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Discrete Dynamics in Nature and Society
Volume 2012 (2012), Article ID 357697, 14 pages
http://dx.doi.org/10.1155/2012/357697
Research Article

Existence of Unbounded Solutions for a Third-Order Boundary Value Problem on Infinite Intervals

School of Sciences, China University of Geosciences, Beijing 100083, China

Received 20 June 2012; Accepted 31 July 2012

Academic Editor: Zengji Du

Copyright © 2012 Hairong Lian and Junfang Zhao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We generalize the unbounded upper and lower solution method to a third-order ordinary differential equation on the half line subject to the Sturm-Liouville boundary conditions. By using such techniques and the Schäuder fixed point theorem, some criteria are presented for the existence of solutions and positive ones to the problem discussed.

1. Introduction

Boundary value problems on infinite intervals, arising from the study of radially symmetric solutions of nonlinear elliptic equation [1], have received much attention in recent years. Because the infinite interval is noncompact, the discussion about BVPs on the half-line is more complicated. There have been many existence results for some boundary value problems of differential equations on the half line. The main methods are the extension of continuous solutions on the corresponding finite intervals under a diagonalization process, fixed point theorems in special Banach space or in special Fréchet space; see [112] and the references therein.

The method of upper and lower solutions is a powerful technique to deal with the existence of boundary value problems (BVPs). In many cases, when given one pair of well-ordered lower and upper solution, nonlinear BVPs always have at least one solution in the closed interval. To obtain this kind of result, we can employ topological degree theory, the monotone iterative technique, or critical theory. For details, we refer the reader to see [14, 7, 9, 1214] and therein.

When the method of upper and lower solution is applied to the infinite interval problems, diagonalization process is always used; see [2, 3, 7]. For example, in [3], Agarwal and O’Regan discussed a Sturm-Liouville boundary value problem of second-order differential equation: 1𝑝(𝑡)𝑝(𝑡)𝑦(𝑡)=𝑞(𝑡)𝑓(𝑡,𝑦(𝑡)),𝑡(0,+),𝑎0𝑦(0)+𝑏0lim𝑡0+𝑝(𝑡)𝑦(𝑡)=𝑐0,orlim𝑡0+𝑝(𝑡)𝑦[(𝑡)=0,𝑦(𝑡)boundedon0,+),orlim𝑡+𝑦(𝑡)=0,(1.1) where 𝑎0>0,𝑏00. General existence criteria were obtained to guarantee the existence of bounded solutions. The methods used therein were based on a diagonalization arguments and existence results of appropriate boundary value problems on finite intervals.

In [12], Yan et al. investigated the boundary value problem 𝑦(𝑡)+Φ(𝑡)𝑓𝑡,𝑦(𝑡),𝑦(𝑡)=0,𝑡(0,+),𝑎𝑦(0)𝑏𝑦(0)=𝑦00,lim𝑡+𝑦(𝑡)=𝑘>0,(1.2) where 𝑎>0,𝑏>0. By using the upper and lower solutions method and the fixed point theorem, the authors presented sufficient conditions for the existence of unbounded positive solutions. In [9], Lian and the coauthors discussed further the existence of the unbounded solutions.

There are many results of third-order boundary value problems on finite interval; see [14, 15] and the references therein. However, there has been few papers concerned with the upper and lower solutions technique for the boundary value problems of third-order differential equation on infinite intervals. In this paper, we aim to investigate a general Sturm-Liouville boundary value problem for third-order differential equation on the half line 𝑢(𝑡)+𝜙(𝑡)𝑓𝑡,𝑢(𝑡),𝑢(𝑡),𝑢(𝑡)=0,𝑡(0,+),𝑢(0)=𝐴,𝑢(0)𝑎𝑢(0)=𝐵,𝑢(+)=𝐶,(1.3) where 𝜙(0,+)(0,+), 𝑓[0,+)×3 are continuous, 𝑎>0, 𝐴,𝐵,𝐶. The methods mainly depend on the unbounded upper and lower solutions method and topological degree theory. The nonlinear is admitted to involve in the high-order derivatives under the considerations of the Nagumo condition. The solutions obtained can be unbounded in this paper. The results obtained in this paper generalize those in [4].

2. Preliminaries

We present here some definitions and lemmas which are essential in the proof of the main results.

Definition 2.1. A function 𝛼𝐶2[0,+)𝐶3(0,+) is called a lower solution of BVP (1.3) if 𝛼(𝑡)+𝜙(𝑡)𝑓𝑡,𝛼(𝑡),𝛼(𝑡),𝛼(𝑡)0,𝑡(0,+),𝛼(0)𝐴,𝛼(0)𝑎𝛼(0)𝐵,𝛼(+)<𝐶.(2.1) Similarly, a function 𝛽𝐶2[0,+)𝐶3(0,+) is called an upper solution of BVP (1.3) if 𝛽(𝑡)+𝜙(𝑡)𝑓𝑡,𝛽(𝑡),𝛽(𝑡),𝛽(𝑡)0,𝑡(0,+),𝛽(0)𝐴,𝛽(0)𝑎𝛽(0)𝐵,𝛽(+)>𝐶.(2.2)

Definition 2.2. Given a positive function 𝜙𝐶(0,+) and a pair of functions 𝛼,𝛽𝐶1[0,+) satisfying 𝛼(0)𝛽(0) and 𝛼(𝑡)𝛽(𝑡),𝑡[0,+); a function 𝑓[0,+)×𝑅3𝑅 is said to satisfy the Nagumo condition with respect to the pair of functions 𝛼,𝛽, if there exist positive functions 𝜓,𝐶[0,+) satisfying 0+𝜓(𝑠)𝜙(𝑠)𝑑𝑠<+,+𝑠/(𝑠)𝑑𝑠=+ such that ||||𝑓(𝑡,𝑥,𝑦,𝑧)𝜓(𝑡)(|𝑧|)(2.3) holds for all 0𝑡<+,𝛼(𝑡)𝑥𝛽(𝑡),𝛼(𝑡)𝑦𝛽(𝑡), and 𝑧.
Let 𝑣0(𝑡)=1+𝑡2,𝑣1(𝑡)=1+𝑡,𝑣2(𝑡)=1 and consider the space 𝑋 defined by 𝑋=𝑥𝐶2[0,+),lim𝑡+𝑥(𝑖)(𝑡)𝑣𝑖(,𝑡)exist,𝑖=0,1,2(2.4) with the norm 𝑥=max{𝑥0,𝑥1,𝑥2}, where 𝑥𝑖=sup𝑡[0,+)|𝑥(𝑡)/𝑣𝑖(𝑡)|. By the standard arguments, we can prove that (𝑋,) is a Banach space.

Lemma 2.3. If 𝑒𝐿1[0,+), then the BVP of third-order linear differential equation 𝑢(𝑡)+𝑒(𝑡)=0,𝑡(0,+),𝑢(0)=𝐴,𝑢(0)𝑎𝑢(0)=𝐵,𝑢(+)=𝐶,(2.5) has a unique solution in 𝑋. Moreover this solution can be expressed as 𝑢(𝑡)=𝑝(𝑡)+0+𝐺(𝑡,𝑠)𝑒(𝑠)𝑑𝑠,(2.6) where 𝐶𝑝(𝑡)=𝐴+(𝑎𝐶+𝐵)𝑡+2𝑡2,1𝐺(𝑡,𝑠)=𝑎𝑡+𝑠𝑡2𝑠21,0𝑠𝑡<+,2𝑡2+𝑎𝑡,0𝑡𝑠<+.(2.7)

Proof. It is easy to verify that (2.6) satisfies BVP (2.5). Now we show the uniqueness. Suppose 𝑢 is a solution of (2.5). Let 𝑣=𝑢, then we have 𝑣(𝑡)+𝑒(𝑡)=0,𝑡(0,+),𝑣(0)𝑎𝑣(0)=𝐵,𝑣(+)=𝐶.(2.8) By a direct calculation, we obtain the general solution of the above equation: 𝑣(𝑡)=𝑐1+𝑐2𝑡+𝑡0𝜏+𝑒(𝑠)𝑑𝑠𝑑𝜏=𝑐1+𝑐2𝑡+𝑡0𝑠𝑒(𝑠)𝑑𝑠+𝑡𝑡+𝑒(𝑠)𝑑𝑠.(2.9) Substituting this to the boundary condition, we arrive at 𝑐1=𝑎𝐶+𝐵+𝑎0+𝑒𝑐(𝑠)𝑑𝑠,2=𝐶.(2.10) Therefore, (2.8) has a unique solution 𝑣(𝑡)=𝑎𝐶+𝐵+𝐶𝑡+0+𝑔(𝑡,𝑠)𝑒(𝑠)𝑑𝑠,(2.11) where 𝑔(𝑡,𝑠)=𝑎+𝑠,0𝑠𝑡<+,𝑎+𝑡,0𝑡𝑠<+.(2.12) Furthermore, 𝑢=𝑣,𝑢(0)=𝐴, so 𝑢(𝑡)=𝑢(0)+𝑡0𝑣(𝜏)𝑑𝜏=𝐴+𝑡0𝑎𝐶+𝐵+𝐶𝜏+0+𝑔(𝜏,𝑠)𝑒(𝑠)𝑑𝑠𝑑𝜏=𝑝(𝑡)+0+𝑡0𝑔(𝜏,𝑠)𝑑𝜏𝑒(𝑠)𝑑𝑠=𝑝(𝑡)+0+𝐺(𝑡,𝑠)𝑒(𝑠)𝑑𝑠.(2.13) The proof is complete.

Theorem 2.4 (see [1]). Let 𝑀𝐶={𝑥𝐶[0,+),lim𝑡+𝑥(𝑡)exists}. Then 𝑀 is relatively compact if the following conditions hold:(a)all functions from 𝑀 are uniformly bounded;(b)all functions from 𝑀 are equicontinuous on any compact interval of [0,+);(c)all functions from 𝑀 are equiconvergent at infinity; that is, for any given 𝜖>0, there exists a 𝑇=𝑇(𝜖)>0 such that |𝑓(𝑡)𝑓(+)|<𝜖, for all 𝑡>𝑇 and 𝑓𝑀.
From the above results, we can obtain the following general criteria for the relative compactness of subsets in 𝐶[0,+).

Theorem 2.5. Given 𝑛+1 continuous functions 𝜌𝑖 satisfying 𝜌𝑖𝜀>0,𝑖=0,1,,𝑛 with 𝜀 a positive constant. Let 𝑀𝐶𝑛={𝑥𝐶𝑛[0,+),lim𝑡+𝜌𝑖(𝑡)𝑥(𝑖)(𝑡)exists,𝑖=0,1,2,,𝑛}. Then 𝑀 is relatively compact if the following conditions hold:(a)all functions from 𝑀 are uniformly bounded;(b)the functions from {𝑦𝑖𝑦𝑖=𝜌𝑖𝑥(𝑖),𝑥𝑀} are equicontinuous on any compact interval of [0,+), 𝑖=0,1,2,,𝑛;(c)the functions from {𝑦𝑖𝑦𝑖=𝜌𝑖𝑥(𝑖),𝑥𝑀} are equiconvergent at infinity, 𝑖=0,1,2,,𝑛.

Proof. Set 𝑀𝑖={𝑦𝑖𝑦𝑖=𝜌𝑖𝑥(𝑖),𝑥(𝑖)𝑀}, then 𝑀𝑖𝐶,𝑖=0,1,2,𝑛. From conditions (a)–(c), we have 𝑀𝑖 is relatively compact in 𝐶. Therefore, for any sequence {𝑦𝑖,𝑚}𝑚=1𝑀𝑖, it has a convergent subsequence. Without loss of generality, we denote it this sequence. Then there exists 𝑦𝑖,0𝑀𝑖 such that 𝑦𝑖,𝑚=𝜌𝑖𝑥𝑚(𝑖)𝑦𝑖,0,𝑚+,𝑖=0,1,2,,𝑛.(2.14)
Set 𝑥𝑖,0=(1/𝜌𝑖)𝑦𝑖,0, then 𝑥𝑚(𝑖)𝑥𝑖,0,𝑖=0,1,2,,𝑛. Noticing that all functions from 𝑀 are uniformly continuous, we can obtain that 𝑥𝑖,0=𝑥(𝑖)0,0,𝑖=1,2,,𝑛. So 𝑀 is relatively compact.

3. Main Results

In this section, we present the existence criteria for the existence of solutions and positive solutions of BVP (1.3). We first cite conditions (H1) and (H2) here.(H1):(1) BVP (1.3) has a pair of upper and lower solutions 𝛽,𝛼 in 𝑋 with 𝛼(𝑡)𝛽(𝑡),𝑡[0,+);(2)𝑓𝐶([0,+)×3,) satisfies the Nagumo condition with respect to 𝛼 and 𝛽.(H2): For any 0𝑡<+,𝛼(𝑡)𝑦𝛽(𝑡) and 𝑧, it holds 𝑓(𝑡,𝛼(𝑡),𝑦,𝑧)𝑓(𝑡,𝑥,𝑦,𝑧)𝑓(𝑡,𝛽(𝑡),𝑦,𝑧),as𝛼(𝑡)𝑥𝛽(𝑡).(3.1)

Lemma 3.1. Suppose condition (H1) holds. And suppose further that the following condition holds:(H3) there exists a constant 𝛾>1 such that sup0𝑡<+(1+𝑡)𝛾𝜙(𝑡)𝜓(𝑡)<+.
If 𝑢 is a solution of (1.3) satisfying 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡),𝛼(𝑡)𝑢(𝑡)𝛽[(𝑡),𝑡0,+),(3.2) then there exists a constant 𝑅>0 (without relations to 𝑢) such that 𝑢2𝑅.

Proof. Let 𝛿>0 and 𝑅>𝐶, 𝜂maxsup𝑡[𝛿,+)𝛽(𝑡)𝛼(0)𝑡,sup𝑡[𝛿,+)𝛽(0)𝛼(𝑡)𝑡(3.3) such that 𝑅𝜂𝑠(𝑠)𝑑𝑠𝑀sup𝑡[0,+)𝛽(𝑡)(1+𝑡)𝛾inf𝑡[0,+)𝛼(𝑡)(1+𝑡)𝛾+𝛾𝛾1sup𝑡[0,+)𝛽(𝑡)1+𝑡,(3.4) where 𝐶 is the nonhomogeneous boundary value, 𝑀=sup0𝑡<+(1+𝑡)𝛾𝜙(𝑡)𝜓(𝑡), then |𝑢(𝑡)|𝑅,𝑡[0,+). If it is untrue, we have the following three cases.

Case 1. Consider ||𝑢||[(𝑡)>𝜂,𝑡0,+).(3.5) Without loss of generality, we suppose 𝑢(𝑡)>𝜂,𝑡[0,+). While for any 𝑇𝛿, 𝛽(𝑇)𝛼(0)𝑇𝑢(𝑇)𝑢(0)𝑇=1𝑇𝑇0𝑢𝛽(𝑠)𝑑𝑠>𝜂(𝑇)𝛼(0)𝑇,(3.6) which is a contraction. So there must exist 𝑡0[0,+) such that |𝑢(𝑡0)|𝜂.

Case 2. Consider ||𝑢||[(𝑡)𝜂,𝑡0,+).(3.7) Just take 𝑅=𝜂 and we can complete the proof.

Case 3. There exists [𝑡1,𝑡2][0,+) such that |𝑢(𝑡1)|=𝜂,|𝑢(𝑡)|>𝜂,𝑡(𝑡1,𝑡2] or |𝑢(𝑡2)|=𝜂,|𝑢(𝑡)|>𝜂,𝑡[𝑡1,𝑡2).
Suppose that 𝑢(𝑡1)=𝜂,𝑢(𝑡)>𝜂,𝑡(𝑡1,𝑡2]. Obviously, 𝑢(𝑡2)𝑢(𝑡1)𝑠(𝑠)𝑑𝑠=𝑡2𝑡1𝑢(𝑠)(𝑢(𝑢𝑠))=(𝑠)𝑑𝑠𝑡2𝑡1𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢𝑢(𝑠)(𝑠)(𝑢(𝑠))𝑑𝑠𝑡2𝑡1𝑢(𝑠)𝜙(𝑠)𝜓(𝑠)𝑑𝑠𝑀𝑡2𝑡1𝑢(𝑠)(1+𝑠)𝛾𝑑𝑠=𝑀𝑡2𝑡1𝑢(𝑠)(1+𝑠)𝛾𝑑𝑠+𝑡2𝑡1𝛾𝑢(𝑠)(1+𝑠)1+𝛾𝑑𝑠𝑀sup𝑡[0,+)𝛽(𝑡)(1+𝑡)𝛾inf𝑡[0,+)𝛼(𝑡)(1+𝑡)𝛾+sup𝑡[0,+)𝛽(𝑡)1+𝑡0+𝛾(1+𝑠)𝛾𝑑𝑠𝑅𝜂𝑠(𝑠)𝑑𝑠(3.8) concludes that 𝑢(𝑡2)𝑅. For 𝑡1 and 𝑡2 are arbitrary, we have 𝑢(𝑡)max{𝑅,𝜂}=𝑅,𝑡[0,+).
Similarly if 𝑢(𝑡1)=𝜂,𝑢(𝑡)<𝜂,𝑡(𝑡1,𝑡2], we can also obtain that 𝑢(𝑡)>𝑅,𝑡[0,+).
Thus there exists 𝑅>0, just related with 𝛼,𝛽, and , such that 𝑢2𝑅.

Remark 3.2. Condition (H3) is necessary for an a priori estimation of 𝑢 in Lemma 3.1. Because the upper and lower solutions are in 𝑋, 𝛽(𝑡) and 𝛼(𝑡) are at most linearly increasing, especially at infinity. Otherwise, sup𝑡[0,+)𝛼(𝑡) and sup𝑡[0,+)𝛽(𝑡) may be equal to infinity.

Theorem 3.3. Suppose 𝜙𝐿1[0,+) and the conditions (H1)–(H3) hold. Then BVP (1.3) has at least one solution 𝑢𝐶2[0,+)𝐶3(0,+) such that 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡),𝛼(𝑡)𝑢(𝑡)𝛽[(𝑡),𝑡0,+).(3.9)

Proof. Let 𝑅>0 be the same definition in Lemma 3.1 and consider the boundary value problem 𝑢(𝑡)+𝜙(𝑡)𝑓𝑡,𝑢(𝑡),𝑢(𝑡),𝑢(𝑡)=0,𝑡(0,+),𝑢(0)=𝐴,𝑢(0)𝑎𝑢(0)=𝐵,𝑢(+)=𝐶,(3.10) where 𝑓(𝐹𝑡,𝑥,𝑦,𝑧)=𝑅𝑡,𝑥,𝛼+(𝑡),𝑧𝑦𝛼(𝑡)||1+𝑦𝛼(||𝑡),𝑦<𝛼𝐹(𝑡),𝑅(𝑡,𝑥,𝑦,𝑧),𝛼(𝑡)𝑦𝛽𝐹(𝑡),𝑅𝑡,𝑥,𝛽(𝑡),𝑧𝑦𝛽(𝑡)||1+𝑦𝛽||(𝑡),𝑦>𝛽𝐹(𝑡),𝑅𝑓(𝑡,𝑥,𝑦,𝑧)=𝑅𝑓(𝑡,𝛼(𝑡),𝑦,𝑧),𝑥<𝛼(𝑡),𝑅𝑓(𝑡,𝑥,𝑦,𝑧),𝛼(𝑡)𝑥𝛽(𝑡),𝑅𝑓(𝑡,𝛽(𝑡),𝑦,𝑧),𝑥>𝛽(𝑡),𝑅(𝑡,𝑥,𝑦,𝑧)=𝑓(𝑡,𝑥,𝑦,𝑅),𝑧<𝑅,𝑓(𝑡,𝑥,𝑦,𝑧),𝑅𝑧𝑅,𝑓(𝑡,𝑥,𝑦,𝑅),𝑧>𝑅.(3.11) Firstly we prove that BVP (3.10) has at least one solution 𝑢. To this end, define the operator 𝑇𝑋𝑋 by (𝑇𝑢)(𝑡)=𝑝(𝑡)+0+𝐺(𝑡,𝑠)𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(𝑠)𝑑𝑠.(3.12) By Lemma 2.3, we can see that the fixed points of 𝑇 coincide with the solutions of BVP (3.10). So it is enough to prove that 𝑇 has a fixed point.
We claim that 𝑇𝑋𝑋 is completely continuous.
(1)𝑇𝑋𝑋 is well defined. For any 𝑢𝑋,𝑢<+ and it holds 0+𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(𝑠)𝑑𝑠0+𝜙𝐻(𝑠)0𝜓(𝑠)+1𝑑𝑠<+,(3.13) where 𝐻0=max0𝑠𝑢(𝑠). By the Lebesgue-dominated convergence theorem, we have lim𝑡+(𝑇𝑢)(𝑡)𝑣0(𝑡)=lim𝑡+𝑙(𝑡)𝑣0+(𝑡)0+𝐺(𝑡,𝑠)𝑣0𝜙(𝑡)(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢=𝐶(𝑠)𝑑𝑠2+120+𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(𝑠)𝑑𝑠<+,lim𝑡+(𝑇𝑢)(𝑡)𝑣1(𝑡)=lim𝑡+(𝑎𝐶+𝐵)+𝐶𝑡𝑣1+(𝑡)0+𝑔(𝑡,𝑠)𝑣1(𝑡)𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(𝑠)𝑑𝑠=𝐶+0+𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(𝑠)𝑑𝑠<+,lim𝑡+(𝑇𝑢)(𝑡)=lim𝑡+𝐶+𝑡+𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(𝑠)𝑑𝑠=𝐶<+,(3.14) so 𝑇𝑢𝑋.
(2)𝑇𝑋𝑋 is continuous. For any convergent sequence 𝑢𝑛𝑢 in 𝑋, there exists 𝑟1>0 such that sup𝑛𝑁𝑢𝑛𝑟1. Similarly, we have 𝑇𝑢𝑛𝑇𝑢=max𝑇𝑢𝑛𝑇𝑢0,𝑇𝑢𝑛(𝑇𝑢)1,𝑇𝑢𝑛(𝑇𝑢)20+maxsup0𝑡<+||||𝐺(𝑡,𝑠)𝑣0||||(𝑡),sup0𝑡<+||||𝑔(𝑡,𝑠)𝑣1||||||𝑓(𝑡),1𝜙(𝑠)𝑠,𝑢𝑛(𝑠),𝑢𝑛(𝑠),𝑢𝑛(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢||(𝑠)𝑑𝑠0+||𝑓𝜙(𝑠)𝑠,𝑢𝑛(𝑠),𝑢𝑛(𝑠),𝑢𝑛(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢||(𝑠)𝑑𝑠0,as𝑛+,(3.15) so 𝑇𝑋𝑋 is continuous.
(3)𝑇𝑋𝑋 is compact. Let 𝐵 be any bounded subset of 𝑋, then there exists 𝑟>0 such that 𝑢𝑟,forall𝑢𝐵. For any 𝑢𝐵, one has 𝑇𝑢=max𝑇𝑢0,(𝑇𝑢)1,(𝑇𝑢)20+||𝑓𝜙(𝑠)𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(||𝑠)𝑑𝑠0+𝐻𝜙(𝑠)𝑟𝜓(𝑠)+1𝑑𝑠<+,(3.16) where 𝐻𝑟=max0𝑠𝑟(𝑠), so 𝑇𝐵 is uniformly bounded. Meanwhile, for any 𝑇>0, if 𝑡1,𝑡2[0,𝑇], we have ||||𝑡𝑇𝑢1𝑣0𝑡1𝑡𝑇𝑢2𝑣0𝑡2||||=|||||0+𝐺𝑡1,𝑠𝑣0𝑡1𝐺𝑡2,𝑠𝑣0𝑡2𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢|||||(𝑠)𝑑𝑠0+||||𝐺𝑡1,𝑠𝑣0𝑡1𝐺𝑡2,𝑠𝑣0𝑡2||||𝐻𝜙(𝑠)𝑟𝜓(𝑠)+1𝑑𝑠0,as𝑡1𝑡2,||||(𝑇𝑢)𝑡1𝑣1𝑡1(𝑇𝑢)𝑡2𝑣1𝑡1||||=|||||0+𝑔𝑡1,𝑠𝑣1𝑡1𝑔𝑡2,𝑠𝑣1𝑡2𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢|||||(𝑠)𝑑𝑠0+||||𝑔𝑡1,𝑠𝑣1𝑡1𝑔𝑡2,𝑠𝑣1𝑡2||||𝐻𝜙(𝑠)𝑟𝜓(𝑠)+1𝑑𝑠0,as𝑡1𝑡2,||||(𝑇𝑢)𝑡1𝑣2𝑡1(𝑇𝑢)𝑡2𝑣2𝑡1||||=||||𝑡2𝑡1𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢||||(𝑠)𝑑𝑠𝑡2𝑡1𝐻𝜙(𝑠)𝑟𝜓(𝑠)+1𝑑𝑠0,as𝑡1𝑡2;(3.17) that is, 𝑇𝐵 is equicontinuous. From Theorem 2.5, we can see that if 𝑇𝐵 is equiconvergent at infinity, then 𝑇𝐵 is relatively compact. In fact, ||||𝑇𝑢(𝑡)𝑣0(𝑡)lim𝑡+𝑇𝑢(𝑡)𝑣0||||=||||(𝑡)𝑙(𝑡)𝑣0𝐶(𝑡)2+0+𝐺(𝑡,𝑠)𝑣01(𝑡)2𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢(||||||||𝑠)𝑑𝑠𝑙(𝑡)𝑣0𝐶(𝑡)2||||+0+||||𝐺(𝑡,𝑠)𝑣01(𝑡)2||||𝜙𝐻(𝑠)𝑟𝜓||||(𝑠)+1𝑑𝑠0,as𝑡+,(𝑇𝑢)(𝑡)𝑣1(𝑡)lim𝑡+(𝑇𝑢)(𝑡)𝑣1||||=||||(𝑡)(𝑎𝐶+𝐵)+𝐶𝑡𝑣1(𝑡)𝐶+0+𝑔(𝑡,𝑠)𝑣1(𝑡)1𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢||||||||(𝑠)𝑑𝑠(𝑎𝐶+𝐵)+𝐶𝑡𝑣1||||+(𝑡)𝐶0+||||𝑔(𝑡,𝑠)𝑣1||||𝐻(𝑡)1𝜙(𝑠)𝑟||||𝜓(𝑠)+1𝑑𝑠0,as𝑡+,(𝑇𝑢)(𝑡)𝑣2||||=||||(𝑡)𝑡+𝜙(𝑠)𝑓𝑠,𝑢(𝑠),𝑢(𝑠),𝑢||||(𝑠)𝑑𝑠𝑡+𝜙𝐻(𝑠)𝑟𝜓(𝑠)+1𝑑𝑠0,as𝑡+.(3.18) Then we can obtain that 𝑇𝑋𝑋 is completely continuous.
By the Schäuder fixed point theorem, 𝑇 has at least one fixed point 𝑢𝑋. Next we will prove 𝑢 satisfying 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡),𝑡[0,+). If 𝑢(𝑡)𝛽(𝑡),𝑡[0,+) does not hold, then, sup0𝑡<+𝑢(𝑡)𝛽(𝑡)>0.(3.19) Because 𝑢(+)𝛽(+)<0, so there are two cases.

Case 1. Consider lim𝑡0+𝑢(𝑡)𝛽(𝑡)=sup0𝑡<+𝑢(𝑡)𝛽(𝑡)>0.(3.20) Easily, 𝑢(0+)𝛽(0+)0. While by the boundary condition, we have 𝑎𝑢(0)𝛽(0)𝑢(0)𝛽(0)>0,(3.21) which is a contraction.

Case 2. There exists 𝑡(0,+) such that 𝑢𝑡𝛽𝑡=sup0𝑡<+𝑢(𝑡)𝛽(𝑡)>0.(3.22) So, 𝑢(𝑡)𝛽(𝑡)=0,𝑢(𝑡)𝛽(𝑡)0. Unfortunately, 𝑢𝑡𝛽𝑡𝑡𝜙𝑓𝑡𝑡,𝛽,𝛽𝑡,𝛽𝑡𝑓𝑡𝑡,𝑢,𝑢𝑡,𝑢𝑡𝑡=𝜙𝑓𝑡𝑡,𝛽,𝛽𝑡,𝛽𝑡𝑓𝑡𝑡,𝑢,𝛽𝑡,𝛽𝑡𝑡+𝜙𝑢𝑡𝛽𝑡||𝑢1+(𝑡)𝛽(𝑡)||𝑡𝜙𝑢𝑡𝛽𝑡||𝑢1+(𝑡)𝛽(𝑡)||>0.(3.23)
Consequently, 𝑢(𝑡)𝛽(𝑡) holds for all 𝑡[0,+). Similarly, we can show that 𝛼(𝑡)𝑢(𝑡) for all 𝑡[0,+). Noticing that 𝛼(0)𝐴𝛽(0), from the inequality 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡), we can obtain that 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡). Lemma 3.1 guarantee that 𝑢𝑅. So, 𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢(𝑡),𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢(𝑡),𝑢(𝑡);(3.24) that is, 𝑢 is a solution of BVP (1.3).

Remark 3.4. For finite interval problem, it is sharp to define the lower and upper solutions satisfying 𝛼(𝑏)𝐶 and 𝛽(𝑏)𝐶; see [15].
If 𝑓[0,+)4[0,+), we can establish a criteria for the existence of positive solutions.

Theorem 3.5. Let 𝑓[0,+)4[0,+) be continuous and 𝜙𝐿1[0,+). Suppose the condition (H2) holds and the following conditions hold.(P1) BVP (1.3) has a pair of positive upper and lower solutions 𝛼,𝛽𝑋 satisfying 𝛼(𝑡)𝛽[(𝑡),𝑡0,+).(3.25)(P2) For any 𝑟>0, there exists 𝜑𝑟 satisfying 0+𝜙(𝑠)𝜑𝑟(𝑠)𝑑𝑠<+ such that 𝑓(𝑡,𝑥,𝑦,𝑧)𝜑𝑟(𝑡)(3.26) holds for all 𝑡[0,+),𝛼(𝑡)𝑥𝛽(𝑡),𝛼(𝑡)𝑦𝛽(𝑡), and 0𝑧𝑟.
Then BVP (1.3) with 𝐴,𝐵,𝐶0 has at least one solution such that 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡),𝛼(𝑡)𝑢(𝑡)𝛽[(𝑡),𝑡0,+).(3.27)

Proof. Choose 𝑅=(1/𝑎)(𝐵+𝛽(0)) and consider the boundary value problem (3.10) except 𝑓𝑅 substituting by 𝑓𝑅(𝑓𝑡,𝑥,𝑦,𝑧)=𝑓(𝑡,𝑥,𝑦,0),𝑧<0,(𝑡,𝑥,𝑦,𝑧),0𝑧𝑅,𝑓(𝑡,𝑥,𝑦,𝑅),𝑧>𝑅.(3.28) Similarly, we can obtain that (3.10) has at least one solution 𝑢 satisfying 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡) and 𝛼(𝑡)𝑢(𝑡)𝛽(𝑡),𝑡[0,+). Because 𝑢(𝑡)=𝜙(𝑡)𝑓𝑡,𝑢(𝑡),𝑢(𝑡),𝑢(𝑡)0(3.29) and 𝑢(+)=𝐶0, we have 0𝑢(𝑡)𝑢1(0)=𝑎𝐵+𝑢(0)𝑅.(3.30) Consequently, the solution 𝑢 is a positive solution of (1.3).

Funding

This research is supported by the National Natural Science Foundation of China (nos. 11101385 and 60974145) and by the Fundamental Research Funds for the Central Universities.

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