Existence of Solutions for Two-Point Boundary Value Problem of Fractional Differential Equations at Resonance

Abstract

We establish the existence results for two-point boundary value problem of fractional differential equations at resonance by means of the coincidence degree theory. Furthermore, a result on the uniqueness of solution is obtained. We give an example to demonstrate our results.

1. Introduction

Fractional differential equations have been studied extensively. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications such as physics, chemistry, phenomena arising in engineering, economy, and science; see, for example, [1–5].

Recently, more and more authors have paid their attentions to the boundary value problems of fractional differential equations; see [6–21]. Moreover, there have been many works related to the existence of solutions for boundary value problems at resonance; see [12–21]. It is considerable that there are many papers that have dealt with the solutions of multipoint boundary value problems of fractional differential equations at resonance (see, e.g., [12, 16]).

In [12], Bai and Zhang considered a three-point boundary value problem of fractional differential equations with nonlinear growth given by 𝐷^𝛼₀⁺𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷₀^𝛼−1+𝑢(𝑡𝑢(0)(1)𝜂,),0<𝑡<1,=0,𝑢=𝜎𝑢 (1) where 1<𝛼≤2, 0<𝜂, 𝜎<1>0, 𝜎𝜂^𝛼−1=1, 𝐷^𝛼₀⁺ is Riemann-Liouville fractional derivative, and 𝑓,𝑔∶[0,1]×ℝ²→ℝ are given functions.

In [13], Hu et al. have studied a two-point boundary value problem for fractional differential equation at resonance 𝐷^𝛼₀⁺𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑥(0),0≤𝑡≤1,=0,𝑥^′(0)=𝑥^′(1), (2) where 1<𝛼≤2, 𝐷^𝛼₀⁺ is Caputo fractional derivative, and 𝑓∶[0,1]×ℝ²→ℝ satisfies Carathéodory conditions.

As far as we know, there are few works on the existence of two-point boundary value problems of the fractional differential equations at resonance. Motivated by the works above, we discuss the existence and uniqueness of solutions for the following nonlinear fractional differential equation: 𝐷^𝛼₀⁺𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷₀^𝛼−1+𝑢(𝑡),𝐷₀^𝛼−2+𝑢(𝑡),…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢(𝑡),𝑢(0)=𝐷₀^𝛼−2+𝑢(0)=⋯=𝐷₀^{𝛼−(𝑁−1)}₊𝑢(0)𝐷=0,_0𝛼−1+𝑢(0)=𝐷₀^𝛼−1+𝑢(1), (3) where 0<𝑡<1, 𝑁−1<𝛼<𝑁, 𝐷^𝛼₀⁺ is Riemann-Liouville fractional derivative, and 𝑓∶[0,1]×ℝ²→ℝ is continuous function.

More precisely, we use the coincidence degree theorem due to Mawhin [22]. The rest of this paper is organized as follows. In Section 2, we give some necessary notations, definitions, and lemmas. In Section 3, we study the existence of solutions of (3) by the coincidence degree theory. Finally, an example is given to illustrate our results in Section 4.

The two-point boundary value problem (3) happens to be at resonance in the sense that the associated linear homogeneous boundary value problem 𝐷^𝛼₀⁺𝑢(𝑡)𝑢(0)=0,=𝐷₀^𝛼−2+𝑢(0)=⋯=𝐷₀^{𝛼−(𝑁−1)}₊𝑢(0)𝐷=0,_0𝛼−1+𝑢(0)=𝐷₀^𝛼−1+𝑢(1), (4) has 𝑢(𝑡)=𝑐₁𝑡^𝛼−1 as a nontrivial solution.

2. Preliminaries

In this section, we present the necessary definitions and lemmas from fractional calculus theory. These definitions and properties can be found in the literature. For more details see [1–3].

Definition 1 (see [1]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑓∶(0,∞)→ℝ is given by 𝐼^𝛼₀₊𝑓(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠, (5) provided that the right-hand side is pointwise defined on (0,∞).

Definition 2 (see [1]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a continuous function 𝑓∶(0,∞)→ℝ is given by 𝐷^𝛼₀⁺𝑓(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑛𝑡₀𝑓(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠, (6) where 𝑛−1<𝛼≤𝑛, provided that the right-hand side is pointwise defined on (0,∞).

Lemma 3 (see [1]). Let 𝑛−1<𝛼≤𝑛, 𝑢∈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, (7) where 𝐶_𝑖∈ℝ, 𝑖=1,2,…𝑛.

Lemma 4 (see [1]). If 𝛼>0, 𝑚∈ℕ and 𝐷=𝑑/𝑑𝑥. If the fractional derivatives 𝐷^𝛼₀₊𝑢(𝑡) and 𝐷^𝛼+𝑚₀₊𝑢(𝑡) exist, then 𝐷^𝑚𝐷^𝛼₀₊𝑢(𝑡)=𝐷^𝛼+𝑚₀₊𝑢(𝑡). (8)

Lemma 5 (see [1]). The relation 𝐼^𝛼_𝑎+𝐼^𝛽_𝑎+𝑓(𝑥)=𝐼^𝛼+𝛽_𝑎+𝑓(𝑥) (9) is valid in following cases 𝛽>0, 𝛼+𝛽>0, and 𝑓(𝑥)∈𝐿₁(𝑎,𝑏).

Now let us recall some notations about the coincidence degree continuation theorem.

Let 𝑌, 𝑍 be real Banach spaces, let 𝐿∶dom𝐿⊂𝑌→𝑍 be a Fredholm map of index zero, and let 𝑃∶𝑌→𝑌, 𝑄∶𝑍→𝑍 be continuous projectors such that ker𝐿=Im𝑃, Im𝐿=ker𝑄, and 𝑌=ker𝐿⊕ker𝑃, 𝑍=Im𝐿⊕Im𝑄. It follows that 𝐿|_{dom𝐿∩ker𝑃}∶dom𝐿∩ker𝑃→Im𝐿 is invertible. We denote the inverse of this map by 𝐾_𝑃. If Ω is an open bounded subset of 𝑌, the map 𝑁 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾_𝑃,𝑄𝑁=𝐾_𝑃(𝐼−𝑄)𝑁∶Ω→𝑌 is compact.

Theorem 6. Let 𝐿 be a Fredholm operator of index zero and 𝑁 be 𝐿-compact on Ω. Suppose that the following conditions are satisfied:(1)𝐿𝑥≠𝜆𝑁𝑥 for each (𝑥,𝜆)∈[(dom𝐿⧵ker𝐿)∩𝜕Ω]×(0,1);(2)𝑁𝑥∉Im𝐿 for each 𝑥∈ker𝐿∩𝜕Ω;(3)deg(𝐽𝑄𝑁|_ker𝐿,Ω∩ker𝐿,0)≠0, where 𝑄∶𝑍→𝑍 is a continuous projection as above with Im𝐿=ker𝑄 and 𝐽∶Im𝑄→ker𝐿 is any isomorphism.Then the equation 𝐿𝑥=𝑁𝑥 has at least one solution in dom𝐿∩Ω.

3. Main Results

In this section, we will prove the existence results for (3).

We use the Banach space 𝐸=𝐶[0,1] with the norm ‖𝑢‖_∞=max_0≤𝑡≤1|𝑢(𝑡)|. For 𝛼>0, 𝑁=[𝛼]+1, we define a linear space 𝑋=𝑢∣𝑢,𝐷^𝛼−𝑖₀₊𝑢∈𝐸,𝑖=1,2,…,𝑁−1. (10) By means of the functional analysis theory, we can prove that 𝑋 is a Banach space with the norm ‖𝑢‖_𝑋=‖𝐷^𝛼−1₀₊𝑢‖_∞+⋯+‖𝐷^{𝛼−(𝑁−1)}₀₊𝑢‖_∞+‖𝑢‖_∞.

Define 𝐿 to be the linear operator from dom(𝐿)⋂𝑋 to 𝐸 with dom(𝐿)={𝑢∈𝑋∣𝐷^𝛼₀₊𝑢(𝑡)∈𝐸,𝑢(0)=𝐷₀^𝛼−2+𝑢(0)=⋯=𝐷₀^{𝛼−(𝑁−1)}₊𝑢(0)=0,𝐷^𝛼−1₀₊𝑢(0)=𝐷^𝛼−1₀₊𝑢(1)} and 𝐿𝑢=𝐷^𝛼₀₊𝑢,𝑢∈dom(𝐿). (11) We define 𝑁∶𝑋→𝐸 by (𝑡)(𝑡)𝑁𝑢=𝑓𝑡,𝑢,𝐷₀^𝛼−1+𝑢(𝑡),𝐷₀^𝛼−2+𝑢(𝑡),…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢(𝑡). (12) Then the problem (3) can be written by 𝐿𝑢=𝑁𝑢.

Lemma 7. The mapping 𝐿∶dom(𝐿)⊂𝐸 is a Fredholm operator of index zero.

Proof. It is clear that ker(𝐿)=𝑐₁𝑡^𝛼−1≅ℝ¹. (13) Let 𝑥∈Im𝐿, so there exists a function 𝑢∈dom𝐿 which satisfies 𝐿𝑢=𝑥. By (11) and Lemma 3, we have 𝑢(𝑡)=𝐼^𝛼₀₊𝑥(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑁𝑡^𝛼−𝑁. (14) By 𝑢(0)=𝐷₀^𝛼−2+𝑢(0)=⋯=𝐷₀^{𝛼−(𝑁−1)}₊𝑢(0)=0, we can obtain 𝑐₂=⋯=𝑐_𝑁=0. Hence 𝑢(𝑡)=𝐼^𝛼₀₊𝑥(𝑡)+𝑐₁𝑡^𝛼−1. (15) Then, we have 𝐷^𝛼−1₀₊𝑢(𝑡)=𝐷^𝛼−1₀₊𝐼^𝛼₀₊𝑥(𝑡)+𝑐₁𝑡^𝛼−1=𝐷^𝛼−1₀₊𝐼^𝛼₀₊𝑥(𝑡)+𝑐₁Γ(𝛼)Γ(1)=^𝑡₀𝑥(𝑠)𝑑𝑠+𝑐₁Γ(𝛼). (16) Taking into account 𝐷₀^𝛼−1+𝑢(0)=𝐷₀^𝛼−1+𝑢(1), we obtain ¹₀𝑥(𝑠)𝑑𝑠=0. (17) On the other hand, suppose 𝑥 satisfy ∫¹₀𝑥(𝑠)𝑑𝑠=0. Let 𝑢(𝑡)=𝐼^𝛼₀₊𝑥(𝑡), we can easily prove 𝑢(𝑡)∈dom(𝐿).
Thus, we conclude that Im(𝐿)=𝑥∶¹₀𝑥(𝑠)𝑑𝑠=0. (18) Consider the linear operators 𝑄∶𝐸→𝐸 defined by 𝑄𝑥(𝑡)=¹₀𝑥(𝑠)𝑑𝑠. (19) Take 𝑥(𝑡)∈𝐸; then 𝑄(𝑄𝑥(𝑡))=𝑄¹₀𝑥(𝑠)=𝑑𝑠₁₀₁₀𝑥(𝑡)𝑑𝑡𝑑𝑠=¹₀𝑥(𝑠)𝑑𝑠=𝑄𝑥(𝑡). (20) We can see 𝑄²=𝑄.
For 𝑥(𝑡)∈𝐸 in the type 𝑥(𝑡)=𝑥(𝑡)−𝑄𝑥(𝑡)+𝑄𝑥(𝑡), obviously, 𝑥(𝑡)−𝑄𝑥(𝑡)∈Ker(𝑄)=Im(𝐿) and 𝑄𝑥(𝑡)∈Im(𝑄). That is to say, 𝐸=Im(𝐿)+Im(𝑄). If 𝑢∈Im(𝐿)⋂Im(𝑄), we have 𝑢=𝑐₁; then ∫¹₀𝑐₁𝑑𝑠=0. As a result 𝑐₁=0, and we get 𝐸=Im(𝐿)⊕Im(𝑄).
Note that Ind𝐿=dimker𝐿−codimIm𝐿=0. Then 𝐿 is a Fredholm mapping of index zero.

We can define the operators 𝑃∶𝑋→𝑋, where 1𝑃𝑢=Γ(𝛼)𝐷^𝛼−1₀₊𝑢(0)𝑡^𝛼−1. (21) For 𝑢∈𝑋, 𝑃()1𝑃𝑢=𝑃Γ(𝛼)𝐷^𝛼−1₀₊𝑢(0)𝑡^𝛼−1=1Γ(𝛼)𝐷^𝛼−1₀₊𝑢(0)𝑡^𝛼−1=𝑃𝑢. (22) So we have 𝑃²=𝑃.

Note that Ker(𝑃)=𝑢∶𝐷^𝛼−1₀₊𝑢(0)=0. (23) Since 𝑢=𝑢−𝑃𝑢+𝑃𝑢, it is easy to say that 𝑢−𝑃𝑢∈Ker(𝑃) and 𝑃𝑢∈Ker(𝐿). So we have 𝑋=Ker(𝑃)+Ker(𝐿). If 𝑢∈Ker(𝐿)⋂Ker(𝑃), then 𝑢=𝑐₁𝑡^𝛼−1. We can derive 𝑐₁=0 from 𝐷^𝛼−1₀₊𝑐₁𝑡^𝛼−1|_𝑡=0=0. Then 𝑋=Ker(𝐿)⊕Ker(𝑃). (24) For 𝑢∈𝑋, ‖𝑃𝑢‖_𝑋=1Γ(𝛼)𝐷^𝛼−1₀₊𝑢(0)⋅𝑡^𝛼−1_𝑋=1Γ(𝛼)𝐷^𝛼−1₀₊𝑢(0)⋅𝑡^𝛼−1_∞+𝐷^𝛼−1₀₊𝑡^𝛼−1_∞+⋯+𝐷^{𝛼−(𝑁−1)}₀₊𝑡^𝛼−1_∞=^𝑁−1_𝑖=11Γ(𝑖)+1Γ(𝛼)𝐷^𝛼−1₀₊𝑢(0𝐷)=𝑎^𝛼−1₀₊𝑢(0), (25) where 𝑎=1/Γ(𝛼)+∑^𝑁−1_𝑖=1(1/Γ(𝑖)).

We define 𝐾_𝑃∶Im𝐿→dom𝐿∩ker𝑃 by 𝐾_𝑝𝑥=𝐼^𝛼₀₊𝑥.

For 𝑥∈Im(𝐿), we have 𝐿𝐾_𝑝𝑥=𝐿𝐼^𝛼₀₊𝑥=𝐷^𝛼₀₊𝐼^𝛼₀₊𝑥=𝑥. (26) For 𝑢∈dom(𝐿)⋂Ker(𝑃), we have 𝐷^𝛼−1₀₊𝑢(0)=0. And for 𝑢∈dom(𝐿), the coefficients 𝑐₁,…,𝑐_𝑁 in the expressions 𝑢=𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑁𝑡^𝛼−𝑁 (27) are all equal to zero. Thus, we obtain 𝐾_𝑝𝐿𝑢=𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢=𝑢. (28) This shows that 𝐾_𝑝=(𝐿_{⋂dom(𝐿)Ker(𝑃)})⁻¹. Again for each 𝑥∈Im(𝐿), 𝐾_𝑝𝑥_𝑋=𝐼^𝛼₀₊𝑥_𝑋=𝐼^𝛼₀₊𝑥_∞+𝐷^𝛼−1₀₊𝐼^𝛼₀₊𝑥_∞+⋯+𝐷^{𝛼−(𝑁−1)}₀₊𝐼^𝛼₀₊𝑥_∞;≤^𝑁−1_𝑖=11Γ()+1𝑖+1Γ()‖𝑥‖𝛼+1^∞=𝑏‖𝑥‖_∞, (29) where 𝑏=1/Γ(𝛼+1)+∑^𝑁−1_𝑖=1(1/Γ(𝑖+1)).

Lemma 8. Assume Ω⊂𝑌 is an open bounded subset such that dom𝐿⋂𝑌≠∅; then map 𝑁 is 𝐿-compact on Ω

Proof. By the continuity of 𝑓, we can get that 𝑄𝑁(Ω) and 𝐾_𝑃(𝐼−𝑄)𝑁(Ω) are bounded. So, in view of the Arzela-Ascoli theorem, we need only to prove that 𝐾_𝑃(𝐼−𝑄)𝑁(Ω) is equicontinuous. From the continuity of 𝑓, there exists a constant 𝑟>0, such that |(𝐼−𝑄)𝑁(𝑢(𝑡))|≤𝑟, for all 𝑢∈Ω, 𝑡∈[0,1].
For 0≤𝑡₁≤𝑡₂≤1, 𝑢∈Ω, we have 𝐾_𝑃,𝑄𝑁𝑢𝑡₂−𝐾_𝑃,𝑄𝑁𝑢𝑡₁=1Γ(𝛼)^𝑡2₀𝑡₂−𝑠^𝛼−1()𝑁(𝑢(𝑠−𝐼−𝑄))𝑑𝑠_𝑡10𝑡₁−𝑠^𝛼−1()𝑁(𝑢(𝑠≤𝑟𝐼−𝑄))𝑑𝑠Γ(𝛼)^𝑡1₀𝑡₁−𝑠^𝛼−1−𝑡₂−𝑠^𝛼−1+𝑟𝑑𝑠Γ(𝛼)^𝑡2_𝑡1𝑡₂−𝑠^𝛼−1=𝑟𝑑𝑠Γ()𝑡𝛼+1^𝛼2−𝑡^𝛼₁. (30) Furthermore, we have 𝐷^𝛼−𝑖₀₊𝐾_𝑃,𝑄𝑁𝑢𝑡₂−𝐷^𝛼−𝑖₀₊𝐾_𝑃,𝑄𝑁𝑢𝑡₁=1Γ(𝑖)^𝑡2₀𝑡₂−𝑠^𝑖−1()𝑁(𝑢(𝑠−𝐼−𝑄))𝑑𝑠_𝑡10𝑡₁−𝑠^𝑖−1()𝑁(𝑢(𝑠≤𝑟𝐼−𝑄))𝑑𝑠Γ(𝑖)^𝑡1₀𝑡₁−𝑠^𝑖−1−𝑡₂−𝑠^𝑖−1+𝑟𝑑𝑠Γ(𝑖)^𝑡2_𝑡1𝑡₂−𝑠^𝑖−1=𝑟𝑑𝑠Γ()𝑡𝑖+1^𝑖2−𝑡^𝑖₁, (31) where 𝑖=1,2,…,𝑁−1. Since 𝑡^𝛼 and 𝑡^𝑖 are uniformly continuous on [0,1], we can get that 𝐾_𝑃(𝐼−𝑄)𝑁∶Ω→𝑌 is compact. The proof is completed.

To obtain our main results, we need the following conditions.(H₁)There exist functions 𝜑,𝜓_𝑖∈𝐿¹[0,1], 𝑖=1,𝑁, such that for all 𝑢∈ℝ², 𝑡∈[0,1], 𝑓𝑡,𝑥₁,𝑥₂,…,𝑥_𝑁≤𝜑+𝜓₁𝑥₁+𝜓₂𝑥₂+⋯+𝜓_𝑁𝑥_𝑁. (32)(H₂)There exists a constant 𝐴>0 such that for every 𝑦∈ℝ, if |𝑥₂|>𝐴 for all 𝑡∈[0,1], then 𝑓𝑡,𝑥₁,𝑥₂,…,𝑥_𝑁≠0. (33)(H₃)There exists a constant 𝐷>0 such that, for each 𝑐_𝑖, 𝑖=1,2 satisfying min{|𝑐₁|,|𝑐₂|}>𝐷. We have either at least one of the following: 𝑐₁𝑁𝑐₁𝑡^𝛼−1>0 (34) or 𝑐₁𝑁𝑐₁𝑡^𝛼−1<0. (35)(H₄)∑^𝑁_𝑖=2𝜌_𝑖<1, where 𝜌_𝑖+1=(𝑎+𝑏)‖𝜓_𝑖‖₁, 𝑖=1,2,…,𝑁.

Lemma 9. Ω₁={𝑢∈dom(𝐿)⧵Ker(𝐿)∣𝐿𝑢=𝜆𝑁𝑢,𝜆∈[0,1]} is bounded.

Proof. For 𝑢∈Ω₁, 𝜆≠0 and 𝐿𝑢=𝜆𝑁𝑢. By (12), 𝐿𝑢=𝜆𝑁𝑢∈Im(𝐿)=Ker(𝑄); that is, 𝜆¹₀𝑓𝑡,𝑢(𝑡),𝐷₀^𝛼−1+𝑢(𝑡),𝐷₀^𝛼−2+𝑢(𝑡),…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢(𝑡)𝑑𝑡=0. (36) By the integral mean value theorem, there exits a constant 𝑡₀∈[0,1] such that 𝑓𝑡₀,𝑢𝑡₀,𝐷₀^𝛼−1+𝑢𝑡₀,𝐷₀^𝛼−2+𝑢𝑡₀,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢𝑡₀=0. (37) Form (H₂), we can get |𝐷^𝛼−1₀₊𝑢(𝑡₀)|≤𝐴.
Again for 𝑢∈Ω₁, (𝐼−𝑃)𝑢∈dom(𝐿)⧵Ker(𝐿) and 𝐿𝑃𝑢=0. From (29), we have ‖(𝐼−𝑃)𝑢‖_𝑋=𝐾_𝑝𝐿(𝐼−𝑃)𝑢_𝑋=𝐾_𝑝𝐿𝑢_𝑋≤𝑏‖𝑁𝑢‖_∞. (38) Now by Lemma 4𝐷^𝛼−1₀₊𝑢(0)≤𝐷^𝛼−1₀₊𝑢𝑡₀+^𝑡0₀𝐷^𝛼₀₊𝑢(𝑠)≤𝐷𝑑𝑠_𝛼−10+𝑢𝑡₀+𝑡₀max_{0≤𝑡≤𝑡0}𝐷^𝛼₀₊𝑢(𝑡≤𝐷_𝛼−10+)𝑢𝑡₀+𝐷^𝛼₀₊𝑢(𝑡)_∞≤𝐴+‖𝐿𝑢‖_∞=𝐴+‖𝑁𝑢‖_∞. (39) That is, 𝐷^𝛼−1₀₊𝑢(0)≤𝐴+‖𝑁𝑢‖_∞. (40) From (25) and (38), we have ‖𝑢‖_𝑋=‖𝑃𝑢+(𝐼−𝑃)𝑢‖_𝑋≤‖𝑃𝑢‖_𝑋+‖(𝐼−𝑃)𝑢‖_𝑋≤𝑎𝐷^𝛼−1₀₊𝑢(0)+𝑏‖𝑁𝑢‖_∞. (41) Furthermore, it follows from (40) and (H₁) that ‖𝑢‖_𝑋≤𝑎𝐷^𝛼−1₀₊𝑢(0)+𝑏‖𝑁𝑢‖_∞≤𝑎𝐴+‖𝑁𝑢‖_∞+𝑏‖𝑁𝑢‖_∞=𝑎𝐴+(𝑎+𝑏)‖𝑁𝑢‖_∞()×𝑓(𝑡)≤𝑎𝐴+𝑎+𝑏𝑡,𝑢,𝐷₀^𝛼−1+𝑢(𝑡),𝐷₀^𝛼−2+𝑢(𝑡),…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢(𝑡)_∞≤𝑎𝐴+(𝑎+𝑏)𝜑₁+𝜓₁‖𝑢‖_∞+𝜓₂₁𝐷^𝛼−1₀₊𝑢_∞+⋯+𝜓_𝑁1𝐷^{𝛼−(𝑁−1)}₀₊𝑢_∞=𝑎𝐴+(𝑎+𝑏)𝜑₁+𝜌₂‖𝑢‖_∞+𝜌₃𝐷^𝛼−1₀₊𝑢_∞+𝜌₄𝐷^𝛼−2₀₊𝑢_∞+⋯+𝜌_𝑁+1𝐷^{𝛼−(𝑁−1)}₀₊𝑢_∞. (42) By the definition ‖𝑢‖_𝑋 and (H₄), it is easy to see that ‖𝐷^𝛼−1₀₊𝑢‖_∞,…,‖𝐷^{𝛼−(𝑁−1)}₀₊𝑢‖_∞ and ‖𝑢‖_∞ are bounded. So, Ω₁ is bounded.

Lemma 10. Ω₂={𝑢∈Ker(𝐿)∶𝑁𝑢∈Im(𝐿)} is bounded.

Proof. Let 𝑢∈Ker(𝐿), so we have 𝑢=𝑐₁𝑡^𝛼−1, 𝑐₁∈ℝ. For 𝑁𝑢∈Im(𝐿)=Ker(𝑄), ¹₀𝑓𝑡,𝑐₁𝑡^𝛼−1,𝑐₁Γ(𝛼)Γ(𝛼),…,Γ()𝑐𝑁−1¹𝑡^𝑁−2𝑑𝑡=0. (43) By the integral mean value theorem, there exits a constant 𝑡₁∈[0,1] such that 𝑓𝑡₁,𝑐₁𝑡₁^𝛼−1,𝑐₁Γ(𝛼)Γ(𝛼),…,Γ()𝑐𝑁−1¹𝑡₁^𝑁−2=0. (44) From (H₂), it follows that |𝑐₁|≤𝐴/Γ(𝛼). Hence, Ω₂ is bounded.

Lemma 11. Ω₃={𝑢∈Ker(𝐿)∶𝜆𝑢+(1−𝜆)𝑄𝑁𝑢=0,𝜆∈[0,1]} is bounded.

Proof. Let 𝑢∈Ker(𝐿), so we have 𝑢=𝑐₁𝑡^𝛼−1, 𝑐₁∈ℝ. If 𝜆=0, then |𝑐₁|≤𝐷. If 𝜆=1, we have 𝑐₁=0.
If 𝜆≠0 and 𝜆≠1, then 𝜆𝑐₁𝑡^𝛼−1+(1−𝜆)𝑄𝑁(𝑢)=0. (45) It follows that 𝜆𝑐₁𝑡^𝛽−1+()×1−𝜆₁₀𝑓𝑡,𝑐₁𝑡^𝛼−1,𝑐₁Γ(𝛼)Γ(𝛼),…,Γ()𝑐𝑁−1¹𝑡^𝑁−2𝑑𝑡=0. (46) Then we get 𝜆𝑐²₁𝑡^𝛼−1+(1−𝜆)¹₀𝑐₁𝑓𝑡,𝑐₁𝑡^𝛼−1,…,𝑐₂Γ(𝛼)𝑑𝑡=0, (47) which, together with (H₃), implies |𝑐₁|≤𝐷. Here, Ω₃ is bounded.

Remark 12. If the other parts of (H₃) hold, then the set Ω^′₃={𝑢∈Ker(𝐿)∶−𝜆𝑢+(1−𝜆)𝑄𝑁𝑢=(0,0),𝜆∈[0,1]} is bounded.

Theorem 13. Suppose (H₁)–(H₄) hold; then the problem (3) has at least one solution in 𝑌.

Proof. Let Ω be a bounded open set of 𝑌, such that ⋃³_𝑖=1Ω_𝑖⊂Ω. It follows from Lemma 8, 𝑁 is 𝐿-compact on Ω. By Lemmas 9, 10, and 11, we get the following:(1)𝐿𝑢≠𝜆𝑁𝑢, for every 𝑢∈[(dom𝐿⧵Ker𝐿)⋂𝜕Ω]×(0,1);(2)𝑁𝑢∉Im𝐿 for every 𝑢∈Ker𝐿⋂𝜕Ω;(3)let 𝐻(𝑢,𝜆)=±𝜆𝐼𝑢+(1−𝜆)𝐽𝑄𝑁𝑢, where 𝐼 is the identical operator. Via the homotopy property of degree, we obtain that deg𝐽𝑄𝑁|_ker𝐿(𝐻())(𝐻())(),Ω∩ker𝐿,0=deg⋅,0,Ω∩ker𝐿,0=deg⋅,1,Ω∩ker𝐿,0=deg𝐼,Ω∩ker𝐿,0=1≠0. (48)Applying Theorem 6, we conclude that 𝐿𝑢=𝑁𝑢 has at least one solution in dom𝐿∩Ω.

Under the stronger conditions imposed on 𝑓, we can prove the uniqueness of solutions to the (3) studied above.

Theorem 14. Suppose the conditions (H₁) in the theorem are replaced by the following conditions.(H₁)′There exist positive constants 𝑎_𝑖, 𝑖=0,1,…,𝑁−1, such that, for all (𝑥₁,𝑥₂,…,𝑥_𝑁), (𝑦₁,𝑦₂,…,𝑦_𝑁)∈ℝ^𝑁, one has 𝑓𝑡,𝑥₁,𝑥₂,…,𝑥_𝑁−𝑓𝑡,𝑦₁,𝑦₂,…,𝑦_𝑁≤𝑎₀𝑥₁−𝑦₁+⋯+𝑎_𝑁−1𝑥_𝑁−𝑦_𝑁. (49)(H₁)′′There exist constants 𝑙_𝑖, 𝑖=1,2,…,𝑁−1, such that for all (𝑥₁,𝑥₂,…,𝑥_𝑁), (𝑦₁,𝑦₂,…,𝑦_𝑁)∈ℝ^𝑁, one has 𝑓𝑡,𝑥₁,𝑥₂,…,𝑥_𝑁−𝑓𝑡,𝑦₁,𝑦₂,…,𝑦_𝑁≥−𝑙₀𝑥₁−𝑦₁+𝑙₁𝑥₂−𝑦₂−𝑙₂𝑥₃−𝑦₃−⋯−𝑙_𝑁−1𝑥_𝑁−𝑦_𝑁. (50)Then, the BVP (3) has a unique solution, provided that 𝑎𝑙₀𝑙₁+𝑎𝑎₀+𝑎₀𝑐+^𝑁−1_𝑖=2𝑎𝑙_𝑖𝑙₁+(𝑎+𝑐)^𝑁−1_𝑖=1𝑎_𝑖<1. (51)

Proof. Let 𝑦_𝑖=0, 𝑖=1,2,…,𝑁, and 𝜑₁=|𝑓(𝑡,0,…,0)|; then the condition (H₁) is satisfied. According to Theorem 13, BVP (3) has at least one solution. Suppose 𝑢_𝑖∈𝑌, 𝑖=1,2 are two solutions of (3); then 𝐷^𝛼₀⁺𝑢_𝑖(𝑡)=𝑓𝑡,𝑢_𝑖(𝑡),𝐷₀^𝛼−1+𝑢_𝑖(𝑡),𝐷₀^𝛼−2+𝑢_𝑖(𝑡),…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢_𝑖(𝑡),𝑖=1,2. (52) Note that 𝑢=𝑢₁−𝑢₂, so 𝑢 satisfy the equation 𝐷^𝛼₀⁺𝑢=𝑓𝑡,𝑢₁,𝐷₀^𝛼−1+𝑢₁,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₁−𝑓𝑡,𝑢₂,𝐷₀^𝛼−1+𝑢₂,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₂. (53) According to Im(𝐿)=Ker(𝑄), we have ¹₀𝑓𝑡,𝑢₁,𝐷₀^𝛼−1+𝑢₁,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₁−𝑓𝑡,𝑢₂,𝐷₀^𝛼−1+𝑢₂,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₂𝑑𝑡=0. (54) By the integral mean value theorem, there exists 𝜂∈[0,1], such that 𝑓𝜂,𝑢₁𝜂,𝐷₀^𝛼−1+𝑢₁𝜂,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₁𝜂−𝑓𝜂,𝑢₂𝜂,𝐷₀^𝛼−1+𝑢₂𝜂,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₂𝜂=0. (55) By (H₁)′′, we have 0=𝑓𝜂,𝑢₁𝜂,𝐷₀^𝛼−1+𝑢₁𝜂,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₁𝜂−𝑓𝜂,𝑢₂𝜂,𝐷₀^𝛼−1+𝑢₂𝜂,…,𝐷₀^{𝛼−(𝑁−1)}₊𝑢₂𝜂≥−𝑙₀𝑢𝜂+𝑙₁𝐷₀^𝛼−1+𝑢𝜂−𝑙₂𝐷₀^𝛼−2+𝑢𝜂−⋯−𝑙_𝑁−1𝐷₀^{𝛼−(𝑁−1)}₊𝑢𝜂. (56) We can have 𝐷₀^𝛼−1+𝑢𝜂≤𝑙₀𝑙₁𝑢𝜂+𝑙₂𝑙₁𝐷₀^𝛼−2+𝑢𝜂𝑙+⋯+^𝑁−1𝑙₁𝐷₀^{𝛼−(𝑁−1)}₊𝑢𝜂≤𝑙₀𝑙₁‖𝑢‖_∞+^𝑁−1_𝑖=2𝑙_𝑖𝑙₁𝐷₀^𝛼−𝑖+𝑢_∞. (57) Thus, we can obtain 𝐷^𝛼−1₀₊𝑢(0)≤𝐷^𝛼−1₀₊𝑢𝜂+^𝜂₀𝐷^𝛼₀₊𝑢(𝑠)≤𝐷𝑑𝑠_𝛼−10+𝑢𝜂+𝜂max_{0≤𝑡≤𝜂}𝐷^𝛼₀₊𝑢(𝑡≤𝑙₀)𝑙₁‖𝑢‖_∞+^𝑁−1_𝑖=2𝑙_𝑖𝑙₁𝐷₀^𝛼−𝑖+𝑢_∞+𝐷^𝛼₀₊𝑢(𝑡)_∞=𝑙₀𝑙₁‖𝑢‖_∞+^𝑁−1_𝑖=2𝑙_𝑖𝑙₁𝐷₀^𝛼−𝑖+𝑢_∞+‖𝐿𝑢‖_∞. (58) According to (25), (38), and (58), we have ‖𝑢‖_𝑋=‖𝑃𝑢+(𝐼−𝑃)𝑢‖_𝑋≤‖𝑃𝑢‖_𝑋+‖(𝐼−𝑃)𝑢‖_𝑋=𝑎𝑙₀𝑙₁‖𝑢‖_∞+^𝑁−1_𝑖=2𝑎𝑙_𝑖𝑙₁𝐷₀^𝛼−𝑖+𝑢_∞+𝑎‖𝐿𝑢‖_∞+𝑐‖𝐿𝑢‖_∞;≤𝑎𝑙₀𝑙₁‖𝑢‖_∞+^𝑁−1_𝑖=2𝑎𝑙_𝑖𝑙₁𝐷₀^𝛼−𝑖+𝑢_∞+(𝑎+𝑐)𝑎₀‖𝑢‖_∞+^𝑁−1_𝑖=1𝑎_𝑖𝐷₀^𝛽−𝑖₊𝑢_∞. (59) From the definition of ‖𝑢‖_𝑋 and the assumption (51), we have ‖𝑢‖=0, so that 𝑢₁=𝑢₂.

4. Example

Let us consider the following boundary value problems: 𝐷₀^2.5+𝑢(𝑡)=𝑡5+19𝐷₀^1.5+𝑢(𝑡)+sin²𝐷₀^0.5+𝑢(𝑡(𝑡),𝑢(0))+arctan𝑢0<𝑡<1,=𝐷₀^0.5+𝑢(0)=0,𝐷₀^1.5+𝑢(0)=𝐷₀^1.5+𝑢(1). (60) Corresponding to the problem (3), we have that 𝛼=2.5 and 𝑓=𝑡𝑡,𝑥,𝑦,𝑧51+arctan𝑥+9𝑦+sin²(𝑧). (61) Moreover, 𝑓≤1𝑡,𝑥,𝑦,𝑧5+𝜋2+19𝑦+1. (62) We can get that the condition (H₁) holds; that is, 𝜑=(12+5𝜋)/10, 𝜓₁=𝜓₃=0, and 𝜓₂=1/9. Taking 𝐴=25, 𝐷=19, we can calculate that (H₂)–(H₄) hold.

Hence, by Theorem 13, we obtain that (60) has at least one solution.

Conflict of Interests

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

Authors’ Contribution

All authors typed, read, and approved the final paper.

Acknowledgments

Research was supported by the National Natural Science Foundation of China (11371364) and 2013 Science and Technology Research Project of Beijing Municipal Education Commission (KM201310016001).