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

Authors: Dandan Yang; Chuanzhi Bai
Journal: Journal of Applied Mathematics
 (2013)

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## 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.


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## Body

## 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:
(1)x′′(t)+f(t,x(t),x′(t))=0,t∈(0,+∞),x(0)=αx(η),limt→∞x′(t)=0,
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:
(2)D0+αu(t)=f(t,u(t)),t∈[0,∞),u(0)=u′(0)=u′′(0)=0,D0+α-1u(t)=limt→∞D0+α-1u(t),
where3<α<4 and D0+α 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:
(3)D0+αx(t)=f(t,x(t)),t∈(0,∞),0<α≤1,limt→∞t1-αx(t)=u0,
where D0+α 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:
(4)D0+αu(t)+a(t)f(t,u(t),D0+α-1u(t))=0,t∈[0,∞),u(0)=0,limt→∞D0+α-1u(t)=u∞,
where1<α≤2, f∈C([0,∞)×ℝ2,ℝ), D0+α-1 and D0+α 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:(5)D0+αu(t)+a(t)f(t,u(t),D0+α-2u(t),D0+α-1u(t))=0,t∈(0,∞),2<α≤3,u(0)=D0+α-2u(0)=0,limt→∞D0+α-1u(t)=0,
wherea(t):[0,∞)→[0,∞), and D0+α 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(6)I0+αf(t)=1Γ(α)∫0t‍f(s)(t-s)1-αds,t>0,
whereα>0, is called the Riemann-Liouville fractional integral of order α and Γ(α) is the Euler gamma function defined by Γ(α)=∫0∞‍tα-1e-tdt, α>0.Definition 2 (see [17]).
A functionf(x) given in the interval [0,∞), the expression
(7)D0+αf(t)=1Γ(n-α)(ddt)n∫0t‍f(s)(t-s)α-n+1ds,
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
(8)D0+αu(t)=0
has
(9)u(t)=c1tα-1+c2tα-2+⋯+cntα-n,ci∈R,i=0,1,2,3,…,n,n=[α]+1
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
(10)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cntα-n,
for someci∈R,i=0,1,2,3,…,n, n=[α]+1.Denote byC[0,∞)the space of all continuous real functions defined on[0,∞)and Lloc(0,∞) the space of all real functions which are Lebesgue integrable on every bounded subinterval of (0,∞). Denote
(11)E={x∈C[0,∞):sup0≤t<∞|x(t)|(1+t)α-1<∞,sup0≤t<∞|D0+α-2x(t)|<∞,limt→∞D0+α-1x(t)existsx∈C[0,∞):sup0≤t<∞|x(t)|(1+t)α-1<∞}
with the norm ∥x∥=max{∥x∥1,∥D0+α-2x∥∞,∥D0+α-1x∥∞}, where
(12)∥x∥1=supt∈[0,∞)|x(t)(1+t)α-1|,∥D0+α-2x∥∞=supt∈[0,∞)|D0+α-2x(t)|,∥D0+α-1x∥∞=supt∈[0,∞)|D0+α-1x(t)|.
By standard arguments, it is easy to prove that(E,∥·∥)is a Banach space.Proposition 5 (see [32]).
Assume thatfis inC(0,∞)⋂‍Lloc(0,∞)with a fractional derivative of order 0<α<1 that belongs to C[0,∞)⋂‍Lloc[0,∞). Then
(13)IαDαf(x)=f(x)+cxα-1,
for somec∈ℝ. When the function f is in C[0,∞), then c=0.Definition 6.
A functionγ∈E⋂‍Lloc(0,∞)is called a lower solution of (5) if
(14)D0+αγ(t)+a(t)f(t,γ(t),D0+α-2γ(t),D0+α-1γ(t))≥0,t∈(0,∞),γ(0)≤0,D0+α-2γ(0)≤0,limt→∞D0+α-1γ(t)≤0.
Similarly, we can define an upper solutionβ∈E⋂‍Lloc(0,∞)of (5) by reversing the above inequalities.Remark 7.
If(15)D0+α-2γ(t)≤D0+α-2β(t),t∈[0,∞),
andγ(0)≤0, β(0)≥0, then we haveγ(t)≤β(t)for allt∈[0,∞).We consider the following two cases.Case 1.
Considerα=3. The inequality (15) reduces toγ′(t)≤β′(t). Then
(16)∫0t‍γ′(s)ds≤∫0t‍β′(s)ds,t∈(0,∞);
that is,
(17)γ(t)-γ(0)≤β(t)-β(0),t∈(0,∞).
From the boundary conditionγ(0)≤0andβ(0)≥0 we obtainγ(0)≤β(0), and
(18)γ(t)≤β(t)+γ(0)-β(0)≤β(t),t∈(0,∞).
Thus,γ(t)≤β(t)fort∈[0,∞).Case 2.
Consider2<α<3. If (15) holds, then
(19)I0+α-2D0+α-2γ(t)≤I0+α-2D0+α-2β(t),t∈[0,∞).
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)=tα-2,β(t)=2tα-2. Obviously,D0+α-2γ(t)=Γ(α+1), D0+α-2β(t)=2Γ(α+1). Thusγ(0)=β(0)=0and
(20)D0+α-2γ(t)≤D0+α-2β(t),t∈[0,∞).
We also getγ(t)≤β(t)for eacht∈[0,∞).Definition 8.
Letγ,β∈E⋂‍Lloc(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
(21)|f(t,x,y,z)|≤ϕ(t)h(|z|),
for all0≤t<∞, γ(t)≤x≤β(t), D0+α-2γ(t)≤y≤D0+α-2β(t),z∈ℝ, and
(22)∫0∞‍sh(s)ds=∞.
We list some assumptions related toa(t),ϕ(t), andf as follows.(H1) Consider
(23)∫0∞‍max{s,1}a(s)ds<∞,∫0∞‍max{s,1}a(s)ϕ(s)ds<∞.(H2) Letf:[0,∞)×ℝ3→ℝ be a continuous function satisfying the like Nagumo condition with respect to the pair of functionsγ, β, such that
(24)f(t,γ(t),y,z)≤f(t,x,y,z)≤f(t,β(t),y,z),for(t,x,y,z)∈[0,∞)×[γ(t),β(t)]×ℝ2.Lemma 9.
Letσ∈C(0,∞)⋂‍L(0,∞), then the fractional boundary value problem
(25)D0+αu(t)+σ(t)=0,t∈(0,∞),2<α≤3,u(0)=D0+α-2u(0)=0,limt→∞D0+α-1u(t)=0
has a unique solutionu(t)=∫0∞‍G(t,s)σ(s)ds, where
(26)G(t,s)=1Γ(α){tα-1-(t-s)α-1,0≤s≤t<∞,tα-1,0≤t≤s<∞.Proof.
By Lemma4, the solution of (25) has the following form:
(27)u(t)=-1Γ(α)∫0t‍(t-s)α-1σ(s)ds+c1tα-1+c2tα-2+c3tα-3.
By boundary value conditionu(0)=0, one hasc3=0. Thus, we have by (27) that
(28)D0+α-2u(t)=-D0+α-2I0+ασ(t)+c1D0+α-2tα-1+c2D0+α-2tα-2=-∫0t‍(t-s)σ(s)ds+c1Γ(α)t+c2Γ(α-1).
Substituting the conditionD0+α-2u(0)=0into (28), we obtain c2=0. Together with limt→∞D0+α-1u(t)=0, we get
(29)c1=1Γ(α)∫0∞‍σ(s)ds.
Hence, we have
(30)u(t)=-1Γ(α)∫0t‍(t-s)α-1σ(s)ds+tα-1Γ(α)∫0∞‍σ(s)ds=-1Γ(α)∫0t‍(t-s)α-1σ(s)ds+1Γ(α)∫0t‍tα-1σ(s)ds+1Γ(α)∫t∞‍tα-1σ(s)ds=1Γ(α)∫0t‍[tα-1-(t-s)α-1]σ(s)ds+1Γ(α)∫t∞‍tα-1σ(s)ds=∫0∞‍G(t,s)σ(s)ds.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)≤tα-1/Γ(α), for (t,s)∈[0,∞)×[0,∞).The proof is easy, so we omit it here.SetCl:={y∈C[0,∞):limt→∞y(t)exists}.Fory∈Cl, define∥y∥:=supt∈[0,∞)|y(t)|.ThenClis a Banach space.Lemma 11 (see [33]).
LetM⊂Cl.ThenMis relatively compact if the following conditions hold:(a)
Mis bounded inCl;(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 Lemma11, 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-2x(t),x∈M}, and {w:w=D0+α-1x(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-2x(t),x∈M}, and {w:w=D0+α-1x(t), x∈M} are equiconvergent at∞.Define the auxiliary functions(31)w(t,x)={γ(t),x<γ(t),x(t),γ(t)≤x≤β(t),β(t),x>β(t),w1(t,y)={D0+α-2γ(t),y<D0+α-2γ(t),y(t),D0+α-2γ(t)≤y≤D0+α-2β(t),D0+α-2β(t),y>D0+α-2β(t).
Consider the following boundary value problem:
(32)D0+αu(t)+a(t)f*(t,u(t),D0+α-2u(t),D0+α-1u(t))=0,t∈(0,∞),2<α≤3,u(0)=D0+α-2u(0)=0,limt→∞D0+α-1u(t)=0,
where
(33)f*(t,x,y,z)=f(t,w0(t,x),w1(t,y),z)+w1(t,y)-y1+|w1(t,y)-y|.For eachu∈E, it follows from (21), (23), and (33) that
(34)|∫0∞‍a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|≤∫0∞‍a(s)[ϕ(s)h(|D0+α-1u(s)|)+1]ds≤∫0∞‍a(s)(H0ϕ(s)+1)ds≤∫0∞‍max{s,1}a(s)(H0ϕ(s)+1)ds<∞,
whereH0=max0≤t≤∥D0+α-1u∥∞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
(35)Tu(t)=∫0∞‍G(t,s)a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds,u∈E,t∈[0,∞).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
(36)∫0∞‍sa(s)(H0ϕ(s)+1)ds≤∫0∞‍max{s,1}a(s)(H0ϕ(s)+1)ds<∞,
which implies
(37)limt→∞ta(t)(H0ϕ(t)+1)=0.
We also have
(38)∫t∞‍a(s)(H0ϕ(s)+1)ds≤∫t∞‍sa(s)(H0ϕ(s)+1)ds,t≥1.
Combining (36) with (38), one has
(39)limt→∞∫t∞‍a(s)(H0ϕ(s)+1)ds=0.
If we apply Lebesgue dominated convergence theorem with (37) and (39), then
(40)limt→∞|Tu(t)|(1+t)α-1≤limt→∞∫0∞‍G(t,s)(1+t)α-1a(s)(H0ϕ(s)+1)ds=0,
which yields
(41)limt→∞Tu(t)(1+t)α-1=0;
that is,
(42)sup0≤t<∞|Tu(t)|(1+t)α-1<∞.
By virtue of (34), we have
(43)sup0≤t<∞|D0+α-2Tu(t)|=sup0≤t<∞|D0+α-2∫t∞‍G(t,s)a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))dsD0+α-2∫t∞‍G(t,s)a(s)f*|=sup0≤t<∞|∫0tsa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds+∫t∞‍ta(s)f*(D0+α-1u(s)s,u(s),D0+α-2u(s),D0+α-1u(s))ds|∫0t‍sa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|≤sup0≤t<∞∫0t‍sa(s)(H0ϕ(s)+1)ds+∫t∞‍ta(s)(H0ϕ(s)+1)ds≤∫0∞‍sa(s)(H0ϕ(s)+1)ds≤∫0∞‍max{s,1}a(s)(H0ϕ(s)+1)ds<∞.
It follows from (39) that
(44)|∫t∞‍a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|≤∫t∞‍a(s)(H0ϕ(s)+1)ds⟶0,t⟶∞.
Therefore, we have
(45)limt→∞D0+α-1Tu(t)=limt→∞∫t∞‍a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds=0.
Thus, we conclude thatTu∈E.
Step
2.
T
:
E
→
Eis continuous. For any convergent sequence un→u in E, we find
(46)un(t)⟶u(t),D0+α-2un⟶D0+α-2u,D0+α-1un⟶D0+α-1u,asn⟶∞,t∈[0,∞).
From continuity off*, we obtain
(47)|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|⟶0,n⟶∞,for  anyt∈[0,∞).
Sinceun→u, we have supn∈N∥D0+α-1u∥∞<∞.Set
(48)Hp=max0≤t≤max{∥D0+α-1u∥∞,supn∈N∥D0+α-1u∥∞}h(t),
which follows that
(49)∫0∞‍sa(s)|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds≤2∫0∞‍sa(s)(Hpϕ(s)+1)ds<∞.
Thus, applying the Lebesgue dominated convergence theorem and then using (49), we have
(50)∥Tun-Tu∥1=sup0≤t<∞|Tun(t)-Tu(t)|(1+t)α-1=sup0≤t<∞|∫0∞‍G(t,s)(1+t)α-1a(s)×((s,u(s),D0+α-2u(s),D0+α-1u(s))f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s)))∫0∞‍G(t,s)(1+t)α-1a(s)|≤1Γ(α)∫0∞‍a(s)|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|⟶0,
asn→∞. On the other hand, we have
(51)∥D0+α-2Tun-D0+α-2Tu∥∞=sup0≤t<∞|D0+α-2Tun(t)-D0+α-2Tu(t)|=sup0≤t<∞|∫0t‍sa(s)×(f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s)))ds+∫t∞‍ta(s)×(f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s)))ds∫0t‍sa(s)|≤sup0≤t<∞[∫0t‍sa(s)×|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))ds-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds+∫t∞‍sa(s)×|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds∫0t‍sa(s)]=∫0∞‍sa(s)×|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds⟶0,n⟶∞.
From (49), it is clear that
(52)∫0∞‍a(s)|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|<+∞.
Moreover, by (52), we have
(53)∥D0+α-1Tun-D0+α-1Tu∥∞=sup0≤t<∞|D0+α-1Tun(t)-D0+α-1Tu(t)|=sup0≤t<∞|∫t∞‍a(s)(f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s)))|∫t∞‍a(s)(f*(s,un(s),D0+α-2un(s),D0+α-1un(s))|≤∫0∞‍a(s)|f*(s,un(s),D0+α-2un(s),D0+α-1un(s))-f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|⟶0,n⟶∞.
Therefore, combining (50), (51), and (53), we get∥Tun-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
(54)Hq=sup0≤t≤∥D0+α-1∥∞,u∈Ah(t)<∞,
in a similar manner as (50), (51), and (53); we have by (21) and (23) that
(55)∥Tu∥1=sup0≤t<∞|Tu(t)|(1+t)α-1≤sup0≤t<∞∫0∞‍G(t,s)(1+t)α-1a(s)×|f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds≤1Γ(α)∫0∞‍a(s)×|f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds<∞,∥D0+α-2Tu∥∞≤∫0∞‍sa(s)×|f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds≤∫0∞‍sa(s)(Hqϕ(s)+1)ds<∞,∥D0+α-1Tu∥∞≤∫0∞‍a(s)|f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|ds≤∫0∞‍a(s)(Hqϕ(s)+1)ds<∞,
which implies that∥Tu∥<∞. Hence,TA is uniformly bounded. Meanwhile, for any B>0, for t1,t2∈[0,B], we have
(56)|Tu(t1)(1+t1)α-1-Tu(t2)(1+t2)α-1|=|∫0∞‍(G(t1,s)(1+t1)α-1-G(t2,s)(1+t2)α-1)×a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))∫0∞‍(G(t1,s)(1+t1)α-1-G(t2,s)(1+t2)α-1)|≤∫0∞‍|G(t1,s)(1+t1)α-1-G(t2,s)(1+t2)α-1|×a(s)(Hqϕ(s)+1)ds⟶0,t1⟶t2.
We also have(57)|D0+α-2Tu(t1)-D0+α-2Tu(t2)|=|∫0t1‍sa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds+∫t1∞‍t1a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds-∫0t2‍sa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds-∫t2∞‍t2a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|≤|∫t1t2‍sa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|+|t1∫t1t2‍a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|+|(t2-t1)∫t2∞‍a(s)f*(D0+α-1u(s)s,u(s),D0+α-2u(s),D0+α-1u(s))ds(t2-t1)∫t2∞‍a(s)|≤∫t1t2‍sa(s)(Hqϕ(s+1))ds+t1∫t1t2‍a(s)(Hqϕ(s+1))ds+|t2-t1|∫t2∞‍a(s)(Hqϕ(s+1))ds,
which approaches0, ast1→t2.Furthermore, we have
(58)|D0+α-1Tu(t1)-D0+α-1Tu(t2)|=|∫t1t2‍a(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|≤∫t1t2‍a(s)(Hqϕ(s)+1)ds,
which approaches0, ast1→t2.Hence, we get that TA is equicontinuous. From (41), we have
(59)|Tu(t)(1+t)α-1-limt→∞Tu(t)(1+t)α-1|=|Tu(t)(1+t)α-1|⟶0,ast⟶∞.
Since
(60)|∫t∞‍ta(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))|≤∫t∞‍ta(s)(H0ϕ(s)+1)ds≤∫t∞‍sa(s)(H0ϕ(s)+1)ds⟶0,t⟶∞,limt→∞D0+α-2Tu(t)=limt→∞[∫0tsa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds+∫t∞‍ta(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds∫0tsa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds]=∫0∞‍sa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds.
Thus, we have
(61)|D0+α-2Tu(t)-limt→∞D0+α-2Tu(t)|=|∫t∞‍ta(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds-∫t∞‍sa(s)f*(s,u(s),D0+α-2u(s),D0+α-1u(s))ds|,
which approaches0, ast→∞.We also have by (45) that
(62)|D0+α-1Tu(t)-limt→∞D0+α-1Tu(t)|=|D0+α-1Tu(t)|=∫t∞‍a(s)f*(s,u(s),D0+α-1u(s),D0+α-2u(s))ds,
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⋂‍Lloc(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⋂‍Lloc(0,∞)satisfying
(63)γ(t)≤u(t)≤β(t),D0+α-2γ(t)≤D0+α-2u(t)≤D0+α-2β(t),|D0+α-1u(t)|≤N,∀t∈[0,∞),
whereN is a constant dependent only onγ,β, andϕ.Proof.
By Lemma13, 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
(64)γ(t)≤u(t)≤β(t),D0+α-2γ(t)≤D0+α-2u(t)≤D0+α-2β(t),t∈[0,∞),
which implies that u is a solution of (5). First of all, we will show that D0+α-2u(t)≤D0+α-2β(t), for all t∈[0,∞).
If not, then(65)sup0≤t<∞(D0+α-2u(t)-D0+α-2β(t))>0.
Note that limt→∞(D0+α-1u(t)-D0+α-1β(t))<0; then there are two cases.
Case
1. There exists at0∈[0,∞)such that
(66)D0+α-2u(t0)-D0+α-2β(t0)=supt∈[0,∞)(D0+α-2u(t)-D0+α-2β(t))>0.
ThenD0+α-1u(t0)=D0+α-1β(t0)and
(67)D0+αu(t0)≤D0+αβ(t0).
On the other hand, in view of (32), (33), and(H2), we have
(68)D0+αu(t0)=-a(t0)[+w1(t0,D0+α-2u)-D0+α-2u(t0)1+|w1(t0,D0+α-2u)-D0+α-2u(t0)|]f(t0,w0(t0,u),w1(t0,D0+α-2u),D0+α-1u)+w1(t0,D0+α-2u)-D0+α-2u(t0)1+|w1(t0,D0+α-2u)-D0+α-2u(t0)|]=-a(t0)[D0+α-2β(t0)-D0+α-2u(t0)1+|D0+α-2β(t0)-D0+α-2u(t0)|f(w1(t0,D0+α-2β(t0))D0+α-1β(t0)t0,w0(t0,u),w1(t0,D0+α-2β(t0)),D0+α-1β(t0))+D0+α-2β(t0)-D0+α-2u(t0)1+|D0+α-2β(t0)-D0+α-2u(t0)|]≥-a(t0)f(t0,β(t0),D0+α-2β(t0),D0+α-1β(t0))+a(t0)D0+α-2u(t0)-D0+α-2β(t0)1+|D0+α-2u(t0)-D0+α-2β(t0)|≥-a(t0)f(t0,β(t0),D0+α-2β(t0),D0+α-1β(t0))≥D0+α-1β(t0),
which contradicts (67).
Case
2. Consider D0+α-2u(0)-D0+α-2β(0)=limt→0+(D0+α-2u(t)-D0+α-2β(t))=supt∈[0,∞)(D0+α-2u(t)−D0+α-2β(t))>0. By the boundary condition, we have the contradiction D0+α-2u(0)−D0+α-2β(0)≤0. Consequently, D0+α-2u(t)≤D0+α-2β(t) holds for all t∈[0,∞). By using the similar method, we also can prove that D0+α-2γ(t)≤D0+α-2u(t) for all t∈[0,∞). By Remark 7, we have
(69)γ(t)≤u(t)≤β(t),for  anyt∈[0,∞).
Ifσ>0, we choose(70)r≥max{supt∈[σ,∞)D0+α-2β(t)-D0+α-2γ(0)t,supt∈[σ,∞)D0+α-2β(0)-D0+α-2γ(t)t},
and N>r such that
(71)∫rN‍sh(s)ds≥m(sup0≤t<∞D0+α-2β(t)-inf0≤t<∞D0+α-2γ(t)),
where m=supt∈[0,∞)a(t)ϕ(t). From (23), it is clear that ∫0∞‍a(s)ϕ(s)ds<∞. Hence,m<∞.
Finally, we will check that|D0+α-1u(t)|<Nfort∈[0,∞).If|D0+α-1u(t)|≤r, for every t∈[0,∞), then |D0+α-1u(t)|<N. IfD0+α-1u(t)>r, for all t∈[0,∞), then for any R≥σ; using (70), we obtain
(72)D0+α-2β(R)-D0+α-2γ(0)R≥D0+α-2u(R)-D0+α-2u(0)R=∫0R‍(d/ds)(D0+α-2u(s))dsR=∫0R‍D0+α-1u(s)dsR>r≥D0+α-2β(R)-D0+α-2γ(0)R,
which is a contradiction. If D0+α-1u(t)<-r, for every t∈[0,∞), we can also have a similar contradiction. There exists t0∈[0,∞) such that |D0+α-2u(t0)|≤r. Thus, there exists [t1,t2]⊂[0,∞) such that |D0+α-1u(t1)|=r,  |D0+α-1u(t)|>r,  t∈(t1,t2]. Otherwise, |D0+α-1u(t2)|=r, |D0+α-1u(t)|>r, t∈[t1,t2). Without loss of generality, we assume that D0+α-1u(t1)=r, D0+α-1u(t)>r, t∈(t1,t2]. Therefore, by a convenient change of variable and using (21) and (71), we get
(73)∫D0+α-1u(t1)D0+α-1u(t2)‍sh(s)ds=∫t1t2‍D0+α-1u(t)h(D0+α-1u(t))D0+αu(t)dt=∫t1t2‍-a(t)f(t,u(t),D0+α-2u(t),D0+α-1u(t))×D0+α-1u(t)h(D0+α-1u(t))dt≤∫t1t2‍a(t)ϕ(t)D0+α-1u(t)dt≤m∫t1t2‍D0+α-1u(t)dt=m(D0+α-2u(t2)-D0+α-2u(t1))≤m(supt∈[0,∞)D0+α-2β(t)-inft∈[0,∞)D0+α-2γ(t))≤∫rN‍sh(s)ds,
which implies that D0+α-1u(t2)≤N, since t2 can be arbitrarily as long as D0+α-1u(t)>r; we have D0+α-1u(t)>r, for any t∈[0,∞), which follows that D0+α-1u(t)≤N. By a similar analysis, we can also obtain that if D0+α-1u(t1)=-r,  D0+α-1u(t)<-r,  t∈(t1,t2], then D0+α-1u(t)>-N,  t∈[0,∞). Therefore,
(74)D0+αu(t)=-a(t)f*(t,u(t),D0+α-2u(t),D0+α-1u(t))=-a(t)f(t,u(t),D0+α-2u(t),D0+α-1u(t));
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(75)D0+8/3u(t)+e-t(t2+u2(t))(Γ(53)-D0+2/3u(t))×(1+arctan(D0+5/3u(t))2)=0,t∈(0,∞),u(0)=D0+2/3u(0)=0,D0+5/3u(∞)=0.
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)=-t2/3 and β(t)=t2/3 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<∞, -t2/3≤x≤t2/3, -Γ(5/3)≤y≤Γ(5/3), we have
(76)|f(t,x,y,z)|≤ϕ(t)h(|z|),
whereϕ(t)=4Γ(5/3)(1+t2) and h(z)=1+z2; we have
(77)∫0∞‍sh(s)ds=∫0∞‍s1+s2ds=∞,
andfsatisfies the like Nagumo condition with respect to -t2/3,t2/3. Furthermore, we have
(78)∫0∞‍max{s,1}a(s)ds=∫01‍e-sds+∫1∞‍se-sds<∞,∫0∞‍max{s,1}a(s)ϕ(s)ds=4Γ(53)∫01‍(1+s2)e-sds+4Γ(53)∫1∞‍s(1+s2)e-sds<∞.
Thus, we conclude by Theorem 14 that there exists at least one solution u(t) to boundary value problem (75) such that
(79)-t2/3≤u(t)≤t2/3,-Γ(53)≤D0+α-2u(t)≤Γ(53),t∈[0,∞).



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*Source: 349025-2013-12-10.xml*