Open Access

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

Advances in Difference Equations20132013:233

https://doi.org/10.1186/1687-1847-2013-233

Received: 11 January 2013

Accepted: 16 July 2013

Published: 7 August 2013

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 dim Ker L = 1 and dim Ker L = 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 dim Ker L = 1 :
D 0 + α u ( t ) = f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = 0 , β u ( η ) = u ( 1 ) , 1 < α 2 .
Recently, Jiang [26] studied the existence of a solution for the following fractional differential equation at resonance under the case dim Ker L = 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 f C ( [ 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 dim Ker L = 1 and dim Ker L = 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 : Im L dom L Ker P 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 ) d s ,

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 d t .

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 d s ,

where n 1 α < 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 ) | d t . 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 n α 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 k N . 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 1 p + ;
  1. (iii)

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

     
  2. (iv)
    Note that for λ > 1 , λ α 1 , α 2 , , α n , 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 u F ,
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 , u F , 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 Ker L = { θ } . It is the so-called nonresonance case. If (A2) holds, then Ker L = { a t α 1 : a R } . If (A3) or (A4) holds, then Ker L = { a t α 2 : a R } . If (A5) holds, then Ker L = { a t α 1 + b t α 2 : a , b R } .

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

Proof If u is a solution to T u = 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 ( f u ) ( 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 ( f u ) ( s ) d s + c 1 Γ ( α ) ,
(2.2)
D 0 + α 2 u ( t ) = 0 t ( t s ) ( f u ) ( s ) d s + c 1 Γ ( α ) t + c 2 Γ ( α 1 ) .
(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 : dom L Y Z 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)

    dim Ker L = codim Im L < + .

     

Let Y, Z be real Banach spaces and L : dom L Y Z be a Fredholm operator of index zero. P : Y Y , Q : Z Z are continuous projectors such that Im P = Ker L , Ker Q = Im L , Y = Ker L Ker P and 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 mapping by K P (generalized inverse operator of L). If Ω is an open bounded subset of Y such that dom L Ω , the mapping N : Y Z will be called L-compact on Ω ¯ , if Q N ( Ω ¯ ) is bounded and K P ( I Q ) 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)

    L x λ N x for every ( x , λ ) [ ( dom L Ker L ) Ω ] × ( 0 , 1 ) .

     
  2. (ii)

    N x Im L for every x Ker L Ω .

     
  3. (iii)

    deg ( Q N | Ker L , Ker L Ω , 0 ) 0 , where Q : Z Z is a projector as above with Im L = Ker Q .

     

Then the equation L x = N x has at least one solution in dom L Ω ¯ .

Let Y = C α 1 [ 0 , 1 ] , Z = L 1 [ 0 , 1 ] . Let the linear operator L : Y dom L Z 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 L u = D 0 + α u ( t ) . Let the nonlinear operator N : Y Z be defined by
( N u ) ( 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
L u = N u .

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

Lemma 2.5 Let L be the linear operator defined as above. If (A2) holds, then
Ker L = { u dom L : u = a t α 1 , a R , t [ 0 , 1 ] }
and
Im L = { 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 ) ] = a Γ ( 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 ] } Ker L .

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

We now show that
Im L = { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } .
If v Im L , then there exists u dom L such that D 0 + α u ( t ) = v ( t ) . Hence,
u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s + a t α 1 + b t α 2
for some a , b R . It yields
Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] + a Γ ( α ) Φ 1 [ 1 ] = Φ 1 [ 0 t v ( s ) d s ] = 0 .
Therefore
Im L { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } .
On the other hand, suppose v Z satisfies
Φ 1 [ 0 t v ( s ) d s ] = 0 .
Let
u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s 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 } Im L .

The proof is complete. □

Lemma 2.6 If Φ 1 [ t ] 0 , then:
  1. (i)

    L is a Fredholm operator of index zero and dim Ker L = codim Im L = 1 .

     
  2. (ii)
    The linear operator K p : Im L dom L Ker P 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 : Im L dom L Ker P C α 1 [ 0 , 1 ] is completely continuous.

     
Proof Firstly, we construct the mapping Q : Z Z defined by
Q y = 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 ] ( Q y ) = Q y ,

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

Now, it is obvious that Im L = Ker Q . Noting that Q is a linear projector, we have Z = Im Q Ker Q . Hence, Z = Im Q Im L and dim Ker L = codim Im L = 1. This means L is a Fredholm mapping of index zero. Taking P : Y Y as
( P u ) ( t ) = D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 ,
then the generalized inverse K p : Im L dom L Ker P 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 v Im L , 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 v Im L , we have
( L K p ) v ( t ) = D 0 + α I 0 + α v ( t ) = v ( t )
and for v dom L Ker P , 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 ,
v dom L Ker P 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 : Im L dom L Ker P C α 1 [ 0 , 1 ] is completely continuous. Let V Im L 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 v V , 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 : Im L dom L Ker P 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 Ker L = { a t α 2 : a R } and
Im L = { 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 dim Ker L = codim Im L = 1 . Here, the projectors P : Y Y , Q : Z Z 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 : Im L dom L Ker P 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 Ker L = { a t α 2 : a R } and
Im L = { 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 dim Ker L = codim Im L = 1 . Here, the projectors P : Y Y , Q : Z Z 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 : Im L dom L Ker P 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
Ker L = { u dom L : u = a t α 1 + b t α 2 , a , b R , t [ 0 , 1 ] }
and
Im L = { 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 ) ] = a Γ ( α ) Φ 2 [ t ] + b Γ ( α 1 ) Φ 2 [ 1 ] = 0 . So,
{ u dom L : u = a t α 1 + b t α 2 , a , b R , t [ 0 , 1 ] } Ker L .
If L u = 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
Ker L { u dom L : u = a t α 1 + b t α 2 , a , b R , t [ 0 , 1 ] } .
We now show that
Im L = { v Z : Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } .
If v Im L , then there exists u dom L such that D 0 + α u ( t ) = v ( t ) . Hence,
u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s + a t α 1 + b t α 2
for some a , b R . It yields
Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] + a Γ ( 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,
Im L { 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 v Z 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 ) d s .
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 } Im L .

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 dim Ker L = codim Im L = 2 . Furthermore, the linear operator K p : Im L dom L Ker P 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 : Z Z 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
Q v = 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 v Z , 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 = Q v .

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

Now we will show that Ker Q = Im L . If v Ker Q , from Q v = 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 ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 , which yields v Im L . On the other hand, if v Im L , from Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 and the definition of Q, it is obvious that Q v = 0 , thus v Ker Q . Hence, Ker Q = Im L .

For v Z , from v = ( v Q v ) + Q v , v Q v Ker Q = Im L , Q v Im Q , we have Z = Im L + Im Q . And for any v Im L Im Q , from v Im Q , there exist constants a , b R such that v ( t ) = a + b t . From v Im L , 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 Im L Im Q = { θ } and Z = Im L Im Q . Since dim Ker L = dim Im Q = codim Im L = 2 , thus L is a Fredholm map of index zero. Let P : Y Y be defined by
( P v ) ( 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 : Im L dom L Ker P of L can be rewritten
( K p v ) ( t ) = I 0 + α v ( t ) .
In fact, for v Im L , 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 v Im L , we have
( L K p ) v ( t ) = D 0 + α I 0 + α v ( t ) = v ( t )
and for v dom L Ker P , 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 ,
v dom L Ker P 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 T x = 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, T x = 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 dim Ker L = 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 u dom L , 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 ) ) d s ] 0 .

(H3) There exists a constant B > 0 such that either for each a R : | a | > B ,
a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] > 0
or for each a R : | 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 L u = λ N u , so λ 0 , N u Im L = Ker Q , 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 ) d s ,
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 , u dom L Ker L , then ( I P ) u dom L Ker P and L P u = 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 = { u Ker L : N u Im L } .
For u Ω 2 , there is u Ker L = { u dom L u = a t α 1 , t [ 0 , 1 ] , a R } , and N u Im L , 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 a R , 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 : Ker L Im Q is the linear isomorphism given by J ( a t α 1 ) = a Φ 1 [ t ] , a R , t [ 0 , 1 ] . For u = a t α 1 Ω 3 ,
λ a = ( 1 λ ) 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 ( 1 λ ) 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 { u Ker L u = 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 ( I Q ) N : Ω Y is compact, thus N is L-compact on Ω ¯ . Then, by the above argument, we have
  1. (i)

    L u λ N u , for every ( u , λ ) [ ( dom L Ker L ) Ω ] × ( 0 , 1 ) ,

     
  2. (ii)

    N u Im L for u Ker L Ω .

     
Finally, we will prove that (iii) of Theorem 2.1 is satisfied. Let H ( u , λ ) = ± λ J u + ( 1 λ ) Q N u . According to the above argument, we know
H ( u , λ ) 0 for  u Ker L Ω .
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, L u = N u has at least one solution in dom L 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 u dom L , 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 .

(H5) There exists a constant B > 0 such that either for each a R : | a | > B ,
a Φ 1 [ 0 t f ( s , a s α 2 , 0 , a Γ ( α 1 ) ) d s ] > 0
or for each a R : | 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 u dom L , if | D 0 + α 1 u ( t ) | + | D 0 + α 2 u ( t ) | > A for all t [ 0 , 1 ] , then 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 .

(H7) There exists a constant B > 0 such that either for each a R : | 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 a R : | 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 dim Ker L = 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 dim Ker L = 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 u dom L , 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 L u = λ N u , so λ 0 , N u Im L = Ker Q , 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 , u dom L Ker L , then ( I P ) u dom L Ker P and L P u = 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 = { u Ker L : N u Im L } .
For u Ω 2 , there is u Ker L = { u dom L u = a t α 1 + b t α 2 , t [ 0 , 1 ] , a , b R } , and N u Im L , 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 2 Γ ( α ) | a | + Γ ( α 1 ) | 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 : Ker L Im Q 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 λ J u + ( 1 λ ) Q N u = 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 ( s ) d s ] = 0 .
If λ = 1 , then a 1 = a 2 = 0 . Otherwise, if | a 1 | + | a 2 | > B , considering the above equalities and (3.5), we have
λ ( a 1 2 + a 2 2 ) = ( 1 λ ) [ a 1 Φ 1 [ 0 t N u ( s ) d s ] + a 2 Φ 2 [ 0 t ( t s ) N u ( s ) d s ] ] < 0 ,
which contradicts λ ( a 1 2 + a 2 2 ) 0 . If (3.6) holds, then we take
Ω 3 = { u Ker L : λ J u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] }
and, again, obtain a contradiction. Thus, in either case,
u C α 1 = a 1 t α 1 + a 2 t α 2 C α 1 | a 1 | ( 1 + 2 Γ ( α ) ) + | a 2 | ( 1 + Γ ( α 1 ) ) B ( 2 + 2 Γ ( α ) + Γ ( α 1 ) )

for all u Ω 3 , that is, Ω 3 is bounded.

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 2.2 and standard arguments, we can prove that K P ( I Q ) N : Ω Y is compact, thus N is L-compact on Ω ¯ . Then, by the above argument, we have
  1. (i)

    L u λ N u for every ( u , λ ) [ ( dom L Ker L ) Ω ] × ( 0 , 1 ) ,

     
  2. (ii)

    N u Im L for u Ker L Ω .

     
Finally, we will prove that (iii) of Theorem 2.1 is satisfied. Let H ( u , λ ) = ± λ J u + ( 1 λ ) Q N u . According to the above argument, we know
H ( u , λ ) 0 for  u Ker L Ω .
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<