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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 528719, 13 pages
Research Article

Positive Solutions for Nonlinear Differential Equations with Periodic Boundary Condition

1College of Information Sciences and Technology, Hainan University, Haikou 570228, China
2College of Science and Information Science, Qingdao Agricultural University, Qingdao 266109, China

Received 22 February 2012; Accepted 23 March 2012

Academic Editor: Yeong-Cheng Liou

Copyright © 2012 Shengjun Li et al. 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.


We study the existence of positive solutions for second-order nonlinear differential equations with nonseparated boundary conditions. Our nonlinearity may be singular in its dependent variable. The proof of the main result relies on a nonlinear alternative principle of Leray-Schauder. Recent results in the literature are generalized and significantly improved.

1. Introduction

In this paper, we establish the positive periodic solutions for the following singular differential equation:𝑝(𝑥)𝑦+𝑞(𝑥)𝑦=𝑓𝑥,𝑦,𝑦,0𝑥𝑇,(1.1) and boundary conditions𝑦(0)=𝑦(𝑇),𝑦[1](0)=𝑦[1](𝑇),(1.2) where 𝑝, 𝑞(/𝑇), the nonlinearity 𝑓((/𝑇)×(0,)×,), and 𝑦[1](𝑥)=𝑝(𝑥)𝑦(𝑥) denotes the quasi-derivative of 𝑦(𝑥). We call boundary conditions (1.2) the periodic boundary conditions which are important representatives of nonseparated boundary conditions. In particular, the nonlinearity may have a repulsive singularity at 𝑦=0, which means thatlim𝑦0+𝑓(𝑥,𝑦,𝑧)=+,uniformlyin(𝑥,𝑧)2.(1.3) Also 𝑓 may take on negative values. Electrostatic or gravitational forces are the most important examples of singular interactions.

During the last few decades, the study of the existence of periodic solutions for singular differential equations have deserved the attention of many researchers [18]. Some classical tools have been used to study singular differential equations in the literature, including the degree theory [79], the method of upper and lower solutions [5, 10], Schauder’s fixed point theorem [2, 11, 12], some fixed point theorems in cones for completely continuous operators [1315], and a nonlinear Leray-Schauder alternative principle [1618].

However, the singular differential equation (1.1), in which the nonlinearity is dependent on the derivative and does not require 𝑓 to be nonnegative, has not attracted much attention in the literature. There are not so many existence results for (1.1) even when the nonlinearity is independent of the derivative. In this paper, we try to fill this gap and establish the existence of positive 𝑇-periodic solutions of (1.1); proof of the existence of positive solutions is based on an application of a nonlinear alternative of Leray-Schauder, which has been used by many authors [1618].

The rest of this paper is organized as follows. In Section 2, some preliminary results will be given, including a famous nonlinear alternative of Leray-Schauder type. In Section 3, we will state and prove the main results.

2. Preliminaries

Let us denote 𝑢(𝑡) and 𝑣(𝑡) by the solutions of the following homogeneous equations:𝑝(𝑥)𝑦+𝑞(𝑥)𝑦=0,0𝑥𝑇,(2.1) satisfying the initial conditions𝑢(0)=1,𝑢[1](0)=0,𝑣(0)=0,𝑣[1](0)=1,(2.2) and set𝐷=𝑢(𝑇)+𝑣[1](𝑇)2.(2.3) Throughout this paper, we assume that (1.1) satisfies the following condition (2.4):𝑝(𝑥)>0,𝑞(𝑥)>0,𝑇01𝑝(𝑥)𝑑𝑥<,𝑇0𝑞(𝑥)𝑑𝑥<.(2.4)

Lemma 2.1 (see [19]). For the solution 𝑦(𝑥) of the boundary value problem 𝑝(𝑥)𝑦+𝑞(𝑥)𝑦=(𝑥),0𝑥𝑇,𝑦(0)=𝑦(𝑇),𝑦[1](0)=𝑦[1](𝑇),(2.5) the formula 𝑦(𝑥)=𝑇0[],𝐺(𝑥,𝑠)(𝑠)𝑑𝑠,𝑥0,𝑇(2.6) holds, where 𝐺(𝑥,𝑠)=𝑣(𝑇)𝐷𝑢𝑢(𝑥)𝑢(𝑠)[1](𝑇)𝐷+𝑣𝑣(𝑥)𝑣(𝑠)[1](𝑇)1𝐷𝑢(𝑥)𝑣(𝑠)𝑢(𝑇)1𝐷𝑢𝑣(𝑠)𝑣(𝑥),0𝑠𝑥𝑇,[1](𝑇)1𝐷𝑢(𝑠)𝑣(𝑥)𝑢(𝑇)1𝐷𝑢(𝑥)𝑣(𝑠),0𝑥𝑠𝑇,(2.7) is the Green's function, and the number 𝐷 is defined by (2.3).

Lemma 2.2 (see [19]). Under condition (2.4), the Green’s function 𝐺(𝑥,𝑠) of the boundary value problem (2.5) is positive, that is, 𝐺(𝑥,𝑠)>0, for 𝑥,𝑠[0,𝑇].
One denotes 𝐴=min0𝑠,𝑥𝑇𝐺(𝑥,𝑠),𝐵=max0𝑠,𝑥𝑇𝐴𝐺(𝑥,𝑠),𝜎=𝐵.(2.8) Thus 𝐵>𝐴>0 and 0<𝜎<1.

Remark 2.3. If 𝑝(𝑥)=1, 𝑞(𝑥)=𝑚2>0, then the Green’s function 𝐺(𝑥,𝑠) of the boundary value problem (2.5) has the form 𝑒𝐺(𝑥,𝑠)=𝑚(𝑥𝑠)+𝑒𝑚(𝑇𝑥+𝑠)𝑒2𝑚𝑚𝑇𝑒1,0𝑠𝑥𝑇,𝑚(𝑠𝑥)+𝑒𝑚(𝑇+𝑥𝑠)𝑒2𝑚𝑚𝑇1,0𝑥𝑠𝑇.(2.9) It is obvious that 𝐺(𝑥,𝑠)>0 for 0𝑠,𝑥𝑇, and a direct calculation shows that 𝑒𝐴=𝑚𝑇/2𝑚𝑒𝑚𝑇1,𝐵=1+𝑒𝑚𝑇𝑒2𝑚𝑚𝑇1,𝜎=2𝑒𝑚𝑇/21+𝑒𝑚𝑇<1.(2.10)
Let 𝑋=[0,𝑇], and we suppose that 𝐹[0,𝑇]××[0,) is a continuous function. Define an operator: (𝑇𝑦)(𝑥)=𝑇0𝐺(𝑥,𝑠)𝐹(𝑠,𝑦(𝑠),𝑧(𝑠))𝑑𝑠(2.11) for 𝑦𝑋 and 𝑥[0,𝑇]. It is easy to prove that 𝑇 is continuous and completely continuous.

3. Main Results

In this section, we state and prove the new existence results for (1.1). In order to prove our main results, the following nonlinear alternative of Leray-Schauder is needed, which can be found in [20]. Let us define the function 𝜔(𝑥)=𝑇0𝐺(𝑥,𝑠)𝑑𝑠. The usual 𝐿1-norm over (0,𝑇) is denoted by 1,and the supremum norm of [0,𝑇] is denoted by .

Lemma 3.1. Assume Ω is a relatively compact subset of a convex set 𝐸 in a normed space 𝑋. Let 𝑇Ω𝐸 be a compact map with 0Ω. Then one of the following two conclusions holds:(i)𝑇 has at least one fixed point in Ω;(ii)there exist 𝑢𝜕Ω and 0<𝜆<1 such that 𝑢=𝜆𝑇𝑢.

Now we present our main existence result of positive solution to problem (1.1).

Theorem 3.2. Suppose that (1.1) satisfies (2.4). Furthermore, assume that there exists a constant 𝑟>0 such that(H1) there exists a constant 𝑀>0 such that 𝐹(𝑥,𝑦,𝑧)=𝑓(𝑥,𝑦,𝑧)+𝑀0 for all (𝑥,𝑦,𝑧)[0,𝑇]×(0,𝑟]×;(H2) there exist continuous, nonnegative functions 𝑔(𝑦), (𝑦), and 𝜚(𝑦) such that []]𝐹(𝑥,𝑦,𝑧)(𝑔(𝑦)+(𝑦))𝜚(|𝑧|),(𝑥,𝑦,𝑧)0,𝑇×(0,𝑟×,(3.1) where 𝑔(𝑦)>0 is nonincreasing, (𝑦)/𝑔(𝑦) is nondecreasing in (0,𝑟] and 𝜚(𝑦) is nondecreasing in (0,);(H3) there exist a nonincreasing positive continuous function 𝑔0(𝑦) on (0,), and a constant 𝑅0>0 such that 𝐹(𝑥,𝑦,𝑧)𝑔0(𝑦) for (𝑥,𝑦,𝑧)[0,𝑇]×(0,𝑅0]×, where 𝑔0(𝑦) satisfies lim𝑦0+𝑔0(𝑦)=+ and lim𝑦0+𝑅0𝑦𝑔0(𝑢)𝑑𝑢=+;(H4) the following inequalities hold: 𝑟𝜎𝑟>𝑀𝜔,𝐿𝑔(𝜎𝑟𝑀𝜔){1+(𝑟)/𝑔(𝑟)}𝜚1𝑟+𝐿2𝑇>𝜔,(3.2) where𝐿1=2𝑞1/min0𝑥𝑇𝑝(𝑥),𝐿2=2𝑀/min0𝑥𝑇𝑝(𝑥).

Then (1.1) has at least one positive 𝑇-periodic solution 𝑦 with 0<𝑦+𝑀𝜔𝑟.

Proof. Since (H4) holds, let 𝑁0={𝑛0,𝑛0+1,}, and we can choose 𝑛0{1,2,} such that 1/𝑛0<𝜎𝑟𝑀𝜔 and )𝜔𝑔(𝜎𝑟𝑀𝜔1+(𝑟)𝜚𝐿𝑔(𝑟)1𝑟+𝐿2𝑇+1𝑛0<𝑟.(3.3)
To show (1.1) has a positive solution, we should only show that 𝑝(𝑥)𝑦+𝑞(𝑥)𝑦=𝐹𝑥,𝑦(𝑥)𝑀𝜔(𝑥),𝑦(𝑥)𝑀𝜔(𝑥)(3.4) has a positive solution 𝑦 satisfying (1.2). If it is right, then 𝑘(𝑥)=𝑦(𝑥)𝑀𝜔(𝑥) is a solution of (1.1) since 𝑝(𝑥)𝑘+𝑞(𝑥)𝑘=𝑝(𝑥)(𝑦(𝑥)𝑀𝜔(𝑥))+𝑞(𝑥)(𝑦(𝑥)𝑀𝜔(𝑥))=𝐹𝑥,𝑦(𝑥)𝑀𝜔(𝑥),𝑦(𝑥)𝑀𝜔(𝑥)𝑀=𝑓𝑥,𝑦(𝑥)𝑀𝜔(𝑥),𝑦(𝑥)𝑀𝜔(𝑥)=𝑓𝑥,𝑘(𝑥),𝑘,(𝑥)(3.5) where [𝑝(𝑥)𝜔(𝑥)]+𝑞(𝑥)𝜔(𝑥)=1 is used.
Consider the family of equations: 𝑝(𝑥)𝑦+𝑞(𝑥)𝑦=𝜆𝐹𝑛𝑥,𝑦(𝑥)𝑀𝜔(𝑥),𝑦(𝑥)𝑀𝜔(+𝑥)𝑞(𝑥)𝑛,(3.6) where 𝜆[0,1],𝑛𝑁0, and 𝐹𝑛1(𝑥,𝑦,𝑧)=𝐹(𝑥,𝑦,𝑧)if𝑦𝑛,𝐹1𝑥,𝑛1,𝑧if𝑦𝑛.(3.7) Problem (3.6)−(1.2) is equivalent to the following fixed point of the operator equation: 𝑦=𝑇𝑛𝑦,(3.8) where 𝑇𝑛 is a continuous and completely continuous operator defined by 𝑇𝑛𝑦(𝑥)=𝜆𝑇0𝐺(𝑥,𝑠)𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔1(𝑠)𝑑𝑠+𝑛,(3.9) and we used the fact 𝑇0𝐺(𝑥,𝑠)𝑞(𝑠)𝑑𝑠1(seeLemma2.1with=𝑞).(3.10)
Now we show 𝑦𝑟 for any fixed point 𝑦 of (3.8). If not, assume that 𝑦 is a fixed point of (3.8) such that 𝑦=𝑟. Note that 1𝑦(𝑥)𝑛=𝜆𝑇0𝐺(𝑥,𝑠)𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔(𝑠)𝑑𝑠𝜆𝐴𝑇0𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔(𝑠)𝑑𝑠=𝜎𝐵𝜆𝑇0𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔(𝑠)𝑑𝑠𝜎max[]𝑥0,𝑇𝜆𝑇0𝐺(𝑥,𝑠)𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔(1𝑠)𝑑𝑠=𝜎𝑦𝑛.(3.11) So we have 1𝑦(𝑥)𝜎𝑦𝑛+1𝑛1𝜎𝑦𝑛+1𝑛𝜎𝑟,for0𝑥𝑇.(3.12) In order to pass the solutions of the truncation equation (3.6) (with 𝜆=1) to that of the original equation (3.4), we need the fact that 𝑦 is bounded. Now we show that 𝑦𝐿1𝑟(3.13) for a solution 𝑦(𝑥) of (3.6).
Integrating (3.6) from 0 to 𝑇 (with 𝜆=1), we obtain 𝑇0𝑞(𝑥)𝑦(𝑥)𝑑𝑥=𝑇0𝐹𝑛𝑥,𝑦(𝑥)𝑀𝜔(𝑥),𝑦(𝑥)𝑀𝜔+(𝑥)𝑞(𝑥)𝑛𝑑𝑥.(3.14) Since 𝑦(0)=𝑦(𝑇), there exists 𝑥0[0,𝑇] such that 𝑦(𝑥0)=0; therefore, ||𝑝(𝑥)𝑦||=||||(𝑥)𝑥𝑥0𝑝(𝑠)𝑦||||=||||(𝑠)𝑑𝑠𝑥𝑥0𝑞(𝑠)𝑦(𝑠)𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔(𝑠)𝑞(𝑠)𝑛||||𝑑𝑠𝑇0𝑞(𝑠)𝑦(𝑠)+𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔+(𝑠)𝑞(𝑠)𝑛𝑑𝑠=2𝑇0𝑞(𝑠)𝑦(𝑠)𝑑𝑠2𝑟𝑇0𝑞(𝑠)𝑑𝑠.(3.15) So, ||𝑦(||𝑥)2𝑟𝑇0𝑞(𝑠)𝑑𝑠𝑝(𝑥)2𝑟𝑞1min0𝑥𝑇𝑝(𝑥)=𝐿1𝑟.(3.16) Similarly we have ||𝜔||(𝑥)2𝑇min0𝑥𝑇𝑝(𝑥).(3.17)
By (3.12), we obtain 𝑦(𝑥)𝑀𝜔(𝑥)𝜎𝑟𝑀𝜔(𝑥)𝜎𝑟𝑀𝜔>1/𝑛01/𝑛. Thus from condition (H2) 𝑦(𝑥)=𝜆𝑇0𝐺(𝑥,𝑠)𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔1(𝑠)𝑑𝑠+𝑛=𝜆𝑇0𝐺(𝑥,𝑠)𝐹𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔1(𝑠)𝑑𝑠+𝑛𝑇0𝐺(𝑥,𝑠)𝐹𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔1(𝑠)𝑑𝑠+𝑛𝑇0𝐺(𝑥,𝑠)𝑔(𝑦(𝑠)𝑀𝜔(𝑠))1+(𝑦(𝑠)𝑀𝜔(𝑠))𝜚||𝑦𝑔(𝑦(𝑠)𝑀𝜔(𝑠))(||||𝜔𝑠)+𝑀(||1𝑠)𝑑𝑠+𝑛𝑔(𝜎𝑟𝑀𝜔))1+(𝑟)𝑔𝜚𝐿(𝑟)1𝑟+𝐿2𝑇𝑇01𝐺(𝑡,𝑠)𝑑𝑠+𝑛𝑔(𝜎𝑟𝑀𝜔))1+(𝑟)𝜚𝐿𝑔(𝑟)1𝑟+𝐿2𝑇1𝜔+𝑛0.(3.18) Therefore, )𝑟=𝑦𝑔(𝜎𝑟𝑀𝜔1+(𝑟)𝜚𝐿𝑔(𝑟)1𝑟+𝐿2𝑇1𝜔+𝑛0.(3.19) This is a contradiction, so 𝑦𝑟.
Using Lemma 3.1, we know that 𝑦=𝑇𝑛𝑦(3.20) has a fixed point, denoted by 𝑦𝑛, that is, equation 𝑝(𝑥)𝑦+𝑞(𝑥)𝑦=𝐹𝑛𝑥,𝑦(𝑥)𝑀𝜔(𝑥),𝑦(𝑥)𝑀𝜔(+𝑥)𝑞(𝑥)𝑛(3.21) has a periodic solution 𝑦𝑛 with 𝑦𝑛<𝑟. Using similar procedure to that of the proof of (3.13), we can prove that 𝑦𝑛𝐿1𝑟.(3.22) In the next lemma, we will show that 𝑦𝑛(𝑥)𝑀𝜔(𝑥) have a uniform positive lower bound, that is, there exists a constant 𝛿>0, independent of 𝑛𝑁0, such that 𝑦𝑛(𝑥)𝑀𝜔(𝑥)𝛿(3.23) for all 𝑛𝑁0.
The fact 𝑦𝑛<𝑟 and 𝑦𝑛𝐿1𝑟 shows that {𝑦𝑛}𝑛𝑁0 is a bounded and equi-continuous family on [0,𝑇]. Thus the Arzela–Ascoli Theorem guarantees that {𝑦𝑛}𝑛𝑁0 has a subsequence,{𝑦𝑛𝑖}𝑖 converging uniformly on [0,𝑇] to a function 𝑦𝑋. 𝐹 is uniformly continuous since 𝑦𝑛 satisfies 𝛿+𝑀𝜔(𝑥)𝑦𝑛(𝑥)𝑟 for all 𝑥[0,𝑇]. Moreover, 𝑦𝑛𝑖 satisfies the integral equation 𝑦𝑛𝑖(𝑥)=𝑇0𝐺(𝑥,𝑠)𝐹𝑛𝑠,𝑦𝑛𝑖(𝑠)𝑀𝜔(𝑠),𝑦𝑛𝑖(𝑠)𝑀𝜔1(𝑠)𝑑𝑠+𝑛𝑖.(3.24) Letting 𝑖, we arrive at 𝑦(𝑥)=𝑇0𝐺(𝑥,𝑠)𝐹𝑛𝑠,𝑦(𝑠)𝑀𝜔(𝑠),𝑦(𝑠)𝑀𝜔(𝑠)𝑑𝑠.(3.25) Therefore, 𝑦 is a positive periodic solution of (1.1) and satisfies 0<𝑦+𝑀𝜔𝑟.

Lemma 3.3. There exists a constant 𝛿>0 such that any solution 𝑦𝑛 of (3.6) (with 𝜆=1) satisfies (3.23) for all 𝑛 large enough.

Proof. By condition (H3), there exist 𝑅1(0,𝑅0) and a continuous function ̃𝑔0 such that 𝐹𝑥,𝑦,𝑦𝑞(𝑥)𝑦̃𝑔0}(𝑦)max{𝑀,𝑟𝑞(3.26) for all (𝑥,𝑦)[0,𝑇]×(0,𝑅1], where ̃𝑔0 satisfies condition also like in (H3).
Choose 𝑛1𝑁0 such that 1/𝑛1𝑅1, and let 𝑁1={𝑛1,𝑛1+1,}. For 𝑛𝑁1, let 𝛼𝑛=min0𝑥𝑇𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝛽𝑛=max0𝑥𝑇𝑦𝑛.(𝑥)𝑀𝜔(𝑥)(3.27) We first show that 𝛽𝑛>𝑅1 for all 𝑛𝑁1. If not, assume that 𝛽𝑛𝑅1 for some 𝑛𝑁1.
If 1/𝑛𝑦𝑛(𝑥)𝑀𝜔(𝑥)𝑅1, we obtain from (3.26) 𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔(𝑥)=𝐹𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔𝑦(𝑥)𝑞(𝑥)𝑛(𝑥)𝑀𝜔(𝑥)+̃𝑔0𝑦𝑛(𝑥)𝑀𝜔(𝑥)̃𝑔0𝑦𝑛(𝑥)𝑀𝜔(𝑥)>𝑟𝑞,(3.28) and, if 𝑦𝑛(𝑥)𝑀𝜔(𝑥)1/𝑛, we obtain 𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(1𝑥)=𝐹𝑥,𝑛,𝑦𝑛(𝑥)𝑞(𝑥)𝑛+̃𝑔01𝑛̃𝑔01𝑛>𝑟𝑞.(3.29) So we have 𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔(𝑥)>𝑟𝑞,for𝛽𝑛𝑅1.(3.30) Integrating (3.6) (with 𝜆=1) from 0 to 𝑇, we deduce that0=𝑇0𝑝(𝑥)𝑦𝑛(𝑥)+𝑞(𝑥)𝑦𝑛(𝑥)𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔(𝑥)𝑞(𝑥)𝑛=𝑑𝑥𝑇0𝑦𝑞(𝑥)𝑛1(𝑥)𝑑𝑥𝑛𝑇0𝑞(𝑥)𝑑𝑥𝑇0𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔<(𝑥)𝑑𝑥𝑇0𝑞(𝑥)𝑦𝑛(𝑥)𝑑𝑥𝑟𝑞𝑇0.(3.31) This is a contradiction. Thus 𝛽𝑛>𝑅1, and we have 𝑦𝑛𝑀𝜔>𝑅1,𝑛𝑁1.(3.32)
To prove (3.23), we first show 𝑦𝑛1(𝑥)𝑀𝜔(𝑥)+𝑛,0𝑥𝑇for𝑛𝑁1.(3.33)
Let 𝑁1=𝑃𝑄; here 𝛼𝑛𝑅1 if 𝑛𝑃, and 𝛼𝑛<𝑅1 if 𝑛𝑄. If 𝑛𝑃, it is easy to verify (3.33) is satisfied. We now show (3.33) holds if 𝑛𝑄. If not, suppose there exists 𝑛𝑄 with 𝛼𝑛=min0𝑥𝑇𝑦𝑛(𝑥)𝑀𝜔(𝑥)=𝑦𝑛𝑐𝑛𝑐𝑀𝜔𝑛<1𝑛(3.34) for some 𝑐𝑛[0,𝑇]. As 𝛼𝑛=𝑦𝑛(𝑐𝑛)𝑀𝜔(𝑐𝑛)<𝑅1, by 𝛽𝑛>𝑅1, there exists 𝑐𝑛[0,𝑇] (without loss of generality, we assume 𝑎𝑛<𝑐𝑛) such that 𝑦𝑛(𝑎𝑛)=𝑀𝜔(𝑎𝑛)+𝑅1 and 𝑦𝑛(𝑥)𝑀𝜔(𝑥)+𝑅1 for 𝑎𝑛𝑥𝑐𝑛.
From (3.26), we easily show that 𝐹𝑛𝑥,𝑦(𝑥)𝑀𝜔(𝑥),𝑦(𝑥)𝑀𝜔𝑦(𝑥)>𝑞(𝑥)𝑛𝑎(𝑥)𝑀𝜔(𝑥)+𝑀for𝑥𝑛,𝑐𝑛.(3.35)
Using (3.6) (with 𝜆=1) for 𝑦𝑛(𝑥), we have, for 𝑥[𝑎𝑛,𝑐𝑛], 𝑦𝑝(𝑥)𝑛(𝑥)𝑀𝜔𝑝𝑦(𝑥)=(𝑥)𝑛𝑝𝜔(𝑥)+𝑀(𝑥)(𝑥)=𝑞(𝑥)𝑦𝑛(𝑥)+𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔(+𝑥)𝑞(𝑥)𝑛[]𝑀1𝑞(𝑥)𝜔(𝑥)>𝑞(𝑥)𝑦𝑛𝑦(𝑥)+𝑞(𝑥)𝑛+(𝑥)𝑀𝜔(𝑥)+𝑀𝑞(𝑥)𝑛[]=𝑀1𝑞(𝑥)𝜔(𝑥)𝑞(𝑥)𝑛0.(3.36)
As 𝑦𝑛(𝑐𝑛)𝑀𝜔(𝑐𝑛)=0, 𝑝(𝑥)>0, so 𝑦𝑛(𝑥)𝑀𝜔(𝑥)<0 for all 𝑥[𝑎𝑛,𝑐𝑛), and the function 𝜈𝑛=𝑦𝑛𝑀𝜔 is strictly decreasing on [𝑎𝑛,𝑐𝑛]. We use 𝜂𝑛 to denote the inverse function of 𝑦𝑛 restricted to [𝑎𝑛,𝑐𝑛]. Thus there exists 𝑏𝑛(𝑎𝑛,𝑐𝑛) such that 𝑦𝑛(𝑏𝑛)𝑀𝜔(𝑏𝑛)=1/𝑛 and 𝑦𝑛1(𝑥)𝑀𝜔(𝑥)𝑛for𝑐𝑛𝑥𝑏𝑛,1𝑛𝑦𝑛(𝑥)𝑀𝜔(𝑥)𝑅1for𝑏𝑛𝑥𝑎𝑛.(3.37) By using the method of substitution, we obtain 𝑅11/𝑛𝐹𝜂𝑛(𝜈),𝜈,𝜈𝑑𝜈=𝑎𝑛𝑏𝑛𝐹𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔𝑦(𝑥)𝑛(𝑥)𝑀𝜔=(𝑥)𝑑𝑥𝑎𝑛𝑏𝑛𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔𝑦(𝑥)𝑛(𝑥)𝑀𝜔=(𝑥)𝑑𝑥𝑎𝑛𝑏𝑛𝑦𝑝(𝑥)𝑛(𝑥)+𝑞(𝑥)𝑦𝑛(𝑥)𝑞(𝑥)𝑛𝑦𝑛(𝑥)𝑀𝜔=(𝑥)𝑑𝑥𝑎𝑛𝑏𝑛𝑦𝑝(𝑥)𝑛(𝑥)𝑦𝑛(𝑥)𝑀𝜔+(𝑥)𝑑𝑥𝑎𝑛𝑏𝑛𝑞(𝑥)𝑦𝑛(𝑥)𝑞(𝑥)𝑛𝑦𝑛(𝑥)𝑀𝜔(𝑥)𝑑𝑥.(3.38) By the facts 𝑦𝑛<𝑟, 𝑦𝑛 and 𝜔 are bounded, one can easily obtain that the second term is bounded. The first term is 𝑝𝑏𝑛𝑦𝑛𝑏𝑛2𝑎𝑝𝑛𝑦𝑛𝑎𝑛2𝑎+𝑀𝑝𝑛𝑦𝑛𝑎𝑛𝜔𝑎𝑛𝑏𝑀𝑝𝑛𝑦𝑛𝑏𝑛𝜔𝑏𝑛+𝑎𝑛𝑏𝑛𝑝(𝑥)𝑦𝑛(𝑥)𝑦𝑛(𝑥)𝑑𝑥𝑀𝑎𝑛𝑏𝑛𝑝(𝑥)𝑦𝑛(𝑥)𝜔(𝑥)𝑑𝑥,(3.39) which is also bounded. As a consequence, there exists 𝐿>0 such that 𝑅11/𝑛𝐹𝜂𝑛(𝑦),𝑦,𝑦𝑑𝑦𝐿.(3.40)
On the other hand, by (H3), we can choose 𝑛2𝑁1 large enough such that 𝑅11/𝑛𝐹𝜂𝑛(𝑦),𝑦,𝑦𝑑𝑦𝑅11/𝑛𝑔0(𝑦)𝑑𝑦>𝐿(3.41) for all 𝑛𝑁2={𝑛2,𝑛2+1,}. This is a contradiction. So (3.33) holds.
Finally, we will show that (3.23) is right in 𝑛𝑄. Noticing estimate (3.33) and employing the method of substitution, we obtain 𝑅1𝛼𝑛𝐹𝜂𝑛(𝑦),𝑦,𝑦𝑑𝑦=𝑎𝑛𝑐𝑛𝐹𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔𝑦(𝑥)𝑛(𝑥)𝑀𝜔=(𝑥)𝑑𝑥𝑎𝑛𝑐𝑛𝐹𝑛𝑥,𝑦𝑛(𝑥)𝑀𝜔(𝑥),𝑦𝑛(𝑥)𝑀𝜔𝑦(𝑥)𝑛(𝑥)𝑀𝜔=(𝑥)𝑑𝑥𝑎𝑛𝑐𝑛𝑝(𝑥)𝑦𝑛(𝑥)+𝑞(𝑥)𝑦𝑛(𝑥)𝑞(𝑥)𝑛𝑦𝑛(𝑥)𝑀𝜔(𝑥)𝑑𝑥.(3.42) Obviously, the right-hand side of the above equality is bounded. On the other hand, by (H3), 𝑅1𝛼𝑛𝐹𝜂𝑛(𝑦),𝑦,𝑦𝑑𝑦𝑅1𝛼𝑛𝑔0(𝑦)𝑑𝑦+(3.43) if 𝛼𝑛0+. Thus we know that 𝛼𝑛𝛿 for some constant 𝛿>0; the proof is completed.

Corollary 3.4. Let the nonlinearity in (1.1) be 𝑓(𝑥,𝑦,𝑧)=(1+|𝑧|𝛾)𝑦𝛼+𝜇𝑦𝛽+𝑒(𝑥),0𝑥𝑇,(3.44) where 𝛼>0, 𝛽, 𝛾0, 𝑒(𝑥)[0,𝑇], and 𝜇>0 is a positive parameter,(i)if 𝛽+𝛾<1, then (1.1) has at least one positive periodic solution for each 𝜇>0;(ii)if 𝛽+𝛾1, then (1.1) has at least one positive periodic solution for each 0<𝜇<𝜇, where 𝜇 is some positive constant.

Proof. We will apply Theorem 3.2. Take 𝑀=𝑒0=max0𝑥𝑇||||𝑒(𝑥),𝑔(𝑦)=𝑦𝛼,(𝑦)=𝜇𝑦𝛽+2𝑒0,𝜚(𝑧)=1+|𝑧|𝛾.(3.45) Then conditions (H1)–(H3) are satisfied and the existence condition (H4) becomes 𝜇<𝑟(𝜎𝑟𝑀𝜔)𝛼𝐿𝜔1+1𝑟+𝐿2𝑇𝛾2𝑒0𝜔𝑟𝛼𝐿1+1𝑟+𝐿2𝑇𝛾𝑟𝛼+𝛽𝐿𝜔1+1𝑟+𝐿2𝑇𝛾(3.46) for some 𝑟>0. So (1.1) has at least one positive periodic solution for 0<𝜇<𝜇=sup𝑟>0𝑟(𝜎𝑟𝑀𝜔)𝛼𝐿𝜔1+1𝑟+𝐿2𝑇𝛾2𝑒0𝜔𝑟𝛼𝐿1+1𝑟+𝐿2𝑇𝛾𝑟𝛼+𝛽𝐿𝜔1+1𝑟+𝐿2𝑇𝛾.(3.47) Note that 𝜇= if 𝛽+𝛾<1 and 𝜇< if 𝛽+𝛾1. We have (i) and (ii).


This work is supported by the National Natural Science Foundation of China (Grant no. 11161017) and Hainan Natural Science Foundation (Grant no. 111002).


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