The Existence of Solutions for a Fractional 2𝑚-Point Boundary Value Problems

Abstract

By using the coincidence degree theory, we consider the following 2𝑚-point boundary value problem for fractional differential equation 𝐷^𝛼0+𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡),𝐷^𝛼−10+𝑢(𝑡),𝐷^𝛼−20+𝑢(𝑡))+𝑒(𝑡), 0<𝑡<1, 𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0=0,𝐷^𝛼−20+𝑢(1)=∑^{𝑚−2𝑖=1}𝑎_𝑖𝐷^𝛼−20+𝑢(𝜉_𝑖),𝑢(1)=∑^{𝑚−2𝑖=1}𝑏_𝑖𝑢(𝜂_𝑖), where 2<𝛼≤3,𝐷^𝛼0+ and 𝐼^𝛼0+ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively. A new result on the existence of solutions for above fractional boundary value problem is obtained.

1. Introduction

Fractional differential equations have been of great interest recently. This is because of the intensive development of the theory of fractional calculus itself as well as its applications. Apart from diverse areas of mathematics, fractional differential equations arise in a variety of different areas such as rheology, fluid flows, electrical networks, viscoelasticity, chemical physics, and many other branches of science (see [1–4] and references cited therein). The research of fractional differential equations on boundary value problems, as one of the focal topics has attained a great deal of attention from many researchers (see [5–13]).

However, there are few papers which consider the boundary value problem at resonance for nonlinear ordinary differential equations of fractional order. In [14], Hu and Liu studied the following BVP of fractional equation at resonance: 𝐷^𝛼0+𝑥𝑥(𝑡()0=)=𝑓𝑥𝑡(,1𝑥),(𝑡),𝑥^𝑥(_𝑡),𝑥^(𝑡),0≤𝑡≤1,(0)=𝑥^(0)=0,(1.1) where 1<𝛼≤2, 𝐷^𝛼0₊ is the standard Caputo fractional derivative.

In [15], Zhang and Bai investigated the nonlinear nonlocal problem 𝐷^𝛼0+𝑢𝑢(0(𝑡))==0𝑓,(𝑡,𝑢𝛽(𝑢𝑡)()𝜂,)=0𝑢<(1𝑡)<,1,(1.2) where 1<𝛼≤2, they consider the case 𝛽𝜂^𝛼−1=1, that is, the resonance case.

In [16], Bai investigated the boundary value problem at resonance 𝐷𝐼^𝛼0+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷^𝛼−10+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,_2−𝛼0+𝑢(𝑡)|_𝑡=0=0,𝐷^𝛼−10+𝑢(1)=^𝑚−2_𝑖=0𝛽_𝑖𝐷^𝛼−10+𝑢𝜂_𝑖(1.3) is considered, where 1<𝛼≤2 is a real number, 𝐷^𝛼0₊and𝐼^𝛼0₊ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and 𝑓∶[0,1]×𝑅²→𝑅 is continuous and 𝑒(𝑡)∈𝐿¹[0,1],𝑚≥2,0<𝜉_𝑖<1,𝛽_𝑖∈𝑅,𝑖=1,2,…,𝑚−2 are given constants such that ∑^{𝑚−2𝑖=1}𝛽_𝑖=1.

In this paper, we study the 2𝑚-point boundary value problem𝐷^𝛼0+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷^𝛼−10+𝑢(𝑡),𝐷^𝛼−20+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.4)𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0=0,𝐷^𝛼−20+𝑢(1)=^𝑚−2_𝑖=1𝑎_𝑖𝐷^𝛼−20+𝑢𝜉_𝑖,𝑢(1)=^𝑚−2_𝑖=1𝑏_𝑖𝑢𝜂_𝑖,(1.5) where 2<𝛼≤3, 𝑚≥2,0<𝜉₁<⋯<𝜉_𝑚<1,0<𝜂₁<⋯<𝜂_𝑚<1,𝑎_𝑖,𝑏_𝑖∈𝑅,𝑓∶[0,1]×𝑅³→𝑅, 𝑓 satisfies Carathéodory conditions, 𝐷^𝛼0₊ and 𝐼^𝛼0₊ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively.

Setting Λ₁=1𝛼(𝛼+1)1−^𝑚−2_𝑖=1𝑎_𝑖𝜉^𝑖𝛼+1,Λ²=1𝛼(𝛼−1)1−^𝑚−2_𝑖=1𝑎_𝑖𝜉^𝛼𝑖Λ₃,=(Γ(𝛼))²Γ(2𝛼)1−^𝑚−2_𝑖=1𝑏_𝑖𝜂^{𝑖2𝛼−1},Λ⁴=Γ(𝛼)Γ(𝛼−1)Γ(2𝛼−1)1−^𝑚−2_𝑖=1𝑏_𝑖𝜂^{𝑖2𝛼−2}.(1.6)

In this paper, we will always suppose that the following conditions hold:(C1): ^𝑚⁻²_𝑖=1𝑎_𝑖𝜉_𝑖=^𝑚−2_𝑖=1𝑎_𝑖=1,^𝑚−2_𝑖=1𝑏_𝑖𝜂^𝑖𝛼−1=^𝑚−2_𝑖=1𝑏_𝑖𝜂^𝑖𝛼−2=1,(1.7)(C2): Λ=Λ₁Λ₄−Λ₂Λ₃≠0.(1.8)

We say that boundary value problem (1.4) and (1.5) is at resonance, if BVP 𝐼𝐷^𝛼0+𝑢(𝑡)=0,_3−𝛼0+𝑢(𝑡)|_𝑡=0=0,D^𝛼−20+𝑢(1)=^𝑚−2_𝑖=1𝑎_𝑖𝐷^𝛼−20+𝑢𝜉_𝑖,𝑢(1)=^𝑚−2_𝑖=1𝑏_𝑖𝑢𝜂_𝑖(1.9) has 𝑢(𝑡)=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2, 𝑎,𝑏∈𝑅 as a nontrivial solution.

The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions for BVP (1.4)-(1.5) under nonlinear growth restriction of 𝑓, basing on the coincidence degree theory due to Mawhin (see [17]).

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

Let 𝑌,𝑍 be real Banach spaces, 𝐿∶dom𝐿⊂𝑌→𝑍 a Fredholm map of index zero,s and 𝑃∶𝑌→𝑌,𝑄∶𝑍→𝑍 continuous projectors such that 𝑌=Ker𝐿⊕Ker𝑃,𝑍=Im𝐿⊕𝑄,Im𝑃=Ker𝐿,Ker𝑄=Im𝐿.(1.10) It follows that 𝐿|_{dom𝐿∩Ker𝑃}∶dom𝐿∩Ker𝑃→Im𝐿 is invertible. We denote the inverse of the map by 𝐾_𝑝. If Ω is an open-bounded subset of 𝑌 such that dom𝐿∩Ω≠∅, the map 𝑁∶𝑌→𝑍 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾_𝑝(𝐼−𝑄)𝑁∶Ω→𝑌 is compact.

The lemma that we used is [17, Theorem 2.4].

Lemma 1.1. Let 𝐿 be a Fredholm operator of index zero and let 𝑁 be L-compact on Ω. Assume that the following conditions are satisfied: (i)𝐿𝑥≠𝜆𝑁𝑥, for all (𝑥,𝜆)∈[dom𝐿⧵Ker𝐿∩𝜕Ω]×[0,1]; (ii)𝑁𝑥∉Im𝐿, for all 𝑥∈Ker𝐿∩𝜕Ω; (iii)deg(𝐽𝑄𝑁|_Ker𝐿,Ker𝐿∩Ω,0)≠0,where 𝑄∶𝑍→𝑍 is a projection as above with Ker𝑄=Im𝐿 and 𝐽∶Im𝑄→Ker𝐿 is any isomorphism. Then the equation 𝐿𝑥=𝑁𝑥 has at least one solution in dom𝐿∩Ω.

2. Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory. These definitions can be found in the recent literature [1–16, 18].

Definition 2.1. The fractional integral of order 𝛼>0 of a function 𝑦∶(0,∞)→𝑅 is given by 𝐼^𝛼0+𝑦(𝑡)=1Γ(𝛼)^𝑡0(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠,(2.1) provided the right side is pointwise defined on (0,∞), where Γ(⋅) is the Gamma function.

Definition 2.2. The fractional derivative of order 𝛼>0 of a function 𝑦∶(0,∞)→𝑅 is given by 𝐷^𝛼0+𝑦(𝑡)=1Γ(𝑛−𝛼)𝑑𝑑𝑡^𝑛^𝑡0𝑦(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, provided the right side is pointwise defined on (0,∞).

Definition 2.3. We say that the map 𝑓∶[0,1]×𝑅^𝑛→𝑅 satisfies Carathéodory conditions with respect to 𝐿¹[0,1] if the following conditions are satisfied: (i)for each 𝑧∈𝑅^𝑛, the mapping 𝑡→𝑓(𝑡,𝑧) is Lebesgue measurable; (ii)for almost every 𝑡∈[0,1], the mapping 𝑡→𝑓(𝑡,𝑧) is continuous on 𝑅^𝑛; (iii) for each 𝑟>0, there exists 𝜌_𝑟∈𝐿¹([0,1],𝑅) such that for a.e. 𝑡∈[0,1] and every |𝑧|≤𝑟, we have 𝑓(𝑡,𝑧)≤𝜌_𝑟(𝑡).

Lemma 2.4 (see [15]). Assume that 𝑢∈𝐶(0,1)∩𝐿¹(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)∩𝐿¹(0,1). Then 𝐼^𝛼0+𝐷^𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₁𝑡^𝛼−2+⋯+𝑐_𝑁𝑡^𝛼−𝑁(2.3) for some 𝑐_𝑖∈𝑅,𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

We use the classical Banach space 𝐶[0,1] with the norm ‖𝑥‖_∞=max_[]𝑡∈0,1||𝑥(𝑡)||,(2.4)𝐿[0,1] with the norm ‖𝑥‖₁=¹⁰||𝑥(𝑡)||𝑑𝑡.(2.5)

Definition 2.5. For 𝑛∈𝑁, we denote by 𝐴𝐶^𝑛[0,1] the space of functions 𝑢(𝑡) which have continuous derivatives up to order 𝑛−1 on [0,1] such that 𝑢^(𝑛−1)(𝑡) is absolutely continuous: 𝐴𝐶^𝑛[0,1]={𝑢∣[0,1]→𝑅 and (𝐷^(𝑛−1))𝑢(𝑡) is absolutely continuous in [0,1]}.

Lemma 2.6 (see [15]). Given 𝜇>0 and 𝑁=[𝜇]+1 we can define a linear space 𝐶^𝜇[0,1]=𝑢(𝑡)∣𝑢(𝑡)=𝐼^𝛼0₊𝑥(𝑡)+𝑐₁𝑡^𝜇−1+𝑐₂𝑡^𝜇−2+⋯+𝑐_𝑁𝑡^{𝜇−(𝑁−1)},𝑡∈[0,1],(2.6) where 𝑥∈[0,1],𝑐_𝑖∈𝑅,𝑖=1,2,…,𝑁−1. By means of the linear functional analysis theory, we can prove that with the ‖𝑢‖_𝐶^𝜇=‖‖𝐷^𝜇0₊𝑢‖‖_∞+⋯+‖‖𝐷^{0𝜇−(𝑁−1)}₊𝑢‖‖_∞+‖𝑢‖_∞,(2.7)𝐶^𝜇[0,1] is a Banach space.

Remark 2.7. If 𝜇 is a natural number, then 𝐶^𝜇[0,1] is in accordance with the classical Banach space 𝐶^𝑛[0,1].

Lemma 2.8 (see [15]). 𝑓⊂𝐶^𝜇[0,1] is a sequentially compact set if and only if 𝑓 is uniformly bounded and equicontinuous. Here uniformly bounded means there exists 𝑀>0, such that for every 𝑢∈𝑓‖𝑢‖_𝐶^𝜇=‖‖𝐷^𝜇0+𝑢‖‖_∞+⋯+‖‖𝐷^{𝜇−(𝑁−1)0+}𝑢‖‖_∞+‖𝑢‖_∞<𝑀,(2.8) and equicontinuous means that ∀𝜀>0, ∃𝛿>0, such that ||𝐷||𝑢𝑡₁−𝑢𝑡₂||<𝜀,∀𝑡₁,𝑡₂∈[0,1],||𝑡₁−𝑡₂||<𝛿,∀𝑢∈𝑓_{𝛼−𝑖0+},𝑢𝑡₁−𝐷^{𝛼−𝑖0+}𝑢𝑡₂||<𝜀,∀𝑡₁,𝑡₂∈[0,1],||𝑡₁−𝑡₂||<𝛿,∀𝑢∈𝑓,∀𝑖=1,2,…,𝑁−1.(2.9)

Lemma 2.9 (see [1]). Let 𝛼>0,𝑛=[𝛼]+1. Assume that 𝑢∈𝐿¹(0,1) with a fractional integration of order 𝑛−𝛼 that belongs to 𝐴𝐶^𝑛[0,1]. Then the equality 𝐼^𝛼0+𝐷^𝛼0+𝑢(𝑡)=𝑢(𝑡)−^𝑛_𝑖=1𝐼^{𝑛−𝛼0+}𝑢(𝑡)^𝑛−𝑖|_𝑡=0Γ(𝛼−𝑖+1)𝑡^𝛼−𝑖(2.10) holds almost everywhere on [0,1].

Definition 2.10 (see [16]). Let 𝐼^𝛼0+(𝐿¹(0,1)),𝛼>0 denote the space of functions 𝑢(𝑡), represented by fractional integral of order 𝛼 of a summable function: 𝑢=𝐼^𝛼0+𝑣,𝑣∈𝐿¹(0,1).

Let 𝑍=𝐿¹[0,1], with the norm ‖𝑦‖=∫¹⁰|𝑦(𝑠)|𝑑𝑠, 𝑌=𝐶^𝛼−1[0,1] defined by Lemma 2.6, with the norm ‖𝑢‖_𝐶^𝛼−1=‖𝐷^0𝛼−1₊𝑢‖_∞+‖𝐷^0𝛼−2₊𝑢‖_∞+‖𝑢‖_∞, where 𝑌 is a Banach space.

Define 𝐿 to be the linear operator from dom𝐿⊂𝑌 to 𝑍 with dom𝐿=𝑢∈𝐶^𝛼−1[0,1]∣𝐷^𝛼0+𝑢∈𝐿¹[0,1],𝑢satisfies(1.5),(2.11)𝐿𝑢=𝐷^𝛼0₊𝑢,𝑢∈dom𝐿,(2.12) we define 𝑁∶𝑌→𝑍 by setting 𝑁𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷^𝛼−10+𝑢(𝑡)𝐷^𝛼−20+𝑢(𝑡)+𝑒(𝑡).(2.13) Then boundary value problem (1.4) and (1.5) can be written as 𝐿𝑢=𝑁𝑢.

3. Main Results

Lemma 3.1. Let L be defined by (2.12), then Ker𝐿=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2∣𝑎,𝑏∈𝑅≅𝑅²,Im𝐿=𝑦∈𝑍∣¹⁰(1−𝑠)𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑦(𝑠)𝑑𝑠=0,₁₀(1−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1b_𝑖^𝜂𝑖₀𝜂_𝑖−𝑠^𝛼−1𝑦(𝑠)𝑑𝑠=0.(3.1)

Proof. In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that 𝐼^𝛼0+=𝐷^−𝛼0+ for 𝛼<0.
Let 𝐿𝑢=𝐷^𝛼0₊𝑢, by Lemma 2.9, 𝐷^𝛼0+𝑢(𝑡)=0 has solution 𝑢(𝑡)=³_𝑖=1𝐼^3−𝛼0+𝑢(𝑡)^3−𝑖|_𝑡=0Γ(𝛼−𝑖+1)𝑡^𝛼−𝑖=𝐼^3−𝛼0+𝑢(𝑡)^|_𝑡=0Γ(𝛼)𝑡^𝛼−1+𝐼^3−𝛼0+𝑢(𝑡)^|_𝑡=0Γ(𝛼−1)𝑡^𝛼−2+𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0Γ(𝛼−2)𝑡^𝛼−3=𝐷^𝛼−10+𝑢(𝑡)|_𝑡=0Γ(𝛼)𝑡^𝛼−1+𝐷^𝛼−20+𝑢(𝑡)|_𝑡=0Γ(𝛼−1)𝑡^𝛼−2+𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0Γ(𝛼−2)𝑡^𝛼−3.(3.2) Combine with (1.5), So, Ker𝐿=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2∣𝑎,𝑏∈𝑅≅𝑅².(3.3) Let 𝑦∈𝑍 and let 𝑢_𝑡=1Γ(𝛼)^𝑡0(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+𝑐₃𝑡^𝛼−3.(3.4)
Then 𝐷^𝛼0+𝑢(𝑡)=𝑦(𝑡) a.e. 𝑡∈[0,1] and, if ¹⁰(1−𝑠)𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑦(𝑠)𝑑𝑠=0,₁₀(1−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜂𝑖₀𝜂_𝑖−𝑠^𝛼−1𝑦(𝑠)𝑑𝑠=0(3.5) hold, then 𝑢(𝑡) satisfies the boundary conditions (1.5). That is, 𝑢∈dom𝐿 and we have {𝑦∈𝑍∣𝑦satisfies(3.4)}⊆Im𝐿.(3.6)
Let 𝑢∈dom𝐿. Then for 𝐷^𝛼0+𝑢∈Im𝐿, we have 𝐼^𝛼0+𝑦(𝑡)=𝑢(𝑡)−𝑐₁𝑡^𝛼−1−𝑐₂𝑡^𝛼−2−𝑐₃𝑡^𝛼−3,(3.7) where 𝑐₁=𝐷^𝛼−10+𝑢(𝑡)|_𝑡=0Γ(𝛼),𝑐²=𝐷^𝛼−20+𝑢(𝑡)|_𝑡=0Γ(𝛼−1),c₃=𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0Γ(𝛼−2),(3.8) which, due to the boundary value condition (1.5), implies that satisfies (3.5). In fact, from 𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0=0 we have 𝑐₃=0, from 𝐷^𝛼−20+𝑢(1)=∑^{𝑚−2𝑖=1}𝑎_𝑖𝐷^𝛼−20+𝑢(𝜉_𝑖), 𝑢(1)=∑^{𝑚−2𝑖=1}𝑏_𝑖𝑢(𝜂_𝑖), we have ¹⁰(1−𝑠)𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑦(𝑠)𝑑𝑠=0,₁₀(1−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜂𝑖₀𝜂_𝑖−𝑠^𝛼−1𝑦(𝑠)𝑑𝑠=0.(3.9) Hence, {𝑦∈𝑍∣𝑦satisfies(3.4)}⊇Im𝐿.(3.10) Therefore, {𝑦∈𝑍∣𝑦satisfies(3.4)}=Im𝐿.(3.11) The proof is complete.

Lemma 3.2. The mapping 𝐿∶dom𝐿⊂𝑌→𝑍 is a Fredholm operator of index zero, and 𝑄𝑦(𝑡)=𝑇₁𝑦(𝑡)𝑡^𝛼−1+𝑇₂𝑦(𝑡)𝑡^𝛼−2,(3.12) where 𝑇₁𝑦=1ΛΛ⁴𝑄₁𝑦−Λ₂𝑄₂𝑦,𝑇₂𝑦=1ΛΛ³𝑄₁𝑦−Λ₁𝑄₂𝑦,(3.13) define by 𝐾_𝑝∶Im𝐿→dom𝐿∩Ker𝑃 by 𝐾_𝑝𝑦(𝑡)=1Γ(𝛼)^𝑡0(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠=𝐼^𝛼0+𝑦(𝑡),𝑦∈Im𝐿,(3.14) and forall𝑦∈Im𝐿,‖𝐾_𝑝𝑦‖_𝐶^𝛼−1≤((1/Γ(𝛼))+2)‖𝑦‖₁.

Proof. Consider the continuous linear mapping 𝑄₁∶𝑍→𝑍 and 𝑄₂∶𝑍→𝑍 defined by 𝑄₁𝑦=¹⁰(1−𝑠)𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖₀𝜉_𝑖𝑄−𝑠𝑦(𝑠)𝑑𝑠,₂𝑦=¹⁰(1−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜂𝑖₀𝜂_𝑖−𝑠^𝛼−1𝑦(𝑠)𝑑𝑠.(3.15) Using the above definitions, we construct the following auxiliary maps 𝑇₁∶𝑍→𝑍 and 𝑇₂∶𝑍→𝑍: 𝑇₁𝑦=1ΛΛ⁴𝑇₂𝑄₁𝑦−Λ₂𝑄₂𝑦,𝑦=1ΛΛ³𝑄₁𝑦−Λ₁Q₂𝑦.(3.16) Since the condition (C2) holds, the mapping defined by 𝑄𝑦(𝑡)=𝑇₁𝑦(𝑡)𝑡^𝛼−1+𝑇₂𝑦(𝑡)𝑡^𝛼−2(3.17) is well defined.
Recall (C2) and note that 𝑇₁𝑇₁𝑦𝑡^𝛼−1=1ΛΛ⁴=1𝑄₁𝑇₁𝑦𝑡^𝛼−1−Λ₂𝑄₂𝑇₁𝑦𝑡^𝛼−1ΛΛ⁴Λ⁴Λ₁Λ𝑄¹𝑦−Λ¹Λ₂Λ𝑄²𝑦−Λ₂Λ⁴Λ₃Λ𝑄¹𝑦−Λ²Λ₃Λ𝑄²=𝑇₁𝑦𝑦,(3.18) and similarly we can derive that 𝑇₁𝑇₂𝑦𝑡^𝛼−2=0,𝑇₂𝑇₁𝑦𝑡^𝛼−1=0,𝑇₂𝑇₂𝑦𝑡^𝛼−2=𝑇₂𝑦.(3.19) So, for 𝑦∈𝑍, it follows from the four relations above that 𝑄²𝑦==𝑄𝑇₁𝑇₁𝑦𝑡^𝛼−1+𝑇₂𝑦𝑡^𝛼−2𝑇₁𝑦𝑡^𝛼−1+𝑇₂𝑦𝑡^𝛼−2𝑡^𝛼−1+𝑇₂𝑇₁𝑦𝑡^𝛼−1+𝑇₂𝑦𝑡^𝛼−2𝑡^𝛼−2=𝑇₁𝑦𝑡^𝛼−1+𝑇₂𝑦𝑡^𝛼−2=𝑄𝑦,(3.20) that is, the map 𝑄 is idempotent. In fact 𝑄 is a continuous linear projector.
Note that 𝑦∈Im𝐿 implies 𝑄𝑦=0. Conversely, if 𝑄𝑦=0, so ΛΛ₄𝑄₁𝑦−Λ₂𝑄₂𝑦=0,₁𝑄₂𝑦−Λ₃𝑄₁𝑦=0,(3.21) but ||||||Λ₄−Λ₂−Λ₃Λ₁||||||=Λ₄Λ₁−Λ₂Λ₃≠0,(3.22) then we must have 𝑄₁𝑦=𝑄₂𝑦=0; since the condition (C2) holds, this can only be the case if 𝑄₁𝑦=𝑄₂𝑦=0, that is, 𝑦∈Im𝐿. In fact Ker𝑄=Im𝐿, take 𝑦∈𝑍 in the form 𝑦=(𝑦−𝑄𝑦)+𝑄𝑦 so that 𝑦−𝑄𝑦∈Ker𝑄=Im𝐿,𝑄𝑦∈Im𝑄, thus, 𝑍=Im𝐿+Im𝑄, Let 𝑦∈Im𝐿∩Im𝑄 and assume that 𝑦=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2 is not identically zero on [0,1]. Then, since 𝑦∈Im𝐿, from (3.5) and the condition (C2), we have 𝑄₁𝑦=¹⁰(1−𝑠)𝑎𝑠^𝛼−1+𝑏𝑠^𝛼−2𝑑𝑠−^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖₀𝜉_𝑖𝑄−𝑠𝑎𝑠^𝛼−1+𝑏𝑠^𝛼−2𝑑𝑠=0,₂𝑦=¹⁰(1−𝑠)^𝛼−1𝑎𝑠^𝛼−1+𝑏𝑠^𝛼−2𝑑𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜂𝑖₀𝜂_𝑖−𝑠^𝛼−1𝑎𝑠^𝛼−1+𝑏𝑠^𝛼−2𝑑𝑠=0.(3.23) So 𝑎𝑎ΛΛ₁₃+𝑏Λ₂=0,+𝑏Λ₄=0,(3.24) but ||||||Λ₁Λ₂Λ₃Λ₄||||||=Λ₁Λ₄−Λ₂Λ₃≠0,(3.25) we derive 𝑎=𝑏=0, which is a contradiction. Hence, Im𝐿∩Im𝑄={0}; thus 𝑍=Im𝐿⊕Im𝑄.
Now, dimKer𝐿=2=codimIm𝐿 and so 𝐿 is a Fredholm operator of index zero.
Let 𝑃∶𝑌→𝑌 be defined by 𝑃𝑢(𝑡)=1Γ(𝛼)𝐷^𝛼−10+𝑢(𝑡)|_𝑡=0𝑡^𝛼−1+1Γ(𝛼−1)𝐷^𝛼−20+𝑢(𝑡)|_𝑡=0𝑡^𝛼−2,𝑡∈[0,1].(3.26) Note that 𝑃 is a continuous linear projector and Ker𝑃=𝑢∈𝑌∣𝐷^𝛼−10+𝑢(0)=𝐷^𝛼−20+𝑢(0)=0.(3.27) It is clear that 𝑌=Ker𝑃⊕Ker𝐿.
Note that the projectors 𝑃 and 𝑄 are exact. Define by 𝐾_𝑝∶Im𝐿→dom𝐿∩Ker𝑃 by 𝐾_𝑝𝑦(𝑡)=1Γ(𝛼)^𝑡0(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠=𝐼^𝛼0+𝑦(𝑡),𝑦∈Im𝐿.(3.28) Hence we have 𝐷^𝛼−10+𝐾_𝑝𝑦𝑡=^𝑡0𝑦(𝑠)𝑑𝑠,𝐷^𝛼−20+𝐾_𝑝𝑦𝑡=^𝑡0(𝑡−𝑠)𝑦(𝑠)𝑑𝑠,(3.29) then ‖‖𝐾_𝑝𝑦‖‖_∞≤1Γ(𝛼)‖𝑦‖¹,‖‖𝐷^𝛼−10+𝐾_𝑝𝑦‖‖_∞≤‖𝑦‖₁,‖‖𝐷^𝛼−20+𝐾_𝑝𝑦‖‖_∞≤‖𝑦‖₁,(3.30) and thus ‖‖𝐾_𝑝𝑦‖‖_𝐶𝛼−1≤1Γ(𝛼)+2‖𝑦‖¹.(3.31)
In fact, if 𝑦∈Im𝐿, then 𝐿𝐾_𝑝𝑦(𝑡)=𝐷^𝛼0+𝐼^𝛼0+𝑦(𝑡)=𝑦(𝑡).(3.32) Also, if 𝑢∈dom𝐿∩Ker𝑃, then 𝐾_𝑝𝐿𝑢(𝑡)=𝐼^𝛼0+𝐷^𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+𝑐₃𝑡^𝛼−3,(3.33) where 𝑐₁=𝐷^𝛼−10+𝑢(𝑡)|_𝑡=0Γ(𝛼),𝑐²=𝐷^𝛼−20+𝑢(𝑡)|_𝑡=0Γ(𝛼−1),𝑐³=𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0Γ(𝛼−2),(3.34) and from the boundary value condition (1.5) and the fact that 𝑢∈dom𝐿∩Ker𝑃, 𝑃𝑢=0, 𝐷^𝛼−10+𝑢(𝑡)|_𝑡=0=𝐷^𝛼−20+𝑢(𝑡)|_𝑡=0=𝐼^3−𝛼0+𝑢(𝑡)|_𝑡=0=0, we have 𝑐₁=𝑐₂=𝑐₃=0, thus 𝐾_𝑝𝐿𝑢(𝑡)=𝑢(𝑡).(3.35) This shows that 𝐾_𝑝=[𝐿|_{dom𝐿∩Ker𝑃}]⁻¹. The proof is complete. Using (3.16), we write 𝐾_𝑝𝑄𝑁𝑢(𝑡)=𝑇₁𝑁𝑢𝑡^𝛼−1+𝑇₂𝑁𝑢𝑡^𝛼−2,(𝐼−𝑄)𝑁𝑢(𝑡)=1Γ(𝛼)^𝑡0(𝑡−𝑠)^𝛼−1[𝑁𝑢(𝑠)−𝑄𝑁𝑢(𝑠)]𝑑𝑠.(3.36)

By Lemma 2.8 and a standard method, we obtain the following lemma.

Lemma 3.3 (see [16]). For every given 𝑒∈𝐿¹[0,1],𝐾_𝑝(𝐼−𝑄)𝑁∶𝑌→𝑌 is completely continuous.

Assume that the following conditions on the function 𝑓(𝑡,𝑥,𝑦,𝑧) are satisfied.(H1)There exist functions 𝑎(𝑡),𝑏(𝑡),𝑐(𝑡),𝑑(𝑡),𝑟(𝑡)∈𝐿¹[0,1], and a constant 𝜃∈[0,1) such that for all (𝑥,𝑦,𝑧)∈𝑅³, 𝑡∈[0,1], one of the following inequalities is satisfied:||𝑓(𝑡,𝑥,𝑦,𝑧)||≤𝑎(𝑡)|𝑥|+𝑏(𝑡)||𝑦||+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑥|^𝜃+𝑟(𝑡),(3.37)||𝑓(𝑡,𝑥,𝑦,𝑧)||≤𝑎(𝑡)|𝑥|+𝑏(𝑡)||𝑦||+𝑐(𝑡)|𝑧|+𝑑(𝑡)||𝑦||^𝜃+𝑟(𝑡),(3.38)||𝑓(𝑡,𝑥,𝑦,𝑧)||≤𝑎(𝑡)|𝑥|+𝑏(𝑡)||𝑦||+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑧|^𝜃+𝑟(𝑡).(3.39)(H2)There exists a constant 𝐴>0, such that for 𝑥∈dom𝐿⧵Ker𝐿 satisfying |𝐷^𝛼−10+𝑥(𝑡)|+|𝐷^𝛼−20+𝑥(𝑡)|>𝐴 for all 𝑡∈[0,1], we have 𝑄₁𝑁𝑥(𝑡)≠0,or𝑄₂𝑁𝑥(𝑡)≠0.(3.40)(H3)There exists a constant 𝐵>0 such that for every 𝑎,𝑏∈𝑅 satisfying 𝑎²+𝑏²>𝐵 then either 𝑎𝑇₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2+𝑏𝑇₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2<0,(3.41)

or 𝑎𝑇₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2+𝑏𝑇₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2>0.(3.42)

Remark 3.4. 𝑇₁𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2) and 𝑇₂𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2) from (H3) stand for the images of 𝑢(𝑡)=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2 under the maps 𝑇₁𝑁 and 𝑇₂𝑁, respectively.

Lemma 3.5. Suppose (H1)-(H2) hold, then the set Ω₁={𝑥∈dom𝐿⧵Ker𝐿∶𝐿𝑥=𝜆𝑁𝑥,𝜆∈[0,1]}(3.43) is bounded.

Proof. Take Ω₁={𝑥∈dom𝐿⧵Ker𝐿∶𝐿𝑥=𝜆𝑁𝑥,𝜆∈[0,1]}.(3.44) Then for 𝑥∈Ω₁, 𝐿𝑥=𝜆𝑁𝑥 thus 𝜆≠0,𝑁𝑥∈Im𝐿=Ker𝑄, and hence 𝑄𝑁𝑥(𝑡) for all 𝑡∈[0,1]. By the definition of 𝑄, we have 𝑄₁𝑁𝑥(𝑡)=𝑄₂𝑁𝑥(𝑡)=0. It follows from (H2) that there exists 𝑡₀∈[0,1], such that |𝐷^𝛼−10+𝑢(𝑡₀)|+|𝐷^𝛼−20+𝑢(𝑡₀)|≤𝐴.
Now 𝐷𝐷^𝛼−10+𝑥(𝑡)=𝐷^𝛼−10+𝑥𝑡₀+^𝑡𝑡₀𝐷^𝛼0+𝑥(𝑠)𝑑𝑠,_𝛼−20+𝑥(𝑡)=𝐷^𝛼−20+𝑥𝑡₀+^𝑡𝑡₀𝐷^𝛼−10+𝑥(𝑠)𝑑𝑠,(3.45) and so ||𝐷^𝛼−10+𝑥(0)||≤‖‖𝐷^𝛼−10+𝑥(𝑡)‖‖_∞≤||𝐷^𝛼−10+𝑥𝑡₀||+‖‖𝐷^𝛼−10+𝑥‖‖₁||𝐷_𝛼−20+≤𝐴+‖𝐿𝑥‖₁≤𝐴+‖𝑁𝑥‖₁,𝑥(0)||≤‖‖𝐷^𝛼−20+𝑥(𝑡)‖‖_∞≤||𝐷^𝛼−20+𝑥𝑡₀||+‖‖𝐷^𝛼−10+𝑥‖‖_∞≤||𝐷^𝛼−20+𝑥𝑡₀||+||𝐷^𝛼−10+𝑥𝑡₀||+‖‖𝐷^𝛼0+𝑥‖‖₁≤𝐴+‖𝐿𝑥‖₁≤𝐴+‖𝑁𝑥‖₁.(3.46) Therefore, we have ‖𝑃𝑥‖_𝐶^𝛼−1=‖‖‖1Γ(𝛼)𝐷^𝛼−10+𝑥(0)𝑡^𝛼−1+1Γ(𝛼−1)𝐷^𝛼−20+𝑥(0)𝑡^𝛼−2‖‖‖_𝐶𝛼−1=‖‖‖1Γ(𝛼)𝐷^𝛼−10+𝑥(0)𝑡^𝛼−1+1Γ(𝛼−1)𝐷^𝛼−20+𝑥(0)𝑡^𝛼−2‖‖‖_∞+‖‖𝐷^𝛼−10+𝑥(0)‖‖_∞+‖‖𝐷^𝛼−10+𝑥(0)𝑡+𝐷^𝛼−20+𝑥(0)‖‖_∞≤2+1Γ(𝛼)||𝐷^𝛼−10+𝑥(0)||+1+1Γ(𝛼−1)||𝐷^𝛼−20+≤2+1𝑥(0)||Γ(𝛼)𝐴+‖𝑁𝑥‖¹+1+Γ(𝛼1−1)𝐴+‖𝑁𝑥‖¹.(3.47) Note that (𝐼−𝑃)𝑥∈dom𝐿∩Ker𝑃 for all 𝑥∈Ω₁. Then, by Lemma 3.2, we have ‖(𝐼−𝑃)𝑥‖_𝐶^𝛼−1=‖‖𝐾_𝑝𝐿(𝐼−𝑃)𝑥‖‖_𝐶𝛼−1≤2+1Γ(𝛼)‖𝑁𝑥‖¹,(3.48) so, we have ‖𝑥‖_𝐶^𝛼−1≤‖(𝐼−𝑃)𝑥‖_𝐶^𝛼−1+‖𝑃𝑥‖_𝐶^𝛼−1≤2+1+2+Γ(𝛼1)𝐴+‖𝑁𝑥‖¹+1+Γ(𝛼1−1)𝐴+‖𝑁𝑥‖¹Γ(𝛼)‖𝑁𝑥‖¹=2Γ(𝛼)+Γ(𝛼1−1)+5‖𝑁𝑥‖¹+1≤𝑚‖𝑁Γ𝑥(𝛼‖)₁+Γ(𝛼1−1)+Γ(𝛼1−2)+3𝐴+𝑛𝐴,(3.49) where 𝑚=((2/Γ(𝛼))+(1/Γ(𝛼−1))+5),𝑛=((1/Γ(𝛼))+(1/Γ(𝛼−1))+(1/(Γ(𝛼−2))+3),𝐴 is a constant. This is for all 𝑥∈Ω₁. If the first condition of (H1) is satisfied, then, we have ‖𝑥‖_𝐶^𝛼−1=≤m𝑚ax‖𝑎‖‖₁𝑥‖_∞,‖‖𝐷^𝛼−10+𝑥‖‖_∞,‖‖𝐷^𝛼−20+𝑥‖‖_∞‖𝑥‖_∞+‖𝑏‖₁‖‖𝐷^𝛼−10+𝑥‖‖_∞+‖𝑐‖₁‖‖𝐷^𝛼−20+𝑥‖‖_∞+‖𝑑‖₁‖‖𝐷^𝛼−20+𝑥‖‖^𝜃∞+𝐷,(3.50) where 𝐷=‖𝑟‖₁+‖𝑒‖₁+𝑛/𝑚, and consequently, for ‖𝑥‖_∞≤‖𝑥‖_𝐶^𝛼−1,‖‖𝐷^𝛼−10+𝑥‖‖_∞≤‖𝑥‖_𝐶^𝛼−1,‖‖𝐷^𝛼−20+𝑥‖‖_∞≤‖𝑥‖_𝐶^𝛼−1,(3.51) so ‖𝑥‖_∞≤𝑚1−𝑚‖𝑎‖₁‖𝑏‖¹‖‖𝐷‖‖𝐷^𝛼−10+𝑥‖‖_∞+‖𝑐‖₁‖‖𝐷^𝛼−20+𝑥‖‖_∞+‖𝑑‖₁‖‖𝐷^𝛼−20+𝑥‖‖^𝜃∞+𝐷_𝛼−10+,𝑥‖‖_∞≤𝑚1−𝑚‖𝑎‖₁−𝑚‖𝑏‖₁‖𝑐‖¹‖‖𝐷‖‖𝐷^𝛼−20+𝑥‖‖_∞+‖𝑑‖₁‖‖𝐷^𝛼−20+𝑥‖‖^𝜃∞+𝐷_𝛼−20+,𝑥‖‖_∞≤𝑚1−𝑚‖𝑎‖₁−𝑚‖𝑏‖₁−𝑚‖𝑐‖₁‖𝑑‖¹‖‖𝐷^𝛼−20+𝑥‖‖^𝜃∞+𝐷.(3.52) But 𝜃∈[0,1) and ‖𝑎‖₁+‖𝑏‖₁+‖𝑐‖₁≤1/𝑚, so there exists 𝐴₁,𝐴₂,𝐴₃>0 such that ‖‖𝐷^𝛼−20+𝑥‖‖_∞≤𝐴₁,‖‖𝐷^𝛼−10+𝑥‖‖_∞≤𝐴₂,‖𝑥‖_∞≤𝐴₃.(3.53) Therefore, for all 𝑥∈Ω₁, ‖𝑥‖_𝐶^𝛼−1=max‖𝑥‖_∞,‖‖𝐷^𝛼−10+𝑥‖‖_∞,‖‖𝐷^𝛼−20+𝑥‖‖_∞≤max𝐴₁,𝐴₂,𝐴₃,(3.54) we can prove that Ω₁ is also bounded.
If (3.38) or (3.39) holds, similar to the above argument, we can prove that Ω₁ is bounded too.

Lemma 3.6. Suppose (H3) holds, then the set Ω₂={𝑥∈Ker𝐿∶𝑁𝑥∈Im𝐿}(3.55) is bounded.

Proof. Let Ω₂={𝑥∈Ker𝐿∶𝑁𝑥∈Im𝐿},(3.56) for 𝑥∈Ω₂,𝑥∈Ker𝐿={𝑥∈dom𝐿∶𝑥=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2,𝑎,𝑏∈𝑅,𝑡∈[0,1]} and 𝑄𝑁𝑥(𝑡)=0; thus 𝑇₁𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2)=𝑇₂𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2)=0. By (H3), 𝑎²+𝑏²≤𝐵, that is, Ω₂ is bounded.

Lemma 3.7. Suppose (H3) holds, then the set Ω₃={𝑥∈Ker𝐿∶−𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈[0,1]}(3.57) is bounded.

Proof. We define the isomorphism 𝐽∶Ker𝐿→Im𝑄 by 𝐽𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2.(3.58)
If the first part of (H3) is satisfied, let Ω₃={𝑥∈Ker𝐿∶−𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈[0,1]}.(3.59) For every 𝑥=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2∈Ω₃, 𝜆𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2=(1−𝜆)𝑇₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2𝑡^𝛼−1+𝑇₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2𝑡^𝛼−1.(3.60) If 𝜆=1, then 𝑎=𝑏=0, and if 𝑎²+𝑏²>𝐵, then by (H3) 𝜆𝑎²+𝑏²=(1−𝜆)𝑎𝑇₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2+𝑏𝑇₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2<0,(3.61) which, in either case, is a contradiction. If the other part of (H3) is satisfied, then we take Ω₃={𝑥∈Ker𝐿∶𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈[0,1]},(3.62) and, again, obtain a contradiction. Thus, in either case ‖𝑥‖_𝐶^𝛼−1=‖‖𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2‖‖_𝐶𝛼−1=‖‖𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2‖‖_∞+‖𝑎Γ(𝛼)‖_∞+‖𝑎Γ(𝛼)𝑡+𝑏Γ(𝛼−1)‖_∞≤≤((21++22ΓΓ((𝛼𝛼)))+|𝑎Γ|+(𝛼(1−+1)Γ)𝐵(𝛼,−1))||𝑏||(3.63) for all 𝑥∈Ω₃, that is, Ω₃ is bounded.

Remark 3.8. Suppose the second part of (H3) holds, then the set Ω^3={𝑥∈Ker𝐿∶𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈[0,1]}(3.64) is bounded.

Theorem 3.9. If (C1)-(C2) and (H1)–(H3) hold, then the boundary value problem (1.4)-(1.5) has at least one solution.

Proof. Set Ω to be a bounded open set of 𝑌 such that ∪^3𝑖=1Ω⊂Ω. It follows from Lemmas 3.2 and 3.3 that 𝐿 is a Fredholm operator of index zero, and the operator 𝐾_𝑝(𝐼−𝑄)𝑁∶Ω→𝑌 is compact 𝑁, thus, is L-compact on Ω. By Lemmas 3.5 and 3.6, we get that the following two conditions are satisfied: (i)𝐿𝑥≠𝜆𝑁𝑥 for every (𝑥,𝜆)∈[dom𝐿⧵Ker𝐿∩𝜕Ω]×[0,1]; (ii)𝑁𝑥∉Im𝐿, for every 𝑥∈Ker𝐿∩𝜕Ω.
Finally, we will prove that (iii) of Lemma 1.1 is satisfied. Let 𝐻(𝑥,𝜆)=±𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥, where 𝐼 is the identity operator in the Banach space 𝑌. According to Lemma 3.7 (or Remark 3.8), we know that 𝐻(𝑥,𝜆)≠0, for all 𝑥∈𝜕Ω∩Ker𝐿, and thus, by the homotopy property of degree, deg𝑄𝑁|_Ker𝐿,Ker𝐿∩Ω,0====dddsgeeenggg⎡⎢⎢⎢⎣(((𝐻±𝐻±𝐼||||||((,⋅⋅ΛK,,10₄e))r,,𝐿KK∩eerrΩ𝐿𝐿,∩∩0)ΩΩ,,00))−ΛΛ₃−ΛΛ²ΛΛΛ¹||||||⎤⎥⎥⎥⎦=sgn±Λ¹Λ₄−Λ₂Λ₃=±1≠0.Λ(3.65) Then by Lemma 1.1, 𝐿𝑥=𝑁𝑥 has at least one solution in dom𝐿∩Ω, so the boundary value problem (1.4) and (1.5) has at least one solution in the space 𝐶^𝛼−1[0,1]. The proof is finished.

4. An Example

Let us consider the following boundary value problem:𝐼𝐷^5/20+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷^𝛼−10+𝑢(𝑡),𝐷^𝛼−20+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,_3−𝛼0+𝑢(𝑡)|_𝑡=0=0,𝐷^1/20+𝑢(1)=2𝐷^1/20+𝑢23−𝐷^1/20+𝑢13,𝑢(1)=654𝑢14−851𝑢19,(4.1) where 𝑓𝑡,𝑢(𝑡),𝐷^𝛼−10+𝑢(𝑡),𝐷^𝛼−20+𝑢(𝑡)=140sin(𝑢(𝑡))+210𝐷^3/20+𝑢(𝑡)+120𝐷^1/20++5cos𝐷_1/20+𝑢(𝑡)𝑢(𝑡)^1/5.(4.2) Corresponding to the problem (1.4)-(1.5), we have that 𝑒(𝑡)=1+3sin²𝑡, 𝛼=5/2, 𝑎₁=−1, 𝑎₂=2, 𝜉₁=1/3, 𝜉₂=2/3, 𝑏₁=64/5, 𝑏₂=−81/5, 𝜂₁=1/4, 𝜂₂=1/9 and 𝑓(𝑡,𝑥,𝑦,𝑧)=140sin𝑥+210𝑦+210𝑧+5cos(𝑧)^1/5,(4.3) then there is 𝑏𝑎₁+𝑎₂=1,𝑎₁𝜉₁+𝑎₂𝜉₂=1,₁Λ𝜂^13/2+𝑏₂𝜂^23/2=1,𝑏₁𝜂^11/2+𝑏₂𝜂^21/2=1,₁=4351−²_𝑖=1𝑎_𝑖𝜉^𝑖7/2,Λ²=4151−²_𝑖=1𝑎_𝑖𝜉^𝑖5/2Λ₃,=(Γ(5/2))²241−²_𝑖=1𝑏_𝑖𝜂^4𝑖,Λ⁴=16Γ(5/22)4Γ(3/2)1−²_𝑖=1𝑏_𝑖𝜂^3𝑖Λ=Λ₁,||𝑓(𝑡,𝑥,𝑦,𝑧)||≤1Λ₄−Λ₂Λ₃≠0,40|𝑥|+210||𝑦||+210|𝑧|+5|𝑧|^1/5.(4.4) Again, taking 𝑎=1/40, 𝑏=𝑐=1/20, then 1‖𝑎‖₁+‖𝑏‖₁+‖𝑐‖₁=1/8,𝑚=(2/Γ(𝛼))+(11/Γ(𝛼−1))+5≈0.131,(4.5) therefore ‖𝑎‖₁+‖𝑏‖₁+‖𝑐‖₁<1𝑚.(4.6) Take 𝐴=181, 𝐵=81. By simple calculation, we can get that (C1)-(C2) and (H1)–(H3) hold. By Lemma 1.1, we obtain that (4.1) has at least one solution.

Acknowledgments

This work is sponsored by NNSF of China (10771212) and the Fundamental Research Funds for the Central Universities (2010LKSX09).