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),𝐷𝛼−10+𝑢(𝑡),𝐷𝛼−20+𝐼𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.1)3−𝛼0+𝑢(0)=0,𝐷𝛼−20+𝑢(0)=𝑛𝑗=1𝛽𝑗𝐷𝛼−20+𝑢𝜉𝑗,𝑢(1)=𝑚𝑖=1𝛼𝑖𝑢𝜂𝑖,(1.2) where 2<𝛼≤3,0<𝜉1<𝜉2<⋯<𝜉𝑛<1,𝑛≥1,0<𝜂1<⋯<𝜂𝑚<1,𝑚≥2,𝛼𝑖,𝛽𝑗∈ℝ,𝑚𝑖=1𝛼𝑖𝜂𝑖𝛼−1=𝑚𝑖=1𝛼𝑖𝜂𝑖𝛼−2=1,𝑛𝑗=1𝛽𝑗𝜉𝑗=0,𝑛𝑗=1𝛽𝑗=1,(1.3)𝑓∶[0,1]×ℝ3→ℝ satisfying the Carathéodory conditions, 𝑒∈𝐿1[0,1]. 𝐷𝛼0+ and 𝐼𝛼0+ are the standard Riemann-Liouville derivative and integral, respectively. We assume, in addition, that 𝑅=Γ(𝛼)2Γ(𝛼−1)Γ(2𝛼)Γ(𝛼+1)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗1−𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼−1−Γ(𝛼)2Γ(𝛼−1)Γ(𝛼+2)Γ(2𝛼−1)𝑛𝑗=1𝛽𝑗𝜉𝑗𝛼+11−𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼−2≠0,(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, [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+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼−(𝑛−1)0+𝑢(𝑡),…,𝐷𝛼−10+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.5) subject to the following boundary value conditions: 𝐼𝑛−𝛼0+𝑢(0)=𝐷𝛼−(𝑛−1)0+𝑢(0)=⋯=𝐷𝛼−20+𝑢(0)=0,𝑢(1)=ğœŽğ‘¢(𝜂),(1.6) where 𝑛>2 is a natural number, 𝑛−1<𝛼≤𝑛 is a real number, 𝑓∶[0,1]×ℝ𝑛→ℝ is continuous, and 𝑒∈𝐿1[0,1],ğœŽâˆˆ(0,∞), and 𝜂∈(0,1) are given constants such that ğœŽğœ‚ğ›¼âˆ’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 𝑌,𝑍 be real Banach spaces, let 𝐿∶dom(𝐿)⊂𝑌→𝑍 be a Fredholm map of index zero, and let 𝑃∶𝑌→𝑌,𝑄∶𝑍→𝑍 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 𝑁∶𝑌→𝑍 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾𝑃(𝐼−𝑄)𝑁∶Ω→𝑌 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)𝐿𝑥≠𝜆𝑁𝑥 for every (𝑥,𝜆)∈[(dom(𝐿)⧵Ker(𝐿))∩𝜕Ω]×(0,1),(ii)𝑁𝑥∉Im(𝐿) for every 𝑥∈Ker(𝐿)∩𝜕Ω,(iii)deg(𝐽𝑄𝑁|Ker(𝐿),Ω∩Ker(𝐿),0)≠0, where 𝑄∶𝑍→𝑍 is a projection as above with Im(𝐿)=Ker(𝑄), and 𝐽∶Im(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,∞)→ℝ is given by 𝐼𝛼0+1𝑦(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑦(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,∞)→ℝ.

Definition 2.2. The fractional derivative of order 𝛼>0 of a function 𝑦∶(0,∞)→ℝ is given by 𝐷𝛼0+1𝑦(𝑡)=𝑑Γ(𝑛−𝛼)𝑑𝑡𝑛𝑡0𝑦(𝑠)(𝑡−𝑠)𝛼−𝑛+1𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, provided the right side is pointwise defined on (0,∞).

Lemma 2.3 (see [3]). 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 𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝐶1𝑡𝛼−1+𝐶2𝑡𝛼−2+⋯+𝐶𝑁𝑡𝛼−𝑁,(2.3) for some 𝐶𝑖∈ℝ,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 â€–ğ‘¥â€–âˆž=max𝑡∈[0,1]|𝑥(𝑡)|. Given 𝜇>0 and 𝑁=[𝜇]+1, one can define a linear space 𝐶𝜇[]0,1∶=𝑢∣𝑢(𝑡)=𝐼𝜇0+𝑥(𝑡)+𝑐1𝑡𝜇−1+𝑐2𝑡𝜇−2+⋯+𝑐𝑁−1𝑡𝜇−(𝑁−1)[],𝑡∈0,1,(2.4) where 𝑥∈𝐶[0,1] and 𝑐𝑖∈ℝ,𝑖=1,2,…,𝑁−1. By means of the linear function analysis theory, one can prove that with the norm ‖𝑢‖𝐶𝜇=‖𝐷𝜇0+ğ‘¢â€–âˆž+⋯+‖𝐷𝜇−(𝑁−1)0+ğ‘¢â€–âˆž+â€–ğ‘¢â€–âˆž,𝐶𝜇[0,1] is a Banach space.

Lemma 2.4 (see [7]). 𝐹⊂𝐶𝜇[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 𝑢∈𝐹, ‖𝑢‖𝐶𝜇=‖‖𝐷𝜇0+ğ‘¢â€–â€–âˆžâ€–â€–ğ·+⋯+𝜇−(𝑁−1)0+ğ‘¢â€–â€–âˆž+â€–ğ‘¢â€–âˆž<𝑀,(2.5) and equicontinuous means that forall𝜀>0,∃𝛿>0 such that ||𝑢𝑡1𝑡−𝑢2||<𝜀,∀𝑡1,𝑡2∈[],||𝑡0,11−𝑡2||,||𝐷<𝛿,∀𝑢∈𝐹𝛼−𝑖0+𝑢𝑡1−𝐷𝛼−𝑖0+𝑢𝑡2||𝑡<𝜀,1,𝑡2∈[],||𝑡0,11−𝑡2||.<𝛿,∀𝑢∈𝐹,∀𝑖∈{0,…,𝑁−1}(2.6)

Let 𝑍=𝐿1[0,1] with the norm ‖𝑔‖1=∫10|𝑔(𝑠)|𝑑𝑠. 𝑌=𝐶𝛼−1[0,1]={𝑢∣𝑢(𝑡)=𝐼𝛼−10+𝑥(𝑡)+𝑐𝑡𝛼−2,𝑡∈[0,1]}, where 𝑥∈𝐶[0,1],𝑐∈ℝ, with the norm ‖𝑢‖𝐶𝛼−1=‖𝐷𝛼−10+ğ‘¢â€–âˆž+‖𝐷𝛼−20+ğ‘¢â€–âˆž+â€–ğ‘¢â€–âˆž, 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 𝑢∈𝐶𝛼−1[0,1] such that 𝐷𝛼−10+𝑢 is absolutely continuous on (0,1) and satisfies (1.1), (1.2).

Definition 2.6. We say that the map 𝑓∶[0,1]×ℝ→ℝ satisfies the Carathéodory conditions with respect to 𝐿1[0,1] if the following conditions are satisfied:(i)for each 𝑧∈ℝ, the mapping 𝑡→𝑓(𝑡,𝑧) is Lebesgue measurable,(ii)for almost every 𝑡∈[0,1], the mapping 𝑧→𝑓(𝑡,𝑧) is continuous on ℝ,(iii)for each 𝑟>0, there exists 𝜌𝑟∈𝐿1([0,1],ℝ) 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)=𝑢∈𝐶𝛼−1[]0,1∣𝐷𝛼0+𝑢∈𝐿1[],0,1,𝑢satisfies(1.2)𝐿𝑢=𝐷𝛼0+𝑢,𝑢∈dom(𝐿).(2.7) We define 𝑁∶𝑌→𝑍 by setting 𝑁𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼−10+𝑢(𝑡),𝐷𝛼−20+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 𝐿∶dom(𝐿)∩𝑌→𝑍 is a Fredholm map of index zero.

Proof. It is clear that Ker(𝐿)={ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2âˆ£ğ‘Ž,𝑏∈ℝ}≅ℝ2.
Let 𝑔∈𝑍 and 1𝑢(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑔(𝑠)𝑑𝑠+𝑐1𝑡𝛼−1+𝑐2𝑡𝛼−2,(2.10) then 𝐷𝛼0+𝑢(𝑡)=𝑔(𝑡)a.e.,𝑡∈(0,1) and, if 10(1−𝑠)𝛼−1𝑔(𝑠)𝑑𝑠−𝑚𝑖=1𝛼𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1𝑔(𝑠)𝑑𝑠=0,𝑛𝑗=1𝛽𝑗𝜉𝑗0𝜉𝑗−𝑠𝑔(𝑠)𝑑𝑠=0(2.11) hold. Then 𝑢(𝑡) satisfies the boundary conditions (1.2), that is, 𝑢∈dom(𝐿), and we have {𝑔∈𝑍∣𝑔satisfies(2.11)}⊆Im(𝐿).(2.12) Let 𝑢∈dom(𝐿), then for 𝐷𝛼0+𝑢∈Im(𝐿), we have 𝑢(𝑡)=𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)+𝑐1𝑡𝛼−1+𝑐2𝑡𝛼−2+𝑐3𝑡𝛼−3,(2.13) which, due to the boundary value condition (1.2), implies that 𝐷𝛼0+𝑢 satisfies (2.11). In fact, from 𝐼3−𝛼0+𝑢(0)=0, we have 𝑐3=0, from ∑𝑢(1)=𝑚𝑖=1𝛼𝑖𝑢(𝜂𝑖), we have 10(1−𝑠)𝛼−1𝐷𝛼0+𝑢(𝑠)𝑑𝑠−𝑚𝑖=1𝛼𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1𝐷𝛼0+𝑢(𝑠)𝑑𝑠=0,(2.14) and from 𝐷𝛼−20+∑𝑢(0)=𝑛𝑗=1𝛽𝑗𝐷𝛼−20+𝑢(𝜉𝑗), we have 𝑛𝑗=1𝛽𝑗𝜉𝑗0𝜉𝑗𝐷−𝑠𝛼0+𝑢(𝑠)𝑑𝑠=0.(2.15) Hence, Im(𝐿)⊆{𝑔∈𝑍∣𝑔satisfies(2.11)}.(2.16) Therefore, Im(𝐿)={𝑔∈𝑍∣𝑔satisfies(2.11)}.(2.17) Consider the continuous linear mapping 𝑄1∶𝑍→𝑍 and 𝑄2∶𝑍→𝑍 defined by 𝑄1𝑔=10(1−𝑠)𝛼−1𝑔(𝑠)𝑑𝑠−𝑚𝑖=1𝛼𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1𝑄𝑔(𝑠)𝑑𝑠,2𝑔=𝑛𝑗=1𝛽𝑗𝜉𝑗0𝜉𝑗−𝑠𝑔(𝑠)𝑑𝑠.(2.18) Using the above definitions, we construct the following auxiliary maps 𝑅1,𝑅2∶𝑍→𝑍: 𝑅11𝑔=𝑅Γ(𝛼−1)Γ(𝛼+1)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗𝑄1𝑔(𝑡)−Γ(𝛼)Γ(𝛼−1)Γ(2𝛼−1)1−𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼−2𝑄2,𝑅𝑔(𝑡)21𝑔=−𝑅Γ(𝛼)Γ(𝛼+2)𝑛𝑗=1𝛽𝑗𝜉𝑗𝛼+1𝑄1𝑔(𝑡)−(Γ(𝛼))2Γ(2𝛼)1−𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼−1𝑄2.𝑔(𝑡)(2.19) Since the condition (1.4) holds, the mapping 𝑄∶𝑍→𝑍 defined by 𝑅(𝑄𝑦)(𝑡)=1𝑡𝑔(𝑡)𝛼−1+𝑅2𝑡𝑔(𝑡)𝛼−2(2.20) is well defined.
Recall (1.4) and note that 𝑅1𝑅1𝑔𝑡𝛼−1=1𝑅Γ(𝛼−1)Γ(𝛼+1)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗𝑄1𝑅1𝑔𝑡𝛼−1−Γ(𝛼)Γ(𝛼−1)Γ(2𝛼−1)1−𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼−2𝑄2𝑅1𝑔𝑡𝛼−1=𝑅1𝑔1𝑅𝛼Γ(𝛼−1)Γ2Γ(𝛼+1)Γ(2𝛼)𝑛𝑗=1𝛽𝑗𝜉𝛼𝑗1−𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼−1−𝛼Γ(𝛼−1)Γ2Γ(2𝛼−1)Γ(𝛼+2)1−𝑚𝑖=1𝛼𝑖𝜂𝑖2𝛼−2𝑛𝑗=1𝛽𝑗𝜉𝑗𝛼+1=𝑅1𝑔,(2.21) and similarly we can derive that 𝑅1𝑅2𝑔𝑡𝛼−2𝑅=0,2𝑅1𝑔𝑡𝛼−1𝑅=0,2𝑅2𝑔𝑡𝛼−2=𝑅2𝑔.(2.22) So, for 𝑔∈𝑍, it follows from the four relations above that 𝑄2𝑔=𝑅1𝑅1𝑔𝑡𝛼−1+𝑅2𝑔𝑡𝛼−2𝑡𝛼−1+𝑅2𝑅1𝑔𝑡𝛼−1+𝑅2𝑔𝑡𝛼−2𝑡𝛼−2=𝑅1𝑅1𝑔𝑡𝛼−1𝑡𝛼−1+𝑅1𝑅2𝑔𝑡𝛼−2𝑡𝛼−1+𝑅2𝑅1𝑔𝑡𝛼−1𝑡𝛼−2+𝑅2𝑅2𝑔𝑡𝛼−2𝑡𝛼−2=𝑅1𝑔𝑡𝛼−1+𝑅2𝑔𝑡𝛼−2=𝑄𝑔,(2.23) that is, the map 𝑄 is idempotent. In fact, 𝑄 is a continuous linear projector.
Note that 𝑔∈Im(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, 𝑔∈Im(𝐿). In fact, Im(𝐿)=Ker(𝑄).
Take 𝑔∈𝑍 in the form 𝑔=(𝑔−𝑄𝑔)+𝑄𝑔, so that 𝑔−𝑄𝑔∈Im(𝐿)=Ker(𝑄) and 𝑄𝑔∈Im(𝑄). Thus, 𝑍=Im(𝐿)+Im(𝑄). Let 𝑔∈Im(𝐿)∩Im(𝑄) and assume that 𝑔(𝑠)=ğ‘Žğ‘ ğ›¼âˆ’1+𝑏𝑠𝛼−2 is not identically zero on [0,1], then, since 𝑔∈Im(𝐿), from (2.11) and the condition (1.4), we derive ğ‘Ž=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 𝑃∶𝑌→𝑌 be defined by 1𝑃𝑢(𝑡)=Γ𝐷(𝛼)𝛼−10+𝑢(0)𝑡𝛼−1+1Γ𝐷(𝛼−1)𝛼−20+𝑢(0)𝑡𝛼−2[],𝑡∈0,1.(2.24) Note that 𝑃 is a continuous linear projector and Ker(𝑃)=𝑢∈𝑌∣𝐷𝛼−10+𝑢(0)=𝐷𝛼−20+𝑢(0)=0.(2.25) It is clear that 𝑌=Ker(𝐿)⊕Ker(𝑃).

Note that the projectors 𝑃 and 𝑄 are exact. Define 𝐾𝑃∶Im(𝐿)→dom(𝐿)∩Ker(𝑃) by 𝐾𝑃1𝑔(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑔(𝑠)𝑑𝑠=𝐼𝛼0+𝑔(𝑡).(2.26) Hence, we have 𝐷𝛼−10+𝐾𝑃𝑔(𝑡)=𝑡0𝑔(𝑠)𝑑𝑠,𝐷𝛼−20+𝐾𝑃𝑔(𝑡)=𝑡0(𝑡−𝑠)𝑔(𝑠)𝑑𝑠,(2.27) then â€–ğ¾ğ‘ƒğ‘”â€–âˆžâ‰¤(1/Γ(𝛼))‖𝑔‖1,‖𝐷𝛼−10+(𝐾𝑃𝑔)â€–âˆžâ‰¤â€–ğ‘”â€–1,‖𝐷𝛼−20+(𝐾𝑃𝑔)â€–âˆžâ‰¤â€–ğ‘”â€–1, and thus ‖‖𝐾𝑃𝑔‖‖𝐶𝛼−1≤12+Γ(𝛼)‖𝑔‖1.(2.28) In fact, if 𝑔∈Im(𝐿), then (𝐿𝐾𝑃)𝑔=𝐷𝛼0+𝐼𝛼0+𝑔=𝑔. Also, if 𝑢∈dom(𝐿)∩Ker(𝑃), then 𝐾𝑃𝐿𝑔(𝑡)=𝐼𝛼0+𝐷𝛼0+𝑔(𝑡)=𝑔(𝑡)+𝑐1𝑡𝛼−1+𝑐2𝑡𝛼−2+𝑐3𝑡𝛼−3,(2.29) from boundary value condition (1.2) and the fact that 𝑢∈dom(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)=1𝑡𝑁𝑢𝛼−1+𝑅2𝑡𝑁𝑢𝛼−2,𝐾𝑃1(𝐼−𝑄)𝑁𝑢(𝑡)=Γ(𝛼)10(𝑡−𝑠)𝛼−1[]𝑁𝑢(𝑠)−𝑄𝑁𝑢(𝑠)𝑑𝑠.(2.31) With arguments similar to those of [7], we obtain the following Lemma.

Lemma 2.8. 𝐾𝑃(𝐼−𝑄)𝑁∶𝑌→𝑌 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 𝑢∈dom(𝐿)⧵Ker(𝐿) satisfying |𝐷𝛼−10+𝑢(𝑡)|+|𝐷𝛼−20+𝑢(𝑡)|>𝐴 for all 𝑡∈[0,1], we have 𝑄1𝑁𝑢(𝑡)≠0or𝑄2𝑁𝑢(𝑡)≠0,(3.1)

(H2) there exist functions ğ‘Ž,𝑏,𝑐,𝑑,𝑟∈𝐿1[0,1] and a constant 𝜃∈[0,1] such that for all (𝑥,𝑦,𝑧)∈ℝ3 and a.e., 𝑡∈[0,1], one of the following inequalities is satisfied:||||||𝑦||𝑓(𝑡,𝑥,𝑦,𝑧)â‰¤ğ‘Ž(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑧|𝜃||||||𝑦||||𝑦||+𝑟(𝑡),𝑓(𝑡,𝑥,𝑦,𝑧)â‰¤ğ‘Ž(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)𝜃||||||𝑦||+𝑟(𝑡),𝑓(𝑡,𝑥,𝑦,𝑧)â‰¤ğ‘Ž(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 ğ‘Ž,𝑏∈ℝ satisfying ğ‘Ž2+𝑏2>𝐵, then either ğ‘Žğ‘…1ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2+𝑏𝑅2ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2<0,(3.3) or else ğ‘Žğ‘…1ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2+𝑏𝑅2ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2>0.(3.4)

Remark 3.1. 𝑅1𝑁(ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2) and 𝑅2𝑁(ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2) from (H3) stand for the images of 𝑢(𝑡)=ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−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 â€–ğ‘Žâ€–1+‖𝑏‖1+‖𝑐‖1<1𝜏,(3.5) where 𝜏=5+2/Γ(𝛼)+1/Γ(𝛼−1).

Proof. Set Ω1[]={𝑢∈dom(𝐿)⧵Ker(𝐿)∣𝐿𝑢=𝜆𝑁𝑢forsome𝜆∈0,1},(3.6) then for 𝑢∈Ω1,𝐿𝑢=𝜆𝑁𝑢; thus, 𝜆≠0,𝑁𝑢∈Im(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 |𝐷𝛼−10+𝑢(𝑡0)|+|𝐷𝛼−20+𝑢(𝑡0)|≤𝐴. Now, 𝐷𝛼−10+𝑢(𝑡)=𝐷𝛼−10+𝑢𝑡0+𝑡𝑡0𝐷𝛼0+𝐷𝑢(𝑠)𝑑𝑠,𝛼−20+𝑢(𝑡)=𝐷𝛼−20+𝑢𝑡0+𝑡𝑡0𝐷𝛼−10+𝑢(𝑠)𝑑𝑠,(3.7) so ||𝐷𝛼−10+||≤‖‖𝐷𝑢(0)𝛼−10+‖‖𝑢(𝑡)∞≤||𝐷𝛼−10+𝑢𝑡0||+‖‖𝐷𝛼0+𝑢‖‖1≤𝐴+‖𝐿𝑢‖1≤𝐴+‖𝑁𝑢‖1,||𝐷𝛼−20+||≤‖‖𝐷𝑢(0)𝛼−20+‖‖𝑢(𝑡)∞≤||𝐷𝛼−20+𝑢𝑡0||+‖‖𝐷𝛼−10+ğ‘¢â€–â€–âˆžâ‰¤||𝐷𝛼−20+𝑢𝑡0||+||𝐷𝛼−10+𝑢𝑡0||+‖‖𝐷𝛼0+𝑢‖‖1≤𝐴+‖𝐿𝑢‖1≤𝐴+‖𝑁𝑢‖1.(3.8) Now by (3.8), we have ‖𝑃𝑢‖𝐶𝛼−1=‖‖‖1𝐷Γ(𝛼)𝛼−10+𝑢(0)𝑡𝛼−1+1𝐷Γ(𝛼−1)𝛼−20+𝑢(0)𝑡𝛼−2‖‖‖𝐶𝛼−1=‖‖‖1𝐷Γ(𝛼)0𝛼−1𝑢(0)𝑡𝛼−1+1𝐷Γ(𝛼−1)0𝛼−2𝑢(0)𝑡𝛼−2‖‖‖∞+‖‖𝐷𝛼−10+‖‖𝑢(0)∞+‖‖𝐷𝛼−10+𝑢(0)𝑡+𝐷𝛼−20+‖‖𝑢(0)∞≤12+||𝐷Γ(𝛼)𝛼−10+||+1𝑢(0)1+||𝐷Γ(𝛼−1)𝛼−20+||≤1𝑢(0)2+Γ(𝛼)𝐴+‖𝑁𝑢‖1+11+Γ(𝛼−1)𝐴+‖𝑁𝑢‖1.(3.9) Note that (𝐼−𝑃)𝑢∈Im(𝐾𝑃)=dom(𝐿)∩Ker(𝑃) for 𝑢∈Ω1, then, by (2.28) and (2.30), ‖‖(𝐼−𝑃)𝑢𝐶𝛼−1=‖‖𝐾𝑃‖‖𝐿(𝐼−𝑃)𝐶𝛼−1≤12−Γ(𝛼)‖𝐿(𝐼−𝑃)𝑢‖1=12−Γ(𝛼)‖𝐿𝑢‖1≤12−Γ(𝛼)‖𝑁𝑢‖1.(3.10) Using (3.9) and (3.10), we obtain ‖𝑢‖𝐶𝛼−1=‖𝑃𝑢+(𝐼−𝑃)𝑢‖𝐶𝛼−1≤‖𝑃𝑢‖𝐶𝛼−1‖+‖(𝐼−𝑃)𝑢𝐶𝛼−1≤12+Γ(𝛼)𝐴+‖𝑁𝑢‖1+11+Γ(𝛼−1)𝐴+‖𝑁𝑢‖1+12+Γ(𝛼)‖𝑁𝑢‖1=25++1Γ(𝛼)Γ(𝛼−1)‖𝑁𝑢‖1+13++1Γ(𝛼)𝐴Γ(𝛼−1)=𝜏‖𝑁𝑢‖1+𝐶1,(3.11) where 𝐶1=(3+1/Γ(𝛼)+1/Γ(𝛼−1))𝐴 is a constant. This is for all 𝑢∈Ω1, ‖𝑢‖𝐶𝛼−1≤𝜏‖𝑁𝑢‖1+𝐶1.(3.12) If the first condition of (H2) is satisfied, then we have maxâ€–ğ‘¢â€–âˆž,‖‖𝐷𝛼−10+ğ‘¢â€–â€–âˆž,‖‖𝐷𝛼−20+ğ‘¢â€–â€–âˆžî€¾â‰¤â€–ğ‘¢â€–ğ¶ğ›¼âˆ’1î‚€â‰¤ğœâ€–ğ‘Žâ€–1â€–ğ‘¢â€–âˆž+‖𝑏‖1‖‖𝐷𝛼−10+ğ‘¢â€–â€–âˆž+‖𝑐‖1‖‖𝐷𝛼−20+ğ‘¢â€–â€–âˆž+‖𝑑‖1‖‖𝐷𝛼−20+ğ‘¢â€–â€–ğœƒâˆž+‖𝑟‖1+‖𝑒‖1+𝐶1,(3.13) and consequently, â€–ğ‘¢â€–âˆžâ‰¤ğœ1âˆ’â€–ğ‘Žâ€–1𝜏‖𝑏‖1‖‖𝐷𝛼−10+ğ‘¢â€–â€–âˆž+‖𝑐‖1‖‖𝐷𝛼−20+ğ‘¢â€–â€–âˆž+‖𝑑‖1‖‖𝐷𝛼−20+ğ‘¢â€–â€–ğœƒâˆž+‖𝑟‖1+‖𝑒‖1+𝐶11âˆ’â€–ğ‘Žâ€–1𝜏,(3.14)‖‖𝐷𝛼−10+ğ‘¢â€–â€–âˆžâ‰¤ğœ1âˆ’â€–ğ‘Žâ€–1𝜏−‖𝑏‖1𝜏‖𝑐‖1‖‖𝐷𝛼−20+ğ‘¢â€–â€–âˆž+‖𝑑‖1‖‖𝐷𝛼−20+ğ‘¢â€–â€–ğœƒâˆž+‖𝑟‖1+‖𝑒‖1+𝐶11âˆ’â€–ğ‘Žâ€–1𝜏−‖𝑏‖1𝜏,(3.15)‖‖𝐷𝛼−10+ğ‘¢â€–â€–âˆžâ‰¤ğœâ€–ğ‘‘â€–1‖‖𝐷𝛼−20+ğ‘¢â€–â€–ğœƒâˆž1âˆ’â€–ğ‘Žâ€–1𝜏−‖𝑏‖1𝜏−‖𝑐‖1𝜏+𝜏‖𝑟‖1+‖𝑒‖1+𝐶11âˆ’â€–ğ‘Žâ€–1𝜏−‖𝑏‖1𝜏−‖𝑐‖1𝜏.(3.16) Note that 𝜃∈[0,1) and â€–ğ‘Žâ€–1+‖𝑏‖1+‖𝑐‖1<1/𝜏, so there exists 𝑀1>0 such that ‖𝐷𝛼−10+ğ‘¢â€–âˆžâ‰¤ğ‘€1 for all 𝑢∈Ω1. The inequalities (3.14) and (3.15) show that there exist 𝑀2,𝑀3>0 such that ‖𝐷𝛼−10+ğ‘¢â€–âˆžâ‰¤ğ‘€2,â€–ğ‘¢â€–âˆžâ‰¤ğ‘€3 for all 𝑢∈Ω1. Therefore, for all 𝑢∈Ω1,‖𝑢‖𝐶𝛼−1=â€–ğ‘¢â€–âˆž+‖𝐷𝛼−10+ğ‘¢â€–âˆž+‖𝐷𝛼−20+ğ‘¢â€–âˆžâ‰¤ğ‘€1+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={𝑢∈Ker(𝐿)∣𝑁𝑢∈Im(𝐿)}.(3.17) For 𝑢∈Ω2,𝑢∈Ker(𝐿)={𝑢∈dom(𝐿)∣𝑢=ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2,ğ‘Ž,𝑏∈ℝ,𝑡∈[0,1]}, and 𝑄𝑁(ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2)=0; thus, 𝑅1𝑁(ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2)=𝑅2𝑁(ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2)=0. By (H3), ğ‘Ž2+𝑏2≤𝐵, that is, Ω2 is bounded.
We define the isomorphism 𝐽∶Im(𝑄)→Ker(𝐿) by ğ½î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2=ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2,ğ‘Ž,𝑏∈ℝ.(3.18)
If the first part of (H3) is satisfied, let Ω3=𝑢∈Ker𝐿∶−𝜆𝐽−1[]𝑢+(1−𝜆)𝑄𝑁𝑢=0,𝜆∈0,1.(3.19) For every ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2∈Ω3, ğœ†î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2𝑅=(1−𝜆)1ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2𝑡𝛼−1+𝑅2ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2𝑡𝛼−2.(3.20) If 𝜆=1, then ğ‘Ž=𝑏=0, and if ğ‘Ž2+𝑏2>𝐵, then by (H3), ğœ†î€·ğ‘Ž2+𝑏2=(1−𝜆)ğ‘Žğ‘…1ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2+𝑏𝑅2ğ‘î€·ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2<0,(3.21) which, in either case, obtain a contradiction. If the other part of (H3) is satisfied, then we take Ω3=𝑢∈Ker𝐿∶𝜆𝐽−1[]𝑢+(1−𝜆)𝑄𝑁𝑢=0,𝜆∈0,1,(3.22) and, again, obtain a contradiction. Thus, in either case, ‖𝑢‖𝐶𝛼−1=â€–ğ‘¢â€–âˆž+‖‖𝐷𝛼−10+ğ‘¢â€–â€–âˆž+‖‖𝐷𝛼−20+ğ‘¢â€–â€–âˆž=â€–â€–ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2‖‖𝐶𝛼−1=â€–â€–ğ‘Žğ‘¡ğ›¼âˆ’1+𝑏𝑡𝛼−2‖‖∞+â€–ğ‘ŽÎ“(𝛼)‖∞+â€–ğ‘ŽÎ“(𝛼)𝑡+𝑏Γ(𝛼−1)‖∞||𝑏||≤(1+2Γ(𝛼))|ğ‘Ž|+(1+Γ(𝛼−1))≤(2+2Γ(𝛼)+Γ(𝛼−1))|ğ‘Ž|,(3.23) for all 𝑢∈Ω3, 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=1Ω𝑖⊂Ω. by Lemma 2.8, the operator 𝐾𝑃(𝐼−𝑄)𝑁∶Ω→𝑌 is compact; thus, 𝑁 is 𝐿-compact on Ω, then by the above argument, we have(i)𝐿𝑢≠𝜆𝑁𝑥, for every (𝑢,𝜆)∈[(dom(𝐿)⧵Ker𝐿)∩𝜕Ω]×(0,1),(ii)𝑁𝑢∉Im(𝐿), for every 𝑢∈Ker(𝐿)∩𝜕Ω.Finally, we will prove that (iii) of Theorem 1.1 is satisfied. Let 𝐻(𝑢,𝜆)=±𝐼𝑢+(1−𝜆)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∈𝜕Ω∩Ker(𝐿),(3.24) and thus, by the homotopy property of degree, ||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)=±1≠0,(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 𝐶𝛼−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.