Existence Results for Nonlinear Impulsive System with Causal Operators

Abstract

In this paper, we establish sufficient conditions for some existence results for nonlinear impulsive differential equations involving causal operators. Our method is based on the monotone iterative technique, a new differential inequality, and the Schauder fixed point theorem. Moreover, we consider three impulsive differential equations as applications to verify our theoretical results.

1. Introduction

Impulsive systems turned out to be the most effective tools for describing many evolutionary progresses that experience instantaneous changes of state at certain moments. Pioneering studies on impulsive differential equations and their dynamics are detailed in [1], establishing a fundamental theoretical framework for impulsive systems. Additionally, a comprehensive analysis of the system properties within impulsive systems is provided in the monographs [2]. Beyond the theory [3,4,5,6,7,8,9], impulsive systems are widely used in biological systems, control systems, ecological systems, and neural network systems [10,11,12]. As a result of these successful applications in various fields, impulsive differential equations have attracted considerable attention.

The theory of differential equations with causal operators is experiencing an important development because it is a framework richer than the corresponding theory of ordinary differential equations. A causal operator is a non-anticipative operator and has been adopted from the engineering literature; some results have been introduced in the monograph [13]. Recently, various types of causal differential equations have been studied widely, such as ordinal differential equations [14,15], functional differential equations [16], differential equations in Banach spaces [17], difference equations [18], and integral differential equations [19]. In addition, Jabeen et al. [20] investigated impulsive differential equations with causal operators and gave the existence of an optimal solution for the control problem. Inspired by the above results, the aim of this paper is to study the impulsive differential equations further enriched by the causal operators, while subject to nonlinear periodic boundary conditions. Therefore, we discuss the following impulsive differential equations involving a causal operator with nonlinear periodic boundary conditions:

$$\left\{\begin{array}{c}{x}^{\prime }\left(t\right)=\left(\mathfrak{R}x\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }x\left(t_k\right)=I_k\left(x\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(x\right(0),x(T\left)\right)=0,\hfill \end{array}\right.$$

(1)

where $J=[0,T](T>0)$, E is a real separable Banach space of continuous functions from $[0,T]$ to the set of real numbers $\mathbb{R}$, $\mathfrak{R}\in C(E,E)$ is a causal operator, $h\in C(\mathbb{R}\times \mathbb{R},\mathbb{R})$, $0=t_0<t_1<t_2<\dots <t_m<t_{m+1}=T$, $I_k\in C(\mathbb{R},\mathbb{R})$, and $\mathsf{\Delta }x\left(t_k\right)=x\left(t_k^{+}\right)-x\left(t_k^{-}\right)$, where $x\left(t_k^{+}\right)$ and $x\left(t_k^{-}\right)$ denote the right and left limits of $x\left(t\right)$ at $t=t_k(k=1,2,\dots ,m)$.

One point of interest of our study lies in the fact that the periodic boundary conditions are nonlinear and encompass the usual linear boundary conditions (such as initial, periodic, and anti-periodic) and other general conditions, such as $e^x\left(0\right)-x\left(T\right)=0$ and ${\displaystyle y\left(0\right)-{\int }_0^{T}y\left(t\right)dt}$ = C (C is a constant). In addition, note that the impulsive differential equations (1) reduce to ordinary differential equations for $I_k\equiv 0$, which has been studied in [14], initial boundary problems for $x\left(0\right)=\xi$, which has been studied in [20], and other differential equations, such as $\left(\mathfrak{R}x\right)\left(t\right)=f(t,x(t),(Tx\left)\right(t),(Sx\left)\right(t\left)\right)$, which has been studied in [21]. Thus, our work includes more types of differential equations.

In this paper, we extend the notion of causal operators to the nonlinear periodic boundary value problems for impulsive differential equations. The rest of this paper is organized as follows. Section 2 establishes a new differential inequality. Section 3 presents the existence of extremal solutions, following the definition of upper and lower solutions. Section 4 gives the existence of extremal quasi-solutions after the definition of coupled lower and upper solutions. In Section 5, a weakly coupled extremal quasi-solution is studied, which is dependent on the introduction of weakly coupled lower and upper solutions. In addition, examples are added in each section to verify the theoretical results.

2. Preliminaries

Let $C(\mathbb{R},\mathbb{R})$ denote the set of real valued continuous functions and $J_0=J\setminus \{t_1,t_2,\dots ,t_m\}$. Let us introduce the space:

$$PC(J,\mathbb{R})=\left\{\begin{array}{c}x:J\to \mathbb{R};x\left(t\right)\mathrm{is}\mathrm{continuous}\mathrm{everywhere}\mathrm{except}\mathrm{for}\mathrm{some}{t}_k\\ \mathrm{at}\mathrm{which}x\left(t_k^{-}\right)\mathrm{and}x\left(t_k^{+}\right)\mathrm{exist}\mathrm{and}x\left(t_k^{-}\right)=x\left(t_k\right),k=1,2,\cdots ,m\end{array}\right\}$$

$$P{C}^1(J,\mathbb{R})=\left\{\begin{array}{c}x\in PC(J,\mathbb{R});x\mathrm{is}\mathrm{continuously}\mathrm{differentiable}\mathrm{for}\mathrm{any}t\in J_0,\mathrm{where}\\ x^{\prime }\left(0^{+}\right),x^{\prime }\left(T^{-}\right),x^{\prime }\left(t_k^{+}\right)\mathrm{and}{x}^{\prime }\left(t_k^{-}\right)\mathrm{exist},k=1,2,\cdots ,p\end{array}\right\}.$$

Clearly, $PC(J,\mathbb{R})$, $P{C}^1(J,\mathbb{R})$ are Banach spaces with the following respective norms:

${\displaystyle {\parallel x\parallel }_{PC(J,\mathbb{R})}=\underset{t\in J}{sup}\left|x\left(t\right)\right|}$, ${\parallel x\parallel }_{P{C}^1(J,\mathbb{R})}={\parallel x\parallel }_{PC(J,\mathbb{R})}+{\parallel x^{\prime }\parallel }_{PC(J,\mathbb{R})}$.

Let $\mathsf{\Omega }=P{C}^1(J,\mathbb{R})$. A function $x\in \mathsf{\Omega }$ is called a solution of (1) if it satisfies (1).

Definition 1.

An operator $\mathfrak{R}\in C(E,E)$ is a causal operator if for all $\zeta ,\xi \in E$, with $\zeta \left(t\right)=\xi \left(t\right)$, such that for every $t\in [0,T]$, we have $\left(\mathfrak{R}\zeta \right)\left(t\right)=\left(\mathfrak{R}\xi \right)\left(t\right)$.

Lemma 1.

Suppose that $p\in \mathsf{\Omega }$ satisfies

$$\left\{\begin{array}{c}{p}^{\prime }\left(t\right)\le -M\left(t\right)p\left(t\right)-\left(\mathcal{L}p\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }p\left(t_k\right)\le -{\mathfrak{L}}_{k}p\left(t_k\right),k=1,2,\dots ,m,\hfill \\ p\left(0\right)\le \lambda p\left(T\right),\hfill \end{array}\right.$$

(2)

where $M\in C(J,[0,+\infty \left)\right)$, $0\le {\mathfrak{L}}_k<1$, $k=1,2,\dots ,m$, and $\mathcal{L}\in C(E,E)$ is a positive linear operator, i.e., $\mathcal{L}z\ge 0$ wherever $z\ge 0$, satisfies

$${\displaystyle {\int }_0^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{k=1}^m{\mathfrak{L}}_k\le \lambda ,where1\left(t\right)=1,0<\lambda \le 1.}$$

(3)

Then $p\left(t\right)\le 0$ for $t\in J$.

Proof. 

If the assumption $p\left(t\right)\le 0,\forall t\in J$ does not hold, two cases arise:

Case 1: There exists $\overline{t}\in J$ such that $p\left(\overline{t}\right)>0$ and $p\left(t\right)\ge 0$ for all $t\in J$.

Then, from (2), we have $p^{\prime }\left(t\right)\le 0$ for $t\ne t_k$ and $\mathsf{\Delta }p\left(t_k\right)\le 0(k=1,2,\dots ,m)$, hence $p\left(t\right)$ is nonincreasing on J. If $\lambda =1$, $p\left(0\right)\le p\left(T\right)$ shows that $p\left(t\right)\equiv c$ (c is a constant), then $p^{\prime }\left(t\right)=0$. Notice that $p\left(\overline{t}\right)>0$, we have $0\equiv p^{\prime }\left(\overline{t}\right)\le -Mp\left(\overline{t}\right)-\left(\mathcal{L}p\right)\left(\overline{t}\right)<0$, which is a contradiction. For $0<\lambda <1$, it follows that $p\left(T\right)\le p\left(0\right)\le \lambda p\left(T\right)$, presenting another contradiction.

Case 2: There exist $t_{*}$ and $t^{*}$, such that $p\left(t_{*}\right)<0$ and $p\left(t^{*}\right)>0$.

Given ${\displaystyle \underset{t\in J}{inf}p\left(t\right)=-r}$, $r>0$, and for a specific $i\in \{1,2,\dots ,m\}$, there is a $t_{*}\in (t_i,t_{i+1}]$, such that $p\left(t_{*}\right)=-r$ or $p\left(t_{*}^{+}\right)=-r$. We only focus on the case where $p\left(t_{*}\right)=-r$, as the proof for the case $p\left(t_{*}^{+}\right)=-r$ is analogous.

Subcase (1): If $t_{*}<t^{*}$, from (2), we have

$$\begin{array}{cc}p\left(t^{*}\right)-p\left(t_{*}\right)\hfill & {\displaystyle ={\int }_{t_{*}}^{t^{*}}{p}^{\prime }\left(t\right)dt+\sum _{t_{*}<t_k<t^{*}}\mathsf{\Delta }p\left(t_k\right)}\hfill \\ & {\displaystyle \le -{\int }_{t_{*}}^{t^{*}}[M\left(t\right)p\left(t\right)+\left(\mathcal{L}p\right)\left(t\right)]dt-\sum _{t_{*}<t_k<t^{*}}{\mathfrak{L}}_{k}p\left(t_k\right)}\hfill \\ & \le r\left\{{\displaystyle {\int }_0^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{k=1}^m{\mathfrak{L}}_k}\right\},\hfill \end{array}$$

and hence

$$\begin{array}{cc}0<p\left(t^{*}\right)\hfill & \le -r+r\left\{{\displaystyle {\int }_0^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{k=1}^m{\mathfrak{L}}_k}\right\}\hfill \\ & =r\left\{{\displaystyle {\int }_0^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{k=1}^m{\mathfrak{L}}_k-1}\right\},\hfill \end{array}$$

which yields

$${\displaystyle {\int }_0^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{k=1}^m{\mathfrak{L}}_k>1,}$$

which contradicts (3).

Subcase (2): If $t_{*}>t^{*}$, from (2), we get

$$\begin{array}{cc}p\left(T\right)\hfill & {\displaystyle =p\left(t_{*}\right)+{\int }_{t_{*}}^T{p}^{\prime }\left(t\right)dt+\sum _{t_{*}<t_k<T}\mathsf{\Delta }p\left(t_k\right)}\hfill \\ & {\displaystyle \le p\left(t_{*}\right)-{\int }_{t_{*}}^T[M\left(t\right)p\left(t\right)+\left(\mathcal{L}p\right)]dt-\sum _{t_{*}<t_k<T}{\mathfrak{L}}_{k}p\left(t_k\right)}\hfill \\ & {\displaystyle \le p\left(t_{*}\right)+r{\int }_{t_{*}}^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+r\sum _{t_{*}<t_k<T}{\mathfrak{L}}_k,}\hfill \end{array}$$

and

$$\begin{array}{cc}p\left(t^{*}\right)\hfill & {\displaystyle =p\left(0\right)+{\int }_0^{t^{*}}{p}^{\prime }\left(t\right)dt+\sum _{0<t_k<t^{*}}\mathsf{\Delta }p\left(t_k\right)}\hfill \\ & {\displaystyle \le p\left(0\right)-{\int }_0^{t^{*}}[M\left(t\right)p\left(t\right)+\left(\mathcal{L}p\right)]dt-\sum _{0<t_k<t^{*}}{\mathfrak{L}}_{k}p\left(t_k\right)}\hfill \\ & {\displaystyle \le p\left(0\right)+r{\int }_0^{t^{*}}[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+r\sum _{0<t_k<t^{*}}{\mathfrak{L}}_k.}\hfill \end{array}$$

Using $p\left(0\right)\le \lambda p\left(T\right)$, $0<\lambda \le 1$, we get

$$\begin{array}{cc}0<p\left(t^{*}\right)\hfill & {\displaystyle \le \lambda \left(p\left(t_{*}\right)+r{\int }_{t_{*}}^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{t_{*}<t_k<T}{\mathfrak{L}}_k\right)}\hfill \\ & {\displaystyle +r{\int }_0^{t^{*}}[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{0<t_k<t^{*}}{\mathfrak{L}}_k}\hfill \\ & \le -r\lambda +r\left\{{\displaystyle {\int }_0^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{k=1}^m{\mathfrak{L}}_k}\right\}\hfill \\ & =r\left\{{\displaystyle {\int }_0^T[M\left(t\right)+\left(\mathcal{L}1\right)\left(t\right)]dt+\sum _{k=1}^m{\mathfrak{L}}_k-\lambda }\right\},\hfill \end{array}$$

which is a contradiction. Then, we get $p\left(t\right)\le 0$, $t\in J$. The proof is complete. □

Next, we give the following linear problems and lemmas, which help to validate our main results.

$$\left\{\begin{array}{c}{x}^{\prime }\left(t\right)=-M\left(t\right)x\left(t\right)-\left(\mathcal{L}x\right)\left(t\right)+{\sigma }_{\eta }\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }x\left(t_k\right)=-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right),k=1,2,\dots ,m,\hfill \\ h(\eta \left(0\right),\eta \left(T\right))+M_1(x\left(0\right)-\eta \left(0\right))-M_2(x\left(t\right)-\eta \left(T\right))=0,\hfill \end{array}\right.$$

(4)

where $M\in C(J,[0,+\infty \left)\right)$, $0\le {\mathfrak{L}}_k<1$, $k=1,2,\dots ,m$, $\eta \in \mathsf{\Omega }$, ${\sigma }_{\eta }\left(t\right)=\left(\mathfrak{R}\eta \right)\left(t\right)+M\left(t\right)\eta \left(t\right)+\left(\mathcal{L}\eta \right)\left(t\right)$.

Lemma 2.

A function $x\in \mathsf{\Omega }$ is a solution to (4) if and only if $x\in E$ satisfies the following integral equation:

$$\begin{array}{cc}x\left(t\right)=\hfill & {\displaystyle \frac{A_{\eta }{e}^{-{\int }_0^{t}M\left(r\right)dr}}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}+{\int }_0^{T}G(t,s)[{\sigma }_{\eta }\left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds}\hfill \\ \hfill & {\displaystyle +\sum _{0<t_k<T}G(t,t_k)[-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right)],}\hfill \end{array}$$

(5)

where $A_{\eta }=-h(\eta \left(0\right),\eta \left(T\right))+M_1\eta \left(0\right)-M_2\eta \left(T\right)$, $M\in C(J,[0,+\infty \left)\right)$, $M_1,M_2$ are constants with $M_1\ne M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}$ and

$$G(t,s)=\left\{\begin{array}{cc}{\displaystyle \frac{M_2}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}}{e}^{-{\int }_0^{t}M\left(r\right)dr}{e}^{-{\int }_s^{T}M\left(r\right)dr}+e^{-{\int }_s^{t}M\left(r\right)dr},\hfill & 0\le s<t\le T,\hfill \\ {\displaystyle \frac{M_2}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}}{e}^{-{\int }_0^{t}M\left(r\right)dr}{e}^{-{\int }_s^{T}M\left(r\right)dr},\hfill & 0\le t\le s\le T.\hfill \end{array}\right.$$

Proof. 

Assume $x\in E$ is a solution to (4); we have

$$x^{\prime }\left(t\right)+M\left(t\right)x\left(t\right)=-\left(\mathcal{L}x\right)\left(t\right)+{\sigma }_{\eta }\left(t\right),t\ne t_k,t\in J,$$

$$\mathsf{\Delta }x\left(t_k\right)=-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right),k=1,2,\dots ,m.$$

It is easy to obtain the following formula:

$$\begin{array}{cc}x\left(t\right)=\hfill & {\displaystyle x\left(0\right)e^{-{\int }_0^{t}M\left(r\right)dr}+{\int }_0^t{e}^{{\int }_t^{s}M\left(r\right)dr}[{\sigma }_{\eta }\left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds}\hfill \\ & {\displaystyle +\sum _{0<t_k<t}{e}^{{\int }_t^{t_k}M\left(r\right)dr}[-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right)].}\hfill \end{array}$$

Let $t=T$; we get

$$\begin{array}{cc}x\left(T\right)=\hfill & {\displaystyle x\left(0\right)e^{-{\int }_0^{T}M\left(r\right)dr}+{\int }_0^T{e}^{{\int }_T^{s}M\left(r\right)dr}[{\sigma }_{\eta }\left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds}\hfill \\ & {\displaystyle +\sum _{0<t_k<T}{e}^{{\int }_T^{t_k}M\left(r\right)dr}[-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right)].}\hfill \end{array}$$

Since $M_{1}x\left(0\right)-M_{2}x\left(t\right)=A_{\eta }$, we obtain

$$\begin{array}{cc}x\left(0\right)\hfill & {\displaystyle =\frac{A_{\eta }}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}+\frac{M_2}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}{\int }_0^T{e}^{{\int }_T^{s}M\left(r\right)dr}[{\sigma }_{\eta }\left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds}\hfill \\ & {\displaystyle +\frac{M_2}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}\sum _{0<t_k<T}{e}^{{\int }_T^{t_k}M\left(r\right)dr}[-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right)].}\hfill \end{array}$$

Then

$$\begin{array}{cc}x\left(t\right)=\hfill & {\displaystyle \frac{A_{\eta }}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}+{\int }_0^{T}G(t,s)[{\sigma }_{\eta }\left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds}\hfill \\ & {\displaystyle +\sum _{0<t_k<T}G(t,t_k)[-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right)],}\hfill \end{array}$$

where

$$G(t,s)=\left\{\begin{array}{cc}{\displaystyle \frac{M_2}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}}{e}^{-{\int }_0^{t}M\left(r\right)dr}{e}^{-{\int }_s^{T}M\left(r\right)dr}+e^{-{\int }_s^{t}M\left(r\right)dr},\hfill & 0\le s<t\le T,\hfill \\ {\displaystyle \frac{M_2}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}}{e}^{-{\int }_0^{t}M\left(r\right)dr}{e}^{-{\int }_s^{T}M\left(r\right)dr},\hfill & 0\le t\le s\le T.\hfill \end{array}\right.$$

We see that if $x\left(t\right)$ is a solution to (4), then $x\left(t\right)$ is also a solution to (5). The proof is complete. □

Apparently, $\parallel G(t,s)\parallel =max\left\{\left|{\displaystyle \frac{M_1}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}}\right|,\left|{\displaystyle \frac{M_2}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}}\right|\right\}$. In the remainder of the paper, we denote $\parallel G(t,s)\parallel =\tau$.

Lemma 3.

Let $\mathcal{L}\in C(E,E)$ be a positive linear operator; $M\in C(J,[0,+\infty \left)\right)$, $M_1\ne h{e}^{-{\int }_0^{T}M\left(r\right)dr}$ and

$$\tau \left(\parallel \mathcal{L}\parallel T+\sum _{k=1}^m\left|{\mathfrak{L}}_k\right|\right)<1.$$

(6)

Then, problem (4) has a unique solution.

Proof. 

For any $x\in E_0$, let us define

$$\begin{array}{cc}\left(Fx\right)\left(t\right)=\hfill & {\displaystyle \frac{A_{\eta }{e}^{-{\int }_0^{t}M\left(r\right)dr}}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}+{\int }_0^{T}G(t,s)[{\sigma }_{\eta }\left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds}\hfill \\ & {\displaystyle +\sum _{0<t_k<t}G(t,t_k)[-{\mathfrak{L}}_{k}x\left(t_k\right)+I_k\left(\eta \left(t_k\right)\right)+{\mathfrak{L}}_k\eta \left(t_k\right)],}\hfill \end{array}$$

and $G(t,s)$, as defined in Lemma 4. For any $x_1,x_2\in E_0$, we have

$$\parallel F{x}_1-F{x}_2{\parallel }_{E_0}\le \tau \left(\parallel \mathcal{L}\parallel T+\sum _{k=1}^m\left|{\mathfrak{L}}_k\right|\right)\parallel x_2-x_1\parallel .$$

Hence, by the Banach contraction principle, there exists a function $x\in E_0$ such that $x=Fx$. Apparently, x is also the unique solution to (4), thus concluding the proof. □

Remark 1.

If $M\in C(J,[0,+\infty \left)\right)$, $M_1\ge M_2>0$, $\mathcal{L}\in C(E,E)$ is a positive linear operator, and

$${\displaystyle \frac{M_1}{M_1-M_2{e}^{-{\int }_0^{T}M\left(r\right)dr}}}\left(\parallel \mathcal{L}\parallel T+\sum _{k=1}^m\left|{\mathfrak{L}}_k\right|\right)<1.$$

(7)

Then, problem (4) has a unique solution.

3. Extremal Solutions of Problem (1)

A function $\mathfrak{m}\in \mathsf{\Omega }$ is called a lower solution of (1) if

$${\mathfrak{m}}^{\prime }\left(t\right)\le \left(\mathfrak{R}\mathfrak{m}\right)\left(t\right),t\ne t_k,\mathsf{\Delta }\mathfrak{m}\left(t_k\right)\le I_k\left(\mathfrak{m}\left(t_k\right)\right),h(\mathfrak{m}\left(0\right),\mathfrak{m}\left(T\right))\le 0,$$

where $t\in J$, $k=1,2,\dots ,m$.

A function $\mathfrak{n}\in \mathsf{\Omega }$ is defined as an upper solution of problem (1) if the above inequalities are reversed.

Theorem 1.

Suppose that (3), (6) hold and $\mathfrak{R}\in C(E,E)$. We also assume that

$\left(H_1\right)$

$\mathfrak{m},\mathfrak{n}\in \mathsf{\Omega }$ are the lower and upper solutions of problem (1), respectively, and $\mathfrak{m}\left(t\right)\le \mathfrak{n}\left(t\right)$ on J;

$\left(H_2\right)$

there exist $M\in C(J,[0,+\infty \left)\right)$, $0\le {\mathfrak{L}}_k<1$, and $k=1,2,\dots ,m$, and $\mathcal{L}\in C(E,E)$ is a positive linear operator that satisfies

$$\left(\mathfrak{R}{x}_1\right)\left(t\right)-\left(\mathfrak{R}{x}_2\right)\left(t\right)\le -M\left(t\right)(x_1\left(t\right)-x_2\left(t\right))-\left(\mathcal{L}(x_1-x_2)\right)\left(t\right),$$

for $\mathfrak{m}\left(t\right)\le x_1\left(t\right)\le x_2\left(t\right)\le \mathfrak{n}\left(t\right)$, $\forall t\in J,t\ne t_k$;

$\left(H_3\right)$

the function $I_k\in C(\mathbb{R},\mathbb{R})$ satisfies

$$I_k\left(x_1\left(t_k\right)\right)-I_k\left(x_2\left(t_k\right)\right)\le -{\mathfrak{L}}_k(x_1\left(t_k\right)-x_2\left(t_k\right))$$

for $\mathfrak{m}\left(t_k\right)\le x_1\left(t_k\right)\le x_2\left(t_k\right)\le \mathfrak{n}\left(t_k\right),k=1,2,\dots ,m$;

$\left(H_4\right)$

the function $h(x,z)\in C(\mathbb{R}\times \mathbb{R},\mathbb{R})$ satisfies

$$h(\overline{x},\overline{z})-h(x,z)\le M_1(\overline{x}-x)-M_2(\overline{z}-z),$$

wherever $\mathfrak{m}\left(0\right)\le x\le \overline{x}\le \mathfrak{n}\left(0\right)$, $\mathfrak{m}\left(T\right)\le z\le \overline{z}\le \mathfrak{n}\left(T\right)$, where $M_1\ge M_2>0$ and $\lambda ={\displaystyle \frac{M_2}{M_1}}$.

Then problem (1) has minimal and maximal solutions within the interval $[\mathfrak{m},\mathfrak{n}]=\{x\in \mathsf{\Omega }:\mathfrak{m}(t)\le x(t)\le \mathfrak{n}(t),t\in J\}$.

Proof. 

For $\eta \in [\mathfrak{m},\mathfrak{n}]$, ${\sigma }_{\eta }\left(t\right)=\left(\mathfrak{R}\eta \right)\left(t\right)+M\left(t\right)\eta \left(t\right)+\left(\mathcal{L}\eta \right)\left(t\right)$. By Lemma 3, problem (4) has exactly one solution $x\in \mathsf{\Omega }$. When $\mathcal{F}\eta =x$ is used to define an operator $\mathcal{F}$, it possesses the following properties:

(a) $\mathfrak{m}\le \mathcal{F}\mathfrak{m}$, $\mathfrak{n}\ge \mathcal{F}\mathfrak{n}$.

Set $p\left(t\right)=\mathfrak{m}\left(t\right)-{\mathfrak{m}}_1\left(t\right)$, where ${\mathfrak{m}}_1=\mathcal{F}\mathfrak{m}$. Employing $\left(H_1\right)$, we have

$$\begin{array}{cc}{p}^{\prime }\left(t\right)\hfill & ={\mathfrak{m}}^{\prime }\left(t\right)-{\mathfrak{m}}_1^{\prime }\left(t\right)\hfill \\ & \le \left(\mathfrak{R}\mathfrak{m}\right)\left(t\right)-\left[\left(\mathfrak{R}\mathfrak{m}\right)\left(t\right)+M\left(t\right)\mathfrak{m}\left(t\right)+\left(\mathcal{L}\mathfrak{m}\right)\left(t\right)-M\left(t\right){\mathfrak{m}}_1\left(t\right)-\left(\mathcal{L}{\mathfrak{m}}_1\right)\left(t\right)\right]\hfill \\ & =-M\left(t\right)p\left(t\right)-\left(\mathcal{L}p\right)\left(t\right),t\ne t_k,t\in J,\hfill \end{array}$$

$$\begin{array}{cc}\mathsf{\Delta }p\left(t_k\right)\hfill & =\mathsf{\Delta }\mathfrak{m}\left(t_k\right)-\mathsf{\Delta }{\mathfrak{m}}_1\left(t_k\right)\hfill \\ & \le I_k\left(\mathfrak{m}\left(t_k\right)\right)-\left[-{\mathfrak{L}}_k{\mathfrak{m}}_1\left(t_k\right)+I_k\left({\mathfrak{m}}_1\left(t_k\right)\right)+{\mathfrak{L}}_k\mathfrak{m}\left(t_k\right)\right]\hfill \\ & =-{\mathfrak{L}}_{k}p\left(t_k\right),k=1,2,\dots ,m,\hfill \end{array}$$

and

$$\begin{array}{cc}p\left(0\right)\hfill & =\mathfrak{m}\left(0\right)-{\mathfrak{m}}_1\left(0\right)\hfill \\ & =\mathfrak{m}\left(0\right)-\left[{\displaystyle -\frac{1}{M_1}h(\mathfrak{m}\left(0\right),\mathfrak{m}\left(T\right))+\frac{M_2}{M_1}({\mathfrak{m}}_1\left(T\right)-\mathfrak{m}\left(T\right))+\mathfrak{m}\left(0\right)}\right]\hfill \\ & {\displaystyle =\frac{1}{M_1}h(\mathfrak{m}\left(0\right),\mathfrak{m}\left(T\right))+\frac{M_2}{M_1}(\mathfrak{m}\left(T\right)-{\mathfrak{m}}_1\left(T\right))}\hfill \\ & {\displaystyle \le \frac{M_2}{M_1}p\left(T\right).}\hfill \end{array}$$

From Lemma 1 and $M_1\ge M_2>0$, we get $p\le 0$, so $\mathfrak{m}\le {\mathfrak{m}}_1$. Similarly, we have $\mathcal{F}\mathfrak{n}\le \mathfrak{n}$.

(b) Operator $\mathcal{F}$ is nondecreasing monotonically.

Let ${\eta }_1,{\eta }_2\in [\mathfrak{m},\mathfrak{n}]$, such that ${\eta }_1\le {\eta }_2$. Assume that $u_1=\mathcal{F}{\eta }_1$, $u_2=\mathcal{F}{\eta }_2$ and $p\left(t\right)=u_1\left(t\right)-u_2\left(t\right)$. Applying the fact of ${\eta }_1\le {\eta }_2$, $\left(H_2\right)$, $\left(H_3\right)$ and $\left(H_4\right)$, we have

$$\begin{array}{cc}{p}^{\prime }\left(t\right)\hfill & =u_1^{\prime }\left(t\right)-u_2^{\prime }\left(t\right)\hfill \\ & =-M\left(t\right)p\left(t\right)-\left(\mathcal{L}p\right)\left(t\right)\hfill \\ & +\left[\left(\mathfrak{R}{\eta }_1\right)\left(t\right)-\left(\mathfrak{R}{\eta }_2\right)\left(t\right)+M\left(t\right)({\eta }_1\left(t\right)-{\eta }_2\left(t\right))+\left(\mathcal{L}({\eta }_1-{\eta }_2)\right)\left(t\right)\right]\hfill \\ & \le -M\left(t\right)p\left(t\right)-\left(\mathcal{L}p\right)\left(t\right),t\ne t_k,t\in J,\hfill \end{array}$$

$$\begin{array}{cc}\mathsf{\Delta }p\left(t_k\right)\hfill & =\mathsf{\Delta }{u}_1\left(t_k\right)-\mathsf{\Delta }{u}_2\left(t_k\right)\hfill \\ & =-{\mathfrak{L}}_{k}p\left(t_k\right)+I_k\left({\eta }_1\left(t_k\right)\right)+{\mathfrak{L}}_k{\eta }_1\left(t_k\right)-I_k\left({\eta }_2\left(t_k\right)\right)+{\mathfrak{L}}_k{\eta }_2\left(t_k\right)\hfill \\ & \le -{\mathfrak{L}}_{k}p\left(t_k\right),k=1,2,\dots ,m,\hfill \end{array}$$

and

$$\begin{array}{cc}p\left(0\right)\hfill & =u_1\left(0\right)-u_2\left(0\right)\hfill \\ & {\displaystyle =-\frac{1}{M_1}h({\eta }_1\left(0\right),{\eta }_1\left(T\right))+\frac{M_2}{M_1}(u_1\left(T\right)-{\eta }_1\left(T\right))+{\eta }_1\left(0\right)}\hfill \\ & -\left[{\displaystyle -\frac{1}{M_1}h({\eta }_2\left(0\right),{\eta }_2\left(T\right))+\frac{M_2}{M_1}(u_2\left(T\right)-{\eta }_2\left(T\right))+{\eta }_2\left(0\right)}\right]\hfill \\ & {\displaystyle \le \frac{M_2}{M_1}p\left(T\right).}\hfill \end{array}$$

From Lemma 1, $p\le 0$ implies $\mathcal{F}{\eta }_1\le \mathcal{F}{\eta }_2$.

Now, we can define the sequences $\left\{{\mathfrak{m}}_n\left(t\right)\right\}$, $\left\{{\mathfrak{n}}_n\left(t\right)\right\}$ with ${\mathfrak{m}}_0=\mathfrak{m}$, ${\mathfrak{n}}_0=\mathfrak{n}$, such that ${\mathfrak{m}}_{n+1}=\mathcal{F}{\mathfrak{m}}_n$, ${\mathfrak{n}}_{n+1}=\mathcal{F}{\mathfrak{n}}_n$. Due to (a) and (b), we achieve

$${\mathfrak{m}}_0\le {\mathfrak{m}}_1\le {\mathfrak{m}}_2\le \cdots \le {\mathfrak{m}}_n\le \cdots \le {\mathfrak{n}}_n\le \cdots \le {\mathfrak{n}}_2\le {\mathfrak{n}}_1\le {\mathfrak{n}}_0.$$

Obviously, the sequence $\left\{{\mathfrak{m}}_n\right\}$ is increasing and bounded, as one has $\left\{{\mathfrak{m}}_n\right\}$ converging $\varrho$ uniformly on J. Similarly, we know that there is ${\displaystyle \underset{n\to \infty }{lim}{\mathfrak{n}}_n\left(t\right)=\varsigma \left(t\right)}$ uniformly on J. Passing to the limit when $n\to \infty$, we find that $\varrho \left(t\right),\varsigma \left(t\right)$ are solutions to (1).

To testify that $\varrho$ and $\varsigma$ are extremal solutions of (1), consider an arbitrary solution x of (1) satisfying $x\in [\mathfrak{m},\mathfrak{n}]$. Now, if ${\mathfrak{m}}_n\le x$, it is easy to obtain ${\mathfrak{m}}_{n+1}=\mathcal{F}{\mathfrak{m}}_n\le \mathcal{F}x=x$ based on the property of $\mathcal{F}$, and by the induction, one reaches ${\mathfrak{m}}_n\le x$ for every $n\in \mathbb{N}$. Thus, passing to the limit, $\varrho \le x$.

The same arguments demonstrate that $x\le \varsigma$. □

Example 1.

Consider the following problem:

$$\left\{\begin{array}{c}{\displaystyle x^{\prime }\left(t\right)=-{\displaystyle \frac{t}{3}}x\left(t\right)-{\displaystyle \frac{1}{10t}}{\int }_0^{t}x\left(s\right)ds,t\in J=[0,1],t\ne {\displaystyle \frac{1}{3}}}\hfill \\ \mathsf{\Delta }x\left(t_k\right)=-{\displaystyle \frac{1}{15}}{x}^2\left({\displaystyle \frac{1}{3}}\right),\hfill \\ x\left(0\right)-x\left(t\right)+{\displaystyle \frac{1}{6}}{x}^2\left(0\right)=0.\hfill \end{array}\right.$$

(8)

Setting $\mathfrak{m}=0,\mathfrak{n}=10$. Evidently, $\mathfrak{m}$ and $\mathfrak{n}$ are lower and upper solutions with $\mathfrak{m}\le \mathfrak{n}$. Taking $\left(\mathfrak{R}x\right)\left(t\right)=-{\displaystyle \frac{t}{3}}x\left(t\right)-{\displaystyle \frac{1}{10t}}{\int }_0^{t}x\left(s\right)ds$ and $h(x,y)=x-y+{\displaystyle \frac{1}{6}}{x}^2$, then assumption (3), (6), $\left(H_1\right)$, $\left(H_2\right)$, $\left(H_3\right)$ and $\left(H_4\right)$ hold with $M\left(t\right)={\displaystyle \frac{t}{3}}$, ${\mathfrak{L}}_k={\displaystyle \frac{1}{6}}$, $\left(\mathcal{L}1\right)\left(t\right)={\displaystyle \frac{1}{10}}$, $M_1=2$, $M_2=1$, $\lambda ={\displaystyle \frac{1}{2}}$, $T=1$. By Theorem (1), the problem (1) has extremal solutions within the interval $[\mathfrak{m},\mathfrak{n}]$.

4. Extremal Quasi-Solutions of Problem (1)

This part aims to verify that the monotone iterative technique remains applicable. It will be used for establishing the existence of extremal quasi-solutions.

Definition 2.

Functions $\mathfrak{x},\mathfrak{z}\in \mathsf{\Omega }$ are called coupled lower and upper solutions of (1) if

$$\left\{\begin{array}{c}{\mathfrak{x}}^{\prime }\left(t\right)\le \left(\mathfrak{R}\mathfrak{x}\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\mathfrak{x}\left(t_k\right)\le I_k\left(\mathfrak{x}\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\mathfrak{x}\right(0),\mathfrak{z}(T\left)\right)\le 0,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}{\mathfrak{z}}^{\prime }\left(t\right)\ge \left(\mathfrak{R}\mathfrak{z}\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\mathfrak{z}\left(t_k\right)\ge I_k\left(\mathfrak{z}\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\mathfrak{z}\right(0),\mathfrak{x}(T\left)\right)\ge 0.\hfill \end{array}\right.$$

Definition 3.

Functions $U,V\in \mathsf{\Omega }$ are called quasi-solutions of (1) if

$$\left\{\begin{array}{c}{U}^{\prime }\left(t\right)=\left(\mathfrak{R}U\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }U\left(t_k\right)=I_k\left(U\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(U\right(0),V(T\left)\right)=0\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}{V}^{\prime }\left(t\right)=\left(\mathfrak{R}V\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }V\left(t_k\right)=I_k\left(V\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(V\right(0),U(T\left)\right)=0.\hfill \end{array}\right.$$

Definition 4.

A paired quasi-solution $(U,V)$ of problems (1) is said to be the minimal and maximal quasi-solution of (1) if $\varrho \le U,V\le \varsigma$. If both the minimal and maximal solutions are present, they are referred to as the extremal quasi-solutions of problem (1).

Theorem 2.

Assume that $\left(H_2\right)$, $\left(H_3\right)$, (3) and (6) hold. Moreover,

$\left(A_1\right)$$\mathfrak{R}\in C(E,E)$ is a causal operator, and $\mathfrak{x},\mathfrak{z}\in \mathsf{\Omega }$ are coupled lower and upper solutions of (1) with $\mathfrak{x}\left(t\right)\le \mathfrak{z}\left(t\right),t\in J$;

$\left(A_2\right)$ there exist $M_1,M_2$, such that $M_1\ge M_2>0$, $h(x,z)\in C(\mathbb{R}\times \mathbb{R},\mathbb{R})$ is nonincreasing in the first variable, satisfying

$$h(x,\overline{z})-h(x,z)\ge M_2(\overline{z}-z),\mathrm{if}\mathfrak{x}\left(T\right)\le z\le \overline{z}\le \mathfrak{z}\left(T\right).$$

Then, problem (1) has extremal quasi-solutions within the internal $[\mathfrak{x},\mathfrak{z}]=\{x\in \mathsf{\Omega }:\mathfrak{x}(t)\le x(t)\le \mathfrak{z}(t),t\in J\}$.

Proof. 

Let

$$\left\{\begin{array}{c}{\mathfrak{x}}_n^{\prime }\left(t\right)+M\left(t\right){\mathfrak{x}}_n\left(t\right)+\left(\mathcal{L}{\mathfrak{x}}_n\right)\left(t\right)=\left(\mathfrak{R}{\mathfrak{x}}_{n-1}\right)\left(t\right)+M\left(t\right){\mathfrak{x}}_{n-1}\left(t\right)+\left(\mathcal{L}{\mathfrak{x}}_{n-1}\right)\left(t\right),t\in J_0,\hfill \\ \mathsf{\Delta }{\mathfrak{x}}_n\left(t_k\right)+{\mathfrak{L}}_k{\mathfrak{x}}_n\left(t_k\right)=I_k\left({\mathfrak{x}}_{n-1}\left(t_k\right)\right)+{\mathfrak{L}}_k{\mathfrak{x}}_{n-1}\left(t_k\right),k=1,2,\dots ,m,\hfill \\ h({\mathfrak{x}}_{n-1}\left(0\right),{\mathfrak{z}}_{n-1}\left(T\right))+M_1({\mathfrak{x}}_n\left(0\right)-{\mathfrak{x}}_{n-1}\left(0\right))-M_2({\mathfrak{x}}_n\left(T\right)-{\mathfrak{x}}_{n-1}\left(T\right))=0,\hfill \end{array}\right.$$

(9)

and

$$\left\{\begin{array}{c}{\mathfrak{z}}_n^{\prime }\left(t\right)+M\left(t\right){\mathfrak{z}}_n\left(t\right)+\left(\mathcal{L}{\mathfrak{z}}_n\right)\left(t\right)=\left(\mathfrak{R}{\mathfrak{z}}_{n-1}\right)\left(t\right)+M\left(t\right){\mathfrak{z}}_{n-1}\left(t\right)+\left(\mathcal{L}{\mathfrak{z}}_{n-1}\right)\left(t\right),t\in J_0,\hfill \\ \mathsf{\Delta }{\mathfrak{z}}_n\left(t_k\right)+{\mathfrak{L}}_k{\mathfrak{z}}_n\left(t_k\right)=I_k\left({\mathfrak{z}}_{n-1}\left(t_k\right)\right)+{\mathfrak{L}}_k{z}_{n-1}\left(t_k\right),k=1,2,\dots ,m,\hfill \\ h({\mathfrak{z}}_{n-1}\left(0\right),{\mathfrak{x}}_{n-1}\left(T\right))+M_1({\mathfrak{z}}_n\left(0\right)-{\mathfrak{z}}_{n-1}\left(0\right))-M_2({\mathfrak{z}}_n\left(T\right)-{\mathfrak{z}}_{n-1}\left(T\right))=0,\hfill \end{array}\right.$$

(10)

for $n=1,2,\dots$, where ${\mathfrak{x}}_0=\mathfrak{x}$, ${\mathfrak{z}}_0=\mathfrak{z}$.

It follows from Lemma 3 that both (9) and (10) have a unique solution, respectively. Now, we conclude the proof through three steps.

Step 1: One proves that ${\mathfrak{x}}_{n-1}\le {\mathfrak{x}}_n$ and ${\mathfrak{z}}_n\le {\mathfrak{z}}_{n-1}$, $n=1,2,\dots$.

Set $p\left(t\right)=\mathfrak{x}\left(t\right)-{\mathfrak{x}}_1\left(t\right)$. Employing $\left(A_1\right)$, we have

$$\begin{array}{cc}{p}^{\prime }\left(t\right)\hfill & ={\mathfrak{x}}^{\prime }\left(t\right)-{\mathfrak{x}}_1^{\prime }\left(t\right)\hfill \\ & \le \left(\mathfrak{R}\mathfrak{x}\right)\left(t\right)+M\left(t\right){\mathfrak{x}}_1\left(t\right)+\left(\mathcal{L}{\mathfrak{x}}_1\right)\left(t\right)-\left(\mathfrak{R}\mathfrak{x}\right)\left(t\right)-M\left(t\right)\mathfrak{x}\left(t\right)-\left(\mathcal{L}\mathfrak{x}\right)\left(t\right)\hfill \\ & =-M\left(t\right)p\left(t\right)-\left(\mathcal{L}p\right)\left(t\right),t\ne t_k,t\in J,\hfill \end{array}$$

$$\begin{array}{cc}\mathsf{\Delta }p\left(t_k\right)\hfill & =\mathsf{\Delta }\mathfrak{x}\left(t_k\right)-\mathsf{\Delta }{\mathfrak{x}}_1\left(t_k\right)\hfill \\ & \le I_k\left(\mathfrak{x}\left(t_k\right)\right)+{\mathfrak{L}}_k{\mathfrak{x}}_1\left(t_k\right)-I_k\left(\mathfrak{x}\left(t_k\right)\right)-{\mathfrak{L}}_k\mathfrak{x}\left(t_k\right),\hfill \\ & =-{\mathfrak{L}}_{k}p\left(t_k\right),k=1,2,\dots ,m,\hfill \end{array}$$

and

$$\begin{array}{cc}p\left(0\right)\hfill & {\displaystyle =\mathfrak{x}\left(0\right)-{\mathfrak{x}}_1\left(0\right)=\frac{1}{M_1}h(\mathfrak{x}\left(0\right),\mathfrak{z}\left(T\right))-\frac{M_2}{M_1}({\mathfrak{x}}_1\left(T\right)-\mathfrak{x}\left(T\right))}\hfill \\ & {\displaystyle \le \frac{M_2}{M_1}p\left(T\right).}\hfill \end{array}$$

From Lemma 1 and $M_2\ge M_1>0$, we get $p\le 0$, so $\mathfrak{x}\le {\mathfrak{x}}_1$.

Utilizing the induction, we deduce that the sequence $\left\{{\mathfrak{x}}_n\right\}$ is monotonically nondecreasing. Analogously, one finds that $\left\{{\mathfrak{z}}_n\right\}$ is monotonically nonincreasing.

Step 2: We demonstrate that ${\mathfrak{x}}_1\le {\mathfrak{z}}_1$ if $\mathfrak{x}\le \mathfrak{z}$.

Let $p\left(t\right)={\mathfrak{x}}_1\left(t\right)-{\mathfrak{z}}_1\left(t\right)$. Using $\left(H_2\right),\left(H_3\right)$ and $\left(A_2\right)$, we get

$$\begin{array}{cc}{p}^{\prime }\left(t\right)\hfill & ={\mathfrak{x}}_1^{\prime }\left(t\right)-{\mathfrak{z}}_1^{\prime }\left(t\right)\hfill \\ & =-M\left(t\right){\mathfrak{x}}_1\left(t\right)-\left(\mathcal{L}{\mathfrak{x}}_1\right)\left(t\right)+\left(\mathfrak{R}\mathfrak{x}\right)\left(k\right)+M\left(t\right)\mathfrak{x}\left(t\right)+\left(\mathcal{L}\mathfrak{x}\right)\left(t\right)\hfill \\ & +M\left(t\right){\mathfrak{z}}_1\left(t\right)+\left(\mathcal{L}{\mathfrak{z}}_1\right)\left(t\right)-\left(\mathfrak{R}\mathfrak{z}\right)\left(t\right)-M\left(t\right)\mathfrak{z}\left(t\right)-\left(\mathcal{L}\mathfrak{z}\right)\left(t\right)\hfill \\ & \le -M\left(t\right)p\left(t\right)-\left(\mathcal{L}p\right)\left(t\right),t\ne t_k,t\in J,\hfill \end{array}$$

$$\begin{array}{cc}\mathsf{\Delta }p\left(t_k\right)\hfill & =\mathsf{\Delta }{\mathfrak{x}}_1\left(t_k\right)-\mathsf{\Delta }{\mathfrak{z}}_1\left(t_k\right)\hfill \\ & =-{\mathfrak{L}}_k{\mathfrak{x}}_1\left(t_k\right)+I_k\left(\mathfrak{x}\left(t_k\right)\right)+{\mathfrak{L}}_k\mathfrak{x}\left(t_k\right)+{\mathfrak{L}}_k{\mathfrak{z}}_1\left(t_k\right)-I_k\left(\mathfrak{z}\left(t_k\right)\right)-{\mathfrak{L}}_k\mathfrak{z}\left(t_k\right)\hfill \\ & \le -{\mathfrak{L}}_{k}p\left(t_k\right),k=1,2,\dots ,m,\hfill \end{array}$$

and

$$\begin{array}{cc}p\left(0\right)\hfill & ={\mathfrak{x}}_1\left(0\right)-{\mathfrak{z}}_1\left(0\right)\hfill \\ & {\displaystyle =-\frac{1}{M_1}h(\mathfrak{x}\left(0\right),\mathfrak{z}\left(T\right))+\frac{M_2}{M_1}({\mathfrak{x}}_1\left(t\right)-\mathfrak{x}\left(T\right))+\mathfrak{x}\left(0\right)}\hfill \\ & -\left[{\displaystyle -\frac{1}{M_1}h(\mathfrak{z}\left(0\right),\mathfrak{x}\left(T\right))+\frac{M_2}{M_1}({\mathfrak{z}}_1\left(T\right)-\mathfrak{z}\left(T\right))+\mathfrak{z}\left(0\right)}\right]\hfill \\ & {\displaystyle \le \frac{M_2}{M_1}p\left(t\right).}\hfill \end{array}$$

Subsequently, by Lemma 1, one has $p\le 0$, which indicates ${\mathfrak{x}}_1\le {\mathfrak{z}}_1$.

Next, we will prove that ${\mathfrak{x}}_1,{\mathfrak{z}}_1$ are coupled lower and upper solutions for (1). Using the assumptions $\left(H_2\right),\left(H_3\right)$, $\left(A_2\right)$, and $\mathfrak{x}\le {\mathfrak{x}}_1$, ${\mathfrak{z}}_1\le \mathfrak{z}$, we obtain

$$\begin{array}{c}{\mathfrak{x}}_1^{\prime }\left(t\right)=\left(\mathfrak{R}\mathfrak{x}\right)\left(t\right)+M\left(t\right)(\mathfrak{x}\left(t\right)-{\mathfrak{x}}_1\left(t\right))+\left(\mathcal{L}(\mathfrak{x}-{\mathfrak{x}}_1)\right)\left(t\right)\le \left(\mathfrak{R}{\mathfrak{x}}_1\right)\left(t\right),\hfill \\ \mathsf{\Delta }{\mathfrak{x}}_1\left(t_k\right)={\mathfrak{L}}_k(\mathfrak{x}\left(t_k\right)-{\mathfrak{x}}_1\left(t_k\right))+I_k\left(\mathfrak{x}\left(t_k\right)\right)\le I_k\left({\mathfrak{x}}_1\left(t_k\right)\right),\hfill \\ h({\mathfrak{x}}_1\left(0\right),{\mathfrak{z}}_1\left(T\right))\le h(\mathfrak{x}\left(0\right),{\mathfrak{z}}_1\left(T\right))\le h(\mathfrak{x}\left(0\right),\mathfrak{z}\left(T\right))+M_2({\mathfrak{z}}_1\left(T\right)-\mathfrak{z}\left(T\right))\le 0.\hfill \end{array}$$

It is evident that ${\mathfrak{x}}_1$ is a coupled lower solution to (1). Analogously, ${\mathfrak{z}}_1\left(t\right)$ is a coupled upper solution to (1). Using the induction, one has ${\mathfrak{x}}_n\le {\mathfrak{z}}_n$, $n=1,2,\dots$.

Step 3: Based on the above two steps, it can be seen that

$${\mathfrak{x}}_0\le {\mathfrak{x}}_1\le {\mathfrak{x}}_2\le \cdots \le {\mathfrak{x}}_n\le \cdots \le {\mathfrak{z}}_n\le \cdots \le {\mathfrak{z}}_2\le {\mathfrak{z}}_1\le {\mathfrak{z}}_0,$$

and each ${\mathfrak{x}}_n,{\mathfrak{z}}_n$ satisfy (9) and (10). Apparently, $\left\{{\mathfrak{x}}_n\right\},\left\{{\mathfrak{x}}_n\right\}$ are uniformly bounded and equicontinuous, using the Ascoli–Arzela theorem and passing to the limit when $n\to \infty$; we find that $\varrho ,\varsigma$ satisfy the following equations

$$\left\{\begin{array}{c}{\varrho }^{\prime }\left(t\right)=\left(\mathfrak{R}\varrho \right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\varrho \left(t_k\right)=I_k\left(\varrho \left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\varrho \right(0),\varsigma (T\left)\right)=0,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}{\varsigma }^{\prime }\left(t\right)=\left(\mathfrak{R}\varsigma \right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\varsigma \left(t_k\right)=I_k\left(\varsigma \left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\varsigma \right(0),\varrho (T\left)\right)=0.\hfill \end{array}\right.$$

It proves that $\varrho ,\varsigma$ are the extremal quasi-solutions of problem (1). This ends the proof. □

Example 2.

Consider the problem of

$$\left\{\begin{array}{c}{\displaystyle x^{\prime }\left(t\right)=-{\displaystyle \frac{t}{2}}x\left(t\right)-{\displaystyle \frac{1}{10t^2}}{\int }_0^{t}sx\left(s\right)ds,t\in J=[0,1],t\ne {\displaystyle \frac{1}{3}}}\hfill \\ \mathsf{\Delta }x\left(t_k\right)=-{\displaystyle \frac{1}{15}}{x}^2\left({\displaystyle \frac{1}{3}}\right),\hfill \\ x\left(0\right)+x\left(t\right)-1=0.\hfill \end{array}\right.$$

(11)

Setting $\mathfrak{x}=0,\mathfrak{z}=\left\{\begin{array}{c}{\displaystyle \frac{1}{2}}t+1,t\in [0,{\displaystyle \frac{1}{3}}),\hfill \\ {\displaystyle \frac{1}{2}}t+{\displaystyle \frac{1}{4}},t\in ({\displaystyle \frac{1}{3}},1].\hfill \end{array}\right.$ Clearly, $\mathfrak{x}$ is a coupled lower solution, and $\mathfrak{z}$ is a coupled upper solution with $\mathfrak{x}\le \mathfrak{z}$. It is easy to see that (3), (6), $\left(A_1\right)$, $\left(H_2\right)$, $\left(H_3\right)$ and $\left(A_2\right)$ hold with $M={\displaystyle \frac{t}{2}}$, ${\mathfrak{L}}_k={\displaystyle \frac{1}{15}}$, $\left(\mathcal{L}x\right)\left(t\right)={\displaystyle \frac{1}{10t^2}}{\int }_0^{t}sx\left(s\right)ds$, $\left(\mathcal{L}1\right)\left(t\right)={\displaystyle \frac{1}{20}}$, $M_1=2$, $M_2=1$, $\lambda ={\displaystyle \frac{1}{2}}$, $T=1$. From Theorem (2), the problem (11) has extremal quasi-solutions within the interval $[\mathfrak{x},\mathfrak{z}]$.

5. Weakly Coupled Quasi-Solutions of Problem (1)

This section establishes the existence of the weakly coupled extremal quasi-solutions of (1).

Definition 5.

$\mathfrak{u},\mathfrak{w}\in \mathsf{\Omega }$ are said to be weakly coupled lower and upper solutions of (1) if

$$\left\{\begin{array}{c}{\mathfrak{u}}^{\prime }\left(t\right)\le \left(\mathfrak{R}\mathfrak{u}\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\mathfrak{u}\left(t_k\right)\le I_k\left(\mathfrak{u}\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\mathfrak{w}\right(0),\mathfrak{u}(T\left)\right)\le 0,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}{\mathfrak{w}}^{\prime }\left(t\right)\ge \left(\mathfrak{R}\mathfrak{w}\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\mathfrak{w}\left(t_k\right)\ge I_k\left(\mathfrak{w}\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\mathfrak{u}\right(0),\mathfrak{w}(T\left)\right)\ge 0.\hfill \end{array}\right.$$

Definition 6.

A pair $(\mathsf{\Phi },\mathsf{\Psi })$, $\mathsf{\Phi },\mathsf{\Psi }\in \mathsf{\Omega }$, is called a weakly coupled quasi-solution of (1) if

$$\left\{\begin{array}{c}{\mathsf{\Phi }}^{\prime }\left(t\right)=\left(\mathfrak{R}\mathsf{\Phi }\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\mathsf{\Phi }\left(t_k\right)=I_k\left(\mathsf{\Phi }\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\mathsf{\Psi }\right(0),\mathsf{\Phi }(T\left)\right)=0,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}{\mathsf{\Psi }}^{\prime }\left(t\right)=\left(\mathfrak{R}\mathsf{\Psi }\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }\mathsf{\Psi }\left(t_k\right)=I_k\left(\mathsf{\Psi }\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h\left(\mathsf{\Phi }\right(0),\mathsf{\Psi }(T\left)\right)=0.\hfill \end{array}\right.$$

Definition 7.

For every weakly coupled quasi-solution $(\mathsf{\Phi },\mathsf{\Psi })$ of (1), a weakly coupled quasi-solution $(\varrho ,\varsigma )$ is said to be weakly coupled minimal and maximal quasi-solutions of (1) if $\varrho \left(t\right)\le \mathsf{\Phi }\left(t\right),\mathsf{\Psi }\left(t\right)\le \varsigma \left(t\right)$ for all $t\in J$. If both minimal and maximal solutions are present, they are called weakly coupled extremal quasi-solutions of problem (1).

Theorem 3.

 Suppose that $\left(H_2\right)$, $\left(H_3\right)$, (3), (6) hold. Furthermore, it is assumed that 

• $\left(B_1\right)$ $\mathfrak{u},\mathfrak{w}\in \mathsf{\Omega }$ are weakly coupled lower and upper solutions of (1) with $\mathfrak{u}\left(t\right)\le \mathfrak{w}\left(t\right)$ on J;
• $\left(B_2\right)$ there exist $M_1,M_2$, such that $M_1\ge M_2>0$. Moreover, $h(x,z)\in C(\mathbb{R}\times \mathbb{R},\mathbb{R})$ is nondecreasing with respect to the first variable and satisfies

  $$h(x,\overline{z})-h(x,z)\le -M_2(\overline{z}-z),if\mathfrak{u}\left(T\right)\le z\le \overline{z}\le \mathfrak{w}\left(T\right).$$

Then, (1) has weakly coupled extremal quasi-solutions within the interval $[\mathfrak{u},\mathfrak{w}]=\{x\in \mathsf{\Omega }:\mathfrak{u}(t)\le x(t)\le \mathfrak{w}(t),t\in J\}$.

Proof. 

Let

$$\left\{\begin{array}{c}{\mathfrak{u}}_n^{\prime }\left(t\right)+M\left(t\right){\mathfrak{u}}_n\left(t\right)+\left(\mathcal{L}{\mathfrak{u}}_n\right)\left(t\right)=\left(\mathfrak{R}{\mathfrak{u}}_{n-1}\right)\left(t\right)+M\left(t\right){\mathfrak{u}}_{n-1}\left(t\right)+\left(\mathcal{L}{\mathfrak{u}}_{n-1}\right)\left(t\right),t\in J_0,\hfill \\ \mathsf{\Delta }{\mathfrak{u}}_n\left(t_k\right)+{\mathfrak{L}}_k{\mathfrak{u}}_n\left(t_k\right)=I_k\left({\mathfrak{u}}_{n-1}\left(t_k\right)\right)+{\mathfrak{L}}_k{\mathfrak{u}}_{n-1}\left(t_k\right),k=1,2,\dots ,m,\hfill \\ h({\mathfrak{w}}_{n-1}\left(0\right),{\mathfrak{u}}_{n-1}\left(T\right))+M_1({\mathfrak{u}}_n\left(0\right)-{\mathfrak{u}}_{n-1}\left(0\right))-M_2({\mathfrak{u}}_n\left(T\right)-{\mathfrak{u}}_{n-1}\left(T\right))=0,\hfill \end{array}\right.$$

(12)

and

$$\left\{\begin{array}{c}{\mathfrak{w}}_n^{\prime }\left(t\right)+M\left(t\right){\mathfrak{w}}_n\left(t\right)+\left(\mathcal{L}{\mathfrak{w}}_n\right)\left(t\right)=\left(\mathfrak{R}{\mathfrak{w}}_{n-1}\right)\left(t\right)+M\left(t\right){\mathfrak{w}}_{n-1}\left(t\right)+\left(\mathcal{L}{\mathfrak{w}}_{n-1}\right)\left(t\right),t\in J_0,\hfill \\ \mathsf{\Delta }{\mathfrak{w}}_n\left(t_k\right)+{\mathfrak{L}}_k{\mathfrak{w}}_n\left(t_k\right)=I_k\left({\mathfrak{w}}_{n-1}\left(t_k\right)\right)+{\mathfrak{L}}_k{\mathfrak{w}}_{n-1}\left(t_k\right),k=1,2,\dots ,m,\hfill \\ h({\mathfrak{u}}_{n-1}\left(0\right),{\mathfrak{w}}_{n-1}\left(T\right))+M_1({\mathfrak{w}}_n\left(0\right)-{\mathfrak{w}}_{n-1}\left(0\right))-M_2({\mathfrak{w}}_n\left(T\right)-{\mathfrak{w}}_{n-1}\left(T\right))=0,\hfill \end{array}\right.$$

(13)

for $n=1,2,\dots$, where ${\mathfrak{u}}_0=\mathfrak{u}$, ${\mathfrak{w}}_0=\mathfrak{w}$.

With regard to Lemma 3, $\mathfrak{u},\mathfrak{w}$ are well defined. Firstly, prove that ${\mathfrak{u}}_0\le {\mathfrak{u}}_1\le {\mathfrak{w}}_1\le {\mathfrak{w}}_0$.

Let $p\left(t\right)={\mathfrak{u}}_0\left(t\right)-{\mathfrak{u}}_1\left(t\right)$, applying $\left(B_1\right)$; we have

$$\begin{array}{cc}{p}^{\prime }\left(t\right)\hfill & ={\mathfrak{u}}_0^{\prime }\left(t\right)-{\mathfrak{u}}_1^{\prime }\left(t\right)\hfill \\ & \le \left(\mathfrak{R}{\mathfrak{u}}_0\right)\left(t\right)-\left[\left(\mathfrak{R}{\mathfrak{u}}_0\right)\left(t\right)+M\left(t\right){\mathfrak{u}}_0\left(t\right)+\left(\mathcal{L}{\mathfrak{u}}_0\right)\left(t\right)-M\left(t\right){\mathfrak{u}}_1\left(t\right)-\left(\mathcal{L}{\mathfrak{u}}_1\right)\left(t\right)\right]\hfill \\ & =-M\left(t\right)p\left(t\right)-\left(\mathcal{L}\right)p\left(t\right),t\ne t_k,t\in J,\hfill \end{array}$$

$$\begin{array}{cc}\mathsf{\Delta }p\left(t_k\right)\hfill & =\mathsf{\Delta }{\mathfrak{u}}_0\left(t_k\right)-\mathsf{\Delta }{\mathfrak{u}}_1\left(t_k\right)\hfill \\ & \le I_k\left({\mathfrak{u}}_0\left(t_k\right)\right)+{\mathfrak{L}}_k{\mathfrak{u}}_1\left(t_k\right)-I_k\left({\mathfrak{u}}_0\left(t_k\right)\right)-{\mathfrak{L}}_k{\mathfrak{u}}_0\left(t_k\right),\hfill \\ & =-{\mathfrak{L}}_{k}p\left(t_k\right),k=1,2,\dots ,m,\hfill \end{array}$$

and

$$\begin{array}{cc}p\left(0\right)\hfill & ={\mathfrak{u}}_0\left(0\right)-\left[{\displaystyle -\frac{1}{M_1}h({\mathfrak{w}}_0\left(0\right),{\mathfrak{u}}_0\left(T\right))+\frac{M_2}{M_1}({\mathfrak{u}}_1\left(T\right)-{\mathfrak{u}}_0\left(T\right))+{\mathfrak{u}}_0\left(0\right)}\right]\hfill \\ & {\displaystyle =\frac{1}{M_1}h({\mathfrak{w}}_0\left(0\right),{\mathfrak{u}}_0\left(T\right))+\frac{M_2}{M_1}({\mathfrak{u}}_0\left(T\right)-{\mathfrak{u}}_1\left(T\right))}\hfill \\ & {\displaystyle \le \frac{M_2}{M_1}p\left(T\right).}\hfill \end{array}$$

By Lemma 1, we obtain $p\left(t\right)\le 0$ with $t\in J$, that is, ${\mathfrak{u}}_0\le {\mathfrak{u}}_1$. Similar arguments prove that ${\mathfrak{w}}_1\le {\mathfrak{w}}_0$.

Now, set $p\left(t\right)={\mathfrak{u}}_1\left(t\right)-{\mathfrak{w}}_1\left(t\right)$. Using $\left(H_2\right),\left(H_3\right)$, we get

$$\begin{array}{cc}{p}^{\prime }\left(t\right)\hfill & ={\mathfrak{u}}_1^{\prime }\left(t\right)-{\mathfrak{w}}_1^{\prime }\left(t\right)\hfill \\ & =-M\left(t\right){\mathfrak{u}}_1\left(t\right)-\left(\mathcal{L}{\mathfrak{u}}_1\right)\left(t\right)+\left(\mathfrak{R}{\mathfrak{u}}_0\right)\left(t\right)+M\left(t\right){\mathfrak{u}}_0\left(t\right)+\left(\mathcal{L}{\mathfrak{u}}_0\right)\left(t\right)\hfill \\ & -\left[-M\left(t\right){\mathfrak{w}}_1\left(k\right)-\left(\mathcal{L}{\mathfrak{w}}_1\right)\left(t\right)+\left(\mathfrak{R}{\mathfrak{w}}_0\right)\left(t\right)+M\left(t\right){\mathfrak{w}}_0\left(t\right)+\left(\mathcal{L}{\mathfrak{w}}_0\right)\left(t\right)\right]\hfill \\ & \le -M\left(t\right)p\left(t\right)-\left(\mathcal{L}p\right)\left(t\right),t\ne t_k,t\in J,\hfill \end{array}$$

and

$$\begin{array}{cc}\mathsf{\Delta }p\left(t_k\right)\hfill & =\mathsf{\Delta }{\mathfrak{u}}_1\left(t_k\right)-\mathsf{\Delta }{\mathfrak{w}}_1\left(t_k\right)\hfill \\ & =-{\mathfrak{L}}_k{\mathfrak{u}}_1\left(t_k\right)+I_k\left(\mathfrak{u}\left(t_k\right)\right)+{\mathfrak{L}}_k\mathfrak{u}\left(t_k\right)+{\mathfrak{L}}_k{\mathfrak{w}}_1\left(t_k\right)-I_k\left(\mathfrak{w}\left(t_k\right)\right)-{\mathfrak{L}}_k\mathfrak{w}\left(t_k\right)\hfill \\ & \le -{\mathfrak{L}}_{k}p\left(t_k\right),k=1,2,\dots ,m.\hfill \end{array}$$

Noticing ${\mathfrak{u}}_0\le {\mathfrak{w}}_0$ and $\left(B_2\right)$, we obtain

$$\begin{array}{cc}p\left(0\right)\hfill & ={\mathfrak{u}}_1\left(0\right)-{\mathfrak{w}}_1\left(0\right)\hfill \\ & {\displaystyle =-\frac{1}{M_1}h({\mathfrak{w}}_0\left(0\right),{\mathfrak{u}}_0\left(T\right))+\frac{M_2}{M_1}({\mathfrak{u}}_1\left(T\right)-{\mathfrak{u}}_0\left(T\right))+{\mathfrak{u}}_0\left(0\right)}\hfill \\ & {\displaystyle -[-\frac{1}{M_1}h({\mathfrak{u}}_0\left(0\right),{\mathfrak{w}}_0\left(T\right))+\frac{M_2}{M_1}({\mathfrak{w}}_1\left(T\right)-{\mathfrak{w}}_0\left(T\right))+{\mathfrak{w}}_0\left(0\right)]}\hfill \\ & {\displaystyle \le \frac{M_2}{M_1}p\left(T\right).}\hfill \end{array}$$

Lemma 1 implies that $p\left(t\right)\le 0$, $t\in J$, i.e., ${\mathfrak{u}}_1\le {\mathfrak{w}}_1$.

Using $H_2$, $H_3$, $B_2$ and the fact of ${\mathfrak{u}}_0\le {\mathfrak{u}}_1$, ${\mathfrak{w}}_1\le {\mathfrak{w}}_0$, we obtain

$$\begin{array}{c}{\mathfrak{u}}_1^{\prime }\left(t\right)=\left(\mathfrak{R}{\mathfrak{u}}_0\right)\left(t\right)+M\left(t\right)({\mathfrak{u}}_0\left(t\right)-{\mathfrak{u}}_1\left(t\right))+\left(\mathcal{L}({\mathfrak{u}}_0-{\mathfrak{u}}_1)\right)\left(t\right)\le \left(\mathfrak{R}{\mathfrak{u}}_1\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }{\mathfrak{u}}_1\left(t_k\right)={\mathfrak{L}}_k({\mathfrak{u}}_0\left(t_k\right)-{\mathfrak{u}}_1\left(t_k\right))+I_k\left({\mathfrak{u}}_0\left(t_k\right)\right)\le I_k\left({\mathfrak{u}}_1\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h({\mathfrak{w}}_1\left(0\right),{\mathfrak{u}}_1\left(T\right))\le h({\mathfrak{w}}_0\left(0\right),{\mathfrak{u}}_1\left(T\right))\le h({\mathfrak{w}}_0\left(0\right),{\mathfrak{u}}_0\left(T\right))-M_2({\mathfrak{u}}_1\left(T\right)-{\mathfrak{u}}_0\left(T\right))\le 0,\hfill \end{array}$$

and ${\mathfrak{u}}_1$ is a weakly coupled lower solution of (1). Similarly,

$$\begin{array}{c}{\mathfrak{w}}_1^{\prime }\left(t\right)=\left(\mathfrak{R}{\mathfrak{w}}_0\right)\left(t\right)+M\left(t\right)({\mathfrak{w}}_0\left(t\right)-{\mathfrak{w}}_1\left(t\right))+\left(\mathcal{L}({\mathfrak{w}}_0-{\mathfrak{w}}_1)\right)\left(t\right)\ge \left(\mathfrak{R}{\mathfrak{w}}_1\right)\left(t\right),t\ne t_k,t\in J,\hfill \\ \mathsf{\Delta }{\mathfrak{w}}_1\left(t_k\right)={\mathfrak{L}}_k({\mathfrak{w}}_0\left(t_k\right)-{\mathfrak{w}}_1\left(t_k\right))+I_k\left({\mathfrak{w}}_0\left(t_k\right)\right)\ge I_k\left({\mathfrak{w}}_1\left(t_k\right)\right),k=1,2,\dots ,m,\hfill \\ h({\mathfrak{u}}_1\left(0\right),{\mathfrak{w}}_1\left(T\right))\ge h({\mathfrak{u}}_0\left(0\right),{\mathfrak{w}}_1\left(T\right))\ge h({\mathfrak{u}}_0\left(0\right),{\mathfrak{w}}_0\left(T\right))-M_2({\mathfrak{w}}_1\left(T\right)-{\mathfrak{w}}_0\left(T\right))\ge 0.\hfill \end{array}$$

and ${\mathfrak{w}}_1$ is a weakly coupled upper solution of (1).

Now, when we define the sequences $\left\{{\mathfrak{u}}_n\right\},\left\{{\mathfrak{w}}_n\right\}$ by (12) and (13), we have

$${\mathfrak{u}}_0\le {\mathfrak{u}}_1\le {\mathfrak{u}}_2\le \cdots \le {\mathfrak{u}}_n\le \cdots \le {\mathfrak{w}}_n\le \cdots \le {\mathfrak{w}}_2\le {\mathfrak{w}}_1\le {\mathfrak{w}}_0.$$

It is easy to see that there exist $\varrho ,\varsigma$, such that ${\displaystyle \underset{n\to \infty }{lim}\mathfrak{u}\left(t\right)=\varrho \left(t\right)}$, ${\displaystyle \underset{n\to \infty }{lim}\mathfrak{w}\left(t\right)=\varsigma \left(t\right)}$ uniformly on J by the induction, and $\varrho ,\varsigma$ are weakly coupled quasi-solutions of problem (1).

Next, we prove that $\varrho ,\varsigma$ are weakly coupled extremal quasi-solutions of (1). Let $x_1,x_2$ be any weakly coupled quasi-solution of (1) on $[{\mathfrak{u}}_0,{\mathfrak{w}}_0]$. Apparently, if ${\mathfrak{u}}_n\le u_1,u_2\le {\mathfrak{w}}_n$, we can see that ${\mathfrak{u}}_{n+1}\le x_1$ by considering $p={\mathfrak{u}}_{n+1}-x_1$, and, employing Lemma 1 and the induction, one reaches ${\mathfrak{u}}_n\le x_1$ for all n. Taking the limit as $n\to \infty$, we can conclude that $\varrho \le x_1$. A similar argument leads to $x_1\le \varsigma$ and $\varrho \le x_2\le \varsigma$. This ends the proof. □

Example 3.

Consider the problem of

$$\left\{\begin{array}{c}{\displaystyle x^{\prime }\left(t\right)=-{\displaystyle \frac{t}{6}}x\left(t\right)-{\displaystyle \frac{1}{3t}}{\int }_0^{t^2}x\left(s\right)ds,t\in J=[0,1],t\ne \frac{1}{2}}\hfill \\ \mathsf{\Delta }x\left(t_k\right)=-{\displaystyle \frac{1}{4}}x\left({\displaystyle \frac{1}{2}}\right),\hfill \\ x^2\left(0\right)+x\left(0\right)-x\left(1\right)=0.\hfill \end{array}\right.$$

(14)

Let $\mathfrak{u}=0,\mathfrak{w}=\left\{\begin{array}{c}t,t\in [0,{\displaystyle \frac{1}{2}}),\hfill \\ 0,t\in ({\displaystyle \frac{1}{2}},1].\hfill \end{array}\right.$ Clearly, $\mathfrak{u}$ is a weakly coupled lower solution and $\mathfrak{w}$ is a weakly coupled upper solution with $v\le w$. It is easy to see that (3), (6), $\left(B_1\right)$, $\left(H_2\right)$, $\left(H_3\right)$ and $\left(B_2\right)$ hold with $M\left(t\right)={\displaystyle \frac{t}{6}}$, ${\mathfrak{L}}_k={\displaystyle \frac{1}{4}}$, $\left(\mathcal{L}x\right)\left(t\right)={\displaystyle \frac{1}{3t}}{\int }_0^{t^2}x\left(s\right)ds$, $\left(\mathcal{L}1\right)\left(t\right)={\displaystyle \frac{t}{3}}$, $M_1=2$, $M_2=1$, $\lambda ={\displaystyle \frac{1}{2}}$, $T=1$. From Theorem (3), the problem (11) has weakly coupled extremal quasi-solutions in $[\mathfrak{u},\mathfrak{w}]$.

6. Conclusions

This study extends the notion of casual operators to impulsive differential equations with nonlinear periodic boundary conditions. Under the assumption of the existence of (coupled or weakly coupled) upper and lower solutions, we applied the monotone iterative technique to prove the existence of extremal solutions, extremal quasi-solutions, and weakly coupled extremal quasi-solutions for impulsive differential equations.

To validate and illustrate the theoretical results obtained, we present three different examples that summarise the essence of our findings. These examples are carefully designed to demonstrate the applicability of our theoretical framework and the effectiveness of the monotone iterative technique in the context of nonlinear impulsive differential equations with causal operators. Each example is analyzed in detail, and the results are shown to be consistent with our theoretical investigation. Through these examples, we highlight the significance of our work in advancing the study of impulsive differential equations and providing new insights into their solution behaviour.