Solutions to Dirichlet-Type Boundary Value Problems of Fractional Order in Banach Spaces

Abstract

We consider the boundary value problems with Dirichlet-type boundary conditions of nonlinear fractional differential equation in Banach space. The existence of the solution to the boundary value problems is established. Our analysis relies on the Sadovskii fixed point theorem. As an application, we give an example to demonstrate our results.

1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications in various sciences, such as physics, mechanics, chemistry, and engineering, (e.g., [1–9]).

Consequently, the fractional calculus and its applications in various fields of science and engineering have received much attention and have developed very rapidly. Jiang and Yuan [10], by using fixed point theorem on the cone, discussed the existence and multiplicity of solutions of the nonlinear fractional differential equation boundary value problem as follows: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)(1)+𝑓𝑡,𝑢))=0,0<𝑡<1,=𝑢=0, (1) where1<𝛼≤2is a real number and𝐷^𝛼₀₊is standard Riemann-Liouville fractional derivative. The authors in [11] consider the same boundary value problem. They derived the corresponding Green function and obtained the existence of solutions of this problem.

As far as we know, the nonlinear integer order differential equation for the Dirichlet boundary value problem has been studied extensively (e.g., [12–17]). However, only a few papers have dealt with the boundary value problem for fractional differential equation, especially, in Banach spaces. The authors in [18] studied the existence of positive solutions of second-order two-point boundary value problem as follows: 𝑢^′′(𝑡)((𝑡𝑢(0)(1)+𝑓𝑡,𝑢))=𝜃,0<𝑡<1,=𝑢=𝜃, (2) in Banach spaces. The authors in [19], by using the Mönch fixed point theorems, obtained the same results.

Motivated by the results mentioned above, we discuss the following boundary value problem (BVP for short): 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)(1)+𝑓𝑡,𝑢))=𝜃,0<𝑡<1,=𝑢=𝜃, (3) in Banach space𝐸, where𝜃is the zero element of𝐸,1<𝛼≤2is a real number,𝐷^𝛼₀₊is standard Riemann-Liouville fractional derivative,𝐼=[0,1], and𝑓∶𝐼×𝐸→𝐸is continuous. We establish an existence result of BVP in Banach spaces. The technique relies on the properties of the Kuratowski noncompactness measure and and Sadovskii fixed point theorem. To the best of our knowledge, this is the first paper considering the existence of solutions to Dirichlet-type value problems of fractional order in Banach spaces.

2. Preliminaries

For the convenience of the reader, we present here the necessary definitions and preliminary facts which are used throughout this paper.

Definition 1 (see [1]). The Riemann-Liouville fractional integral of order𝛼>0of a function𝑓∶(0,∞)→𝑅is given by 𝐼^𝛼₀₊𝑓(𝑥)=1Γ(𝛼)^𝑥₀𝑓(𝑡)(𝑥−𝑡)^1−𝛼𝑑𝑡, (4) provided that the right side is pointwise defined on(0,∞), whereΓ(𝛼)=∫^∞₀𝑡^𝛼−1𝑒^−𝑡𝑑𝑡.

Definition 2 (see [1]). The Riemann-liouville fractional derivative of order𝛼>0of a continuous function𝑓∶(0,∞)→𝑅is given by 𝐷^𝛼₀₊𝑓(𝑥)=1Γ()𝑑𝑛−𝛼𝑑𝑥^𝑛𝑥₀𝑓(𝑡)(𝑥−𝑡)^{𝛼−𝑛−1}𝑑𝑡, (5) where𝑛=[𝛼]+1,[𝛼]denotes the integer part of the number𝛼, provided that the right side is pointwise defined on(0,∞).

Lemma 3 (see [2]). Let𝛼>0. If we assume that𝑢∈𝐶(0,1)∩𝐿(0,1), then the fractional differential equation 𝐷^𝛼₀₊𝑢(𝑡)=0 (6) has𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑁(𝑐_𝑖∈𝑅,𝑖=1,2,3,…,𝑁)as unique solutions, where𝑁is the smallest integer greater than or equal to𝛼.

Lemma 4 (see [2]). Assume that𝑢∈𝐶(0,1)∩𝐿(0,1)with a fractional derivative of order𝛼>0that belongs to𝐶(0,1)∩𝐿(0,1). Then 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑁𝑡^𝛼−𝑁, (7) for some𝑐_𝑖∈𝑅,𝑖=1,2,3,…,𝑁, where𝑁is the smallest integer greater than or equal to𝛼.

Definition 5 (Kuratowski noncompactness measure). Let 𝐸 be a real Banach space, let 𝑆be a bounded set in𝐸. We denote 𝛼(𝑆)=inf𝛿>0∶𝑆=^𝑚_𝑖=1𝑆_𝑖,diam𝑆_𝑖<𝛿,𝑖=1,2,…,𝑚. (8)𝛼(𝑆)is called Kuratowski noncompactness measure of𝑆, wherediam(𝑆_𝑖)denote the diameters of𝑆_𝑖. Obviously0≤𝛼(𝑆)<∞.

Definition 6. Let𝐸₁and𝐸₂be real Banach spaces,𝐷⊂𝐸₁, and𝐴∶𝐷→𝐸₂be a continuous and bounded operator.
(1)𝐴is called a𝑘-set contraction operator if there exists a constant𝑘, such that𝛼(𝐴(𝑆))≤𝑘𝛼(𝑆)for any bounded set𝑆in𝐸. When𝑘<1,𝐴is called a strict set contraction operator.
(2)Let𝑆be a nonrelative compact, bounded subset in𝐸.𝐴is called a condensation if𝛼(𝐴(𝑆))<𝛼(𝑆).

Remark 7. A strict set contraction operator is condensation.
Now, we denote the Banach space of continuous functions𝑢∶𝐼→𝐸by𝐶[𝐼,𝐸]with the maximal norm‖𝑢‖=max_𝑡∈𝐼‖𝑢(𝑡)‖.The basic space used in this paper is: 𝑋[𝐼,𝐸]=𝑢∈𝐶[𝐼,𝐸]∶sup_𝑡∈𝐼‖𝑢(𝑡)‖1+𝑡^𝛼−1<+∞ (9) equipped with norm‖𝑢‖_𝑋=sup_𝑡∈𝐼(‖𝑢(𝑡)‖/(1+𝑡^𝛼−1)). It is easy to see that𝑋[𝐼,𝐸]is a Banach space.

A map𝑥∈𝑋[𝐼,𝐸]∩𝐶¹[𝐼,𝐸]is called a solution of BVP if it satisfies (3). For a bounded subset𝐷of Banach space𝐸, let𝛼(𝐷)be the Kuratowski noncompactness measure of𝐷. In this paper, the Kuratowski noncompactness measure in𝐸, 𝐶[𝐼,𝐸], and 𝑋[𝐼,𝐸] is denoted by𝛼_𝐸(⋅),𝛼_𝐶(⋅),  and  𝛼_𝑋(⋅), respectively. The following properties of the Kuratowski noncompactness measure and Sadovskii fixed point theorem are needed for our discussion.

Lemma 8 (see [20]). If𝐻⊂𝐶[𝐼,𝐸]is bounded and equicontinuous. Then𝛼_𝐸(𝐻(𝑡))is continuous on𝐼and𝛼_𝐶(𝐻)=max_𝑡∈𝐼𝛼_𝐸(𝐻(𝑡)),𝛼_𝐸(∫_𝐼𝑥(𝑡)𝑑𝑡∶𝑥∈𝐻)≤∫_𝐼𝛼_𝐸(𝐻(𝑡))𝑑𝑡, where𝐻(𝑡)={𝑥(𝑡)∶𝑥∈𝐻}for each  𝑡∈𝐼.

Lemma 9 (see [21] (Sadovskii)). Let𝐷be a bounded, closed, and convex subset of the Banach space𝐸. If the operator𝐴∶𝐷→𝐷is condensing, then𝐴has a fixed point in𝐷.

3. Main Result

In order to discuss the BVP, the preliminary lemmas are given in this section.

For convenience, let us list some conditions.(𝐻₁) There exist nonnegative functions𝑎,𝑏∈𝐶[0,1]such that 𝑓(𝑡,𝑥)≤𝑎(𝑡)‖𝑥‖+𝑏(𝑡),∀𝑡∈𝐼,𝑥∈𝐸,₁₀1+𝑡^𝛼−1𝑎(𝑡)Γ(𝛼)𝑑𝑡<2₁₀,𝑏(𝑡)𝑑𝑡<+∞. (10)(𝐻₂)For any𝑟>0,[𝛼,𝛽]⊂𝐼,𝑓(𝑡,𝑥)is uniformly continuous on[𝛼,𝛽]×𝐵_𝐸[𝜃,𝑟], where𝜃is the zero element of𝐸and𝐵_𝐸[𝜃,𝑟]={𝑥∈𝐸∶‖𝑥‖≤𝑟}.(𝐻₃)There exists a positive constant𝑘with(8/Γ(𝛼)(1+𝑡^𝛼−1))𝑘<1,  such that𝛼_𝐸(𝑓(𝑡,𝑊))≤𝑘𝛼_𝐸(𝑊)for all𝑡∈[0,1]and all bounded subset𝑊in𝐸.

Lemma 10 (see [2]). Given𝑦∈𝐶[0,1]and1<𝛼≤2, then the problem 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)(1)+𝑦=0,0<𝑡<1,=𝑢=0, (11) has a unique solution satisfying 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠+¹₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠. (12)

Lemma 11 (see [2]). Suppose that condition(𝐻₁)is satisfied, then BVP is equivalent to integral equation: 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓((𝑠+𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑓(𝑠,𝑢(𝑠))𝑑𝑠. (13) For any𝑢∈𝑋[𝐼,𝐸], we define the operator𝑇by (𝑇𝑢)(𝑡)=−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓((𝑠+𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑓(𝑠,𝑢(𝑠))𝑑𝑠. (14)

Remark 12. Lemma 11 indicates that the solution of BVP coincides with the fixed point of the operator𝑇.𝑇𝑢is well defined and𝑇𝑢∈𝑋[𝐼,𝐸].Indeed, from condition(𝐻₁), we have ¹₀𝑓((𝑡≤𝑡,𝑢))𝑑𝑡₁₀[𝑎(𝑡𝑢‖(𝑡=)‖+𝑏)]𝑑𝑡₁₀𝑎(𝑡)1+𝑡^𝛼−1‖𝑢‖1+𝑡^𝛼−1(𝑡≤+𝑏)𝑑𝑡₁₀1+𝑡^𝛼−1𝑎(𝑡)‖𝑢‖_𝑋(𝑡=‖𝑢‖+𝑏)𝑑𝑡_𝑋101+𝑡^𝛼−1𝑎(𝑡)𝑑𝑡+¹₀𝑏(𝑡)𝑑𝑡<∞. (15) By (14) and (15), we get that (𝑇𝑢)(𝑡)1+𝑡^𝛼−1=11+𝑡^𝛼−1×−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓((𝑠+𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑓((𝑠≤2𝑠,𝑢))𝑑𝑠Γ(𝛼)¹₀𝑓(𝑠,𝑢(𝑠))𝑑𝑠<∞. (16)

Lemma 13. Suppose conditions(𝐻₁)and(𝐻₂)are satisfied, then𝑇∶𝑋[𝐼,𝐸]→𝑋[𝐼,𝐸]is continuous and bounded.

Proof. First, by (14) and (16), we get(𝑇𝑢)(𝑡)∈𝑋[𝐼,𝐸]for any𝑢(𝑡)∈𝑋[𝐼,𝐸]. Thus,𝑇∶𝑋[𝐼,𝐸]→𝑋[𝐼,𝐸]is bounded. Next, we will prove that𝑇is continuous on𝑋[𝐼,𝐸]. Take{𝑢_𝑛},{𝑢}∈𝑋[𝐼,𝐸]and‖𝑢_𝑛−𝑢‖_𝑋→0(𝑛→+∞).Hence,{𝑢_𝑛}is a bounded subset of𝑋[𝐼,𝐸], that is, there exists𝑀>0such that‖𝑢_𝑛‖_𝑋≤𝑀for all𝑛≥1. Taking the limit, we have‖𝑢‖_𝑋≤𝑀.
In addition, by (14) we have 𝑇𝑢_𝑛(𝑡)1+𝑡^𝛼−1−(𝑇𝑢)(𝑡)1+𝑡^𝛼−1=11+𝑡^𝛼−1−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓𝑠,𝑢_𝑛(𝑠+)𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑓𝑠,𝑢_𝑛(𝑠−1)𝑑𝑠1+𝑡^𝛼−1−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓((𝑠+𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑓((𝑠≤1𝑠,𝑢))𝑑𝑠1+𝑡^𝛼−1−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓𝑠,𝑢_𝑛(𝑠)+𝑑𝑠_𝑡0(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓((𝑠+1𝑠,𝑢))𝑑𝑠1+𝑡^𝛼−11₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑓𝑠,𝑢_𝑛(𝑠)−𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)𝑓((𝑠≤2𝑠,𝑢))𝑑𝑠Γ(𝛼)1+𝑡^𝛼−11₀𝑓𝑠,𝑢_𝑛(𝑠)−𝑓(𝑠,𝑢(𝑠))𝑑𝑠. (17) It follows from condition(𝐻₂)that for any𝜀>0, there exists𝑁>0such that 𝑓𝑠,𝑢_𝑛(𝑠)((𝑠≤𝜀−𝑓𝑠,𝑢))3Γ(𝛼),∀𝑛≥𝑁,𝑠∈𝐼. (18) Therefore, for any𝜀>0,𝑡∈[0,1],  and𝑛>𝑁, by (17) and (18), we have 𝑇𝑢_𝑛(𝑡)1+𝑡^𝛼−1−(𝑇𝑢)(𝑡)1+𝑡^𝛼−1≤2Γ(𝛼)1+𝑡^𝛼−11₀𝜀3Γ(𝛼)𝑑𝑠<𝜀. (19) Thus, we conclude that‖𝑇𝑢_𝑛−𝑇𝑢‖_𝑋<𝜀; namely,𝑇∶𝑋[𝐼,𝐸]→𝑋[𝐼,𝐸]is continuous, and the conclusion of lemma follows.

Lemma 14. Let condition(𝐻₁)be satisfied and let𝑉be a bounded subset of𝑋[𝐼,𝐸]. Then(𝑇𝑉)(𝑡)/(1+𝑡^𝛼−1)is equicontinuous on[0,1].

Proof. In order to show that(𝑇𝑉)(𝑡)/(1+𝑡^𝛼−1)is equicontinuous on[0,1], we only need to testify the following conclusion.
For any𝜀>0,𝑢∈𝑉, there exists a𝛿>0such that (𝑇𝑢)𝑡₂1+𝑡₂^𝛼−1−(𝑇𝑢)𝑡₁1+𝑡₁^𝛼−1<𝜀,∀𝑡₁,𝑡₂∈[],𝑡0,1₁−𝑡₂<𝛿. (20) In fact, from condition(𝐻₁), it follows that(𝑎₁) nonnegative functions𝑎(𝑡),𝑏(𝑡)∈𝐶[0,1]are bounded in[0,1], that is, there exist positive constants𝐴and𝐵, such thatmax_𝑡∈𝐼𝑎(𝑡)≤𝐴andmax_𝑡∈𝐼𝑏(𝑡)≤𝐵,(𝑎₂)‖𝑓(𝑡,𝑢(𝑡))‖≤(1+𝑡^𝛼−1)𝑎(𝑡)‖𝑢‖_𝑋+𝑏(𝑡).
In view of the boundedness of𝑉, there exists𝑅>0such that‖𝑢‖_𝑋≤𝑅for any𝑢∈𝑉. Without loss of generality, for any𝑢∈𝑉,𝑡₁,𝑡₂∈𝐼with𝑡₁<𝑡₂, by (14), we can find that (𝑇𝑢)𝑡₂1+𝑡₂^𝛼−1−(𝑇𝑢)𝑡₁1+𝑡₁^𝛼−1=−^𝑡2₀𝑡₂−𝑠^𝛼−11+𝑡₂^𝛼−1Γ(𝛼)𝑓((𝑠+𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡₂^𝛼−11+𝑡₂^𝛼−1Γ(𝛼)𝑓((𝑠−−𝑠,𝑢))𝑑𝑠_𝑡10𝑡₁−𝑠^𝛼−11+𝑡₁^𝛼−1Γ(𝛼)𝑓((𝑠+𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡₁^𝛼−11+𝑡₁^𝛼−1Γ(𝛼)𝑓((𝑠≤𝑠,𝑢))𝑑𝑠_𝑡20𝑡₂−𝑠^𝛼−11+𝑡₂^𝛼−1Γ(𝛼)𝑓(𝑠,𝑢(𝑠))𝑑𝑠−^𝑡1₀𝑡₁−𝑠^𝛼−11+𝑡₁^𝛼−1Γ(𝛼)𝑓((𝑠+𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡₂^𝛼−11+𝑡₂^𝛼−1Γ(𝛼)𝑓((𝑠−𝑠,𝑢))𝑑𝑠₁₀(1−𝑠)^𝛼−1𝑡₁^𝛼−11+𝑡₁^𝛼−1Γ(𝛼)𝑓((𝑠≤1𝑠,𝑢))𝑑𝑠Γ(𝛼)^𝑡2₀𝑡₂−𝑠^𝛼−11+𝑡₂^𝛼−1𝑓(𝑠,𝑢(𝑠))𝑑𝑠−^𝑡1₀𝑡₂−𝑠^𝛼−11+𝑡₂^𝛼−1𝑓((𝑠+1𝑠,𝑢))𝑑𝑠Γ(𝛼)^𝑡1₀𝑡₂−𝑠^𝛼−11+𝑡₂^𝛼−1𝑓((𝑠−𝑠,𝑢))𝑑𝑠_𝑡10𝑡₁−𝑠^𝛼−11+𝑡₁^𝛼−1𝑓((𝑠+1𝑠,𝑢))𝑑𝑠Γ(𝛼)¹₀𝑡₂^𝛼−11+𝑡₂^𝛼−1−𝑡₁^𝛼−11+𝑡₁^𝛼−1×(1−𝑠)^𝛼−1𝑓((𝑠≤1𝑠,𝑢))𝑑𝑠Γ(𝛼)^𝑡2_𝑡1𝑡₂−𝑠^𝛼−11+𝑡₂^𝛼−1𝑓((𝑠+1𝑠,𝑢))𝑑𝑠Γ(𝛼)^𝑡1₀𝑡₂−𝑠^𝛼−11+𝑡₂^𝛼−1−𝑡₁−𝑠^𝛼−11+𝑡₁^𝛼−1×𝑓((𝑠+1𝑠,𝑢))𝑑𝑠Γ(𝛼)¹₀𝑡₂^𝛼−11+𝑡₂^𝛼−1−𝑡₁^𝛼−11+𝑡₁^𝛼−1(1−𝑠)^𝛼−1×𝑓((𝑠≤𝑠,𝑢))𝑑𝑠2𝐴𝑅+𝐵Γ(𝛼)𝑡₂−𝑡₁+2𝐴𝑅+𝐵Γ()𝑡𝛼+11+^𝛼2𝑡₂^𝛼−1−𝑡^𝛼₁1+𝑡₁^𝛼−1−𝑡₂−𝑡₁^𝛼2+2𝐴𝑅+𝐵Γ(𝛼)𝑡₂^𝛼−11+𝑡₂^𝛼−1−𝑡₁^𝛼−11+𝑡₁^𝛼−1. (21) This ensures that(𝑇𝑉)(𝑡)/(1+𝑡^𝛼−1)is equicontinuous on[0,1]. If𝑡₁≥𝑡₂, we can also get the same result. Thus,(𝑇𝑉)(𝑡)/(1+𝑡^𝛼−1)is equicontinuous on[0,1].

The main result of this paper is as follows.

Theorem 15. Let conditions(𝐻₁)–(𝐻₃)be satisfied. Then the BVP has at least one solution belonging to𝑋[𝐼,𝐸].

Proof. We only need to prove that the the operator𝑇has a fixed point in𝑋[𝐼,𝐸]. By condition(𝐻₁), we can choose a real number𝑅such that 𝑅>¹₀𝑏(𝑠)Γ(𝛼)𝑑𝑠2−¹₀1+𝑠^𝛼−1𝑎(𝑠)𝑑𝑠⁻¹, (22) and let 𝐵=∶𝐵_𝑋(𝜃,𝑅)=𝑢∈𝑋[𝐼,𝐸]∶‖𝑢‖_𝑋≤𝑅. (23) Frist we prove that𝑇𝐵⊂𝐵. In fact, for any𝑢∈𝐵, by (16), we have (𝑇𝑢)(𝑡)1+𝑡^𝛼−1≤2Γ(𝛼)¹₀𝑓((𝑠≤2𝑠,𝑢))𝑑𝑠Γ(𝛼)¹₀1+𝑠^𝛼−1𝑎(𝑠)𝑑𝑠‖𝑢‖_𝑋+¹₀𝑏(𝑠)≤2𝑑𝑠Γ(𝛼)¹₀1+𝑠^𝛼−1𝑎(𝑠Γ(𝛼))𝑑𝑠𝑅+𝑅2−¹₀1+𝑠^𝛼−1𝑎(𝑠)().𝑑𝑠<𝑅,∀𝑡∈𝐼 (24) From Lemma 13, it follows that𝑇𝐵⊂𝐵.
Choose𝐷=co_𝑋(𝑇𝐵), that is,𝐷is the convex closure of𝑇𝐵in𝑋[𝐼,𝐸]. Clearly,𝐷is nonempty, bounded, convex, and closed subset of𝐵. By Lemma 14, it follows that(𝑇𝐵)(𝑡)/(1+𝑡^𝛼−1)is equicontinuous on[0,1], together with the definition of 𝐷, it follows that𝐷(𝑡)/(1+𝑡^𝛼−1)are equicontinuous on[0,1].
Now we show that𝑇is a strict set contraction operator from𝐷to𝐷.
Observing that𝐷⊂𝐵and𝑇𝐵⊂𝐷, together with Lemma 13 we know that𝑇∶𝐷→𝐷is bounded and continuous. Finally, we prove that there exists a constant0≤𝑎<1such that𝛼_𝑋(𝑇𝑉)≤𝑎𝛼_𝑋(𝑉), for𝑉⊂𝐷.
In fact, by (14), condition(𝐻₂)and Lemma 14, applying [22, Lemma 2.6], we have 𝛼_𝑋(𝑇𝑉)=sup_𝑡∈𝐼𝛼_𝐸(𝑇𝑉)(𝑡)1+𝑡^𝛼−1,∀𝑉⊂𝐷, (25) where(𝑇𝑉)(𝑡)/(1+𝑡^𝛼−1)={(𝑇𝑢)(𝑡)/(1+𝑡^𝛼−1)∶𝑢∈𝑉,𝑡isfixed}⊂𝐷for each𝑡∈𝐼. Thus, by (25), we need only to prove thatsup_𝑡∈𝐼𝛼_𝐸((𝑇𝑢)(𝑡)/(1+𝑡^𝛼−1))≤𝑎𝛼_𝑋(𝑉), for𝑉⊂𝐷.By condition(𝐻₂)and the definition of𝐷, we know{𝑓(𝑠,𝑢(𝑠))∶𝑢∈𝐷}are equicontinuous on[0,1]. Thus, by virtue of Lemma 8 and condition(𝐻₃), we get 𝛼_𝐸(𝑇𝑉)(𝑡)1+𝑡^𝛼−11≤−1+𝑡^𝛼−1𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝛼_𝐸𝑓((𝑠+1𝑠,𝑢))∶𝑢∈𝑉𝑑𝑠1+𝑡^𝛼−11₀(1−𝑠)^𝛼−1𝑡^𝛼−1Γ(𝛼)×𝛼_𝐸𝑓((𝑠≤2𝑠,𝑢))∶𝑢∈𝑉𝑑𝑠Γ(𝛼)1+𝑡^𝛼−11₀𝛼_𝐸𝑓((𝑠=2𝑠,𝑢))∶𝑢∈𝑉𝑑𝑠Γ(𝛼)1+𝑡^𝛼−11₀𝛼_𝐸𝑓()≤2𝑠,𝑊𝑑𝑠Γ(𝛼)1+𝑡^𝛼−1𝑘𝛼_𝐸(𝑊), (26) where𝑊={𝑢(𝑠)∶𝑠∈𝐼,𝑢∈𝑉}. For any given𝜀>0, there exists a partition𝑉=⋃^𝑛_𝑗=1𝑉_𝑗 with diam𝑉_𝑗<𝛼_𝑋(𝑉)+𝜀5,𝑗=1,2,…,𝑛. (27) Now for𝑗=1,2,…,𝑛, choose𝑢_𝑗∈𝑉_𝑗and a partition0=𝑡₀<𝑡₁<𝑡₂<⋯<𝑡_𝑚=1such that 𝑢_𝑗(𝑡)−𝑢_𝑗𝑡<𝜀5,∀𝑗=1,2,…,𝑛,𝑡,𝑡∈𝑡_𝑖−1,𝑡_𝑖,∀𝑖=1,2,…,𝑚. (28) Clearly,𝑊=⋃^𝑚_𝑖=1⋃^𝑛_𝑗=1𝑊_𝑖𝑗, where𝑊_𝑖𝑗={𝑢(𝑡)∶𝑡∈[𝑡_𝑖−1,𝑡_𝑖],𝑢∈𝑉_𝑗}. For any𝑢(𝑡),𝑢(𝑡)∈𝑊_𝑖𝑗(𝑡,𝑡∈[𝑡_𝑖−1,𝑡_𝑖],𝑢,𝑢∈𝑉_𝑗), by (27) and (28), we have 𝑢(𝑡)−𝑢𝑡≤𝑢(𝑡)−𝑢_𝑗(𝑡)+𝑢_𝑗(𝑡)−𝑢_𝑗𝑡+𝑢_𝑗𝑡−𝑢𝑡=1+𝑡^𝛼−1𝑢(𝑡)−𝑢_𝑗(𝑡)1+𝑡^𝛼−1+𝑢_𝑗(𝑡)−𝑢_𝑗𝑡+1+𝑡^𝛼−1𝑢_𝑗𝑡−𝑢𝑡1+𝑡^𝛼−1≤1+𝑡^𝛼−1𝑢−𝑢_𝑗𝑋+𝜀5+1+𝑡^𝛼−1𝑢_𝑗−𝑢_𝑋≤4diam𝑉_𝑗+𝜀5<4𝛼_𝑋(𝑉)+𝜀, (29) which impliesdiam𝑊_𝑖𝑗≤4𝛼_𝑋(𝑉)+𝜀and, thus,𝛼_𝐸(𝑊)≤4𝛼_𝑋(𝑉)+𝜀.Since𝜀is arbitrary, we get 𝛼_𝐸(𝑊)≤4𝛼_𝑋(𝑉). (30) It follows from (25), (26), and (30) that 𝛼_𝐸(𝑇𝑉)(𝑡)1+𝑡^𝛼−1≤8Γ(𝛼)1+𝑡^𝛼−1𝑘𝛼_𝑋(𝑉),∀𝑉⊂𝐷. (31) Thus sup_𝑡∈𝐼𝛼_𝐸(𝑇𝑉)(𝑡)1+𝑡^𝛼−1≤8Γ(𝛼)1+𝑡^𝛼−1𝑘𝛼_𝑋(𝑉),∀𝑉⊂𝐷. (32) Consequently 𝛼_𝑋(𝑇𝑉)≤𝑎𝛼_𝑋(𝑉), (33) where𝑎=(8/Γ(𝛼)(1+𝑡^𝛼−1))𝑘. From condition(𝐻₃), it follows that0≤𝑎<1. Therefore,𝑇is a strict set contraction operator from𝐷to𝐷, obviously𝑇is condensing too. It follows from Lemma 9 that𝑇has at least one fixed point in𝐷, that is, the BVP has at least one solution in𝑋[𝐼,𝐸].

4. Example

Now we consider the system of scalar nonlinear fractional differential equations to illustrate our results.

Let 𝐸=𝑙_∞=𝑢=𝑢₁,𝑢₂,…,𝑢_𝑛,…,sup_𝑛𝑢_𝑛<+∞,𝑡∈[0,1], (34) with the norm‖𝑢‖=sup_𝑛|𝑢_𝑛|.Evidently𝐸is a Banach space. Consider the boundary value problem: 𝐷^3/2𝑢_𝑛=𝑢_𝑛161+√𝑡1+𝑡²+sin𝑡+sin𝑢²_𝑛+142𝑛²𝑒^√𝑡,(),𝑢𝑡∈0,1_𝑛(0)=𝑢_𝑛(1)=0. (35) System (35) can be regard as a boundary value problem of the form (3), where 3𝛼=2,𝑓(𝑡,𝑢)=𝑓₁(𝑡,𝑢),𝑓₂(𝑡,𝑢),…,𝑓_𝑛(𝑡,𝑢),…with𝑓_𝑛()=𝑢𝑡,𝑢^𝑛161+√𝑡1+𝑡²+sin𝑡+sin𝑢²_𝑛+142𝑛²𝑒^√𝑡. (36)

Next, we show that conditions(𝐻₁)–(𝐻₃)are satisfied. Clearly,𝑓(𝑡,𝑢)∈𝐶(𝐼×𝐸,𝐸)and 𝑓(≤1𝑡,𝑢)161+√𝑡1+𝑡²+142𝑒^√𝑡𝑢‖+1‖42𝑒^√𝑡. (37) With the aid of simple computation, we have 1+√¹₀𝑡1161+√𝑡1+𝑡²+142𝑒^√𝑡=𝜋𝑑𝑡+1647−13𝑒≈0.0693,₁₀142𝑒^√𝑡𝑑𝑡≈0.0126. (38) Note thatΓ(3/2)≈0.8862.Take𝑎(𝑡)=(1/16(1+√𝑡)(1+𝑡²))+(1/42𝑒^√𝑡)and𝑏(𝑡)=1/42𝑒^√𝑡. Clearly, condition(𝐻₁)is satisfied. It is easy to see that condition(𝐻₂)is also satisfied.

Finally, we verify condition(𝐻₃). Denote𝑓=𝑓⁽¹⁾+𝑓⁽²⁾, where 𝑓_𝑛⁽¹⁾=𝑢_𝑛(𝑡)161+√𝑡1+𝑡²,𝑓_𝑛⁽²⁾=sin𝑡+sin𝑢²_𝑛+1(𝑡)42𝑛²𝑒^√𝑡. (39) Then we can obtain that𝛼(𝑓⁽²⁾(𝑡,𝐷))=0for any bounded set𝐷⊂𝐸. Indeed, let{𝑢^(𝑚)}⊂𝐸be bounded, that is, there exists𝑀>0, such that‖𝑢^(𝑚)‖≤𝑀,𝑚=1,2,3,…, where𝑢^(𝑚)=(𝑢₁^(𝑚),…,𝑢_𝑛^(𝑚),…). Then we have, for each𝑡∈𝐼, 𝑓_𝑛⁽²⁾𝑡,𝑢^(𝑚)≤1+𝑀42𝑛²,𝑛,𝑚=1,2,3,… (40) which implied that{𝑓_𝑛⁽²⁾(𝑡,𝑢^(𝑚))}is bounded. And by the diagonal method, we can choose a subsequence{𝑢^(𝑚𝑘)}⊂{𝑢^(𝑚)}such that 𝑓_𝑛⁽²⁾𝑡,𝑢^(𝑚𝑘)⟶𝑣_𝑛,𝑘⟶∞,𝑛=1,2,3,…. (41) From (40), it follows that 𝑣_𝑛≤1+𝑀42𝑛²,𝑛=1,2,3,… (42) that is,𝑣={𝑣₁,…,𝑣_𝑛,…}∈𝐸.

For any𝜀>0, (40) and (42) imply that there exists𝑁>0such that 𝑓_𝑛⁽²⁾𝑡,𝑢^(𝑚𝑘)<𝜀2,𝑣_𝑛<𝜀2,𝑛>𝑁,𝑘=1,2,3,…. (43) On the other hand, from (41) we obtain that there exists𝐾>0such that 𝑓_𝑛⁽²⁾𝑡,𝑢^(𝑚𝑘)−𝑣_𝑛<𝜀,𝑘>𝐾,𝑛=1,2,…𝑁. (44) It follows from (43), (44), and the definition of the norm in𝐸that 𝑓⁽²⁾𝑡,𝑢^(𝑚𝑘)−𝑣<𝜀,𝑘>𝐾. (45) This means that‖𝑓⁽²⁾(𝑡,𝑢^(𝑚𝑘))−𝑣‖→0as𝑘→∞ and so𝑓⁽²⁾(𝑡,𝐷)is relatively compact for any bounded𝐷⊂𝐸. Hence 𝛼𝑓⁽²⁾(𝑡,𝐷)=0,∀𝑡∈𝐼,𝐷⊂𝐸. (46) Consequently, we arrive at 𝛼𝑓(𝑡,𝐷)≤𝛼𝑓⁽¹⁾()=𝛼(𝐷)𝑡,𝐷161+√𝑡1+𝑡²≤1𝛼(𝐷)16,∀𝑡∈𝐼,𝐷⊂𝐸, (47) since 81+√𝑡Γ()12/316<1, (48) we conclude that condition(𝐻₃)is satisfied for𝑘=1/16.

Therefore, all the conditions of Theorem 15 are satisfied. An application of Theorem 15 implies that problem (35) has a solution.

5. Conclusion

In this paper, we present some sufficient conditions which ensure the existence of solutions to fractional differential equation for Dirichlet-type boundary value problems. Applying the Sadovskii fixed point theorem, we establish some new existence criteria for boundary value problems (3) in Banach space. Although, for the fractional differential equation for the Dirichlet boundary value problem (3), only a few papers have dealt with the boundary value problem for fractional differential equations, especially in Banach space. In this aspect, our work fills up the deficiency. As applications, examples are presented to illustrate the main results.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.