And condition (H5) is also satisfied for
(4.11)c2(t)=(1+t)3960,h2,k(u,v)=∥u∥+∥v∥,u*=(1,…,1k3,…),
in this situation,
(4.12)h2,k=lim∥u∥+∥v∥→0∥u∥+∥v∥∥u∥+∥v∥=∞.
Finally, choose β=1, it is easy to check that condition (H6) is satisfied. In this case, Mβ≈0.4162, and so
(4.13)NMβ∫01‍G(s)a(s)ds≈0.7287<β=1.
Hence, our conclusion follows from Theorem 3.6.



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*Source: 172963-2012-12-02.xml*