International Journal of Differential Equations

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Fractional Differential Equations 2011

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Volume 2011 |Article ID 401803 | https://doi.org/10.1155/2011/401803

Xiaoling Han, Ting Wang, "The Existence of Solutions for a Nonlinear Fractional Multi-Point Boundary Value Problem at Resonance", International Journal of Differential Equations, vol. 2011, Article ID 401803, 14 pages, 2011. https://doi.org/10.1155/2011/401803

The Existence of Solutions for a Nonlinear Fractional Multi-Point Boundary Value Problem at Resonance

Academic Editor: Nikolai Leonenko
Received16 May 2011
Accepted16 Jun 2011
Published18 Aug 2011

Abstract

We discuss the existence of solution for a multipoint boundary value problem of fractional differential equation. An existence result is obtained with the use of the coincidence degree theory.

1. Introduction

In this paper, we study the multipoint boundary value problem 𝐷𝛼0+𝑒(𝑑)=𝑓𝑑,𝑒(𝑑),π·π›Όβˆ’10+𝑒(𝑑),π·π›Όβˆ’20+𝐼𝑒(𝑑)+𝑒(𝑑),0<𝑑<1,(1.1)3βˆ’π›Ό0+𝑒(0)=0,π·π›Όβˆ’20+𝑒(0)=𝑛𝑗=1π›½π‘—π·π›Όβˆ’20+π‘’ξ€·πœ‰π‘—ξ€Έ,𝑒(1)=π‘šξ“π‘–=1π›Όπ‘–π‘’ξ€·πœ‚π‘–ξ€Έ,(1.2) where 2<𝛼≀3,0<πœ‰1<πœ‰2<β‹―<πœ‰π‘›<1,𝑛β‰₯1,0<πœ‚1<β‹―<πœ‚π‘š<1,π‘šβ‰₯2,𝛼𝑖,π›½π‘—βˆˆβ„,π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–π›Όβˆ’1=π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–π›Όβˆ’2=1,𝑛𝑗=1π›½π‘—πœ‰π‘—=0,𝑛𝑗=1𝛽𝑗=1,(1.3)π‘“βˆΆ[0,1]×ℝ3→ℝ satisfying the CarathΓ©odory conditions, π‘’βˆˆπΏ1[0,1]. 𝐷𝛼0+ and 𝐼𝛼0+ are the standard Riemann-Liouville derivative and integral, respectively. We assume, in addition, that 𝑅=Ξ“(𝛼)2Ξ“(π›Όβˆ’1)Ξ“(2𝛼)Ξ“(𝛼+1)𝑛𝑗=1π›½π‘—πœ‰π›Όπ‘—ξƒ©1βˆ’π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–2π›Όβˆ’1ξƒͺβˆ’Ξ“(𝛼)2Ξ“(π›Όβˆ’1)Ξ“(𝛼+2)Ξ“(2π›Όβˆ’1)𝑛𝑗=1π›½π‘—πœ‰π‘—π›Ό+11βˆ’π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–2π›Όβˆ’2ξƒͺβ‰ 0,(1.4) where Ξ“ is the Gamma function. Due to condition (1.3), the fractional differential operator in (1.1), (1.2) is not invertible.

Fractional differential equation can describe many phenomena in various fields of science and engineering. Many methods have been introduced for solving fractional differential equations, such as the popular Laplace transform method, the iteration method. For details, see [1, 2] and the references therein.

Recently, there are some papers dealing with the solvability of nonlinear boundary value problems of fractional differential equation, by use of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, etc.), see, for example, [3–6]. But there are few papers that consider the fractional-order boundary problems at resonance. Very recently [7], Y. H. Zhang and Z. B. Bai considered the existence of solutions for the fractional ordinary differential equation𝐷𝛼0+𝑒(𝑑)=𝑓𝑑,𝑒(𝑑),π·π›Όβˆ’(π‘›βˆ’1)0+𝑒(𝑑),…,π·π›Όβˆ’10+𝑒(𝑑)+𝑒(𝑑),0<𝑑<1,(1.5) subject to the following boundary value conditions: πΌπ‘›βˆ’π›Ό0+𝑒(0)=π·π›Όβˆ’(π‘›βˆ’1)0+𝑒(0)=β‹―=π·π›Όβˆ’20+𝑒(0)=0,𝑒(1)=πœŽπ‘’(πœ‚),(1.6) where 𝑛>2 is a natural number, π‘›βˆ’1<𝛼≀𝑛 is a real number, π‘“βˆΆ[0,1]×ℝ𝑛→ℝ is continuous, and π‘’βˆˆπΏ1[0,1],𝜎∈(0,∞), and πœ‚βˆˆ(0,1) are given constants such that πœŽπœ‚π›Όβˆ’1=1. 𝐷𝛼0+ and 𝐼𝛼0+ are the standard Riemann-Liouville derivative and integral, respectively. By the conditions, the kernel of the linear operator is one dimensional.

Motivated by the above work and recent studies on fractional differential equations [8–18], in this paper, we consider the existence of solutions for multipoint boundary value problem (1.1), (1.2) at resonance. Note that under condition (1.3), the kernel of the linear operator in (1.1), (1.2) is two dimensional. Our method is based upon the coincidence degree theory of Mawhin [18].

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

Let π‘Œ,𝑍 be real Banach spaces, let 𝐿∢dom(𝐿)βŠ‚π‘Œβ†’π‘ be a Fredholm map of index zero, and let π‘ƒβˆΆπ‘Œβ†’π‘Œ,π‘„βˆΆπ‘β†’π‘ be continuous projectors such that Im(𝑃)=Ker(𝑃),Ker(𝑄)=Im(𝐿), and π‘Œ=Ker(𝐿)βŠ•Ker(𝑃),𝑍=Im(𝐿)βŠ•Im(𝑄). It follows that 𝐿|dom(𝐿)∩Ker(𝑃)∢dom(𝐿)∩Ker(𝑃)β†’Im(𝐿) is invertible. We denote the inverse of the map by 𝐾𝑃. If Ξ© is an open-bounded subset of π‘Œ such that dom(𝐿)βˆ©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 theorem that we used is Theorem  2.4 of [18].

Theorem 1.1. Let 𝐿 be a Fredholm operator of index zero and 𝑁 be 𝐿-compact on Ξ©. Assume that the following conditions are satisfied: (i)𝐿π‘₯β‰ πœ†π‘π‘₯ for every (π‘₯,πœ†)∈[(dom(𝐿)⧡Ker(𝐿))βˆ©πœ•Ξ©]Γ—(0,1),(ii)𝑁π‘₯βˆ‰Im(𝐿) for every π‘₯∈Ker(𝐿)βˆ©πœ•Ξ©,(iii)deg(𝐽𝑄𝑁|Ker(𝐿),Ω∩Ker(𝐿),0)β‰ 0, where π‘„βˆΆπ‘β†’π‘ is a projection as above with Im(𝐿)=Ker(𝑄), and 𝐽∢Im(𝑄)β†’Ker(𝐿) is any isomorphism,
then the equation 𝐿π‘₯=𝑁π‘₯ has at least one solution in dom(𝐿)∩Ω.

The rest of this paper is organized as follows. In Section 2, we give some notation and Lemmas. In Section 3, we establish an existence theorem of a solution for the problem (1.1), (1.2).

2. Background Materials and Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory, and these definitions can be found in the recent literature [1, 2].

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’1𝑦(𝑠)𝑑𝑠,(2.1) provided the right side is pointwise defined on (0,∞). And we let 𝐼00+𝑦(𝑑)=𝑦(𝑑) for every continuous π‘¦βˆΆ(0,∞)→ℝ.

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,∞).

Lemma 2.3 (see [3]). 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)+𝐢1π‘‘π›Όβˆ’1+𝐢2π‘‘π›Όβˆ’2+β‹―+πΆπ‘π‘‘π›Όβˆ’π‘,(2.3) for some πΆπ‘–βˆˆβ„,𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

We use the classical space 𝐢[0,1] with the norm β€–π‘₯β€–βˆž=maxπ‘‘βˆˆ[0,1]|π‘₯(𝑑)|. Given πœ‡>0 and 𝑁=[πœ‡]+1, one can define a linear space πΆπœ‡[]ξ€½0,1∢=π‘’βˆ£π‘’(𝑑)=πΌπœ‡0+π‘₯(𝑑)+𝑐1π‘‘πœ‡βˆ’1+𝑐2π‘‘πœ‡βˆ’2+β‹―+π‘π‘βˆ’1π‘‘πœ‡βˆ’(π‘βˆ’1)[]ξ€Ύ,π‘‘βˆˆ0,1,(2.4) where π‘₯∈𝐢[0,1] and π‘π‘–βˆˆβ„,𝑖=1,2,…,π‘βˆ’1. By means of the linear function analysis theory, one can prove that with the norm β€–π‘’β€–πΆπœ‡=β€–π·πœ‡0+π‘’β€–βˆž+β‹―+β€–π·πœ‡βˆ’(π‘βˆ’1)0+π‘’β€–βˆž+β€–π‘’β€–βˆž,πΆπœ‡[0,1] is a Banach space.

Lemma 2.4 (see [7]). πΉβŠ‚πΆπœ‡[0,1] is a sequentially compact set if and only if 𝐹 is uniformly bounded and equicontinuous. Here, uniformly bounded means that there exists 𝑀>0 such that for every π‘’βˆˆπΉ, 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,(2.5) and equicontinuous means that forallπ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Ήπ›Όβˆ’π‘–0+𝑒𝑑1ξ€Έβˆ’π·π›Όβˆ’π‘–0+𝑒𝑑2ξ€Έ||𝑑<πœ€,1,𝑑2∈[],||𝑑0,11βˆ’π‘‘2||ξ€Έ.<𝛿,βˆ€π‘’βˆˆπΉ,βˆ€π‘–βˆˆ{0,…,π‘βˆ’1}(2.6)

Let 𝑍=𝐿1[0,1] with the norm ‖𝑔‖1=∫10|𝑔(𝑠)|𝑑𝑠. π‘Œ=πΆπ›Όβˆ’1[0,1]={π‘’βˆ£π‘’(𝑑)=πΌπ›Όβˆ’10+π‘₯(𝑑)+π‘π‘‘π›Όβˆ’2,π‘‘βˆˆ[0,1]}, where π‘₯∈𝐢[0,1],π‘βˆˆβ„, with the norm β€–π‘’β€–πΆπ›Όβˆ’1=β€–π·π›Όβˆ’10+π‘’β€–βˆž+β€–π·π›Όβˆ’20+π‘’β€–βˆž+β€–π‘’β€–βˆž, and π‘Œ is a Banach space.

Definition 2.5. By a solution of the boundary value problem (1.1), (1.2), we understand a function π‘’βˆˆπΆπ›Όβˆ’1[0,1] such that π·π›Όβˆ’10+𝑒 is absolutely continuous on (0,1) and satisfies (1.1), (1.2).

Definition 2.6. We say that the map π‘“βˆΆ[0,1]×ℝ→ℝ satisfies the CarathΓ©odory conditions with respect to 𝐿1[0,1] if the following conditions are satisfied:(i)for each π‘§βˆˆβ„, the mapping 𝑑→𝑓(𝑑,𝑧) is Lebesgue measurable,(ii)for almost every π‘‘βˆˆ[0,1], the mapping 𝑧→𝑓(𝑑,𝑧) is continuous on ℝ,(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).

Define 𝐿 to be the linear operator from dom(𝐿)βˆ©π‘Œ to 𝑍 with ξ€½dom(𝐿)=π‘’βˆˆπΆπ›Όβˆ’1[]0,1βˆ£π·π›Ό0+π‘’βˆˆπΏ1[]ξ€Ύ,0,1,𝑒satisfies(1.2)𝐿𝑒=𝐷𝛼0+𝑒,π‘’βˆˆdom(𝐿).(2.7) 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).(2.8) Then boundary value problem (1.1), (1.2) can be written as 𝐿𝑒=𝑁𝑒.(2.9)

Lemma 2.7. Let condition (1.3) and (1.4) hold, then 𝐿∢dom(𝐿)βˆ©π‘Œβ†’π‘ is a Fredholm map of index zero.

Proof. It is clear that Ker(𝐿)={π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2βˆ£π‘Ž,π‘βˆˆβ„}≅ℝ2.
Let π‘”βˆˆπ‘ and 1𝑒(𝑑)=ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1𝑔(𝑠)𝑑𝑠+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2,(2.10) then 𝐷𝛼0+𝑒(𝑑)=𝑔(𝑑)a.e.,π‘‘βˆˆ(0,1) and, if ξ€œ10(1βˆ’π‘ )π›Όβˆ’1𝑔(𝑠)π‘‘π‘ βˆ’π‘šξ“π‘–=1π›Όπ‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1𝑔(𝑠)𝑑𝑠=0,𝑛𝑗=1π›½π‘—ξ€œπœ‰π‘—0ξ€·πœ‰π‘—ξ€Έβˆ’π‘ π‘”(𝑠)𝑑𝑠=0(2.11) hold. Then 𝑒(𝑑) satisfies the boundary conditions (1.2), that is, π‘’βˆˆdom(𝐿), and we have {π‘”βˆˆπ‘βˆ£π‘”satisfies(2.11)}βŠ†Im(𝐿).(2.12) Let π‘’βˆˆdom(𝐿), then for 𝐷𝛼0+π‘’βˆˆIm(𝐿), we have 𝑒(𝑑)=𝐼𝛼0+𝐷𝛼0+𝑒(𝑑)+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+𝑐3π‘‘π›Όβˆ’3,(2.13) which, due to the boundary value condition (1.2), implies that 𝐷𝛼0+𝑒 satisfies (2.11). In fact, from 𝐼3βˆ’π›Ό0+𝑒(0)=0, we have 𝑐3=0, from βˆ‘π‘’(1)=π‘šπ‘–=1𝛼𝑖𝑒(πœ‚π‘–), we have ξ€œ10(1βˆ’π‘ )π›Όβˆ’1𝐷𝛼0+𝑒(𝑠)π‘‘π‘ βˆ’π‘šξ“π‘–=1π›Όπ‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1𝐷𝛼0+𝑒(𝑠)𝑑𝑠=0,(2.14) and from π·π›Όβˆ’20+βˆ‘π‘’(0)=𝑛𝑗=1π›½π‘—π·π›Όβˆ’20+𝑒(πœ‰π‘—), we have 𝑛𝑗=1π›½π‘—ξ€œπœ‰π‘—0ξ€·πœ‰π‘—ξ€Έπ·βˆ’π‘ π›Ό0+𝑒(𝑠)𝑑𝑠=0.(2.15) Hence, Im(𝐿)βŠ†{π‘”βˆˆπ‘βˆ£π‘”satisfies(2.11)}.(2.16) Therefore, Im(𝐿)={π‘”βˆˆπ‘βˆ£π‘”satisfies(2.11)}.(2.17) Consider the continuous linear mapping 𝑄1βˆΆπ‘β†’π‘ and 𝑄2βˆΆπ‘β†’π‘ defined by 𝑄1ξ€œπ‘”=10(1βˆ’π‘ )π›Όβˆ’1𝑔(𝑠)π‘‘π‘ βˆ’π‘šξ“π‘–=1π›Όπ‘–ξ€œπœ‚π‘–0ξ€·πœ‚π‘–ξ€Έβˆ’π‘ π›Όβˆ’1𝑄𝑔(𝑠)𝑑𝑠,2𝑔=𝑛𝑗=1π›½π‘—ξ€œπœ‰π‘—0ξ€·πœ‰π‘—ξ€Έβˆ’π‘ π‘”(𝑠)𝑑𝑠.(2.18) Using the above definitions, we construct the following auxiliary maps 𝑅1,𝑅2βˆΆπ‘β†’π‘: 𝑅11𝑔=𝑅Γ(π›Όβˆ’1)Ξ“(𝛼+1)𝑛𝑗=1π›½π‘—πœ‰π›Όπ‘—π‘„1𝑔(𝑑)βˆ’Ξ“(𝛼)Ξ“(π›Όβˆ’1)Γ(2π›Όβˆ’1)1βˆ’π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–2π›Όβˆ’2ξƒͺ𝑄2ξƒ­,𝑅𝑔(𝑑)21𝑔=βˆ’π‘…ξƒ¬Ξ“(𝛼)Ξ“(𝛼+2)𝑛𝑗=1π›½π‘—πœ‰π‘—π›Ό+1𝑄1𝑔(𝑑)βˆ’(Ξ“(𝛼))2Γ(2𝛼)1βˆ’π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–2π›Όβˆ’1ξƒͺ𝑄2ξƒ­.𝑔(𝑑)(2.19) Since the condition (1.4) holds, the mapping π‘„βˆΆπ‘β†’π‘ defined by 𝑅(𝑄𝑦)(𝑑)=1𝑑𝑔(𝑑)π›Όβˆ’1+𝑅2𝑑𝑔(𝑑)π›Όβˆ’2(2.20) is well defined.
Recall (1.4) and note that 𝑅1𝑅1π‘”π‘‘π›Όβˆ’1ξ€Έ=1𝑅Γ(π›Όβˆ’1)Ξ“(𝛼+1)𝑛𝑗=1π›½π‘—πœ‰π›Όπ‘—π‘„1𝑅1π‘”π‘‘π›Όβˆ’1ξ€Έβˆ’Ξ“(𝛼)Ξ“(π›Όβˆ’1)Γ(2π›Όβˆ’1)1βˆ’π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–2π›Όβˆ’2ξƒͺ𝑄2𝑅1π‘”π‘‘π›Όβˆ’1ξ€Έξƒ­=𝑅1𝑔1𝑅𝛼Γ(π›Όβˆ’1)Ξ“2ξ€ΈΞ“(𝛼+1)Ξ“(2𝛼)𝑛𝑗=1π›½π‘—πœ‰π›Όπ‘—ξƒ©1βˆ’π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–2π›Όβˆ’1ξƒͺβˆ’ξ€·π›ΌΞ“(π›Όβˆ’1)Ξ“2Γ(2π›Όβˆ’1)Ξ“(𝛼+2)1βˆ’π‘šξ“π‘–=1π›Όπ‘–πœ‚π‘–2π›Όβˆ’2ξƒͺ𝑛𝑗=1π›½π‘—πœ‰π‘—π›Ό+1ξƒ­=𝑅1𝑔,(2.21) and similarly we can derive that 𝑅1𝑅2π‘”π‘‘π›Όβˆ’2𝑅=0,2𝑅1π‘”π‘‘π›Όβˆ’1𝑅=0,2𝑅2π‘”π‘‘π›Όβˆ’2ξ€Έ=𝑅2𝑔.(2.22) So, for π‘”βˆˆπ‘, it follows from the four relations above that 𝑄2𝑔=𝑅1𝑅1π‘”π‘‘π›Όβˆ’1+𝑅2π‘”π‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’1+𝑅2𝑅1π‘”π‘‘π›Όβˆ’1+𝑅2π‘”π‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’2=𝑅1𝑅1π‘”π‘‘π›Όβˆ’1ξ€Έπ‘‘π›Όβˆ’1+𝑅1𝑅2π‘”π‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’1+𝑅2𝑅1π‘”π‘‘π›Όβˆ’1ξ€Έπ‘‘π›Όβˆ’2+𝑅2𝑅2π‘”π‘‘π›Όβˆ’2ξ€Έπ‘‘π›Όβˆ’2=𝑅1π‘”π‘‘π›Όβˆ’1+𝑅2π‘”π‘‘π›Όβˆ’2=𝑄𝑔,(2.23) that is, the map 𝑄 is idempotent. In fact, 𝑄 is a continuous linear projector.
Note that π‘”βˆˆIm(𝐿) implies 𝑄𝑔=0. Conversely, if 𝑄𝑔=0, then we must have 𝑅1𝑔=𝑅2𝑔=0; since the condition (1.4) holds, this can only be the case if 𝑄1𝑔=𝑄2𝑔=0, that is, π‘”βˆˆIm(𝐿). In fact, Im(𝐿)=Ker(𝑄).
Take π‘”βˆˆπ‘ in the form 𝑔=(π‘”βˆ’π‘„π‘”)+𝑄𝑔, so that π‘”βˆ’π‘„π‘”βˆˆIm(𝐿)=Ker(𝑄) and π‘„π‘”βˆˆIm(𝑄). Thus, 𝑍=Im(𝐿)+Im(𝑄). Let π‘”βˆˆIm(𝐿)∩Im(𝑄) and assume that 𝑔(𝑠)=π‘Žπ‘ π›Όβˆ’1+π‘π‘ π›Όβˆ’2 is not identically zero on [0,1], then, since π‘”βˆˆIm(𝐿), from (2.11) and the condition (1.4), we derive π‘Ž=𝑏=0, which is a contradiction. Hence, Im(𝐿)∩Im(𝑄)={0}; thus, 𝑍=Im(𝐿)βŠ•Im(𝑄).
Now, dimKer(𝐿)=2=codimIm(𝐿), 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(0)π‘‘π›Όβˆ’1+1Γ𝐷(π›Όβˆ’1)π›Όβˆ’20+𝑒(0)π‘‘π›Όβˆ’2[],π‘‘βˆˆ0,1.(2.24) Note that 𝑃 is a continuous linear projector and ξ€½Ker(𝑃)=π‘’βˆˆπ‘Œβˆ£π·π›Όβˆ’10+𝑒(0)=π·π›Όβˆ’20+𝑒(0)=0.(2.25) It is clear that π‘Œ=Ker(𝐿)βŠ•Ker(𝑃).

Note that the projectors 𝑃 and 𝑄 are exact. Define πΎπ‘ƒβˆΆIm(𝐿)β†’dom(𝐿)∩Ker(𝑃) by 𝐾𝑃1𝑔(𝑑)=ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1𝑔(𝑠)𝑑𝑠=𝐼𝛼0+𝑔(𝑑).(2.26) Hence, we have π·π›Όβˆ’10+𝐾𝑃𝑔(ξ€œπ‘‘)=𝑑0𝑔(𝑠)𝑑𝑠,π·π›Όβˆ’20+𝐾𝑃𝑔(ξ€œπ‘‘)=𝑑0(π‘‘βˆ’π‘ )𝑔(𝑠)𝑑𝑠,(2.27) then β€–πΎπ‘ƒπ‘”β€–βˆžβ‰€(1/Ξ“(𝛼))‖𝑔‖1,β€–π·π›Όβˆ’10+(𝐾𝑃𝑔)β€–βˆžβ‰€β€–π‘”β€–1,β€–π·π›Όβˆ’20+(𝐾𝑃𝑔)β€–βˆžβ‰€β€–π‘”β€–1, and thus β€–β€–πΎπ‘ƒπ‘”β€–β€–πΆπ›Όβˆ’1≀12+ξ‚ΆΞ“(𝛼)‖𝑔‖1.(2.28) In fact, if π‘”βˆˆIm(𝐿), then (𝐿𝐾𝑃)𝑔=𝐷𝛼0+𝐼𝛼0+𝑔=𝑔. Also, if π‘’βˆˆdom(𝐿)∩Ker(𝑃), then 𝐾𝑃𝐿𝑔(𝑑)=𝐼𝛼0+𝐷𝛼0+𝑔(𝑑)=𝑔(𝑑)+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+𝑐3π‘‘π›Όβˆ’3,(2.29) from boundary value condition (1.2) and the fact that π‘’βˆˆdom(𝐿)∩Ker(𝑃), we have 𝑐1=𝑐2=𝑐3=0. Thus, 𝐾𝑃=𝐿||dom(𝐿)∩Ker(𝑃)ξ‚βˆ’1.(2.30) Using (2.19), we write 𝑅𝑄𝑁𝑒(𝑑)=1ξ€Έπ‘‘π‘π‘’π›Όβˆ’1+𝑅2ξ€Έπ‘‘π‘π‘’π›Όβˆ’2,𝐾𝑃1(πΌβˆ’π‘„)𝑁𝑒(𝑑)=ξ€œΞ“(𝛼)10(π‘‘βˆ’π‘ )π›Όβˆ’1[]𝑁𝑒(𝑠)βˆ’π‘„π‘π‘’(𝑠)𝑑𝑠.(2.31) With arguments similar to those of [7], we obtain the following Lemma.

Lemma 2.8. 𝐾𝑃(πΌβˆ’π‘„)π‘βˆΆπ‘Œβ†’π‘Œ is completely continuous.

3. The Main Results

Assume that the following conditions on the function 𝑓(𝑑,π‘₯,𝑦,𝑧) are satisfied:

(H1) there exists a constant 𝐴>0, such that for π‘’βˆˆdom(𝐿)⧡Ker(𝐿) satisfying |π·π›Όβˆ’10+𝑒(𝑑)|+|π·π›Όβˆ’20+𝑒(𝑑)|>𝐴 for all π‘‘βˆˆ[0,1], we have 𝑄1𝑁𝑒(𝑑)β‰ 0or𝑄2𝑁𝑒(𝑑)β‰ 0,(3.1)

(H2) there exist functions π‘Ž,𝑏,𝑐,𝑑,π‘ŸβˆˆπΏ1[0,1] and a constant πœƒβˆˆ[0,1] such that for all (π‘₯,𝑦,𝑧)βˆˆβ„3 and a.e., 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)|𝑧|πœƒ||||||𝑦||||𝑦||+π‘Ÿ(𝑑),𝑓(𝑑,π‘₯,𝑦,𝑧)β‰€π‘Ž(𝑑)|π‘₯|+𝑏(𝑑)+𝑐(𝑑)|𝑧|+𝑑(𝑑)πœƒ||||||𝑦||+π‘Ÿ(𝑑),𝑓(𝑑,π‘₯,𝑦,𝑧)β‰€π‘Ž(𝑑)|π‘₯|+𝑏(𝑑)+𝑐(𝑑)|𝑧|+𝑑(𝑑)|π‘₯|πœƒ+π‘Ÿ(𝑑),(3.2)

(H3) there exists a constant 𝐡>0 such that for every π‘Ž,π‘βˆˆβ„ satisfying π‘Ž2+𝑏2>𝐡, then either π‘Žπ‘…1π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ+𝑏𝑅2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ<0,(3.3) or else π‘Žπ‘…1π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ+𝑏𝑅2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ>0.(3.4)

Remark 3.1. 𝑅1𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2) and 𝑅2𝑁(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 𝑅1𝑁 and 𝑅2𝑁, respectively.

Theorem 3.2. If (H1)–(H3) hold, then boundary value problem (1.1)-(1.2) has at least one solution provided that β€–π‘Žβ€–1+‖𝑏‖1+‖𝑐‖1<1𝜏,(3.5) where 𝜏=5+2/Ξ“(𝛼)+1/Ξ“(π›Όβˆ’1).

Proof. Set Ξ©1[]={π‘’βˆˆdom(𝐿)⧡Ker(𝐿)βˆ£πΏπ‘’=πœ†π‘π‘’forsomeπœ†βˆˆ0,1},(3.6) then for π‘’βˆˆΞ©1,𝐿𝑒=πœ†π‘π‘’; thus, πœ†β‰ 0,π‘π‘’βˆˆIm(𝐿)=Ker(𝑄), and hence 𝑄𝑁𝑒(𝑑)=0 for all π‘‘βˆˆ[0,1]. By the definition of 𝑄, we have 𝑄1𝑁𝑒(𝑑)=𝑄2𝑁𝑒(𝑑)=0. It follows from (H1) 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+𝑒(𝑑)=π·π›Όβˆ’10+𝑒𝑑0ξ€Έ+ξ€œπ‘‘π‘‘0𝐷𝛼0+𝐷𝑒(𝑠)𝑑𝑠,π›Όβˆ’20+𝑒(𝑑)=π·π›Όβˆ’20+𝑒𝑑0ξ€Έ+ξ€œπ‘‘π‘‘0π·π›Όβˆ’10+𝑒(𝑠)𝑑𝑠,(3.7) so ||π·π›Όβˆ’10+||≀‖‖𝐷𝑒(0)π›Όβˆ’10+‖‖𝑒(𝑑)βˆžβ‰€||π·π›Όβˆ’10+𝑒𝑑0ξ€Έ||+‖‖𝐷𝛼0+𝑒‖‖1≀𝐴+‖𝐿𝑒‖1≀𝐴+‖𝑁𝑒‖1,||π·π›Όβˆ’20+||≀‖‖𝐷𝑒(0)π›Όβˆ’20+‖‖𝑒(𝑑)βˆžβ‰€||π·π›Όβˆ’20+𝑒𝑑0ξ€Έ||+β€–β€–π·π›Όβˆ’10+π‘’β€–β€–βˆžβ‰€||π·π›Όβˆ’20+𝑒𝑑0ξ€Έ||+||π·π›Όβˆ’10+𝑒𝑑0ξ€Έ||+‖‖𝐷𝛼0+𝑒‖‖1≀𝐴+‖𝐿𝑒‖1≀𝐴+‖𝑁𝑒‖1.(3.8) Now by (3.8), we have β€–π‘ƒπ‘’β€–πΆπ›Όβˆ’1=β€–β€–β€–1𝐷Γ(𝛼)π›Όβˆ’10+𝑒(0)π‘‘π›Όβˆ’1+1𝐷Γ(π›Όβˆ’1)π›Όβˆ’20+𝑒(0)π‘‘π›Όβˆ’2β€–β€–β€–πΆπ›Όβˆ’1=β€–β€–β€–1𝐷Γ(𝛼)0π›Όβˆ’1𝑒(0)π‘‘π›Όβˆ’1+1𝐷Γ(π›Όβˆ’1)0π›Όβˆ’2𝑒(0)π‘‘π›Όβˆ’2β€–β€–β€–βˆž+β€–β€–π·π›Όβˆ’10+‖‖𝑒(0)∞+β€–β€–π·π›Όβˆ’10+𝑒(0)𝑑+π·π›Όβˆ’20+‖‖𝑒(0)βˆžβ‰€ξ‚΅12+ξ‚Ά||𝐷Γ(𝛼)π›Όβˆ’10+||+ξ‚΅1𝑒(0)1+ξ‚Ά||𝐷Γ(π›Όβˆ’1)π›Όβˆ’20+||≀1𝑒(0)2+ξ‚Άξ€·Ξ“(𝛼)𝐴+‖𝑁𝑒‖1ξ€Έ+ξ‚΅11+ξ‚Άξ€·Ξ“(π›Όβˆ’1)𝐴+‖𝑁𝑒‖1ξ€Έ.(3.9) Note that (πΌβˆ’π‘ƒ)π‘’βˆˆIm(𝐾𝑃)=dom(𝐿)∩Ker(𝑃) for π‘’βˆˆΞ©1, then, by (2.28) and (2.30), β€–β€–(πΌβˆ’π‘ƒ)π‘’πΆπ›Όβˆ’1=‖‖𝐾𝑃‖‖𝐿(πΌβˆ’π‘ƒ)πΆπ›Όβˆ’1≀12βˆ’ξ‚ΆΞ“(𝛼)‖𝐿(πΌβˆ’π‘ƒ)𝑒‖1=ξ‚΅12βˆ’ξ‚ΆΞ“(𝛼)‖𝐿𝑒‖1≀12βˆ’ξ‚ΆΞ“(𝛼)‖𝑁𝑒‖1.(3.10) Using (3.9) and (3.10), we obtain β€–π‘’β€–πΆπ›Όβˆ’1=‖𝑃𝑒+(πΌβˆ’π‘ƒ)π‘’β€–πΆπ›Όβˆ’1β‰€β€–π‘ƒπ‘’β€–πΆπ›Όβˆ’1β€–+β€–(πΌβˆ’π‘ƒ)π‘’πΆπ›Όβˆ’1≀12+ξ‚Άξ€·Ξ“(𝛼)𝐴+‖𝑁𝑒‖1ξ€Έ+ξ‚΅11+ξ‚Άξ€·Ξ“(π›Όβˆ’1)𝐴+‖𝑁𝑒‖1ξ€Έ+ξ‚΅12+ξ‚ΆΞ“(𝛼)‖𝑁𝑒‖1=ξ‚΅25++1Ξ“(𝛼)ξ‚ΆΞ“(π›Όβˆ’1)‖𝑁𝑒‖1+ξ‚΅13++1Ξ“(𝛼)𝐴Γ(π›Όβˆ’1)=πœβ€–π‘π‘’β€–1+𝐢1,(3.11) where 𝐢1=(3+1/Ξ“(𝛼)+1/Ξ“(π›Όβˆ’1))𝐴 is a constant. This is for all π‘’βˆˆΞ©1, β€–π‘’β€–πΆπ›Όβˆ’1β‰€πœβ€–π‘π‘’β€–1+𝐢1.(3.12) If the first condition of (H2) is satisfied, then we have ξ€½maxβ€–π‘’β€–βˆž,β€–β€–π·π›Όβˆ’10+π‘’β€–β€–βˆž,β€–β€–π·π›Όβˆ’20+π‘’β€–β€–βˆžξ€Ύβ‰€β€–π‘’β€–πΆπ›Όβˆ’1ξ‚€β‰€πœβ€–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œƒβˆž+β€–π‘Ÿβ€–1+‖𝑒‖1+𝐢1,(3.13) and consequently, β€–π‘’β€–βˆžβ‰€πœ1βˆ’β€–π‘Žβ€–1πœξ‚€β€–π‘β€–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œƒβˆž+β€–π‘Ÿβ€–1+‖𝑒‖1+𝐢11βˆ’β€–π‘Žβ€–1𝜏,(3.14)β€–β€–π·π›Όβˆ’10+π‘’β€–β€–βˆžβ‰€πœ1βˆ’β€–π‘Žβ€–1πœβˆ’β€–π‘β€–1πœξ‚€β€–π‘β€–1β€–β€–π·π›Όβˆ’20+π‘’β€–β€–βˆž+‖𝑑‖1β€–β€–π·π›Όβˆ’20+π‘’β€–β€–πœƒβˆž+β€–π‘Ÿβ€–1+‖𝑒‖1+𝐢11βˆ’β€–π‘Žβ€–1πœβˆ’β€–π‘β€–1𝜏,(3.15)β€–β€–π·π›Όβˆ’10+π‘’β€–β€–βˆžβ‰€πœβ€–π‘‘β€–1β€–β€–π·π›Όβˆ’20+π‘’β€–β€–πœƒβˆž1βˆ’β€–π‘Žβ€–1πœβˆ’β€–π‘β€–1πœβˆ’β€–π‘β€–1𝜏+πœξ€·β€–π‘Ÿβ€–1+‖𝑒‖1ξ€Έ+𝐢11βˆ’β€–π‘Žβ€–1πœβˆ’β€–π‘β€–1πœβˆ’β€–π‘β€–1𝜏.(3.16) Note that πœƒβˆˆ[0,1) and β€–π‘Žβ€–1+‖𝑏‖1+‖𝑐‖1<1/𝜏, so there exists 𝑀1>0 such that β€–π·π›Όβˆ’10+π‘’β€–βˆžβ‰€π‘€1 for all π‘’βˆˆΞ©1. The inequalities (3.14) and (3.15) show that there exist 𝑀2,𝑀3>0 such that β€–π·π›Όβˆ’10+π‘’β€–βˆžβ‰€π‘€2,β€–π‘’β€–βˆžβ‰€π‘€3 for all π‘’βˆˆΞ©1. Therefore, for all π‘’βˆˆΞ©1,β€–π‘’β€–πΆπ›Όβˆ’1=β€–π‘’β€–βˆž+β€–π·π›Όβˆ’10+π‘’β€–βˆž+β€–π·π›Όβˆ’20+π‘’β€–βˆžβ‰€π‘€1+𝑀2+𝑀3, that is, Ξ©1 is bounded given the first condition of (H2). If the other conditions of (H2) hold, by using an argument similar to the above, we can prove that Ξ©1 is also bounded.
Let Ξ©2={π‘’βˆˆKer(𝐿)βˆ£π‘π‘’βˆˆIm(𝐿)}.(3.17) For π‘’βˆˆΞ©2,π‘’βˆˆKer(𝐿)={π‘’βˆˆdom(𝐿)βˆ£π‘’=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2,π‘Ž,π‘βˆˆβ„,π‘‘βˆˆ[0,1]}, and 𝑄𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2)=0; thus, 𝑅1𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2)=𝑅2𝑁(π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2)=0. By (H3), π‘Ž2+𝑏2≀𝐡, that is, Ξ©2 is bounded.
We define the isomorphism 𝐽∢Im(𝑄)β†’Ker(𝐿) by π½ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έ=π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2,π‘Ž,π‘βˆˆβ„.(3.18)
If the first part of (H3) is satisfied, let Ξ©3=ξ€½π‘’βˆˆKerπΏβˆΆβˆ’πœ†π½βˆ’1[]𝑒+(1βˆ’πœ†)𝑄𝑁𝑒=0,πœ†βˆˆ0,1.(3.19) For every π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2∈Ω3, πœ†ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2𝑅=(1βˆ’πœ†)ξ€Ίξ€·1π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2π‘‘ξ€Έξ€Έπ›Όβˆ’1+𝑅2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2π‘‘ξ€Έξ€Έπ›Όβˆ’2ξ€».(3.20) 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+𝑏𝑅2π‘ξ€·π‘Žπ‘‘π›Όβˆ’1+π‘π‘‘π›Όβˆ’2ξ€Έξ€»<0,(3.21) which, in either case, obtain a contradiction. If the other part of (H3) is satisfied, then we take Ξ©3=ξ€½π‘’βˆˆKerπΏβˆΆπœ†π½βˆ’1[]𝑒+(1βˆ’πœ†)𝑄𝑁𝑒=0,πœ†βˆˆ0,1,(3.22) and, again, obtain a contradiction. Thus, in either case, β€–π‘’β€–πΆπ›Όβˆ’1=β€–π‘’β€–βˆž+β€–β€–π·π›Όβˆ’10+π‘’β€–β€–βˆž+β€–β€–π·π›Όβˆ’20+π‘’β€–β€–βˆž=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€–βˆž||𝑏||≀(1+2Ξ“(𝛼))|π‘Ž|+(1+Ξ“(π›Όβˆ’1))≀(2+2Ξ“(𝛼)+Ξ“(π›Όβˆ’1))|π‘Ž|,(3.23) for all π‘’βˆˆΞ©3, that is, Ξ©3 is bounded.
In the following, we will prove that all the conditions of Theorem 1.1 are satisfied. Set Ξ© to be a bounded open set of π‘Œ such that π‘ˆ3𝑖=1Ξ©π‘–βŠ‚Ξ©. by Lemma 2.8, the operator 𝐾𝑃(πΌβˆ’π‘„)π‘βˆΆΞ©β†’π‘Œ is compact; thus, 𝑁 is 𝐿-compact on Ξ©, then by the above argument, we have(i)πΏπ‘’β‰ πœ†π‘π‘₯, for every (𝑒,πœ†)∈[(dom(𝐿)⧡Ker𝐿)βˆ©πœ•Ξ©]Γ—(0,1),(ii)π‘π‘’βˆ‰Im(𝐿), for every π‘’βˆˆKer(𝐿)βˆ©πœ•Ξ©.Finally, we will prove that (iii) of Theorem 1.1 is satisfied. Let 𝐻(𝑒,πœ†)=±𝐼𝑒+(1βˆ’πœ†)𝐽𝑄𝑁𝑒, where 𝐼 is the identity operator in the Banach space π‘Œ. According to the above argument, we know that 𝐻(𝑒,πœ†)β‰ 0,βˆ€π‘’βˆˆπœ•Ξ©βˆ©Ker(𝐿),(3.24) and thus, by the homotopy property of degree, ξ‚€||deg𝐽𝑄𝑁Ker(𝐿),Ω∩Ker(𝐿),0=deg(𝐻(…,0),Ω∩Ker(𝐿),0)=deg(𝐻(…,1),Ω∩Ker(𝐿),0)=deg(±𝐼,Ω∩Ker(𝐿),0)=Β±1β‰ 0,(3.25) then by Theorem 1.1, 𝐿𝑒=𝑁𝑒 has at least one solution in dom(𝐿)∩Ω, so boundary problem (1.1), (1.2) has at least one solution in the space πΆπ›Όβˆ’1[0,1]. The proof is finished.

Acknowledgments

This work is supported by the NSFC (no. 11061030, no. 11026060), the Fundamental Research Funds for the Gansu Universities, the nwnu-kjcxgc-03-69, nwnu-kjcxgc-03-61.

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Copyright Β© 2011 Xiaoling Han and Ting Wang. 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.


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