We discuss the existence of solution for a multipoint boundary value problem of fractional differential equation. An existence result is obtained with the use of the coincidence degree theory.

1. Introduction

In this paper, we study the multipoint boundary value problem D0+αu(t)=f(t,u(t),D0+α-1u(t),D0+α-2u(t))+e(t),0<t<1,I0+3-αu(0)=0,D0+α-2u(0)=∑j=1nβjD0+α-2u(ξj),u(1)=∑i=1mαiu(ηi),
where 2<α≤3,0<ξ1<ξ2<⋯<ξn<1,n≥1,0<η1<⋯<ηm<1,m≥2,αi,βj∈ℝ,∑i=1mαiηiα-1=∑i=1mαiηiα-2=1,∑j=1nβjξj=0,∑j=1nβj=1,f:[0,1]×ℝ3→ℝ satisfying the Carathéodory conditions, e∈L1[0,1]. D0+α and I0+α are the standard Riemann-Liouville derivative and integral, respectively. We assume, in addition, that R=Γ(α)2Γ(α-1)Γ(2α)Γ(α+1)∑j=1nβjξjα(1-∑i=1mαiηi2α-1)-Γ(α)2Γ(α-1)Γ(α+2)Γ(2α-1)∑j=1nβjξjα+1(1-∑i=1mαiηi2α-2)≠0,
where Γ is the Gamma function. Due to condition (1.3), the fractional differential operator in (1.1), (1.2) is not invertible.

Fractional differential equation can describe many phenomena in various fields of science and engineering. Many methods have been introduced for solving fractional differential equations, such as the popular Laplace transform method, the iteration method. For details, see [1, 2] and the references therein.

Recently, there are some papers dealing with the solvability of nonlinear boundary value problems of fractional differential equation, by use of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, etc.), see, for example, [3–6]. But there are few papers that consider the fractional-order boundary problems at resonance. Very recently [7], Y. H. Zhang and Z. B. Bai considered the existence of solutions for the fractional ordinary differential equationD0+αu(t)=f(t,u(t),D0+α-(n-1)u(t),…,D0+α-1u(t))+e(t),0<t<1,
subject to the following boundary value conditions: I0+n-αu(0)=D0+α-(n-1)u(0)=⋯=D0+α-2u(0)=0,u(1)=σu(η),
where n>2 is a natural number, n-1<α≤n is a real number, f:[0,1]×ℝn→ℝ is continuous, and e∈L1[0,1],σ∈(0,∞), and η∈(0,1) are given constants such that σηα-1=1. D0+α and I0+α are the standard Riemann-Liouville derivative and integral, respectively. By the conditions, the kernel of the linear operator is one dimensional.

Motivated by the above work and recent studies on fractional differential equations [8–18], in this paper, we consider the existence of solutions for multipoint boundary value problem (1.1), (1.2) at resonance. Note that under condition (1.3), the kernel of the linear operator in (1.1), (1.2) is two dimensional. Our method is based upon the coincidence degree theory of Mawhin [18].

Now, we will briefly recall some notation and abstract existence result.

Let Y,Z be real Banach spaces, let L:dom(L)⊂Y→Z be a Fredholm map of index zero, and let P:Y→Y,Q:Z→Z be continuous projectors such that Im(P)=Ker(P),Ker(Q)=Im(L), and Y=Ker(L)⊕Ker(P),Z=Im(L)⊕Im(Q). It follows that L|dom(L)∩Ker(P):dom(L)∩Ker(P)→Im(L) is invertible. We denote the inverse of the map by KP. If Ω is an open-bounded subset of Y such that dom(L)∩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 theorem that we used is Theorem 2.4 of [18].

Theorem 1.1.

Let L be a Fredholm operator of index zero and N be L-compact on Ω¯. Assume that the following conditions are satisfied:

Lx≠λNx for every (x,λ)∈[(dom(L)∖Ker(L))∩∂Ω]×(0,1),

Nx∉Im(L) for every x∈Ker(L)∩∂Ω,

deg(JQN|Ker(L),Ω∩Ker(L),0)≠0, where Q:Z→Z is a projection as above with Im(L)=Ker(Q), and J:Im(Q)→Ker(L) is any isomorphism,

then the equation Lx=Nx has at least one solution in dom(L)∩Ω¯.

The rest of this paper is organized as follows. In Section 2, we give some notation and Lemmas. In Section 3, we establish an existence theorem of a solution for the problem (1.1), (1.2).

2. Background Materials and Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory, and these definitions can be found in the recent literature [1, 2].

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,∞). And we let I0+0y(t)=y(t) for every continuous y:(0,∞)→ℝ.

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,∞).

Lemma 2.3 (see [<xref ref-type="bibr" rid="B3">3</xref>]).

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+C2tα-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 space C[0,1] with the norm ∥x∥∞=maxt∈[0,1]|x(t)|. Given μ>0 and N=[μ]+1, one can define a linear space Cμ[0,1]:={u∣u(t)=I0+μx(t)+c1tμ-1+c2tμ-2+⋯+cN-1tμ-(N-1),t∈[0,1]},
where x∈C[0,1] and ci∈ℝ,i=1,2,…,N-1. By means of the linear function analysis theory, one can prove that with the norm ∥u∥Cμ=∥D0+μu∥∞+⋯+∥D0+μ-(N-1)u∥∞+∥u∥∞,Cμ[0,1] is a Banach space.

Lemma 2.4 (see [<xref ref-type="bibr" rid="B7">7</xref>]).

F⊂Cμ[0,1] is a sequentially compact set if and only if F is uniformly bounded and equicontinuous. Here, uniformly bounded means that 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 forallε>0,∃δ>0 such that
|u(t1)-u(t2)|<ε,(∀t1,t2∈[0,1],|t1-t2|<δ,∀u∈F),|D0+α-iu(t1)-D0+α-iu(t2)|<ε,(t1,t2∈[0,1],|t1-t2|<δ,∀u∈F,∀i∈{0,…,N-1}).

Let Z=L1[0,1] with the norm ∥g∥1=∫01|g(s)|ds. Y=Cα-1[0,1]={u∣u(t)=I0+α-1x(t)+ctα-2,t∈[0,1]}, where x∈C[0,1],c∈ℝ, with the norm ∥u∥Cα-1=∥D0+α-1u∥∞+∥D0+α-2u∥∞+∥u∥∞, and Y is a Banach space.

Definition 2.5.

By a solution of the boundary value problem (1.1), (1.2), we understand a function u∈Cα-1[0,1] such that D0+α-1u is absolutely continuous on (0,1) and satisfies (1.1), (1.2).

Definition 2.6.

We say that the map f:[0,1]×ℝ→ℝ satisfies the Carathéodory conditions with respect to L1[0,1] if the following conditions are satisfied:

for each z∈ℝ, the mapping t→f(t,z) is Lebesgue measurable,

for almost every t∈[0,1], the mapping z→f(t,z) is continuous on ℝ,

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).

Define L to be the linear operator from dom(L)∩Y to Z with dom(L)={u∈Cα-1[0,1]∣D0+αu∈L1[0,1],usatisfies(1.2)},Lu=D0+αu,u∈dom(L).
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 (1.1), (1.2) can be written as Lu=Nu.

Lemma 2.7.

Let condition (1.3) and (1.4) hold, then L:dom(L)∩Y→Z is a Fredholm map of index zero.

Proof.

It is clear that Ker(L)={atα-1+btα-2∣a,b∈ℝ}≅ℝ2.

Let g∈Z and
u(t)=1Γ(α)∫0t(t-s)α-1g(s)ds+c1tα-1+c2tα-2,
then D0+αu(t)=g(t)a.e.,t∈(0,1) and, if
∫01(1-s)α-1g(s)ds-∑i=1mαi∫0ηi(ηi-s)α-1g(s)ds=0,∑j=1nβj∫0ξj(ξj-s)g(s)ds=0
hold. Then u(t) satisfies the boundary conditions (1.2), that is, u∈dom(L), and we have
{g∈Z∣gsatisfies(2.11)}⊆Im(L).
Let u∈dom(L), then for D0+αu∈Im(L), we have
u(t)=I0+αD0+αu(t)+c1tα-1+c2tα-2+c3tα-3,
which, due to the boundary value condition (1.2), implies that D0+αu satisfies (2.11). In fact, from I0+3-αu(0)=0, we have c3=0, from u(1)=∑i=1mαiu(ηi), we have
∫01(1-s)α-1D0+αu(s)ds-∑i=1mαi∫0ηi(ηi-s)α-1D0+αu(s)ds=0,
and from D0+α-2u(0)=∑j=1nβjD0+α-2u(ξj), we have
∑j=1nβj∫0ξj(ξj-s)D0+αu(s)ds=0.
Hence,
Im(L)⊆{g∈Z∣gsatisfies(2.11)}.
Therefore,
Im(L)={g∈Z∣gsatisfies(2.11)}.
Consider the continuous linear mapping Q1:Z→Z and Q2:Z→Z defined by
Q1g=∫01(1-s)α-1g(s)ds-∑i=1mαi∫0ηi(ηi-s)α-1g(s)ds,Q2g=∑j=1nβj∫0ξj(ξj-s)g(s)ds.
Using the above definitions, we construct the following auxiliary maps R1,R2:Z→Z:
R1g=1R[Γ(α-1)Γ(α+1)∑j=1nβjξjαQ1g(t)-Γ(α)Γ(α-1)Γ(2α-1)(1-∑i=1mαiηi2α-2)Q2g(t)],R2g=-1R[Γ(α)Γ(α+2)∑j=1nβjξjα+1Q1g(t)-(Γ(α))2Γ(2α)(1-∑i=1mαiηi2α-1)Q2g(t)].
Since the condition (1.4) holds, the mapping Q:Z→Z defined by
(Qy)(t)=(R1g(t))tα-1+(R2g(t))tα-2
is well defined.

Recall (1.4) and note that
R1(R1gtα-1)=1R[Γ(α-1)Γ(α+1)∑j=1nβjξjαQ1(R1gtα-1)-Γ(α)Γ(α-1)Γ(2α-1)(1-∑i=1mαiηi2α-2)Q2(R1gtα-1)Γ(α-1)Γ(α+1)∑j=1nβjξjαQ1(R1gtα-1)]=R1g1R[Γ(α-1)Γ(α2)Γ(α+1)Γ(2α)∑j=1nβjξjα(1-∑i=1mαiηi2α-1)-Γ(α-1)Γ(α2)Γ(2α-1)Γ(α+2)(1-∑i=1mαiηi2α-2)∑j=1nβjξjα+1]=R1g,
and similarly we can derive that
R1(R2gtα-2)=0,R2(R1gtα-1)=0,R2(R2gtα-2)=R2g.
So, for g∈Z, it follows from the four relations above that
Q2g=R1(R1gtα-1+R2gtα-2)tα-1+R2(R1gtα-1+R2gtα-2)tα-2=R1(R1gtα-1)tα-1+R1(R2gtα-2)tα-1+R2(R1gtα-1)tα-2+R2(R2gtα-2)tα-2=R1gtα-1+R2gtα-2=Qg,
that is, the map Q is idempotent. In fact, Q is a continuous linear projector.

Note that g∈Im(L) implies Qg=0. Conversely, if Qg=0, then we must have R1g=R2g=0; since the condition (1.4) holds, this can only be the case if Q1g=Q2g=0, that is, g∈Im(L). In fact, Im(L)=Ker(Q).

Take g∈Z in the form g=(g-Qg)+Qg, so that g-Qg∈Im(L)=Ker(Q) and Qg∈Im(Q). Thus, Z=Im(L)+Im(Q). Let g∈Im(L)∩Im(Q) and assume that g(s)=asα-1+bsα-2 is not identically zero on [0,1], then, since g∈Im(L), from (2.11) and the condition (1.4), we derive a=b=0, which is a contradiction. Hence, Im(L)∩Im(Q)={0}; thus, Z=Im(L)⊕Im(Q).

Now, dimKer(L)=2=codimIm(L), and so L is a Fredholm operator of index zero.

Let P:Y→Y be defined by Pu(t)=1Γ(α)D0+α-1u(0)tα-1+1Γ(α-1)D0+α-2u(0)tα-2,t∈[0,1].
Note that P is a continuous linear projector and Ker(P)={u∈Y∣D0+α-1u(0)=D0+α-2u(0)=0}.
It is clear that Y=Ker(L)⊕Ker(P).

Note that the projectors P and Q are exact. Define KP:Im(L)→dom(L)∩Ker(P) by KPg(t)=1Γ(α)∫0t(t-s)α-1g(s)ds=I0+αg(t).
Hence, we have D0+α-1(KPg)(t)=∫0tg(s)ds,D0+α-2(KPg)(t)=∫0t(t-s)g(s)ds,
then ∥KPg∥∞≤(1/Γ(α))∥g∥1,∥D0+α-1(KPg)∥∞≤∥g∥1,∥D0+α-2(KPg)∥∞≤∥g∥1, and thus ‖KPg‖Cα-1≤(2+1Γ(α))‖g‖1.
In fact, if g∈Im(L), then (LKP)g=D0+αI0+αg=g. Also, if u∈dom(L)∩Ker(P), then (KPLg)(t)=I0+αD0+αg(t)=g(t)+c1tα-1+c2tα-2+c3tα-3,
from boundary value condition (1.2) and the fact that u∈dom(L)∩Ker(P), we have c1=c2=c3=0. Thus, KP=(L|dom(L)∩Ker(P))-1.
Using (2.19), we write QNu(t)=(R1Nu)tα-1+(R2Nu)tα-2,KP(I-Q)Nu(t)=1Γ(α)∫01(t-s)α-1[Nu(s)-QNu(s)]ds.
With arguments similar to those of [7], we obtain the following Lemma.

Lemma 2.8.

KP(I-Q)N:Y→Y is completely continuous.

3. The Main Results

Assume that the following conditions on the function f(t,x,y,z) are satisfied:

(H1) there exists a constant A>0, such that for u∈dom(L)∖Ker(L) satisfying |D0+α-1u(t)|+|D0+α-2u(t)|>A for all t∈[0,1], we have Q1Nu(t)≠0orQ2Nu(t)≠0,

(H2) there exist functions a,b,c,d,r∈L1[0,1] and a constant θ∈[0,1] such that for all (x,y,z)∈ℝ3 and a.e., t∈[0,1], one of the following inequalities is satisfied:|f(t,x,y,z)|≤a(t)|x|+b(t)|y|+c(t)|z|+d(t)|z|θ+r(t),|f(t,x,y,z)|≤a(t)|x|+b(t)|y|+c(t)|z|+d(t)|y|θ+r(t),|f(t,x,y,z)|≤a(t)|x|+b(t)|y|+c(t)|z|+d(t)|x|θ+r(t),

(H3) there exists a constant B>0 such that for every a,b∈ℝ satisfying a2+b2>B, then either
aR1N(atα-1+btα-2)+bR2N(atα-1+btα-2)<0,
or else
aR1N(atα-1+btα-2)+bR2N(atα-1+btα-2)>0.

Remark 3.1.

R1N(atα-1+btα-2) and R2N(atα-1+btα-2) from (H3) stand for the images of u(t)=atα-1+btα-2 under the maps R1N and R2N, respectively.

Theorem 3.2.

If (H1)–(H3) hold, then boundary value problem (1.1)-(1.2) has at least one solution provided that
‖a‖1+‖b‖1+‖c‖1<1τ,
where τ=5+2/Γ(α)+1/Γ(α-1).

Proof.

Set
Ω1={u∈dom(L)∖Ker(L)∣Lu=λNuforsomeλ∈[0,1]},
then for u∈Ω1,Lu=λNu; thus, λ≠0,Nu∈Im(L)=Ker(Q), and hence QNu(t)=0 for all t∈[0,1]. By the definition of Q, we have Q1Nu(t)=Q2Nu(t)=0. It follows from (H1) that there exists t0∈[0,1] such that |D0+α-1u(t0)|+|D0+α-2u(t0)|≤A. Now,
D0+α-1u(t)=D0+α-1u(t0)+∫t0tD0+αu(s)ds,D0+α-2u(t)=D0+α-2u(t0)+∫t0tD0+α-1u(s)ds,
so
|D0+α-1u(0)|≤‖D0+α-1u(t)‖∞≤|D0+α-1u(t0)|+‖D0+αu‖1≤A+‖Lu‖1≤A+‖Nu‖1,|D0+α-2u(0)|≤‖D0+α-2u(t)‖∞≤|D0+α-2u(t0)|+‖D0+α-1u‖∞≤|D0+α-2u(t0)|+|D0+α-1u(t0)|+‖D0+αu‖1≤A+‖Lu‖1≤A+‖Nu‖1.
Now by (3.8), we have
‖Pu‖Cα-1=‖1Γ(α)D0+α-1u(0)tα-1+1Γ(α-1)D0+α-2u(0)tα-2‖Cα-1=‖1Γ(α)D0α-1u(0)tα-1+1Γ(α-1)D0α-2u(0)tα-2‖∞+‖D0+α-1u(0)‖∞+‖D0+α-1u(0)t+D0+α-2u(0)‖∞≤(2+1Γ(α))|D0+α-1u(0)|+(1+1Γ(α-1))|D0+α-2u(0)|≤(2+1Γ(α))(A+‖Nu‖1)+(1+1Γ(α-1))(A+‖Nu‖1).
Note that (I-P)u∈Im(KP)=dom(L)∩Ker(P) for u∈Ω1, then, by (2.28) and (2.30),
‖(I-P)u‖Cα-1=‖KPL(I-P)‖Cα-1≤(2-1Γ(α))‖L(I-P)u‖1=(2-1Γ(α))‖Lu‖1≤(2-1Γ(α))‖Nu‖1.
Using (3.9) and (3.10), we obtain
‖u‖Cα-1=‖Pu+(I-P)u‖Cα-1≤‖Pu‖Cα-1+‖(I-P)u‖Cα-1≤(2+1Γ(α))(A+‖Nu‖1)+(1+1Γ(α-1))(A+‖Nu‖1)+(2+1Γ(α))‖Nu‖1=(5+2Γ(α)+1Γ(α-1))‖Nu‖1+(3+1Γ(α)+1Γ(α-1))A=τ∥Nu∥1+C1,
where C1=(3+1/Γ(α)+1/Γ(α-1))A is a constant. This is for all u∈Ω1,
‖u‖Cα-1≤τ‖Nu‖1+C1.
If the first condition of (H2) is satisfied, then we have
max{‖u‖∞,‖D0+α-1u‖∞,‖D0+α-2u‖∞}≤‖u‖Cα-1≤τ(‖D0+α-2u‖∞θ‖a‖1‖u‖∞+‖b‖1‖D0+α-1u‖∞+‖c‖1‖D0+α-2u‖∞+‖d‖1‖D0+α-2u‖∞θ+‖r‖1+‖e‖1)+C1,
and consequently,
‖u‖∞≤τ1-‖a‖1τ(‖b‖1‖D0+α-1u‖∞+‖c‖1‖D0+α-2u‖∞+‖d‖1‖D0+α-2u‖∞θ+‖r‖1+‖e‖1)+C11-‖a‖1τ,‖D0+α-1u‖∞≤τ1-‖a‖1τ-∥b∥1τ(‖c‖1‖D0+α-2u‖∞+‖d‖1‖D0+α-2u‖∞θ+‖r‖1+‖e‖1‖D0+α-2u‖∞θ)+C11-‖a‖1τ-‖b‖1τ,‖D0+α-1u‖∞≤τ‖d‖1‖D0+α-2u‖∞θ1-‖a‖1τ-∥b∥1τ-‖c‖1τ+τ(‖r‖1+‖e‖1)+C11-‖a‖1τ-∥b∥1τ-‖c‖1τ.
Note that θ∈[0,1) and ∥a∥1+∥b∥1+∥c∥1<1/τ, so there exists M1>0 such that ∥D0+α-1u∥∞≤M1 for all u∈Ω1. The inequalities (3.14) and (3.15) show that there exist M2,M3>0 such that ∥D0+α-1u∥∞≤M2,∥u∥∞≤M3 for all u∈Ω1. Therefore, for all u∈Ω1,∥u∥Cα-1=∥u∥∞+∥D0+α-1u∥∞+∥D0+α-2u∥∞≤M1+M2+M3, that is, Ω1 is bounded given the first condition of (H2). If the other conditions of (H2) hold, by using an argument similar to the above, we can prove that Ω1 is also bounded.

Let
Ω2={u∈Ker(L)∣Nu∈Im(L)}.
For u∈Ω2,u∈Ker(L)={u∈dom(L)∣u=atα-1+btα-2,a,b∈ℝ,t∈[0,1]}, and QN(atα-1+btα-2)=0; thus, R1N(atα-1+btα-2)=R2N(atα-1+btα-2)=0. By (H3), a2+b2≤B, that is, Ω2 is bounded.

We define the isomorphism J:Im(Q)→Ker(L) by
J(atα-1+btα-2)=atα-1+btα-2,a,b∈R.

If the first part of (H3) is satisfied, let
Ω3={u∈KerL:-λJ-1u+(1-λ)QNu=0,λ∈[0,1]}.
For every atα-1+btα-2∈Ω3,
λ(atα-1+btα-2)=(1-λ)[(R1N(atα-1+btα-2))tα-1+(R2N(atα-1+btα-2))tα-2].
If λ=1, then a=b=0, and if a2+b2>B, then by (H3),
λ(a2+b2)=(1-λ)[aR1N(atα-1+btα-2)+bR2N(atα-1+btα-2)]<0,
which, in either case, obtain a contradiction. If the other part of (H3) is satisfied, then we take
Ω3={u∈KerL:λJ-1u+(1-λ)QNu=0,λ∈[0,1]},
and, again, obtain a contradiction. Thus, in either case,
‖u‖Cα-1=‖u‖∞+‖D0+α-1u‖∞+‖D0+α-2u‖∞=‖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))|a|,
for all u∈Ω3, that is, Ω3 is bounded.

In the following, we will prove that all the conditions of Theorem 1.1 are satisfied. Set Ω to be a bounded open set of Y such that Ui=13Ωi¯⊂Ω. by Lemma 2.8, the operator KP(I-Q)N:Ω¯→Y is compact; thus, N is L-compact on Ω¯, then by the above argument, we have

Lu≠λNx, for every (u,λ)∈[(dom(L)∖KerL)∩∂Ω]×(0,1),

Nu∉Im(L), for every u∈Ker(L)∩∂Ω.

Finally, we will prove that (iii) of Theorem 1.1 is satisfied. Let H(u,λ)=±Iu+(1-λ)JQNu, where I is the identity operator in the Banach space Y. According to the above argument, we know that
H(u,λ)≠0,∀u∈∂Ω∩Ker(L),
and thus, by the homotopy property of degree,
deg(JQN|Ker(L),Ω∩Ker(L),0)=deg(H(…,0),Ω∩Ker(L),0)=deg(H(…,1),Ω∩Ker(L),0)=deg(±I,Ω∩Ker(L),0)=±1≠0,
then by Theorem 1.1, Lu=Nu has at least one solution in dom(L)∩Ω¯, so boundary problem (1.1), (1.2) has at least one solution in the space Cα-1[0,1]. The proof is finished. Acknowledgments

This work is supported by the NSFC (no. 11061030, no. 11026060), the Fundamental Research Funds for the Gansu Universities, the nwnu-kjcxgc-03-69, nwnu-kjcxgc-03-61.

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