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Existence results for a functional boundary value problem of fractional differential equations

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Abstract

In this paper, a functional boundary value problem of fractional differential equations is studied. Based on Mawhin’s coincidence degree theory, some existence theorems are obtained in the case of nonresonance and the cases of dimKerL=1 and dimKerL=2 at resonance.

1 Introduction

The subject of fractional calculus has gained considerable popularity and importance because of its frequent appearance in various fields such as physics, chemistry, and engineering. In consequence, the subject of fractional differential equations has attracted much attention. Many methods have been introduced to solve fractional differential equations, such as the popular Laplace transform method, the iteration method, the Fourier transform method and the operational method. For details, see [13] and the references therein. Recently, there have been some papers dealing with the basic theory for initial value problems of nonlinear fractional differential equations; for example, see [4, 5]. Also, there are some articles which deal with the existence and multiplicity of solutions for nonlinear boundary value problems of fractional order differential equations using techniques of topological degree theory. We refer the reader to [616] for some recent results at nonresonance and to [1726] at resonance.

In [18], by making use of the coincidence degree theory of Mawhin, Zhang and Bai discussed the existence results for the following nonlinear nonlocal problem at resonance under the case dimKerL=1:

D 0 + α u(t)=f ( t , u ( t ) ) ,0<t<1,u(0)=0,βu(η)=u(1),1<α2.

Recently, Jiang [26] studied the existence of a solution for the following fractional differential equation at resonance under the case dimKerL=2:

D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) ) , u ( 0 ) = 0 , D 0 + α 1 u ( 0 ) = i = 1 m a i D 0 + α 1 u ( ξ i ) , D 0 + α 2 u ( 1 ) = j = 1 n b j D 0 + α 2 u ( η j ) .

Being directly inspired by [18, 20, 26], we intend in this paper to study the following functional boundary value problems (FBVP) of fractional order differential equation:

D 0 + α u(t)=f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) ,
(1.1)
I 0 + 3 α u(t) | t = 0 =0, Φ 1 [ D 0 + α 1 u ( t ) ] =0, Φ 2 [ D 0 + α 2 u ( t ) ] =0,
(1.2)

where 2<α<3, D 0 + α and I 0 + α are the standard Riemann-Liouville differentiation and integration, and fC([0,1]× R 3 ,R); Φ 1 , Φ 2 :C[0,1]R are continuous linear functionals.

In this paper, we shall give some sufficient conditions to construct the existence theorems for FBVP (1.1), (1.2) at nonresonance and resonance (both cases of dimKerL=1 and dimKerL=2), respectively. To the best of our knowledge, the method of Mawhin’s theorem has not been developed for fractional order differential equation with functional boundary value problems at resonance. So, it is interesting and important to discuss the existence of a solution for FBVP (1.1), (1.2). Many difficulties occur when we deal with them. For example, the construction of the generalized inverse K p :ImLdomLKerP of L. So, we need to introduce some new tools and methods to investigate the existence of a solution for FBVP (1.1), (1.2).

The rest of this paper is organized as follows. In Section 2, we give some notations and lemmas. In Section 3, we establish the existence results of a solution for functional boundary value problem (1.1), (1.2).

2 Preliminaries and lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions and properties can be found in the literature. The readers who are unfamiliar with this area can consult, for example, [1, 2, 4] for details.

Definition 2.1 [1, 2]

The Riemann-Liouville fractional integral of order α>0 of a function u:(0,)R is given by

I 0 + α u(t)= 1 Γ ( α ) 0 t ( t s ) α 1 u(s)ds,

provided that the right-hand side is pointwise defined on (0,). Here Γ(α) is the Gamma function given by Γ(α)= 0 + t α 1 e t dt.

Definition 2.2 [1, 2]

The Riemann-Liouville fractional derivative of order α>0 of a continuous function u:(0,)R is given by

D 0 + α u(t)= 1 Γ ( n α ) ( d d t ) n 0 t u ( s ) ( t s ) α n + 1 ds,

where n1α<n, provided that the right-hand side is pointwise defined on (0,).

We use the classical spaces C[0,1] with the norm u = max t [ 0 , 1 ] |u(t)|, L 1 [0,1] with the norm u 1 = 0 1 |u(t)|dt. We also use the space A C n [0,1] defined by

A C n [0,1]= { u : [ 0 , 1 ] R u ( n 1 )  are absolutely continuous on  [ 0 , 1 ] }

and the Banach space C μ [0,1] (μ>0)

C μ [ 0 , 1 ] = { u ( t ) u ( t ) = I 0 + μ x ( t ) + c 1 t μ 1 + c 2 t μ 2 + + c N 1 t μ ( N 1 ) , x C [ 0 , 1 ] , t [ 0 , 1 ] , c i R , i = 1 , 2 , , N = [ μ ] + 1 }

with the norm u C μ = D 0 + μ u ++ D 0 + μ ( N 1 ) u + u .

Lemma 2.1 [2]

Let α>0, n=[α]+1. Assume that u L 1 (0,1) with a fractional integration of order nα that belongs to A C n [0,1]. Then the equality

( I 0 + α D 0 + α u ) (t)=u(t) i = 1 n ( ( I 0 + n α u ) ( t ) ) ( n i ) | t = 0 Γ ( α i + 1 ) t α i

holds almost everywhere on [0,1].

Remark 2.1 If u satisfies D 0 + α u=f(t) L 1 (0,1) and I 0 + 3 α u | t = 0 =0, then u C α 1 [0,1]. In fact, with Lemma 2.1, one has

u(t)= I 0 + α f(t)+ c 1 t α 1 + c 2 t α 2 + c 3 t α 3 .

Combine with I 0 + 3 α u | t = 0 =0, there is c 3 =0. So,

u(t)= I 0 + α f(t)+ c 1 t α 1 + c 2 t α 2 = I 0 + α 1 [ I 0 + 1 f ( t ) + c 1 Γ ( α ) ] + c 2 t ( α 1 ) 1 .

In the following lemma, we use the unified notation both for fractional integrals and fractional derivatives assuming that I 0 + α = D 0 + α for α<0.

Lemma 2.2 [2]

Assume α>0, then:

  1. (i)

    Let kN. If D a + α u(t) and ( D a + α + k u)(t) exist, then

    ( D k D a + α ) u(t)= ( D a + α + k u ) (t);
  2. (ii)

    If β>0, α+β>1, then

    ( I a + α I a + β ) u(t)= ( I a + α + β u ) (t)

is satisfied at any point on [a,b] for u L p (a,b) and 1p+;

  1. (iii)

    Let uC[a,b]. Then ( D a + α I a + α )u(t)=u(t) holds on [a,b];

  2. (iv)

    Note that for λ>1, λα1,α2,,αn, we have

    D α t λ = Γ ( λ + 1 ) Γ ( λ α + 1 ) t λ α , D α t α i =0,i=1,2,,n.

Lemma 2.3 [18]

F C μ [0,1] is a sequentially compact set if and only if F is uniformly bounded and equicontinuous. Here ‘F is uniformly bounded and equicontinuous’ means that there exists M>0 such that for every uF,

u C μ = D 0 + μ u ++ D 0 + μ [ μ ] u + u <M

and that ϵ>0, N>0, for all t 1 , t 2 [0,1], | t 1 t 2 |<δ, uF, i{0,1,,[μ]}, there hold

| u ( t 1 ) u ( t 2 ) | <ε, | D 0 + μ i u ( t 1 ) D 0 + μ i u ( t 2 ) | <ε,

respectively.

Next, consider the following conditions:

(A1) Φ 1 [1] Φ 2 [1]0.

(A2) Φ 1 [1]=0, Φ 2 [1]0, Φ 2 [t]=0.

(A3) Φ 1 [1]=0, Φ 2 [1]=0, Φ 2 [t]0.

(A4) Φ 1 [1]0, Φ 2 [1]=0, Φ 2 [t]=0.

(A5) Φ 1 [1]=0, Φ 2 [1]=0, Φ 2 [t]=0.

We shall prove that: If (A1) holds, then KerL={θ}. It is the so-called nonresonance case. If (A2) holds, then KerL={a t α 1 :aR}. If (A3) or (A4) holds, then KerL={a t α 2 :aR}. If (A5) holds, then KerL={a t α 1 +b t α 2 :a,bR}.

In the nonresonance case, FBVP (1.1), (1.2) can be transformed into an operator equation.

Lemma 2.4 Assume that (A1) holds. Then functional boundary value problem (1.1) and (1.2) has a solution if and only if the operator T: C α 1 [0,1] C α 1 [0,1], defined by

( T u ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 ( f u ) ( s ) d s Φ 1 [ 0 t ( f u ) ( s ) d s ] Γ ( α ) Φ 1 [ 1 ] t α 1 Φ 1 [ 1 ] Φ 2 [ 0 t ( t s ) ( f u ) ( s ) d s ] Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 2 [ t ] Γ ( α 1 ) Φ 1 [ 1 ] Φ 2 [ 1 ] t α 2 ,

has a fixed point, where (fu)(t)=f(t,u(t), D 0 + α 1 u(t), D 0 + α 2 u(t)).

Proof If u is a solution to Tu=u, by Lemma 2.2, we get

D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) , D 0 + α 1 u ( t ) = 0 t ( f u ) ( s ) d s Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 1 [ 1 ]

and

D 0 + α 2 u ( t ) = 0 t ( t s ) ( f u ) ( s ) d s Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 1 [ 1 ] t Φ 1 [ 1 ] Φ 2 [ 0 t ( t s ) ( f u ) ( s ) d s ] Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 2 [ t ] Φ 1 [ 1 ] Φ 2 [ 1 ] .

Considering the linearity of Φ i (i=1,2), we have

I 0 + 3 α u ( t ) | t = 0 = 0 , Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 1 [ 1 ] Φ 1 [ 1 ] = 0 , Φ 2 [ D 0 + α 2 u ( t ) ] = Φ 2 [ 0 t ( t s ) ( f u ) ( s ) d s ] Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 1 [ 1 ] Φ 2 [ t ] Φ 1 [ 1 ] Φ 2 [ 0 t ( t s ) ( f u ) ( s ) d s ] Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 2 [ t ] Φ 1 [ 1 ] Φ 2 [ 1 ] Φ 2 [ 1 ] = 0 .

So, u is a solution to FBVP (1.1), (1.2).

If u is a solution to (1.1), by Lemma 2.1, we can reduce (1.1) to an equivalent integral equation

u(t)= I 0 + α (fu)(t)+ c 1 t α 1 + c 2 t α 2 + c 3 t α 3 .
(2.1)

By I 0 + 3 α u(t) | t = 0 =0, there is c 3 =0, and

D 0 + α 1 u(t)= 0 t (fu)(s)ds+ c 1 Γ(α),
(2.2)
D 0 + α 2 u(t)= 0 t (ts)(fu)(s)ds+ c 1 Γ(α)t+ c 2 Γ(α1).
(2.3)

Applying Φ 1 and Φ 2 to (2.2) and (2.3), respectively, we obtain

0 = Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t ( f u ) ( s ) d s ] + c 1 Γ ( α ) Φ 1 [ 1 ] , 0 = Φ 2 [ D 0 + α 2 u ( t ) ] = Φ 2 [ 0 t ( t s ) ( f u ) ( s ) d s ] + c 1 Γ ( α ) Φ 2 [ t ] + c 2 Γ ( α 1 ) Φ 2 [ 1 ] .

Thus,

c 1 = Φ 1 [ 0 t ( f u ) ( s ) d s ] Γ ( α ) Φ 1 [ 1 ] ,
(2.4)
c 2 = Φ 1 [ 1 ] Φ 2 [ 0 t ( t s ) ( f u ) ( s ) d s ] Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 2 [ t ] Γ ( α 1 ) Φ 1 [ 1 ] Φ 2 [ 1 ] .
(2.5)

Substituting (2.4) and (2.5) into (2.1), we obtain

u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 ( f u ) ( s ) d s Φ 1 [ 0 t ( f u ) ( s ) d s ] Γ ( α ) Φ 1 [ 1 ] t α 1 Φ 1 [ 1 ] Φ 2 [ 0 t ( t s ) ( f u ) ( s ) d s ] Φ 1 [ 0 t ( f u ) ( s ) d s ] Φ 2 [ t ] Γ ( α 1 ) Φ 1 [ 1 ] Φ 2 [ 1 ] t α 2 .

The proof is complete. □

The following definitions and lemmas are a preparation for the existence of solutions to (1.1), (1.2) at resonance.

Definition 2.3 Let Y, Z be real Banach spaces, let L:domLYZ be a linear operator. L is said to be a Fredholm operator of index zero provided that:

  1. (i)

    ImL is a closed subset of Z,

  2. (ii)

    dimKerL=codimImL<+.

Let Y, Z be real Banach spaces and L:domLYZ be a Fredholm operator of index zero. P:YY, Q:ZZ are continuous projectors such that ImP=KerL, KerQ=ImL, Y=KerLKerP and Z=ImLImQ. It follows that L | dom L Ker P :domLKerPImL is invertible. We denote the inverse of the mapping by K P (generalized inverse operator of L). If Ω is an open bounded subset of Y such that domLΩ, the mapping N:YZ will be called L-compact on Ω ¯ , if QN( Ω ¯ ) is bounded and K P (IQ)N: Ω ¯ Y is compact.

We need the following known result for the sequel (Theorem 2.4 [27]).

Theorem 2.1 Let L be a Fredholm operator of index zero, and let N be L-compact on Ω ¯ . Assume that the following conditions are satisfied:

  1. (i)

    LxλNx for every (x,λ)[(domLKerL)Ω]×(0,1).

  2. (ii)

    NxImL for every xKerLΩ.

  3. (iii)

    deg(QN | Ker L ,KerLΩ,0)0, where Q:ZZ is a projector as above with ImL=KerQ.

Then the equation Lx=Nx has at least one solution in domL Ω ¯ .

Let Y= C α 1 [0,1], Z= L 1 [0,1]. Let the linear operator L:YdomLZ with

dom L = { u C α 1 [ 0 , 1 ] : D 0 + α u ( t ) Z , I 0 + 3 α u ( t ) | t = 0 = 0 , Φ 1 [ D 0 + α 1 u ( t ) ] = 0 , Φ 2 [ D 0 + α 2 u ( t ) ] = 0 }

be defined by Lu= D 0 + α u(t). Let the nonlinear operator N:YZ be defined by

(Nu)(t)=f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) .

Then (1.1), (1.2) can be written as

Lu=Nu.

Now, we give KerL, ImL and some necessary operators at dimKerL=1 and dimKerL=2, respectively.

Lemma 2.5 Let L be the linear operator defined as above. If (A2) holds, then

KerL= { u dom L : u = a t α 1 , a R , t [ 0 , 1 ] }

and

ImL= { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } .

Proof Let u(t)=a t α 1 . Clearly, D 0 + α u(t)=0 and I 0 + 3 α u(t) | t = 0 =0. Considering (A2), Φ 1 [ D 0 + α 1 u(t)]=a Φ 1 [Γ(α)]=aΓ(α) Φ 1 [1]=0 and Φ 2 [ D 0 + α 2 u(t)]=Γ(α) Φ 2 [t]=0. So,

{ u dom L : u = a t α 1 , a R , t [ 0 , 1 ] } KerL.

If Lu= D 0 + α u(t)=0, then u(t)=a t α 1 +b t α 2 +c t α 3 . Considering I 0 + 3 α u(t) | t = 0 =0 and (A2), we can obtain that b=c=0. It yields u(t)=a t α 1 and KerL{udomL:u=a t α 1 ,aR,t[0,1]}.

We now show that

ImL= { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } .

If vImL, then there exists udomL such that D 0 + α u(t)=v(t). Hence,

u(t)= 1 Γ ( α ) 0 t ( t s ) α 1 v(s)ds+a t α 1 +b t α 2

for some a,bR. It yields

Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] +aΓ(α) Φ 1 [1]= Φ 1 [ 0 t v ( s ) d s ] =0.

Therefore

ImL { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } .

On the other hand, suppose vZ satisfies

Φ 1 [ 0 t v ( s ) d s ] =0.

Let

u(t)= 1 Γ ( α ) 0 t ( t s ) α 1 v(s)ds t α 2 Γ ( α 1 ) Φ 2 [ 1 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] .

Obviously, D 0 + α u(t)=v(t) and I 0 + 3 α u(t) | t = 0 =0. Considering (A2) and the linearity of Φ i (i=1,2), we have

Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] =0

and

Φ 2 [ D 0 + α 2 u ( t ) ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] Φ 2 [ 1 Φ 2 [ 1 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] ] =0.

It yields

{ v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } ImL.

The proof is complete. □

Lemma 2.6 If Φ 1 [t]0, then:

  1. (i)

    L is a Fredholm operator of index zero and dimKerL=codimImL=1.

  2. (ii)

    The linear operator K p :ImLdomLKerP can be defined by

    ( K p v)(t)= I 0 + α v(t) t α 2 Γ ( α 1 ) Φ 2 [ 1 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] .
  3. (iii)

    K p v C α 1 Δ 1 v 1 , where Δ 1 =2+ 1 Γ ( α ) + ( 1 + Γ ( α 1 ) ) Φ 2 Γ ( α 1 ) | Φ 2 [ 1 ] | and Φ 2 is the norm of a continuous linear functional Φ 2 .

  4. (iv)

    The linear operator K p :ImLdomLKerP C α 1 [0,1] is completely continuous.

Proof Firstly, we construct the mapping Q:ZZ defined by

Qy= 1 Φ 1 [ t ] Φ 1 [ 0 t y ( s ) d s ] .
(2.6)

Noting that

Q 2 y= 1 Φ 1 [ t ] Φ 1 [ 0 t ( Q y ) d s ] = 1 Φ 1 [ t ] Φ 1 [ 0 t d s ] (Qy)=Qy,

we get Q:ZZ is a well-defined projector.

Now, it is obvious that ImL=KerQ. Noting that Q is a linear projector, we have Z=ImQKerQ. Hence, Z=ImQImL and dimKerL=codimImL=1. This means L is a Fredholm mapping of index zero. Taking P:YY as

(Pu)(t)= D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 ,

then the generalized inverse K p :ImLdomLKerP of L can be rewritten

( K p v)(t)= I 0 + α v(t) t α 2 Γ ( α 1 ) Φ 2 [ 1 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] .

In fact, for vImL, we have

I 0 + 3 α ( K p v ) ( t ) | t = 0 = 0 , Φ 1 [ D 0 + α 1 ( K p v ) ( t ) ] = Φ 1 [ D 0 + α 1 I 0 + α v ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] = 0

and

Φ 2 [ D 0 + α 2 ( K p v ) ( t ) ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] 1 Φ 2 [ 1 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] Φ 2 [1]=0,

which implies that K p is well defined on ImL. Moreover, for vImL, we have

(L K p )v(t)= D 0 + α I 0 + α v(t)=v(t)

and for vdomLKerP, we know

I 0 + α D 0 + α v(t)=v(t) D 0 + α 1 v ( t ) | t = 0 Γ ( α ) t α 1 D 0 + α 2 v ( t ) | t = 0 Γ ( α 1 ) t α 2 I 0 + 3 α v ( t ) | t = 0 Γ ( α ) t α 3 ,

vdomLKerP means that I 0 + 3 α v(t) | t = 0 = D 0 + α 1 v(t) | t = 0 = Φ 2 [ D 0 + α 2 v(t)]=0. So,

( K p L ) v ( t ) = I 0 + α D 0 + α v ( t ) t α 2 Γ ( α 1 ) Φ 2 [ 1 ] Φ 2 [ D 0 + α 2 v ( t ) D 0 + α 2 v ( t ) | t = 0 ] = v ( t ) D 0 + α 2 v ( t ) | t = 0 Γ ( α 1 ) t α 2 + t α 2 Γ ( α 1 ) Φ 2 [ 1 ] Φ 2 [ D 0 + α 2 v ( t ) | t = 0 ] = v ( t ) .

That is, K p = ( L | dom L Ker P ) 1 . Since

D 0 + α 1 ( K p v ) ( t ) = 0 t v ( s ) d s , D 0 + α 2 ( K p v ) ( t ) = 0 t ( t s ) v ( s ) d s 1 Φ 2 [ 1 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] ,

then

K p v ( 1 Γ ( α ) + Φ 2 Γ ( α 1 ) | Φ 2 [ 1 ] | ) v 1 , D 0 + α 1 ( K p v ) v 1 , D 0 + α 2 ( K p v ) ( 1 + Φ 2 | Φ 2 [ 1 ] | ) v 1 .

It follows that

K p v C α 1 ( 2 + 1 Γ ( α ) + ( 1 + Γ ( α 1 ) ) Φ 2 Γ ( α 1 ) | Φ 2 [ 1 ] | ) v 1 .

Finally, we prove that K p :ImLdomLKerP C α 1 [0,1] is completely continuous. Let VImL L 1 [0,1] be a bounded set. From the above discussion, we only need to prove that K p V is equicontinuous on [0,1]. For vV, t 1 , t 2 [0,1] with t 1 < t 2 , we have

| D 0 + α 1 ( K p v ) ( t 1 ) D 0 + α 1 ( K p v ) ( t 2 ) | = | t 2 t 1 v ( s ) d s | t 2 t 1 | v ( s ) | d s , | D 0 + α 2 ( K p v ) ( t 1 ) D 0 + α 2 ( K p v ) ( t 2 ) | = | 0 t 1 ( t 1 s ) v ( s ) d s 0 t 2 ( t 2 s ) v ( s ) d s | | t 2 t 1 ( t 1 s ) v ( s ) d s | + | 0 t 2 ( t 1 t 2 ) v ( s ) d s | t 2 t 1 | v ( s ) | d s + ( t 1 t 2 ) v 1

and

| ( K p v ) ( t 1 ) ( K p v ) ( t 2 ) | 1 Γ ( α ) | 0 t 1 ( t 1 s ) α 1 v ( s ) d s 0 t 2 ( t 2 s ) α 1 v ( s ) d s | + | Φ 2 [ 0 t ( t s ) v ( s ) d s ] | Γ ( α 1 ) | Φ 2 [ 1 ] | | t 1 α 2 t 2 α 2 | 1 Γ ( α ) | t 2 t 1 ( t 1 s ) α 1 v ( s ) d s | + 1 Γ ( α ) | 0 t 2 ( ( t 1 s ) α 1 ( t 2 s ) α 1 ) v ( s ) d s | + Φ 2 v 1 Γ ( α 1 ) | Φ 2 [ 1 ] | | t 1 α 2 t 2 α 2 | 1 Γ ( α ) t 2 t 1 | v ( s ) | d s + α 1 Γ ( α ) v 1 ( t 1 t 2 ) + Φ 2 v 1 Γ ( α 1 ) | Φ 2 [ 1 ] | | t 1 α 2 t 2 α 2 | .

Therefore, K p (V) is equicontinuous. Thus, the operator K p :ImLdomLKerP is completely continuous. The proof is complete. □

Similar to Lemmas 2.5 and 2.6, we can obtain the following lemma.

Lemma 2.7 If (A3) holds, then KerL={a t α 2 :aR} and

ImL= { v : Φ 1 [ 0 t v ( s ) d s ] = 0 } .

Furthermore, if Φ 1 [t]0 also holds, then L is a Fredholm operator of index zero and dimKerL=codimImL=1. Here, the projectors P:YY, Q:ZZ can be defined as follows:

( P v ) ( t ) = D 0 + α 2 v ( t ) | t = 0 Γ ( α 1 ) t α 2 , ( Q v ) ( t ) = Φ 1 [ 0 t v ( s ) d s ] Φ 1 [ t ] .

The generalized inverse operator of L, K P :ImLdomLKerP can be defined by

( K p v)(t)= I 0 + α v(t) Φ 2 [ 0 t ( t s ) v ( s ) d s ] Γ ( α ) Φ 2 [ t ] t α 1 .

Also,

K p v C α 1 Δ 2 v 1 ,

where Δ 2 =2+ 1 Γ ( α ) + ( 1 + 2 Γ ( α ) ) Φ 2 Γ ( α ) | Φ 2 [ t ] | .

Lemma 2.8 If (A4) holds, then KerL={a t α 2 :aR} and

ImL= { v : Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } .

Furthermore, if Φ 2 [ t 2 ]0 also holds, then L is a Fredholm operator of index zero and dimKerL=codimImL=1. Here, the projectors P:YY, Q:ZZ can be defined as follows:

( P v ) ( t ) = D 0 + α 2 v | t = 0 Γ ( α 1 ) t α 2 , ( Q v ) ( t ) = 2 Φ 2 [ 0 t ( t s ) v ( s ) d s ] Φ 2 [ t 2 ] .

The generalized inverse operator of L, K P :ImLdomLKerP can be defined by

( K p v)(t)= I 0 + α v(t) Φ 1 [ 0 t v ( s ) d s ] Γ ( α ) Φ 1 [ 1 ] t α 1 .

Also,

K p v C α 1 Δ 3 v 1 ,

where Δ 3 =2+ 1 Γ ( α ) + ( 1 + 2 Γ ( α ) ) Φ 1 Γ ( α ) | Φ 1 [ 1 ] | and Φ 1 is the norm of the continuous linear functional  Φ 1 .

Lemma 2.9 If (A5) holds, then

KerL= { u dom L : u = a t α 1 + b t α 2 , a , b R , t [ 0 , 1 ] }

and

ImL= { v Z : Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } .

Proof Let u(t)=a t α 1 +b t α 2 . Clearly, D 0 + α u(t)=0 and I 0 + 3 α u(t) | t = 0 =0. Considering (A5), Φ 1 [ D 0 + α 1 u(t)]= Φ 1 [Γ(α)]=Γ(α) Φ 1 [1]=0 and Φ 2 [ D 0 + α 2 u(t)]=aΓ(α) Φ 2 [t]+bΓ(α1) Φ 2 [1]=0. So,

{ u dom L : u = a t α 1 + b t α 2 , a , b R , t [ 0 , 1 ] } KerL.

If Lu= D 0 + α u(t)=0, then u(t)=a t α 1 +b t α 2 +c t α 3 . Considering D 0 + α u(t)=0 and (A5), we can obtain that

KerL { u dom L : u = a t α 1 + b t α 2 , a , b R , t [ 0 , 1 ] } .

We now show that

ImL= { v Z : Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } .

If vImL, then there exists udomL such that D 0 + α u(t)=v(t). Hence,

u(t)= 1 Γ ( α ) 0 t ( t s ) α 1 v(s)ds+a t α 1 +b t α 2

for some a,bR. It yields

Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] +aΓ(α) Φ 1 [1]= Φ 1 [ 0 t v ( s ) d s ] =0

and

Φ 2 [ D 0 + α 2 u ( t ) ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] + a Γ ( α ) Φ 2 [ t ] + b Γ ( α 1 ) Φ 2 [ 1 ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 .

Therefore,

ImL { v Z : Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } .

On the other hand, suppose vZ satisfies

Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] =0.

Let

u(t)= I 0 + α v(t)= 1 Γ ( α ) 0 t ( t s ) α 1 v(s)ds.

Obviously, D 0 + α u(t)=v(t) and I 0 + 3 α u(t) | t = 0 =0. Considering (A5) and the linearity of Φ i (i=1,2), we have

Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] =0

and

Φ 2 [ D 0 + α 2 u ( t ) ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] =0.

It yields

{ v Z : Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } ImL.

The proof is complete. □

Lemma 2.10 If 2 Φ 1 [t] Φ 2 [ t 3 ]3 Φ 1 [ t 2 ] Φ 2 [ t 2 ]0, then L is a Fredholm operator of index zero and dimKerL=codimImL=2. Furthermore, the linear operator K p :ImLdomLKerP can be defined by

( K p v)(t)= I 0 + α v(t).

Also,

K p v C α 1 ( 2 + 1 Γ ( α ) ) v 1 .

Proof Firstly, we construct the mapping Q:ZZ defined by

( Q v ) ( t ) = 2 Φ 2 [ t 3 ] Φ 1 [ 0 t v ( s ) d s ] 6 Φ 1 [ t 2 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] 6 Φ 2 [ t 2 ] Φ 1 [ 0 t v ( s ) d s ] 12 Φ 1 [ t ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] t .

Let

T 1 v= 2 Φ 2 [ t 3 ] Φ 1 [ 0 t v ( s ) d s ] 6 Φ 1 [ t 2 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ]

and

T 2 v= 6 Φ 2 [ t 2 ] Φ 1 [ 0 t v ( s ) d s ] 12 Φ 1 [ t ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] .

We have

Qv= T 1 v+( T 2 v)t.
(2.7)

Noting that

T 1 ( T 1 v ) = 2 Φ 2 [ t 3 ] Φ 1 [ 0 t ( T 1 v ) d s ] 6 Φ 1 [ t 2 ] Φ 2 [ 0 t ( t s ) ( T 1 v ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] T 1 ( T 1 v ) = 2 Φ 2 [ t 3 ] Φ 1 [ 0 t d s ] 6 Φ 1 [ t 2 ] Φ 2 [ 0 t ( t s ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 1 v ) T 1 ( T 1 v ) = 2 Φ 2 [ t 3 ] Φ 1 [ t ] 6 Φ 1 [ t 2 ] Φ 2 [ t 2 2 ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 1 v ) T 1 ( T 1 v ) = T 1 v , T 1 ( T 2 v ) = 2 Φ 2 [ t 3 ] Φ 1 [ 0 t ( T 2 v ) s d s ] 6 Φ 1 [ t 2 ] Φ 2 [ 0 t ( t s ) ( T 2 v ) s d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] T 1 ( T 2 v ) = 2 Φ 2 [ t 3 ] Φ 1 [ 0 t s d s ] 6 Φ 1 [ t 2 ] Φ 2 [ 0 t ( t s ) s d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 2 v ) T 1 ( T 2 v ) = 2 Φ 2 [ t 3 ] Φ 1 [ t 2 2 ] 6 Φ 1 [ t 2 ] Φ 2 [ t 3 6 ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 2 v ) T 1 ( T 2 v ) = 0 , T 2 ( T 1 v ) = 6 Φ 2 [ t 2 ] Φ 1 [ 0 t ( T 1 v ) d s ] 12 Φ 1 [ t ] Φ 2 [ 0 t ( t s ) ( T 1 v ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] T 2 ( T 1 v ) = 6 Φ 2 [ t 2 ] Φ 1 [ 0 t d s ] 12 Φ 1 [ t ] Φ 2 [ 0 t ( t s ) d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 1 v ) T 2 ( T 1 v ) = 6 Φ 2 [ t 2 ] Φ 1 [ t ] 12 Φ 1 [ t ] Φ 2 [ t 2 2 ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 1 v ) T 2 ( T 1 v ) = 0

and

T 2 ( T 2 v ) = 6 Φ 2 [ t 2 ] Φ 1 [ 0 t ( T 2 v ) s d s ] 12 Φ 1 [ t ] Φ 2 [ 0 t ( t s ) ( T 2 v ) s d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] = 6 Φ 2 [ t 2 ] Φ 1 [ 0 t s d s ] 12 Φ 1 [ t ] Φ 2 [ 0 t ( t s ) s d s ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 2 v ) = 6 Φ 2 [ t 2 ] Φ 1 [ t 2 2 ] 12 Φ 1 [ t ] Φ 2 [ t 3 6 ] 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] ( T 2 v ) = T 2 v ,

we have, for each vZ, that

Q 2 v= T 1 ( T 1 v + ( T 2 v ) t ) + T 2 ( T 1 v + ( T 2 v ) t ) t= T 1 v+( T 2 v)t=Qv.

So, Q:ZZ is a well-defined projector.

Now we will show that KerQ=ImL. If vKerQ, from Qv=0, we have T 1 v=0 and T 2 v=0. Considering the definitions of T 1 and T 2 , we have

{ Φ 2 [ t 3 ] Φ 1 [ 0 t v ( s ) d s ] 3 Φ 1 [ t 2 ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 , Φ 2 [ t 2 ] Φ 1 [ 0 t v ( s ) d s ] 2 Φ 1 [ t ] Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 .

Since

| Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] 2 Φ 1 [ t ] | =2 Φ 1 [t] Φ 2 [ t 3 ] +3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] 0,

so Φ 1 [ 0 t v(s)ds]= Φ 2 [ 0 t (ts)v(s)ds]=0, which yields vImL. On the other hand, if vImL, from Φ 1 [ 0 t v(s)ds]= Φ 2 [ 0 t (ts)v(s)ds]=0 and the definition of Q, it is obvious that Qv=0, thus vKerQ. Hence, KerQ=ImL.

For vZ, from v=(vQv)+Qv, vQvKerQ=ImL, QvImQ, we have Z=ImL+ImQ. And for any vImLImQ, from vImQ, there exist constants a,bR such that v(t)=a+bt. From vImL, we obtain

{ Φ 1 [ t ] a + Φ 1 [ t 2 2 ] b = 0 , Φ 2 [ t 2 2 ] a + Φ 2 [ t 3 6 ] b = 0 .
(2.8)

In view of

| Φ 1 [ t ] Φ 1 [ t 2 2 ] Φ 2 [ t 2 2 ] Φ 2 [ t 3 6 ] | = 1 6 Φ 1 [t] Φ 2 [ t 3 ] 1 4 Φ 1 [ t 2 ] Φ 2 [ t 2 ] 0,

therefore (2.8) has a unique solution a=b=0, which implies ImLImQ={θ} and Z=ImLImQ. Since dimKerL=dimImQ=codimImL=2, thus L is a Fredholm map of index zero. Let P:YY be defined by

(Pv)(t)= D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 + D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) t α 2 .

Then the generalized inverse K p :ImLdomLKerP of L can be rewritten

( K p v)(t)= I 0 + α v(t).

In fact, for vImL, we have

I 0 + 3 α ( K p v ) ( t ) | t = 0 = 0 , Φ 1 [ D 0 + α 1 ( K p v ) ( t ) ] = Φ 1 [ D 0 + α 1 I 0 + α v ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] = 0

and

Φ 2 [ D 0 + α 2 ( K p v ) ( t ) ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] =0,

which implies that K p is well defined on ImL. Moreover, for vImL, we have

(L K p )v(t)= D 0 + α I 0 + α v(t)=v(t)

and for vdomLKerP, we know

I 0 + α D 0 + α v(t)=v(t) D 0 + α 1 v ( t ) | t = 0 Γ ( α ) t α 1 D 0 + α 2 v ( t ) | t = 0 Γ ( α 1 ) t α 2 I 0 + 3 α v ( t ) | t = 0 Γ ( α ) t α 3 ,

vdomLKerP means that I 0 + 3 α v(t) | t = 0 = D 0 + α 1 v(t) | t = 0 = D 0 + α 2 v(t) | t = 0 =0. So,

( K p L)v(t)= I 0 + α D 0 + α v(t)=v(t).

That is, K p = ( L | dom L Ker P ) 1 . Since

D 0 + α 1 ( K p v ) ( t ) = 0 t v ( s ) d s , D 0 + α 2 ( K p v ) ( t ) = 0 t ( t s ) v ( s ) d s ,

then

K p v 1 Γ ( α ) v 1 , D 0 + α 1 ( K p v ) v 1 , D 0 + α 2 ( K p v ) v 1 .

It follows that

K p v C α 1 ( 2 + 1 Γ ( α ) ) v 1 .

The proof is complete. □

3 Main results

From Lemma 2.4, we can obtain the existence theorem for FBVP (1.1), (1.2).

Theorem 3.1 Assume that (A1) and the following conditions hold:

| f ( t , x 1 , x 2 , x 3 ) f ( t , y 1 , y 2 , y 3 ) | β ( | x 1 y 1 | + | x 2 y 2 | + | x 3 y 3 | ) .

Then FBVP (1.1), (1.2) has a unique solution in C α 1 [0,1] provided that

β ( 2 + Φ 2 | Φ 2 [ 1 ] | + 1 Γ ( α ) + Φ 2 Γ ( α 1 ) | Φ 2 [ 1 ] | ) ( 1 + Φ 1 | Φ 1 [ 1 ] | ) <1.

Proof We shall prove that Tx=x has a unique solution in C α 1 [0,1]. For each u,v C α 1 [0,1], considering the linearity of Φ i (i=1,2), we have

( T u ) ( t ) ( T v ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 ( ( f u ) ( s ) ( f u ) ( s ) ) d s Φ 1 [ 0 t ( ( f u ) ( s ) ( f u ) ( s ) ) d s ] Γ ( α ) Φ 1 [ 1 ] t α 1 Φ 2 [ 0 t ( t s ) ( ( f u ) ( s ) ( f u ) ( s ) ) d s ] Γ ( α 1 ) Φ 2 [ 1 ] t α 2 + Φ 1 [ 0 t ( ( f u ) ( s ) ( f u ) ( s ) ) d s ] Φ 2 [ t ] Γ ( α 1 ) Φ 1 [ 1 ] Φ 2 [ 1 ] t α 2 .

Then

| ( T u ) ( t ) ( T v ) ( t ) | β u v C α 1 ( 1 Γ ( α ) + Φ 2 Γ ( α 1 ) | Φ 2 [ 1 ] | ) ( 1 + Φ 1 | Φ 1 [ 1 ] | ) , | D 0 + α 1 ( T u ) ( t ) D 0 + α 1 ( T v ) ( t ) | β u v C α 1 ( 1 + Φ 1 | Φ 1 [ 1 ] | )

and

| D 0 + α 2 ( T u ) ( t ) D 0 + α 2 ( T v ) ( t ) | β u v C α 1 ( 1 + Φ 2 | Φ 2 [ 1 ] | ) ( 1 + Φ 1 | Φ 1 [ 1 ] | ) .

So,

T u T v C α 1 β u v C α 1 ( 2 + Φ 2 | Φ 2 [ 1 ] | + 1 Γ ( α ) + Φ 2 Γ ( α 1 ) | Φ 2 [ 1 ] | ) ( 1 + Φ 1 | Φ 1 [ 1 ] | ) .

The above inequality implies that T is a contraction. By using Banach’s contraction principle, Tx=x has a unique solution in C α 1 [0,1]. From Lemma 2.4, FBVP (1.1), (1.2) has a unique solution in C α 1 [0,1]. The proof is complete. □

From Lemmas 2.5-2.8 and Theorem 2.1, we can obtain the existence theorem for FBVP (1.1), (1.2) in the case of dimKerL=1.

Theorem 3.2 Let f:[0,1]× R 3 R be a continuous function. Assume that Φ 1 [t]0, (A2) and the following conditions (H1)-(H3) hold:

(H1) There exist functions α,β,γ,ω L 1 [0,1] such that for all (x,y,z) R 3 , t[0,1],

| f ( t , x , y ) | ω(t)+α(t)|x|+β(t)|y|+γ(t)|z|.

(H2) There exists a constant A>0 such that for udomL, if | D 0 + α 1 u(t)|>A for all t[0,1], then Φ 1 [ 0 t f(s,u(s), D 0 + α 1 u(s), D 0 + α 2 u(s))ds]0.

(H3) There exists a constant B>0 such that either for each aR:|a|>B,

a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] >0

or for each aR:|a|>B,

a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] <0.

Then FBVP (1.1), (1.2) has at least one solution in C α 1 [0,1] provided

( 1 Γ ( α ) + 2 + Δ 1 ) ( α 1 + β 1 + γ 1 ) <1,

where Δ 1 is the same as in Lemma  2.6.

Proof Set

Ω 1 = { u dom L Ker L : L u = λ N u  for some  λ [ 0 , 1 ] } .

Then, for u Ω 1 , since Lu=λNu, so λ0, NuImL=KerQ, hence

Φ 1 [ 0 t f ( s , u ( s ) , D 0 + α 1 u ( s ) , D 0 + α 2 u ( s ) ) d s ] =0.

Thus, from (H2), there exists t 0 [0,1] such that

| D 0 + α 1 u ( t 0 ) | A.

Now,

D 0 + α 1 u(t)= D 0 + α 1 u( t 0 )+ t 0 t D 0 + α u(s)ds,

and so

| D 0 + α 1 u ( 0 ) | D 0 + α 1 u ( t ) | D 0 + α 2 u ( t 0 ) | + D 0 + α u ( t ) 1 A + L u 1 A + N u 1 .
(3.1)

Again, for u Ω 1 , udomLKerL, then (IP)udomLKerP and LPu=0. Thus, from Lemma 2.6, we have

( I P ) u C α 1 = K P L ( I P ) u C α 1 Δ 1 L ( I P ) u 1 Δ 1 N u 1 .
(3.2)

From (3.1), (3.2), we have

u C α 1 P u C α 1 + ( I P ) u C α 1 = ( 1 Γ ( α ) + 2 ) | D 0 + α 1 u ( 0 ) | + ( I P ) u C α 1 A ( 1 Γ ( α ) + 2 ) + ( 1 Γ ( α ) + 2 + Δ 1 ) N u 1 .

By this and (H1), we have

u C α 1 A ( 1 Γ ( α ) + 2 ) + ( 1 Γ ( α ) + 2 + Δ 1 ) ω 1 + ( 1 Γ ( α ) + 2 + Δ 1 ) ( α 1 + β 1 + γ 1 ) u C α 1

and

u C α 1 A ( 1 Γ ( α ) + 2 ) + ( 1 Γ ( α ) + 2 + Δ 1 ) ω 1 1 ( 1 Γ ( α ) + 2 + Δ 1 ) ( α 1 + β 1 + γ 1 ) .

Therefore, Ω 1 is bounded. Let

Ω 2 ={uKerL:NuImL}.

For u Ω 2 , there is uKerL={udomLu=a t α 1 ,t[0,1],aR}, and NuImL, thus

Φ 1 [ 0 t f ( s , a t α 1 , a Γ ( α ) , a Γ ( α ) s ) d s ] =0.

From (H2), we get |a| A Γ ( α ) , thus Ω 2 is bounded.

Next, according to the condition (H3), for any aR, if |a|>B, then either

a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] <0
(3.3)

or else

a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] >0.
(3.4)

If (3.3) holds, set

Ω 3 = { u Ker L : λ J u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] } ,

here Q is given by (2.6) and J:KerLImQ is the linear isomorphism given by J(a t α 1 )= a Φ 1 [ t ] , aR, t[0,1]. For u=a t α 1 Ω 3 ,

λa=(1λ) Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] .

If λ=1, then a=0. Otherwise, if |a|>B, in view of (3.3), one has

a(1λ) Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] <0,

which contradicts λ a 2 0. Thus, Ω 3 {uKerLu=a t α 1 ,|a|B} is bounded.

If (3.4) holds, then define the set

Ω 3 = { x Ker L : λ J u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] } ,

here J is as above. Similar to the above argument, we can show that Ω 3 is bounded too.

In the following, we shall prove that all the conditions of Theorem 2.1 are satisfied. Let Ω be a bounded open subset of Y such that i = 1 3 Ω i ¯ Ω. By Lemma 2.6 and standard arguments, we can prove that K P (IQ)N:ΩY is compact, thus N is L-compact on Ω ¯ . Then, by the above argument, we have

  1. (i)

    LuλNu, for every (u,λ)[(domLKerL)Ω]×(0,1),

  2. (ii)

    NuImL for uKerLΩ.

Finally, we will prove that (iii) of Theorem 2.1 is satisfied. Let H(u,λ)=±λJu+(1λ)QNu. According to the above argument, we know

H(u,λ)0for uKerLΩ.

Thus, by the homotopy property of degree, we have

deg ( Q N | Ker L , Ker L Ω , 0 ) = deg ( H ( , 0 ) , Ker L Ω , 0 ) = deg ( H ( , 1 ) , Ker L Ω , 0 ) = deg ( J , Ker L Ω , 0 ) 0 .

Then, by Theorem 2.1, Lu=Nu has at least one solution in domL Ω ¯ , so that FBVP (1.1), (1.2) has a solution in C α 1 [0,1]. The proof is complete. □

Theorem 3.3 Let f:[0,1]× R 3 R be a continuous function. Assume that Φ 1 [t]0, (A3), (H1) and the following conditions (H4), (H5) hold:

(H4) There exists a constant A>0 such that for udomL, if | D 0 + α 1 u(t)|+| D 0 + α 2 u(t)|>A for all t[0,1], then Φ 1 [ 0 t f(s,u(s), D 0 + α 1 u(s), D 0 + α 2 u(s))ds]0.

(H5) There exists a constant B>0 such that either for each aR:|a|>B,

a Φ 1 [ 0 t f ( s , a s α 2 , 0 , a Γ ( α 1 ) ) d s ] >0

or for each aR:|a|>B,

a Φ 1 [ 0 t f ( s , a s α 2 , 0 , a Γ ( α 1 ) ) d s ] <0.

Then FBVP (1.1), (1.2) has at least one solution in C α 1 [0,1] provided

( 1 Γ ( α 1 ) + 1 + Δ 2 ) ( α 1 + β 1 + γ 1 ) <1,

where Δ 2 is the same as in Lemma  2.7.

Theorem 3.4 Let f:[0,1]× R 3 R be a continuous function. Assume that Φ 2 [ t 2 ]0, (A4), (H1) and the following conditions (H6), (H7) hold:

(H6) There exists a constant A>0 such that for udomL, if | D 0 + α 1 u(t)|+| D 0 + α 2 u(t)|>A for all t[0,1], then Φ 2 [ 0 t (ts)f(s,u(s), D 0 + α 1 u(s), D 0 + α 2 u(s))ds]0.

(H7) There exists a constant B>0 such that either for each aR:|a|>B,

a Φ 2 [ 0 t ( t s ) f ( s , a s α 2 , 0 , a Γ ( α 1 ) ) d s ] >0

or for each aR:|a|>B,

a Φ 2 [ 0 t ( t s ) f ( s , a s α 2 , 0 , a Γ ( α 1 ) ) d s ] <0.

Then FBVP (1.1), (1.2) has at least one solution in C α 1 [0,1] provided

( 1 Γ ( α 1 ) + 1 + Δ 3 ) ( α 1 + β 1 + γ 1 ) <1,

where Δ 3 is the same as in Lemma  2.8.

The proofs of Theorem 3.3 and Theorem 3.4 are similar to that of Theorem 3.2. So, we omit them.

The above Theorem 3.2, Theorem 3.3 and Theorem 3.4 are the existence of solutions to FBVP (1.1), (1.2) in the case of dimKerL=1. By making use of Theorem 2.1, Lemma 2.9 and Lemma 2.10, we obtain the existence of solutions for FBVP (1.1), (1.2) in the case of dimKerL=2.

Theorem 3.5 Let f:[0,1]× R 3 R be a continuous function. Assume that 2 Φ 1 [t] Φ 2 [ t 3 ]3 Φ 1 [ t 2 ] Φ 2 [ t 2 ]0, (A5), (H1) and the following conditions (H8), (H9) hold:

(H8) There exists a constant A>0 such that for udomL, if | D 0 + α 1 u(t)|+| D 0 + α 2 u(t)|>A for all t[0,1], then

Φ 1 [ 0 t f ( s , u ( s ) , D 0 + α 1 u ( s ) , D 0 + α 2 u ( s ) ) d s ] 0 or Φ 2 [ 0 t ( t s ) f ( s , u ( s ) , D 0 + α 1 u ( s ) , D 0 + α 2 u ( s ) ) d s ] 0 .

(H9) There exists a constant B>0 such that for a 1 , a 2 R satisfying | a 1 |+| a 2 |>B, either

a 1 Φ 1 [ 0 t N ( a 1 t α 1 + a 2 t α 2 ) d s ] > 0 , a 2 Φ 2 [ 0 t ( t s ) N ( a 1 t α 1 + a 2 t α 2 ) d s ] > 0
(3.5)

or

a 1 Φ 1 [ 0 t N ( a 1 t α 1 + a 2 t α 2 ) d s ] < 0 , a 2 Φ 2 [ 0 t ( t s ) N ( a 1 t α 1 + a 2 t α 2 ) d s ] < 0 .
(3.6)

Then FBVP (1.1), (1.2) has at least one solution in C α 1 [0,1] provided

( 2 Γ ( α ) + 5 + 1 Γ ( α 1 ) ) ( α 1 + β 1 + γ 1 ) <1.

Proof Set

Ω 1 = { u dom L Ker L : L u = λ N u  for some  λ [ 0 , 1 ] } .

Then, for u Ω 1 , since Lu=λNu, so λ0, NuImL=KerQ, hence

Φ 1 [ 0 t f ( s , u ( s ) , D 0 + α 1 u ( s ) , D 0 + α 2 u ( s ) ) d s ] =0

and

Φ 2 [ 0 t ( t s ) f ( s , u ( s ) , D 0 + α 1 u ( s ) , D 0 + α 2 u ( s ) ) d s ] =0.

Thus, from (H8), there exists t 0 [0,1] such that

| D 0 + α 1 u ( t 0 ) | + | D 0 + α 2 u ( t 0 ) | A.

Now,

D 0 + α 1 u ( t ) = D 0 + α 1 u ( t 0 ) + t 0 t D 0 + α u ( s ) d s , D 0 + α 2 u ( t ) = D 0 + α 2 u ( t 0 ) + t 0 t D 0 + α 1 u ( s ) d s ,

and so

| D 0 + α 1 u ( 0 ) | D 0 + α 1 u ( t ) | D 0 + α 1 u ( t 0 ) | + D 0 + α u ( t ) 1 A + L u 1 A + N u 1 ,
(3.7)
| D 0 + α 2 u ( 0 ) | D 0 + α 2 u ( t ) | D 0 + α 2 u ( t 0 ) | + D 0 + α 1 u ( t ) 1 | D 0 + α 2 u ( t 0 ) | + D 0 + α 1 u ( t ) | D 0 + α 2 u ( t 0 ) | + | D 0 + α 1 u ( t 0 ) | + D 0 + α u ( t ) 1 A + N u 1 .
(3.8)

Again, for u Ω 1 , udomLKerL, then (IP)udomLKerP and LPu=0. Thus, from Lemma 2.10, we have

( I P ) u C α 1 = K P L ( I P ) u C α 1 ( 2 + 1 Γ ( α ) ) L ( I P ) u 1 ( 2 + 1 Γ ( α ) ) N u 1 .
(3.9)

From (3.7), (3.8) and (3.9), we have

u C α 1 P u C α 1 + ( I P ) u C α 1 = ( 1 Γ ( α ) + 2 ) | D 0 + α 1 u ( 0 ) | + ( 1 Γ ( α 1 ) + 1 ) | D 0 + α 2 u ( 0 ) | + ( I P ) u C α 1 A ( 1 Γ ( α ) + 1 Γ ( α 1 ) + 3 ) + ( 2 Γ ( α ) + 5 + 1 Γ ( α 1 ) ) N u 1 .

By this and (H1), we have

u C α 1 A ( 1 Γ ( α ) + 1 Γ ( α 1 ) + 3 ) + ( 2 Γ ( α ) + 5 + 1 Γ ( α 1 ) ) ω 1 + ( 2 Γ ( α ) + 5 + 1 Γ ( α 1 ) ) ( α 1 + β 1 + γ 1 ) u C α 1

and

u C α 1 A ( 1 Γ ( α ) + 1 Γ ( α 1 ) + 3 ) + ( 2 Γ ( α ) + 5 + 1 Γ ( α 1 ) ) ω 1 1 ( 2 Γ ( α ) + 5 + 1 Γ ( α 1 ) ) ( α 1 + β 1 + γ 1 ) .

Therefore, Ω 1 is bounded. Let

Ω 2 ={uKerL:NuImL}.

For u Ω 2 , there is uKerL={udomLu=a t α 1 +b t α 2 ,t[0,1],a,bR}, and NuImL, thus

Φ 1 [ 0 t f ( s , a s α 1 + b s α 2 , a Γ ( α ) , a Γ ( α ) s + b Γ ( α 1 ) ) d s ] =0

and

Φ 2 [ 0 t ( t s ) f ( s , a s α 1 + b s α 2 , a Γ ( α ) , a Γ ( α ) s + b Γ ( α 1 ) ) d s ] =0.

From (H8), we get 2Γ(α)|a|+Γ(α1)|b|A. Then, for u Ω 2 , we have

u C α 1 ( 2 Γ ( α ) + 1 ) |a|+ ( Γ ( α 1 ) + 1 ) |b| ( 1 + 1 Γ ( α 1 ) ) A,

thus Ω 2 is bounded.

Next, for any a 1 , a 2 R, define a linear isomorphism J:KerLImQ by

J ( a 1 t α 1 + a 2 t α 2 ) = 2 a 1 Φ 2 [ t 3 ] 6 a 2 Φ 1 [ t 2 ] ( 6 a 1 Φ 2 [ t 2 ] 12 a 2 Φ 1 [ t ] ) t 2 Φ 1 [ t ] Φ 2 [ t 3 ] 3 Φ 1 [ t 2 ] Φ 2 [ t 2 ] .

If (3.5) holds, set

Ω 3 = { u Ker L : λ J u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] } ,

where Q is given by (2.7). For u= a 1 t α 1 + a 2 t α 2 Ω 3 , from λJu+(1λ)QNu=0, we obtain

Φ 2 [ t 3 ] ( a 1 λ + ( 1 λ ) Φ 1 [ 0 t N u ( s ) d s ] ) 3 Φ 1 [ t 2 ] ( a 2 λ + ( 1 λ ) Φ 2 [ 0 t ( t s ) N u ( s ) d s ] ) = 0

and

Φ 2 [ t 2 ] ( a 1 λ + ( 1 λ ) Φ 1 [ 0 t N u ( s ) d s ] ) 2 Φ 1 [ t ] ( a 2 λ + ( 1 λ ) Φ 2 [ 0 t ( t s ) N u ( s ) d s ] ) = 0 .

By 2 Φ 1 [t] Φ 2 [ t 3 ]3 Φ 1 [ t 2 ] Φ 2 [ t 2 ]0, it yields

{ a 1 λ + ( 1 λ ) Φ 1 [ 0 t N u ( s ) d s ] = 0 , a 2 λ + ( 1 λ ) Φ 2 [ 0 t ( t s ) N u