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Journal of Applied Mathematics
VolumeΒ 2012, Article IDΒ 841349, 18 pages
http://dx.doi.org/10.1155/2012/841349
Research Article

The Existence of Solutions for a Fractional 2π‘š-Point Boundary Value Problems

1Department of Mathematics, China University of Mining and Technology, Xuzhou 221008, China
2School of Mathematics and Physical Sciences, Xuzhou Institute of Technology, Xuzhou 221008, China

Received 24 June 2011; Revised 9 October 2011; Accepted 11 October 2011

Academic Editor: YongkunΒ Li

Copyright Β© 2012 Gang Wang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

By using the coincidence degree theory, we consider the following 2π‘š-point boundary value problem for fractional differential equation 𝐷𝛼0+𝑒(𝑑)=𝑓(𝑑,𝑒(𝑑),π·π›Όβˆ’10+𝑒(𝑑),π·π›Όβˆ’20+𝑒(𝑑))+𝑒(𝑑), 0<𝑑<1, 𝐼3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0=0,π·π›Όβˆ’20+βˆ‘π‘’(1)=π‘šβˆ’2𝑖=1π‘Žπ‘–π·π›Όβˆ’20+𝑒(πœ‰π‘–βˆ‘),𝑒(1)=π‘šβˆ’2𝑖=1𝑏𝑖𝑒(πœ‚π‘–), where 2<𝛼≀3,𝐷𝛼0+ and 𝐼𝛼0+ 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: 𝐷𝛼0+π‘₯ξ€·(𝑑)=𝑓𝑑,π‘₯(𝑑),π‘₯ξ…ž(𝑑),π‘₯ξ…žξ…žξ€Έ(𝑑),0≀𝑑≀1,π‘₯(0)=π‘₯(1),π‘₯ξ…ž(0)=π‘₯ξ…žξ…ž(0)=0,(1.1) where 1<𝛼≀2, 𝐷𝛼0+ is the standard Caputo fractional derivative.

In [15], Zhang and Bai investigated the nonlinear nonlocal problem 𝐷𝛼0+𝑒(𝑑)=𝑓(𝑑,𝑒(𝑑)),0<𝑑<1,𝑒(0)=0,𝛽𝑒(πœ‚)=𝑒(1),(1.2) where 1<𝛼≀2, they consider the case π›½πœ‚π›Όβˆ’1=1, that is, the resonance case.

In [16], Bai investigated the boundary value problem at resonance 𝐷𝛼0+𝑒(𝑑)=𝑓𝑑,𝑒(𝑑),π·π›Όβˆ’10+𝐼𝑒(𝑑)+𝑒(𝑑),0<𝑑<1,2βˆ’π›Ό0+𝑒(𝑑)|𝑑=0=0,π·π›Όβˆ’10+𝑒(1)=π‘šβˆ’2𝑖=0π›½π‘–π·π›Όβˆ’10+π‘’ξ€·πœ‚π‘–ξ€Έ(1.3) is considered, where 1<𝛼≀2 is a real number, 𝐷𝛼0+and𝐼𝛼0+ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and π‘“βˆΆ[0,1]×𝑅2→𝑅 is continuous and 𝑒(𝑑)∈𝐿1[0,1],π‘šβ‰₯2,0<πœ‰π‘–<1,π›½π‘–βˆˆπ‘…,𝑖=1,2,…,π‘šβˆ’2 are given constants such that βˆ‘π‘šβˆ’2𝑖=1𝛽𝑖=1.

In this paper, we study the 2π‘š-point boundary value problem𝐷𝛼0+𝑒(𝑑)=𝑓𝑑,𝑒(𝑑),π·π›Όβˆ’10+𝑒(𝑑),π·π›Όβˆ’20+𝑒(𝑑)+𝑒(𝑑),0<𝑑<1,(1.4)𝐼3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0=0,π·π›Όβˆ’20+𝑒(1)=π‘šβˆ’2𝑖=1π‘Žπ‘–π·π›Όβˆ’20+π‘’ξ€·πœ‰π‘–ξ€Έ,𝑒(1)=π‘šβˆ’2𝑖=1π‘π‘–π‘’ξ€·πœ‚π‘–ξ€Έ,(1.5) where 2<𝛼≀3, π‘šβ‰₯2,0<πœ‰1<β‹―<πœ‰π‘š<1,0<πœ‚1<β‹―<πœ‚π‘š<1,π‘Žπ‘–,π‘π‘–βˆˆπ‘…,π‘“βˆΆ[0,1]×𝑅3→𝑅, 𝑓 satisfies CarathΓ©odory conditions, 𝐷𝛼0+ and 𝐼𝛼0+ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively.

Setting Ξ›1=1𝛼(𝛼+1)1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–πœ‰π‘–π›Ό+1ξƒͺ,Ξ›2=1𝛼(π›Όβˆ’1)1βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–πœ‰π›Όπ‘–ξƒͺ,Ξ›3=(Ξ“(𝛼))2Γ(2𝛼)1βˆ’π‘šβˆ’2𝑖=1π‘π‘–πœ‚π‘–2π›Όβˆ’1ξƒ­,Ξ›4=Ξ“(𝛼)Ξ“(π›Όβˆ’1)Γ(2π›Όβˆ’1)1βˆ’π‘šβˆ’2𝑖=1π‘π‘–πœ‚π‘–2π›Όβˆ’2ξƒ­.(1.6)

In this paper, we will always suppose that the following conditions hold:(C1):β€‰π‘šβˆ’2𝑖=1π‘Žπ‘–πœ‰π‘–=π‘šβˆ’2𝑖=1π‘Žπ‘–=1,π‘šβˆ’2𝑖=1π‘π‘–πœ‚π‘–π›Όβˆ’1=π‘šβˆ’2𝑖=1π‘π‘–πœ‚π‘–π›Όβˆ’2=1,(1.7)(C2): Λ=Ξ›1Ξ›4βˆ’Ξ›2Ξ›3β‰ 0.(1.8)

We say that boundary value problem (1.4) and (1.5) is at resonance, if BVP 𝐷𝛼0+𝐼𝑒(𝑑)=0,3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0=0,Dπ›Όβˆ’20+𝑒(1)=π‘šβˆ’2𝑖=1π‘Žπ‘–π·π›Όβˆ’20+π‘’ξ€·πœ‰π‘–ξ€Έ,𝑒(1)=π‘šβˆ’2𝑖=1π‘π‘–π‘’ξ€·πœ‚π‘–ξ€Έ(1.9) has 𝑒(𝑑)=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2, π‘Ž,π‘βˆˆπ‘… 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 π‘Œ,𝑍 be real Banach spaces, 𝐿∢domπΏβŠ‚π‘Œβ†’π‘ a Fredholm map of index zero,s and π‘ƒβˆΆπ‘Œβ†’π‘Œ,π‘„βˆΆπ‘β†’π‘ continuous projectors such that π‘Œ=KerπΏβŠ•Ker𝑃,𝑍=ImπΏβŠ•π‘„,Im𝑃=Ker𝐿,Ker𝑄=Im𝐿.(1.10) It follows that 𝐿|dom𝐿∩Kerπ‘ƒβˆΆdom𝐿∩Ker𝑃→Im𝐿 is invertible. We denote the inverse of the map by 𝐾𝑝. If Ξ© is an open-bounded subset of π‘Œ such that domπΏβˆ©Ξ©β‰ βˆ…, the map π‘βˆΆπ‘Œβ†’π‘ will be called 𝐿-compact on Ξ© if 𝑄𝑁(Ξ©) is bounded and 𝐾𝑝(πΌβˆ’π‘„)π‘βˆΆΞ©β†’π‘Œ 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 Ξ©. Assume that the following conditions are satisfied: (i)𝐿π‘₯β‰ πœ†π‘π‘₯, for all (π‘₯,πœ†)∈[dom𝐿⧡KerπΏβˆ©πœ•Ξ©]Γ—[0,1]; (ii)𝑁π‘₯βˆ‰Im𝐿, for all π‘₯∈KerπΏβˆ©πœ•Ξ©; (iii)deg(𝐽𝑄𝑁|Ker𝐿,Ker𝐿∩Ω,0)β‰ 0,where π‘„βˆΆπ‘β†’π‘ is a projection as above with Ker𝑄=Im𝐿 and 𝐽∢Im𝑄→Ker𝐿 is any isomorphism. Then the equation 𝐿π‘₯=𝑁π‘₯ 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 π‘¦βˆΆ(0,∞)→𝑅 is given by 𝐼𝛼0+1𝑦(𝑑)=ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1,𝑦(𝑠)𝑑𝑠(2.1) provided the right side is pointwise defined on (0,∞), where Ξ“(β‹…) is the Gamma function.

Definition 2.2. The fractional derivative of order 𝛼>0 of a function π‘¦βˆΆ(0,∞)→𝑅 is given by 𝐷𝛼0+1𝑦(𝑑)=𝑑Γ(π‘›βˆ’π›Ό)ξ‚π‘‘π‘‘π‘›ξ€œπ‘‘0𝑦(𝑠)(π‘‘βˆ’π‘ )π›Όβˆ’π‘›+1𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, provided the right side is pointwise defined on (0,∞).

Definition 2.3. We say that the map π‘“βˆΆ[0,1]×𝑅𝑛→𝑅 satisfies CarathΓ©odory conditions with respect to 𝐿1[0,1] if the following conditions are satisfied: (i)for each π‘§βˆˆπ‘…π‘›, the mapping 𝑑→𝑓(𝑑,𝑧) is Lebesgue measurable; (ii)for almost every π‘‘βˆˆ[0,1], the mapping 𝑑→𝑓(𝑑,𝑧) is continuous on 𝑅𝑛; (iii) for each π‘Ÿ>0, there exists πœŒπ‘ŸβˆˆπΏ1([0,1],𝑅) such that for a.e. π‘‘βˆˆ[0,1] and every |𝑧|β‰€π‘Ÿ, we have 𝑓(𝑑,𝑧)β‰€πœŒπ‘Ÿ(𝑑).

Lemma 2.4 (see [15]). Assume that π‘’βˆˆπΆ(0,1)∩𝐿1(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐢(0,1)∩𝐿1(0,1). Then 𝐼𝛼0+𝐷𝛼0+𝑒(𝑑)=𝑒(𝑑)+𝑐1π‘‘π›Όβˆ’1+𝑐1π‘‘π›Όβˆ’2+β‹―+π‘π‘π‘‘π›Όβˆ’π‘(2.3) for some π‘π‘–βˆˆπ‘…,𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

We use the classical Banach space 𝐢[0,1] with the norm β€–π‘₯β€–βˆž=max[]π‘‘βˆˆ0,1||||,π‘₯(𝑑)(2.4)𝐿[0,1] with the norm β€–π‘₯β€–1=ξ€œ10||||π‘₯(𝑑)𝑑𝑑.(2.5)

Definition 2.5. For π‘›βˆˆπ‘, we denote by 𝐴𝐢𝑛[0,1] the space of functions 𝑒(𝑑) which have continuous derivatives up to order π‘›βˆ’1 on [0,1] such that 𝑒(π‘›βˆ’1)(𝑑) is absolutely continuous: 𝐴𝐢𝑛[0,1]={π‘’βˆ£[0,1]→𝑅 and (𝐷(π‘›βˆ’1))𝑒(𝑑) is absolutely continuous in [0,1]}.

Lemma 2.6 (see [15]). Given πœ‡>0 and 𝑁=[πœ‡]+1 we can define a linear space πΆπœ‡[]=0,1𝑒(𝑑)βˆ£π‘’(𝑑)=𝐼𝛼0+π‘₯(𝑑)+𝑐1π‘‘πœ‡βˆ’1+𝑐2π‘‘πœ‡βˆ’2+β‹―+π‘π‘π‘‘πœ‡βˆ’(π‘βˆ’1)[],,π‘‘βˆˆ0,1(2.6) where π‘₯∈[0,1],π‘π‘–βˆˆπ‘…,𝑖=1,2,…,π‘βˆ’1. By means of the linear functional analysis theory, we can prove that with the β€–π‘’β€–πΆπœ‡=β€–β€–π·πœ‡0+π‘’β€–β€–βˆžβ€–β€–π·+β‹―+0πœ‡βˆ’(π‘βˆ’1)+π‘’β€–β€–βˆž+β€–π‘’β€–βˆž,(2.7)πΆπœ‡[0,1] is a Banach space.

Remark 2.7. If πœ‡ is a natural number, then πΆπœ‡[0,1] is in accordance with the classical Banach space 𝐢𝑛[0,1].

Lemma 2.8 (see [15]). π‘“βŠ‚πΆπœ‡[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 π‘’βˆˆπ‘“β€–π‘’β€–πΆπœ‡=β€–β€–π·πœ‡0+π‘’β€–β€–βˆžβ€–β€–π·+β‹―+πœ‡βˆ’(π‘βˆ’1)0+π‘’β€–β€–βˆž+β€–π‘’β€–βˆž<𝑀,(2.8) and equicontinuous means that βˆ€πœ€>0, βˆƒπ›Ώ>0, such that ||𝑒𝑑1ξ€Έξ€·π‘‘βˆ’π‘’2ξ€Έ||ξ€·<πœ€,βˆ€π‘‘1,𝑑2∈[],||𝑑0,11βˆ’π‘‘2||ξ€Έ,||𝐷<𝛿,βˆ€π‘’βˆˆπ‘“π›Όβˆ’π‘–0+𝑒𝑑1ξ€Έβˆ’π·π›Όβˆ’π‘–0+𝑒𝑑2ξ€Έ||ξ€·<πœ€,βˆ€π‘‘1,𝑑2∈[],||𝑑0,11βˆ’π‘‘2||ξ€Έ.<𝛿,βˆ€π‘’βˆˆπ‘“,βˆ€π‘–=1,2,…,π‘βˆ’1(2.9)

Lemma 2.9 (see [1]). Let 𝛼>0,𝑛=[𝛼]+1. Assume that π‘’βˆˆπΏ1(0,1) with a fractional integration of order π‘›βˆ’π›Ό that belongs to 𝐴𝐢𝑛[0,1]. Then the equality 𝐼𝛼0+𝐷𝛼0+𝑒(𝑑)=𝑒(𝑑)βˆ’π‘›ξ“π‘–=1πΌξ€·ξ€·π‘›βˆ’π›Ό0+𝑒(𝑑)π‘›βˆ’π‘–|𝑑=0𝑑Γ(π›Όβˆ’π‘–+1)π›Όβˆ’π‘–(2.10) holds almost everywhere on [0,1].

Definition 2.10 (see [16]). Let 𝐼𝛼0+(𝐿1(0,1)),𝛼>0 denote the space of functions 𝑒(𝑑), represented by fractional integral of order 𝛼 of a summable function: 𝑒=𝐼𝛼0+𝑣,π‘£βˆˆπΏ1(0,1).

Let 𝑍=𝐿1[0,1], with the norm βˆ«β€–π‘¦β€–=10|𝑦(𝑠)|𝑑𝑠, π‘Œ=πΆπ›Όβˆ’1[0,1] defined by Lemma 2.6, with the norm β€–π‘’β€–πΆπ›Όβˆ’1=‖𝐷0π›Όβˆ’1+π‘’β€–βˆž+‖𝐷0π›Όβˆ’2+π‘’β€–βˆž+β€–π‘’β€–βˆž, where π‘Œ is a Banach space.

Define 𝐿 to be the linear operator from domπΏβŠ‚π‘Œ to 𝑍 with ξ€½dom𝐿=π‘’βˆˆπΆπ›Όβˆ’1[]0,1βˆ£π·π›Ό0+π‘’βˆˆπΏ1[]0,1,𝑒satisfiesξ€Ύ,(1.5)(2.11)𝐿𝑒=𝐷𝛼0+𝑒,π‘’βˆˆdom𝐿,(2.12) we define π‘βˆΆπ‘Œβ†’π‘ by setting 𝑁𝑒(𝑑)=𝑓𝑑,𝑒(𝑑),π·π›Όβˆ’10+𝑒(𝑑)π·π›Όβˆ’20+𝑒(𝑑)+𝑒(𝑑).(2.13) Then boundary value problem (1.4) and (1.5) can be written as 𝐿𝑒=𝑁𝑒.

3. Main Results

Lemma 3.1. Let L be defined by (2.12), then Ker𝐿=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Ύβˆ£π‘Ž,π‘βˆˆπ‘…β‰…π‘…2,Imξƒ―ξ€œπΏ=π‘¦βˆˆπ‘βˆ£10(1βˆ’π‘ )𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ€œπœ‰π‘–0ξ€·πœ‰π‘–ξ€Έξ€œβˆ’π‘ π‘¦(𝑠)𝑑𝑠=0,10(1βˆ’π‘ )π›Όβˆ’1𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1bπ‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1ξƒ°.𝑦(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 𝐼𝛼0+=π·βˆ’π›Ό0+ for 𝛼<0.
Let 𝐿𝑒=𝐷𝛼0+𝑒, by Lemma 2.9, 𝐷𝛼0+𝑒(𝑑)=0 has solution 𝑒(𝑑)=3𝑖=1𝐼3βˆ’π›Ό0+𝑒(𝑑)3βˆ’π‘–|𝑑=0Γ𝑑(π›Όβˆ’π‘–+1)π›Όβˆ’π‘–=𝐼3βˆ’π›Ό0+𝑒(𝑑)ξ…žξ…ž|𝑑=0𝑑Γ(𝛼)π›Όβˆ’1+𝐼3βˆ’π›Ό0+𝑒(𝑑)ξ…ž|𝑑=0𝑑Γ(π›Όβˆ’1)π›Όβˆ’2+𝐼3βˆ’π›Ό0+𝑒|(𝑑)𝑑=0𝑑Γ(π›Όβˆ’2)π›Όβˆ’3=π·π›Όβˆ’10+𝑒(𝑑)|𝑑=0𝑑Γ(𝛼)π›Όβˆ’1+π·π›Όβˆ’20+𝑒(𝑑)|𝑑=0𝑑Γ(π›Όβˆ’1)π›Όβˆ’2+𝐼3βˆ’π›Ό0+𝑒|(𝑑)𝑑=0𝑑Γ(π›Όβˆ’2)π›Όβˆ’3.(3.2) Combine with (1.5), So, ξ€½Ker𝐿=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Ύβˆ£π‘Ž,π‘βˆˆπ‘…β‰…π‘…2.(3.3) Let π‘¦βˆˆπ‘ and let 𝑒𝑑=1ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1𝑦(𝑠)𝑑𝑠+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+𝑐3π‘‘π›Όβˆ’3.(3.4)
Then 𝐷𝛼0+𝑒(𝑑)=𝑦(𝑑) a.e. π‘‘βˆˆ[0,1] and, if ξ€œ10(1βˆ’π‘ )𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ€œπœ‰π‘–0ξ€·πœ‰π‘–ξ€Έξ€œβˆ’π‘ π‘¦(𝑠)𝑑𝑠=0,10(1βˆ’π‘ )π›Όβˆ’1𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1𝑦(𝑠)𝑑𝑠=0(3.5) hold, then 𝑒(𝑑) satisfies the boundary conditions (1.5). That is, π‘’βˆˆdom𝐿 and we have {π‘¦βˆˆπ‘βˆ£π‘¦satisfies(3.4)}βŠ†Im𝐿.(3.6)
Let π‘’βˆˆdom𝐿. Then for 𝐷𝛼0+π‘’βˆˆIm𝐿, we have 𝐼𝛼0+𝑦(𝑑)=𝑒(𝑑)βˆ’π‘1π‘‘π›Όβˆ’1βˆ’π‘2π‘‘π›Όβˆ’2βˆ’π‘3π‘‘π›Όβˆ’3,(3.7) where 𝑐1=π·π›Όβˆ’10+𝑒(𝑑)|𝑑=0Ξ“(𝛼),𝑐2=π·π›Όβˆ’20+𝑒(𝑑)|𝑑=0,Ξ“(π›Όβˆ’1)c3=𝐼3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0,Ξ“(π›Όβˆ’2)(3.8) which, due to the boundary value condition (1.5), implies that satisfies (3.5). In fact, from 𝐼3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0=0 we have 𝑐3=0, from π·π›Όβˆ’20+βˆ‘π‘’(1)=π‘šβˆ’2𝑖=1π‘Žπ‘–π·π›Όβˆ’20+𝑒(πœ‰π‘–), βˆ‘π‘’(1)=π‘šβˆ’2𝑖=1𝑏𝑖𝑒(πœ‚π‘–), we have ξ€œ10(1βˆ’π‘ )𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ€œπœ‰π‘–0ξ€·πœ‰π‘–ξ€Έξ€œβˆ’π‘ π‘¦(𝑠)𝑑𝑠=0,10(1βˆ’π‘ )π›Όβˆ’1𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1𝑦(𝑠)𝑑𝑠=0.(3.9) Hence, {π‘¦βˆˆπ‘βˆ£π‘¦satisfies(3.4)}βŠ‡Im𝐿.(3.10) Therefore, {π‘¦βˆˆπ‘βˆ£π‘¦satisfies(3.4)}=Im𝐿.(3.11) The proof is complete.

Lemma 3.2. The mapping 𝐿∢domπΏβŠ‚π‘Œβ†’π‘ is a Fredholm operator of index zero, and 𝑇𝑄𝑦(𝑑)=1𝑑𝑦(𝑑)π›Όβˆ’1+𝑇2𝑑𝑦(𝑑)π›Όβˆ’2,(3.12) where 𝑇11𝑦=Ξ›ξ€·Ξ›4𝑄1π‘¦βˆ’Ξ›2𝑄2𝑦,𝑇21𝑦=Ξ›ξ€·Ξ›3𝑄1π‘¦βˆ’Ξ›1𝑄2𝑦,(3.13) define by πΎπ‘βˆΆIm𝐿→dom𝐿∩Ker𝑃 by 𝐾𝑝1𝑦(𝑑)=ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1𝑦(𝑠)𝑑𝑠=𝐼𝛼0+𝑦(𝑑),π‘¦βˆˆIm𝐿,(3.14) and forallπ‘¦βˆˆIm𝐿,β€–πΎπ‘π‘¦β€–πΆπ›Όβˆ’1≀((1/Ξ“(𝛼))+2)‖𝑦‖1.

Proof. Consider the continuous linear mapping 𝑄1βˆΆπ‘β†’π‘ and 𝑄2βˆΆπ‘β†’π‘ defined by 𝑄1ξ€œπ‘¦=10(1βˆ’π‘ )𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ€œπœ‰π‘–0ξ€·πœ‰π‘–ξ€Έπ‘„βˆ’π‘ π‘¦(𝑠)𝑑𝑠,2ξ€œπ‘¦=10(1βˆ’π‘ )π›Όβˆ’1𝑦(𝑠)π‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1𝑦(𝑠)𝑑𝑠.(3.15) Using the above definitions, we construct the following auxiliary maps 𝑇1βˆΆπ‘β†’π‘ and 𝑇2βˆΆπ‘β†’π‘: 𝑇11𝑦=Ξ›ξ€·Ξ›4𝑄1π‘¦βˆ’Ξ›2𝑄2𝑦,𝑇21𝑦=Ξ›ξ€·Ξ›3𝑄1π‘¦βˆ’Ξ›1Q2𝑦.(3.16) Since the condition (C2) holds, the mapping defined by 𝑇𝑄𝑦(𝑑)=1𝑑𝑦(𝑑)π›Όβˆ’1+𝑇2𝑑𝑦(𝑑)π›Όβˆ’2(3.17) is well defined.
Recall (C2) and note that 𝑇1𝑇1π‘¦π‘‘π›Όβˆ’1ξ€Έ=1Ξ›ξ€·Ξ›4𝑄1𝑇1π‘¦π‘‘π›Όβˆ’1ξ€Έβˆ’Ξ›2𝑄2𝑇1π‘¦π‘‘π›Όβˆ’1=1ξ€Έξ€ΈΞ›ξ‚ΈΞ›4ξ‚΅Ξ›4Ξ›1Λ𝑄1Ξ›π‘¦βˆ’1Ξ›2Λ𝑄2π‘¦ξ‚Άβˆ’Ξ›2ξ‚΅Ξ›4Ξ›3Λ𝑄1Ξ›π‘¦βˆ’2Ξ›3Λ𝑄2𝑦=𝑇1𝑦,(3.18) and similarly we can derive that 𝑇1𝑇2π‘¦π‘‘π›Όβˆ’2ξ€Έ=0,𝑇2𝑇1π‘¦π‘‘π›Όβˆ’1ξ€Έ=0,𝑇2𝑇2π‘¦π‘‘π›Όβˆ’2ξ€Έ=𝑇2𝑦.(3.19) So, for π‘¦βˆˆπ‘, it follows from the four relations above that 𝑄2𝑇𝑦=𝑄1π‘¦ξ€Έπ‘‘π›Όβˆ’1+𝑇2π‘¦ξ€Έπ‘‘π›Όβˆ’2ξ€Έ=𝑇1𝑇1π‘¦ξ€Έπ‘‘π›Όβˆ’1+𝑇2π‘¦ξ€Έπ‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’1+𝑇2𝑇1π‘¦ξ€Έπ‘‘π›Όβˆ’1+𝑇2π‘¦ξ€Έπ‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’2=𝑇1π‘¦ξ€Έπ‘‘π›Όβˆ’1+𝑇2π‘¦ξ€Έπ‘‘π›Όβˆ’2=𝑄𝑦,(3.20) that is, the map 𝑄 is idempotent. In fact 𝑄 is a continuous linear projector.
Note that π‘¦βˆˆIm𝐿 implies 𝑄𝑦=0. Conversely, if 𝑄𝑦=0, so Ξ›4𝑄1π‘¦βˆ’Ξ›2𝑄2Λ𝑦=0,1𝑄2π‘¦βˆ’Ξ›3𝑄1𝑦=0,(3.21) but ||||||Ξ›4βˆ’Ξ›2βˆ’Ξ›3Ξ›1||||||=Ξ›4Ξ›1βˆ’Ξ›2Ξ›3β‰ 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, π‘¦βˆˆIm𝐿. In fact Ker𝑄=Im𝐿, take π‘¦βˆˆπ‘ in the form 𝑦=(π‘¦βˆ’π‘„π‘¦)+𝑄𝑦 so that π‘¦βˆ’π‘„π‘¦βˆˆKer𝑄=Im𝐿,π‘„π‘¦βˆˆIm𝑄, thus, 𝑍=Im𝐿+Im𝑄, Let π‘¦βˆˆIm𝐿∩Im𝑄 and assume that 𝑦=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2 is not identically zero on [0,1]. Then, since π‘¦βˆˆIm𝐿, from (3.5) and the condition (C2), we have 𝑄1ξ€œπ‘¦=10ξ€·(1βˆ’π‘ )π‘Žπ‘ π›Όβˆ’1+π‘π‘ π›Όβˆ’2ξ€Έπ‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘Žπ‘–ξ€œπœ‰π‘–0ξ€·πœ‰π‘–βˆ’π‘ ξ€Έξ€·π‘Žπ‘ π›Όβˆ’1+π‘π‘ π›Όβˆ’2𝑄𝑑𝑠=0,2ξ€œπ‘¦=10(1βˆ’π‘ )π›Όβˆ’1ξ€·π‘Žπ‘ π›Όβˆ’1+π‘π‘ π›Όβˆ’2ξ€Έπ‘‘π‘ βˆ’π‘šβˆ’2𝑖=1π‘π‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1ξ€·π‘Žπ‘ π›Όβˆ’1+π‘π‘ π›Όβˆ’2𝑑𝑠=0.(3.23) So π‘ŽΞ›1+𝑏Λ2=0,π‘ŽΞ›3+𝑏Λ4=0,(3.24) but ||||||Ξ›1Ξ›2Ξ›3Ξ›4||||||=Ξ›1Ξ›4βˆ’Ξ›2Ξ›3β‰ 0,(3.25) we derive π‘Ž=𝑏=0, which is a contradiction. Hence, Im𝐿∩Im𝑄={0}; thus 𝑍=ImπΏβŠ•Im𝑄.
Now, dimKer𝐿=2=codimIm𝐿 and so 𝐿 is a Fredholm operator of index zero.
Let π‘ƒβˆΆπ‘Œβ†’π‘Œ be defined by 1𝑃𝑒(𝑑)=Γ𝐷(𝛼)π›Όβˆ’10+𝑒(𝑑)|𝑑=0π‘‘π›Όβˆ’1+1Γ𝐷(π›Όβˆ’1)π›Όβˆ’20+𝑒(𝑑)|𝑑=0π‘‘π›Όβˆ’2[].,π‘‘βˆˆ0,1(3.26) Note that 𝑃 is a continuous linear projector and ξ€½Ker𝑃=π‘’βˆˆπ‘Œβˆ£π·π›Όβˆ’10+𝑒(0)=π·π›Όβˆ’20+ξ€Ύ.𝑒(0)=0(3.27) It is clear that π‘Œ=Kerπ‘ƒβŠ•Ker𝐿.
Note that the projectors 𝑃 and 𝑄 are exact. Define by πΎπ‘βˆΆIm𝐿→dom𝐿∩Ker𝑃 by 𝐾𝑝1𝑦(𝑑)=ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1𝑦(𝑠)𝑑𝑠=𝐼𝛼0+𝑦(𝑑),π‘¦βˆˆIm𝐿.(3.28) Hence we have π·π›Όβˆ’10+ξ€·πΎπ‘π‘¦ξ€Έξ€œπ‘‘=𝑑0𝑦(𝑠)𝑑𝑠,π·π›Όβˆ’20+ξ€·πΎπ‘π‘¦ξ€Έξ€œπ‘‘=𝑑0(π‘‘βˆ’π‘ )𝑦(𝑠)𝑑𝑠,(3.29) then β€–β€–πΎπ‘π‘¦β€–β€–βˆžβ‰€1Ξ“(𝛼)‖𝑦‖1,β€–β€–π·π›Όβˆ’10+ξ€·πΎπ‘π‘¦ξ€Έβ€–β€–βˆžβ‰€β€–π‘¦β€–1,β€–β€–π·π›Όβˆ’20+ξ€·πΎπ‘π‘¦ξ€Έβ€–β€–βˆžβ‰€β€–π‘¦β€–1,(3.30) and thus β€–β€–πΎπ‘π‘¦β€–β€–πΆπ›Όβˆ’1≀1ξ‚ΆΞ“(𝛼)+2‖𝑦‖1.(3.31)
In fact, if π‘¦βˆˆIm𝐿, then 𝐿𝐾𝑝𝑦(𝑑)=𝐷𝛼0+𝐼𝛼0+𝑦(𝑑)=𝑦(𝑑).(3.32) Also, if π‘’βˆˆdom𝐿∩Ker𝑃, then 𝐾𝑝𝐿𝑒(𝑑)=𝐼𝛼0+𝐷𝛼0+𝑒(𝑑)=𝑒(𝑑)+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+𝑐3π‘‘π›Όβˆ’3,(3.33) where 𝑐1=π·π›Όβˆ’10+𝑒(𝑑)|𝑑=0Ξ“(𝛼),𝑐2=π·π›Όβˆ’20+𝑒(𝑑)|𝑑=0Ξ“(π›Όβˆ’1),𝑐3=𝐼3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0,Ξ“(π›Όβˆ’2)(3.34) and from the boundary value condition (1.5) and the fact that π‘’βˆˆdom𝐿∩Ker𝑃, 𝑃𝑒=0, π·π›Όβˆ’10+𝑒(𝑑)|𝑑=0=π·π›Όβˆ’20+𝑒(𝑑)|𝑑=0=𝐼3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0=0, we have 𝑐1=𝑐2=𝑐3=0, thus 𝐾𝑝𝐿𝑒(𝑑)=𝑒(𝑑).(3.35) This shows that 𝐾𝑝=[𝐿|dom𝐿∩Ker𝑃]βˆ’1. The proof is complete. Using (3.16), we write 𝑇𝑄𝑁𝑒(𝑑)=1ξ€Έπ‘‘π‘π‘’π›Όβˆ’1+𝑇2ξ€Έπ‘‘π‘π‘’π›Όβˆ’2,𝐾𝑝1(πΌβˆ’π‘„)𝑁𝑒(𝑑)=ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1[]𝑁𝑒(𝑠)βˆ’π‘„π‘π‘’(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 π‘’βˆˆπΏ1[0,1],𝐾𝑝(πΌβˆ’π‘„)π‘βˆΆπ‘Œβ†’π‘Œ is completely continuous.

Assume that the following conditions on the function 𝑓(𝑑,π‘₯,𝑦,𝑧) are satisfied.(H1)There exist functions π‘Ž(𝑑),𝑏(𝑑),𝑐(𝑑),𝑑(𝑑),π‘Ÿ(𝑑)∈𝐿1[0,1], and a constant πœƒβˆˆ[0,1) such that for all (π‘₯,𝑦,𝑧)βˆˆπ‘…3, π‘‘βˆˆ[0,1], one of the following inequalities is satisfied:||||||𝑦||𝑓(𝑑,π‘₯,𝑦,𝑧)β‰€π‘Ž(𝑑)|π‘₯|+𝑏(𝑑)+𝑐(𝑑)|𝑧|+𝑑(𝑑)|π‘₯|πœƒ+π‘Ÿ(𝑑),(3.37)||||||𝑦||||𝑦||𝑓(𝑑,π‘₯,𝑦,𝑧)β‰€π‘Ž(𝑑)|π‘₯|+𝑏(𝑑)+𝑐(𝑑)|𝑧|+𝑑(𝑑)πœƒ+π‘Ÿ(𝑑),(3.38)||||||𝑦||𝑓(𝑑,π‘₯,𝑦,𝑧)β‰€π‘Ž(𝑑)|π‘₯|+𝑏(𝑑)+𝑐(𝑑)|𝑧|+𝑑(𝑑)|𝑧|πœƒ+π‘Ÿ(𝑑).(3.39)(H2)There exists a constant 𝐴>0, such that for π‘₯∈dom𝐿⧡Ker𝐿 satisfying |π·π›Όβˆ’10+π‘₯(𝑑)|+|π·π›Όβˆ’20+π‘₯(𝑑)|>𝐴 for all π‘‘βˆˆ[0,1], we have 𝑄1𝑁π‘₯(𝑑)β‰ 0,or𝑄2𝑁π‘₯(𝑑)β‰ 0.(3.40)(H3)There exists a constant 𝐡>0 such that for every π‘Ž,π‘βˆˆπ‘… satisfying π‘Ž2+𝑏2>𝐡 then either π‘Žπ‘‡1π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ+𝑏𝑇2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ<0,(3.41)

or π‘Žπ‘‡1π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ+𝑏𝑇2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ>0.(3.42)

Remark 3.4. 𝑇1𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2) and 𝑇2𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2) from (H3) stand for the images of 𝑒(𝑑)=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2 under the maps 𝑇1𝑁 and 𝑇2𝑁, respectively.

Lemma 3.5. Suppose (H1)-(H2) hold, then the set Ξ©1={π‘₯∈dom𝐿⧡Ker[]}𝐿∢𝐿π‘₯=πœ†π‘π‘₯,πœ†βˆˆ0,1(3.43) is bounded.

Proof. Take Ξ©1[]={π‘₯∈dom𝐿⧡Ker𝐿∢𝐿π‘₯=πœ†π‘π‘₯,πœ†βˆˆ0,1}.(3.44) Then for π‘₯∈Ω1, 𝐿π‘₯=πœ†π‘π‘₯ thus πœ†β‰ 0,𝑁π‘₯∈Im𝐿=Ker𝑄, and hence 𝑄𝑁π‘₯(𝑑) for all π‘‘βˆˆ[0,1]. By the definition of 𝑄, we have 𝑄1𝑁π‘₯(𝑑)=𝑄2𝑁π‘₯(𝑑)=0. It follows from (H2) that there exists 𝑑0∈[0,1], such that |π·π›Όβˆ’10+𝑒(𝑑0)|+|π·π›Όβˆ’20+𝑒(𝑑0)|≀𝐴.
Now π·π›Όβˆ’10+π‘₯(𝑑)=π·π›Όβˆ’10+π‘₯𝑑0ξ€Έ+ξ€œπ‘‘π‘‘0𝐷𝛼0+𝐷π‘₯(𝑠)𝑑𝑠,π›Όβˆ’20+π‘₯(𝑑)=π·π›Όβˆ’20+π‘₯𝑑0ξ€Έ+ξ€œπ‘‘π‘‘0π·π›Όβˆ’10+π‘₯(𝑠)𝑑𝑠,(3.45) and so ||π·π›Όβˆ’10+||≀‖‖𝐷π‘₯(0)π›Όβˆ’10+β€–β€–π‘₯(𝑑)βˆžβ‰€||π·π›Όβˆ’10+π‘₯𝑑0ξ€Έ||+β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–1≀𝐴+‖𝐿π‘₯β€–1≀𝐴+‖𝑁π‘₯β€–1,||π·π›Όβˆ’20+π‘₯||≀‖‖𝐷(0)π›Όβˆ’20+π‘₯β€–β€–(𝑑)βˆžβ‰€||π·π›Όβˆ’20+π‘₯𝑑0ξ€Έ||+β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–βˆžβ‰€||π·π›Όβˆ’20+π‘₯𝑑0ξ€Έ||+||π·π›Όβˆ’10+π‘₯𝑑0ξ€Έ||+‖‖𝐷𝛼0+π‘₯β€–β€–1≀𝐴+‖𝐿π‘₯β€–1≀𝐴+‖𝑁π‘₯β€–1.(3.46) Therefore, we have ‖𝑃π‘₯β€–πΆπ›Όβˆ’1=β€–β€–β€–1𝐷Γ(𝛼)π›Όβˆ’10+π‘₯(0)π‘‘π›Όβˆ’1+1𝐷Γ(π›Όβˆ’1)π›Όβˆ’20+π‘₯(0)π‘‘π›Όβˆ’2β€–β€–β€–πΆπ›Όβˆ’1=β€–β€–β€–1𝐷Γ(𝛼)π›Όβˆ’10+π‘₯(0)π‘‘π›Όβˆ’1+1𝐷Γ(π›Όβˆ’1)π›Όβˆ’20+π‘₯(0)π‘‘π›Όβˆ’2β€–β€–β€–βˆž+β€–β€–π·π›Όβˆ’10+β€–β€–π‘₯(0)∞+β€–β€–π·π›Όβˆ’10+π‘₯(0)𝑑+π·π›Όβˆ’20+β€–β€–π‘₯(0)βˆžβ‰€ξ‚΅12+ξ‚Ά||𝐷Γ(𝛼)π›Όβˆ’10+||+ξ‚΅1π‘₯(0)1+ξ‚Ά||𝐷Γ(π›Όβˆ’1)π›Όβˆ’20+||≀1π‘₯(0)2+ξ‚Άξ€·Ξ“(𝛼)𝐴+‖𝑁π‘₯β€–1ξ€Έ+ξ‚΅11+ξ‚Άξ€·Ξ“(π›Όβˆ’1)𝐴+‖𝑁π‘₯β€–1ξ€Έ.(3.47) Note that (πΌβˆ’π‘ƒ)π‘₯∈dom𝐿∩Ker𝑃 for all π‘₯∈Ω1. Then, by Lemma 3.2, we have β€–β€–(πΌβˆ’π‘ƒ)π‘₯πΆπ›Όβˆ’1=‖‖𝐾𝑝‖‖𝐿(πΌβˆ’π‘ƒ)π‘₯πΆπ›Όβˆ’1≀12+ξ‚ΆΞ“(𝛼)‖𝑁π‘₯β€–1,(3.48) so, we have β€–π‘₯β€–πΆπ›Όβˆ’1≀‖(πΌβˆ’π‘ƒ)π‘₯β€–πΆπ›Όβˆ’1+‖𝑃π‘₯β€–πΆπ›Όβˆ’1≀12+Ξ“ξ‚Άξ€·(𝛼)𝐴+‖𝑁π‘₯β€–1ξ€Έ+ξ‚΅11+Ξ“ξ‚Άξ€·(π›Όβˆ’1)𝐴+‖𝑁π‘₯β€–1ξ€Έ+ξ‚΅12+ξ‚ΆΞ“(𝛼)‖𝑁π‘₯β€–1=ξ‚΅2+1Ξ“(𝛼)ξ‚ΆΞ“(π›Όβˆ’1)+5‖𝑁π‘₯β€–1+ξ‚΅1+1Ξ“(𝛼)+1Ξ“(π›Όβˆ’1)𝐴Γ(π›Όβˆ’2)+3β‰€π‘šβ€–π‘π‘₯β€–1+𝑛𝐴,(3.49) where π‘š=((2/Ξ“(𝛼))+(1/Ξ“(π›Όβˆ’1))+5),𝑛=((1/Ξ“(𝛼))+(1/Ξ“(π›Όβˆ’1))+(1/(Ξ“(π›Όβˆ’2))+3),𝐴 is a constant. This is for all π‘₯∈Ω1. If the first condition of (H1) is satisfied, then, we have β€–π‘₯β€–πΆπ›Όβˆ’1ξ€½=maxβ€–π‘₯β€–βˆž,β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–βˆž,β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–βˆžξ€Ύξ‚ƒβ‰€π‘šβ€–π‘Žβ€–1β€–π‘₯β€–βˆž+‖𝑏‖1β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–βˆž+‖𝑐‖1β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–βˆž+‖𝑑‖1β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–πœƒβˆžξ‚„,+𝐷(3.50) where 𝐷=β€–π‘Ÿβ€–1+‖𝑒‖1+𝑛/π‘š, and consequently, for β€–π‘₯β€–βˆžβ‰€β€–π‘₯β€–πΆπ›Όβˆ’1,β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–βˆžβ‰€β€–π‘₯β€–πΆπ›Όβˆ’1,β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–βˆžβ‰€β€–π‘₯β€–πΆπ›Όβˆ’1,(3.51) so β€–π‘₯β€–βˆžβ‰€π‘š1βˆ’π‘šβ€–π‘Žβ€–1‖𝑏‖1β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–βˆž+‖𝑐‖1β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–βˆž+‖𝑑‖1β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–πœƒβˆžξ‚„,‖‖𝐷+π·π›Όβˆ’10+π‘₯β€–β€–βˆžβ‰€π‘š1βˆ’π‘šβ€–π‘Žβ€–1βˆ’π‘šβ€–π‘β€–1‖𝑐‖1β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–βˆž+‖𝑑‖1β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–πœƒβˆžξ‚„,‖‖𝐷+π·π›Όβˆ’20+π‘₯β€–β€–βˆžβ‰€π‘š1βˆ’π‘šβ€–π‘Žβ€–1βˆ’π‘šβ€–π‘β€–1βˆ’π‘šβ€–π‘β€–1‖𝑑‖1β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–πœƒβˆžξ‚.+𝐷(3.52) But πœƒβˆˆ[0,1) and β€–π‘Žβ€–1+‖𝑏‖1+‖𝑐‖1≀1/π‘š, so there exists 𝐴1,𝐴2,𝐴3>0 such that β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–βˆžβ‰€π΄1,β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–βˆžβ‰€π΄2,β€–π‘₯β€–βˆžβ‰€π΄3.(3.53) Therefore, for all π‘₯∈Ω1, β€–π‘₯β€–πΆπ›Όβˆ’1ξ€½=maxβ€–π‘₯β€–βˆž,β€–β€–π·π›Όβˆ’10+π‘₯β€–β€–βˆž,β€–β€–π·π›Όβˆ’20+π‘₯β€–β€–βˆžξ€Ύξ€½π΄β‰€max1,𝐴2,𝐴3ξ€Ύ,(3.54) we can prove that Ξ©1 is also bounded.
If (3.38) or (3.39) holds, similar to the above argument, we can prove that Ξ©1 is bounded too.

Lemma 3.6. Suppose (H3) holds, then the set Ξ©2={π‘₯∈KerπΏβˆΆπ‘π‘₯∈Im𝐿}(3.55) is bounded.

Proof. Let Ξ©2={π‘₯∈KerπΏβˆΆπ‘π‘₯∈Im𝐿},(3.56) for π‘₯∈Ω2,π‘₯∈Ker𝐿={π‘₯∈dom𝐿∢π‘₯=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2,π‘Ž,π‘βˆˆπ‘…,π‘‘βˆˆ[0,1]} and 𝑄𝑁π‘₯(𝑑)=0; thus 𝑇1𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2)=𝑇2𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2)=0. By (H3), π‘Ž2+𝑏2≀𝐡, that is, Ξ©2 is bounded.

Lemma 3.7. Suppose (H3) holds, then the set Ξ©3={π‘₯∈Ker[]}πΏβˆΆβˆ’πœ†π½π‘₯+(1βˆ’πœ†)𝑄𝑁π‘₯=0,πœ†βˆˆ0,1(3.57) is bounded.

Proof. We define the isomorphism 𝐽∢Ker𝐿→Im𝑄 by π½ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2.(3.58)
If the first part of (H3) is satisfied, let Ξ©3[]={π‘₯∈KerπΏβˆΆβˆ’πœ†π½π‘₯+(1βˆ’πœ†)𝑄𝑁π‘₯=0,πœ†βˆˆ0,1}.(3.59) For every π‘₯=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2∈Ω3, πœ†ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2𝑇=(1βˆ’πœ†)1π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’1+𝑇2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’1ξ€».(3.60) If πœ†=1, then π‘Ž=𝑏=0, and if π‘Ž2+𝑏2>𝐡, then by (H3) πœ†ξ€·π‘Ž2+𝑏2ξ€Έ=ξ€Ί(1βˆ’πœ†)π‘Žπ‘‡1π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ+𝑏𝑇2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έξ€»<0,(3.61) which, in either case, is a contradiction. If the other part of (H3) is satisfied, then we take Ξ©3[]={π‘₯∈KerπΏβˆΆπœ†π½π‘₯+(1βˆ’πœ†)𝑄𝑁π‘₯=0,πœ†βˆˆ0,1},(3.62) and, again, obtain a contradiction. Thus, in either case β€–π‘₯β€–πΆπ›Όβˆ’1=β€–β€–π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2β€–β€–πΆπ›Όβˆ’1=β€–β€–π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2β€–β€–βˆž+β€–π‘ŽΞ“(𝛼)β€–βˆž+β€–π‘ŽΞ“(𝛼)𝑑+𝑏Γ(π›Όβˆ’1)β€–βˆžβ‰€||𝑏||(1+2Ξ“(𝛼))|π‘Ž|+(1+Ξ“(π›Όβˆ’1))≀(2+2Ξ“(𝛼)+Ξ“(π›Όβˆ’1))𝐡,(3.63) for all π‘₯∈Ω3, that is, Ξ©3 is bounded.

Remark 3.8. Suppose the second part of (H3) holds, then the set Ξ©ξ…ž3[]}={π‘₯∈KerπΏβˆΆπœ†π½π‘₯+(1βˆ’πœ†)𝑄𝑁π‘₯=0,πœ†βˆˆ0,1(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 Ξ© to be a bounded open set of π‘Œ such that βˆͺ3𝑖=1Ξ©βŠ‚Ξ©. It follows from Lemmas 3.2 and 3.3 that 𝐿 is a Fredholm operator of index zero, and the operator 𝐾𝑝(πΌβˆ’π‘„)π‘βˆΆΞ©β†’π‘Œ is compact 𝑁, thus, is L-compact on Ξ©. By Lemmas 3.5 and 3.6, we get that the following two conditions are satisfied: (i)𝐿π‘₯β‰ πœ†π‘π‘₯ for every (π‘₯,πœ†)∈[dom𝐿⧡KerπΏβˆ©πœ•Ξ©]Γ—[0,1]; (ii)𝑁π‘₯βˆ‰Im𝐿, for every π‘₯∈KerπΏβˆ©πœ•Ξ©.
Finally, we will prove that (iii) of Lemma 1.1 is satisfied. Let 𝐻(π‘₯,πœ†)=Β±πœ†π½π‘₯+(1βˆ’πœ†)𝑄𝑁π‘₯, where 𝐼 is the identity operator in the Banach space π‘Œ. According to Lemma 3.7 (or Remark 3.8), we know that 𝐻(π‘₯,πœ†)β‰ 0, for all π‘₯βˆˆπœ•Ξ©βˆ©Ker𝐿, and thus, by the homotopy property of degree, ξ€·deg𝑄𝑁|KerπΏξ€ΈβŽ‘βŽ’βŽ’βŽ’βŽ£Β±||||||Ξ›,Ker𝐿∩Ω,0=deg(𝐻(β‹…,0),Ker𝐿∩Ω,0)=deg(𝐻(β‹…,1),Ker𝐿∩Ω,0)=deg(±𝐼,Ker𝐿∩Ω,0)=sgn4Ξ›βˆ’Ξ›2Ξ›βˆ’Ξ›3ΛΛ1Ξ›||||||⎀βŽ₯βŽ₯βŽ₯βŽ¦ξ‚΅Β±Ξ›=sgn1Ξ›4βˆ’Ξ›2Ξ›3Ξ›ξ‚Ά=Β±1β‰ 0.(3.65) Then by Lemma 1.1, 𝐿π‘₯=𝑁π‘₯ 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 πΆπ›Όβˆ’1[0,1]. The proof is finished.

4. An Example

Let us consider the following boundary value problem:𝐷5/20+𝑒(𝑑)=𝑓𝑑,𝑒(𝑑),π·π›Όβˆ’10+𝑒(𝑑),π·π›Όβˆ’20+𝐼𝑒(𝑑)+𝑒(𝑑),0<𝑑<1,3βˆ’π›Ό0+𝑒(𝑑)|𝑑=0=0,𝐷1/20+𝑒(1)=2𝐷1/20+𝑒23ξ‚βˆ’π·1/20+𝑒13,𝑒(1)=645𝑒14ξ‚βˆ’815𝑒19,(4.1) where 𝑓𝑑,𝑒(𝑑),π·π›Όβˆ’10+𝑒(𝑑),π·π›Όβˆ’20+ξ€Έ=1𝑒(𝑑)140sin(𝑒(𝑑))+𝐷203/20+1𝑒(𝑑)+𝐷201/20+𝐷𝑒(𝑑)+5cos1/20+𝑒(𝑑)1/5.(4.2) Corresponding to the problem (1.4)-(1.5), we have that 𝑒(𝑑)=1+3sin2𝑑, 𝛼=5/2, π‘Ž1=βˆ’1, π‘Ž2=2, πœ‰1=1/3, πœ‰2=2/3, 𝑏1=64/5, 𝑏2=βˆ’81/5, πœ‚1=1/4, πœ‚2=1/9 and 1𝑓(𝑑,π‘₯,𝑦,𝑧)=140sinπ‘₯+120𝑦+20𝑧+5cos(𝑧)1/5,(4.3) then there is π‘Ž1+π‘Ž2=1,π‘Ž1πœ‰1+π‘Ž2πœ‰2𝑏=1,1πœ‚13/2+𝑏2πœ‚23/2=1,𝑏1πœ‚11/2+𝑏2πœ‚21/2Ξ›=1,1=4351βˆ’2𝑖=1π‘Žπ‘–πœ‰π‘–7/2ξƒͺ,Ξ›2=4151βˆ’2𝑖=1π‘Žπ‘–πœ‰π‘–5/2ξƒͺ,Ξ›3=(Ξ“(5/2))2241βˆ’2𝑖=1π‘π‘–πœ‚4𝑖,Ξ›4=16Ξ“(5/2)Ξ“(3/2)241βˆ’2𝑖=1π‘π‘–πœ‚3𝑖,Ξ›=Ξ›1Ξ›4βˆ’Ξ›2Ξ›3||||≀1β‰ 0,𝑓(𝑑,π‘₯,𝑦,𝑧)|140π‘₯|+||𝑦||+120|20𝑧|+5|𝑧|1/5.(4.4) Again, taking π‘Ž=1/40, 𝑏=𝑐=1/20, then β€–π‘Žβ€–1+‖𝑏‖1+‖𝑐‖11=1/8,π‘š=1(2/Ξ“(𝛼))+(1/Ξ“(π›Όβˆ’1))+5β‰ˆ0.131,(4.5) therefore β€–π‘Žβ€–1+‖𝑏‖1+‖𝑐‖1<1π‘š.(4.6) Take 𝐴=181, 𝐡=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).

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