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.
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:
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 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 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 problemD0+αu(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+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.
We say that boundary value problem (1.4) and (1.5) is at resonance, if BVP 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. Section 2 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.
Let Y,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 Y=KerL⊕KerP,Z=ImL⊕Q,ImP=KerL,KerQ=ImL.
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.
Let L be a Fredholm operator of index zero and let N be L-compact on Ω¯. Assume that the following conditions are satisfied:
Lx≠λNx, for all (x,λ)∈[domL∖KerL∩∂Ω]×[0,1];
Nx∉ImL, for all x∈KerL∩∂Ω;
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 order α>0 of a function y:(0,∞)→R is given by
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 order α>0 of a function y:(0,∞)→R is given by
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 map f:[0,1]×Rn→R satisfies Carathéodory conditions with respect to L1[0,1] if the following conditions are satisfied:
for each z∈Rn, the mapping t→f(t,z) is Lebesgue measurable;
for almost every t∈[0,1], the mapping t→f(t,z) is continuous on Rn;
for each r>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 that u∈C(0,1)∩L1(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L1(0,1). Then
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 space C[0,1] with the norm ‖x‖∞=maxt∈[0,1]|x(t)|,L[0,1] with the norm ‖x‖1=∫01|x(t)|dt.
Definition 2.5.
For n∈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]).
Given μ>0 and N=[μ]+1 we can define a linear space
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
‖u‖Cμ=‖D0+μu‖∞+⋯+‖D0+μ-(N-1)u‖∞+‖u‖∞,Cμ[0,1] is a Banach space.
Remark 2.7.
If μ 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‖u‖Cμ=‖D0+μu‖∞+⋯+‖D0+μ-(N-1)u‖∞+‖u‖∞<M,
and equicontinuous means that ∀ɛ>0, ∃δ>0, such that
|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]).
Let α>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
(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]).
Let I0+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).
Let Z=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.
Define L to be the linear operator from domL⊂Y to Z with domL={u∈Cα-1[0,1]∣D0+αu∈L1[0,1],usatisfies(1.5)},Lu=D0+αu,u∈domL,
we define N:Y→Z by setting 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 ResultsLemma 3.1.
Let L be defined by (2.12), then
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 that I0+α=D0+-α for α<0.
Let Lu=D0+αu, by Lemma 2.9, D0+αu(t)=0 has solution
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,
KerL={atα-1+btα-2∣a,b∈R}≅R2.
Let y∈Z and let
ut=1Γ(α)∫0t(t-s)α-1y(s)ds+c1tα-1+c2tα-2+c3tα-3.
Then D0+αu(t)=y(t) a.e. t∈[0,1] and, if
∫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
{y∈Z∣ysatisfies(3.4)}⊆ImL.
Let u∈domL. Then for D0+αu∈ImL, we have
I0+αy(t)=u(t)-c1tα-1-c2tα-2-c3tα-3,
where
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
∫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,
{y∈Z∣ysatisfies(3.4)}⊇ImL.
Therefore,
{y∈Z∣ysatisfies(3.4)}=ImL.
The proof is complete.
Lemma 3.2.
The mapping L:domL⊂Y→Z is a Fredholm operator of index zero, and
Qy(t)=(T1y(t))tα-1+(T2y(t))tα-2,
where
T1y=1Λ(Λ4Q1y-Λ2Q2y),T2y=1Λ(Λ3Q1y-Λ1Q2y),
define by Kp:ImL→domL∩KerP by
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 mapping Q1:Z→Z and Q2:Z→Z defined by
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:
T1y=1Λ(Λ4Q1y-Λ2Q2y),T2y=1Λ(Λ3Q1y-Λ1Q2y).
Since the condition (C2) holds, the mapping defined by
Qy(t)=(T1y(t))tα-1+(T2y(t))tα-2
is well defined.
Recall (C2) and note that
T1(T1ytα-1)=1Λ(Λ4Q1(T1ytα-1)-Λ2Q2(T1ytα-1))=1Λ[Λ4(Λ4Λ1ΛQ1y-Λ1Λ2ΛQ2y)-Λ2(Λ4Λ3ΛQ1y-Λ2Λ3ΛQ2y)]=T1y,
and similarly we can derive that
T1(T2ytα-2)=0,T2(T1ytα-1)=0,T2(T2ytα-2)=T2y.
So, for y∈Z, it follows from the four relations above that
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 that y∈ImL implies Qy=0. Conversely, if Qy=0, so
Λ4Q1y-Λ2Q2y=0,Λ1Q2y-Λ3Q1y=0,
but
|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
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
aΛ1+bΛ2=0,aΛ3+bΛ4=0,
but
|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.
Let P:Y→Y be defined by
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
KerP={u∈Y∣D0+α-1u(0)=D0+α-2u(0)=0}.
It is clear that Y=KerP⊕KerL.
Note that the projectors P and Q are exact. Define by Kp:ImL→domL∩KerP by
Kpy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds=I0+αy(t),y∈ImL.
Hence we have
D0+α-1(Kpy)t=∫0ty(s)ds,D0+α-2(Kpy)t=∫0t(t-s)y(s)ds,
then
‖Kpy‖∞≤1Γ(α)‖y‖1,‖D0+α-1(Kpy)‖∞≤‖y‖1,‖D0+α-2(Kpy)‖∞≤‖y‖1,
and thus
‖Kpy‖Cα-1≤(1Γ(α)+2)‖y‖1.
In fact, if y∈ImL, then
(LKp)y(t)=D0+αI0+αy(t)=y(t).
Also, if u∈domL∩KerP, then
(KpL)u(t)=I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+c3tα-3,
where
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
(KpL)u(t)=u(t).
This shows that Kp=[L|domL∩KerP]-1. The proof is complete. Using (3.16), we write
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 Lemma 2.8 and a standard method, we obtain the following lemma.
Lemma 3.3 (see [16]).
For every given e∈L1[0,1],Kp(I-Q)N:Y→Y is completely continuous.
Assume that the following conditions on the function f(t,x,y,z) are satisfied.
There exist functions a(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:
There exists a constant A>0, such that for x∈domL∖KerL satisfying |D0+α-1x(t)|+|D0+α-2x(t)|>A for all t∈[0,1], we have
Q1Nx(t)≠0,orQ2Nx(t)≠0.
There exists a constant B>0 such that for every a,b∈R satisfying a2+b2>B then either
aT1N(atα-1+btα-2)+bT2N(atα-1+btα-2)<0,
or
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
Ω1={x∈domL∖KerL:Lx=λNx,λ∈[0,1]}
is bounded.
Proof.
Take
Ω1={x∈domL∖KerL: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
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
|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
‖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
‖(I-P)x‖Cα-1=‖KpL(I-P)x‖Cα-1≤(2+1Γ(α))‖Nx‖1,
so, we have
‖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
‖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
‖x‖∞≤‖x‖Cα-1,‖D0+α-1x‖∞≤‖x‖Cα-1,‖D0+α-2x‖∞≤‖x‖Cα-1,
so
‖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
‖D0+α-2x‖∞≤A1,‖D0+α-1x‖∞≤A2,‖x‖∞≤A3.
Therefore, for all x∈Ω1,
‖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
Ω2={x∈KerL:Nx∈ImL}
is bounded.
Proof.
Let
Ω2={x∈KerL:Nx∈ImL},
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
Ω3={x∈KerL:-λJx+(1-λ)QNx=0,λ∈[0,1]}
is bounded.
Proof.
We define the isomorphism J:KerL→ImQ by
J(atα-1+btα-2)=atα-1+btα-2.
If the first part of (H3) is satisfied, let
Ω3={x∈KerL:-λJx+(1-λ)QNx=0,λ∈[0,1]}.
For every x=atα-1+btα-2∈Ω3,
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)
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
Ω3={x∈KerL:λJx+(1-λ)QNx=0,λ∈[0,1]},
and, again, obtain a contradiction. Thus, in either case
‖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
Ω3′={x∈KerL: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.
Set Ω 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:
Lx≠λNx for every (x,λ)∈[domL∖KerL∩∂Ω]×[0,1];
Nx∉ImL, for every x∈KerL∩∂Ω.
Finally, we will prove that (iii) of Lemma 1.1 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,
deg(QN|KerL,KerL∩Ω,0)=deg(H(⋅,0),KerL∩Ω,0)=deg(H(⋅,1),KerL∩Ω,0)=deg(±I,KerL∩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: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 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 f(t,x,y,z)=140sinx+120y+120z+5cos(z)1/5,
then there is 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 ‖a‖1+‖b‖1+‖c‖1=1/8,1m=1(2/Γ(α))+(1/Γ(α-1))+5≈0.131,
therefore
‖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.
Acknowledgments
This work is sponsored by NNSF of China (10771212) and the Fundamental Research Funds for the Central Universities (2010LKSX09).
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