MultiPoint BVPs for Second-Order Functional Differential Equations with Impulses

Abstract

This paper is concerned about the existence of extreme solutions of multipoint boundary value problem for a class of second-order impulsive functional differential equations. We introduce a new concept of lower and upper solutions. Then, by using the method of upper and lower solutions introduced and monotone iterative technique, we obtain the existence results of extreme solutions.

1. Introduction

In this paper, we consider the multipoint boundary value problems for the impulsive functional differential equation: −𝑢^(𝑡)Δ=𝑢_𝑓(𝑡,𝑢(𝑡),𝑢(𝜃(𝑡))),𝑡∈𝐽=[0,1],𝑡≠𝑡_𝑘,𝑢(0)−𝑎𝑢_𝑡_𝑘=𝐼_𝑘𝑢𝑡_𝑘,𝑘=1,…,𝑚,(0)=𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)=𝑑𝑢(𝜉),(1.1) where 𝑓∈𝐶(𝐽×𝐑²,𝐑),    0≤𝜃(𝑡)≤𝑡,𝑡∈𝐽,𝜃∈𝐶(𝐽),𝑎≥0,𝑏≥0,0≤𝑐≤1,0≤𝑑≤1, 0<𝜂,𝜉<1. 0<𝑡₁<𝑡₂<⋯<𝑡_𝑚<1, 𝑓 is continuous everywhere except at {𝑡_𝑘}×𝑅²; 𝑓(𝑡^+𝑘,⋅,⋅), and 𝑓(𝑡^−𝑘,⋅,⋅) exist with 𝑓(𝑡^−𝑘,⋅,⋅)=𝑓(𝑡_𝑘,⋅,⋅); 𝐼_𝑘∈𝐶(𝑅,𝑅),Δ𝑢^(𝑡_𝑘)=𝑢^(𝑡^+𝑘)−𝑢^(𝑡^−𝑘). Denote 𝐽⁻=𝐽⧵{𝑡_𝑖,𝑖=1,2,…,𝑚}. Let 𝑃𝐶(𝐽,𝑅)={𝑢∶𝐽→𝑅;𝑢(𝑡)|_𝐽⁻ is continuous, 𝑢(𝑡^+𝑘) and 𝑢(𝑡^−𝑘) exist with 𝑢(𝑡^−𝑘)=𝑢(𝑡_𝑘),𝑘=1,2,…,𝑚}; 𝑃𝐶¹(𝐽,𝑅)={𝑢∶𝐽→𝑅;𝑢(𝑡)|_𝐽⁻ is continuous differentiable, 𝑢^(𝑡^+𝑘) and 𝑢^(𝑡^−𝑘) exist with 𝑢^(𝑡^−𝑘)=𝑢^(𝑡_𝑘),𝑘=1,2,…,𝑚}. Let 𝐸=𝑃𝐶¹(𝐽,𝑅)∩𝐶²(𝐽,𝑅). A function 𝑢∈𝐸 is called a solution of BVP(1.1) if it satisfies (1.1).

The method of upper and lower solutions combining monotone iterative technique offers an approach for obtaining approximate solutions of nonlinear differential equations [1–3]. There exist much literature devoted to the applications of this technique to general boundary value problems and periodic boundary value problems, for example, see [1, 4–6] for ordinary differential equations, [7–11] for functional differential equations, and [12] for differential equations with piecewise constant arguments. For the studies about some special boundary value problems, for example, Lidston boundary value problems and antiperiodic boundary value problems, one may see [13, 14] and the references cited therein.

Here, we hope to mention some papers where existence results of solutions of certain boundary value problems of impulsive differential equations were studied [11, 15] and certain multipoint boundary value problems also were studied [6, 16–21]. These works motivate that we study the multipoint boundary value problems for the impulsive functional differential equation (1.1).

We also note that when 𝐼_𝑘=0 and 𝜃(𝑡)=𝑡, the boundary value problem (1.1) reduces to multi-point boundary value problems for ordinary differential equations which have been studied in many papers, see, for example, [6, 16–18] and the references cited therein. To our knowledge, only a few papers paid attention to multi-point boundary value problems for impulsive functional differential equations.

In this paper, we are concerned with the existence of extreme solutions for the boundary value problem (1.1). The paper is organized as follows. In Section 2, we establish two comparison principles. In Section 3, we consider a linear problem associated to (1.1) and then give a proof for the existence theorem. In Section 4, we first introduce a new concept of lower and upper solutions. By using the method of upper and lower solutions with a monotone iterative technique, we obtain the existence of extreme solutions for the boundary value problem (1.1).

2. Comparison Principles

In the following, we always assume that the following condition (𝐻) is satisfied:

(𝐻)  𝑎≥0,𝑏≥0,0≤𝑐≤1,0≤𝑑≤1,0<𝜂,𝜉<1,𝑎+𝑐>0,𝑏+𝑑>0.

For any given function 𝑔∈𝐸, we denote 𝐴_𝑔=max𝑔(0)−𝑎𝑔^(0)−𝑐𝑔(𝜂)𝑎𝜋+𝑐sin𝜋𝜂,𝑔(1)+𝑏𝑔^(1)−𝑑𝑔(𝜉)𝐵𝑏𝜋+𝑑sin𝜋𝜉_𝑔,=max𝐴_𝑔,0,𝑐_𝑔(𝑡)=𝐵_𝑔sin(𝜋𝑡),𝑟=𝜋².(2.1) We now present main results of this section.

Theorem 2.1. Assume that  𝑢∈𝐸 satisfies −𝑢^(𝑡)Δ+𝑢_𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))≤0,𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑢(0)−𝑎𝑢_𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘,𝑘=1,…,𝑚,(0)≤𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)≤𝑑𝑢(𝜉),(2.2) where 𝑎≥0,𝑏≥0, 0≤𝑐≤1,0≤𝑑≤1,0<𝜂,  𝜉<1, 𝐿_𝑘≥0 and constants 𝑀,𝑁 satisfy 𝑀>0,𝑁≥0,𝑀+𝑁2+^𝑚_𝑘=1𝐿_𝑘≤1.(2.3) Then 𝑢(𝑡)≤0 for 𝑡∈𝐽.

Proof. Suppose, to the contrary, that 𝑢(𝑡)>0 for some 𝑡∈𝐽.
If 𝑢(1)=max_𝑡∈𝐽𝑢(𝑡)>0, then 𝑢^(1)≥0, 𝑢(1)≥𝑢(𝜉), and 𝑑𝑢(𝜉)≤𝑢(1)≤𝑢(1)+𝑏𝑢^(1)≤𝑑𝑢(𝜉).(2.4) So 𝑑=1 and 𝑢(𝜉) is a maximum value.
If 𝑢(0)=max_𝑡∈𝐽𝑢(𝑡)>0, then 𝑢^(0)≤0, 𝑢(0)≥𝑢(𝜂), and 𝑐𝑢(𝜂)≤𝑢(0)≤𝑢(0)−𝑏𝑢^(0)≤𝑐𝑢(𝜂).(2.5) So 𝑐=1 and 𝑢(𝜂) is a maximum value.
Therefore, there is a 𝜌∈(0,1) such that 𝑢(𝜌)=max_𝑡∈𝐽𝑢(𝑡)>0,𝑢^𝜌⁺≤0.(2.6)
Suppose that 𝑢(𝑡)≥0 for 𝑡∈𝐽. From the first inequality of (2.2), we obtain that 𝑢^(𝑡)≥0 for 𝑡∈𝐽. Hence 𝑢(0)=max_𝑡∈𝐽𝑢(𝑡)or𝑢(1)=max_𝑡∈𝐽𝑢(𝑡).(2.7)
If 𝑢^(0)≥0, then 𝑢^(𝑡)≥0, 𝑡∈(𝑡_𝑖,𝑡_𝑖+1], it is easy to obtain that 𝑢(𝑡) is nondecreasing. Since 𝑢(1)≤𝑑𝑢(𝜉)≤𝑢(1), it follows that 𝑢(𝑡)≡𝐾 (𝐾>0) for 𝑡∈[𝜉,1]. From the first inequality of (2.2), we have that when 𝑡∈[𝜉,1],0<𝑀𝐾≤𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))≤𝑢^(𝑡)=0,(2.8) which is a contradiction.
If 𝑢^(0)≤0, then 𝑢(0)=max_𝑡∈𝐽𝑢(𝑡)>0, or 𝑢(1)=max_𝑡∈𝐽𝑢(𝑡)>0. If 𝑢(0)=max_𝑡∈𝐽𝑢(𝑡)>0, then 𝑢(𝑡)≡𝐾 (𝐾>0) for 𝑡∈[0,𝜂]. If 𝑢(1)=max_𝑡∈𝐽𝑢(𝑡)>0, then 𝑢(𝑡)≡𝐾 for 𝑡∈[𝜉,1].
From the first inequality of (2.2), we have that when 𝑡∈[𝜉,1],0<𝑀𝐾≤𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))≤𝑢^(𝑡)=0,(2.9) which is a contradiction.
Suppose that there exist 𝑡₁,𝑡₂∈𝐽 such that 𝑢(𝑡₁)>0 and 𝑢(𝑡₂)<0. We consider two possible cases.Case 1 (𝑢(0)>0). Since 𝑢(𝑡₂)<0, there is 𝜅>0,𝜀>0 such that 𝑢(𝜅)=0, 𝑢(𝑡)≥0 for 𝑡∈[0,𝜅) and 𝑢(𝑡)<0 for all 𝑡∈(𝜅,𝜅+𝜀]. It is easy to obtain that 𝑢^(𝑡)≥0 for 𝑡∈[0,𝜅]. If 𝑡^∗<𝜅, then 0<𝑀𝑢(𝑡^∗)≤𝑢^(𝑡^∗)≤0, a contradiction. Hence 𝑡^∗>𝜅+𝜀. Let 𝑡_∗∈[0,𝑡^∗) such that 𝑢(𝑡_∗)=min_{𝑡∈[0,𝑡∗)}𝑢(𝑡), then 𝑢(𝑡_∗)<0. From the first inequality of (2.2), we have 𝑢^(𝑡)Δ≥𝑢_(𝑀+𝑁)𝑢𝑡_∗,𝑡∈0,𝑡^∗,𝑡≠𝑡_𝑘,𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘,𝑘=1,…,𝑚.(2.10) Integrating the above inequality from 𝑠(𝑡_∗≤𝑠≤𝑡^∗) to 𝑡^∗, we obtain 𝑢^𝑡^∗−𝑢^(𝑠)≥≥𝑡𝑡^∗_∗−𝑠(𝑀+𝑁)𝑢𝑡_∗+_{𝑠<𝑡𝑘<𝑡∗}𝐿_𝑘𝑢𝑡_𝑘−𝑠(𝑀+𝑁)𝑢𝑡_∗+^𝑚_𝑘=1𝐿_𝑘𝑢𝑡_∗.(2.11) Hence −𝑢^(𝑠)≥[𝑡^∗−𝑠(𝑀+𝑁)+^𝑚_𝑘=1𝐿_𝑘𝑢𝑡_∗,𝑡_∗≤𝑠≤𝑡^∗,(2.12) and then integrate from 𝑡_∗ to 𝑡^∗ to obtain −𝑢𝑡_∗≤<𝑢_{𝑡∗𝑡∗}𝑡^∗−𝑢𝑡_∗≤−𝑠𝑀−+𝑡^∗𝑁(𝑀+𝑁)𝑢𝑡_∗𝑑𝑠−^𝑚_𝑘=1𝐿_𝑘𝑢𝑡_∗2𝑡^∗−𝑡_∗²+^𝑚_𝑘=1𝐿_𝑘𝑢𝑡^∗≤−𝑀+𝑁2+^𝑚_𝑘=1𝐿_𝑘𝑢𝑡^∗.(2.13) From (2.3), we have that 𝑢(𝑡_∗)>0. This is a contradiction.Case 2 (𝑢(0)≤0). Let 𝑡_∗∈[0,𝑡^∗) such that 𝑢(𝑡_∗)=min_{𝑡∈[0,𝑡∗)}𝑢(𝑡)≤0. From the first inequality of (2.2), we have 𝑢^(𝑡)Δ≥𝑢_(𝑀+𝑁)𝑢𝑡_∗,𝑡∈0,𝑡^∗,𝑡≠𝑡_𝑘,𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘,𝑘=1,…,𝑚.(2.14) The rest proof is similar to that of Case 1. The proof is complete.

Theorem 2.2. Assume that (𝐻) holds and 𝑢∈𝐸 satisfies −𝑢^(𝑡)+𝑀𝑢(𝑡Δ)+𝑢_𝑁𝑢(𝜃(𝑡))+(𝑀+𝑟)𝑐_𝑢(𝑡)+𝑁𝑐_𝑢(𝜃(𝑡))≤0,𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘+𝐿_𝑘𝑐_𝑢𝑡_𝑘,𝑘=1,…,𝑚,(2.15) where constants 𝑀,𝑁 satisfy (2.3), and 𝐿_𝑘≥0, then 𝑢(𝑡)≤0 for 𝑡∈𝐽.

Proof. Assume that 𝑢(0)−𝑎𝑢^(0)≤𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)≤𝑑𝑢(𝜉), then 𝑐_𝑢(𝑡)≡0. By Theorem 2.1, 𝑢(𝑡)≤0.
Assume that 𝑢(0)−𝑎𝑢^(0)≤𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)>𝑑𝑢(𝜉), then
𝑐_𝑢(𝑡)=sin(𝜋𝑡)𝑏𝜋+𝑑sin(𝜋𝜉)𝑢(1)+𝑏𝑢^(1)−𝑑𝑢(𝜉).(2.16)
Put 𝑦(𝑡)=𝑢(𝑡)+𝑐_𝑢(𝑡),𝑡∈𝐽, then 𝑦(𝑡)≥𝑢(𝑡) for all 𝑡∈𝐽, and
𝑦^(𝑡)=𝑢^(𝑡)+𝜋cos(𝜋𝑡)𝑏𝜋+𝑑sin(𝜋𝜉)𝑢(1)+𝑏𝑢^𝑦(1)−𝑑𝑢(𝜉),𝑡∈𝐽,_(𝑡)=𝑢^(𝑡)−𝑟𝑐_𝑢(𝑡),𝑡∈𝐽.(2.17) Hence 𝑦𝑦((0𝜉))==𝑢𝑢((𝜉0)),+𝑦s(i1n)(=𝜋𝜉𝑢)(1),𝑏𝜋+𝑑sin(𝜋𝜉)𝑢(1)+𝑏𝑢^𝑦(1)−𝑑𝑢(𝜉)_,(0)=𝑢^(0)+𝜋𝑏𝜋+𝑑sin(𝜋𝜉)𝑢(1)+𝑏𝑢^𝑦(1)−𝑑𝑢(𝜉)_,(1)=𝑢^(1)−𝜋𝑏𝜋+𝑑sin(𝜋𝜉)𝑢(1)+𝑏𝑢^−𝑦_(1)−𝑑𝑢(𝜉),𝑦(0)−(𝑡)𝑎𝑦+_𝑀𝑦(𝑡)+𝑁𝑦(𝜃(𝑡))=−𝑢^(𝑡)+𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))+(𝑀+𝑟)𝑐_𝑢(𝑡)+𝑁𝑐_𝑢(𝜃(𝑡))≤0,(0)=𝑢(0)−𝑎𝑢^(0)−𝑎𝜋𝑏𝜋+𝑑sin(𝜋𝜉)𝑢(1)+𝑏𝑢^𝑦(1)+𝑏𝑦_(1)−𝑑𝑢(𝜉)≤𝑐𝑢(𝜂)≤𝑐𝑦(𝜂),=𝑢(1)(1+)−𝑏𝑢𝑑_𝑦(𝜉)(1)−𝑑𝑢(𝜉)−𝑏𝜋𝑏𝜋+𝑑sin(𝜋𝜉)𝑢(1)+𝑏𝑢^−𝑑sin(𝜋𝜉)(1)−𝑑𝑢(𝜉)𝑏𝜋+𝑑sin𝜋𝜉𝑢(1)+𝑏𝑢^Δ𝑦_(1)−𝑑𝑢(𝜉)≤0,𝑡_𝑘=Δ𝑢^𝑡_𝑘Δ𝑐^𝑢𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘+𝐿_𝑘𝑐_𝑢𝑡_𝑘=𝐿_𝑘𝑦𝑡_𝑘.(2.18) By Theorem 2.1, 𝑦(𝑡)≤0 for all 𝑡∈𝐽, which implies that 𝑢(𝑡)≤0 for 𝑡∈𝐽.
Assume that 𝑢(0)−𝑎𝑢^(0)>𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)≤𝑑𝑢(𝜉), then
𝑐_𝑢(𝑡)=sin𝜋𝑡𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^(0)−𝑐𝑢(𝜂).(2.19) Put 𝑦(𝑡)=𝑢(𝑡)+𝑐_𝑢(𝑡),𝑡∈𝐽, then 𝑦(𝑡)≥𝑢(𝑡) for all 𝑡∈𝐽, and 𝑦^(𝑡)=𝑢^(𝑡)+𝜋cos(𝜋𝑡)𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^𝑦(0)−𝑐𝑢(𝜂),𝑡∈𝐽,_(𝑡)=𝑢^(𝑡)−𝑟𝑐_𝑢(𝑡),𝑡∈𝐽.(2.20) Hence 𝑦𝑦((0𝜂))==𝑢𝑢((0𝜂)),+𝑦(s1i)n=(𝜋𝑢𝜂()1),𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^𝑦(0)−𝑐𝑢(𝜂)_,(0)=𝑢^(0)+𝜋𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^𝑦(0)−𝑐𝑢(𝜂)_,(1)=𝑢^(1)−𝜋𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^−𝑦_(0)−𝑐𝑢(𝜂),𝑦(0)−(𝑡)𝑎𝑦+_𝑀𝑦(𝑡)+𝑁𝑦(𝜃(𝑡))=−𝑢^(𝑡)+𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))+(𝑀+𝑟)𝑐_𝑢(𝑡)+𝑁𝑐_𝑢(𝜃(𝑡))≤0,=𝑢(0)(0−)𝑎−𝑢𝑐_𝑦(𝜂)(0)−𝑐𝑢(𝜂)−𝑎𝜋𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^−𝑐sin(𝜋𝜂)(0)−𝑐𝑢(𝜂)𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^𝑦(1)+𝑏𝑦_(0)−𝑐𝑢(𝜂)≤0,(1)=𝑢(1)+𝑏𝑢^(1)−𝑏𝜋𝑎𝜋+𝑐sin(𝜋𝜂)𝑢(0)−𝑎𝑢^Δ𝑦_(0)−𝑐𝑢(𝜂)≤𝑑𝑢(𝜉)≤𝑑𝑦(𝜉),𝑡_𝑘=Δ𝑢^𝑡_𝑘+Δ𝑐^𝑢𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘+𝐿_𝑘𝑐_𝑢𝑡_𝑘=𝐿_𝑘𝑦𝑡_𝑘.(2.21)
By Theorem 2.1, 𝑦(𝑡)≤0 for all 𝑡∈𝐽, which implies that 𝑢(𝑡)≤0 for 𝑡∈𝐽.
Assume that 𝑢(0)−𝑎𝑢^(0)>𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)>𝑑𝑢(𝜉), then 𝑐_𝑢(𝑡)=𝐴_𝑢sin(𝜋𝑡).
Put 𝑦(𝑡)=𝑢(𝑡)+𝑐_𝑢(𝑡),𝑡∈𝐽, then 𝑦(𝑡)≥𝑢(𝑡) for all 𝑡∈𝐽, and 𝑦^(𝑦𝑡)=𝑢^(𝑡)+𝐴_𝑢𝜋cos(𝜋𝑡),𝑡∈𝐽,_(𝑡)=𝑢^(𝑡)−𝑟𝑐_𝑢(𝑡),𝑡∈𝐽.(2.22) Hence 𝑦𝑦((0𝜂))==𝑢𝑢((0𝜂)),+𝐴_𝑢𝑦(1)=𝑢(1),𝑦sin(𝜋𝜂),𝑦(𝜉)=𝑢(𝜉)+𝐴_𝑢sin(𝜋𝜉),_−(𝑦0)_=𝑢^(0)+𝐴_𝑢𝜋,𝑦^(1)=𝑢^(1)−𝐴_𝑢𝜋,=(𝑡−)+𝑢_𝑀𝑦(𝑡)+𝑁𝑦(𝜃(𝑡))𝑦(0)−𝑎𝑦_(𝑡)+𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))+(𝑀+𝑟)𝑐_𝑢(𝑡)+𝑁𝑐_𝑢(𝜃(𝑡))≤0,𝑦(1)+𝑏𝑦_(0)−𝑐𝑦(𝜂)=𝑢(0)−𝑎𝑢^(0)−𝑐𝑢(𝜂)−𝑎𝐴_𝑢𝜋−𝑐𝐴_𝑢sin(𝜋𝜂)≤0,Δ𝑦_(1)−𝑑𝑦(𝜉)=𝑢(1)+𝑏𝑢^(1)−𝑑𝑢(𝜉)−𝑏𝐴_𝑢𝜋−𝑑𝐴_𝑢sin(𝜋𝜉)≤0,𝑡_𝑘=Δ𝑢^𝑡_𝑘+Δ𝑐^𝑢𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘+𝐿_𝑘𝑐_𝑢𝑡_𝑘=𝐿_𝑘𝑦𝑡_𝑘.(2.23)
By Theorem 2.1, 𝑦(𝑡)≤0 for all 𝑡∈𝐽, which implies that 𝑢(𝑡)≤0 for 𝑡∈𝐽. The proof is complete.

3. Linear Problem

In this section, we consider the linear boundary value problem −𝑢^(Δ𝑡)𝑢+_𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))=𝜎(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑢(0)−𝑎𝑢_𝑡_𝑘=𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚,(0)=𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)=𝑑𝑢(𝜉).(3.1)

Theorem 3.1. Assume that (𝐻) holds, 𝜎∈𝐶(𝐽), 𝑒_𝑘∈𝑅, and constants 𝑀,𝑁 satisfy (2.3) with 𝜇=𝑎(1+2𝑏)2(𝑎+𝑏+1)+18𝑎1++𝑏2+𝑏1²(𝑀+𝑁)+1+𝑎(1++𝑏𝑏+)²1^𝑚_𝑘=1𝐿_𝑘<1.(3.2) Further suppose that there exist 𝛼,𝛽∈𝐸 such that (ℎ₁)𝛼≤𝛽 on 𝐽,(ℎ₂)−𝛼^(𝑡)+𝑀𝛼(𝑡Δ)𝛼+_𝑁𝛼(𝜃(𝑡))+(𝑀+𝑟)𝑐_𝛼(𝑡)+𝑁𝑐_𝛼(𝜃(𝑡))≤𝜎(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘≥𝐿_𝑘𝛼𝑡_𝑘+𝐿_𝑘𝑐_𝛼𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚,(3.3)(ℎ₃)−𝛽^(𝑡)+𝑀𝛽(𝑡Δ)+𝛽_𝑁𝛽(𝜃(𝑡))−(𝑀+𝑟)𝑐_−𝛽(𝑡)+𝑁𝑐_−𝛽(𝜃(𝑡))≥𝜎(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘≤𝐿_𝑘𝛽𝑡_𝑘−𝐿_𝑘𝑐_−𝛽𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚.(3.4)Then the boundary value problem (3.1) has one unique solution 𝑢(𝑡) and 𝛼≤𝑢≤𝛽 for 𝑡∈𝐽.

Proof. We first show that the solution of (3.1) is unique. Let 𝑢₁,𝑢₂ be the solution of (3.1) and set 𝑣=𝑢₁−𝑢₂. Thus, −𝑣^(𝑡Δ)+𝑣_𝑀𝑣(𝑡)+𝑁𝑣(𝜃(𝑡))=0,𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑣(0)−𝑎𝑣_𝑡_𝑘=𝐿_𝑘𝑣𝑡_𝑘,𝑘=1,…,𝑚,(0)=𝑐𝑣(𝜂),𝑣(1)+𝑏𝑣^(1)=𝑑𝑣(𝜉).(3.5) By Theorem 2.1, we have that 𝑣≤0 for 𝑡∈𝐽, that is, 𝑢₁≤𝑢₂ on 𝐽. Similarly, one can obtain that 𝑢₂≤𝑢₁ on 𝐽. Hence 𝑢₁=𝑢₂.
Next, we prove that if 𝑢 is a solution of (3.1), then 𝛼≤𝑢≤𝛽. Let 𝑝=𝛼−𝑢. From boundary conditions, we have that 𝑐_𝛼(𝑡)=𝑐_𝑝(𝑡) for all 𝑡∈𝐽. From (ℎ₂) and (3.1), we have
−𝑝^(𝑡)+𝑀𝑝(𝑡Δ)+𝑝_𝑁𝑝(𝜃(𝑡))+(𝑀+𝑟)𝑐_𝑝(𝑡)+𝑁𝑐_𝑝(𝜃(𝑡))≤0,𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘≥𝐿_𝑘𝑝𝑡_𝑘+𝐿_𝑘𝑐_𝑝𝑡_𝑘,𝑘=1,…,𝑚.(3.6) By Theorem 2.1, we have that 𝑝=𝛼−𝑢≤0 on 𝐽. Analogously, 𝑢≤𝛽 on 𝐽.
Finally, we show that the boundary value problem (3.1) has a solution by five steps as follows.
Step 1. Let 𝛼(𝑡)=𝛼(𝑡)+𝑐_𝛼(𝑡),𝛽(𝑡)=𝛽(𝑡)−𝑐_−𝛽(𝑡). We claim that(1)−𝛼^(𝑡)+𝑀𝛼(𝑡)+𝑁𝛼Δ(𝜃(𝑡))+(𝑀+𝑟)𝑐_𝛼(𝑡)+𝑁𝑐_𝛼(𝜃(𝑡))≤𝜎(𝑡)for𝑡∈𝐽,𝑡≠𝑡_𝑘,𝛼^𝑡_𝑘≥𝐿_𝑘𝛼𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚,(3.7)(2)−𝛽^(𝑡)+𝑀𝛽(𝑡)+𝑁𝛽(𝜃Δ(𝑡))−(𝑀+𝑟)𝑐_−𝛽(𝑡)+𝑁𝑐_−𝛽(𝜃(𝑡))≥𝜎(𝑡)for𝑡∈𝐽,𝑡≠𝑡_𝑘,𝛽^𝑡_𝑘≤𝐿_𝑘𝛽𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚,(3.8)(3)𝛼(𝑡)≤𝛼(𝑡)≤𝛽(𝑡)≤𝛽(𝑡) for 𝑡∈𝐽.
From (ℎ₂) and (ℎ₃), we have
−𝛼^(𝑡Δ)+𝑀𝛼(𝑡)+𝑁𝛼(𝜃(𝑡))≤𝜎(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,−𝛼^𝑡_𝑘≥𝐿_𝑘𝛼𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚.(3.9)𝛽^(𝑡Δ)+𝑀𝛽(𝑡)+𝑁𝛽(𝜃(𝑡))≥𝜎(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝛽^𝑡_𝑘≤𝐿_𝑘𝛽𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚,(3.10)𝛼(0)−𝑎𝛼^(0)−𝑐𝛼(𝜂)=𝛼(0)−𝑎𝛼^(0)−𝑐𝛼(𝜂)−(𝑎𝜋+𝑐sin(𝜋𝜂))𝐵_𝛼≤0,(3.11)−𝛼(1)+𝑏𝛼^(1)−𝑑𝛼(𝜉)=𝛼(1)+𝑏𝛼^(0)−𝑑𝛼(𝜉)−(𝑏𝜋+𝑑sin(𝜋𝜉))𝐵_𝛼≤0,(3.12)−𝛽(0)−𝑎𝛽^(0)−𝑐𝛽(𝜂)=−𝛽(0)+𝑎𝛽^(0)+𝑐𝛽(𝜂)−(𝑎𝜋+𝑐sin(𝜋𝜂))𝐵_−𝛽≤0,(3.13)𝛽(1)+𝑏𝛽^(1)−𝑑𝛽(𝜉)=−𝛽(1)−𝑏𝛽^(0)+𝑑𝛽(𝜉)−(𝑏𝜋+𝑑sin(𝜋𝜉))𝐵_−𝛽≤0.(3.14) From (3.9)–(3.14), we obtain that 𝑐_𝛼(𝑡)=𝑐_−𝛽(𝑡)≡0,𝑡∈𝐽. Combining (3.9) and (3.10), we obtain that ( 1) and ( 2) hold.
It is easy to see that 𝛼≤𝛼,𝛽≤𝛽 on 𝐽. We show that 𝛼≤𝛽 on 𝐽. Let 𝑝=𝛼−𝛽, then 𝑝(𝑡)=𝛼(𝑡)−𝛽(𝑡)+𝑐_𝛼(𝑡)+𝑐_−𝛽(𝑡). From (3.9)–(3.14), we have
Δ−𝑝𝑝^_(𝑡)+𝑀𝑝(𝑡)+𝑁𝑝(𝜃(𝑡))≤0,𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑝(0)−𝑡_𝑘𝑎𝑝≥_𝐿_𝑘𝑝𝑡_𝑘,𝑘=1,…,𝑚,(0)−𝑐𝑝(𝜂)=𝛼(0)−𝑎𝛼^(0)−𝑐𝛼(𝜂)−(𝑎𝜋+𝑐sin𝐼(𝜋𝜂))𝐵_𝛼𝑝(1)+𝑏𝑝_−𝛽(0)+𝑎𝛽^(0)+𝑐𝛽(𝜂)−(𝑎𝜋+𝑐sin(𝜋𝜂))𝐵_−𝛽≤0,(1)−𝑑𝑝(𝜉)=𝛼(1)+𝑏𝛼^(1)−𝑑𝛼(𝜉)−(𝑏𝜋+𝑑sin(𝜋𝜉))𝐵_𝛼Δ𝑝_−𝛽(1)−𝑏𝛽^(1)+𝑑𝛽(𝜂)−(𝑏𝜋+𝑑sin(𝜋𝜉))𝐵_−𝛽≤0,𝑡_𝑘+=𝐿Δ_𝑘𝛼^𝑡_𝑘−Δ𝛽^𝑡_𝑘+Δ𝑐^𝑎𝑡_𝑘+Δ𝑐^−𝛽𝑡_𝑘≥𝐿_𝑘𝛼𝑡_𝑘−𝛽𝑡_𝑘𝑐_𝛼𝑡_𝑘+𝑐_−𝛽𝑡_𝑘=𝐿_𝑘𝑝𝑡_𝑘.(3.15) By Theorem 2.1, we have that 𝑝≤0 on 𝐽, that is, 𝛼≤𝛽 on 𝐽. So ( 3) holds.Step 2. Consider the boundary value problem −𝑢^(𝑡Δ)+𝑢_𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))=𝜎(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑢(0)−𝑡_𝑘𝑎𝑢=_𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚,(0)=𝜆,𝑢(1)+𝑏𝑢^(1)=𝛿,(3.16) where 𝜆∈𝐑,𝛿∈𝐑. We show that the boundary value problem (3.16) has one unique solution 𝑢(𝑡,𝜆,𝛿).
It is easy to check that the boundary value problem (3.16) is equivalent to the integral equation:
𝑢(𝑡)=𝛿𝑡+(1−𝑡)𝜆+𝑏𝜆+𝑎𝛿𝑎+𝑏+1+¹⁰+𝐺(𝑡,𝑠)[𝜎(𝑠)−𝑀𝑢(𝑠)−𝑁𝑢(𝜃(𝑠))]𝑑𝑠_{0<𝑡𝑘<𝑡}𝑡−𝑡_𝑘𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘−𝑎+1𝑏+1(𝑡+𝑏)^𝑚_𝑘=11−𝑡_𝑘+𝑏𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘,(3.17) where 𝐺(𝑡,𝑠)=1𝑎+𝑏+1((𝑎𝑎++𝑡𝑠))((11++𝑏𝑏−−𝑠𝑡)),,00≤≤𝑡𝑠≤≤𝑠𝑡≤≤11,.(3.18)
It is easy to see that 𝑃𝐶(𝐽,𝑅) with norm ‖𝑢‖=max_𝑡∈𝐽|𝑢(𝑡)| is a Banach space. Define a mapping Φ∶𝑃𝐶(𝐽,𝑅)→𝑃𝐶(𝐽,𝑅) by
(Φ𝑢)(𝑡)=𝛿𝑡+(1−𝑡)𝜆+𝑏𝜆+𝑎𝛿𝑎+𝑏+1+¹⁰+𝐺(𝑡,𝑠)[𝜎(𝑠)−𝑀𝑢(𝑠)−𝑁𝑢(𝜃(𝑠))]𝑑𝑠_{0<𝑡𝑘<𝑡}𝑡−𝑡_𝑘𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘−𝑎+1𝑏+1(𝑡+𝑏)^𝑚_𝑘=11−𝑡_𝑘+𝑏𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘.(3.19) For any 𝑥,𝑦∈𝑃𝐶(𝐽,𝑅), we have ||(Φ𝑥≤)(𝑡)−(Φ𝑦)(𝑡)₁₀||𝐺(𝑡,𝑠)[𝑀(𝑦(𝑠)−𝑥(𝑠))+𝑁(𝑦(𝜃(𝑠))−𝑥(𝜃(𝑠)))]𝑑𝑠+1+(1+𝑏)²𝑎+𝑏+1^𝑚_𝑘=1≤𝐿_𝑘‖𝑥−𝑦‖₁₀𝐺(𝑡,𝑠)𝑑𝑠‖𝑥−𝑦‖(𝑀+𝑁)+1+(1+𝑏)²𝑎+𝑏+1^𝑚_𝑘=1𝐿_𝑘‖𝑥−𝑦‖.(3.20) Since max_𝑡∈𝐽¹⁰𝐺(𝑡,𝑠)𝑑𝑠=𝑎(1+2𝑏)2(𝑎+𝑏+1)+81𝑎1++𝑏2+𝑏1²,(3.21) the inequality (3.2) implies that Φ∶𝑃𝐶(𝐽)→𝑃𝐶(𝐽) is a contraction mapping. Thus there exists a unique 𝑢∈𝑃𝐶(𝐽) such that Φ𝑢=𝑢. The boundary value problem (3.16) has a unique solution.Step 3. We show that for any 𝑡∈𝐽, the unique solution 𝑢(𝑡,𝜆,𝛿) of the boundary value problem (3.16) is continuous in 𝜆 and 𝛿. Let 𝑢(𝑡,𝜆_𝑖,𝛿_𝑖),𝑖=1,2, be the solution of −𝑢^(𝑡Δ)𝑢+_𝑀𝑢(𝑡)+𝑁𝑢(𝜃(𝑡))=𝜎(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑢(0)−𝑎𝑢_𝑡_𝑘=𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘,𝑘=1,…,𝑚,(0)=𝜆_𝑖,𝑢(1)+𝑏𝑢^(1)=𝛿_𝑖,𝑖=1,2.(3.22) Then 𝑢𝑡,𝜆_𝑖,𝛿_𝑖=𝛿^𝑖𝑡+(1−𝑡)𝜆_𝑖+𝑏𝜆_𝑖+𝑎𝛿_𝑖𝑎+𝑏+1+¹⁰+𝐺(𝑡,𝑠)𝜎(𝑠)−𝑀𝑢𝑠,𝜆_𝑖,𝛿_𝑖−𝑁𝑢𝜃(𝑠),𝜆_𝑖,𝛿_𝑖𝑑𝑠_{0<𝑡𝑘<𝑡}×𝑡−𝑡_𝑘𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘−𝑎+1𝑏+1(𝑡+𝑏)_𝑚_𝑘=11−𝑡_𝑘+𝑏𝐿_𝑘𝑢𝑡_𝑘+𝑒_𝑘,𝑖=1,2.(3.23)
From (3.23), we have that
‖‖𝑢𝑡,𝜆₁,𝛿₁−𝑢𝑡,𝜆₂,𝛿₂‖‖≤+||𝜆(₁𝑀−+𝜆₂𝑁||+)‖‖||𝛿𝑢₁−𝑡,𝛿𝜆₂₁||,𝛿₁−𝑢𝑡,𝜆₂,𝛿₂‖‖max_𝑡∈𝐽¹⁰+‖𝑢𝑡,𝜆₁𝐺(𝑡,𝑠)𝑑𝑠,𝛿₁−𝑢𝑡,𝜆₂,𝛿₂‖1+(1+𝑏)²𝑎+𝑏+1^𝑚_𝑘=1≤||𝜆𝐿_𝑘‖𝑥−𝑦‖₁−𝜆₂||+||𝛿₁−𝛿₂||+𝜇‖‖𝑢𝑡,𝜆₁,𝛿₁−𝑢𝑡,𝜆₂,𝛿₂‖‖.(3.24)
Hence
‖‖𝑢(𝑡,𝜆₁,𝛿₁)−𝑢𝑡,𝜆₂,𝛿₂‖‖₀≤11−𝜇||𝜆¹−𝜆₂||+||𝛿₁−𝛿₂||.(3.25)Step 4. We show that 𝛼(𝑡)≤𝑢(𝑡,𝜆,𝛿)≤𝛽(𝑡)(3.26) for any 𝑡∈𝐽, 𝜆∈[𝑐𝛼(𝜂),𝑐𝛽(𝜂)], and 𝛿∈[𝑑𝛼(𝜉),𝑑𝛽(𝜉)], where 𝑢(𝑡,𝜆,𝛿) is unique solution of the boundary value problem (3.16).
Let 𝑚(𝑡)=𝛼(𝑡)−𝑢(𝑡,𝜆,𝛿). From (3.9), (3.11), (3.12), and (3.16), we have that 𝑚(0)−𝑎𝑚^(0)≤𝑐𝑚(𝜂),𝑚(1)+𝑏𝑚^(1)≤𝑑𝑚(𝜉), and
−=𝑚−^(𝑡)+𝑀𝑚(𝑡)+𝑁𝑚(𝜃(𝑡))Δ𝑚_𝛼^(𝑡)+𝑀𝛼(𝑡)+𝑁𝛼(𝜃(𝑡))+𝑢^(𝑡,𝜆)−𝑀𝑢(𝑡,𝜆,𝛿)−𝑁𝑢(𝜃(𝑡),𝜆,𝛿)≤𝜎(𝑡)−𝜎(𝑡)≤0,𝑡_𝑘≥𝐿_𝑘𝑢𝑡_𝑘.(3.27) By Theorem 2.1, we obtain that 𝑚≤0 on 𝐽, that is, 𝛼(𝑡)≤𝑢(𝑡,𝜆,𝛿) on 𝐽. Similarly, 𝑢(𝑡,𝜆,𝛿)≤𝛽(𝑡) on 𝐽.Step 5. Let 𝐷=[𝑐𝛼(𝜂),𝑐𝛽(𝜂)]×[𝑑𝛼(𝜉),𝑑𝛽(𝜉)]. Define a mapping 𝐹∶𝐷→𝐑² by 𝐹(𝜆,𝛿)=(𝑢(𝜂,𝜆,𝛿),𝑢(𝜉,𝜆,𝛿)),(3.28) where 𝑢(𝑡,𝜆,𝛿) is unique solution of the boundary value problem (3.16). From Step 4, we have 𝐹(𝐷)⊂𝐷.(3.29) Since 𝐷 is a compact convex set and 𝐹 is continuous, it follows by Schauder’s fixed point theorem that 𝐹 has a fixed point (𝜆₀,𝛿₀)∈𝐷 such that 𝑢𝜂,𝜆₀,𝛿₀=𝜆₀,𝑢𝜉,𝜆₀,𝛿₀=𝛿₀.(3.30) Obviously, 𝑢(𝑡,𝜆₀,𝛿₀) is unique solution of the boundary value problem (3.1). This completes the proof.

4. Main Results

Let 𝑀∈𝐑, 𝑁∈𝐑. We first give the following definition.

Definition 4.1. A function 𝛼∈𝐸 is called a lower solution of the boundary value problem (1.2) if −𝛼^(𝑡)+(𝑀Δ𝛼+_𝑟)𝑐_𝛼(𝑡)+𝑁𝑐_𝛼(𝜃(𝑡))≤𝑓(𝑡,𝛼(𝑡),𝛼(𝜃(𝑡))),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘≥𝐼_𝑘𝛼𝑡_𝑘+𝐿_𝑘𝑐_𝛼𝑡_𝑘,𝑘=1,…,𝑚.(4.1)

Definition 4.2. A function 𝛽∈𝐸 is called an upper solution of the boundary value problem (1.2) if −𝛽^(𝑡)−(𝑀+𝑟Δ)𝑐𝛽_−𝛽(𝑡)−𝑁𝑐_−𝛽(𝜃(𝑡))≥𝑓(𝑡,𝛽(𝑡),𝛽(𝜃(𝑡)))𝑡∈𝐽,𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘≤𝐼_𝑘𝛽𝑡_𝑘−𝐿_𝑘𝑐_−𝛽𝑡_𝑘,𝑘=1,…,𝑚.(4.2) Our main result is the following theorem.

Theorem 4.3. Assume that (𝐻)  holds. If the following conditions are satisfied: (𝐻₁)𝛼,𝛽 are lower and upper solutions for boundary value problem (1.2) respectively, and 𝛼(𝑡)≤𝛽(𝑡) on 𝐽,(𝐻₂)the constants 𝑀,𝑁 in definition of upper and lower solutions satisfy (2.3), (3.2), and 𝑓(𝑡,𝑥,𝑦𝐼)_𝑘−𝑓𝑡,𝑥,𝑦≥−𝑀𝑥−𝑥−𝑁𝑦−𝑦,(𝑥)−𝐼_𝑘(𝑦)≥𝐿_𝑘(𝑥−𝑦),𝑥≤𝑦,(4.3) for 𝛼(𝑡)≤𝑥≤𝑥≤𝛽(𝑡),𝛼(𝜃(𝑡))≤𝑦≤𝑦≤𝛽(𝜃(𝑡)),𝑡∈𝐽.
Then, there exist monotone sequences {𝛼_𝑛},{𝛽_𝑛} with 𝛼₀=𝛼,𝛽₀=𝛽 such that lim_𝑛→∞𝛼_𝑛(𝑡)=𝜌(𝑡),lim_𝑛→∞𝛽_𝑛(𝑡)=𝜚(𝑡) uniformly on 𝐽, and 𝜌,𝜚 are the minimal and the maximal solutions of (1.2), respectively, such that
𝛼₀≤𝛼₁≤𝛼₂≤⋯𝛼_𝑛≤𝜌≤𝑥≤𝜚≤𝛽_𝑛≤⋯≤𝛽₂≤𝛽₁≤𝛽₀(4.4) on 𝐽, where 𝑥 is any solution of (1.1) such that 𝛼(𝑡)≤𝑥(𝑡)≤𝛽(𝑡) on 𝐽.

Proof. Let [𝛼,𝛽]={𝑢∈𝐸∶𝛼(𝑡)≤𝑢(𝑡)≤𝛽(𝑡),𝑡∈𝐽}. For any 𝛾∈[𝛼,𝛽], we consider the boundary value problem −𝑢^(𝑡)+𝑀Δ𝑢𝑢_(𝑡)+𝑁𝑢(𝜃(𝑡))=𝑓(𝑡,𝛾(𝑡),𝛾(𝜃(𝑡)))+𝑀𝛾(𝑡)+𝑁𝛾(𝜃(𝑡)),𝑡∈𝐽,𝑢𝑡_𝑘(0)=−𝐼𝑎_𝑘𝑥_𝛾𝑡_𝑘−𝐿_𝑘𝑢𝑡_𝑘−𝛾𝑡_𝑘,𝑘=1,…,𝑚.(0)=𝑐𝑢(𝜂),𝑢(1)+𝑏𝑢^(1)=𝑑𝑢(𝜉).(4.5)
Since 𝛼 is a lower solution of (1.2), from (𝐻₂), we have that
−𝛼≤^(𝑡𝑓)+(𝑡,𝑀𝛼(𝛼𝑡)(𝑡,)𝛼+(𝜃𝑁(𝑡𝛼))()𝜃+(𝑡𝑀))𝛼(𝑡)+𝑁𝛼(𝜃(𝑡))−(𝑀+𝑟)𝑐_𝛼≤𝑓(𝑡,𝛾(𝑡),𝛾(𝜃(𝑡)))+𝑀𝛾(𝑡)+𝑁𝛾(𝜃(𝑡))−(𝑀+𝑟)𝑐_𝛼(𝑡)−𝑁𝑐_𝛼(𝜃(𝑡))Δ𝛼_(𝑡)−𝑁𝑐_𝛼(𝜃(𝑡)),𝑡_𝑘≥𝐼_𝑘𝛼𝑡_𝑘+𝐿_𝑘𝑐_𝛼𝑡_𝑘≥𝐼_𝑘𝛾𝑡_𝑘+𝐿_𝑘𝛼𝑡_𝑘−𝐿_𝑘𝛾𝑡_𝑘+𝐿_𝑘𝑐_𝛼𝑡_𝑘.(4.6)
Similarly, we have that
−𝛽≥^(𝑡𝑓)+(𝑡,𝑀𝛾(𝛽𝑡)(𝑡,)𝛾+(𝜃𝑁(𝑡𝛽))()𝜃+(𝑡𝑀))𝛾(𝑡)+𝑁𝛾(𝜃(𝑡))+(𝑀+𝑟)𝑐_−𝛽Δ𝛽_(𝑡)+𝑁𝑐_−𝛽(𝜃(𝑡)),𝑡_𝑘≤𝐼_𝑘𝛽𝑡_𝑘−𝐿_𝑘𝑐_−𝛽𝑡_𝑘≤𝐼_𝑘𝛾𝑡_𝑘+𝐿_𝑘𝛽𝑡_𝑘−𝐿_𝑘𝛾𝑡_𝑘−𝐿_𝑘𝑐_−𝛽𝑡_𝑘.(4.7)
By Theorem 3.1, the boundary value problem (4.5) has a unique solution 𝑢∈[𝛼,𝛽]. We define an operator Ψ by 𝑢=Ψ𝛾, then Ψ is an operator from [𝛼,𝛽] to [𝛼,𝛽].
We will show that
(a)𝛼≤Ψ𝛼,Ψ𝛽≤𝛽,(b)Ψ is nondecreasing in [𝛼,𝛽].
From Ψ𝛼∈[𝛼,𝛽] and Ψ𝛽∈[𝛼,𝛽], we have that (a) holds. To prove (b), we show that Ψ𝜈₁≤Ψ𝜈₂ if 𝛼≤𝜈₁≤𝜈₂≤𝛽.
Let 𝜈^∗1=Ψ𝜈₁,𝜈^∗2=Ψ𝜈₂ ,and 𝑝=𝜈^∗1−𝜈^∗2, then by (𝐻₂) and boundary conditions, we have that
−𝑝^=(𝑡𝑓)+𝑡𝑀,𝜈₁𝑝(𝑡)+𝑁𝑝(𝜃(𝑡))−𝑓𝑡,𝜈(₂𝑡),𝜈₁(𝜃(𝑡))+𝑀𝜈₁(𝑡)+𝑁𝜈₁(𝜃(𝑡))Δ𝑝_(𝑡),𝜈₂(𝜃(𝑡))−𝑀𝜈₂(𝑡)−𝑁𝜈₂(𝜃(𝑡))≤0,𝑝(0)−𝑡_𝑘𝑎𝑝≥_𝐿_𝑘𝑝𝑡_𝑘,(0)=𝑐𝑝(𝜂),𝑝(1)+𝑝𝑢^(1)=𝑑𝑝(𝜉).(4.8) By Theorem 2.1, 𝑝(𝑡)≤0 on 𝐽, which implies that Ψ𝜈₁≤Ψ𝜈₂.
Define the sequences {𝛼_𝑛},{𝛽_𝑛} with 𝛼₀=𝛼,𝛽₀=𝛽 such that 𝛼_𝑛+1=Ψ𝛼_𝑛,𝛽_𝑛+1=Ψ𝛽_𝑛 for 𝑛=0,1,2,… From (a) and (b), we have
𝛼₀≤𝛼₁≤𝛼₂≤⋯≤𝛼_𝑛≤𝛽_𝑛≤⋯≤𝛽₂≤𝛽₁≤𝛽₀(4.9) on 𝑡∈𝐽, and each 𝛼_𝑛,𝛽_𝑛∈𝐸 satisfies −𝛼^𝑛(𝑡)+𝑀𝛼_𝑛(Δ𝑡)𝛼+_𝑛𝑁𝛼_𝑛(𝜃(𝑡))=𝑓𝑡,𝛼_𝑛−1(𝑡),𝛼_𝑛−1(𝜃(𝑡))+𝑀𝛼_𝑛−1(𝑡)+𝑁𝛼_𝑛−1(𝜃(𝑡)),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘𝛼=𝐼_𝑘𝛼_𝑛−1𝑡_𝑘+𝐿_𝑘𝛼_𝑛𝑡_𝑘−𝛼_𝑛−1𝑡_𝑘,𝑘=1,2,…,𝑚,_𝑛−𝛽_𝑛(0)−𝑎𝛼^𝑛(0)=𝑐𝛼_𝑛(𝜂),𝛼_𝑛(1)+𝑏𝛼^𝑛(1)=𝑑𝛼_𝑛(𝜉),(𝑡)+𝑀𝛽_𝑛(Δ𝑡)𝛽+_𝑛𝑁𝛽_𝑛(𝜃(𝑡))=𝑓𝑡,𝛽_𝑛−1(𝑡),𝛽_𝑛−1(𝜃(𝑡))+𝑀𝛽_𝑛−1(𝑡)+𝑁𝛽_𝑛−1(𝜃(𝑡)),𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑡_𝑘𝛽=𝐼_𝑘𝛽_𝑛−1𝑡_𝑘+𝐿_𝑘𝛽_𝑛𝑡_𝑘−𝛽_𝑛−1𝑡_𝑘,𝑘=1,2,…,𝑚,_𝑛(0)−𝑎𝛽^𝑛(0)=𝑐𝛽_𝑛(𝜂),𝛽_𝑛(1)+𝑏𝛽^𝑛(1)=𝑑𝛽_𝑛(𝜉).(4.10) Therefore, there exist 𝜌,𝜚 such that such that lim_𝑛→∞𝛼_𝑛(𝑡)=𝜌(𝑡), lim_𝑛→∞𝛽_𝑛(𝑡)=𝜚(𝑡) uniformly on 𝐽. Clearly, 𝜌,𝜚 are solutions of (1.1).
Finally, we prove that if 𝑥∈[𝛼₀,𝛽₀] is any solution of (1.1), then 𝜌(𝑡)≤𝑥(𝑡)≤𝜚(𝑡) on 𝐽. To this end, we assume, without loss of generality, that 𝛼_𝑛(𝑡)≤𝑥(𝑡)≤𝛽_𝑛(𝑡) for some 𝑛. Since 𝛼₀(𝑡)≤𝑥(𝑡)≤𝛽₀(𝑡), from property (b), we can obtain
𝛼_𝑛+1(𝑡)≤𝑥(𝑡)≤𝛽_𝑛+1(𝑡),𝑡∈𝐽.(4.11) Hence, we can conclude that 𝛼_𝑛(𝑡)≤𝑥(𝑡)≤𝛽_𝑛(𝑡),∀𝑛.(4.12) Passing to the limit as 𝑛→∞, we obtain 𝜌(𝑡)≤𝑥(𝑡)≤𝜚(𝑡),𝑡∈𝐽.(4.13) This completes the proof.

Acknowledgments

This work is supported by the NNSF of China (10571050;10871062) and Hunan Provincial Natural Science Foundation of China (NO:09JJ3010), and Science Research Fund of Hunan provincial Education Department (No: 06C052 ).