Upper and Lower Solution Method for Fractional Boundary Value Problems on the Half-Line

Abstract

We establish the existence of unbounded solutions for nonlinear fractional boundary value problems on the half-line. By the upper and lower solution method technique, sufficient conditions for the existence of solutions for the fractional boundary value problems are established. An example is presented to illustrate our main result.

1. Introduction

Boundary value problems on the half-line arise in the study of radially symmetric solutions of nonlinear elliptic equations and various physical phenomena, such as the theory of drain flows and plasma physics; see [1–10] and the references therein. In 2006, Lian and Ge in [11] investigated the following boundary value problem on the half-line for the second-order differential equation: 𝑥^′′(𝑡)+𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)(),𝑥(0)𝜂=0,𝑡∈0,+∞=𝛼𝑥,lim_𝑡→∞𝑥^′(𝑡)=0, (1) where𝛼∈ℝ, 𝛼≠1, and 𝜂∈(0,∞) are given. Based on Leray-Schauder continuation theorem, some suitable conditions for the existence of solutions to (1) are established.

On the other hand, fractional calculus is a generalization of the ordinary differentiation and integration to arbitrary noninteger order. Fractional calculus is a wonderful technique to understand memory and hereditary properties of materials and processes. Some recent contributions to fractional differential equations are present in the monographs [12–19]. Very recently, Chen and Tang in [20] considered the following fractional differential boundary value problem on the half-line: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡[),𝑢(0)=𝑓𝑡,𝑢)),𝑡∈0,∞=𝑢^′(0)=𝑢^′′(0)=0,𝐷₀^𝛼−1₊𝑢(𝑡)=lim_𝑡→∞𝐷₀^𝛼−1₊𝑢(𝑡), (2) where3<𝛼<4 and 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative. By the recent Leggett-Williams norm-type theorem, the existence of positive solutions is obtained. In 2011, [21] set up the global existence results of solutions of initial value problems on the half-axis as follows: 𝐷^𝛼₀₊𝑥(𝑡)=𝑓(𝑡,𝑥(𝑡)),𝑡∈(0,∞),0<𝛼≤1,lim_𝑡→∞𝑡^1−𝛼𝑥(𝑡)=𝑢₀, (3) where 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative. By constructing a special Banach space and employing fixed point theorems, some sufficient conditions for the existence of solutions are obtained. In [22], the authors studied the following boundary value problem of fractional order on the half-line: 𝐷^𝛼₀₊𝑢(𝑡)+𝑎(𝑡)𝑓𝑡,𝑢(𝑡),𝐷₀^𝛼−1₊𝑢(𝑡[),𝑢(0))=0,𝑡∈0,∞=0,lim_𝑡→∞𝐷₀^𝛼−1₊𝑢(𝑡)=𝑢_∞, (4) where1<𝛼≤2, 𝑓∈𝐶([0,∞)×ℝ²,ℝ), 𝐷₀^𝛼−1₊ and 𝐷^𝛼₀₊ are the standard Riemann-Liouville fractional derivatives. By Schauder’s fixed point theorem on an unbounded domain, they obtain the existence result for (4). Some papers have recently been done for fractional boundary value problem on the half-line or unbounded domain, see [22–31].

Inspired by the above-mentioned works, in this paper, we study the existence of solutions to the following fractional differential equations with boundary value problems: 𝐷^𝛼₀₊𝑢(𝑡)+𝑎(𝑡)𝑓𝑡,𝑢(𝑡),𝐷₀^𝛼−2₊𝑢(𝑡),𝐷₀^𝛼−1₊𝑢(𝑡()𝑢(0))=0,𝑡∈0,∞,2<𝛼≤3,=𝐷₀^𝛼−2₊𝑢(0)=0,lim_𝑡→∞𝐷₀^𝛼−1₊𝑢(𝑡)=0, (5) where𝑎(𝑡)∶[0,∞)→[0,∞), and 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative. By upper and lower solution method techniques, the sufficient conditions for solutions to (5) are obtained. Our main findings given in this paper have some new features. Firstly, the like Nagumo condition defined by us plays an important role in the nonlinear term involving the standard Riemann-Liouville derivatives. Secondly, to the best of our knowledge, no work has been done concerning fractional boundary value problems (5) and our method is different from that of [22, 26, 31]. Thirdly, the nonlinear term 𝑓 may take negative values, and 𝑓(𝑡,𝑥,𝑦,𝑧) depends on 𝑧 allowed to be quadratic, referring to our example. The rest of this paper is organized as follows: in Section 2, we present some preliminaries and some lemmas that will be used in Section 3. The main result and proof will be given in Section 3. In addition, an example is given to demonstrate the application of our main result.

2. Preliminaries

We first present some basic definitions and preliminary results about fractional calculus; we refer the reader to [17, 18] for more details.

Definition 1 (see [18]). The integral 𝐼^𝛼₀₊𝑓(𝑡)=1Γ(𝛼)^𝑡₀𝑓(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠,𝑡>0, (6) where𝛼>0, is called the Riemann-Liouville fractional integral of order 𝛼 and Γ(𝛼) is the Euler gamma function defined by Γ(𝛼)=∫^∞₀𝑡^𝛼−1𝑒^−𝑡𝑑𝑡, 𝛼>0.

Definition 2 (see [17]). A function 𝑓(𝑥) given in the interval [0,∞), the expression 𝐷^𝛼₀₊𝑓(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀𝑓(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠, (7) where𝑛=[𝛼]+1,[𝛼]denotes the integer part of number𝛼, is called the Riemann-Liouville fractional derivative of order𝛼.

Lemma 3 (see [17]). Let𝛼>0and 𝑢(𝑡)∈𝐶(0,∞)⋂𝐿(0,∞).Then the fractional differential equation 𝐷^𝛼₀₊𝑢(𝑡)=0 (8) has 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛,𝑐_𝑖[𝛼]∈𝑅,𝑖=0,1,2,3,…,𝑛,𝑛=+1 (9) as a unique solution.

Lemma 4 (see [17]). Assume that𝑢(𝑡)∈𝐶(0,∞)⋂𝐿(0,∞)with a fractional derivative of order𝛼>0that belongs to𝐶(0,∞)⋂𝐿(0,∞); then 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛, (10) for some𝑐_𝑖∈𝑅,𝑖=0,1,2,3,…,𝑛, 𝑛=[𝛼]+1.

Denote by𝐶[0,∞)the space of all continuous real functions defined on[0,∞)and 𝐿_loc(0,∞) the space of all real functions which are Lebesgue integrable on every bounded subinterval of (0,∞). Denote 𝐸=𝑥∈𝐶[0,∞)∶sup_0≤𝑡<∞|𝑥(𝑡)|(1+𝑡)^𝛼−1<∞,sup_0≤𝑡<∞𝐷₀^𝛼−2₊𝑥(𝑡)<∞,lim_𝑡→∞𝐷₀^𝛼−1₊𝑥(𝑡)exists (11) with the norm ‖𝑥‖=max{‖𝑥‖₁,‖𝐷₀^𝛼−2₊𝑥‖_∞,‖𝐷₀^𝛼−1₊𝑥‖_∞}, where ‖𝑥‖₁=sup_{[)𝑡∈0,∞}𝑥(𝑡)(1+𝑡)^𝛼−1,𝐷₀^𝛼−2₊𝑥_∞=sup_{𝑡∈[0,∞)}𝐷₀^𝛼−2₊𝑥(𝑡,𝐷_0𝛼−1+)𝑥_∞=sup_{𝑡∈[0,∞)}𝐷₀^𝛼−1₊𝑥(𝑡). (12) By standard arguments, it is easy to prove that(𝐸,‖⋅‖)is a Banach space.

Proposition 5 (see [32]). Assume that𝑓is in𝐶(0,∞)⋂𝐿_loc(0,∞)with a fractional derivative of order 0<𝛼<1 that belongs to 𝐶[0,∞)⋂𝐿_loc[0,∞). Then 𝐼^𝛼𝐷^𝛼𝑓(𝑥)=𝑓(𝑥)+𝑐𝑥^𝛼−1, (13) for some𝑐∈ℝ. When the function 𝑓 is in 𝐶[0,∞), then 𝑐=0.

Definition 6. A function𝛾∈𝐸⋂𝐿_loc(0,∞)is called a lower solution of (5) if 𝐷^𝛼₀₊𝛾(𝑡)+𝑎(𝑡)𝑓𝑡,𝛾(𝑡),𝐷₀^𝛼−2₊𝛾(𝑡),𝐷₀^𝛼−1₊𝛾(𝑡(),𝛾(0))≥0,𝑡∈0,∞≤0,𝐷₀^𝛼−2₊𝛾(0)≤0,lim_𝑡→∞𝐷₀^𝛼−1₊𝛾(𝑡)≤0. (14) Similarly, we can define an upper solution𝛽∈𝐸⋂𝐿_loc(0,∞)of (5) by reversing the above inequalities.

Remark 7. If 𝐷₀^𝛼−2₊𝛾(𝑡)≤𝐷₀^𝛼−2₊𝛽(𝑡),𝑡∈[0,∞), (15) and𝛾(0)≤0, 𝛽(0)≥0, then we have𝛾(𝑡)≤𝛽(𝑡)for all𝑡∈[0,∞).

We consider the following two cases.

Case 1. Consider 𝛼=3. The inequality (15) reduces to𝛾^′(𝑡)≤𝛽^′(𝑡). Then ^𝑡₀𝛾^′(𝑠)𝑑𝑠≤^𝑡₀𝛽^′(𝑠)𝑑𝑠,𝑡∈(0,∞); (16) that is, 𝛾(𝑡)−𝛾(0)≤𝛽(𝑡)−𝛽(0),𝑡∈(0,∞). (17) From the boundary condition𝛾(0)≤0and𝛽(0)≥0 we obtain𝛾(0)≤𝛽(0), and 𝛾(𝑡)≤𝛽(𝑡)+𝛾(0)−𝛽(0)≤𝛽(𝑡),𝑡∈(0,∞). (18) Thus,𝛾(𝑡)≤𝛽(𝑡)for𝑡∈[0,∞).

Case 2. Consider2<𝛼<3. If (15) holds, then 𝐼₀^𝛼−2₊𝐷₀^𝛼−2₊𝛾(𝑡)≤𝐼₀^𝛼−2₊𝐷₀^𝛼−2₊𝛽(𝑡),𝑡∈[0,∞). (19) Noting that0<𝛼−2<1and𝛾,𝛽∈𝐶[0,∞), from the above inequality and Proposition 5, we have that𝛾(𝑡)≤𝛽(𝑡)for𝑡∈[0,∞).
For example, consider𝛾(𝑡)=𝑡^𝛼−2,𝛽(𝑡)=2𝑡^𝛼−2. Obviously,𝐷₀^𝛼−2₊𝛾(𝑡)=Γ(𝛼+1), 𝐷₀^𝛼−2₊𝛽(𝑡)=2Γ(𝛼+1). Thus𝛾(0)=𝛽(0)=0and 𝐷₀^𝛼−2₊𝛾(𝑡)≤𝐷₀^𝛼−2₊𝛽(𝑡),𝑡∈[0,∞). (20) We also get𝛾(𝑡)≤𝛽(𝑡)for each𝑡∈[0,∞).

Definition 8. Let 𝛾,𝛽∈𝐸⋂𝐿_loc(0,∞) be lower and upper solutions to (5) and suppose that (15) holds. A continuous function𝑓∶[0,∞)×ℝ³→ℝ is said to satisfy the like Nagumo condition with respect to the pair of functions𝛾,𝛽, if there exist a nonnegative function𝜙∈𝐶[0,∞)and a positive oneℎ∈𝐶[0,∞)such that 𝑓𝑡,𝑥,𝑦,𝑧≤𝜙(𝑡)ℎ(|𝑧|), (21) for all0≤𝑡<∞, 𝛾(𝑡)≤𝑥≤𝛽(𝑡), 𝐷₀^𝛼−2₊𝛾(𝑡)≤𝑦≤𝐷₀^𝛼−2₊𝛽(𝑡),𝑧∈ℝ, and ^∞₀𝑠ℎ(𝑠)𝑑𝑠=∞. (22) We list some assumptions related to𝑎(𝑡),𝜙(𝑡), and𝑓 as follows. (𝐻₁) Consider max{^∞₀𝑠,1}𝑎(𝑠)𝑑𝑠<∞,_∞0max{𝑠,1}𝑎(𝑠)𝜙(𝑠)𝑑𝑠<∞. (23) (𝐻₂) Let𝑓∶[0,∞)×ℝ³→ℝ be a continuous function satisfying the like Nagumo condition with respect to the pair of functions𝛾, 𝛽, such that 𝑓(𝑡)(𝑡),∈[)×𝛾(𝑡)(𝑡𝑡,𝛾,𝑦,𝑧≤𝑓𝑡,𝑥,𝑦,𝑧≤𝑓𝑡,𝛽,𝑦,𝑧for𝑡,𝑥,𝑦,𝑧0,∞,𝛽)×ℝ². (24)

Lemma 9. Let𝜎∈𝐶(0,∞)⋂𝐿(0,∞), then the fractional boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)()𝑢(0)+𝜎=0,𝑡∈0,∞,2<𝛼≤3,=𝐷₀^𝛼−2₊𝑢(0)=0,lim_𝑡→∞𝐷₀^𝛼−1₊𝑢(𝑡)=0 (25) has a unique solution𝑢(𝑡)=∫^∞₀𝐺(𝑡,𝑠)𝜎(𝑠)𝑑𝑠, where 𝐺()=1𝑡,𝑠Γ(𝛼)𝑡^𝛼−1−(𝑡−𝑠)^𝛼−1𝑡,0≤𝑠≤𝑡<∞,_𝛼−1,0≤𝑡≤𝑠<∞. (26)

Proof. By Lemma 4, the solution of (25) has the following form: 𝑢(𝑡)1=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜎(𝑠)𝑑𝑠+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+𝑐₃𝑡^𝛼−3. (27) By boundary value condition𝑢(0)=0, one has𝑐₃=0. Thus, we have by (27) that 𝐷₀^𝛼−2₊𝑢(𝑡)=−𝐷₀^𝛼−2₊𝐼^𝛼₀₊𝜎(𝑡)+𝑐₁𝐷₀^𝛼−2₊𝑡^𝛼−1+𝑐₂𝐷₀^𝛼−2₊𝑡^𝛼−2=−^𝑡₀(𝑡−𝑠)𝜎(𝑠)𝑑𝑠+𝑐₁Γ(𝛼)𝑡+𝑐₂Γ(𝛼−1). (28) Substituting the condition𝐷₀^𝛼−2₊𝑢(0)=0into (28), we obtain 𝑐₂=0. Together with lim_𝑡→∞𝐷₀^𝛼−1₊𝑢(𝑡)=0, we get 𝑐₁=1Γ(𝛼)^∞₀𝜎(𝑠)𝑑𝑠. (29) Hence, we have 𝑢(𝑡)1=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜎(𝑠)𝑡𝑑𝑠+^𝛼−1Γ(𝛼)^∞₀𝜎(𝑠)1𝑑𝑠=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜎(𝑠)+1𝑑𝑠Γ(𝛼)^𝑡₀𝑡^𝛼−1𝜎(𝑠)1𝑑𝑠+Γ(𝛼)^∞_𝑡𝑡^𝛼−1𝜎(𝑠)=1𝑑𝑠Γ(𝛼)^𝑡₀𝑡^𝛼−1−(𝑡−𝑠)^𝛼−1𝜎(𝑠)+1𝑑𝑠Γ(𝛼)^∞_𝑡𝑡^𝛼−1𝜎(𝑠)=𝑑𝑠_∞0𝐺(𝑡,𝑠)𝜎(𝑠)𝑑𝑠. (30)

Lemma 10. The function𝐺(𝑡,𝑠)defined by (26) satisfies the following properties:(1)𝐺(𝑡,𝑠)is a continuous function and𝐺(𝑡,𝑠)≥0 for [0,∞)×[0,∞);(2)𝐺(𝑡,𝑠)≤𝑡^𝛼−1/Γ(𝛼), for (𝑡,𝑠)∈[0,∞)×[0,∞).

The proof is easy, so we omit it here.

Set𝐶_𝑙∶={𝑦∈𝐶[0,∞)∶lim_𝑡→∞𝑦(𝑡)exists}.For𝑦∈𝐶_𝑙, define‖𝑦‖∶=sup_{𝑡∈[0,∞)}|𝑦(𝑡)|.Then𝐶_𝑙is a Banach space.

Lemma 11 (see [33]). Let𝑀⊂𝐶_𝑙.Then𝑀is relatively compact if the following conditions hold:(a)𝑀is bounded in𝐶_𝑙;(b)the functions belonging to𝑀are locally equicontinuous on[0,∞);(c)the functions from𝑀are equiconvergent; that is given𝜖>0, there corresponds𝑇(𝜖)>0such that |𝑥(𝑡)−𝑥(∞)|<𝜖, and𝑥∈𝑀.
By Lemma 11, similar to the proof of Theorem2.3in [34], we can easily have the following lemma.

Lemma 12. Let𝑀⊂𝐸.Then 𝑀 is relatively compact if the following conditions hold:(1)𝑀 is bounded in 𝐸;(2)the functions belonging to{𝑦∶𝑦=𝑥/(1+𝑡)^𝛼−1,𝑥∈𝑀},{𝑧∶𝑧=𝐷₀^𝛼−2₊𝑥(𝑡),𝑥∈𝑀}, and {𝑤∶𝑤=𝐷₀^𝛼−1₊𝑥(𝑡), 𝑥∈𝑀}, are locally equicontinuous on[0,∞);(3)the functions from {𝑦∶𝑦=𝑥/(1+𝑡)^𝛼−1,𝑥∈𝑀}, {𝑧∶𝑧=𝐷₀^𝛼−2₊𝑥(𝑡),𝑥∈𝑀}, and {𝑤∶𝑤=𝐷₀^𝛼−1₊𝑥(𝑡), 𝑥∈𝑀} are equiconvergent at∞.

Define the auxiliary functions 𝑤()=𝛾(𝑡)(𝑡),𝑥(𝑡)(𝑡)(𝑡),𝛽(𝑡)(𝑡),𝑤𝑡,𝑥,𝑥<𝛾,𝛾≤𝑥≤𝛽,𝑥>𝛽₁=𝐷𝑡,𝑦^0𝛼−2+𝛾(𝑡),𝑦<𝐷₀^𝛼−2₊𝛾(𝑡),𝑦(𝑡),𝐷₀^𝛼−2₊𝛾(𝑡)≤𝑦≤𝐷₀^𝛼−2₊𝛽(𝑡),𝐷₀^𝛼−2₊𝛽(𝑡),𝑦>𝐷₀^𝛼−2₊𝛽(𝑡). (31) Consider the following boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)+𝑎(𝑡)𝑓^∗𝑡,𝑢(𝑡),𝐷₀^𝛼−2₊𝑢(𝑡),𝐷₀^𝛼−1₊𝑢(𝑡()𝑢(0))=0,𝑡∈0,∞,2<𝛼≤3,=𝐷₀^𝛼−2₊𝑢(0)=0,lim_𝑡→∞𝐷₀^𝛼−1₊𝑢(𝑡)=0, (32) where 𝑓^∗𝑡,𝑥,𝑦,𝑧=𝑓𝑡,𝑤₀(𝑡,𝑥),𝑤₁+𝑤𝑡,𝑦,𝑧₁𝑡,𝑦−𝑦1+𝑤₁.𝑡,𝑦−𝑦 (33)

For each𝑢∈𝐸, it follows from (21), (23), and (33) that ^∞₀𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)≤𝑑𝑠_∞0𝑎(𝑠)𝜙(𝑠)ℎ𝐷₀^𝛼−1₊𝑢(𝑠≤)+1𝑑𝑠_∞0𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠≤^∞₀max{𝑠,1}𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠<∞, (34) where𝐻₀=max_{0≤𝑡≤‖𝐷0+}^𝛼−1_𝑢‖∞ℎ(𝑡). From (34) and Lemma 9, we know that u is a solution of (32) if and only if𝑢solves the operator equation 𝑇𝑢=𝑢, and 𝑇 is defined by (𝑡)=𝑇𝑢_∞0𝐺(𝑡,𝑠)𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠[).)𝑑𝑠,𝑢∈𝐸,𝑡∈0,∞ (35)

Lemma 13. Suppose that(𝐻₁)-(𝐻₂)hold; then 𝑇∶𝐸→𝐸 (the operator defined in (35)) is completely continuous.

Proof. Consider the following.
Step 1. 𝑇∶𝐸→𝐸 is well defined. For𝑢∈𝐸, it follows from (34) that 𝑠𝑎(^∞₀𝑠)𝐻₀𝜙(𝑠)≤+1𝑑𝑠_∞0max{𝑠,1}𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠<∞, (36) which implies lim_𝑡→∞𝑡𝑎(𝑡)𝐻₀𝜙(𝑡)+1=0. (37) We also have ^∞_𝑡𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠≤^∞_𝑡𝑠𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠,𝑡≥1. (38) Combining (36) with (38), one has lim_𝑡→∞^∞_𝑡𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠=0. (39) If we apply Lebesgue dominated convergence theorem with (37) and (39), then lim_𝑡→∞|𝑇𝑢(𝑡)|(1+𝑡)^𝛼−1≤lim_𝑡→∞^∞₀𝐺(𝑡,𝑠)(1+𝑡)^𝛼−1𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠=0, (40) which yields lim_𝑡→∞𝑇𝑢(𝑡)(1+𝑡)^𝛼−1=0; (41) that is, sup_0≤𝑡<∞|𝑇𝑢(𝑡)|(1+𝑡)^𝛼−1<∞. (42) By virtue of (34), we have sup_0≤𝑡<∞𝐷₀^𝛼−2+𝑇𝑢(𝑡)=sup_0≤𝑡<∞𝐷₀^𝛼−2+∞_𝑡𝐺(𝑡,𝑠)𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)𝑑𝑠=sup_0≤𝑡<∞^𝑡₀𝑠𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)+𝑑𝑠_∞𝑡𝑡𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)𝑑𝑠≤sup_0≤𝑡<∞^𝑡₀𝑠𝑎(𝑠)𝐻₀𝜙(𝑠)++1𝑑𝑠_∞𝑡𝑡𝑎(𝑠)𝐻₀𝜙(𝑠)≤+1𝑑𝑠_∞0𝑠𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠≤^∞₀max{𝑠,1}𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠<∞. (43) It follows from (39) that ^∞_𝑡𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)≤𝑑𝑠_∞𝑡𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠⟶0,𝑡⟶∞. (44) Therefore, we have lim_𝑡→∞𝐷₀^𝛼−1₊𝑇𝑢(𝑡)=lim_𝑡→∞^∞_𝑡𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)𝑑𝑠=0. (45) Thus, we conclude that𝑇𝑢∈𝐸.
Step 2. 𝑇∶𝐸→𝐸is continuous. For any convergent sequence 𝑢_𝑛→𝑢 in 𝐸, we find 𝑢_𝑛(𝑡)⟶𝑢(𝑡),𝐷₀^𝛼−2₊𝑢_𝑛⟶𝐷₀^𝛼−2₊𝐷𝑢,_0𝛼−1+𝑢_𝑛⟶𝐷₀^𝛼−1₊𝑢,as𝑛⟶∞,𝑡∈[0,∞). (46) From continuity of𝑓^∗, we obtain 𝑓^∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)[).⟶0,𝑛⟶∞,forany𝑡∈0,∞ (47) Since𝑢_𝑛→𝑢, we have sup_𝑛∈𝑁‖𝐷₀^𝛼−1₊𝑢‖_∞<∞.Set 𝐻_𝑝=max_{0≤𝑡≤max‖𝐷0+𝛼−1𝑢‖∞,sup𝑛∈𝑁‖𝐷0+𝛼−1𝑢‖∞}ℎ(𝑡), (48) which follows that 𝑠𝑎(^∞₀𝑠)𝑓^∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)𝑑𝑠≤2^∞₀𝑠𝑎(𝑠)𝐻_𝑝𝜙(𝑠)+1𝑑𝑠<∞. (49) Thus, applying the Lebesgue dominated convergence theorem and then using (49), we have 𝑇𝑢_𝑛−𝑇𝑢₁=sup_0≤𝑡<∞𝑇𝑢_𝑛(𝑡)−𝑇𝑢(𝑡)(1+𝑡)^𝛼−1=sup_0≤𝑡<∞^∞₀𝐺(𝑡,𝑠)(1+𝑡)^𝛼−1𝑎(𝑠)×𝑓^∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)≤1Γ(𝛼)^∞₀𝑎(𝑠)𝑓^∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)⟶0, (50) as𝑛→∞. On the other hand, we have 𝐷₀^𝛼−2₊𝑇𝑢_𝑛−𝐷₀^𝛼−2₊𝑇𝑢_∞=sup_0≤𝑡<∞𝐷₀^𝛼−2₊𝑇𝑢_𝑛(𝑡)−𝐷₀^𝛼−2₊𝑇𝑢(𝑡)=sup_0≤𝑡<∞^𝑡₀(𝑠)×𝑓𝑠𝑎_∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠+)𝑑𝑠_∞𝑡(𝑠)×𝑓𝑡𝑎_∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)𝑑𝑠≤sup_0≤𝑡<∞^𝑡₀(𝑠)×𝑓𝑠𝑎_∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)𝑑𝑠−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)+𝑑𝑠_∞𝑡(𝑠)×𝑓𝑠𝑎_∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠=)𝑑𝑠_∞0(𝑠)×𝑓𝑠𝑎_∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)𝑑𝑠⟶0,𝑛⟶∞. (51) From (49), it is clear that ^∞₀𝑎(𝑠)𝑓^∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)<+∞. (52) Moreover, by (52), we have 𝐷₀^𝛼−1₊𝑇𝑢_𝑛−𝐷₀^𝛼−1₊𝑇𝑢_∞=sup_0≤𝑡<∞𝐷₀^𝛼−1₊𝑇𝑢_𝑛(𝑡)−𝐷₀^𝛼−1₊𝑇𝑢(𝑡)=sup_0≤𝑡<∞^∞_𝑡𝑎(𝑠)𝑓^∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)≤^∞₀𝑎(𝑠)𝑓^∗𝑠,𝑢_𝑛(𝑠),𝐷₀^𝛼−2+𝑢_𝑛(𝑠),𝐷₀^𝛼−1+𝑢_𝑛(𝑠)−𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)⟶0,𝑛⟶∞. (53) Therefore, combining (50), (51), and (53), we get‖𝑇𝑢_𝑛−𝑇𝑢‖→0, as𝑛→∞; then we claim that𝑇∶𝐸→𝐸is continuous.
Step 3. 𝑇∶𝐸→𝐸 is compact. Let 𝐴 be any bounded subset of 𝐸; then, for𝑢∈𝐴, set 𝐻_𝑞=sup_{0≤𝑡≤‖𝐷0+𝛼−1‖∞,𝑢∈𝐴}ℎ(𝑡)<∞, (54) in a similar manner as (50), (51), and (53); we have by (21) and (23) that ‖𝑇𝑢‖₁=sup_0≤𝑡<∞|𝑇𝑢(𝑡)|(1+𝑡)^𝛼−1≤sup_0≤𝑡<∞^∞₀𝐺(𝑡,𝑠)(1+𝑡)^𝛼−1𝑎(𝑠)×𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)≤1𝑑𝑠Γ(𝛼)^∞₀𝑎(𝑠)×𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)𝐷𝑑𝑠<∞,_0𝛼−2+𝑇𝑢_∞≤^∞₀(𝑠)×𝑓𝑠𝑎_∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠≤)𝑑𝑠_∞0𝑠𝑎(𝑠)𝐻_𝑞𝜙(𝑠)𝐷+1𝑑𝑠<∞,_0𝛼−1+𝑇𝑢_∞≤^∞₀𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2+𝑢(𝑠),𝐷₀^𝛼−1+𝑢(𝑠)≤𝑑𝑠_∞0𝑎(𝑠)𝐻_𝑞𝜙(𝑠)+1𝑑𝑠<∞, (55) which implies that‖𝑇𝑢‖<∞. Hence,𝑇𝐴 is uniformly bounded. Meanwhile, for any 𝐵>0, for 𝑡₁,𝑡₂∈[0,𝐵], we have 𝑇𝑢𝑡₁1+𝑡₁^𝛼−1−𝑇𝑢𝑡₂1+𝑡₂^𝛼−1=^∞₀𝐺𝑡₁,𝑠1+𝑡₁^𝛼−1−𝐺𝑡₂,𝑠1+𝑡₂^𝛼−1×𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)≤^∞₀𝐺𝑡₁,𝑠1+𝑡₁^𝛼−1−𝐺𝑡₂,𝑠1+𝑡₂^𝛼−1×𝑎(𝑠)𝐻_𝑞𝜙(𝑠)+1𝑑𝑠⟶0,𝑡₁⟶𝑡₂. (56)
We also have 𝐷₀^𝛼−2₊𝑇𝑢𝑡₁−𝐷₀^𝛼−2₊𝑇𝑢𝑡₂=^𝑡1₀𝑠𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)+𝑑𝑠_∞𝑡1𝑡₁𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)−𝑑𝑠_𝑡20𝑠𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)−𝑑𝑠_∞𝑡2𝑡₂𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)≤𝑑𝑠_𝑡2𝑡1𝑠𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)+𝑡𝑑𝑠_1𝑡2𝑡1𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)+𝑡𝑑𝑠₂−𝑡₁^∞_𝑡2𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)𝑑𝑠≤^𝑡2_𝑡1𝑠𝑎(𝑠)𝐻_𝑞𝜙(𝑠+1)𝑑𝑠+𝑡₁^𝑡2_𝑡1𝑎(𝑠)𝐻_𝑞𝜙(+𝑡𝑠+1)𝑑𝑠₂−𝑡₁^∞_𝑡2𝑎(𝑠)𝐻_𝑞𝜙(𝑠+1)𝑑𝑠, (57) which approaches0, as𝑡₁→𝑡₂.Furthermore, we have 𝐷₀^𝛼−1₊𝑇𝑢𝑡₁−𝐷₀^𝛼−1₊𝑇𝑢𝑡₂=^𝑡2_𝑡1𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)≤𝑑𝑠_𝑡2𝑡1𝑎(𝑠)𝐻_𝑞𝜙(𝑠)+1𝑑𝑠, (58) which approaches0, as𝑡₁→𝑡₂.Hence, we get that 𝑇𝐴 is equicontinuous. From (41), we have 𝑇𝑢(𝑡)(1+𝑡)^𝛼−1−lim_𝑡→∞𝑇𝑢(𝑡)(1+𝑡)^𝛼−1=𝑇𝑢(𝑡)(1+𝑡)^𝛼−1⟶0,as𝑡⟶∞. (59) Since 𝑡𝑎(^∞_𝑡𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)≤^∞_𝑡𝑡𝑎(𝑠)𝐻₀𝜙(𝑠)≤+1𝑑𝑠_∞𝑡𝑠𝑎(𝑠)𝐻₀𝜙(𝑠)+1𝑑𝑠⟶0,𝑡⟶∞,lim_𝑡→∞𝐷₀^𝛼−2₊𝑇𝑢(𝑡)=lim_𝑡→∞^𝑡₀𝑠𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠+)𝑑𝑠_∞𝑡𝑡𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)=𝑑𝑠_∞0𝑠𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)𝑑𝑠. (60) Thus, we have 𝐷₀^𝛼−2₊𝑇𝑢(𝑡)−lim_𝑡→∞𝐷₀^𝛼−2₊(𝑡=𝑇𝑢)_∞𝑡𝑡𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)−𝑑𝑠_∞𝑡𝑠𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠)𝑑𝑠, (61) which approaches0, as𝑡→∞.We also have by (45) that 𝐷₀^𝛼−1₊𝑇𝑢(𝑡)−lim_𝑡→∞𝐷₀^𝛼−1₊(𝑡=𝐷𝑇𝑢)_0𝛼−1+(𝑡=𝑇𝑢)_∞𝑡𝑎(𝑠)𝑓^∗𝑠,𝑢(𝑠),𝐷₀^𝛼−1₊𝑢(𝑠),𝐷₀^𝛼−2₊𝑢(𝑠)𝑑𝑠, (62) which approaches0as𝑡→∞.Thus,𝑇𝐴is equiconvergent at∞.Then𝑇𝐴is relatively compact. Therefore,𝑇∶𝐸→𝐸is completely continuous. The proof is complete.

3. Main Result

We are in the position to state the main existence result.

Theorem 14. Let𝛾,𝛽∈𝐸⋂𝐿_loc(0,∞)be lower and upper solutions to (5), and suppose that (15) holds. Moreover,(𝐻₁)-(𝐻₂)hold. Then fractional boundary value problem (5) has at least one solution𝑢(𝑡)∈𝐸⋂𝐿_loc(0,∞)satisfying 𝛾(𝑡)(𝑡)(𝑡),𝐷≤𝑢≤𝛽_0𝛼−2+𝛾(𝑡)≤𝐷₀^𝛼−2₊𝑢(𝑡)≤𝐷₀^𝛼−2₊𝛽(𝑡),𝐷₀^𝛼−1₊𝑢(𝑡)≤𝑁,∀𝑡∈[0,∞), (63) where𝑁 is a constant dependent only on𝛾,𝛽, and𝜙.

Proof. By Lemma 13, we know that 𝑇∶𝐸→𝐸 is completely continuous. By the Schauder fixed point theorem, we can easily obtain that𝑇has at least one fixed point 𝑢∈𝐸. Thus, 𝑢 is a solution of (32). Next, we will show that 𝑢 satisfies the inequalities 𝛾(𝑡)(𝑡)(𝑡),𝐷≤𝑢≤𝛽_0𝛼−2+𝛾(𝑡)≤𝐷₀^𝛼−2₊𝑢(𝑡)≤𝐷₀^𝛼−2₊𝛽(𝑡),𝑡∈[0,∞), (64) which implies that 𝑢 is a solution of (5). First of all, we will show that 𝐷₀^𝛼−2₊𝑢(𝑡)≤𝐷₀^𝛼−2₊𝛽(𝑡), for all 𝑡∈[0,∞).
If not, then sup_0≤𝑡<∞𝐷₀^𝛼−2₊𝑢(𝑡)−𝐷₀^𝛼−2₊𝛽(𝑡)>0. (65) Note that lim_𝑡→∞(𝐷₀^𝛼−1₊𝑢(𝑡)−𝐷₀^𝛼−1₊𝛽(𝑡))<0; then there are two cases.
Case 1. There exists a𝑡₀∈[0,∞)such that 𝐷₀^𝛼−2₊𝑢𝑡₀−𝐷₀^𝛼−2₊𝛽𝑡₀=sup_{[)𝑡∈0,∞}𝐷₀^𝛼−2₊𝑢(𝑡)−𝐷₀^𝛼−2₊𝛽(𝑡)>0. (66) Then𝐷₀^𝛼−1₊𝑢(𝑡₀)=𝐷₀^𝛼−1₊𝛽(𝑡₀)and 𝐷^𝛼₀₊𝑢𝑡₀≤𝐷^𝛼₀₊𝛽𝑡₀. (67) On the other hand, in view of (32), (33), and(𝐻₂), we have 𝐷^𝛼₀₊𝑢𝑡₀=−𝑎𝑡₀𝑓𝑡₀,𝑤₀𝑡₀,𝑢,𝑤₁𝑡₀,𝐷₀^𝛼−2₊𝑢,𝐷₀^𝛼−1₊𝑢+𝑤₁𝑡₀,𝐷₀^𝛼−2₊𝑢−𝐷₀^𝛼−2₊𝑢𝑡₀1+𝑤₁𝑡₀,𝐷₀^𝛼−2₊𝑢−𝐷₀^𝛼−2₊𝑢𝑡₀=−𝑎𝑡₀𝑓𝑡₀,𝑤₀𝑡₀,𝑤,𝑢₁𝑡₀,𝐷₀^𝛼−2₊𝛽𝑡₀,𝐷₀^𝛼−1₊𝛽𝑡₀+𝐷₀^𝛼−2₊𝛽𝑡₀−𝐷₀^𝛼−2₊𝑢𝑡₀1+𝐷₀^𝛼−2₊𝛽𝑡₀−𝐷₀^𝛼−2₊𝑢𝑡₀≥−𝑎𝑡₀𝑓𝑡₀,𝛽𝑡₀,𝐷₀^𝛼−2₊𝛽𝑡₀,𝐷₀^𝛼−1₊𝛽𝑡₀+𝑎𝑡₀𝐷₀^𝛼−2₊𝑢𝑡₀−𝐷₀^𝛼−2₊𝛽𝑡₀1+𝐷₀^𝛼−2₊𝑢𝑡₀−𝐷₀^𝛼−2₊𝛽𝑡₀≥−𝑎𝑡₀𝑓𝑡₀,𝛽𝑡₀,𝐷₀^𝛼−2₊𝛽𝑡₀,𝐷₀^𝛼−1₊𝛽𝑡₀≥𝐷₀^𝛼−1₊𝛽𝑡₀, (68) which contradicts (67).
Case 2. Consider 𝐷₀^𝛼−2₊𝑢(0)−𝐷₀^𝛼−2₊𝛽(0)=lim_𝑡→0⁺(𝐷₀^𝛼−2₊𝑢(𝑡)−𝐷₀^𝛼−2₊𝛽(𝑡))=sup_{𝑡∈[0,∞)}(𝐷₀^𝛼−2₊𝑢(𝑡)−𝐷₀^𝛼−2₊𝛽(𝑡))>0. By the boundary condition, we have the contradiction 𝐷₀^𝛼−2₊𝑢(0)−𝐷₀^𝛼−2₊𝛽(0)≤0. Consequently, 𝐷₀^𝛼−2₊𝑢(𝑡)≤𝐷₀^𝛼−2₊𝛽(𝑡) holds for all 𝑡∈[0,∞). By using the similar method, we also can prove that 𝐷₀^𝛼−2₊𝛾(𝑡)≤𝐷₀^𝛼−2₊𝑢(𝑡) for all 𝑡∈[0,∞). By Remark 7, we have 𝛾(𝑡)≤𝑢(𝑡)≤𝛽(𝑡),forany𝑡∈[0,∞). (69) If𝜎>0, we choose𝑟≥maxsup_{[)𝑡∈𝜎,∞}𝐷₀^𝛼−2₊𝛽(𝑡)−𝐷₀^𝛼−2₊𝛾(0)𝑡,sup_{[)𝑡∈𝜎,∞}𝐷₀^𝛼−2₊𝛽(0)−𝐷₀^𝛼−2₊𝛾(𝑡)𝑡, (70) and 𝑁>𝑟 such that ^𝑁_𝑟𝑠ℎ(𝑠)𝑑𝑠≥𝑚sup_0≤𝑡<∞𝐷₀^𝛼−2₊𝛽(𝑡)−inf_0≤𝑡<∞𝐷₀^𝛼−2₊𝛾(𝑡), (71) where 𝑚=sup_{𝑡∈[0,∞)}𝑎(𝑡)𝜙(𝑡). From (23), it is clear that ∫^∞₀𝑎(𝑠)𝜙(𝑠)𝑑𝑠<∞. Hence,𝑚<∞.
Finally, we will check that|𝐷₀^𝛼−1₊𝑢(𝑡)|<𝑁for𝑡∈[0,∞).If|𝐷₀^𝛼−1₊𝑢(𝑡)|≤𝑟, for every 𝑡∈[0,∞), then |𝐷₀^𝛼−1₊𝑢(𝑡)|<𝑁. If𝐷₀^𝛼−1₊𝑢(𝑡)>𝑟, for all 𝑡∈[0,∞), then for any 𝑅≥𝜎; using (70), we obtain 𝐷₀^𝛼−2₊𝛽(𝑅)−𝐷₀^𝛼−2₊𝛾(0)𝑅≥𝐷₀^𝛼−2₊𝑢(𝑅)−𝐷₀^𝛼−2₊𝑢(0)𝑅=∫^𝑅₀(𝑑/𝑑𝑠)𝐷₀^𝛼−2₊𝑢(𝑠)𝑑𝑠𝑅=∫^𝑅₀𝐷₀^𝛼−1₊𝑢(𝑠)𝑑𝑠𝑅𝐷>𝑟≥^0𝛼−2+𝛽(𝑅)−𝐷₀^𝛼−2₊𝛾(0)𝑅, (72) which is a contradiction. If 𝐷₀^𝛼−1₊𝑢(𝑡)<−𝑟, for every 𝑡∈[0,∞), we can also have a similar contradiction. There exists 𝑡₀∈[0,∞) such that |𝐷₀^𝛼−2₊𝑢(𝑡₀)|≤𝑟. Thus, there exists [𝑡₁,𝑡₂]⊂[0,∞) such that |𝐷₀^𝛼−1₊𝑢(𝑡₁)|=𝑟,  |𝐷₀^𝛼−1₊𝑢(𝑡)|>𝑟,  𝑡∈(𝑡₁,𝑡₂]. Otherwise, |𝐷₀^𝛼−1₊𝑢(𝑡₂)|=𝑟, |𝐷₀^𝛼−1₊𝑢(𝑡)|>𝑟, 𝑡∈[𝑡₁,𝑡₂). Without loss of generality, we assume that 𝐷₀^𝛼−1₊𝑢(𝑡₁)=𝑟, 𝐷₀^𝛼−1₊𝑢(𝑡)>𝑟, 𝑡∈(𝑡₁,𝑡₂]. Therefore, by a convenient change of variable and using (21) and (71), we get ^{𝐷0+𝛼−1𝑢(𝑡2)}_{𝐷0+𝛼−1𝑢𝑡1}𝑠ℎ(𝑠)=𝑑𝑠_𝑡2𝑡1𝐷₀^𝛼−1₊𝑢(𝑡)ℎ𝐷₀^𝛼−1₊𝑢(𝑡𝐷)^𝛼0+𝑢(𝑡)=𝑑𝑡_𝑡2𝑡1−𝑎(𝑡)𝑓𝑡,𝑢(𝑡),𝐷₀^𝛼−2₊𝑢(𝑡),𝐷₀^𝛼−1₊𝑢(𝑡×𝐷_0𝛼−1+)𝑢(𝑡)ℎ𝐷₀^𝛼−1₊𝑢(𝑡)≤𝑑𝑡_𝑡2𝑡1𝑎(𝑡)𝜙(𝑡)𝐷₀^𝛼−1₊𝑢(𝑡)𝑑𝑡≤𝑚^𝑡2_𝑡1𝐷₀^𝛼−1₊𝑢(𝑡)𝐷𝑑𝑡=𝑚₀^𝛼−2₊𝑢𝑡₂−𝐷₀^𝛼−2₊𝑢𝑡₁≤𝑚sup_{[)𝑡∈0,∞}𝐷₀^𝛼−2₊𝛽(𝑡)−inf_{[)𝑡∈0,∞}𝐷₀^𝛼−2₊𝛾(𝑡)≤^𝑁_𝑟𝑠ℎ(𝑠)𝑑𝑠, (73) which implies that 𝐷₀^𝛼−1₊𝑢(𝑡₂)≤𝑁, since 𝑡₂ can be arbitrarily as long as 𝐷₀^𝛼−1₊𝑢(𝑡)>𝑟; we have 𝐷₀^𝛼−1₊𝑢(𝑡)>𝑟, for any 𝑡∈[0,∞), which follows that 𝐷₀^𝛼−1₊𝑢(𝑡)≤𝑁. By a similar analysis, we can also obtain that if 𝐷₀^𝛼−1₊𝑢(𝑡₁)=−𝑟,  𝐷₀^𝛼−1₊𝑢(𝑡)<−𝑟,  𝑡∈(𝑡₁,𝑡₂], then 𝐷₀^𝛼−1₊𝑢(𝑡)>−𝑁,  𝑡∈[0,∞). Therefore, 𝐷^𝛼₀₊𝑢(𝑡)=−𝑎(𝑡)𝑓^∗𝑡,𝑢(𝑡),𝐷₀^𝛼−2₊𝑢(𝑡),𝐷₀^𝛼−1₊𝑢(𝑡(𝑡)𝑓(𝑡))=−𝑎𝑡,𝑢,𝐷₀^𝛼−2₊𝑢(𝑡),𝐷₀^𝛼−1₊𝑢(𝑡); (74) that is, 𝑢is a solution of (5), which completes the proof.

Example 15. Consider the boundary value problem of the fractional differential equation on the half-line 𝐷₀^8/3₊𝑢(𝑡)+𝑒^−𝑡𝑡²+𝑢²(𝑡)Γ53−𝐷₀^2/3₊𝑢(𝑡)×1+arctan𝐷₀^5/3₊𝑢(𝑡)²(),𝑢(0)=0,𝑡∈0,∞=𝐷₀^2/3₊𝑢(0)=0,𝐷₀^5/3₊𝑢(∞)=0. (75) In this case,𝛼=8/3,𝑎(𝑡)=𝑒^−𝑡, 𝑓(𝑡,𝑥,𝑦,𝑧)=(𝑡²+𝑥²)(Γ(5/3)−𝑦)(1+arctan(𝑧²)). Obviously, 𝛾(𝑡)=−𝑡^2/3 and 𝛽(𝑡)=𝑡^2/3 are a pair of lower and upper solutions of (75). Furthermore, we have 𝛾,𝛽∈𝐸, 𝛾(𝑡)≤𝛽(𝑡), for 𝑡∈[0,∞). It is clear that 𝑓 is continuous on [0,∞)×ℝ³. If 0≤𝑡<∞, −𝑡^2/3≤𝑥≤𝑡^2/3, −Γ(5/3)≤𝑦≤Γ(5/3), we have 𝑓𝑡,𝑥,𝑦,𝑧≤𝜙(𝑡)ℎ(|𝑧|), (76) where𝜙(𝑡)=4Γ(5/3)(1+𝑡²) and ℎ(𝑧)=1+𝑧²; we have ^∞₀𝑠ℎ(𝑠)𝑑𝑠=^∞₀𝑠1+𝑠²𝑑𝑠=∞, (77) and𝑓satisfies the like Nagumo condition with respect to −𝑡^2/3,𝑡^2/3. Furthermore, we have max{^∞₀𝑠,1}𝑎(𝑠)𝑑𝑠=¹₀𝑒^−𝑠𝑑𝑠+^∞₁𝑠𝑒^−𝑠𝑑𝑠<∞,_∞0{}𝑎(𝑠)𝜙(𝑠)5max𝑠,1𝑑𝑠=4Γ3¹₀1+𝑠²𝑒^−𝑠5𝑑𝑠+4Γ3^∞₁𝑠1+𝑠²𝑒^−𝑠𝑑𝑠<∞. (78) Thus, we conclude by Theorem 14 that there exists at least one solution 𝑢(𝑡) to boundary value problem (75) such that −𝑡^2/3≤𝑢(𝑡)≤𝑡^2/3,5−Γ3≤𝐷₀^𝛼−2₊𝑢(𝑡)5≤Γ3,𝑡∈[0,∞). (79)

Conflict of Interests

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

Acknowledgments

The authors thank the referees for their valuable and helpful suggestions and comments that improved the paper. This work is supported by the Natural Science Foundation of Jiangsu Province (BK2011407) and the Natural Science Foundation of China (11271364 and 10771212).