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

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 𝐼𝐷^𝛼0+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷^𝛼−10+𝑢(𝑡),𝐷^𝛼−20+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.1)_3−𝛼0+𝑢(0)=0,𝐷^𝛼−20+𝑢(0)=^𝑛_𝑗=1𝛽_𝑗𝐷^𝛼−20+𝑢𝜉_𝑗,𝑢(1)=^𝑚_𝑖=1𝛼_𝑖𝑢𝜂_𝑖,(1.2) where 2<𝛼≤3,0<𝜉₁<𝜉₂<⋯<𝜉_𝑛<1,𝑛≥1,0<𝜂₁<⋯<𝜂_𝑚<1,𝑚≥2,𝛼_𝑖,𝛽_𝑗∈ℝ,^𝑚_𝑖=1𝛼_𝑖𝜂^𝑖𝛼−1=^𝑚_𝑖=1𝛼_𝑖𝜂^𝑖𝛼−2=1,^𝑛_𝑗=1𝛽_𝑗𝜉_𝑗=0,^𝑛_𝑗=1𝛽_𝑗=1,(1.3)𝑓∶[0,1]×ℝ³→ℝ satisfying the Carathéodory conditions, 𝑒∈𝐿¹[0,1]. 𝐷^𝛼0+ and 𝐼^𝛼0+ 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=1𝛽_𝑗𝜉^𝛼𝑗1−^𝑚_𝑖=1𝛼_𝑖𝜂^{𝑖2𝛼−1}−Γ(𝛼)₂Γ(𝛼−1)Γ(𝛼+2)Γ(2𝛼−1)^𝑛_𝑗=1𝛽_𝑗𝜉^𝑗𝛼+11−^𝑚_𝑖=1𝛼_𝑖𝜂^{𝑖2𝛼−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 equation𝐷^𝛼0+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷^{𝛼−(𝑛−1)0+}𝑢(𝑡),…,𝐷^𝛼−10+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.5) subject to the following boundary value conditions: 𝐼^{𝑛−𝛼0+}𝑢(0)=𝐷^{𝛼−(𝑛−1)0+}𝑢(0)=⋯=𝐷^𝛼−20+𝑢(0)=0,𝑢(1)=𝜎𝑢(𝜂),(1.6) where 𝑛>2 is a natural number, 𝑛−1<𝛼≤𝑛 is a real number, 𝑓∶[0,1]×ℝ^𝑛→ℝ is continuous, and 𝑒∈𝐿¹[0,1],𝜎∈(0,∞), and 𝜂∈(0,1) are given constants such that 𝜎𝜂^𝛼−1=1. 𝐷^𝛼0+ and 𝐼^𝛼0+ 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 𝐾_𝑃. 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 𝐾_𝑃(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 𝐼^𝛼0+𝑦(𝑡)=1Γ(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 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,∞).

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 𝐼^𝛼0+𝐷^𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^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 space 𝐶[0,1] with the norm ‖𝑥‖_∞=max_𝑡∈[0,1]|𝑥(𝑡)|. Given 𝜇>0 and 𝑁=[𝜇]+1, one can define a linear space 𝐶^𝜇[0,1]∶=𝑢∣𝑢(𝑡)=𝐼^𝜇0+𝑥(𝑡)+𝑐₁𝑡^𝜇−1+𝑐₂𝑡^𝜇−2+⋯+𝑐_𝑁−1𝑡^{𝜇−(𝑁−1)},𝑡∈[0,1],(2.4) where 𝑥∈𝐶[0,1] and 𝑐_𝑖∈ℝ,𝑖=1,2,…,𝑁−1. By means of the linear function analysis theory, one can prove that with the norm ‖𝑢‖_𝐶^𝜇=‖𝐷^𝜇0+𝑢‖_∞+⋯+‖𝐷^{𝜇−(𝑁−1)0+}𝑢‖_∞+‖𝑢‖_∞,𝐶^𝜇[0,1] is a Banach space.

Lemma 2.4 (see [7]). 𝐹⊂𝐶^𝜇[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‖_𝐶^𝜇=‖‖𝐷^𝜇0+𝑢‖‖_∞+⋯+‖‖𝐷^{𝜇−(𝑁−1)0+}𝑢‖‖_∞+‖𝑢‖_∞<𝑀,(2.5) and equicontinuous means that forall𝜀>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∈{0,…,𝑁−1}.(2.6)

Let 𝑍=𝐿¹[0,1] with the norm ‖𝑔‖₁=∫¹⁰|𝑔(𝑠)|𝑑𝑠. 𝑌=𝐶^𝛼−1[0,1]={𝑢∣𝑢(𝑡)=𝐼^𝛼−10+𝑥(𝑡)+𝑐𝑡^𝛼−2,𝑡∈[0,1]}, where 𝑥∈𝐶[0,1],𝑐∈ℝ, with the norm ‖𝑢‖_𝐶^𝛼−1=‖𝐷^𝛼−10+𝑢‖_∞+‖𝐷^𝛼−20+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 𝑢∈𝐶^𝛼−1[0,1] such that 𝐷^𝛼−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 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).

Define 𝐿 to be the linear operator from dom(𝐿)∩𝑌 to 𝑍 with dom(𝐿)=𝑢∈𝐶𝐿𝑢^𝛼−1[=0,𝐷1]_𝛼0+∣𝐷^𝛼0+𝑢∈𝐿¹[0,1],𝑢satisfies(1.2),𝑢,𝑢∈dom(𝐿).(2.7) 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).(2.8) 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)={𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2∣𝑎,𝑏∈ℝ}≅ℝ².
Let 𝑔∈𝑍 and 𝑢(𝑡)=1Γ(𝛼)^𝑡0(𝑡−𝑠)^𝛼−1𝑔(𝑠)𝑑𝑠+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2,(2.10) then 𝐷^𝛼0+𝑢(𝑡)=𝑔(𝑡)a.e.,𝑡∈(0,1) and, if ¹⁰(1−𝑠)^𝛼−1𝑔(𝑠)𝑑𝑠−^𝑚_𝑖=1𝛼_𝑖^𝜂𝑖₀𝜂_𝑖−𝑠^𝛼−1𝑔(𝑠)𝑑𝑠=0,_𝑛_𝑗=1𝛽_𝑗^𝜉𝑗₀𝜉_𝑗−𝑠𝑔(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 𝐷^𝛼0+𝑢∈Im(𝐿), we have 𝑢(𝑡)=𝐼^𝛼0+𝐷^𝛼0+𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+𝑐₃𝑡^𝛼−3,(2.13) which, due to the boundary value condition (1.2), implies that 𝐷^𝛼0+𝑢 satisfies (2.11). In fact, from 𝐼^3−𝛼0+𝑢(0)=0, we have 𝑐₃=0, from 𝑢(1)=∑^𝑚𝑖=1𝛼_𝑖𝑢(𝜂_𝑖), we have ¹⁰(1−𝑠)^𝛼−1𝐷^𝛼0+𝑢(𝑠)𝑑𝑠−^𝑚_𝑖=1𝛼_𝑖^𝜂𝑖₀𝜂_𝑖−𝑠^𝛼−1𝐷^𝛼0+𝑢(𝑠)𝑑𝑠=0,(2.14) and from 𝐷^𝛼−20+𝑢(0)=∑^𝑛𝑗=1𝛽_𝑗𝐷^𝛼−20+𝑢(𝜉_𝑗), we have ^𝑛_𝑗=1𝛽_𝑗^𝜉𝑗₀𝜉_𝑗−𝑠𝐷^𝛼0+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 𝑄₁𝑔=¹⁰(1−𝑠)^𝛼−1𝑔(𝑠)𝑑𝑠−^𝑚_𝑖=1𝛼_𝑖^𝜂𝑖₀𝜂_𝑖𝑄−𝑠^𝛼−1𝑔(𝑠)𝑑𝑠,₂𝑔=^𝑛_𝑗=1𝛽_𝑗^𝜉𝑗₀𝜉_𝑗−𝑠𝑔(𝑠)𝑑𝑠.(2.18) Using the above definitions, we construct the following auxiliary maps 𝑅₁,𝑅₂∶𝑍→𝑍: 𝑅₁𝑔=1𝑅ΓΓ((𝛼𝛼+−11))^𝑛_𝑗=1𝛽_𝑗𝜉^𝛼𝑗𝑄₁𝑔(𝑡)−Γ(𝛼)Γ(𝛼−1)Γ(2𝛼−1)1−^𝑚_𝑖=1𝛼_𝑖𝜂^{𝑖2𝛼−2}𝑄²𝑅𝑔(𝑡)₂,𝑔=−1𝑅Γ(Γ𝛼(+𝛼)2)^𝑛_𝑗=1𝛽_𝑗𝜉^𝑗𝛼+1𝑄₁𝑔(𝑡)−(Γ(𝛼))²Γ(2𝛼)1−^𝑚_𝑖=1𝛼_𝑖𝜂^{𝑖2𝛼−1}𝑄²𝑔(𝑡).(2.19) Since the condition (1.4) holds, the mapping 𝑄∶𝑍→𝑍 defined by (𝑄𝑦)(𝑡)=𝑅₁𝑔(𝑡)𝑡^𝛼−1+𝑅₂𝑔(𝑡)𝑡^𝛼−2(2.20) is well defined.
Recall (1.4) and note that 𝑅₁𝑅₁𝑔𝑡^𝛼−1=1𝑅ΓΓ((𝛼𝛼−+11))^𝑛_𝑗=1−Γ(𝛼)Γ(𝛼−𝛽1_𝑗𝜉)^𝛼𝑗𝑄₁𝑅₁𝑔𝑡^𝛼−1Γ(2𝛼−1)1−^𝑚_𝑖=1𝛼_𝑖𝜂^{𝑖2𝛼−2}𝑄²=𝑅₁𝑅₁𝑔𝑡^𝛼−1𝑔1𝑅Γ(𝛼−1)Γ𝛼²Γ(𝛼+1)Γ(2𝛼)^𝑛_𝑗=1𝛽_𝑗𝜉^𝛼𝑗1−^𝑚_𝑖=1𝛼_𝑖𝜂^{𝑖2𝛼−1}−Γ(𝛼−1)Γ𝛼₂Γ(2𝛼−1)Γ(𝛼+2)1−^𝑚_𝑖=1𝛼_𝑖𝜂^{𝑖2𝛼−2}^𝑛_𝑗=1𝛽_𝑗𝜉^𝑗𝛼+1=𝑅₁𝑔,(2.21) and similarly we can derive that 𝑅𝑅₁𝑅₂𝑔𝑡^𝛼−2=0,₂𝑅𝑅₁𝑔𝑡^𝛼−1=0,₂𝑅₂𝑔𝑡^𝛼−2=𝑅₂𝑔.(2.22) So, for 𝑔∈𝑍, it follows from the four relations above that 𝑄²𝑔=𝑅₁𝑅₁𝑔𝑡^𝛼−1+𝑅₂𝑔𝑡^𝛼−2𝑡^𝛼−1+𝑅₂𝑅₁𝑔𝑡^𝛼−1+𝑅₂𝑔𝑡^𝛼−2𝑡^𝛼−2=𝑅₁𝑅₁𝑔𝑡^𝛼−1𝑡^𝛼−1+𝑅₁𝑅₂𝑔𝑡^𝛼−2𝑡^𝛼−1+𝑅₂𝑅₁𝑔𝑡^𝛼−1𝑡^𝛼−2+𝑅₂𝑅₂𝑔𝑡^𝛼−2𝑡^𝛼−2=𝑅₁𝑔𝑡^𝛼−1+𝑅₂𝑔𝑡^𝛼−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)=𝑎𝑠^𝛼−1+𝑏𝑠^𝛼−2 is not identically zero on [0,1], then, since 𝑔∈Im(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 𝑃∶𝑌→𝑌 be defined by 𝑃𝑢(𝑡)=1Γ(𝛼)𝐷^𝛼−10+𝑢(0)𝑡^𝛼−1+1Γ(𝛼−1)𝐷^𝛼−20+𝑢(0)𝑡^𝛼−2,𝑡∈[0,1].(2.24) Note that 𝑃 is a continuous linear projector and Ker(𝑃)=𝑢∈𝑌∣𝐷^𝛼−10+𝑢(0)=𝐷^𝛼−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 𝐾_𝑃∶Im(𝐿)→dom(𝐿)∩Ker(𝑃) by 𝐾_𝑃𝑔(𝑡)=1Γ(𝛼)^𝑡0(𝑡−𝑠)^𝛼−1𝑔(𝑠)𝑑𝑠=𝐼^𝛼0+𝑔(𝑡).(2.26) Hence, we have 𝐷^𝛼−10+𝐾_𝑃𝑔(𝑡)=^𝑡0𝑔(𝑠)𝑑𝑠,𝐷^𝛼−20+𝐾_𝑃𝑔(𝑡)=^𝑡0(𝑡−𝑠)𝑔(𝑠)𝑑𝑠,(2.27) then ‖𝐾_𝑃𝑔‖_∞≤(1/Γ(𝛼))‖𝑔‖₁,‖𝐷^𝛼−10+(𝐾_𝑃𝑔)‖_∞≤‖𝑔‖₁,‖𝐷^𝛼−20+(𝐾_𝑃𝑔)‖_∞≤‖𝑔‖₁, and thus ‖‖𝐾_𝑃𝑔‖‖_𝐶𝛼−1≤2+1Γ(𝛼)‖𝑔‖¹.(2.28) In fact, if 𝑔∈Im(𝐿), then (𝐿𝐾_𝑃)𝑔=𝐷^𝛼0+𝐼^𝛼0+𝑔=𝑔. Also, if 𝑢∈dom(𝐿)∩Ker(𝑃), then 𝐾_𝑃𝐿𝑔(𝑡)=𝐼^𝛼0+𝐷^𝛼0+𝑔(𝑡)=𝑔(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+𝑐₃𝑡^𝛼−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, 𝐾_𝑃=𝐿||_{dom(𝐿)∩Ker(𝑃)}⁻¹.(2.30) Using (2.19), we write 𝐾_𝑃𝑄𝑁𝑢(𝑡)=𝑅₁𝑁𝑢𝑡^𝛼−1+𝑅₂𝑁𝑢𝑡^𝛼−2,(𝐼−𝑄)𝑁𝑢(𝑡)=1Γ(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. 𝐾_{𝑃(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 |𝐷^𝛼−10+𝑢(𝑡)|+|𝐷^𝛼−20+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,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:||||𝑓𝑓((𝑡𝑡,,𝑥𝑥,,𝑦𝑦,,𝑧𝑧))||||≤≤𝑎𝑎((𝑡𝑡))||𝑥𝑥||++𝑏𝑏((𝑡𝑡))||||𝑦𝑦||||++𝑐𝑐((𝑡𝑡))||𝑧𝑧||++𝑑𝑑((𝑡𝑡))|||𝑧𝑦|||^𝜃_𝜃+𝑟(𝑡),||𝑓(𝑡,𝑥,𝑦,𝑧)||≤𝑎(𝑡)|𝑥|+𝑏(𝑡)||𝑦||+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑥|_𝜃+𝑟(𝑡),+𝑟(𝑡),(3.2)

(H3) there exists a constant 𝐵>0 such that for every 𝑎,𝑏∈ℝ satisfying 𝑎²+𝑏²>𝐵, then either 𝑎𝑅₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2+𝑏𝑅₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2<0,(3.3) or else 𝑎𝑅₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2+𝑏𝑅₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2>0.(3.4)

Remark 3.1. 𝑅₁𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2) and 𝑅₂𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2) from (H3) stand for the images of 𝑢(𝑡)=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−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 ‖𝑎‖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 |𝐷^𝛼−10+𝑢(𝑡₀)|+|𝐷^𝛼−20+𝑢(𝑡₀)|≤𝐴. Now, 𝐷𝐷^𝛼−10+𝑢(𝑡)=𝐷^𝛼−10+𝑢𝑡₀+^𝑡𝑡₀𝐷^𝛼0+𝑢(𝑠)𝑑𝑠,_𝛼−20+𝑢(𝑡)=𝐷^𝛼−20+𝑢𝑡₀+^𝑡𝑡₀𝐷^𝛼−10+𝑢(𝑠)𝑑𝑠,(3.7) so ||𝐷^𝛼−10+𝑢(0)||≤‖‖𝐷^𝛼−10+𝑢(𝑡)‖‖_∞≤||𝐷^𝛼−10+𝑢𝑡₀||+‖‖𝐷^𝛼0+𝑢‖‖₁≤𝐴+‖𝐿𝑢‖₁||𝐷_𝛼−20+≤𝐴+‖𝑁𝑢‖₁,𝑢(0)||≤‖‖𝐷^𝛼−20+𝑢(𝑡)‖‖_∞≤||𝐷^𝛼−20+𝑢𝑡₀||+‖‖𝐷^𝛼−10+𝑢‖‖_∞≤||𝐷^𝛼−20+𝑢𝑡₀||+||𝐷^𝛼−10+𝑢𝑡₀||+‖‖𝐷^𝛼0+𝑢‖‖₁≤𝐴+‖𝐿𝑢‖₁≤𝐴+‖𝑁𝑢‖₁.(3.8) Now by (3.8), we have ‖𝑃𝑢‖_𝐶^𝛼−1=‖‖‖1Γ(𝛼)𝐷^𝛼−10+𝑢(0)𝑡^𝛼−1+1Γ(𝛼−1)𝐷^𝛼−20+𝑢(0)𝑡^𝛼−2‖‖‖_𝐶𝛼−1=‖‖‖1Γ(𝛼)𝐷^0𝛼−1𝑢(0)𝑡^𝛼−1+1Γ(𝛼−1)𝐷^0𝛼−2𝑢(0)𝑡^𝛼−2‖‖‖_∞+‖‖𝐷^𝛼−10+𝑢(0)‖‖_∞+‖‖𝐷^𝛼−10+𝑢(0)𝑡+𝐷^𝛼−20+𝑢(0)‖‖_∞≤2+1Γ(𝛼)||𝐷^𝛼−10+𝑢(0)||+1+1Γ(𝛼−1)||𝐷^𝛼−20+≤2+1𝑢(0)||Γ(𝛼)𝐴+‖𝑁𝑢‖¹+1+Γ(𝛼1−1)𝐴+‖𝑁𝑢‖¹.(3.9) Note that (𝐼−𝑃)𝑢∈Im(𝐾_𝑃)=dom(𝐿)∩Ker(𝑃) for 𝑢∈Ω₁, then, by (2.28) and (2.30), ‖(𝐼−𝑃)𝑢‖_𝐶^𝛼−1=‖‖𝐾_𝑃𝐿(𝐼−𝑃)‖‖_𝐶𝛼−1≤2−1Γ(𝛼)‖𝐿(𝐼−𝑃)𝑢‖¹=2−1Γ(𝛼)‖𝐿𝑢‖¹≤2−1Γ(𝛼)‖𝑁𝑢‖¹.(3.10) Using (3.9) and (3.10), we obtain ‖𝑢‖_𝐶^𝛼−1=‖𝑃𝑢+(𝐼−𝑃)𝑢‖_𝐶^𝛼−1≤‖𝑃𝑢‖_𝐶^𝛼−1+‖(𝐼−𝑃)𝑢‖_𝐶^𝛼−1≤2+1Γ(𝛼)𝐴+‖𝑁𝑢‖¹+1+Γ(𝛼1−1)𝐴+‖𝑁𝑢‖¹+2+Γ1(𝛼)‖𝑁𝑢‖¹=5+2Γ(𝛼)+Γ(𝛼1−1)‖𝑁𝑢‖¹+3+1=𝜏‖𝑁𝑢‖₁Γ(𝛼)+Γ(𝛼1−1)𝐴+𝐶₁,(3.11) where 𝐶₁=(3+1/Γ(𝛼)+1/Γ(𝛼−1))𝐴 is a constant. This is for all 𝑢∈Ω₁, ‖𝑢‖_𝐶^𝛼−1≤𝜏‖𝑁𝑢‖₁+𝐶₁.(3.12) If the first condition of (H2) is satisfied, then we have max≤‖‖𝑢𝑢‖‖_∞,‖‖_𝐶𝛼−1𝐷^𝛼−10+𝑢‖‖_∞,‖‖𝐷^𝛼−20+𝑢‖‖_∞≤𝜏‖𝑎‖₁‖𝑢‖_∞+‖𝑏‖₁‖‖𝐷^𝛼−10+𝑢‖‖_∞+‖𝑐‖₁‖‖𝐷^𝛼−20+𝑢‖‖_∞+‖𝑑‖₁‖‖𝐷^𝛼−20+𝑢‖‖^𝜃∞+‖𝑟‖₁+‖𝑒‖₁+𝐶¹,(3.13) and consequently, ‖𝑢‖_∞≤𝜏1−‖𝑎‖₁𝜏‖𝑏‖¹‖‖𝐷^𝛼−10+𝑢‖‖_∞+‖𝑐‖₁‖‖𝐷^𝛼−20+𝑢‖‖_∞+‖𝑑‖₁‖‖𝐷^𝛼−20+𝑢‖‖^𝜃∞+‖𝑟‖₁+‖𝑒‖₁+𝐶₁1−‖𝑎‖₁𝜏,(3.14)‖‖𝐷^𝛼−10+𝑢‖‖_∞≤𝜏1−‖𝑎‖₁𝜏−‖𝑏‖₁𝜏‖𝑐‖¹‖‖𝐷^𝛼−20+𝑢‖‖_∞+‖𝑑‖₁‖‖𝐷^𝛼−20+𝑢‖‖^𝜃∞+‖𝑟‖₁+‖𝑒‖₁+𝐶₁1−‖𝑎‖₁𝜏−‖𝑏‖₁𝜏,(3.15)‖‖𝐷^𝛼−10+𝑢‖‖_∞≤𝜏‖𝑑‖¹‖‖𝐷^𝛼−20+𝑢‖‖^𝜃∞1−‖𝑎‖₁𝜏−‖𝑏‖₁𝜏−‖𝑐‖₁𝜏+𝜏‖𝑟‖¹+‖𝑒‖₁+𝐶₁1−‖𝑎‖₁𝜏−‖𝑏‖₁𝜏−‖𝑐‖₁𝜏.(3.16) Note that 𝜃∈[0,1) and ‖𝑎‖₁+‖𝑏‖₁+‖𝑐‖₁<1/𝜏, so there exists 𝑀₁>0 such that ‖𝐷^𝛼−10+𝑢‖_∞≤𝑀₁ for all 𝑢∈Ω₁. The inequalities (3.14) and (3.15) show that there exist 𝑀₂,𝑀₃>0 such that ‖𝐷^𝛼−10+𝑢‖_∞≤𝑀₂,‖𝑢‖_∞≤𝑀₃ for all 𝑢∈Ω₁. Therefore, for all 𝑢∈Ω₁,‖𝑢‖_𝐶^𝛼−1=‖𝑢‖_∞+‖𝐷^𝛼−10+𝑢‖_∞+‖𝐷^𝛼−20+𝑢‖_∞≤𝑀₁+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=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2,𝑎,𝑏∈ℝ,𝑡∈[0,1]}, and 𝑄𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2)=0; thus, 𝑅₁𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2)=𝑅₂𝑁(𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2)=0. By (H3), 𝑎²+𝑏²≤𝐵, that is, Ω₂ is bounded.
We define the isomorphism 𝐽∶Im(𝑄)→Ker(𝐿) by 𝐽𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2=𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2,𝑎,𝑏∈ℝ.(3.18)
If the first part of (H3) is satisfied, let Ω₃=𝑢∈Ker𝐿∶−𝜆𝐽⁻¹𝑢+(1−𝜆)𝑄𝑁𝑢=0,𝜆∈[0,1].(3.19) For every 𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2∈Ω₃, 𝜆𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2=(1−𝜆)𝑅₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2𝑡^𝛼−1+𝑅₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2𝑡^𝛼−2.(3.20) If 𝜆=1, then 𝑎=𝑏=0, and if 𝑎²+𝑏²>𝐵, then by (H3), 𝜆𝑎²+𝑏²=(1−𝜆)𝑎𝑅₁𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2+𝑏𝑅₂𝑁𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−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⁻¹𝑢+(1−𝜆)𝑄𝑁𝑢=0,𝜆∈[0,1],(3.22) and, again, obtain a contradiction. Thus, in either case, ‖𝑢‖_𝐶^𝛼−1=‖𝑢‖_∞+‖‖𝐷^𝛼−10+𝑢‖‖_∞+‖‖𝐷^𝛼−20+𝑢‖‖_∞=‖‖𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2‖‖_𝐶𝛼−1=‖‖𝑎𝑡^𝛼−1+𝑏𝑡^𝛼−2‖‖_∞+‖𝑎Γ(𝛼)‖_∞+‖𝑎Γ(𝛼)𝑡+𝑏Γ(𝛼−1)‖_∞≤≤((12++22ΓΓ((𝛼𝛼)))+|𝑎Γ|+(𝛼(1−+1)Γ)|(𝑎𝛼|,−1))||𝑏||(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 𝑈^3𝑖=1Ω_𝑖⊂Ω. by Lemma 2.8, the operator 𝐾_𝑃(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, deg====𝐽ddd±𝑄eee1ggg𝑁≠(((||𝐻𝐻±0,𝐼((,……Ω_Ker(𝐿),∩,,Ω01))K∩,,eΩΩrK∩∩(e𝐿rKK)(𝐿,ee0rr)),((0𝐿𝐿)),,00))(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 𝐶^𝛼−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.