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International Journal of Differential Equations
Volume 2011 (2011), Article ID 401803, 14 pages
http://dx.doi.org/10.1155/2011/401803
Research Article

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

Department of Mathematics, Northwest Normal University, Lanzhou 730070, China

Received 16 May 2011; Accepted 16 June 2011

Academic Editor: Nikolai Leonenko

Copyright © 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.

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+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼10+𝑢(𝑡),𝐷𝛼20+𝐼𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.1)3𝛼0+𝑢(0)=0,𝐷𝛼20+𝑢(0)=𝑛𝑗=1𝛽𝑗𝐷𝛼20+𝑢𝜉𝑗,𝑢(1)=𝑚𝑖=1𝛼𝑖𝑢𝜂𝑖,(1.2) where 2<𝛼3,0<𝜉1<𝜉2<<𝜉𝑛<1,𝑛1,0<𝜂1<<𝜂𝑚<1,𝑚2,𝛼𝑖,𝛽𝑗,𝑚𝑖=1𝛼𝑖𝜂𝑖𝛼1=𝑚𝑖=1𝛼𝑖𝜂𝑖𝛼2=1,𝑛𝑗=1𝛽𝑗𝜉𝑗=0,𝑛𝑗=1𝛽𝑗=1,(1.3)𝑓[0,1]×3 satisfying the Carathéodory conditions, 𝑒𝐿1[0,1]. 𝐷𝛼0+ and 𝐼𝛼0+ are the standard Riemann-Liouville derivative and integral, respectively. We assume, in addition, that 𝑅=Γ(𝛼)2Γ(𝛼1)Γ(2𝛼)Γ(𝛼+1)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗1𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼1Γ(𝛼)2Γ(𝛼1)Γ(𝛼+2)Γ(2𝛼1)𝑛𝑗=1𝛽𝑗𝜉𝑗𝛼+11𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼20,(1.4) where Γ 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, [36]. 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+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼(𝑛1)0+𝑢(𝑡),,𝐷𝛼10+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.5) subject to the following boundary value conditions: 𝐼𝑛𝛼0+𝑢(0)=𝐷𝛼(𝑛1)0+𝑢(0)==𝐷𝛼20+𝑢(0)=0,𝑢(1)=𝜎𝑢(𝜂),(1.6) where 𝑛>2 is a natural number, 𝑛1<𝛼𝑛 is a real number, 𝑓[0,1]×𝑛 is continuous, and 𝑒𝐿1[0,1],𝜎(0,), and 𝜂(0,1) are given constants such that 𝜎𝜂𝛼1=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 [818], 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 𝑌,𝑍 be real Banach spaces, let 𝐿dom(𝐿)𝑌𝑍 be a Fredholm map of index zero, and let 𝑃𝑌𝑌,𝑄𝑍𝑍 be continuous projectors such that Im(𝑃)=Ker(𝑃),Ker(𝑄)=Im(𝐿), and 𝑌=Ker(𝐿)Ker(𝑃),𝑍=Im(𝐿)Im(𝑄). It follows that 𝐿|dom(𝐿)Ker(𝑃)dom(𝐿)Ker(𝑃)Im(𝐿) is invertible. We denote the inverse of the map by 𝐾𝑃. If Ω is an open-bounded subset of 𝑌 such that dom(𝐿)Ω, the map 𝑁𝑌𝑍 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾𝑃(𝐼𝑄)𝑁Ω𝑌 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 Ω. Assume that the following conditions are satisfied: (i)𝐿𝑥𝜆𝑁𝑥 for every (𝑥,𝜆)[(dom(𝐿)Ker(𝐿))𝜕Ω]×(0,1),(ii)𝑁𝑥Im(𝐿) for every 𝑥Ker(𝐿)𝜕Ω,(iii)deg(𝐽𝑄𝑁|Ker(𝐿),ΩKer(𝐿),0)0, where 𝑄𝑍𝑍 is a projection as above with Im(𝐿)=Ker(𝑄), and 𝐽Im(𝑄)Ker(𝐿) is any isomorphism,
then the equation 𝐿𝑥=𝑁𝑥 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 𝑦(0,) is given by 𝐼𝛼0+1𝑦(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑦(𝑠)𝑑𝑠,(2.1) provided the right side is pointwise defined on (0,). And we let 𝐼00+𝑦(𝑡)=𝑦(𝑡) for every continuous 𝑦(0,).

Definition 2.2. The fractional derivative of order 𝛼>0 of a function 𝑦(0,) is given by 𝐷𝛼0+1𝑦(𝑡)=𝑑Γ(𝑛𝛼)𝑑𝑡𝑛𝑡0𝑦(𝑠)(𝑡𝑠)𝛼𝑛+1𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, provided the right side is pointwise defined on (0,).

Lemma 2.3 (see [3]). Assume that 𝑢𝐶(0,1)𝐿1[0,1] with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)𝐿1[0,1], then 𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝐶1𝑡𝛼1+𝐶2𝑡𝛼2++𝐶𝑁𝑡𝛼𝑁,(2.3) for some 𝐶𝑖,𝑖=1,2,,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

We use the classical space 𝐶[0,1] with the norm 𝑥=max𝑡[0,1]|𝑥(𝑡)|. Given 𝜇>0 and 𝑁=[𝜇]+1, one can define a linear space 𝐶𝜇[]0,1=𝑢𝑢(𝑡)=𝐼𝜇0+𝑥(𝑡)+𝑐1𝑡𝜇1+𝑐2𝑡𝜇2++𝑐𝑁1𝑡𝜇(𝑁1)[],𝑡0,1,(2.4) where 𝑥𝐶[0,1] and 𝑐𝑖,𝑖=1,2,,𝑁1. By means of the linear function analysis theory, one can prove that with the norm 𝑢𝐶𝜇=𝐷𝜇0+𝑢++𝐷𝜇(𝑁1)0+𝑢+𝑢,𝐶𝜇[0,1] is a Banach space.

Lemma 2.4 (see [7]). 𝐹𝐶𝜇[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 𝑢𝐹, 𝑢𝐶𝜇=𝐷𝜇0+𝑢𝐷++𝜇(𝑁1)0+𝑢+𝑢<𝑀,(2.5) and equicontinuous means that forall𝜀>0,𝛿>0 such that ||𝑢𝑡1𝑡𝑢2||<𝜀,𝑡1,𝑡2[],||𝑡0,11𝑡2||,||𝐷<𝛿,𝑢𝐹𝛼𝑖0+𝑢𝑡1𝐷𝛼𝑖0+𝑢𝑡2||𝑡<𝜀,1,𝑡2[],||𝑡0,11𝑡2||.<𝛿,𝑢𝐹,𝑖{0,,𝑁1}(2.6)

Let 𝑍=𝐿1[0,1] with the norm 𝑔1=10|𝑔(𝑠)|𝑑𝑠. 𝑌=𝐶𝛼1[0,1]={𝑢𝑢(𝑡)=𝐼𝛼10+𝑥(𝑡)+𝑐𝑡𝛼2,𝑡[0,1]}, where 𝑥𝐶[0,1],𝑐, with the norm 𝑢𝐶𝛼1=𝐷𝛼10+𝑢+𝐷𝛼20+𝑢+𝑢, and 𝑌 is a Banach space.

Definition 2.5. By a solution of the boundary value problem (1.1), (1.2), we understand a function 𝑢𝐶𝛼1[0,1] such that 𝐷𝛼10+𝑢 is absolutely continuous on (0,1) and satisfies (1.1), (1.2).

Definition 2.6. We say that the map 𝑓[0,1]× satisfies the Carathéodory conditions with respect to 𝐿1[0,1] if the following conditions are satisfied:(i)for each 𝑧, the mapping 𝑡𝑓(𝑡,𝑧) is Lebesgue measurable,(ii)for almost every 𝑡[0,1], the mapping 𝑧𝑓(𝑡,𝑧) is continuous on ,(iii)for each 𝑟>0, there exists 𝜌𝑟𝐿1([0,1],) such that, for a.e., 𝑡[0,1] and every |𝑧|𝑟, we have |𝑓(𝑡,𝑧)|𝜌𝑟(𝑡).

Define 𝐿 to be the linear operator from dom(𝐿)𝑌 to 𝑍 with dom(𝐿)=𝑢𝐶𝛼1[]0,1𝐷𝛼0+𝑢𝐿1[],0,1,𝑢satises(1.2)𝐿𝑢=𝐷𝛼0+𝑢,𝑢dom(𝐿).(2.7) We define 𝑁𝑌𝑍 by setting 𝑁𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼10+𝑢(𝑡),𝐷𝛼20+𝑢(𝑡)+𝑒(𝑡).(2.8) Then boundary value problem (1.1), (1.2) can be written as 𝐿𝑢=𝑁𝑢.(2.9)

Lemma 2.7. Let condition (1.3) and (1.4) hold, then 𝐿dom(𝐿)𝑌𝑍 is a Fredholm map of index zero.

Proof. It is clear that Ker(𝐿)={𝑎𝑡𝛼1+𝑏𝑡𝛼2𝑎,𝑏}2.
Let 𝑔𝑍 and 1𝑢(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔(𝑠)𝑑𝑠+𝑐1𝑡𝛼1+𝑐2𝑡𝛼2,(2.10) then 𝐷𝛼0+𝑢(𝑡)=𝑔(𝑡)a.e.,𝑡(0,1) and, if 10(1𝑠)𝛼1𝑔(𝑠)𝑑𝑠𝑚𝑖=1𝛼𝑖𝜂𝑖0𝜂𝑖𝑠𝛼1𝑔(𝑠)𝑑𝑠=0,𝑛𝑗=1𝛽𝑗𝜉𝑗0𝜉𝑗𝑠𝑔(𝑠)𝑑𝑠=0(2.11) hold. Then 𝑢(𝑡) satisfies the boundary conditions (1.2), that is, 𝑢dom(𝐿), and we have {𝑔𝑍𝑔satises(2.11)}Im(𝐿).(2.12) Let 𝑢dom(𝐿), then for 𝐷𝛼0+𝑢Im(𝐿), we have 𝑢(𝑡)=𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)+𝑐1𝑡𝛼1+𝑐2𝑡𝛼2+𝑐3𝑡𝛼3,(2.13) which, due to the boundary value condition (1.2), implies that 𝐷𝛼0+𝑢 satisfies (2.11). In fact, from 𝐼3𝛼0+𝑢(0)=0, we have 𝑐3=0, from 𝑢(1)=𝑚𝑖=1𝛼𝑖𝑢(𝜂𝑖), we have 10(1𝑠)𝛼1𝐷𝛼0+𝑢(𝑠)𝑑𝑠𝑚𝑖=1𝛼𝑖𝜂𝑖0𝜂𝑖𝑠𝛼1𝐷𝛼0+𝑢(𝑠)𝑑𝑠=0,(2.14) and from 𝐷𝛼20+𝑢(0)=𝑛𝑗=1𝛽𝑗𝐷𝛼20+𝑢(𝜉𝑗), we have 𝑛𝑗=1𝛽𝑗𝜉𝑗0𝜉𝑗𝐷𝑠𝛼0+𝑢(𝑠)𝑑𝑠=0.(2.15) Hence, Im(𝐿){𝑔𝑍𝑔satises(2.11)}.(2.16) Therefore, Im(𝐿)={𝑔𝑍𝑔satises(2.11)}.(2.17) Consider the continuous linear mapping 𝑄1𝑍𝑍 and 𝑄2𝑍𝑍 defined by 𝑄1𝑔=10(1𝑠)𝛼1𝑔(𝑠)𝑑𝑠𝑚𝑖=1𝛼𝑖𝜂𝑖0𝜂𝑖𝑠𝛼1𝑄𝑔(𝑠)𝑑𝑠,2𝑔=𝑛𝑗=1𝛽𝑗𝜉𝑗0𝜉𝑗𝑠𝑔(𝑠)𝑑𝑠.(2.18) Using the above definitions, we construct the following auxiliary maps 𝑅1,𝑅2𝑍𝑍: 𝑅11𝑔=𝑅Γ(𝛼1)Γ(𝛼+1)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗𝑄1𝑔(𝑡)Γ(𝛼)Γ(𝛼1)Γ(2𝛼1)1𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼2𝑄2,𝑅𝑔(𝑡)21𝑔=𝑅Γ(𝛼)Γ(𝛼+2)𝑛𝑗=1𝛽𝑗𝜉𝑗𝛼+1𝑄1𝑔(𝑡)(Γ(𝛼))2Γ(2𝛼)1𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼1𝑄2.𝑔(𝑡)(2.19) Since the condition (1.4) holds, the mapping 𝑄𝑍𝑍 defined by 𝑅(𝑄𝑦)(𝑡)=1𝑡𝑔(𝑡)𝛼1+𝑅2𝑡𝑔(𝑡)𝛼2(2.20) is well defined.
Recall (1.4) and note that 𝑅1𝑅1𝑔𝑡𝛼1=1𝑅Γ(𝛼1)Γ(𝛼+1)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗𝑄1𝑅1𝑔𝑡𝛼1Γ(𝛼)Γ(𝛼1)Γ(2𝛼1)1𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼2𝑄2𝑅1𝑔𝑡𝛼1=𝑅1𝑔1𝑅𝛼Γ(𝛼1)Γ2Γ(𝛼+1)Γ(2𝛼)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗1𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼1𝛼Γ(𝛼1)Γ2Γ(2𝛼1)Γ(𝛼+2)1𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼2𝑛𝑗=1𝛽𝑗𝜉𝑗𝛼+1=𝑅1𝑔,(2.21) and similarly we can derive that 𝑅1𝑅2𝑔𝑡𝛼2𝑅=0,2𝑅1𝑔𝑡𝛼1𝑅=0,2𝑅2𝑔𝑡𝛼2=𝑅2𝑔.(2.22) So, for 𝑔𝑍, it follows from the four relations above that 𝑄2𝑔=𝑅1𝑅1𝑔𝑡𝛼1+𝑅2𝑔𝑡𝛼2𝑡𝛼1+𝑅2𝑅1𝑔𝑡𝛼1+𝑅2𝑔𝑡𝛼2𝑡𝛼2=𝑅1𝑅1𝑔𝑡𝛼1𝑡𝛼1+𝑅1𝑅2𝑔𝑡𝛼2𝑡𝛼1+𝑅2𝑅1𝑔𝑡𝛼1𝑡𝛼2+𝑅2𝑅2𝑔𝑡𝛼2𝑡𝛼2=𝑅1𝑔𝑡𝛼1+𝑅2𝑔𝑡𝛼2=𝑄𝑔,(2.23) that is, the map 𝑄 is idempotent. In fact, 𝑄 is a continuous linear projector.
Note that 𝑔Im(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, 𝑔Im(𝐿). In fact, Im(𝐿)=Ker(𝑄).
Take 𝑔𝑍 in the form 𝑔=(𝑔𝑄𝑔)+𝑄𝑔, so that 𝑔𝑄𝑔Im(𝐿)=Ker(𝑄) and 𝑄𝑔Im(𝑄). Thus, 𝑍=Im(𝐿)+Im(𝑄). Let 𝑔Im(𝐿)Im(𝑄) and assume that 𝑔(𝑠)=𝑎𝑠𝛼1+𝑏𝑠𝛼2 is not identically zero on [0,1], then, since 𝑔Im(𝐿), from (2.11) and the condition (1.4), we derive 𝑎=𝑏=0, which is a contradiction. Hence, Im(𝐿)Im(𝑄)={0}; thus, 𝑍=Im(𝐿)Im(𝑄).
Now, dimKer(𝐿)=2=codimIm(𝐿), and so 𝐿 is a Fredholm operator of index zero.

Let 𝑃𝑌𝑌 be defined by 1𝑃𝑢(𝑡)=Γ𝐷(𝛼)𝛼10+𝑢(0)𝑡𝛼1+1Γ𝐷(𝛼1)𝛼20+𝑢(0)𝑡𝛼2[],𝑡0,1.(2.24) Note that 𝑃 is a continuous linear projector and Ker(𝑃)=𝑢𝑌𝐷𝛼10+𝑢(0)=𝐷𝛼20+𝑢(0)=0.(2.25) It is clear that 𝑌=Ker(𝐿)Ker(𝑃).

Note that the projectors 𝑃 and 𝑄 are exact. Define 𝐾𝑃Im(𝐿)dom(𝐿)Ker(𝑃) by 𝐾𝑃1𝑔(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔(𝑠)𝑑𝑠=𝐼𝛼0+𝑔(𝑡).(2.26) Hence, we have 𝐷𝛼10+𝐾𝑃𝑔(𝑡)=𝑡0𝑔(𝑠)𝑑𝑠,𝐷𝛼20+𝐾𝑃𝑔(𝑡)=𝑡0(𝑡𝑠)𝑔(𝑠)𝑑𝑠,(2.27) then 𝐾𝑃𝑔(1/Γ(𝛼))𝑔1,𝐷𝛼10+(𝐾𝑃𝑔)𝑔1,𝐷𝛼20+(𝐾𝑃𝑔)𝑔1, and thus 𝐾𝑃𝑔𝐶𝛼112+Γ(𝛼)𝑔1.(2.28) In fact, if 𝑔Im(𝐿), then (𝐿𝐾𝑃)𝑔=𝐷𝛼0+𝐼𝛼0+𝑔=𝑔. Also, if 𝑢dom(𝐿)Ker(𝑃), then 𝐾𝑃𝐿𝑔(𝑡)=𝐼𝛼0+𝐷𝛼0+𝑔(𝑡)=𝑔(𝑡)+𝑐1𝑡𝛼1+𝑐2𝑡𝛼2+𝑐3𝑡𝛼3,(2.29) from boundary value condition (1.2) and the fact that 𝑢dom(𝐿)Ker(𝑃), we have 𝑐1=𝑐2=𝑐3=0. Thus, 𝐾𝑃=𝐿||dom(𝐿)Ker(𝑃)1.(2.30) Using (2.19), we write 𝑅𝑄𝑁𝑢(𝑡)=1𝑡𝑁𝑢𝛼1+𝑅2𝑡𝑁𝑢𝛼2,𝐾𝑃1(𝐼𝑄)𝑁𝑢(𝑡)=Γ(𝛼)10(𝑡𝑠)𝛼1[]𝑁𝑢(𝑠)𝑄𝑁𝑢(𝑠)𝑑𝑠.(2.31) With arguments similar to those of [7], we obtain the following Lemma.

Lemma 2.8. 𝐾𝑃(𝐼𝑄)𝑁𝑌𝑌 is completely continuous.

3. The Main Results

Assume that the following conditions on the function 𝑓(𝑡,𝑥,𝑦,𝑧) are satisfied:

(H1) there exists a constant 𝐴>0, such that for 𝑢dom(𝐿)Ker(𝐿) satisfying |𝐷𝛼10+𝑢(𝑡)|+|𝐷𝛼20+𝑢(𝑡)|>𝐴 for all 𝑡[0,1], we have 𝑄1𝑁𝑢(𝑡)0or𝑄2𝑁𝑢(𝑡)0,(3.1)

(H2) there exist functions 𝑎,𝑏,𝑐,𝑑,𝑟𝐿1[0,1] and a constant 𝜃[0,1] such that for all (𝑥,𝑦,𝑧)3 and a.e., 𝑡[0,1], one of the following inequalities is satisfied:||||||𝑦||𝑓(𝑡,𝑥,𝑦,𝑧)𝑎(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑧|𝜃||||||𝑦||||𝑦||+𝑟(𝑡),𝑓(𝑡,𝑥,𝑦,𝑧)𝑎(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)𝜃||||||𝑦||+𝑟(𝑡),𝑓(𝑡,𝑥,𝑦,𝑧)𝑎(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑥|𝜃+𝑟(𝑡),(3.2)

(H3) there exists a constant 𝐵>0 such that for every 𝑎,𝑏 satisfying 𝑎2+𝑏2>𝐵, then either 𝑎𝑅1𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2+𝑏𝑅2𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2<0,(3.3) or else 𝑎𝑅1𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2+𝑏𝑅2𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2>0.(3.4)

Remark 3.1. 𝑅1𝑁(𝑎𝑡𝛼1+𝑏𝑡𝛼2) and 𝑅2𝑁(𝑎𝑡𝛼1+𝑏𝑡𝛼2) from (H3) stand for the images of 𝑢(𝑡)=𝑎𝑡𝛼1+𝑏𝑡𝛼2 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 𝑎1+𝑏1+𝑐1<1𝜏,(3.5) where 𝜏=5+2/Γ(𝛼)+1/Γ(𝛼1).

Proof. Set Ω1[]={𝑢dom(𝐿)Ker(𝐿)𝐿𝑢=𝜆𝑁𝑢forsome𝜆0,1},(3.6) then for 𝑢Ω1,𝐿𝑢=𝜆𝑁𝑢; thus, 𝜆0,𝑁𝑢Im(𝐿)=Ker(𝑄), and hence 𝑄𝑁𝑢(𝑡)=0 for all 𝑡[0,1]. By the definition of 𝑄, we have 𝑄1𝑁𝑢(𝑡)=𝑄2𝑁𝑢(𝑡)=0. It follows from (H1) that there exists 𝑡0[0,1] such that |𝐷𝛼10+𝑢(𝑡0)|+|𝐷𝛼20+𝑢(𝑡0)|𝐴. Now, 𝐷𝛼10+𝑢(𝑡)=𝐷𝛼10+𝑢𝑡0+𝑡𝑡0𝐷𝛼0+𝐷𝑢(𝑠)𝑑𝑠,𝛼20+𝑢(𝑡)=𝐷𝛼20+𝑢𝑡0+𝑡𝑡0𝐷𝛼10+𝑢(𝑠)𝑑𝑠,(3.7) so ||𝐷𝛼10+||𝐷𝑢(0)𝛼10+𝑢(𝑡)||𝐷𝛼10+𝑢𝑡0||+𝐷𝛼0+𝑢1𝐴+𝐿𝑢1𝐴+𝑁𝑢1,||𝐷𝛼20+||𝐷𝑢(0)𝛼20+𝑢(𝑡)||𝐷𝛼20+𝑢𝑡0||+𝐷𝛼10+𝑢||𝐷𝛼20+𝑢𝑡0||+||𝐷𝛼10+𝑢𝑡0||+𝐷𝛼0+𝑢1𝐴+𝐿𝑢1𝐴+𝑁𝑢1.(3.8) Now by (3.8), we have 𝑃𝑢𝐶𝛼1=1𝐷Γ(𝛼)𝛼10+𝑢(0)𝑡𝛼1+1𝐷Γ(𝛼1)𝛼20+𝑢(0)𝑡𝛼2𝐶𝛼1=1𝐷Γ(𝛼)0𝛼1𝑢(0)𝑡𝛼1+1𝐷Γ(𝛼1)0𝛼2𝑢(0)𝑡𝛼2+𝐷𝛼10+𝑢(0)+𝐷𝛼10+𝑢(0)𝑡+𝐷𝛼20+𝑢(0)12+||𝐷Γ(𝛼)𝛼10+||+1𝑢(0)1+||𝐷Γ(𝛼1)𝛼20+||1𝑢(0)2+Γ(𝛼)𝐴+𝑁𝑢1+11+Γ(𝛼1)𝐴+𝑁𝑢1.(3.9) Note that (𝐼𝑃)𝑢Im(𝐾𝑃)=dom(𝐿)Ker(𝑃) for 𝑢Ω1, then, by (2.28) and (2.30), (𝐼𝑃)𝑢𝐶𝛼1=𝐾𝑃𝐿(𝐼𝑃)𝐶𝛼112Γ(𝛼)𝐿(𝐼𝑃)𝑢1=12Γ(𝛼)𝐿𝑢112Γ(𝛼)𝑁𝑢1.(3.10) Using (3.9) and (3.10), we obtain 𝑢𝐶𝛼1=𝑃𝑢+(𝐼𝑃)𝑢𝐶𝛼1𝑃𝑢𝐶𝛼1+(𝐼𝑃)𝑢𝐶𝛼112+Γ(𝛼)𝐴+𝑁𝑢1+11+Γ(𝛼1)𝐴+𝑁𝑢1+12+Γ(𝛼)𝑁𝑢1=25++1Γ(𝛼)Γ(𝛼1)𝑁𝑢1+13++1Γ(𝛼)𝐴Γ(𝛼1)=𝜏𝑁𝑢1+𝐶1,(3.11) where 𝐶1=(3+1/Γ(𝛼)+1/Γ(𝛼1))𝐴 is a constant. This is for all 𝑢Ω1, 𝑢𝐶𝛼1𝜏𝑁𝑢1+𝐶1.(3.12) If the first condition of (H2) is satisfied, then we have max𝑢,𝐷𝛼10+𝑢,𝐷𝛼20+𝑢𝑢𝐶𝛼1𝜏𝑎1𝑢+𝑏1𝐷𝛼10+𝑢+𝑐1𝐷𝛼20+𝑢+𝑑1𝐷𝛼20+𝑢𝜃+𝑟1+𝑒1+𝐶1,(3.13) and consequently, 𝑢𝜏1𝑎1𝜏𝑏1𝐷𝛼10+𝑢+𝑐1𝐷𝛼20+𝑢+𝑑1𝐷𝛼20+𝑢𝜃+𝑟1+𝑒1+𝐶11𝑎1𝜏,(3.14)𝐷𝛼10+𝑢𝜏1𝑎1𝜏𝑏1𝜏𝑐1𝐷𝛼20+𝑢+𝑑1𝐷𝛼20+𝑢𝜃+𝑟1+𝑒1+𝐶11𝑎1𝜏𝑏1𝜏,(3.15)𝐷𝛼10+𝑢𝜏𝑑1𝐷𝛼20+𝑢𝜃1𝑎1𝜏𝑏1𝜏𝑐1𝜏+𝜏𝑟1+𝑒1+𝐶11𝑎1𝜏𝑏1𝜏𝑐1𝜏.(3.16) Note that 𝜃[0,1) and 𝑎1+𝑏1+𝑐1<1/𝜏, so there exists 𝑀1>0 such that 𝐷𝛼10+𝑢𝑀1 for all 𝑢Ω1. The inequalities (3.14) and (3.15) show that there exist 𝑀2,𝑀3>0 such that 𝐷𝛼10+𝑢𝑀2,𝑢𝑀3 for all 𝑢Ω1. Therefore, for all 𝑢Ω1,𝑢𝐶𝛼1=𝑢+𝐷𝛼10+𝑢+𝐷𝛼20+𝑢𝑀1+𝑀2+𝑀3, that is, Ω1 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 Ω1 is also bounded.
Let Ω2={𝑢Ker(𝐿)𝑁𝑢Im(𝐿)}.(3.17) For 𝑢Ω2,𝑢Ker(𝐿)={𝑢dom(𝐿)𝑢=𝑎𝑡𝛼1+𝑏𝑡𝛼2,𝑎,𝑏,𝑡[0,1]}, and 𝑄𝑁(𝑎𝑡𝛼1+𝑏𝑡𝛼2)=0; thus, 𝑅1𝑁(𝑎𝑡𝛼1+𝑏𝑡𝛼2)=𝑅2𝑁(𝑎𝑡𝛼1+𝑏𝑡𝛼2)=0. By (H3), 𝑎2+𝑏2𝐵, that is, Ω2 is bounded.
We define the isomorphism 𝐽Im(𝑄)Ker(𝐿) by 𝐽𝑎𝑡𝛼1+𝑏𝑡𝛼2=𝑎𝑡𝛼1+𝑏𝑡𝛼2,𝑎,𝑏.(3.18)
If the first part of (H3) is satisfied, let Ω3=𝑢Ker𝐿𝜆𝐽1[]𝑢+(1𝜆)𝑄𝑁𝑢=0,𝜆0,1.(3.19) For every 𝑎𝑡𝛼1+𝑏𝑡𝛼2Ω3, 𝜆𝑎𝑡𝛼1+𝑏𝑡𝛼2𝑅=(1𝜆)1𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2𝑡𝛼1+𝑅2𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2𝑡𝛼2.(3.20) If 𝜆=1, then 𝑎=𝑏=0, and if 𝑎2+𝑏2>𝐵, then by (H3), 𝜆𝑎2+𝑏2=(1𝜆)𝑎𝑅1𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2+𝑏𝑅2𝑁𝑎𝑡𝛼1+𝑏𝑡𝛼2<0,(3.21) which, in either case, obtain a contradiction. If the other part of (H3) is satisfied, then we take Ω3=𝑢Ker𝐿𝜆𝐽1[]𝑢+(1𝜆)𝑄𝑁𝑢=0,𝜆0,1,(3.22) and, again, obtain a contradiction. Thus, in either case, 𝑢𝐶𝛼1=𝑢+𝐷𝛼10+𝑢+𝐷𝛼20+𝑢=𝑎𝑡𝛼1+𝑏𝑡𝛼2𝐶𝛼1=𝑎𝑡𝛼1+𝑏𝑡𝛼2+𝑎Γ(𝛼)+𝑎Γ(𝛼)𝑡+𝑏Γ(𝛼1)||𝑏||(1+2Γ(𝛼))|𝑎|+(1+Γ(𝛼1))(2+2Γ(𝛼)+Γ(𝛼1))|𝑎|,(3.23) for all 𝑢Ω3, that is, Ω3 is bounded.
In the following, we will prove that all the conditions of Theorem 1.1 are satisfied. Set Ω to be a bounded open set of 𝑌 such that 𝑈3𝑖=1Ω𝑖Ω. by Lemma 2.8, the operator 𝐾𝑃(𝐼𝑄)𝑁Ω𝑌 is compact; thus, 𝑁 is 𝐿-compact on Ω, then by the above argument, we have(i)𝐿𝑢𝜆𝑁𝑥, for every (𝑢,𝜆)[(dom(𝐿)Ker𝐿)𝜕Ω]×(0,1),(ii)𝑁𝑢Im(𝐿), for every 𝑢Ker(𝐿)𝜕Ω.Finally, we will prove that (iii) of Theorem 1.1 is satisfied. Let 𝐻(𝑢,𝜆)=±𝐼𝑢+(1𝜆)𝐽𝑄𝑁𝑢, where 𝐼 is the identity operator in the Banach space 𝑌. According to the above argument, we know that 𝐻(𝑢,𝜆)0,𝑢𝜕ΩKer(𝐿),(3.24) and thus, by the homotopy property of degree, ||deg𝐽𝑄𝑁Ker(𝐿),ΩKer(𝐿),0=deg(𝐻(,0),ΩKer(𝐿),0)=deg(𝐻(,1),ΩKer(𝐿),0)=deg(±𝐼,ΩKer(𝐿),0)=±10,(3.25) then by Theorem 1.1, 𝐿𝑢=𝑁𝑢 has at least one solution in dom(𝐿)Ω, so boundary problem (1.1), (1.2) has at least one solution in the space 𝐶𝛼1[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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