Integral BVPs for a Class of First-Order Impulsive Functional Differential Equations

Abstract

The methods of lower and upper solutions and monotone iterative technique are employed to the study of integral boundary value problems for a class of first-order impulsive functional differential equations. Sufficient conditions are obtained for the existence of extreme solutions.

1. Introduction and Preliminaries

In this paper, we study the following integral boundary value problems (BVPs for short) of the impulsive functional differential equation

𝑥^′(𝑡)+𝑏(𝑡)𝑥(𝑡)=𝑓(𝑡,𝑥(𝑡),[𝐾𝑥](𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽=[0,𝑇],Δ𝑥(𝑡_𝑘)=𝐼_𝑘(𝑥(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝑥(0)+𝜇^𝑇₀𝑥(𝑠)𝑑𝑠=𝑥(𝑇),𝜇≤0, (1.1) where 𝑓∈𝐶(𝐽×𝑅²,𝑅),𝐼_𝑘∈𝐶(𝑅,𝑅),(1≤𝑘≤𝑚),𝑏(𝑡)∈𝐶(𝑅),𝑏(𝑡)≤0,𝐽=[0,𝑇],0=𝑡₀<𝑡₁<𝑡₂<⋯<𝑡_𝑚<𝑡_𝑚+1=𝑇. 𝐾∶𝑃𝐶(𝐽)→𝑃𝐶(𝐽), where 𝑃𝐶(𝐽)={𝑢∶𝐽→𝑅,𝑢 is continuous for 𝑡∈𝐽,𝑡≠𝑡_𝑘,𝑢(𝑡⁺_𝑖),𝑢(𝑡⁻_𝑖) exist, and 𝑢(𝑡⁻_𝑖)=𝑢(𝑡_𝑖),𝑖=1,2,…,𝑚}. Furthermore, we will assume that 𝐾 is continuous and monotone nondecreasing, and for any bounded set 𝐴⊆𝑃𝐶(𝐽),𝐾𝐴 is bounded. Δ𝑥(𝑡_𝑘)=𝑥(𝑡⁺_𝑘)−𝑥(𝑡⁻_𝑘) denotes the jump of 𝑥(𝑡) at 𝑡=𝑡_𝑘; 𝑥(𝑡⁺_𝑘) and 𝑥(𝑡⁻_𝑘) represent the right and left limits of 𝑥(𝑡) at 𝑡=𝑡_𝑘, respectively. Denote 𝐽^′=𝐽⧵{𝑡₁,𝑡₂,…,𝑡_𝑚}.

Let 𝑃𝐶¹(𝐽)={𝑢∈𝑃𝐶(𝐽)∶𝑢 be continuously differentiable for 𝑡∈𝐽,𝑡≠𝑡_𝑘}. 𝑃𝐶(𝐽) and 𝑃𝐶¹(𝐽) are Banach spaces with the norms

‖𝑢‖_{𝑃𝐶(𝐽)}=sup{|𝑢(𝑡)|∶𝑡∈𝐽},‖𝑢‖_𝑃𝐶¹_(𝐽)=max{‖𝑢‖_{𝑃𝐶(𝐽)},‖𝑢^′‖_{𝑃𝐶(𝐽)}}. (1.2)

By a solution of (1.1) we mean a 𝑢∈𝑃𝐶¹(𝐽) for which problem (1.1) is satisfied.

Note that (1.1) has a very general form, as special instances resulting from (1.1), one can have impulsive differential equations with deviating arguments and impulsive differential equations with the Volterra or Fredholm operators. When 𝜇=0, 𝐼_𝑘≡0, (1.1) reduces to

𝑥^′(𝑡)+𝑏(𝑡)𝑥(𝑡)=𝑓(𝑡,𝑥(𝑡),[𝐾𝑥](𝑡)),𝑡∈𝐽=[0,𝑇],𝑥(0)=𝑥(𝑇). (1.3) In [1], Cao and Li. studied and understood existence and stability of solution of this equation by using fixed theorem and monotone iteration techniques.

When 𝜇=0, 𝑏(𝑡)≡0, (1.1) reduces to

𝑥^′(𝑡)=𝑓(𝑡,𝑥(𝑡),[𝐾𝑥](𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽=[0,𝑇],Δ𝑥(𝑡_𝑘)=𝐼_𝑘(𝑥(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝑥(0)=𝑥(𝑇). (1.4) In [2], Li discussed and built the existence theorem of solutions of this equation by using fixed theorem, upper and lower solutions methods and monotone iterative techniques.

When 𝜇=0, 𝑏(𝑡)≡0, [𝐾𝑥](𝑡)=𝑥(𝑡), the equation (1.1) reduces to the periodic boundary value problem of the impulsive differential equation

𝑥^′(𝑡)=𝑓(𝑡,𝑥(𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽=[0,𝑇],Δ𝑥(𝑡_𝑘)=𝐼_𝑘(𝑥(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝑥(0)=𝑥(𝑇). (1.5) There are plenty of results on studying the periodic boundary value problem of impulsive differential equations (see [3–8]). According to author’s know, there are no dependent references for studying the (1.1) yet. To fill in this void, we try to find the conditions on 𝑓 and 𝐼_𝑘, so that make sure that the (1.1) exists extremal solution.

It is well known that the monotone iterative technique offers an approach for obtaining approximate solutions of nonlinear differential equations, for details, see [4] and the references therein. There also exist several works devoted to the applications of this technique to boundary value problems of impulsive differential equations, see, for example, [1–3, 5–14]. In this paper, we consider (1.1) by using the method of upper and lower solutions combined with monotone iterative technique. This technique plays an important role in constructing monotone sequences which converge to the solutions of our problems. In presence of a lower solution 𝛼 and an upper solution 𝛽 with 𝛼≤𝛽, we show under suitable conditions the sequences converge to the solutions of (1.1) by using the method of upper and lower solutions and monotone iterative technique.

Definition 1.1. The functions 𝛼,𝛽∈𝑃𝐶¹(𝐽) are called lower solution and upper solution of (1.1), respectively, if 𝛼^′(𝑡)+𝑏(𝑡)𝛼(𝑡)≤𝑓(𝑡,𝛼(𝑡),[𝐾𝛼](𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝛼(𝑡_𝑘)≤𝐼_𝑘(𝛼(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝛼(0)+𝜇^𝑇₀𝛽𝛼(𝑠)𝑑𝑠≤𝛼(𝑇),𝜇≤0._′(𝑡)+𝑏(𝑡)𝛽(𝑡)≥𝑓(𝑡,𝛽(𝑡),[𝐾𝛽](𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝛽(𝑡_𝑘)≥𝐼_𝑘(𝛽(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝛽(0)+𝜇^𝑇₀𝛽(𝑠)𝑑𝑠≥𝛽(𝑇),𝜇≤0. (1.6)

In what follows we define the set

[𝛼,𝛽]={𝑤∈𝑃𝐶(𝐽,𝑅)∶𝛼(𝑡)≤𝑤(𝑡)≤𝛽(𝑡),𝑡∈𝐽} (1.7) for 𝛼,𝛽∈𝑃𝐶(𝐽,𝑅) and 𝛼≤𝛽.

We list the following conditions.

(𝐻₁)𝛼(𝑡),𝛽(𝑡) are lower and upper solutions of (1.1) such that 𝛼(𝑡)≤𝛽(𝑡).(𝐻₂) There exists 𝑀≥0 such that 𝑓(𝑡,𝑥,𝑦)−𝑓(𝑡,𝑥,𝑦)≥−𝑀(𝑥−𝑥), (1.8) for 𝛼(𝑡)≤𝑥≤𝑥≤𝛽(𝑡),[𝑇𝛼](𝑡)≤𝑦≤𝑦≤[𝑇𝛽](𝑡),𝑡∈𝐽.(𝐻₃) There exist 0≤𝐿_𝑘<1,𝑘=1,2,…,𝑚 such that 𝐼_𝑘(𝑥)−𝐼_𝑘(𝑦)≥−𝐿_𝑘(𝑥−𝑦),𝑘=1,2,…,𝑚, (1.9) for 𝛼(𝑡)≤𝑦≤𝑥≤𝛽(𝑡),𝑡∈𝐽.

2. Main Results

To obtain our main results, we need the following lemmas.

Lemma 2.1 (see [9]). Suppose that the following conditions are satisfied.(𝐴₀) Sequence {𝑡_𝑘} satisfies 0≤𝑡₀<𝑡₁<𝑡₂⋯, and lim_𝑛→∞𝑡_𝑛=∞.(𝐴₁)𝑚∈𝑃𝐶¹[𝐽,𝑅] and 𝑚(𝑡) is left continuous at 𝑡_𝑘,𝑘=1,2,….(𝐴₂) For 𝑘=1,2,…,𝑡≥𝑡₀, 𝑚^′(𝑡)≤𝑝(𝑡)𝑚(𝑡)+𝑞(𝑡),𝑡≠𝑡_𝑘,𝑡∈𝐽,𝑚(𝑡⁺_𝑘)≤𝑑_𝑘𝑚(𝑡_𝑘)+𝑏_𝑘,𝑘=1,2,…,𝑚, (2.1) where 𝑞,𝑝∈𝐶[𝑅₊,𝑅],𝑏_𝑘,𝑑_𝑘≥0 are constants, then 𝑚(𝑡)≤𝑚(𝑡₀)_{𝑡0<𝑡𝑘≤𝑡}𝑑_𝑘exp(^𝑡_𝑡0𝑝(𝑠)𝑑𝑠)+_{𝑡0<𝑡𝑘≤𝑡}(_{𝑡𝑘<𝑡𝑗≤𝑡}𝑑_𝑗exp(^𝑡_𝑡𝑘𝑝(𝑠)𝑑𝑠))𝑏_𝑘+^𝑡_{𝑡0𝑠<𝑡𝑘<𝑡}𝑑_𝑘exp(^𝑡_𝑠𝑝(𝜎)𝑑𝜎)𝑞(𝑠)𝑑𝑠. (2.2)

Lemma 2.2 (see [12]). If 𝑚∈𝑃𝐶¹(𝐽) and 𝑚^′(𝑡)≤−𝑀𝑚(𝑡),𝑡≠𝑡_𝑘,𝑡∈𝐽=[0,𝑇],Δ𝑚(𝑡_𝑘)≤−𝐿_𝑘𝑚(𝑡_𝑘),𝑘=1,2,…,𝑚,𝑚(0)≤𝑚(𝑇), (2.3) where 𝑀>0,0<𝐿_𝑘≤1, then 𝑚(𝑡)≤0,𝑡∈𝐽.

Lemma 2.3. If 𝑥∈𝑃𝐶(𝐽),𝑀>0,0<𝐿_𝑘≤1,𝑘=1,2,…,𝑚, and (1/(1−𝑒^−𝑀𝑇))∑^𝑚_𝑘=0𝐿_𝑘<1, then the equation 𝑥^′(𝑡)+𝑀𝑥(𝑡)=𝜎(𝑡),𝑡≠𝑡_𝑘,𝑡∈𝐽=[0,𝑇],Δ𝑥(𝑡_𝑘)=−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘,𝑘=1,2,…,𝑚,𝑥(0)+𝑑=𝑥(𝑇),𝑑∈𝑅 (2.4) has one unique solution.

Proof. Firstly, we prove that (2.4) is equivalent to the integral equation 𝑒𝑥(𝑡)=−^−𝑀𝑡1−𝑒^−𝑀𝑇𝑑+^𝑇₀𝐺(𝑡,𝑠)𝜎(𝑠)𝑑𝑠+^𝑚_𝑘=0𝐺(𝑡,𝑡_𝑘)(−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘), (2.5) where 𝑒𝐺(𝑡,𝑠)=1−𝑒^{−𝑀(𝑡−𝑠)−𝑀𝑇}𝑒,0≤𝑠<𝑡≤𝑇,_{−𝑀(𝑇+𝑡−𝑠)}1−𝑒^−𝑀𝑇,0≤𝑡≤𝑠≤𝑇. (2.6)
If 𝑥(𝑡)∈𝑃𝐶¹(𝐽) is solution of (2.4), then, by directly integrating we obtain
𝑒𝑥(𝑡)=−^−𝑀𝑡1−𝑒^−𝑀𝑇𝑑+^𝑇₀𝐺(𝑡,𝑠)𝜎(𝑠)𝑑𝑠+^𝑚_𝑘=0𝐺(𝑡,𝑡_𝑘)(−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘). (2.7)
If 𝑥(𝑡)∈𝑃𝐶¹(𝐽) is solution of the above-mentioned integral equation, then
𝑥^′𝑒(𝑡)=−𝑀[−^−𝑀𝑡1−𝑒^−𝑀𝑇𝑑+^𝑇₀𝐺(𝑡,𝑠)𝜎(𝑠)𝑑𝑠+^𝑚_𝑘=0𝐺(𝑡,𝑡_𝑘)(−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘)]+𝜎(𝑡)=−𝑀𝑥(𝑡)+𝜎(𝑡),𝑡≠𝑡_𝑘,Δ𝑥(𝑡_𝑘)=−𝐿_𝑘(𝑥(𝑡_𝑘))+𝑑_𝑘1,𝑘=1,2,…,𝑚,𝑥(0)=−1−𝑒^−𝑀𝑇𝑑+^𝑇₀𝑒^{−𝑀(𝑇−𝑠)}1−𝑒^−𝑀𝑇𝜎(𝑠)𝑑𝑠+^𝑚_𝑘=0𝑒^{−𝑀(𝑇−𝑡𝑘)}1−𝑒^−𝑀𝑇(−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘𝑒),𝑥(𝑇)=−^−𝑀𝑇1−𝑒^−𝑀𝑇𝑑+^𝑇₀𝑒^{−𝑀(𝑇−𝑠)}1−𝑒^−𝑀𝑇𝜎(𝑠)𝑑𝑠+^𝑚_𝑘=0𝑒^{−𝑀(𝑇−𝑡𝑘)}1−𝑒^−𝑀𝑇(−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘). (2.8)
This yields 𝑥(0)+𝑑=𝑥(𝑇). So (2.4) is equivalent to the integral equation
𝑒𝑥(𝑡)=−^−𝑀𝑡1−𝑒^−𝑀𝑇𝑑+^𝑇₀𝐺(𝑡,𝑠)𝜎(𝑠)𝑑𝑠+^𝑚_𝑘=0𝐺(𝑡,𝑡_𝑘)(−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘). (2.9)
Now, we define operator 𝐴∶𝑃𝐶(𝐽)→𝑃𝐶(𝐽) as
𝑒(𝐴𝑥)(𝑡)=−^−𝑀𝑡1−𝑒^−𝑀𝑇𝑑+^𝑇₀𝐺(𝑡,𝑠)𝜎(𝑠)𝑑𝑠+^𝑚_𝑘=0𝐺(𝑡,𝑡_𝑘)(−𝐿_𝑘𝑥(𝑡_𝑘)+𝑑_𝑘). (2.10)
For each 𝑥,𝑦∈𝑃𝐶(𝐽),
1|(𝐴𝑥)(𝑡)−(𝐴𝑦)(𝑡)|≤1−𝑒^{𝑚−𝑀𝑇}_𝑘=0𝐿_𝑘1|𝑥−𝑦|≤1−𝑒^{𝑚−𝑀𝑇}_𝑘=0𝐿_𝑘‖𝑥(𝑡_𝑘)−𝑦(𝑡_𝑘)‖, (2.11) and so 1‖(𝐴𝑥)(𝑡)−(𝐴𝑦)(𝑡)‖≤1−𝑒^{𝑚−𝑀𝑇}_𝑘=0𝐿_𝑘‖𝑥(𝑡_𝑘)−𝑦(𝑡_𝑘)‖. (2.12)
This indicates that 𝐴∶𝑃𝐶(𝐽)→𝑃𝐶(𝐽) is a contraction mapping. Then there is one unique 𝑥∈𝑃𝐶(𝐽) such that 𝐴𝑥=𝑥, that is, (2.4) has an unique solution 𝑥(𝑡). The proof is complete.

Theorem 2.4. If the conditions (𝐻₁),(𝐻₂),(𝐻₃) are all satisfied, and, in addition, if there exist 𝑀>0,0<𝐿_𝑘≤1,𝑘=1,2,…,𝑚, such that (1/(1−𝑒^−𝑀𝑇))∑^𝑚_𝑘=0𝐿_𝑘<1, then the impulsive equation (1.1) has minimal and maximal solutions 𝜌(𝑡),𝑟(𝑡)∈𝑃𝐶¹(𝐽) in [𝛼,𝛽], and there are monotone sequences {𝛼_𝑛},{𝛽_𝑛} convergeing uniformly to 𝜌(𝑡),𝑟(𝑡) in 𝐽, respectively, where 𝛼₀=𝛼,𝛽₀=𝛽, and 𝛼_𝑛(𝑡),𝛽_𝑛(𝑡) are lower and upper solutions of (1.1), respectively.

Proof. For each 𝜓∈[𝛼,𝛽], we consider the equation 𝑥^′(𝑡)=𝑓(𝑡,𝜓(𝑡),[𝐾𝜓](𝑡))−𝑏(𝑡)𝜓(𝑡)−𝑀(𝑥(𝑡)−𝜓(𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝑥(𝑡_𝑘)=𝐼_𝑘(𝜓(𝑡_𝑘))−𝐿_𝑘(𝑥(𝑡_𝑘)−𝜓(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝑥(0)+𝜇^𝑇₀𝜓(𝑠)𝑑𝑠=𝑥(𝑇),𝜇≤0. (2.13) By Lemma 2.3, we know that (2.13) has a unique solution 𝑥(𝑡)∈𝑃𝐶¹(𝐽). Now, we define operator 𝐴∶𝑃𝐶¹(𝐽)→𝑃𝐶¹(𝐽) as 𝐴𝜓=𝑥.
We will prove that {𝛼_𝑛},{𝛽_𝑛} have the following properties.
(a)𝛼₀≤𝐴𝛼₀,𝐴𝛽₀≤𝛽₀.(b)𝐴 is monotone nondecreasing on [𝛼₀,𝛽₀].
Proofs of properties (a),(b) are divided into three steps to proceed.
Step 1. Suppose that 𝑝=𝛼₀−𝛼₁, then 𝑝^′=𝛼^′₀−𝛼^′₁≤−𝑏(𝑡)𝛼(𝑡)+𝑓(𝑡,𝛼(𝑡),[𝐾𝛼](𝑡))−𝑓(𝑡,𝛼(𝑡),[𝐾𝛼(𝑡)](𝑡))+𝑏(𝑡)𝛼(𝑡)+𝑀(𝛼₁(𝑡)−𝛼(𝑡))=−𝑀𝑝(𝑡),𝑡≠𝑡_𝑘,Δ𝑝(𝑡_𝑘)=Δ𝛼₀−Δ𝛼₁≤𝐼_𝑘(𝛼(𝑡_𝑘))−𝐼_𝑘(𝛼(𝑡_𝑘))+𝐿_𝑘(𝛼₁(𝑡_𝑘)−𝛼(𝑡_𝑘))=−𝐿_𝑘𝑝(𝑡_𝑘),𝑡=𝑡_𝑘,𝑝(0)=𝛼₀(0)−𝛼₁(0)≤𝑝(𝑇). (2.14) By Lemma 2.2, we obtain 𝑝(𝑡)≤0,𝑡∈𝐽, so 𝛼₀(𝑡)≤𝛼₁(𝑡).Step 2. Suppose that 𝑝=𝛽₁−𝛽₀, then 𝑝^′=𝛽^′₁−𝛽^′₀≤𝑏(𝑡)𝛽(𝑡)−𝑓(𝑡,𝛽(𝑡),[𝐾𝛽](𝑡))+𝑓(𝑡,𝛽(𝑡),[𝐾𝛽(𝑡)](𝑡))−𝑏(𝑡)𝛽(𝑡)−𝑀(𝛽₁(𝑡)−𝛽(𝑡))=−𝑀𝑝(𝑡),𝑡=𝑡_𝑘,Δ𝑝(𝑡_𝑘)=Δ𝛽₁−Δ𝛽₀≤−𝐼_𝑘(𝛽(𝑡_𝑘))+𝐼_𝑘(𝛽(𝑡_𝑘))−𝐿_𝑘(𝛽₁(𝑡_𝑘)−𝛽(𝑡_𝑘))=−𝐿_𝑘𝑝(𝑡_𝑘),𝑡=𝑡_𝑘,𝑝(0)=𝛽₁(0)−𝛽₀(0)≤𝑝(𝑇). (2.15) By Lemma 2.2, we obtain 𝑝(𝑡)≤0,𝑡∈𝐽, so 𝛽₁(𝑡)≤𝛽₀(𝑡).
Similary we can show that 𝛼₁(𝑡)≤𝛽₁(𝑡), hence 𝛼₀(𝑡)≤𝛼₁(𝑡)≤𝛽₁(𝑡)≤𝛽₀(𝑡).
Step 3. If 𝑛=𝑚,𝛼_𝑚−1≤𝛼_𝑚≤𝛽_𝑚≤𝛽_𝑚−1, then when 𝑛=𝑚+1, let 𝑝=𝛼_𝑚−𝛼_𝑚+1. Then 𝑝^′=𝛼^′_𝑚−𝛼^′_𝑚+1≤−𝑏(𝑡)𝛼_𝑚−1(𝑡)+𝑓(𝑡,𝛼_𝑚−1(𝑡),[𝐾𝛼_𝑚−1](𝑡))−𝑀(𝛼_𝑚(𝑡)−𝛼_𝑚−1(𝑡))−𝑓(𝑡,𝛼_𝑚(𝑡),[𝐾𝛼_𝑚(𝑡)](𝑡))+𝑏(𝑡)𝛼_𝑚(𝑡)+𝑀(𝛼_𝑚+1(𝑡)−𝛼_𝑚(𝑡))=−𝑏(𝑡)(𝛼_𝑚−1−𝛼_𝑚)−𝑀𝑝(𝑡),𝑡≠𝑡_𝑘. (2.16)Furthermore, 𝑏(𝑡)≤0,𝛼_𝑚−1−𝛼_𝑚≤0, thus 𝑝^′=𝛼^′_𝑚−𝛼^′_𝑚+1≤−𝑀𝑝(𝑡),Δ𝑝(𝑡_𝑘)=Δ𝛼_𝑚−Δ𝛼_𝑚+1=𝐼_𝑘(𝛼_𝑚−1(𝑡_𝑘))−𝐿_𝑘(𝛼_𝑚(𝑡_𝑘)−𝛼_𝑚−1(𝑡_𝑘))−𝐼_𝑘(𝛼_𝑚(𝑡_𝑘))+𝐿_𝑘(𝛼_𝑚+1(𝑡_𝑘)−𝛼_𝑚(𝑡_𝑘))≤−𝐿_𝑘𝑝(𝑡_𝑘),𝑡=𝑡_𝑘,𝑝(0)=𝛼_𝑚(0)−𝛼_𝑚+1(0)=𝜇^𝑇₀(𝛼_𝑚(𝑠)−𝛼_𝑚−1(𝑠))𝑑𝑠+𝑝(𝑇)≤𝑝(𝑇). (2.17) By Lemma 2.2, we obtain 𝑝(𝑡)≤0,𝑡∈𝐽, so 𝛼_𝑚(𝑡)≤𝛼_𝑚+1(𝑡).
Similarly, we can assume that 𝑝=𝛽_𝑚+1−𝛽_𝑚. When 𝑡≠𝑡_𝑘,
𝑝^′=𝛽^′_𝑚+1−𝛽^′_𝑚≤−𝑏(𝑡)𝛽_𝑚(𝑡)+𝑓(𝑡,𝛽_𝑚(𝑡),[𝐾𝛽_𝑚](𝑡))−𝑀(𝛽_𝑚+1(𝑡)−𝛽_𝑚(𝑡))−𝑓(𝑡,𝛽_𝑚−1(𝑡),[𝐾𝛽_𝑚−1(𝑡)](𝑡))+𝑏(𝑡)𝛽_𝑚−1(𝑡)+𝑀(𝛽_𝑚(𝑡)−𝛽_𝑚−1(𝑡))=−𝑏(𝑡)(𝛽_𝑚−𝛽_𝑚−1)−𝑀𝑝(𝑡). (2.18) Furthermore, 𝑏(𝑡)≤0, 𝛽_𝑚−𝛽_𝑚−1≤0, thus 𝑝^′=𝛽^′_𝑚+1−𝛽^′_𝑚≤−𝑀𝑝(𝑡), when 𝑡=𝑡_𝑘, Δ𝑝(𝑡_𝑘)=Δ𝛽_𝑚+1−Δ𝛽_𝑚=𝐼_𝑘(𝛽_𝑚(𝑡_𝑘))−𝐿_𝑘(𝛽_𝑚+1(𝑡_𝑘)−𝛽_𝑚(𝑡_𝑘))−𝐼_𝑘(𝛽_𝑚−1(𝑡_𝑘))+𝐿_𝑘(𝛽_𝑚(𝑡_𝑘)−𝛽_𝑚−1(𝑡_𝑘))≤−𝐿_𝑘𝑝(𝑡_𝑘),𝑝(0)=𝛽_𝑚+1(0)−𝛽_𝑚(0)=𝜇^𝑇₀(𝛽_𝑚−1(𝑠)−𝛽_𝑚(𝑠))𝑑𝑠+𝑝(𝑇)≤𝑝(𝑇), (2.19) hence by Lemma 2.2, we obtain 𝑝(𝑡)≤0,𝑡∈𝐽, so 𝛽_𝑚+1(𝑡)≤𝛽_𝑚(𝑡).
In the same way we can prove that 𝛼_𝑚+1(𝑡)≤𝛽_𝑚+1(𝑡).
Thus by mathematical induction we can know that
𝛼_𝑛−1≤𝛼_𝑛≤𝛽_𝑛≤𝛽_𝑛−1,𝑛=0,1,2,…,𝑡∈𝐽. (2.20) So far, we finish the proof of the properties (𝑎),(𝑏).
Now we prove that 𝛼_𝑛,𝛽_𝑛,𝑛=0,1,2,…, are lower and upper solutions of (1.1). Similarly, we can use mathematical induction to prove this.
When 𝑛=0, 𝛼₀,𝛽₀ are already lower and upper solutions of (1.1).
When 𝑛=1,
𝛼₁(𝑡)=𝑓(𝑡,𝛼(𝑡),[𝐾𝛼](𝑡))−𝑏(𝑡)𝛼(𝑡)−𝑀(𝛼₁(𝑡)−𝛼(𝑡))−𝑓(𝑡,𝛼₁(𝑡),[𝐾𝛼₁](𝑡))+𝑓(𝑡,𝛼₁(𝑡),[𝐾𝛼₁](𝑡))≤𝑓(𝑡,𝛼₁(𝑡),[𝐾𝛼₁](𝑡))−𝑏(𝑡)𝛼₁(𝑡),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝛼₁(𝑡_𝑘)=𝐼_𝑘(𝛼(𝑡_𝑘))−𝐿_𝑘(𝛼₁(𝑡_𝑘)−𝛼(𝑡_𝑘))+𝐼_𝑘(𝛼₁(𝑡_𝑘))−𝐼_𝑘(𝛼₁(𝑡_𝑘))≤𝐼_𝑘(𝛼₁(𝑡_𝑘𝛼)),𝑘=1,2,…,𝑚,₁(0)+𝜇^𝑇₀𝛼₁(𝑠)𝑑𝑠≤𝛼₁(0)+𝜇^𝑇₀𝛼(𝑠)𝑑𝑠=𝛼₁(𝑇),𝜇≤0. (2.21)
Thus 𝛼₁ is lower solution of (1.1).
Suppose that 𝛼_𝑛 is lower solution of (1.1) when 𝑛=𝑚.
Then when 𝑛=𝑚+1,
𝛼_𝑚+1(𝑡)=𝑓(𝑡,𝛼_𝑚(𝑡),[𝐾𝛼_𝑚](𝑡))−𝑏(𝑡)𝛼_𝑚(𝑡)−𝑀(𝛼_𝑚+1(𝑡)−𝛼_𝑚(𝑡))−𝑓(𝑡,𝛼_𝑚+1(𝑡),[𝐾𝛼_𝑚+1](𝑡))+𝑓(𝑡,𝛼_𝑚+1(𝑡),[𝐾𝛼_𝑚+1](𝑡))≤𝑓(𝑡,𝛼_𝑚+1(𝑡),[𝐾𝛼_𝑚+1](𝑡))−𝑏(𝑡)𝛼_𝑚+1(𝑡),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝛼_𝑚+1(𝑡_𝑘)=𝐼_𝑘(𝛼_𝑚(𝑡_𝑘))−𝐿_𝑘(𝛼_𝑚+1(𝑡_𝑘)−𝛼_𝑚(𝑡_𝑘))+𝐼_𝑘(𝛼_𝑚+1(𝑡_𝑘))−𝐼_𝑘(𝛼_𝑚+1(𝑡_𝑘))≤𝐼_𝑘(𝛼_𝑚+1(𝑡_𝑘𝛼)),𝑘=1,2,…,𝑚,_𝑚+1(0)+𝜇^𝑇₀𝛼_𝑚+1(𝑠)𝑑𝑠≤𝛼_𝑚+1(0)+𝜇^𝑇₀𝛼_𝑚(𝑠)𝑑𝑠=𝛼_𝑚+1(𝑇). (2.22)
Thus by mathematical induction we can know that 𝛼_𝑛 is lower solution of (1.1). In the same way we can prove that 𝛽_𝑛 is upper solution of (1.1).
By 𝛼_𝑛−1≤𝛼_𝑛≤𝛽_𝑛≤𝛽_𝑛−1,𝑛=0,1,2,…,𝑡∈𝐽, we can know that when 𝑛→+∞,{𝛼_𝑛},{𝛽_𝑛} have limits 𝜌(𝑡),𝑟(𝑡), respectively. Since they are independent of 𝑡 when 𝑛→+∞{𝛼_𝑛}, {𝛽_𝑛} converge uniformly to 𝜌(𝑡),𝑟(𝑡) and 𝛼_𝑛≤𝜌(𝑡)≤𝑟(𝑡)≤𝛽_𝑛≤𝛽_𝑛−1,𝑛=0,1,2,…,𝑡∈𝐽.
According to 𝛼_𝑛,𝛽_𝑛 satisfying (2.13), that is,
𝛼^′_𝑛(𝑡)=𝑓(𝑡,𝛼_𝑛−1(𝑡),[𝐾𝛼_𝑛−1](𝑡))−𝑏(𝑡)𝛼_𝑛−1(𝑡)−𝑀(𝛼_𝑛(𝑡)−𝛼_𝑛−1(𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝛼_𝑛(𝑡_𝑘)=𝐼_𝑘(𝛼_𝑛−1(𝑡_𝑘))−𝐿_𝑘(𝛼_𝑛(𝑡_𝑘)−𝛼_𝑛−1(𝑡_𝑘𝛼)),𝑘=1,2,…,𝑚,_𝑛(0)+𝜇^𝑇₀𝛼_𝑛−1(𝑠)𝑑𝑠=𝛼_𝑛𝛽(𝑇),𝜇≤0,_′𝑛(𝑡)=𝑓(𝑡,𝛽_𝑛−1(𝑡),[𝐾𝛽_𝑛−1](𝑡))−𝑏(𝑡)𝛽_𝑛−1(𝑡)−𝑀(𝛽_𝑛(𝑡)−𝛽_𝑛−1(𝑡)),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝛽_𝑛(𝑡_𝑘)=𝐼_𝑘(𝛽_𝑛−1(𝑡_𝑘))−𝐿_𝑘(𝛽_𝑛(𝑡_𝑘)−𝛽_𝑛−1(𝑡_𝑘𝛽)),𝑘=1,2,…,𝑚,_𝑛(0)+𝜇^𝑇₀𝛽_𝑛−1(𝑠)𝑑𝑠=𝛽_𝑛(𝑇),𝜇≤0, (2.23) when 𝑛→+∞, we have 𝜌^′(𝑡)=𝑓(𝑡,𝜌(𝑡),[𝐾𝜌](𝑡))−𝑏(𝑡)𝜌(𝑡),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝜌(𝑡_𝑘)=𝐼_𝑘(𝜌(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝜌(0)+𝜇^𝑇₀𝑟𝜌(𝑠)𝑑𝑠=𝜌(𝑇),𝜇≤0._′(𝑡)=𝑓(𝑡,𝑟(𝑡),[𝐾𝑟](𝑡))−𝑏(𝑡)𝑟(𝑡),𝑡≠𝑡_𝑘,𝑡∈𝐽,Δ𝑟(𝑡_𝑘)=𝐼_𝑘(𝑟(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝑟(0)+𝜇^𝑇₀𝑟(𝑠)𝑑𝑠=𝑟(𝑇),𝜇≤0. (2.24) Equation (2.24) indicates that 𝜌(𝑡),𝑟(𝑡) are solutions of (1.1).
Lastly, we prove that 𝜌(𝑡),𝑟(𝑡) are minimal and maximal solutions of the equation (1.1) in [𝛼,𝛽].
Suppose that 𝑥(𝑡) is a solution of the equation and satisfies 𝑥(𝑡)∈[𝛼,𝛽],𝑡∈𝐽, obviously, we can assume that there is an 𝑛 such that 𝛼_𝑛≤𝑥≤𝛽_𝑛.
If 𝑝(𝑡)=𝛼_𝑛+1−𝑥, then
𝑝^′=𝛼^′_𝑛+1−𝑥^′≤−𝑏(𝑡)𝛼_𝑛(𝑡)+𝑓(𝑡,𝛼_𝑛(𝑡),[𝐾𝛼_𝑛](𝑡))−𝑀(𝛼_𝑛+1(𝑡)−𝛼_𝑛(𝑡))−𝑓(𝑡,𝑥(𝑡),[𝐾𝑥(𝑡)](𝑡))+𝑏(𝑡)𝑥(𝑡)≤−𝑏(𝑡)(𝛼_𝑛−𝑥)−𝑀𝑝(𝑡),𝑡≠𝑡_𝑘. (2.25)
And since 𝑏(𝑡)≤0, 𝛼_𝑚−𝑥≤0, 𝑝^′≤−𝑀𝑝(𝑡),
Δ𝑝(𝑡_𝑘)=Δ𝛼_𝑛+1−Δ𝑥=𝐼_𝑘(𝛼_𝑛(𝑡_𝑘))−𝐿_𝑘(𝛼_𝑛+1(𝑡_𝑘)−𝛼_𝑛(𝑡_𝑘))−𝐼_𝑘(𝑥(𝑡_𝑘))≤−𝐿_𝑘𝑝(𝑡_𝑘),𝑡=𝑡_𝑘,𝑝(0)=𝛼_𝑛+1(0)−𝑥(0)=𝜇^𝑇₀(𝑥(𝑠)−𝛼_𝑛(𝑠))𝑑𝑠+𝑝(𝑇)≤𝑝(𝑇). (2.26) Hence by Lemma 2.2, we can obtain 𝑝(𝑡)≤0,𝑡∈𝐽, so 𝛼_𝑛+1(𝑡)≤𝑥(𝑡). Similarly, we can obtain: 𝑥(𝑡)≤𝛽_𝑛+1(𝑡),𝑡∈𝐽. This indicates that 𝛼_𝑛(𝑡)≤𝑥(𝑡)≤𝛽_𝑛+1(𝑡),𝑡∈𝐽,𝑛=0,1,2,…. Hence when 𝑛→+∞, we can obtain that 𝜌(𝑡)≤𝑥(𝑡)≤𝑟(𝑡),𝑡∈𝐽. This ends the proof.

Finally, we give an example to illustrate the efficiency of our results.

Example 2.5. Consider the problem of 𝑥^′(𝑡)−𝑥(𝑡)sin𝑡=−𝑥(𝑡)+𝑡+^𝑡₀𝑥(𝑠)𝑑𝑠,0<𝑡<1,𝑡≠𝑡₁,Δ𝑥(𝑡₁1)=−8𝑥(𝑡₁),𝑡₁=12,𝑥(0)−¹₀𝑥(𝑠)𝑑𝑠=𝑥(1), (2.27) where 𝑏(𝑡)=sin𝑡,𝑓(𝑡,𝑥(𝑡),𝐾𝑥(𝑡))=−𝑥(𝑡)+𝑡+∫^𝑡₀𝑥(𝑠)𝑑𝑠,𝐼₁(𝑥)=−𝑥. Obviously, 𝛼(𝑡)=0,𝛽(𝑡)=1−𝑡 are the lower solution and upper solution for (2.27) with 𝛼(𝑡)≤𝛽(𝑡), respectively. 𝑓(𝑡,𝑥,𝐾𝑥)−𝑓(𝑡,𝑦,𝐾𝑦)=−(𝑥−𝑦)−∫^𝑡₀(𝑥(𝑠)−𝑦(𝑠))𝑑𝑠, 𝐼₁(𝑥)−𝐼₁(𝑦)=−(𝑥−𝑦). Let 𝑇=1,𝐿_𝑘=1/8, the conditions of Theorem 2.4 are all satisfied, so problem (2.27) has the maximal and minimal solutions in the segement [𝛼(𝑡),𝛽(𝑡)].

Acknowledgments

This work was supported by the NNSF of China (10571050; 10871062), the Hunan Provincial Natural Science Foundation of China (09JJ6010), and the Zhejiang Provincial Natural Science Foundation of China (Y6090057).