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Advances in Difference Equations

, 2013:233 | Cite as

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

  • Yumei Zou
  • Yujun CuiEmail author
Open Access
Research
Part of the following topical collections:
  1. Progress in Functional Differential and Difference Equations
  2. Recent Advances in Operator Equations, Boundary Value Problems, Fixed Point Theory and Applications, and General Inequalities

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 Open image in new window and dim Ker L = 2 Open image in new window 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 [1, 2, 3] 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 [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16] for some recent results at nonresonance and to [17, 18, 19, 20, 21, 22, 23, 24, 25, 26] 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 Open image in new window:
D 0 + α u ( t ) = f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = 0 , β u ( η ) = u ( 1 ) , 1 < α 2 . Open image in new window
Recently, Jiang [26] studied the existence of a solution for the following fractional differential equation at resonance under the case dim Ker L = 2 Open image in new window:
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 ) . Open image in new window
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 ) ) , Open image in new window
(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 , Open image in new window
(1.2)

where 2 < α < 3 Open image in new window, D 0 + α Open image in new window and I 0 + α Open image in new window are the standard Riemann-Liouville differentiation and integration, and f C ( [ 0 , 1 ] × R 3 , R ) Open image in new window; Φ 1 , Φ 2 : C [ 0 , 1 ] R Open image in new window 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 Open image in new window and dim Ker L = 2 Open image in new window), 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 Open image in new window 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 Open image in new window of a function u : ( 0 , ) R Open image in new window is given by
I 0 + α u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 u ( s ) d s , Open image in new window

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

Definition 2.2 [1, 2]

The Riemann-Liouville fractional derivative of order α > 0 Open image in new window of a continuous function u : ( 0 , ) R Open image in new window 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 , Open image in new window

where n 1 α < n Open image in new window, provided that the right-hand side is pointwise defined on ( 0 , ) Open image in new window.

We use the classical spaces C [ 0 , 1 ] Open image in new window with the norm u = max t [ 0 , 1 ] | u ( t ) | Open image in new window, L 1 [ 0 , 1 ] Open image in new window with the norm u 1 = 0 1 | u ( t ) | d t Open image in new window. We also use the space A C n [ 0 , 1 ] Open image in new window defined by
A C n [ 0 , 1 ] = { u : [ 0 , 1 ] R u ( n 1 )  are absolutely continuous on  [ 0 , 1 ] } Open image in new window
and the Banach space C μ [ 0 , 1 ] Open image in new window ( μ > 0 Open image in new window)
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 } Open image in new window

with the norm u C μ = D 0 + μ u + + D 0 + μ ( N 1 ) u + u Open image in new window.

Lemma 2.1 [2]

Let α > 0 Open image in new window, n = [ α ] + 1 Open image in new window. Assume that u L 1 ( 0 , 1 ) Open image in new window with a fractional integration of order n α Open image in new window that belongs to A C n [ 0 , 1 ] Open image in new window. 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 Open image in new window

holds almost everywhere on [ 0 , 1 ] Open image in new window.

Remark 2.1 If u satisfies D 0 + α u = f ( t ) L 1 ( 0 , 1 ) Open image in new window and I 0 + 3 α u | t = 0 = 0 Open image in new window, then u C α 1 [ 0 , 1 ] Open image in new window. 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 . Open image in new window
Combine with I 0 + 3 α u | t = 0 = 0 Open image in new window, there is c 3 = 0 Open image in new window. 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 . Open image in new window

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

Lemma 2.2 [2]

Assume α > 0 Open image in new window, then:
  1. (i)
    Let k N Open image in new window. If D a + α u ( t ) Open image in new window and ( D a + α + k u ) ( t ) Open image in new window exist, then
    ( D k D a + α ) u ( t ) = ( D a + α + k u ) ( t ) ; Open image in new window
     
  2. (ii)
    If β > 0 Open image in new window, α + β > 1 Open image in new window, then
    ( I a + α I a + β ) u ( t ) = ( I a + α + β u ) ( t ) Open image in new window
     
is satisfied at any point on [ a , b ] Open image in new window for u L p ( a , b ) Open image in new window and 1 p + Open image in new window;
  1. (iii)

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

     
  2. (iv)
    Note that for λ > 1 Open image in new window, λ α 1 , α 2 , , α n Open image in new window, 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 . Open image in new window
     

Lemma 2.3 [18]

F C μ [ 0 , 1 ] Open image in new window 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 Open image in new window such that for every u F Open image in new window,
u C μ = D 0 + μ u + + D 0 + μ [ μ ] u + u < M Open image in new window
and that ϵ > 0 Open image in new window, N > 0 Open image in new window, for all t 1 , t 2 [ 0 , 1 ] Open image in new window, | t 1 t 2 | < δ Open image in new window, u F Open image in new window, i { 0 , 1 , , [ μ ] } Open image in new window, there hold
| u ( t 1 ) u ( t 2 ) | < ε , | D 0 + μ i u ( t 1 ) D 0 + μ i u ( t 2 ) | < ε , Open image in new window

respectively.

Next, consider the following conditions:

(A1) Φ 1 [ 1 ] Φ 2 [ 1 ] 0 Open image in new window.

(A2) Φ 1 [ 1 ] = 0 Open image in new window, Φ 2 [ 1 ] 0 Open image in new window, Φ 2 [ t ] = 0 Open image in new window.

(A3) Φ 1 [ 1 ] = 0 Open image in new window, Φ 2 [ 1 ] = 0 Open image in new window, Φ 2 [ t ] 0 Open image in new window.

(A4) Φ 1 [ 1 ] 0 Open image in new window, Φ 2 [ 1 ] = 0 Open image in new window, Φ 2 [ t ] = 0 Open image in new window.

(A5) Φ 1 [ 1 ] = 0 Open image in new window, Φ 2 [ 1 ] = 0 Open image in new window, Φ 2 [ t ] = 0 Open image in new window.

We shall prove that: If (A1) holds, then Ker L = { θ } Open image in new window. It is the so-called nonresonance case. If (A2) holds, then Ker L = { a t α 1 : a R } Open image in new window. If (A3) or (A4) holds, then Ker L = { a t α 2 : a R } Open image in new window. If (A5) holds, then Ker L = { a t α 1 + b t α 2 : a , b R } Open image in new window.

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 ] Open image in new window, 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 , Open image in new window

has a fixed point, where ( f u ) ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) Open image in new window.

Proof If u is a solution to T u = u Open image in new window, 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 ] Open image in new window
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 ] . Open image in new window
Considering the linearity of Φ i Open image in new window ( i = 1 , 2 Open image in new window), 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 . Open image in new window

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 . Open image in new window
(2.1)
By I 0 + 3 α u ( t ) | t = 0 = 0 Open image in new window, there is c 3 = 0 Open image in new window, and
D 0 + α 1 u ( t ) = 0 t ( f u ) ( s ) d s + c 1 Γ ( α ) , Open image in new window
(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 ) . Open image in new window
(2.3)
Applying Φ 1 Open image in new window and Φ 2 Open image in new window 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 ] . Open image in new window
Thus,
c 1 = Φ 1 [ 0 t ( f u ) ( s ) d s ] Γ ( α ) Φ 1 [ 1 ] , Open image in new window
(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 ] . Open image in new window
(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 . Open image in new window

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 Open image in new window 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 < + Open image in new window.

     

Let Y, Z be real Banach spaces and L : dom L Y Z Open image in new window be a Fredholm operator of index zero. P : Y Y Open image in new window, Q : Z Z Open image in new window are continuous projectors such that Im P = Ker L Open image in new window, Ker Q = Im L Open image in new window, Y = Ker L Ker P Open image in new window and Z = Im L Im Q Open image in new window. It follows that L | dom L Ker P : dom L Ker P Im L Open image in new window is invertible. We denote the inverse of the mapping by K P Open image in new window (generalized inverse operator of L). If Ω is an open bounded subset of Y such that dom L Ω Open image in new window, the mapping N : Y Z Open image in new window will be called L-compact on Ω ¯ Open image in new window, if Q N ( Ω ¯ ) Open image in new window is bounded and K P ( I Q ) N : Ω ¯ Y Open image in new window 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 ¯ Open image in new window. Assume that the following conditions are satisfied:
  1. (i)

    L x λ N x Open image in new window for every ( x , λ ) [ ( dom L Ker L ) Ω ] × ( 0 , 1 ) Open image in new window.

     
  2. (ii)

    N x Im L Open image in new window for every x Ker L Ω Open image in new window.

     
  3. (iii)

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

     

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

Let Y = C α 1 [ 0 , 1 ] Open image in new window, Z = L 1 [ 0 , 1 ] Open image in new window. Let the linear operator L : Y dom L Z Open image in new window 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 } Open image in new window
be defined by L u = D 0 + α u ( t ) Open image in new window. Let the nonlinear operator N : Y Z Open image in new window be defined by
( N u ) ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) . Open image in new window
Then (1.1), (1.2) can be written as

Now, we give KerL, ImL and some necessary operators at dim Ker L = 1 Open image in new window and dim Ker L = 2 Open image in new window, 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 ] } Open image in new window
and
Im L = { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } . Open image in new window
Proof Let u ( t ) = a t α 1 Open image in new window. Clearly, D 0 + α u ( t ) = 0 Open image in new window and I 0 + 3 α u ( t ) | t = 0 = 0 Open image in new window. Considering (A2), Φ 1 [ D 0 + α 1 u ( t ) ] = a Φ 1 [ Γ ( α ) ] = a Γ ( α ) Φ 1 [ 1 ] = 0 Open image in new window and Φ 2 [ D 0 + α 2 u ( t ) ] = Γ ( α ) Φ 2 [ t ] = 0 Open image in new window. So,
{ u dom L : u = a t α 1 , a R , t [ 0 , 1 ] } Ker L . Open image in new window

If L u = D 0 + α u ( t ) = 0 Open image in new window, then u ( t ) = a t α 1 + b t α 2 + c t α 3 Open image in new window. Considering I 0 + 3 α u ( t ) | t = 0 = 0 Open image in new window and (A2), we can obtain that b = c = 0 Open image in new window. It yields u ( t ) = a t α 1 Open image in new window and Ker L { u dom L : u = a t α 1 , a R , t [ 0 , 1 ] } Open image in new window.

We now show that
Im L = { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } . Open image in new window
If v Im L Open image in new window, then there exists u dom L Open image in new window such that D 0 + α u ( t ) = v ( t ) Open image in new window. Hence,
u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s + a t α 1 + b t α 2 Open image in new window
for some a , b R Open image in new window. 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 . Open image in new window
Therefore
Im L { v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } . Open image in new window
On the other hand, suppose v Z Open image in new window satisfies
Φ 1 [ 0 t v ( s ) d s ] = 0 . Open image in new window
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 ] . Open image in new window
Obviously, D 0 + α u ( t ) = v ( t ) Open image in new window and I 0 + 3 α u ( t ) | t = 0 = 0 Open image in new window. Considering (A2) and the linearity of Φ i Open image in new window ( i = 1 , 2 Open image in new window), we have
Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] = 0 Open image in new window
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 . Open image in new window
It yields
{ v Z : Φ 1 [ 0 t v ( s ) d s ] = 0 } Im L . Open image in new window

The proof is complete. □

Lemma 2.6 If Φ 1 [ t ] 0 Open image in new window, then:
  1. (i)

    L is a Fredholm operator of index zero and dim Ker L = codim Im L = 1 Open image in new window.

     
  2. (ii)
    The linear operator K p : Im L dom L Ker P Open image in new window 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 ] . Open image in new window
     
  3. (iii)

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

     
  4. (iv)

    The linear operator K p : Im L dom L Ker P C α 1 [ 0 , 1 ] Open image in new window is completely continuous.

     
Proof Firstly, we construct the mapping Q : Z Z Open image in new window defined by
Q y = 1 Φ 1 [ t ] Φ 1 [ 0 t y ( s ) d s ] . Open image in new window
(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 , Open image in new window

we get Q : Z Z Open image in new window is a well-defined projector.

Now, it is obvious that Im L = Ker Q Open image in new window. Noting that Q is a linear projector, we have Z = Im Q Ker Q Open image in new window. Hence, Z = Im Q Im L Open image in new window and dim Ker L = codim Im L = Open image in new window1. This means L is a Fredholm mapping of index zero. Taking P : Y Y Open image in new window as
( P u ) ( t ) = D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 , Open image in new window
then the generalized inverse K p : Im L dom L Ker P Open image in new window 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 ] . Open image in new window
In fact, for v Im L Open image in new window, 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 Open image in new window
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 , Open image in new window
which implies that K p Open image in new window is well defined on ImL. Moreover, for v Im L Open image in new window, we have
( L K p ) v ( t ) = D 0 + α I 0 + α v ( t ) = v ( t ) Open image in new window
and for v dom L Ker P Open image in new window, 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 , Open image in new window
v dom L Ker P Open image in new window 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 Open image in new window. 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 ) . Open image in new window
That is, K p = ( L | dom L Ker P ) 1 Open image in new window. 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 ] , Open image in new window
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 . Open image in new window
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 . Open image in new window
Finally, we prove that K p : Im L dom L Ker P C α 1 [ 0 , 1 ] Open image in new window is completely continuous. Let V Im L L 1 [ 0 , 1 ] Open image in new window be a bounded set. From the above discussion, we only need to prove that K p V Open image in new window is equicontinuous on [ 0 , 1 ] Open image in new window. For v V Open image in new window, t 1 , t 2 [ 0 , 1 ] Open image in new window with t 1 < t 2 Open image in new window, 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 Open image in new window
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 | . Open image in new window

Therefore, K p ( V ) Open image in new window is equicontinuous. Thus, the operator K p : Im L dom L Ker P Open image in new window 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 } Open image in new window and
Im L = { v : Φ 1 [ 0 t v ( s ) d s ] = 0 } . Open image in new window
Furthermore, if Φ 1 [ t ] 0 Open image in new window also holds, then L is a Fredholm operator of index zero and dim Ker L = codim Im L = 1 Open image in new window. Here, the projectors P : Y Y Open image in new window, Q : Z Z Open image in new window 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 ] . Open image in new window
The generalized inverse operator of L , K P : Im L dom L Ker P Open image in new window 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 . Open image in new window
Also,
K p v C α 1 Δ 2 v 1 , Open image in new window

where Δ 2 = 2 + 1 Γ ( α ) + ( 1 + 2 Γ ( α ) ) Φ 2 Γ ( α ) | Φ 2 [ t ] | Open image in new window.

Lemma 2.8 If (A4) holds, then Ker L = { a t α 2 : a R } Open image in new window and
Im L = { v : Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } . Open image in new window
Furthermore, if Φ 2 [ t 2 ] 0 Open image in new window also holds, then L is a Fredholm operator of index zero and dim Ker L = codim Im L = 1 Open image in new window. Here, the projectors P : Y Y Open image in new window, Q : Z Z Open image in new window 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 ] . Open image in new window
The generalized inverse operator of L , K P : Im L dom L Ker P Open image in new window 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 . Open image in new window
Also,
K p v C α 1 Δ 3 v 1 , Open image in new window

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

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 ] } Open image in new window
and
Im L = { v Z : Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } . Open image in new window
Proof Let u ( t ) = a t α 1 + b t α 2 Open image in new window. Clearly, D 0 + α u ( t ) = 0 Open image in new window and I 0 + 3 α u ( t ) | t = 0 = 0 Open image in new window. Considering (A5), Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ Γ ( α ) ] = Γ ( α ) Φ 1 [ 1 ] = 0 Open image in new window and Φ 2 [ D 0 + α 2 u ( t ) ] = a Γ ( α ) Φ 2 [ t ] + b Γ ( α 1 ) Φ 2 [ 1 ] = 0 Open image in new window. So,
{ u dom L : u = a t α 1 + b t α 2 , a , b R , t [ 0 , 1 ] } Ker L . Open image in new window
If L u = D 0 + α u ( t ) = 0 Open image in new window, then u ( t ) = a t α 1 + b t α 2 + c t α 3 Open image in new window. Considering D 0 + α u ( t ) = 0 Open image in new window 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 ] } . Open image in new window
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 } . Open image in new window
If v Im L Open image in new window, then there exists u dom L Open image in new window such that D 0 + α u ( t ) = v ( t ) Open image in new window. Hence,
u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s + a t α 1 + b t α 2 Open image in new window
for some a , b R Open image in new window. 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 Open image in new window
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 . Open image in new window
Therefore,
Im L { v Z : Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 } . Open image in new window
On the other hand, suppose v Z Open image in new window satisfies
Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 . Open image in new window
Let
u ( t ) = I 0 + α v ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s . Open image in new window
Obviously, D 0 + α u ( t ) = v ( t ) Open image in new window and I 0 + 3 α u ( t ) | t = 0 = 0 Open image in new window. Considering (A5) and the linearity of Φ i Open image in new window ( i = 1 , 2 Open image in new window), we have
Φ 1 [ D 0 + α 1 u ( t ) ] = Φ 1 [ 0 t v ( s ) d s ] = 0 Open image in new window
and
Φ 2 [ D 0 + α 2 u ( t ) ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 . Open image in new window
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 . Open image in new window

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 Open image in new window, then L is a Fredholm operator of index zero and dim Ker L = codim Im L = 2 Open image in new window. Furthermore, the linear operator K p : Im L dom L Ker P Open image in new window can be defined by
( K p v ) ( t ) = I 0 + α v ( t ) . Open image in new window
Also,
K p v C α 1 ( 2 + 1 Γ ( α ) ) v 1 . Open image in new window
Proof Firstly, we construct the mapping Q : Z Z Open image in new window 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 . Open image in new window
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 ] Open image in new window
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 ] . Open image in new window
We have
Q v = T 1 v + ( T 2 v ) t . Open image in new window
(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 Open image in new window
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 , Open image in new window
we have, for each v Z Open image in new window, 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 . Open image in new window

So, Q : Z Z Open image in new window is a well-defined projector.

Now we will show that Ker Q = Im L Open image in new window. If v Ker Q Open image in new window, from Q v = 0 Open image in new window, we have T 1 v = 0 Open image in new window and T 2 v = 0 Open image in new window. Considering the definitions of T 1 Open image in new window and T 2 Open image in new window, 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 . Open image in new window
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 , Open image in new window

so Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 Open image in new window, which yields v Im L Open image in new window. On the other hand, if v Im L Open image in new window, from Φ 1 [ 0 t v ( s ) d s ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 Open image in new window and the definition of Q, it is obvious that Q v = 0 Open image in new window, thus v Ker Q Open image in new window. Hence, Ker Q = Im L Open image in new window.

For v Z Open image in new window, from v = ( v Q v ) + Q v Open image in new window, v Q v Ker Q = Im L Open image in new window, Q v Im Q Open image in new window, we have Z = Im L + Im Q Open image in new window. And for any v Im L Im Q Open image in new window, from v Im Q Open image in new window, there exist constants a , b R Open image in new window such that v ( t ) = a + b t Open image in new window. From v Im L Open image in new window, 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 . Open image in new window
(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 , Open image in new window
therefore (2.8) has a unique solution a = b = 0 Open image in new window, which implies Im L Im Q = { θ } Open image in new window and Z = Im L Im Q Open image in new window. Since dim Ker L = dim Im Q = codim Im L = 2 Open image in new window, thus L is a Fredholm map of index zero. Let P : Y Y Open image in new window 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 . Open image in new window
Then the generalized inverse K p : Im L dom L Ker P Open image in new window of L can be rewritten
( K p v ) ( t ) = I 0 + α v ( t ) . Open image in new window
In fact, for v Im L Open image in new window, 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 Open image in new window
and
Φ 2 [ D 0 + α 2 ( K p v ) ( t ) ] = Φ 2 [ 0 t ( t s ) v ( s ) d s ] = 0 , Open image in new window
which implies that K p Open image in new window is well defined on ImL. Moreover, for v Im L Open image in new window, we have
( L K p ) v ( t ) = D 0 + α I 0 + α v ( t ) = v ( t ) Open image in new window
and for v dom L Ker P Open image in new window, 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 , Open image in new window
v dom L Ker P Open image in new window 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 Open image in new window. So,
( K p L ) v ( t ) = I 0 + α D 0 + α v ( t ) = v ( t ) . Open image in new window
That is, K p = ( L | dom L Ker P ) 1 Open image in new window. 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 , Open image in new window
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 . Open image in new window
It follows that
K p v C α 1 ( 2 + 1 Γ ( α ) ) v 1 . Open image in new window

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 | ) . Open image in new window
Then FBVP (1.1), (1.2) has a unique solution in C α 1 [ 0 , 1 ] Open image in new window 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 . Open image in new window
Proof We shall prove that T x = x Open image in new window has a unique solution in C α 1 [ 0 , 1 ] Open image in new window. For each u , v C α 1 [ 0 , 1 ] Open image in new window, considering the linearity of Φ i Open image in new window ( i = 1 , 2 Open image in new window), 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 . Open image in new window
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 ] | ) Open image in new window
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 ] | ) . Open image in new window
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 ] | ) . Open image in new window

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

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

(H1) There exist functions α , β , γ , ω L 1 [ 0 , 1 ] Open image in new window such that for all ( x , y , z ) R 3 Open image in new window, t [ 0 , 1 ] Open image in new window,
| f ( t , x , y ) | ω ( t ) + α ( t ) | x | + β ( t ) | y | + γ ( t ) | z | . Open image in new window

(H2) There exists a constant A > 0 Open image in new window such that for u dom L Open image in new window, if | D 0 + α 1 u ( t ) | > A Open image in new window for all t [ 0 , 1 ] Open image in new window, then Φ 1 [ 0 t f ( s , u ( s ) , D 0 + α 1 u ( s ) , D 0 + α 2 u ( s ) ) d s ] 0 Open image in new window.

(H3) There exists a constant B > 0 Open image in new window such that either for each a R : | a | > B Open image in new window,
a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] > 0 Open image in new window
or for each a R : | a | > B Open image in new window,
a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] < 0 . Open image in new window
Then FBVP (1.1), (1.2) has at least one solution in C α 1 [ 0 , 1 ] Open image in new window provided
( 1 Γ ( α ) + 2 + Δ 1 ) ( α 1 + β 1 + γ 1 ) < 1 , Open image in new window

where Δ 1 Open image in new window 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 ] } . Open image in new window
Then, for u Ω 1 Open image in new window, since L u = λ N u Open image in new window, so λ 0 Open image in new window, N u Im L = Ker Q Open image in new window, hence
Φ 1 [ 0 t f ( s , u ( s ) , D 0 + α 1 u ( s ) , D 0 + α 2 u ( s ) ) d s ] = 0 . Open image in new window
Thus, from (H2), there exists t 0 [ 0 , 1 ] Open image in new window such that
| D 0 + α 1 u ( t 0 ) | A . Open image in new window
Now,
D 0 + α 1 u ( t ) = D 0 + α 1 u ( t 0 ) + t 0 t D 0 + α u ( s ) d s , Open image in new window
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 . Open image in new window
(3.1)
Again, for u Ω 1 Open image in new window, u dom L Ker L Open image in new window, then ( I P ) u dom L Ker P Open image in new window and L P u = 0 Open image in new window. 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 . Open image in new window
(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 . Open image in new window
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 Open image in new window
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 ) . Open image in new window
Therefore, Ω 1 Open image in new window is bounded. Let
Ω 2 = { u Ker L : N u Im L } . Open image in new window
For u Ω 2 Open image in new window, there is u Ker L = { u dom L u = a t α 1 , t [ 0 , 1 ] , a R } Open image in new window, and N u Im L Open image in new window, thus
Φ 1 [ 0 t f ( s , a t α 1 , a Γ ( α ) , a Γ ( α ) s ) d s ] = 0 . Open image in new window

From (H2), we get | a | A Γ ( α ) Open image in new window, thus Ω 2 Open image in new window is bounded.

Next, according to the condition (H3), for any a R Open image in new window, if | a | > B Open image in new window, then either
a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] < 0 Open image in new window
(3.3)
or else
a Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] > 0 . Open image in new window
(3.4)
If (3.3) holds, set
Ω 3 = { u Ker L : λ J u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] } , Open image in new window
here Q is given by (2.6) and J : Ker L Im Q Open image in new window is the linear isomorphism given by J ( a t α 1 ) = a Φ 1 [ t ] Open image in new window, a R Open image in new window, t [ 0 , 1 ] Open image in new window. For u = a t α 1 Ω 3 Open image in new window,
λ a = ( 1 λ ) Φ 1 [ 0 t f ( s , a s α 1 , a Γ ( α ) , 0 ) d s ] . Open image in new window
If λ = 1 Open image in new window, then a = 0 Open image in new window. Otherwise, if | a | > B Open image in new window, 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 , Open image in new window

which contradicts λ a 2 0 Open image in new window. Thus, Ω 3 { u Ker L u = a t α 1 , | a | B } Open image in new window 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 ] } , Open image in new window

here J is as above. Similar to the above argument, we can show that Ω 3 Open image in new window 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 Open image in new window. By Lemma 2.6 and standard arguments, we can prove that K P ( I Q ) N : Ω Y Open image in new window is compact, thus N is L-compact on Ω ¯ Open image in new window. Then, by the above argument, we have
  1. (i)

    L u λ N u Open image in new window, for every ( u , λ ) [ ( dom L Ker L ) Ω ] × ( 0 , 1 ) Open image in new window,

     
  2. (ii)

    N u Im L Open image in new window for u Ker L Ω