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ISRN Mathematical Analysis
Volume 2012 (2012), Article ID 830983, 12 pages
http://dx.doi.org/10.5402/2012/830983
Research Article

Positive Solutions to Periodic Boundary Value Problems for Four-Order Differential Equations

1Hunan College of Information, Changsha, Hunan 410200, China
2Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China

Received 26 November 2011; Accepted 9 January 2012

Academic Editor: G. Gripenberg

Copyright © 2012 Huantao Zhu and Zhiguo Luo. 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 apply fixed point theorem in a cone to obtain sufficient conditions for the existence of single and multiple positive solutions of periodic boundary value problems for a class of four-order differential equations.

1. Introduction

In this paper, we investigate the existence of positive solutions of the following periodic boundary value problem:𝑦(4)(𝑡)𝐴𝑦(𝑡)+𝐵𝑦(𝑡)=𝑓𝑡,𝑦(𝑡),𝑦𝑦(𝑡),𝑡𝐼,(𝑖)(0)=𝑦(𝑖)(2𝜋),𝑖=0,1,2,3,(1.1) where 𝐼=[0,2𝜋], 𝐴 and 𝐵 are positive constants with 𝐴2>4𝐵, 𝑓𝐶(𝐼×(0,+)×𝑅,[0,+)).

In recent years, the nonlinear periodic boundary value problems have been widely studied by many authors, for example, see [17] and the references therein. Many theorems and methods of nonlinear functional analysis, for instance, upper and lower solutions method, fixed point theorems, variational method, and critical point theory, and so on, have been applied to their problems. When positive solutions are discussed, it seems that fixed point theorem in cones is quite effective in dealing with the problems with singularity. In [8], Zhang and Wang proved periodic boundary value problems with singularity𝑦(𝑡)+𝑘2𝑦𝑦(𝑡)=𝑔(𝑡,𝑦(𝑡)),𝑡𝐼,(𝑖)(0)=𝑦(𝑖)(2𝜋),𝑖=0,1,(1.2) have multiple positive solutions under some conditions, where 𝑔(𝑡,𝑦)  is singular at 𝑦=0, that is,lim𝑦0+𝑔(𝑡,𝑦)=+.(1.3) Relying on a nonlinear alternative of Leray-Schauder type and fixed point theorem, Chu and Zhou [9] discussed the existence of positive solutions for the third-order periodic boundary value problem𝑦(𝑡)+𝑘3𝑦𝑦(𝑡)=𝑔(𝑡,𝑦(𝑡)),𝑡𝐼,(𝑖)(0)=𝑦(𝑖)(2𝜋),𝑖=0,1,2,(1.4) where 𝑘(0,1/3).  However, relatively few papers have been published on the same problem for four-order differential equations. Recently, by using a maximum principle for operator 𝐿𝑢=𝑢(4)𝛽𝑢+𝛼𝑢  in periodic boundary condition and fixed point index theory in cones, Li [10] considered the existence of positive solution for the fourth-order periodic boundary value problem𝑦(4)(𝑡)𝛽𝑦[],𝑦(𝑡)+𝛼𝑦(𝑡)=𝑔(𝑡,𝑦(𝑡)),𝑡0,1(𝑖)(0)=𝑦(𝑖)(1),𝑖=0,1,2,3,(1.5) where 𝑓[0,1]×𝑅+𝑅+  is continuous, 𝛼, 𝛽𝑅 and satisfy 0<𝛼<(𝛽/2+2𝜋2)2, 𝛽>2𝜋2, 𝛼/𝜋4+𝛽/𝜋2>1. However, since there appears 𝑦  in nonlinear term 𝑔, the method in [9] cannot be directly applied to (1.1). The main aim of this paper is to establish sufficient conditions for the existence of positive solutions to the problem (1.1).

To prove our main results, we present an existence theorem.

Theorem 1.1 (see [11]). Let 𝐸 be a Banach space and 𝑃 a cone in 𝐸. Suppose Λ1 and Λ2 are open subsets of 𝐸 such that 0Λ1Λ1Λ2  and suppose that 𝑇𝑃Λ2Λ1𝑃(1.6) is a completely continuous operator. If one of the following conditions is satisfied:(i)𝑇𝑥𝑥for 𝑥𝑃𝜕Λ1,𝑇𝑥𝑥  for 𝑥𝑃𝜕Λ2,(ii)𝑇𝑥𝑥  for 𝑥𝑃𝜕Λ1,𝑇𝑥𝑥  for 𝑥𝑃𝜕Λ2. Then 𝑇 has a fixed point in 𝑃(Λ2/Λ1).

2. Preliminaries

In this section, we present some preliminary results which will be needed in Section 3.

Let 𝛼>0, and for any function 𝑢𝐶(𝐼),  we defined the operator𝐽𝛼𝑢(𝑡)=02𝜋𝐺𝛼(𝑡,𝑠)𝑢(𝑠)𝑑𝑠,(2.1) where𝐺𝛼𝑒(𝑡,𝑠)=𝛼(𝑡𝑠)+𝑒𝛼(2𝜋𝑡+𝑠)𝑒2𝛼2𝜋𝛼𝑒1,0𝑠𝑡2𝜋,𝛼(𝑠𝑡)+𝑒𝛼(2𝜋𝑠+𝑡)𝑒2𝛼2𝜋𝛼1,0𝑡𝑠2𝜋.(2.2) By a direct calculation, we easily obtain02𝜋𝐺𝛼(𝑡,𝑠)𝑑𝑠=𝛼2,𝑚𝛼=min𝑠,𝑡𝐼𝐺𝛼(𝑡,𝑠)=2𝑒𝛼𝜋𝑒2𝛼2𝜋𝛼1,𝑀𝛼=max𝑠,𝑡𝐼𝐺𝛼(𝑡,𝑠)=1+𝑒2𝛼𝜋𝑒2𝛼2𝜋𝛼.1(2.3) Set𝜌=𝐴+𝐴24𝐵2,𝜆=𝐴𝐴24𝐵2,or𝜌=𝐴𝐴24𝐵2,𝜆=𝐴+𝐴24𝐵2,(2.4) then 𝜌, 𝜆𝑅. Now, we consider the problem𝑢(𝑡)+𝜆2𝑢(𝑡)=𝑓𝑡,𝐽𝜌𝑢,𝜌2𝐽𝜌𝑢𝑢𝑢,𝑡𝐼,(𝑖)(0)=𝑢(𝑖)(2𝜋),𝑖=0,1.(2.5)

Lemma 2.1. If 𝑢𝐶2(𝐼)  is a (positive) solution of problem (2.5), then 𝑦=𝐽𝜌𝑢𝐶4(𝐼)  is a (positive) solution of problem (1.1). Moreover, the problem (2.5) is equivalent to integral equation 𝑢(𝑡)=02𝜋𝐺𝜆𝐽(𝑡,𝑠)𝑓𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠.(2.6)

Proof. If 𝑢𝐶2(𝐼),  then 𝐽𝜌𝑢𝐶4(𝐼)  and 𝐽𝜌𝑢(𝑡)=𝑢(𝑡)+𝜌2𝐽𝜌𝑢𝐽(𝑡),𝑡𝐼,𝜌𝑢(𝑖)𝐽(0)=𝜌𝑢(𝑖)(2𝜋),𝑖=0,1.(2.7) Thus, 𝑦(𝑡)=𝑢(𝑡)+𝜌2𝐽𝜌𝑢(𝑡),𝑦(𝑡)=𝑢(𝑡)+𝜌2𝐽𝜌𝑢𝑦(𝑡),(4)(𝑡)=𝑢(𝑡)+𝜌2𝐽𝜌𝑢(𝑡)=𝑢(𝑡)𝜌2𝑢(𝑡)+𝜌4𝐽𝜌𝑢(𝑡).(2.8) Then, 𝑦(4)(𝑡)𝐴𝑦(𝑡)+𝐵𝑦(𝑡)=𝑢(𝑡)+𝐴𝜌2𝜌𝑢(𝑡)+4𝐴𝜌2𝐽+𝐵𝜌𝑢(𝑡)=𝑢(𝑡)+𝜆2𝑢(𝑡)=𝑓𝑡,𝐽𝜌𝑢,𝜌2𝐽𝜌𝑢𝑢=𝑓𝑡,𝑦(𝑡),𝑦.(𝑡)(2.9) On the other hand, 𝑦(0)=𝑦(2𝜋),𝑦(0)=𝑦(2𝜋), 𝑦(0)=𝑢(0)+𝜌2𝐽𝜌𝑢(0)=𝑢(2𝜋)+𝜌2𝐽𝜌𝑢(2𝜋)=𝑦𝑦(2𝜋),(0)=𝑢(0)+𝜌2𝐽𝜌𝑢(0)=𝑢(2𝜋)+𝜌2𝐽𝜌𝑢(2𝜋)=𝑦(2𝜋).(2.10) Hence, if 𝑢𝐶2(𝐼) is a solution of problem (2.5), then 𝑦=𝐽𝜌𝑢𝐶4(𝐼) is a solution of problem (1.1). And, if 𝑢𝐶2(𝐼) is a positive function, noting that 𝐺𝜌(𝑡,𝑠)>0 for any 𝑠, 𝑡𝐼, we have 𝑦=02𝜋𝐺𝜌(𝑡,𝑠)𝑢(𝑠)𝑑𝑠>0.(2.11) Noting that, for any function 𝐶(𝐼), linear problem 𝑢(𝑡)+𝜆2𝑢𝑢(𝑡)=(𝑡),𝑡𝐼,(𝑖)(0)=𝑢(𝑖)(2𝜋),𝑖=0,1,(2.12) has a unique solution 𝑢(𝑡)=02𝜋𝐺𝜆(𝑡,𝑠)(𝑠)𝑑𝑠,(2.13) one can easily obtain that (2.6) holds. The proof is complete.

In the following application, we take 𝐸=𝐶2(𝐼) with the supremum norm and define𝑃𝜆=𝑢𝐸𝑢(𝑡)0forall𝑡𝐼,min𝑡𝐽𝑢(𝑡)𝛿𝜆𝑢,(2.14) where 𝛿𝜆=𝑚𝜆/𝑀𝜆(0,1).

One easily checks and verifies that 𝑃𝜆  is a cone in 𝐸. For any 𝑟>0, let 𝑃𝜆𝑟={𝑢𝑃𝜆𝑢<𝑟},  then 𝜕𝑃𝜆𝑟={𝑢𝑃𝜆𝑢=𝑟}.  For any 𝑢𝐸, define mapping 𝑇𝜆𝐸𝐸 by𝑇𝜆𝑢(𝑡)=02𝜋𝐺𝜆𝐽(𝑡,𝑠)𝑓𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠,(2.15) then the fixed point of 𝑇𝜆 in 𝐸 is a positive solution of (2.5).

Lemma 2.2. For any 𝜍>𝜏>0, 𝑇𝜆𝑃𝜆𝜍𝑃𝜆𝜏𝑃𝜆  is completely continuous.

Proof. For any 𝑢𝑃𝜆𝜍𝑃𝜆𝜏, 𝜏𝑢𝜍   and 𝛿𝜆𝜏𝑢(𝑡)𝜍 for all 𝑡𝐼. Thus, if 𝑢𝑃𝜆𝜍𝑃𝜆𝜏, 𝐽𝜌𝑢(𝑡)=02𝜋𝐺𝜌(𝑡,𝑠)𝑢(𝑠)𝑑𝑠𝛿𝜆𝜏02𝜋𝐺𝜌𝛿(𝑡,𝑠)𝑑𝑠𝜆𝜏𝜌2.(2.16) It is easy to see that 𝑇𝜆 is continuous and completely continuous since 𝑓𝐼×(0,+)×𝑅[0,+)  is continuous. Next, we show that 𝑇𝜆𝑃𝜆𝜍𝑃𝜆𝜏𝑃𝜆.  Since 𝑓(𝑠,(𝐽𝜌𝑢)(𝑠),𝜌2(𝐽𝜌𝑢)(𝑠)𝑢(𝑠))0  for 𝑢𝑃𝜆𝜍𝑃𝜆𝜏, 𝑇𝜆𝑢0. On the other hand, 𝑇𝜆𝑢=max𝑡𝐼02𝜋𝐺𝜆𝐽(𝑡,𝑠)𝑓𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠𝑀𝜆02𝜋𝑓𝐽𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢𝑇(𝑠)𝑢(𝑠)𝑑𝑠,𝜆𝑢(𝑡)𝑚𝜆02𝜋𝑓𝐽𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠𝛿𝜆𝑇𝜆𝑢.(2.17) The proof is complete.

3. Positive Solutions of (1.1)

In this section, we make the following hypotheses. (𝐻𝜆1) There exist nonnegative functions (𝑢), 𝑔(𝑢)𝐶((0,+)(0,+))  and 𝑝(𝑡,𝑣), 𝑞(𝑡,𝑣)𝐶(𝐽×𝑅) such that 𝑓(𝑡,𝑢,𝑣)𝑝(𝑡,𝑣)(𝑢)+𝑞(𝑡,𝑣)𝑔(𝑢)(3.1) for all (𝑡,𝑢,𝑣)𝐼×(0,)×𝑅  and sup𝑢>0𝜌𝑢𝑔2𝑢max𝛿𝜆1𝑢𝑣1𝛿𝜆𝑢02𝜋𝜌𝑝(𝑡,𝑣)𝑑𝑡2𝑢+02𝜋𝜌𝑞(𝑡,𝑣)𝑑𝑡𝑔2𝑢𝑔𝜌2𝛿𝜆𝑢>𝑀𝜆,(3.2) where 𝑔 is nonincreasing and /𝑔 is nondecreasing on (0,).(𝐻𝜆2) One has liminf𝑢0+min02𝜋𝑓(𝑡,𝑤,𝑣)𝑑𝑡,𝜌2𝛿𝜆𝑢𝑤𝜌2𝛿𝑢,𝜆1𝑢𝑣1𝛿𝜆𝑢𝑢>𝑚𝜆1.(3.3)(𝐻𝜆3) One has liminf𝑢+min02𝜋𝑓(𝑡,𝑤,𝑣)𝑑𝑡,𝜌2𝛿𝜆𝑢𝑤𝜌2𝛿𝑢,𝜆1𝑢𝑣1𝛿𝜆𝑢𝑢>𝑚𝜆1.(3.4)

Under the above hypotheses, we can obtain the following result.

Theorem 3.1. Assume that (𝐻𝜆1) and (𝐻𝜆2)  are satisfied, then there exist two positive constants 𝛼, 𝛽   such that (1.1) has at least positive solution 𝑦 with 𝜌2𝛼<𝑦<𝜌2𝛽.(3.5)
Assume (𝐻𝜆1) and (𝐻𝜆3) are satisfied, then there exist two positive constants 𝛽, 𝛾 such that (1.1) has at least positive solution 𝑦 with 𝜌2𝛽<𝑦<𝜌2𝛾.(3.6)
Assume (𝐻𝜆1), (𝐻𝜆2), and (𝐻𝜆3) are satisfied, then there exist positive constants 𝛼, 𝛽, 𝛾   such that (1.1) has at least two positive solutions 𝑦1, 𝑦2 with 𝜌2𝑦𝛼<1<𝜌2𝑦𝛽<2<𝜌2𝛾.(3.7)

Proof. First, we assume that (𝐻𝜆1) and (𝐻𝜆2) are satisfied. From the condition (𝐻𝜆1),  one can obtain that there exist a 𝛽>0 such that 𝜌𝛽𝑔2𝛽max(𝛿𝜆1)𝛽𝑣(1𝛿𝜆)𝛽02𝜋𝜌𝑝(𝑡,𝑣)𝑑𝑡2𝛽+02𝜋𝜌𝑞(𝑡,𝑣)𝑑𝑡𝑔2𝛽𝑔𝜌2𝛿𝜆𝛽>𝑀𝜆.(3.8) For any 𝑢𝜕𝑃𝜆𝛽, 𝛿𝜆𝛽𝑢(𝑡)𝛽  for all 𝑡𝐼 and 𝛽𝛿𝜆𝜌2𝐽𝜌𝑢(𝑡)=02𝜋𝐺𝜌𝛽(𝑡,𝑠)𝑢(𝑠)𝑑𝑠𝜌2,𝛿𝜆𝛽𝛽𝜌2𝐽𝜌𝑢(𝑡)𝑢(𝑡)𝛽𝛿𝜆𝛽.(3.9) Thus, for 𝑢𝜕𝑃𝜆𝛽, from (𝐻𝜆1),  we have 𝑇𝜆𝑢(𝑡)=02𝜋𝐺𝜆𝐽(𝑡,𝑠)𝑓𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠𝑀𝜆02𝜋𝑝𝑠,𝜌2𝐽𝜌𝑢𝐽(𝑠)𝑢(𝑠)𝜌𝑢(𝑠)+𝑞𝑠,𝜌2𝐽𝜌𝑢𝑔𝐽(𝑠)𝑢(𝑠)𝜌𝑢(𝑠)𝑑𝑠𝑀𝜆02𝜋𝑝𝑠,𝜌2𝐽𝜌𝑢𝐽(𝑠)𝑢(𝑠)𝜌𝑢(𝑠)𝑔𝐽𝜌𝑢(𝑠)+𝑞𝑠,𝜌2𝐽𝜌𝑢𝑔𝐽(𝑠)𝑢(𝑠)𝜌𝑢(𝑠)𝑑𝑠𝑀𝜆02𝜋𝑝𝑠,𝜌2𝐽𝜌𝑢𝜌(𝑠)𝑢(𝑠)2𝛽𝑔𝜌2𝛽+𝑞𝑠,𝜌2𝐽𝜌𝑢𝑔𝜌(𝑠)𝑢(𝑠)2𝛿𝜆𝛽𝑑𝑠𝑀𝜆𝑔𝜌2𝛿𝜆𝛽𝑔𝜌2𝛽max𝛿𝜆1𝛽𝑣1𝛿𝜆𝛽02𝜋𝜌𝑝(𝑡,𝑣)𝑑𝑡2𝛽+02𝜋𝜌𝑞(𝑡,𝑣)𝑑𝑡𝑔2𝛽<𝛽=𝑢,(3.10) which implies that 𝑇𝜆𝑢<𝑢,𝑢𝜕𝑃𝜆𝛽.(3.11)
From (𝐻𝜆2)  is satisfied, there exists a positive constant 𝛼<𝛽  such that min02𝜋𝑓(𝑡,𝑤,𝑣)𝑑𝑡,𝜌2𝛿𝜆𝛼𝑤𝜌2𝛿𝛼,𝜆1𝛼𝑣1𝛿𝜆𝛼>𝑚𝜆1𝛼.(3.12) For any 𝑢𝜕𝑃𝜆𝛼, 𝛿𝜆𝛼𝑢(t)𝛼 for all 𝑡𝐼 and 𝛿𝜆𝛼𝜌2𝐽𝜌𝑢(𝑡)=02𝜋𝐺𝜌𝛼(𝑡,𝑠)𝑢(𝑠)𝑑𝑠𝜌2,𝛿𝜆𝛼𝛼𝜌2𝐽𝜌𝑢(𝑡)𝑢(𝑡)𝛼𝛿𝜆𝛼,(𝑇𝑢)(𝑡)=02𝜋𝐺𝜆𝐽(𝑡,𝑠)𝑓𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠𝑚𝜆02𝜋𝑓𝐽𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠𝑚𝜆min02𝜋𝑓(𝑡,𝑤,𝑣)𝑑𝑡,𝜌2𝛿𝜆𝛼𝑤𝜌2𝛿𝛼,𝜆1𝛼𝑣1𝛿𝜆𝛼>𝛼=𝑢,(3.13) which implies that 𝑇𝜆𝑢>𝑢,𝑢𝜕𝑃𝜆𝛼.(3.14) From (3.11) and (3.14) and Theorem 1.1, one can obtain that 𝑇𝜆 has a fixed point 𝑢 in  𝑃𝜆𝛽𝑃𝜆𝛼 with 𝛼<𝑢<𝛽.  Hence, 𝑦=𝐽𝜌𝑢  is a positive solution of (1.1) with 𝜌2𝛼<𝑦<𝜌2𝛽.
Next, we assume that (𝐻𝜆1) and (𝐻𝜆3) are satisfied. In this case, we have (3.11).
Suppose that (𝐻𝜆3) is satisfied, there exists a positive constant 𝛾>𝛽 such that min02𝜋𝑓(𝑡,𝑤,𝑣)𝑑𝑡,𝜌2𝛿𝜆𝛾𝑤𝜌2𝛿𝛾,𝜆1𝛾𝑣1𝛿𝜆𝛾>𝑚𝜆1𝛾.(3.15) For any 𝑢𝜕𝑃𝜆𝛾, 𝛿𝜆𝛾𝑢(𝑡)𝛾  for all 𝑡𝐼 and 𝛿𝜆𝛾𝜌2𝐽𝜌𝑢(𝑡)=02𝜋𝐺𝜌𝛾(𝑡,𝑠)𝑢(𝑠)𝑑𝑠𝜌2,𝛿𝜆𝛾𝛾𝜌2𝐽𝜌𝑢(𝑡)𝑢(𝑡)𝛾𝛿𝜆𝛾,(𝑇𝑢)(𝑡)=02𝜋𝐺𝜆𝐽(𝑡,𝑠)𝑓𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠𝑚𝜆02𝜋𝑓𝐽𝑠,𝜌𝑢(𝑠),𝜌2𝐽𝜌𝑢(𝑠)𝑢(𝑠)𝑑𝑠𝑚𝜆min02𝜋𝑓(𝑡,𝑤,𝑣)𝑑𝑡,𝜌2𝛿𝜆𝛾𝑤𝜌2𝛿𝛾,𝜆1𝛾𝑣1𝛿𝜆𝛾>𝛾=𝑢,(3.16) which implies that 𝑇𝜆𝑢>𝑢,𝑢𝜕𝑃𝜆𝛾.(3.17) From (3.11) and (3.17) and Theorem 1.1, one can obtain that 𝑇𝜆 has a fixed point 𝑢 in  𝑃𝜆𝛾𝑃𝜆𝛽 with 𝛽<𝑢<𝛾.  Thus, 𝑦=𝐽𝜌𝑢   is a positive solution of (1.1) with 𝜌2𝛽<𝑦<𝜌2𝛾.
Assume that (𝐻𝜆1), (𝐻𝜆2), and (𝐻𝜆3) are satisfied. Repeating the above argument, one can obtain that 𝑇𝜆 has a fixed point 𝑢1 in  𝑃𝜆𝛽P𝜆𝛼  and a fixed point 𝑢2 in  𝑃𝜆𝛾𝑃𝜆𝛽 with 𝑢𝛼<1𝑢<𝛽<2<𝛾.(3.18) Hence, 𝑦1=𝐽𝜌𝑢1, 𝑦2=𝐽𝜌𝑢2  are two positive solutions of (1.1) with 𝜌2𝑦𝛼<1<𝜌2𝑦𝛽<2<𝜌2𝛾.(3.19) The proof is complete.

4. A Similar Problem

In this section, we use the idea in Sections 2 and 3 to consider the following problem:𝑦(4)(𝑡)+𝑘4𝑦(𝑡)=𝑓𝑡,𝑦(𝑡),𝑦1(𝑡),𝑡𝐼,0<𝑘<2,𝑦(𝑖)(0)=𝑦(𝑖)(2𝜋),𝑖=0,1,2,3,(4.1) where 𝐼=[0,2𝜋], 𝑓𝐶(𝐼×(0,+)×𝑅,[0,+)).

Let 𝑢𝐶2(𝐼)  and 𝑦=𝐽𝑢=02𝜋𝐾(𝑡,𝑠)𝑢(𝑠)𝑑𝑠,  where𝐾(𝑡,𝑠)=cos𝑘(𝜋𝑡+𝑠)2𝑘sin𝑘𝜋,0𝑠𝑡2𝜋,cos𝑘(𝜋+𝑡𝑠)2𝑘sin𝑘𝜋,0𝑡𝑠2𝜋.(4.2) Then,02𝜋𝐾(𝑡,𝑠)𝑑𝑠=𝑘2,𝐽𝑢(𝑡)=𝑢(𝑡)𝑘2𝐽𝑢(𝑡),𝑡𝐼,𝐽𝑢(𝑖)(0)=𝐽𝑢(𝑖)(2𝜋),𝑖=0,1,(4.3) and one easily check that (4.1) is equivalent to the problem𝑢(𝑡)+𝑘2𝑢(𝑡)=𝑓𝑡,𝐽𝑢,𝑢𝑘2𝑢𝐽𝑢,𝑡𝐼,(𝑖)(0)=𝑢(𝑖)(2𝜋),𝑖=0,1.(4.4) If 𝑢𝐶2(𝐼) is a (positive) solution of problem (4.4), then 𝑦=𝐽𝑢𝐶4(𝐼)  is a (positive) solution of problem (4.1). Moreover, the problem (4.4) is equivalent to integral equation𝑢(𝑡)=02𝜋𝐺𝑘(𝑡,𝑠)𝑓𝑠,𝐽𝑢(𝑠),𝑢(𝑠)𝑘2𝐽𝑢(𝑠)𝑑𝑠.(4.5) For any 𝑢𝐸, define mapping 𝑇𝐸𝐸 by(𝑇𝑢)(𝑡)=02𝜋𝐺𝑘(𝑡,𝑠)𝑓𝑠,𝐽𝑢(𝑠),𝑢(𝑠)𝑘2𝐽𝑢(𝑠)𝑑𝑠.(4.6) For any 𝜍>𝜏>0, one can obtain that 𝑇𝑃𝑘𝜍𝑃𝑘𝜏𝑃𝑘  is completely continuous.

Similar to the proof of Theorem 3.1, we can obtain the following result.

Theorem 4.1. Assume that (𝐻𝑘1) and (𝐻𝑘2) are satisfied, then there exist two positive constants 𝛼, 𝛽  such that (4.1) has at least positive solution 𝑦 with 𝑘2𝛼<y<𝑘2𝛽.(4.7)
Assume (𝐻𝑘1) and (𝐻𝑘3) are satisfied, then there exist two positive constants 𝛽, 𝛾 such that (4.1) has at least positive solution 𝑦 with 𝑘2𝛽<𝑦<𝑘2𝛾.(4.8)
Assume (𝐻𝑘1), (𝐻𝑘2), and (𝐻𝑘3) are satisfied, then there exist positive constants 𝛼, 𝛽, 𝛾 such that (4.1) has at least two positive solutions 𝑦1, 𝑦2 with 𝑘2𝑦𝛼<1<𝑘2𝑦𝛽<2<𝑘2𝛾,(4.9) where (𝐻𝑘𝑗)(𝑗=1,2,3)  is condition obtained by replacing 𝜆 and 𝜌 by 𝑘 in the condition (𝐻𝜆𝑗) defined in Section 3.

Example 4.2. Consider the differential equation 𝑦(4)(𝑡)𝜇𝑦1(𝑡)+𝑦(𝑡)=𝑒20𝜋𝜇𝑦(𝑡)+1+𝑡𝑦𝜇𝑦(𝑡),𝑡𝐼,(𝑖)(0)=𝑦(𝑖)(2𝜋),𝑖=0,1,2,3,(4.10) where 𝜇>4 is a constant.
Let𝜆=𝜇𝜇242,𝜌=𝜇+𝜇242𝑒,𝑓(𝑡,𝑢,𝑣)=𝑣+(1+𝑡)(𝜇𝑢)1,120𝜋𝜇(𝑢)1,𝑔(𝑢)=𝑢𝑒,𝑝(𝑡,𝑣)=𝑣20𝜋𝜇,𝑞(𝑡,𝑣)=1+𝑡20𝜋𝜇2.(4.11) Then, for all (𝑡,𝑢,𝑣)𝐼×(0,)×𝑅, 0<𝑓(𝑡,𝑢,𝑣)𝑝(𝑡,𝑣)(𝑢)+𝑞(𝑡,𝑣)𝑔(𝑢).(4.12) Noting that sup𝑢>0𝑢max(𝛿𝜆1)𝑢𝑣(1𝛿𝜆)𝑢02𝜋𝜌𝑝(𝑡,𝑣)𝑑𝑡2𝑢𝜌/𝑔2𝑢+02𝜋𝑔𝜌𝑞(𝑡,𝑣)𝑑𝑡2𝛿𝜆𝑢>10𝜇𝛿𝜆𝑒+3𝜋𝜇1𝜌2,lim𝜇+𝑀𝜆𝜇𝛿𝜆=14𝜋,lim𝜇+𝜌2𝜇=1,(4.13) we obtain that (𝐻𝜆1) holds when 𝜇 is sufficiently large. On the other hand, it is easy to check that (𝐻𝜆2) is satisfied since 𝑓(𝑡,𝑢,𝑣)+  as 𝑢0+ for any 𝑡𝐼 and 𝑣𝑅. Hence, (4.10) has at least a positive solution when 𝜇 is sufficiently large.

Acknowledgment

A Project Supported by the NNSF of China (10871063) and ScientificResearch Fund of human Provincial Education Department (10C0258).

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