# The Existence of Solutions for a Fractional 2m-Point Boundary Value Problems

Authors: Gang Wang; Wenbin Liu; Jinyun Yang; Sinian Zhu; Ting Zheng
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

By using the coincidence degree theory, we consider the following 2m-point boundary value problem for fractional differential equation D0+αut=ft,ut,D0+α-1ut,D0+α-2ut+et,
0<t<1,
I0+3-αut|t=0=0,D0+α-2u1=∑i=1m-2aiD0+α-2uξi,u1=∑i=1m-2biuηi, where 2<α≤3,D0+α and I0+α are the standard Riemann-Liouville fractional derivative and fractional integral, respectively. A new result on the existence of solutions for above fractional boundary value problem is obtained.


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. This is because of the intensive development of the theory of fractional calculus itself as well as its applications. Apart from diverse areas of mathematics, fractional differential equations arise in a variety of different areas such as rheology, fluid flows, electrical networks, viscoelasticity, chemical physics, and many other branches of science (see [1–4] and references cited therein). The research of fractional differential equations on boundary value problems, as one of the focal topics has attained a great deal of attention from many researchers (see [5–13]).However, there are few papers which consider the boundary value problem at resonance for nonlinear ordinary differential equations of fractional order. In [14], Hu and Liu studied the following BVP of fractional equation at resonance:
(1.1)D0+αx(t)=f(t,x(t),x′(t),x′′(t)),0≤t≤1,x(0)=x(1),x′(0)=x′′(0)=0,
where 1<α≤2, D0+α is the standard Caputo fractional derivative.In [15], Zhang and Bai investigated the nonlinear nonlocal problem (1.2)D0+αu(t)=f(t,u(t)),0<t<1,u(0)=0,βu(η)=u(1),
where 1<α≤2, they consider the case βηα-1=1, that is, the resonance case.In [16], Bai investigated the boundary value problem at resonance (1.3)D0+αu(t)=f(t,u(t),D0+α-1u(t))+e(t),0<t<1,I0+2-αu(t)|t=0=0,D0+α-1u(1)=∑i=0m-2βiD0+α-1u(ηi)
is considered, where 1<α≤2 is a real number, D0+αandI0+α are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and f:[0,1]×R2→R is continuous and e(t)∈L1[0,1],m≥2,0<ξi<1,βi∈R,i=1,2,…,m-2 are given constants such that ∑i=1m-2βi=1.In this paper, we study the 2m-point boundary value problem(1.4)D0+αu(t)=f(t,u(t),D0+α-1u(t),D0+α-2u(t))+e(t),0<t<1,(1.5)I0+3-αu(t)|t=0=0,D0+α-2u(1)=∑i=1m-2aiD0+α-2u(ξi),u(1)=∑i=1m-2biu(ηi),
where 2<α≤3, m≥2,0<ξ1<⋯<ξm<1,0<η1<⋯<ηm<1,ai,bi∈R,f:[0,1]×R3→R, f satisfies Carathéodory conditions, D0+α and I0+α are the standard Riemann-Liouville fractional derivative and fractional integral, respectively.Setting(1.6)Λ1=1α(α+1)(1-∑i=1m-2aiξiα+1),Λ2=1α(α-1)(1-∑i=1m-2aiξiα),Λ3=(Γ(α))2Γ(2α)[1-∑i=1m-2biηi2α-1],Λ4=Γ(α)Γ(α-1)Γ(2α-1)[1-∑i=1m-2biηi2α-2].In this paper, we will always suppose that the following conditions hold:(C1):(1.7)∑i=1m-2aiξi=∑i=1m-2ai=1,∑i=1m-2biηiα-1=∑i=1m-2biηiα-2=1,(C2):(1.8)Λ=Λ1Λ4-Λ2Λ3≠0.We say that boundary value problem (1.4) and (1.5) is at resonance, if BVP (1.9)D0+αu(t)=0,I0+3-αu(t)|t=0=0,D0+α-2u(1)=∑i=1m-2aiD0+α-2u(ξi),u(1)=∑i=1m-2biu(ηi)
has u(t)=atα-1+btα-2, a,b∈R as a nontrivial solution.The rest of this paper is organized as follows. Section2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions for BVP (1.4)-(1.5) under nonlinear growth restriction of f, basing on the coincidence degree theory due to Mawhin (see [17]).Now, we will briefly recall some notation and an abstract existence result.LetY,Z be real Banach spaces, L:domL⊂Y→Z a Fredholm map of index zero,s and P:Y→Y,Q:Z→Z continuous projectors such that (1.10)Y=Ker⁡L⊕Ker⁡P,Z=Im⁡L⊕Q,Im⁡P=Ker⁡L,Ker⁡Q=Im⁡L.
It follows that L|domL∩KerP:domL∩KerP→ImL is invertible. We denote the inverse of the map by Kp. If Ω is an open-bounded subset of Y such that domL∩Ω≠∅, the map N:Y→Z will be called L-compact on Ω if QN(Ω¯) is bounded and Kp(I-Q)N:Ω¯→Y is compact.The lemma that we used is [17, Theorem 2.4].Lemma 1.1.
LetL be a Fredholm operator of index zero and let N be L-compact on Ω¯. Assume that the following conditions are satisfied: (i)
Lx≠λNx, for all (x,λ)∈[domL∖KerL∩∂Ω]×[0,1];(ii)
Nx∉ImL, for all x∈KerL∩∂Ω;(iii)
deg(JQN|KerL,KerL∩Ω,0)≠0,where Q:Z→Z is a projection as above with KerQ=ImL and J:ImQ→KerL is any isomorphism. Then the equation Lx=Nx has at least one solution in domL∩Ω¯.


## 2. Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory. These definitions can be found in the recent literature [1–16, 18].Definition 2.1.
The fractional integral of orderα>0 of a function y:(0,∞)→R is given by
(2.1)I0+αy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds,
provided the right side is pointwise defined on (0,∞), where Γ(·) is the Gamma function.Definition 2.2.
The fractional derivative of orderα>0 of a function y:(0,∞)→R is given by
(2.2)D0+αy(t)=1Γ(n-α)(ddt)n∫0ty(s)(t-s)α-n+1ds,
where n=[α]+1, provided the right side is pointwise defined on (0,∞).Definition 2.3.
We say that the mapf:[0,1]×Rn→R satisfies Carathéodory conditions with respect to L1[0,1] if the following conditions are satisfied: (i)
for eachz∈Rn, the mapping t→f(t,z) is Lebesgue measurable;(ii)
for almost everyt∈[0,1], the mapping t→f(t,z) is continuous on Rn;(iii)
for eachr>0, there exists ρr∈L1([0,1],R) such that for a.e. t∈[0,1] and every |z|≤r, we have f(t,z)≤ρr(t).Lemma 2.4 (see [15]).
Assume thatu∈C(0,1)∩L1(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L1(0,1). Then
(2.3)I0+αD0+αu(t)=u(t)+c1tα-1+c1tα-2+⋯+cNtα-N
for some ci∈R,i=1,2,…,N, where N is the smallest integer greater than or equal to α.We use the classical Banach spaceC[0,1] with the norm (2.4)‖x‖∞=max⁡t∈[0,1]|x(t)|,L[0,1] with the norm (2.5)‖x‖1=∫01|x(t)|dt.Definition 2.5.
Forn∈N, we denote by ACn[0,1] the space of functions u(t) which have continuous derivatives up to order n-1 on [0,1] such that u(n-1)(t) is absolutely continuous: ACn[0,1]={u∣[0,1]→R and (D(n-1))u(t) is absolutely continuous in [0,1]}.Lemma 2.6 (see [15]).
Givenμ>0 and N=[μ]+1 we can define a linear space
(2.6)Cμ[0,1]={u(t)∣u(t)=I0+αx(t)+c1tμ-1+c2tμ-2+⋯+cNtμ-(N-1),t∈[0,1]},
where x∈[0,1],ci∈R,i=1,2,…,N-1. By means of the linear functional analysis theory, we can prove that with the
(2.7)‖u‖Cμ=‖D0+μu‖∞+⋯+‖D0+μ-(N-1)u‖∞+‖u‖∞,Cμ[0,1] is a Banach space.Remark 2.7.
Ifμ is a natural number, then Cμ[0,1] is in accordance with the classical Banach space Cn[0,1].Lemma 2.8 (see [15]).
f⊂Cμ[0,1] is a sequentially compact set if and only if f is uniformly bounded and equicontinuous. Here uniformly bounded means there exists M>0, such that for every u∈f(2.8)‖u‖Cμ=‖D0+μu‖∞+⋯+‖D0+μ-(N-1)u‖∞+‖u‖∞<M,
and equicontinuous means that ∀ɛ>0, ∃δ>0, such that
(2.9)|u(t1)-u(t2)|<ɛ,(∀t1,t2∈[0,1],|t1-t2|<δ,∀u∈f),|D0+α-iu(t1)-D0+α-iu(t2)|<ɛ,(∀t1,t2∈[0,1],|t1-t2|<δ,∀u∈f,∀i=1,2,…,N-1).Lemma 2.9 (see [1]).
Letα>0,n=[α]+1. Assume that u∈L1(0,1) with a fractional integration of order n-α that belongs to ACn[0,1]. Then the equality
(2.10)(I0+αD0+αu)(t)=u(t)-∑i=1n((I0+n-αu)(t))n-i|t=0Γ(α-i+1)tα-i
holds almost everywhere on [0,1].Definition 2.10 (see [16]).
LetI0+α(L1(0,1)),α>0 denote the space of functions u(t), represented by fractional integral of order α of a summable function: u=I0+αv,v∈L1(0,1).LetZ=L1[0,1], with the norm ∥y∥=∫01|y(s)|ds, Y=Cα-1[0,1] defined by Lemma 2.6, with the norm ∥u∥Cα-1=∥D0+α-1u∥∞+∥D0+α-2u∥∞+∥u∥∞, where Y is a Banach space.DefineL to be the linear operator from domL⊂Y to Z with (2.11)dom⁡L={u∈Cα-1[0,1]∣D0+αu∈L1[0,1],usatisfies(1.5)},(2.12)Lu=D0+αu,u∈dom⁡L,
we define N:Y→Z by setting (2.13)Nu(t)=f(t,u(t),D0+α-1u(t)D0+α-2u(t))+e(t).
Then boundary value problem (1.4) and (1.5) can be written as Lu=Nu.


## 3. Main Results

Lemma 3.1.
Let L be defined by (2.12), then
(3.1)KerL={atα-1+btα-2∣a,b∈R}≅R2,ImL={y∈Z∣∫01(1-s)y(s)ds-∑i=1m-2ai∫0ξi(ξi-s)y(s)ds=0,∫01(1-s)α-1y(s)ds-∑i=1m-2bi∫0ηi(ηi-s)α-1y(s)ds=0}.Proof.
In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming thatI0+α=D0+-α for α<0.
LetLu=D0+αu, by Lemma 2.9, D0+αu(t)=0 has solution
(3.2)u(t)=∑i=13((I0+3-αu)(t))3-i|t=0Γ(α-i+1)tα-i=((I0+3-αu)(t))′′|t=0Γ(α)tα-1+((I0+3-αu)(t))′|t=0Γ(α-1)tα-2+((I0+3-αu)(t))|t=0Γ(α-2)tα-3=D0+α-1u(t)|t=0Γ(α)tα-1+D0+α-2u(t)|t=0Γ(α-1)tα-2+((I0+3-αu)(t))|t=0Γ(α-2)tα-3.
Combine with (1.5), So,
(3.3)Ker⁡L={atα-1+btα-2∣a,b∈R}≅R2.
Let y∈Z and let
(3.4)ut=1Γ(α)∫0t(t-s)α-1y(s)ds+c1tα-1+c2tα-2+c3tα-3.
ThenD0+αu(t)=y(t) a.e. t∈[0,1] and, if
(3.5)∫01(1-s)y(s)ds-∑i=1m-2ai∫0ξi(ξi-s)y(s)ds=0,∫01(1-s)α-1y(s)ds-∑i=1m-2bi∫0ηi(ηi-s)α-1y(s)ds=0
hold, then u(t) satisfies the boundary conditions (1.5). That is, u∈domL and we have
(3.6){y∈Z∣ysatisfies(3.4)}⊆Im⁡L.
Letu∈domL. Then for D0+αu∈ImL, we have
(3.7)I0+αy(t)=u(t)-c1tα-1-c2tα-2-c3tα-3,
where
(3.8)c1=D0+α-1u(t)|t=0Γ(α),c2=D0+α-2u(t)|t=0Γ(α-1),c3=I0+3-αu(t)|t=0Γ(α-2),
which, due to the boundary value condition (1.5), implies that satisfies (3.5). In fact, from I0+3-αu(t)|t=0=0 we have c3=0, from D0+α-2u(1)=∑i=1m-2aiD0+α-2u(ξi), u(1)=∑i=1m-2biu(ηi), we have
(3.9)∫01(1-s)y(s)ds-∑i=1m-2ai∫0ξi(ξi-s)y(s)ds=0,∫01(1-s)α-1y(s)ds-∑i=1m-2bi∫0ηi(ηi-s)α-1y(s)ds=0.
Hence,
(3.10){y∈Z∣ysatisfies(3.4)}⊇Im⁡L.
Therefore,
(3.11){y∈Z∣ysatisfies(3.4)}=Im⁡L.
The proof is complete.Lemma 3.2.
The mappingL:domL⊂Y→Z is a Fredholm operator of index zero, and
(3.12)Qy(t)=(T1y(t))tα-1+(T2y(t))tα-2,
where
(3.13)T1y=1Λ(Λ4Q1y-Λ2Q2y),T2y=1Λ(Λ3Q1y-Λ1Q2y),
define by Kp:ImL→domL∩KerP by
(3.14)Kpy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds=I0+αy(t),y∈ImL,
and forally∈ImL,∥Kpy∥Cα-1≤((1/Γ(α))+2)∥y∥1.Proof.
Consider the continuous linear mappingQ1:Z→Z and Q2:Z→Z defined by
(3.15)Q1y=∫01(1-s)y(s)ds-∑i=1m-2ai∫0ξi(ξi-s)y(s)ds,Q2y=∫01(1-s)α-1y(s)ds-∑i=1m-2bi∫0ηi(ηi-s)α-1y(s)ds.
Using the above definitions, we construct the following auxiliary maps T1:Z→Z and T2:Z→Z:
(3.16)T1y=1Λ(Λ4Q1y-Λ2Q2y),T2y=1Λ(Λ3Q1y-Λ1Q2y).
Since the condition (C2) holds, the mapping defined by
(3.17)Qy(t)=(T1y(t))tα-1+(T2y(t))tα-2
is well defined.
Recall (C2) and note that(3.18)T1(T1ytα-1)=1Λ(Λ4Q1(T1ytα-1)-Λ2Q2(T1ytα-1))=1Λ[Λ4(Λ4Λ1ΛQ1y-Λ1Λ2ΛQ2y)-Λ2(Λ4Λ3ΛQ1y-Λ2Λ3ΛQ2y)]=T1y,
and similarly we can derive that
(3.19)T1(T2ytα-2)=0,T2(T1ytα-1)=0,T2(T2ytα-2)=T2y.
So, for y∈Z, it follows from the four relations above that
(3.20)Q2y=Q((T1y)tα-1+(T2y)tα-2)=T1((T1y)tα-1+(T2y)tα-2)tα-1+T2((T1y)tα-1+(T2y)tα-2)tα-2=(T1y)tα-1+(T2y)tα-2=Qy,
that is, the map Q is idempotent. In fact Q is a continuous linear projector.
Note thaty∈ImL implies Qy=0. Conversely, if Qy=0, so
(3.21)Λ4Q1y-Λ2Q2y=0,Λ1Q2y-Λ3Q1y=0,
but
(3.22)|Λ4-Λ2-Λ3Λ1|=Λ4Λ1-Λ2Λ3≠0,
then we must have Q1y=Q2y=0; since the condition (C2) holds, this can only be the case if Q1y=Q2y=0, that is, y∈ImL. In fact KerQ=ImL, take y∈Z in the form y=(y-Qy)+Qy so that y-Qy∈KerQ=ImL,Qy∈ImQ, thus, Z=ImL+ImQ, Let y∈ImL∩ImQ and assume that y=atα-1+btα-2 is not identically zero on [0,1]. Then, since y∈ImL, from (3.5) and the condition (C2), we have
(3.23)Q1y=∫01(1-s)(asα-1+bsα-2)ds-∑i=1m-2ai∫0ξi(ξi-s)(asα-1+bsα-2)ds=0,Q2y=∫01(1-s)α-1(asα-1+bsα-2)ds-∑i=1m-2bi∫0ηi(ηi-s)α-1(asα-1+bsα-2)ds=0.
So
(3.24)aΛ1+bΛ2=0,aΛ3+bΛ4=0,
but
(3.25)|Λ1Λ2Λ3Λ4|=Λ1Λ4-Λ2Λ3≠0,
we derive a=b=0, which is a contradiction. Hence, ImL∩ImQ={0}; thus Z=ImL⊕ImQ.
Now,dimKerL=2=codimImL and so L is a Fredholm operator of index zero.
LetP:Y→Y be defined by
(3.26)Pu(t)=1Γ(α)D0+α-1u(t)|t=0tα-1+1Γ(α-1)D0+α-2u(t)|t=0tα-2,t∈[0,1].
Note that P is a continuous linear projector and
(3.27)Ker⁡P={u∈Y∣D0+α-1u(0)=D0+α-2u(0)=0}.
It is clear that Y=KerP⊕KerL.
Note that the projectorsP and Q are exact. Define by Kp:ImL→domL∩KerP by
(3.28)Kpy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds=I0+αy(t),y∈Im⁡L.
Hence we have
(3.29)D0+α-1(Kpy)t=∫0ty(s)ds,D0+α-2(Kpy)t=∫0t(t-s)y(s)ds,
then
(3.30)‖Kpy‖∞≤1Γ(α)‖y‖1,‖D0+α-1(Kpy)‖∞≤‖y‖1,‖D0+α-2(Kpy)‖∞≤‖y‖1,
and thus
(3.31)‖Kpy‖Cα-1≤(1Γ(α)+2)‖y‖1.
In fact, ify∈ImL, then
(3.32)(LKp)y(t)=D0+αI0+αy(t)=y(t).
Also, if u∈domL∩KerP, then
(3.33)(KpL)u(t)=I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+c3tα-3,
where
(3.34)c1=D0+α-1u(t)|t=0Γ(α),c2=D0+α-2u(t)|t=0Γ(α-1),c3=I0+3-αu(t)|t=0Γ(α-2),
and from the boundary value condition (1.5) and the fact that u∈domL∩KerP, Pu=0, D0+α-1u(t)|t=0=D0+α-2u(t)|t=0=I0+3-αu(t)|t=0=0, we have c1=c2=c3=0, thus
(3.35)(KpL)u(t)=u(t).
This shows that Kp=[L|domL∩KerP]-1. The proof is complete. Using (3.16), we write
(3.36)QNu(t)=(T1Nu)tα-1+(T2Nu)tα-2,Kp(I-Q)Nu(t)=1Γ(α)∫0t(t-s)α-1[Nu(s)-QNu(s)]ds.By Lemma2.8 and a standard method, we obtain the following lemma.Lemma 3.3 (see [16]).
For every givene∈L1[0,1],Kp(I-Q)N:Y→Y is completely continuous.Assume that the following conditions on the functionf(t,x,y,z) are satisfied.(H1)
There exist functionsa(t),b(t),c(t),d(t),r(t)∈L1[0,1], and a constant θ∈[0,1) such that for all (x,y,z)∈R3, t∈[0,1], one of the following inequalities is satisfied:(3.37)|f(t,x,y,z)|≤a(t)|x|+b(t)|y|+c(t)|z|+d(t)|x|θ+r(t),(3.38)|f(t,x,y,z)|≤a(t)|x|+b(t)|y|+c(t)|z|+d(t)|y|θ+r(t),(3.39)|f(t,x,y,z)|≤a(t)|x|+b(t)|y|+c(t)|z|+d(t)|z|θ+r(t).(H2)
There exists a constantA>0, such that for x∈domL∖KerL satisfying |D0+α-1x(t)|+|D0+α-2x(t)|>A for all t∈[0,1], we have
(3.40)Q1Nx(t)≠0,orQ2Nx(t)≠0.(H3)
There exists a constantB>0 such that for every a,b∈R satisfying a2+b2>B then either
(3.41)aT1N(atα-1+btα-2)+bT2N(atα-1+btα-2)<0,or(3.42)aT1N(atα-1+btα-2)+bT2N(atα-1+btα-2)>0.Remark 3.4.
T1N(atα-1+btα-2) and T2N(atα-1+btα-2) from (H3) stand for the images of u(t)=atα-1+btα-2 under the maps T1N and T2N, respectively.Lemma 3.5.
Suppose (H1)-(H2)  hold, then the set(3.43)Ω1={x∈domL∖KerL:Lx=λNx,λ∈[0,1]}
is bounded.Proof.
Take(3.44)Ω1={x∈dom⁡L∖Ker⁡L:Lx=λNx,λ∈[0,1]}.
Then for x∈Ω1, Lx=λNx thus λ≠0,Nx∈ImL=KerQ, and hence QNx(t) for all t∈[0,1]. By the definition of Q, we have Q1Nx(t)=Q2Nx(t)=0. It follows from (H2) that there exists t0∈[0,1], such that |D0+α-1u(t0)|+|D0+α-2u(t0)|≤A.
Now(3.45)D0+α-1x(t)=D0+α-1x(t0)+∫t0tD0+αx(s)ds,D0+α-2x(t)=D0+α-2x(t0)+∫t0tD0+α-1x(s)ds,
and so
(3.46)|D0+α-1x(0)|≤‖D0+α-1x(t)‖∞≤|D0+α-1x(t0)|+‖D0+α-1x‖1≤A+‖Lx‖1≤A+‖Nx‖1,|D0+α-2x(0)|≤‖D0+α-2x(t)‖∞≤|D0+α-2x(t0)|+‖D0+α-1x‖∞≤|D0+α-2x(t0)|+|D0+α-1x(t0)|+‖D0+αx‖1≤A+‖Lx‖1≤A+‖Nx‖1.
Therefore, we have
(3.47)‖Px‖Cα-1=‖1Γ(α)D0+α-1x(0)tα-1+1Γ(α-1)D0+α-2x(0)tα-2‖Cα-1=‖1Γ(α)D0+α-1x(0)tα-1+1Γ(α-1)D0+α-2x(0)tα-2‖∞+‖D0+α-1x(0)‖∞+‖D0+α-1x(0)t+D0+α-2x(0)‖∞≤(2+1Γ(α))|D0+α-1x(0)|+(1+1Γ(α-1))|D0+α-2x(0)|≤(2+1Γ(α))(A+‖Nx‖1)+(1+1Γ(α-1))(A+‖Nx‖1).
Note that (I-P)x∈domL∩KerP for all x∈Ω1. Then, by Lemma 3.2, we have
(3.48)‖(I-P)x‖Cα-1=‖KpL(I-P)x‖Cα-1≤(2+1Γ(α))‖Nx‖1,
so, we have
(3.49)‖x‖Cα-1≤‖(I-P)x‖Cα-1+‖Px‖Cα-1≤(2+1Γ(α))(A+‖Nx‖1)+(1+1Γ(α-1))(A+‖Nx‖1)+(2+1Γ(α))‖Nx‖1=(2Γ(α)+1Γ(α-1)+5)‖Nx‖1+(1Γ(α)+1Γ(α-1)+1Γ(α-2)+3)A≤m‖Nx‖1+nA,
where m=((2/Γ(α))+(1/Γ(α-1))+5),n=((1/Γ(α))+(1/Γ(α-1))+(1/(Γ(α-2))+3),A is a constant. This is for all x∈Ω1. If the first condition of (H1) is satisfied, then, we have
(3.50)‖x‖Cα-1=max⁡{‖x‖∞,‖D0+α-1x‖∞,‖D0+α-2x‖∞}≤m[+‖d‖1‖D0+α-2x‖∞θ+D]‖a‖1‖x‖∞+‖b‖1‖D0+α-1x‖∞+‖c‖1‖D0+α-2x‖∞+‖d‖1‖D0+α-2x‖∞θ+D],
where D=∥r∥1+∥e∥1+n/m, and consequently, for
(3.51)‖x‖∞≤‖x‖Cα-1,‖D0+α-1x‖∞≤‖x‖Cα-1,‖D0+α-2x‖∞≤‖x‖Cα-1,
so
(3.52)‖x‖∞≤m1-m‖a‖1[‖b‖1‖D0+α-1x‖∞+‖c‖1‖D0+α-2x‖∞+‖d‖1‖D0+α-2x‖∞θ+D],‖D0+α-1x‖∞≤m1-m‖a‖1-m‖b‖1[‖c‖1‖D0+α-2x‖∞+‖d‖1‖D0+α-2x‖∞θ+D],‖D0+α-2x‖∞≤m1-m‖a‖1-m‖b‖1-m‖c‖1(‖d‖1‖D0+α-2x‖∞θ+D).
But θ∈[0,1) and ∥a∥1+∥b∥1+∥c∥1≤1/m, so there exists A1,A2,A3>0 such that
(3.53)‖D0+α-2x‖∞≤A1,‖D0+α-1x‖∞≤A2,‖x‖∞≤A3.
Therefore, for all x∈Ω1,
(3.54)‖x‖Cα-1=max⁡{‖x‖∞,‖D0+α-1x‖∞,‖D0+α-2x‖∞}≤max⁡{A1,A2,A3},
we can prove that Ω1 is also bounded.
If (3.38) or (3.39) holds, similar to the above argument, we can prove that Ω1 is bounded too.Lemma 3.6.
Suppose (H3) holds, then the set(3.55)Ω2={x∈KerL:Nx∈ImL}
is bounded.Proof.
Let(3.56)Ω2={x∈Ker⁡L:Nx∈Im⁡L},
for x∈Ω2,x∈KerL={x∈domL:x=atα-1+btα-2,a,b∈R,t∈[0,1]} and QNx(t)=0; thus T1N(atα-1+btα-2)=T2N(atα-1+btα-2)=0. By (H3), a2+b2≤B, that is, Ω2 is bounded.Lemma 3.7.
Suppose  (H3)  holds, then the set(3.57)Ω3={x∈KerL:-λJx+(1-λ)QNx=0,λ∈[0,1]}
is bounded.Proof.
We define the isomorphismJ:KerL→ImQ by
(3.58)J(atα-1+btα-2)=atα-1+btα-2.
If the first part of (H3) is satisfied, let(3.59)Ω3={x∈Ker⁡L:-λJx+(1-λ)QNx=0,λ∈[0,1]}.
For every x=atα-1+btα-2∈Ω3,
(3.60)λ(atα-1+btα-2)=(1-λ)[T1N(atα-1+btα-2)tα-1+T2N(atα-1+btα-2)tα-1].
If λ=1, then a=b=0, and if a2+b2>B, then by (H3)
(3.61)λ(a2+b2)=(1-λ)[aT1N(atα-1+btα-2)+bT2N(atα-1+btα-2)]<0,
which, in either case, is a contradiction. If the other part of (H3) is satisfied, then we take
(3.62)Ω3={x∈Ker⁡L:λJx+(1-λ)QNx=0,λ∈[0,1]},
and, again, obtain a contradiction. Thus, in either case
(3.63)‖x‖Cα-1=‖atα-1+btα-2‖Cα-1=‖atα-1+btα-2‖∞+‖aΓ(α)‖∞+‖aΓ(α)t+bΓ(α-1)‖∞≤(1+2Γ(α))|a|+(1+Γ(α-1))|b|≤(2+2Γ(α)+Γ(α-1))B,
for all x∈Ω3, that is, Ω3 is bounded.Remark 3.8.
Suppose the second part of (H3) holds, then the set(3.64)Ω3′={x∈Ker⁡L:λJx+(1-λ)QNx=0,λ∈[0,1]}
is bounded.Theorem 3.9.
If (C1)-(C2) and (H1)–(H3) hold, then the boundary value problem (1.4)-(1.5) has at least one solution.Proof.
SetΩ to be a bounded open set of Y such that ∪i=13Ω¯⊂Ω. It follows from Lemmas 3.2 and 3.3 that L is a Fredholm operator of index zero, and the operator Kp(I-Q)N:Ω¯→Y is compact N, thus, is L-compact on Ω¯. By Lemmas 3.5 and 3.6, we get that the following two conditions are satisfied: (i)
Lx≠λNx for every (x,λ)∈[domL∖KerL∩∂Ω]×[0,1];(ii)
Nx∉ImL, for every x∈KerL∩∂Ω.
Finally, we will prove that (iii) of Lemma1.1 is satisfied. Let H(x,λ)=±λJx+(1-λ)QNx, where I is the identity operator in the Banach space Y. According to Lemma 3.7 (or Remark 3.8), we know that H(x,λ)≠0, for all x∈∂Ω∩KerL, and thus, by the homotopy property of degree,
(3.65)deg⁡(QN|Ker⁡L,Ker⁡L∩Ω,0)=deg⁡(H(⋅,0),Ker⁡L∩Ω,0)=deg⁡(H(⋅,1),Ker⁡L∩Ω,0)=deg⁡(±I,Ker⁡L∩Ω,0)=sgn⁡[±|Λ4Λ-Λ2Λ-Λ3ΛΛ1Λ|]=sgn⁡(±Λ1Λ4-Λ2Λ3Λ)=±1≠0.
Then by Lemma 1.1, Lx=Nx has at least one solution in domL∩Ω¯, so the boundary value problem (1.4) and (1.5) has at least one solution in the space Cα-1[0,1]. The proof is finished.


## 4. An Example

Let us consider the following boundary value problem:(4.1)D0+5/2u(t)=f(t,u(t),D0+α-1u(t),D0+α-2u(t))+e(t),0<t<1,I0+3-αu(t)|t=0=0,D0+1/2u(1)=2D0+1/2u(23)-D0+1/2u(13),u(1)=645u(14)-815u(19),
where (4.2)f(t,u(t),D0+α-1u(t),D0+α-2u(t))=140sin⁡(u(t))+120D0+3/2u(t)+120D0+1/2u(t)+5cos⁡(D0+1/2u(t))1/5.
Corresponding to the problem (1.4)-(1.5), we have that e(t)=1+3sin2t, α=5/2, a1=-1, a2=2, ξ1=1/3, ξ2=2/3, b1=64/5, b2=-81/5, η1=1/4, η2=1/9 and (4.3)f(t,x,y,z)=140sin⁡x+120y+120z+5cos⁡(z)1/5,
then there is (4.4)a1+a2=1,a1ξ1+a2ξ2=1,b1η13/2+b2η23/2=1,b1η11/2+b2η21/2=1,Λ1=435(1-∑i=12aiξi7/2),Λ2=415(1-∑i=12aiξi5/2),Λ3=(Γ(5/2))224[1-∑i=12biηi4],Λ4=16Γ(5/2)Γ(3/2)24[1-∑i=12biηi3],Λ=Λ1Λ4-Λ2Λ3≠0,|f(t,x,y,z)|≤140|x|+120|y|+120|z|+5|z|1/5.
Again, taking a=1/40, b=c=1/20, then (4.5)‖a‖1+‖b‖1+‖c‖1=1/8,1m=1(2/Γ(α))+(1/Γ(α-1))+5≈0.131,
therefore
(4.6)‖a‖1+‖b‖1+‖c‖1<1m.
Take A=181, B=81. By simple calculation, we can get that (C1)-(C2) and (H1)–(H3) hold. By Lemma 1.1, we obtain that (4.1) has at least one solution.



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*Source: 841349-2012-01-26.xml*