Abstract

By means of the fixed point theory of strict set contraction operators, we establish a new existence theorem on multiple positive solutions to a singular boundary value problem for second-order impulsive differential equations with periodic boundary conditions in a Banach space. Moreover, an application is given to illustrate the main result.

1. Introduction

The theory of impulsive differential equations describes processes that experience a sudden change of their state at certain moments. In recent years, a great deal of work has been done in the study of the existence of solutions for impulsive boundary value problems, by which a number of chemotherapy, population dynamics, optimal control, ecology, industrial robotics, and physics phenomena are described. For the general aspects of impulsive differential equations, we refer the reader to the classical monograph [1]. For some general and recent works on the theory of impulsive differential equations, we refer the reader to [214]. Meanwhile, the theory of ordinary differential equations in abstract spaces has become a new important branch (see [1518]). So it is interesting and important to discuss the existence of positive solutions for impulsive boundary value problem in a Banach space.

Let (𝐸,) be a real Banach space, 𝐽=[0,2𝜋], 0=𝑡0<𝑡1<𝑡2<<𝑡𝑚<𝑡𝑚+1=2𝜋, 𝐽0=[0,𝑡1], and 𝐽𝑖=(𝑡𝑖,𝑡𝑖+1], 𝑖=1,,𝑚. Note that 𝑃𝐶[𝐽,𝐸]={𝑢𝑢 is a map from𝐽into𝐸such that 𝑢(𝑡)is continuous at 𝑡𝑡𝑘 and left continuous at𝑡=𝑡𝑘 and 𝑢(𝑡+𝑘)exist, 𝑘=1,2,,𝑚}, and it is also a Banach space with norm 𝑢𝑃𝐶=sup𝑡𝐽𝑢(𝑡).(1.1)

Let the Banach space 𝐸 be partially ordered by a cone 𝑃 of 𝐸; that is, 𝑥𝑦 if and only if 𝑦𝑥𝑃, and 𝑃𝐶[𝐽,𝐸] is partially ordered by 𝐾={𝑢𝑃𝐶[𝐽,𝐸]𝑢(𝑡)𝑃,𝑡𝐽}𝑢𝑣 if and only if 𝑣𝑢𝐾; that is, 𝑢(𝑡)𝑣(𝑡) for all 𝑡𝐽.

In this paper, we consider the following singular periodic boundary value problem with impulsive effects in Banach 𝐸𝑢(𝑡)+𝑀2𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),𝑡𝐽,𝑡𝑡𝑘,||Δ𝑢𝑡=𝑡𝑘=𝐼𝑘𝑢𝑡𝑘,Δ𝑢||𝑡=𝑡𝑘=𝐼𝑘𝑢𝑡𝑘𝑢,𝑘=1,2,,𝑚,(0)=𝑢(2𝜋),𝑢(0)=𝑢(2𝜋),(1.2) where 𝑀>0 is constant, 𝑓(𝑡,𝑢) may be singular at 𝑡=0and/or𝑡=2𝜋, 𝑓𝐶[(0,2𝜋)×𝑃,𝑃], 𝐼𝑘,𝐼𝑘𝐶[𝑃,𝑃], Δ𝑢|𝑡=𝑡𝑘=𝑢(𝑡+𝑘)𝑢(𝑡𝑘), Δ𝑢|𝑡=𝑡𝑘=𝑢(𝑡+𝑘)𝑢(𝑡𝑘), 𝑘=1,2,,𝑚, and 𝑢𝑖(𝑡+𝑘) (resp., 𝑢𝑖(𝑡𝑘)) denote the right limit (resp., left limit) of 𝑢𝑖(𝑡) at 𝑡=𝑡𝑘, 𝑖=0,1.

In the special case where 𝐸=+=[0,+), and 𝐼𝑘=𝐼𝑘=0, 𝑘=1,2,,𝑚, problem (1.2) is reduced to the usual second-order periodic boundary value problem. For example, in [19], the periodic boundary value problem: 𝑢(𝑡)+𝑀𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),𝑡(0,2𝜋),𝑢(0)=𝑢(2𝜋),𝑢(0)=𝑢(2𝜋),(1.3) was proved to have at least one positive solution, by Jiang [19].

In [20], the authors studied the multiplicity of positive solutions for IBVP(1.2) in 𝐸=+; the main tool is the theory of fixed point index.

In [21], the author considers the following periodic boundary value problem of second-order integrodifferential equations of mixed type in Banach space: 𝑢||=𝑓(𝑡,𝑢,𝑇𝑢,𝑆𝑢),𝑡(0,2𝜋),Δ𝑢𝑡=𝑡𝑘=𝐼𝑘𝑢𝑡𝑘,Δ𝑢||𝑡=𝑡𝑘=𝐼𝑘𝑢𝑡𝑘,𝑘=1,2,,𝑚,𝑢(0)=𝑢(2𝜋),𝑢(0)=𝑢(2𝜋),(1.4) where 𝑓𝐶[𝐽×𝐸×𝐸×𝐸,𝐸], 𝐼𝑘,𝐼𝑘𝐶[𝐸,𝐸], and the operators 𝑇, 𝑆 are given by 𝑇𝑢(𝑡)=𝑡0𝑘(𝑡,𝑠)𝑢(𝑠)𝑑𝑠,𝑆𝑢(𝑡)=02𝜋𝑘1(𝑡,𝑠)𝑢(𝑠)𝑑𝑠,(1.5) with 𝑘𝐶[𝐷,], 𝐷={(𝑡,𝑠)20𝑠𝑡2𝜋}, 𝑘1𝐶[𝐽×𝐽,]. By applying the monotone iterative technique and cone theory based on a comparison result, the author obtained an existence theorem of minimal and maximal solutions for the IBVP(1.4).

Motivated by the above facts, our aim is to study the multiplicity of positive solutions for IBVP(1.2) in a Banach space. By means of the fixed point index theory of strict set contraction operators, we establish a new existence theorem on multiple positive solutions for IBVP(1.2). Moreover, an application is given to illustrate the main result.

The rest of this paper is organized as follows. In Section 2, we present some basic lemmas and preliminary facts which will be needed in the sequel. Our main result and its proof are arranged in Section 3. An example is given to show the application of the result in Section 4.

2. Preliminaries

Let 𝑇𝑟={𝑥𝐸𝑥𝑟}, 𝐵𝑟={𝑢𝑃𝐶[𝐽,𝐸]𝑢𝑃𝐶𝑟} (𝑟>0); for 𝐷𝑃𝐶[𝐽,𝐸], we denote 𝐷(𝑡)={𝑢(𝑡)𝑢𝐷}𝐸(𝑡𝐽).𝛼 denotes the Kuratowski measure of noncompactness.

Let 𝑃𝐶1[𝐽,𝐸]={𝑢𝑢be a map from𝐽into𝐸such that 𝑢(𝑡)is continuously differentiable at 𝑡𝑡𝑘 and left continuous at𝑡=𝑡𝑘 and𝑢(𝑡+𝑘), 𝑢(𝑡𝑘), 𝑢(𝑡+𝑘)exist, 𝑘=1,2,,𝑚}. Evidently, 𝑃𝐶1[𝐽,𝐸] is a Banach space with norm 𝑢𝑃𝐶1=max𝑢𝑃𝐶,𝑢𝑃𝐶.(2.1)

Let 𝐽=𝐽{𝑡1,𝑡2,,𝑡𝑚}; a map 𝑢𝑃𝐶1[𝐽,𝐸]𝐶2[𝐽,𝐸] is a solution of IBVP(1.2) if it satisfies (1.2).

Now, we first give the following lemmas in order to prove our main result.

Lemma 2.1 (see [17]). Let 𝐾 be a cone in real Banach space 𝐸, and let Ω be a nonempty bounded open convex subset of 𝐾. Suppose that 𝐴Ω𝐾 is a strict set contraction and 𝐴(Ω)𝐾. Then the fixed-point index 𝑖(𝐴,Ω,𝐾)=1.

Lemma 2.2 (see [21]). 𝑢𝑃𝐶1[𝐽,𝐸]𝐶2[𝐽,𝐸] is a solution of IBVP (1.2) if and only if 𝑢𝑃𝐶[𝐽,𝐸] is a solution of the impulsive integral equation: 𝑢(𝑡)=02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝑚𝑘=1𝐺𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘+𝐻𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘,(2.2) where 𝑒𝐺(𝑡,𝑠)=2𝑀2𝜋𝑀11𝑒𝑀(2𝜋𝑡+𝑠)+𝑒𝑀(𝑡𝑠)𝑒,0𝑠𝑡2𝜋,𝑀(2𝜋+𝑡𝑠)+𝑒𝑀(𝑠𝑡)2𝑒,0𝑡𝑠2𝜋,𝐻(𝑡,𝑠)=2𝜋𝑀11𝑒𝑀(2𝜋𝑡+𝑠)𝑒𝑀(𝑡𝑠)𝑒,0𝑠𝑡2𝜋,𝑀(𝑠𝑡)𝑒𝑀(2𝜋+𝑡𝑠),0𝑡<𝑠2𝜋.(2.3) By simple calculations, we obtain that for (𝑡,𝑠)𝐽×𝐽, 𝑙0𝑒=𝜋𝑀𝑀𝑒2𝜋𝑀𝑒1𝐺(𝑡,𝑠)2𝜋𝑀+1𝑒2𝑀2𝜋𝑀1=𝑙1,||||1(2.4)𝐻(𝑡,𝑠)2,𝑀𝐺(𝑡,𝑠)+𝐻(𝑡,𝑠)>0.(2.5)

To establish the existence of multiple positive solutions in 𝑃𝐶1[𝐽,𝐸]𝐶2[𝐽,𝐸] of IBVP(1.2), let us list the following assumptions:(A1)𝑓(𝑡,𝑥)𝑔(𝑡)(𝑥), 𝑡(0,2𝜋), 𝑥𝑃, where 𝑔(0,2𝜋)(0,) is continuous and𝑃𝑃 is bounded and continuous and satisfies 02𝜋𝑔(𝑠)𝑑𝑠<+.(A2)(𝑥) in (A1) satisfies𝑐𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑙1𝑚𝑘=1𝑘+12𝑚𝑘=1𝑐𝑘<1,𝑑𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑙1𝑚𝑘=1𝑒𝑘+12𝑚𝑘=1𝑑𝑘<1,(2.6) where 𝑐=lim𝑥0(𝑥)𝑥,𝑑=lim𝑥+(𝑥),𝑐𝑥𝑘=lim𝑥0𝐼𝑘𝑥,𝑑𝑘=lim𝑥+𝐼𝑘,𝑥𝑘=lim𝑥0𝐼𝑘𝑥,𝑒𝑘=lim𝑥+𝐼𝑘.𝑥(2.7)(A3)For any 𝑟>0 and [𝑎,𝑏](0,2𝜋), 𝑓 is uniformly continuous on [𝑎,𝑏]×𝑇𝑟.(A4)There exist 𝐿,𝐿𝑘,𝐻𝑘0 such that 𝛼(𝑓(𝑡,𝐷))𝐿𝛼(𝐷), 𝛼(𝐼𝑘(𝐷))𝐿𝑘𝛼(𝐷), 𝛼(𝐼𝑘(𝐷))𝐻𝑘𝛼(𝐷) (𝑘=1,,𝑚), and 4𝜋𝐿𝑙1+𝑙1𝑚𝑘=1𝐻𝑘+(1/2)𝑚𝑘=1𝐿𝑘<1,for 𝑡(0,2𝜋), and 𝐷𝑃 is bounded.(A5)For any 𝑥𝑃, 𝐼𝑘(𝑥)𝑀𝐼𝑘(𝑥);(A6)𝑃 is a solid cone, and there exist 𝑢0𝑃, 𝐽0=[𝑎,𝑏]𝐽 such that 𝑡𝐽0, 𝑥𝑢0

imply 𝑓(𝑡,𝑥)(𝑡)𝑢0, 𝐶(𝐽0,[0,+)), and 𝑙=𝑙0𝑏𝑎(𝑠)𝑑𝑠>1.

Define an operator 𝐴 as follows: (𝐴𝑢)(𝑡)=02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝑚𝑘=1𝐺𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘+𝐻𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘,𝑡𝐽.(2.8)

Lemma 2.3. Assuming (A1) and (A4) hold, then, for any 𝑟>0, 𝐴𝑃𝐶[𝐽,𝑃]𝐵𝑟𝑃𝐶[𝐽,𝑃] is bounded and continuous.

Proof. According to (A1) and (A4), we obtain that 𝐴 is a bounded operator. In the following, we will show that 𝐴 is continuous.
Let {𝑢𝑛},{𝑢}𝑃𝐶[𝐽,𝑃]𝐵𝑟, and 𝑢𝑛𝑢𝑃𝐶0. Next we show that 𝐴𝑢𝑛𝐴𝑢𝑃𝐶0. By (A1), {(𝐴𝑢𝑛)(𝑡)} is equicontinuous on each 𝐽𝑖(𝑖=0,,𝑚). By the Lebesgue dominated convergence theorem and (2.4), we have 𝐴𝑢𝑛(𝑡)𝐴𝑢(𝑡)02𝜋𝑓𝐺(𝑡,𝑠)𝑠,𝑢𝑛+(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑚𝑘=1𝐺𝑡,𝑡𝑘𝐼𝑘𝑢𝑛𝑡𝑘𝐼𝑘𝑢𝑡𝑘+𝑚𝑘=1||𝐻𝑡,𝑡𝑘||𝐼𝑘𝑢𝑛𝑡𝑘𝐼𝑘𝑢𝑡𝑘𝑙102𝜋𝑓𝑠,𝑢𝑛(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝑙1𝑚𝑘=1𝐼𝑘𝑢𝑛𝑡𝑘𝐼𝑘𝑢𝑡𝑘+12𝑚𝑘=1𝐼𝑘𝑢𝑛𝑡𝑘𝐼𝑘𝑢𝑡𝑘0(𝑛).(2.9) In view of the Ascoli-Arzela theorem, {𝐴𝑢𝑛} is a relatively compact set in 𝑃𝐶[𝐽,𝐸]. In the following we will verify that 𝐴𝑢𝑛𝐴𝑢𝑃𝐶0(𝑛).
If this is not true, then there are 𝜀0>0 and {𝑢𝑛𝑖}{𝑢𝑛} such that 𝐴𝑢𝑛𝑖𝐴𝑢𝑃𝐶𝜀0(𝑖=1,2,). Since {𝐴𝑢𝑛} is a relatively compact set, there exists a subsequence of {𝐴𝑢𝑛𝑖} which converges to 𝑣𝑃𝐶[𝐽,𝑃], without loss of generality, and we assume that lim𝑖𝐴𝑢𝑛𝑖=𝑣, that is, lim𝑖𝐴𝑢𝑛𝑖𝑣𝑃𝐶=0, so 𝑣=𝐴𝑢, which imply a contradiction. Therefore 𝐴 is continuous.

Lemma 2.4. Assuming (A1), (A3), and (A4) hold, then, for any 𝑅>0, 𝐴𝑃𝐶[𝐽,𝑃]𝐵𝑅𝑃𝐶[𝐽,𝑃] is a strict set contraction operator.

Proof. For any 𝑅>0, 𝑆𝑃𝐶[𝐽,𝑃]𝐵𝑅, by (A1), 𝐴𝑆 is bounded and equicontinuous on each 𝐽𝑖, 𝑖=0,,𝑚, and by [17], 𝛼𝑃𝐶(𝐴𝑆)=sup𝑡𝐽𝛼((𝐴𝑆)(𝑡)),(2.10) where (𝐴𝑆)(𝑡)={𝐴𝑢(𝑡)𝑢𝑆,𝑡𝐽}.
Let 𝐷=02𝜋,𝐷𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑢𝑆𝛿=𝛿2𝜋𝛿𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑢𝑆,0<𝛿<min𝜋,𝑡1,2𝜋𝑡𝑚.(2.11) By (A1) and (2.4), for any 𝑢𝑆, 𝛿2𝜋𝛿𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑙1max𝑥𝑇𝑅(𝑢)𝛿0𝑔(𝑠)𝑑𝑠+𝑙1max𝑢𝑇𝑅(𝑥)2𝜋2𝜋𝛿𝑔(𝑠)𝑑𝑠.(2.12) In view of (2.12) and (A1), we have 𝑑𝐻(𝐷𝛿,𝐷)0(𝛿0+), where 𝑑𝐻(𝐷𝛿,𝐷) denotes the Hausdorff distance of 𝐷 and 𝐷𝛿.
Therefore, lim𝛿0+𝛼𝐷𝛿=𝛼(𝐷).(2.13) Next we will estimate 𝛼(𝐷𝛿). Since 𝛿2𝜋𝛿𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠(2𝜋2𝛿)[]co({𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑠𝛿,2𝜋𝛿}),(2.14) thus 𝛼𝐷𝛿=𝛼𝛿2𝜋𝛿𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑢𝑆2(𝜋𝛿)𝛼[][]co{𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑠𝛿,2𝜋𝛿,𝑢𝑆}2𝜋𝛼({𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑠𝛿,2𝜋𝛿,𝑢𝑆})2𝜋𝑙1𝛼𝑓𝐼𝛿𝐼×𝑆𝛿,(2.15) where 𝐼𝛿=[𝛿,2𝜋𝛿], 𝑆(𝐼𝛿)={𝑢(𝑡)𝑡𝐼𝛿,𝑢𝑆}.
By (A3) and (A4), it is not difficult to prove that 𝛼𝑓𝐼𝛿𝐼×𝑆𝛿=max𝑡𝐼𝛿𝛼𝑓𝐼𝑡,𝑆𝛿𝑆𝐼𝐿𝛼𝛿𝐿𝛼(𝑆(𝐽)).(2.16) By [17], we have 𝐿𝛼(𝑆(𝐽))2𝐿𝛼𝑃𝐶(𝑆).(2.17) Let 𝛿0+, and making use of the fact that lim𝛿0+𝛼(𝐷𝛿)=𝛼(𝐷), we obtain 𝛼(𝐷)2𝜋𝑙12𝐿𝛼𝑃𝐶(𝑆)=4𝜋𝐿𝑙1𝛼𝑃𝐶(𝑆).(2.18) It is clear that 𝛼𝑚𝑘=1𝐺𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘𝑙𝑢𝑆1𝑚𝑘=1𝐻𝑘𝛼𝑃𝐶𝛼(𝑆),(2.19)𝑚𝑘=1𝐻𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘1𝑢𝑆2𝑚𝑘=1𝐿𝑘𝛼𝑃𝐶(𝑆).(2.20) Hence, according to (2.18)–(2.20), we have 𝛼𝑃𝐶(𝐴𝑆)4𝜋𝐿𝑙1+𝑙1𝑚𝑘=1𝐻𝑘+12𝑚𝑘=1𝐿𝑘𝛼𝑃𝐶(𝑆).(2.21)
By (A4) and Lemma 2.3, 𝐴 is a strict set contraction operator from 𝑃𝐶[𝐽,𝑃] into 𝑃𝐶[𝐽,𝑃].

3. Main Result

Theorem 3.1. Assuming that (A1)–(A6) hold, then the IBVP (1.2) has at least two positive solutions 𝑢1 and 𝑢2 satisfying 𝑢1(𝑡)𝑙𝑢0(𝑡),for𝑡𝐽0=𝑎,𝑏,𝑙>1,(3.1) where 𝑙 was specified in (A6).

Proof. First we verify that there exists 𝛿>0 such that 𝑣𝛿 for 𝑣𝑢0. If this is not true, then there exists {𝑣𝑛}𝐸which satisfies 𝑣𝑛𝑢0 and 𝑣𝑛<(1/𝑛)(𝑛=1,2,), so we have 𝑢0𝜃, which is a contradiction with 𝑢0𝑃.
By (A2), there exist 𝑐>𝑐, 𝑐𝑘>𝑐𝑘, 𝑑>𝑑, 𝑑𝑘>𝑑𝑘, 𝑘>𝑘, and 𝑒𝑘>𝑒𝑘, and 0<𝑟1<𝛿,𝑟2𝑢>max𝛿,𝑙0(3.2) satisfy 𝑐𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑙1𝑚𝑘=1𝑘+12𝑚𝑘=1𝑐𝑘<1,(3.3)𝑏=𝑑𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑙1𝑚𝑘=1𝑒𝑘+12𝑚𝑘=1𝑑𝑘<1.(3.4) For 𝑥𝑇𝑟1𝑃, (𝑥)𝑐𝐼𝑥,𝑘(𝑥)𝑐𝑘𝑥,𝐼𝑘(𝑥)𝑘𝑥.(3.5) For 𝑥𝑟2 and 𝑥𝑃, (𝑥)𝑑𝐼𝑥,𝑘(𝑥)𝑑𝑘𝑥,𝐼𝑘(𝑥)𝑒𝑘𝑥.(3.6) Therefore, for any 𝑥𝑃, we have (𝑥)𝑑𝑥+𝑀,𝐼𝑘(𝑥)𝑑𝑘𝑥+𝑀,𝐼𝑘(𝑥)𝑒𝑘𝑥+𝑀,(3.7) where 𝑀𝑀=max0,𝑀1,,𝑀𝑚,𝐾1,,𝐾𝑚,𝑀0=sup(𝑥)𝑥𝑇𝑟2,𝑀𝑃𝑘𝐼=sup𝑘(𝑥)𝑥𝑇𝑟2𝑃,𝐾𝑘=sup𝐼𝑘(𝑥)𝑥𝑇𝑟2𝑃(𝑘=1,2,,𝑚).(3.8)
Let 𝑟3=𝑟2+(1𝑏)1𝐺, 𝐺=𝑀[𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑚𝑙1+𝑚/2], 𝑈1={𝑢𝑃𝐶[𝐽,𝑃]𝑢𝑃𝐶<𝑟1}, 𝑈2={𝑢𝑃𝐶[𝐽,𝑃]𝑢𝑃𝐶<𝑟3}, 𝑈3={𝑢𝑃𝐶[𝐽,𝑃]𝑢𝑃𝐶<𝑟3, 𝑢(𝑡)𝑙𝑢0 for 𝑡𝐽0 and 𝑙>1}. It is clear that 𝑈1,𝑈2,𝑈3 are nonempty, bounded, and convex open sets in 𝑃𝐶[𝐽,𝑃], and 𝑈1=𝑃𝐶[𝐽,𝑃]𝐵𝑟1, 𝑈2=𝑃𝐶[𝐽,𝑃]𝐵𝑟3, and 𝑈3={𝑢𝑈2𝑢(𝑡)𝑙𝑢0,𝑡𝐽0}.
From (3.2), we obtain 𝑈1𝑈2,𝑈3𝑈2,𝑈1𝑈3=.(3.9) According to Lemma 2.4, 𝐴𝑈2𝑃𝐶[𝐽,𝑃] is a strict set contraction operator, and for 𝑢𝑈2, by (2.4) and (3.7), we obtain (𝐴𝑢)(𝑡)=02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝑚𝑘=1𝐺𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘+𝐻𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘𝑙102𝜋𝑔(𝑠)𝑑𝑠(𝑢)+𝑚𝑘=1𝑙1𝐼𝑘+12𝐼𝑘𝑙102𝜋𝑑𝑔(𝑠)𝑑𝑠𝑢+𝑀+𝑙1𝑚𝑘=1𝑒𝑘𝑢+𝑀+12𝑚𝑘=1𝑑𝑘𝑢+𝑀=𝑑𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑙1𝑚𝑘=1𝑒𝑘+12𝑚𝑘=1𝑑𝑘𝑢+𝑀𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑚𝑙1+𝑚2=𝑏𝑢+𝐺𝑏𝑟3+𝐺<𝑟3.(3.10) Hence 𝐴𝑈2𝑈2.(3.11)
Similarly, 𝐴𝑈1𝑃𝐶[𝐽,𝑃] is a strict set contraction operator, and for 𝑢𝑈1, by (3.3) and (3.5), we obtain (𝐴𝑢)(𝑡)=02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝑚𝑘=1𝐺𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘+𝐻𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘𝑐𝑙102𝜋𝑔(𝑠)𝑑𝑠𝑢+𝑚𝑘=1𝑙1𝑘1𝑢+2𝑐𝑘=𝑐𝑢𝑙102𝜋𝑔(𝑠)𝑑𝑠+𝑙1𝑚𝑘=1𝑘+12𝑚𝑘=1𝑐𝑘𝑢<𝑢𝑟1,(3.12) so 𝐴𝑈1𝑈1.(3.13)
Let 𝑢𝑈3, by (3.11), we have 𝐴𝑢𝑃𝐶<𝑟3.
By (2.5), (A5), and (A6), for 𝑡𝐽0, (𝐴𝑢)(𝑡)=02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝑚𝑘=1𝐺𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘+𝐻𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝑚𝑘=1𝑀𝐺𝑡,𝑡𝑘+𝐻𝑡,𝑡𝑘𝐼𝑘𝑢𝑡𝑘02𝜋𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑙0𝑏𝑎𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑙0𝑏𝑎(𝑠)𝑑𝑠𝑢0=𝑙𝑢0.(3.14) So 𝐴𝑢𝑈3, and 𝐴𝑈3𝑈3.(3.15) According to (3.11)–(3.15) and Lemma 2.1, we have 𝑖𝐴,𝑈𝑗[],𝑃𝐶𝐽,𝑃=1(𝑗=1,2,3).(3.16) Hence 𝑖𝐴,𝑈2𝑈1𝑈3[],𝑃𝐶𝐽,𝑃=𝑖𝐴,𝑈2[],𝑃𝐶𝐽,𝑃𝑖𝐴,𝑈1[],𝑃𝐶𝐽,𝑃𝑖𝐴,𝑈3[],𝑃𝐶𝐽,𝑃=1.(3.17)
Thus, 𝐴 has two fixed points 𝑢1 and 𝑢2 in 𝑈3 and 𝑈2(𝑈1𝑈3), respectively, which means 𝑢1(𝑡) and 𝑢2(𝑡) are positive solution of the IBVP (1.2), where 𝑢1(𝑡)𝑙𝑢0, for 𝑡𝐽0 and 𝑙>1.

4. Example

To illustrate how our main result can be used in practice, we present an example.

Example 4.1. Consider the following problem: 𝑥𝑛(𝑡)+4𝑥𝑛=1𝑡32+999𝜋𝑡ln1+𝑥2𝑛+1+𝑥𝑛3,𝑡𝐽,Δ𝑥𝑛||𝑡=1/3=18𝑥𝑛13,Δ𝑥𝑛||𝑡=1/31=4𝑥𝑛13,𝑥𝑛(0)=𝑥𝑛(2𝜋),𝑥𝑛(0)=𝑥𝑛(2𝜋),(4.1) where 𝑥𝑚+𝑛=𝑥𝑛 (𝑛=1,2,,𝑚).

Conclusion
IBVP (4.1) has at least two positive solutions {𝑥1𝑛(𝑡)} and {𝑥2𝑛(𝑡)} such that 𝑥1𝑛(𝑡)>1 for 𝑡[𝜋,2𝜋], 𝑛=1,2,,𝑚.

Proof. Let 𝐽=[0,2𝜋], 𝐸=𝑚={𝑥=(𝑥1,𝑥2,,𝑥𝑚)𝑥𝑛,𝑛=1,2,,𝑚}; then, 𝐸 is a Banach space with norm 𝑥=max1𝑛𝑚|𝑥𝑛|. Let 𝑃={(𝑥1,𝑥2,,𝑥𝑚)𝑥𝑛0,𝑛=1,2,,𝑚}; then, 𝑃 is a solid cone in 𝐸. Compared to IBVP (1.2), 𝑓(𝑡,𝑥)=(𝑓1,𝑓2,,𝑓𝑚), 𝑓𝑛(𝑡,𝑥)=(1/𝑡)[3(2+(999/𝜋)𝑡)ln(1+𝑥2𝑛+1)+𝑥𝑛/3)] is singular at 𝑡=0. 𝐼(𝑥)=(𝐼1(𝑥),𝐼2(𝑥),,𝐼𝑚(𝑥)), and 𝐼𝑛(𝑥)=(1/8)𝑥𝑛(1/3). 𝐼(𝑥)=(𝐼1(𝑥),,𝐼𝑚(𝑥)), and 𝐼𝑛(𝑥)=(1/4)𝑥𝑛(1/3), 𝑛=1,2,,𝑚.
Next we will verify that the conditions in Theorem 3.1 are satisfied.
Let 𝑔(𝑡)=1/𝑡, (𝑥)=(1(𝑥),2(𝑥),,𝑚(𝑥)), and 𝑛(𝑥)=6000ln(1+𝑥2𝑛+1)+𝑥𝑛/3. It is clear that 𝑓(𝑡,𝑥)𝑔(𝑡)(𝑥), for 𝑡(0,2𝜋) and 𝑥𝐸, so (A1) is satisfied.
By simple calculations, we have 𝑀=2, 𝑐=𝑑=1/3, 𝑐1=𝑑1=1/8, 1=𝑒1=1/4, 02𝜋𝑔(𝑠)𝑑𝑠=5.01326, 𝑙0=0.00093, and 𝑙1=0.25. Hence, 𝑙102𝜋𝑔(𝑠)𝑑𝑠𝑐+𝑙11+(1/2)𝑐1<1; that is, (A2) is satisfied.
Since 𝐸 is a finite-dimensional space, it is obvious that (A3) and (A4) are satisfied.
It is clear that 𝐼𝑛(𝑥)=(1/4)𝑥𝑛(1/3) and 𝑀𝐼𝑛(𝑥)=2×(1/8)𝑥𝑛(1/3)=(1/4)𝑥𝑛(1/3), so 𝐼𝑛(𝑥)=𝑀𝐼𝑛(𝑥); that is, (A5) is satisfied.
Let 𝑢0=(1,1,,1)𝑃 and 𝐽0=[𝜋,2𝜋][0,2𝜋]; for 𝑡𝐽0 and 𝑥𝑢0, we have 𝑓𝑛1(𝑡,𝑥)=𝑡32+999𝜋𝑡ln1+𝑥2𝑛+1+𝑥𝑛3>3000ln2𝑡.(4.2) Let (𝑡)=3000ln2/𝑡; then, for 𝑡𝐽0 and 𝑥𝑢0, we obtain that 𝑓(𝑡,𝑥)(𝑡)𝑢0 and 𝑙0𝜋2𝜋(𝑠)𝑑𝑠>1. Therefore (A6) is satisfied.
By Theorem 3.1, IBVP (4.1) has at least two positive solutions {𝑥1𝑛(𝑡)} and {𝑥2𝑛(𝑡)} and satisfies 𝑥1𝑛(𝑡)>1, 𝑛=1,2,,𝑚.

Acknowledgments

This work is supported by the NNSF of China (no. 10871160) and project of NWNU-KJCXGC-3-47.