A Class of Coupled Causal Differential Equations

Abstract

In this thesis, we consider the existence of extreme solutions to a class of coupled causal differential equations. By utilizing two comparison theorems and a monotone iterative technique, we have obtained sufficient conditions under which the equations have extreme solutions. One practical example is presented.

1. Introduction

We first study the following causal differential equations:

$$\left\{\begin{array}{c}{x}^{\prime }\left(t\right)=Q_1(x,y)\left(t\right),t\in I,\hfill \\ y^{\prime }\left(t\right)=Q_2(y,x)\left(t\right),t\in I,\hfill \\ g_1(x\left(0\right),x\left(Y\right),y\left(0\right),y\left(Y\right))=0,\hfill \\ g_2(y\left(0\right),y\left(Y\right),x\left(0\right),x\left(Y\right))=0,\hfill \end{array}\right.$$

(1)

where $I=[0,Y]$, $E=C(I^2,\mathbb{R})$, $Q_1,Q_2\in C(E,E)$ are causal operators and $g_1,g_2\in C({\mathbb{R}}^4,\mathbb{R})$ and $Q_1,Q_2$ are bounded. The term causal operator first appeared in Volterra’s work on the integral equations. It is a non-anticipative operator, and it was first applied by Volterra. The definition of the operator was given by Tonelli and Tychonoff. This theory can unify many types of equations. Peer works were reported by Drici et al. [1] and Geng [2]. Compared with traditional models, causal differential equations are more similar to real-world problems and more widely used. It has been tapped in depth by many authors in these years. Interested readers are referred to the monograph [3] and article [4] for more details.

At the same time, the research of a class of coupled differential equations is also crucial as such a system often occurs in real-life applications; it is a physical model of phenomena, in terms of the fluid dynamic traffic model, oscillations of earthquakes and economics and has attracted much attention; see, e.g., [5,6,7]. It should be noted that, in [8], Jankowski discussed coupled causal difference equations. The author developed the monotone iterative technique and gained the existence of extreme solutions to such equations. Affected by the above results, we study a class of coupled causal differential equations and obtain the extreme solutions.

The following is the frame of this paper. In Section 2, two comparison theorems are presented. In Section 3, we investigate the linear differential equations of (1). In Section 4, we establish the existence of the extremal solutions to (1). In the end, a computed example is given to verify the outcome.

2. Comparison Results

In this part, we set $E=C(I,\mathbb{R})$ and we define:

$$\parallel f-g\parallel :=\underset{t\in I}{\max }|f\left(t\right)-g\left(t\right)|.$$

Let $\mathsf{\Omega }:=E\bigcap C^1(I,\mathbb{R})$. A pair $(x,y)$, x, $y\in \mathsf{\Omega }$ is treated as a solution to Problem (1) if it contains (1).

Definition 1.

Suppose that $Q\in C(E,E)$ is called a causal operator if for every pair of elements in E such that $x_i\left(s\right)=y_i\left(s\right),i=1,2$ for $t_0\le s\le t\le Y$, then:

$$Q(x_1,x_2)\left(s\right)=Q(y_1,y_2)\left(s\right),t_0\le s\le t.$$

Lemma 1.

Suppose that $z\in \mathsf{\Omega }$ satisfies:

$$\left\{\begin{array}{c}{z}^{\prime }\left(t\right)\le -A\left(t\right)z\left(t\right)-\left(\mathcal{L}z\right)\left(t\right),t\in I,\hfill \\ z\left(0\right)\le \lambda z\left(Y\right),\hfill \end{array}\right.$$

(2)

where $A\in C(I,[0,+\infty \left)\right)$ and $\mathcal{L}\in C(E,E)$ is a positive linear operator satisfying:

$${\displaystyle {\int }_0^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt\le ϵ,0<ϵ\le 1.}$$

(3)

then $z\left(t\right)\le 0$, $t\in I$.

Proof. 

Let us say $z\left(t\right)\le 0\forall t\in I$ is not true; one has two situations:

Case 1. There exists $\overline{t}\in I$ that meets $z\left(\overline{t}\right)>0$, $z\left(t\right)\ge 0$, $t\in I$.

Then, from (2), we have $z^{\prime }\left(t\right)\le 0$ for $t\in I$; hence, $z\left(t\right)$ is nonincreasing on I. If $ϵ=1$, $z\left(0\right)\le z\left(Y\right)$ shows that $z\left(t\right)\equiv$ is constant, so $z^{\prime }\left(t\right)\equiv 0$. As $z\left(\overline{t}\right)>0$, one has $0\equiv z^{\prime }\left(\overline{t}\right)\le -A\left(t\right)z\left(\overline{t}\right)-\left(\mathcal{L}z\right)\left(\overline{t}\right)<0$, which is a conflict. If $0<ϵ<1$, then $z\left(Y\right)\le z\left(0\right)\le ϵz\left(Y\right)$, so it is also a conflict.

Case 2. There are $t_m$ and $t_n\in I$ satisfying $z\left(t_m\right)<0$ and $z\left(t_n\right)>0$.

Put ${\displaystyle \underset{t\in I}{\inf }z\left(t\right)=-r}$, $r>0$. With generality, we may assume that $z\left(t_m\right)=-r$.

Subcase (a). If $t_m<t_n$, from (2), we gain:

$$\begin{array}{cc}\hfill z\left(t_n\right)-z\left(t_m\right)& {\displaystyle ={\int }_{t_m}^{t_n}{z}^{\prime }\left(t\right)dt\le -{\int }_{t_m}^{t_n}[A\left(t\right)z\left(t\right)+\left(\mathcal{L}z\right)\left(t\right)]dt}\hfill \\ \hfill & \le r\left\{{\displaystyle {\int }_0^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt}\right\}\hfill \end{array}$$

hence:

$${\displaystyle 0<z\left(t_n\right)\le -r+r{\int }_0^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt]=r[{\int }_0^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt-1],}$$

which yields:

$${\displaystyle {\int }_0^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt>1,}$$

which contradicts (3).

Subcase (b). If $t_m>t_n$, from (2), we get:

$${\displaystyle z\left(Y\right)=z\left(t_m\right)+{\int }_{t_m}^Y{z}^{\prime }\left(t\right)dt\le z\left(t_m\right)+r{\int }_{t_m}^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt}$$

and:

$${\displaystyle z\left(t_n\right)=z\left(0\right)+{\int }_0^{t_n}{z}^{\prime }\left(t\right)dt\le z\left(0\right)+r{\int }_0^{t_n}[A\left(t\right)+\mathcal{L}\left(t\right)]dt.}$$

Using the fact $z\left(0\right)\le ϵz\left(Y\right)$, $0<ϵ\le 1$, we obtain:

$$\begin{array}{cc}0<z\left(t_n\right)\hfill & {\displaystyle \le ϵ\left(z\left(t_m\right)+r{\int }_{t_m}^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt\right)+r{\int }_0^{t_n}[A\left(t\right)+\mathcal{L}\left(t\right)]dt}\hfill \\ \hfill & \le -rϵ+r\left\{{\displaystyle {\int }_0^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt}\right\}\hfill \\ \hfill & =r\left\{{\displaystyle {\int }_0^Y[A\left(t\right)+\mathcal{L}\left(t\right)]dt-ϵ}\right\},\hfill \end{array}$$

which is a conflict with a given condition. Therefore, $z\left(t\right)\le 0$, $t\in J$. The proving process is done. ☐

Lemma 2.

Let $j,k\in \mathsf{\Omega }$ be such that:

$$\left\{\begin{array}{cc}\hfill j^{\prime }\left(t\right)& \le -A_1\left(t\right)j\left(t\right)+A_2\left(t\right)k\left(t\right)-\left({\mathcal{L}}_{1}j\right)\left(t\right)+\left({\mathcal{L}}_{2}k\right)\left(t\right),t\in I,\hfill \\ \hfill k^{\prime }\left(t\right)& \le -A_1\left(t\right)k\left(t\right)+A_2\left(t\right)j\left(t\right)-\left({\mathcal{L}}_{1}k\right)\left(t\right)+\left({\mathcal{L}}_{2}j\right)\left(t\right),t\in I,\hfill \\ \hfill j\left(0\right)& \le {ϵ}_{1}j\left(Y\right)+{ϵ}_{2}k\left(Y\right),\hfill \\ \hfill k\left(0\right)& \le {ϵ}_{1}k\left(Y\right)+{ϵ}_{2}j\left(Y\right),\hfill \end{array}\right.$$

(4)

where $0<{ϵ}_1+{ϵ}_2\le 1,0<{ϵ}_1-{ϵ}_2\le 1$, $A_1,A_2\in C(I,[0,+\infty )),A_1\left(t\right)\ge A_2\left(t\right)$, $t\in I$, ${\mathcal{L}}_1-{\mathcal{L}}_2\in C(E,E)$ and ${\mathcal{L}}_1+{\mathcal{L}}_2\in C(E,E)$ are positive linear operators. Additionally,

$$\begin{array}{c}\begin{array}{c}{\displaystyle \max \left\{\frac{{\int }_0^Y[A_1\left(t\right)+A_2\left(t\right)+{\mathcal{L}}_1\left(t\right)+{\mathcal{L}}_2\left(t\right)]dt}{{ϵ}_1-{ϵ}_2},\frac{{\int }_0^Y[A_1\left(t\right)-A_2\left(t\right)+{\mathcal{L}}_1\left(t\right)-{\mathcal{L}}_2\left(t\right)]dt}{{ϵ}_1+{ϵ}_2}\right\}}\hfill \\ \le 1.\hfill \end{array}\hfill \end{array}$$

(5)

Then, $j\left(t\right)\le 0$ and $k\left(t\right)\le 0$, $t\in I$.

Proof. 

Let $c\left(t\right)=j\left(t\right)+k\left(t\right),t\in I$. Then:

$$\begin{array}{cc}\hfill c^{\prime }\left(t\right)=j^{\prime }\left(t\right)+k^{\prime }\left(t\right)& \le -A_1\left(t\right)c\left(t\right)+A_2\left(t\right)c\left(t\right)-\left[({\mathcal{L}}_1-{\mathcal{L}}_2)c\right]\left(t\right)\hfill \\ \hfill & \le -[A_1\left(t\right)-A_2\left(t\right)]c\left(t\right)-\left[({\mathcal{L}}_1-{\mathcal{L}}_2)c\right]\left(t\right),t\in I\hfill \end{array}$$

and:

$$c\left(0\right)=j\left(0\right)+k\left(0\right)\le {ϵ}_{1}c\left(Y\right)+{ϵ}_{2}c\left(Y\right)=({ϵ}_1+{ϵ}_2)c\left(Y\right).$$

From Lemma (1), $c\left(t\right)\le 0$. Hence, $j\left(t\right)\le -k\left(t\right)$ and $k\left(t\right)\le -j\left(t\right)$, then we have:

$$\left\{\begin{array}{c}{j}^{\prime }\left(t\right)\le -[A_1\left(t\right)+A_2\left(t\right)]j\left(t\right)-\left[({\mathcal{L}}_1+{\mathcal{L}}_2)j\right]\left(t\right),t\in I,\hfill \\ j\left(0\right)\le ({ϵ}_1-{ϵ}_2)j\left(Y\right)\hfill \end{array}\right.$$

and:

$$\left\{\begin{array}{c}{k}^{\prime }\left(t\right)\le -[A_1\left(t\right)+A_2\left(t\right)]k\left(t\right)-\left[({\mathcal{L}}_1+{\mathcal{L}}_2)k\right]\left(t\right),t\in I,\hfill \\ k\left(0\right)\le ({ϵ}_1-{ϵ}_2)k\left(Y\right).\hfill \end{array}\right.$$

From Lemma (1), one obtains $j\left(t\right)\le 0,k\left(t\right)\le 0,t\in I$. This is the end. ☐

3. Existence Results

Let us consider the linear problem:

$$\left\{\begin{array}{c}{x}^{\prime }\left(t\right)=-A\left(t\right)x\left(t\right)-\left(\mathcal{L}x\right)\left(t\right)+\sigma \left(t\right),t\in I,\hfill \\ N_{1}x\left(0\right)-N_{2}x\left(Y\right)=B,\hfill \end{array}\right.$$

(6)

where $N_1,N_2$ are constants, $\sigma \in E$, $A\in C(I,[0,+\infty \left)\right)$, $B\in \mathbb{R}$.

Lemma 3.

$x\in \mathsf{\Omega }$ is one solution of (6) when $x\in E$ is one solution of the equation:

$$\begin{array}{c}{\displaystyle x\left(t\right)=\frac{B{e}^{-{\int }_0^{t}A\left(r\right)dr}}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}+{\int }_0^{Y}D(t,s)[\sigma \left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds,}\hfill \end{array}$$

(7)

where $B=-g(\eta \left(0\right),\eta \left(T\right))+N_1\eta \left(0\right)-N_2\eta \left(Y\right)$, $\eta \in \mathsf{\Omega }$, $A\in C(I,[0,+\infty \left)\right)$, $N_1,N_2$ are constants with $N_1\ne N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}$, and:

$$D(t,s):=\left\{\begin{array}{cc}\frac{N_2}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}{e}^{{\int }_t^{0}A\left(r\right)dr}{e}^{{\int }_Y^{s}A\left(r\right)dr}+e^{{\int }_t^{s}A\left(r\right)dr},\hfill & 0\le s<t\le Y,\hfill \\ \frac{N_2}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}{e}^{{\int }_t^{0}A\left(r\right)dr}{e}^{{\int }_Y^{s}A\left(r\right)dr},\hfill & 0\le t\le s\le Y.\hfill \end{array}\right.$$

Proof. 

Suppose that $x\in \mathsf{\Omega }$ is one solution of Problem (6), one achieves:

$$\begin{array}{cc}x\left(t\right)=\hfill & {\displaystyle x\left(0\right)e^{-{\int }_0^{t}A\left(r\right)dr}+{\int }_0^t{e}^{{\int }_t^{s}A\left(r\right)dr}[\sigma \left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds.}\hfill \end{array}$$

Let $t=Y$; we get:

$$\begin{array}{cc}x\left(Y\right)=\hfill & {\displaystyle x\left(0\right)e^{-{\int }_0^{Y}A\left(r\right)dr}+{\int }_0^Y{e}^{{\int }_Y^{s}A\left(r\right)dr}[\sigma \left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds.}\hfill \end{array}$$

Since $N_{1}x\left(0\right)-N_{2}x\left(T\right)=B$, we obtain:

$$\begin{array}{cc}\hfill x\left(0\right)=& {\displaystyle \frac{B}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}}\hfill \\ \hfill & {\displaystyle +\frac{N_2}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}{\int }_0^Y{e}^{{\int }_Y^{s}A\left(r\right)dr}[\sigma \left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds.}\hfill \end{array}$$

Then:

$$\begin{array}{c}{\displaystyle x\left(t\right)=\frac{B{e}^{-{\int }_0^{t}A\left(r\right)dr}}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}+{\int }_0^{Y}D(t,s)[\sigma \left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds,}\hfill \end{array}$$

where:

$$D(t,s)=\left\{\begin{array}{cc}\frac{N_2}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}{e}^{{\int }_t^{0}A\left(r\right)dr}{e}^{{\int }_Y^{s}A\left(r\right)dr}+e^{{\int }_t^{s}A\left(r\right)dr},\hfill & 0\le s<t\le Y,\hfill \\ \frac{N_2}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}{e}^{{\int }_t^{0}A\left(r\right)dr}{e}^{{\int }_Y^{s}A\left(r\right)dr},\hfill & 0\le t\le s\le Y.\hfill \end{array}\right.$$

Apparently, if $x\left(t\right)$ is a solution of (6), so $x\left(t\right)$ will also be a solution of (7). The proof is complete. ☐

Clearly, $\parallel D(t,s)\parallel =\max \left\{\left|\frac{N_1}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}\right|,\left|\frac{N_2}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}\right|\right\}$. In the next part, we define $\parallel D(t,s)\parallel =d$.

Lemma 4.

$A\in C(I,[0,+\infty \left)\right)$, $N_1\ne N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}$, and:

$$d\parallel \mathcal{L}\parallel Y<1.$$

Then, (6) has only one solution.

Proof. 

Give a mapping $F:E\to E$ by:

$$\begin{array}{cc}\left(Fx\right)\left(t\right):=\hfill & {\displaystyle \frac{B{e}^{-{\int }_0^{t}A\left(r\right)dr}}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}+{\int }_0^{Y}D(t,s)[\sigma \left(s\right)-\left(\mathcal{L}x\right)\left(s\right)]ds.}\hfill \end{array}$$

For any $x_1,x_2\in E$, we have:

$$\parallel F{x}_1-F{x}_2\parallel \le d\parallel \mathcal{L}\parallel Y\parallel x_2-x_1\parallel .$$

From the Banach contraction principle, (6) has only one solution. The proof is then finished.

Remark 1.

If $A\in C(I,[0,+\infty \left)\right)$, $N_1\ge N_2>0$, and:

$$\frac{N_1}{N_1-N_2{e}^{-{\int }_0^{Y}A\left(r\right)dr}}\parallel \mathcal{L}\parallel Y<1.$$

Then, (6) has only one solution.

Let $B(k,h)\left(t\right):=A_1\left(t\right)k\left(t\right)+A_2\left(t\right)h\left(t\right)+\left({\mathcal{L}}_{1}k\right)\left(t\right)+\left({\mathcal{L}}_{2}h\right)\left(t\right)$.

Theorem 1.

Suppose that linear operators ${\mathcal{L}}_1-{\mathcal{L}}_2\in C(E,E)$ and ${\mathcal{L}}_1+{\mathcal{L}}_2\in C(E,E)$ are positive. Let ${\sigma }_1,{\sigma }_2\in E$, $C_1$, $C_2\in \mathbb{R}$, $A_1$, $A_2\in C(I,[0,+\infty ))$ and $A_1\left(t\right)\ge A_2\left(t\right)$ for $t\in I$. Moreover, we assume that (5) holds, $N_1$, $N_2$, $N_3$ are constants with $N_1\ne (N_2-N_3)e^{-{\int }_0^Y[A_1\left(r\right)+A_2\left(r\right)]dr}$, $N_1\ne (N_2+N_3)e^{-{\int }_0^Y[A_1\left(r\right)-A_2\left(r\right)]dr}$, and:

$$\begin{array}{c}\max \left\{\frac{N_1\parallel {\mathcal{L}}_1+{\mathcal{L}}_2\parallel Y}{N_1-(N_2-N_3)e^{-{\int }_0^Y[A_1\left(r\right)+A_2\left(r\right)]dr}},\frac{N_1\parallel {\mathcal{L}}_1-{\mathcal{L}}_2\parallel Y}{N_1-(N_2+N_3)e^{-{\int }_0^Y[A_1\left(r\right)-A_2\left(r\right)]dr}}\right\}<1.\hfill \end{array}$$

(8)

Then, the system:

$$\left\{\begin{array}{c}{k}^{\prime }\left(t\right)=-B(k,h)\left(t\right)+{\sigma }_1\left(t\right),t\in I,\hfill \\ h^{\prime }\left(t\right)=-B(h,k)\left(t\right)+{\sigma }_2\left(t\right),t\in I,\hfill \\ N_{1}x\left(0\right)=N_{2}k\left(Y\right)-N_{3}h\left(Y\right)+A_1,\hfill \\ N_{1}y\left(0\right)=N_{2}h\left(Y\right)-N_{3}k\left(Y\right)+A_2,\hfill \end{array}\right.$$

(9)

has only one solution.

Proof. 

Set $e=k+h,l=k-h$. Then, we have:

$$\left\{\begin{array}{c}{e}^{\prime }\left(t\right)=-[A_1\left(t\right)+A_2\left(t\right)]e\left(t\right)-\left[({\mathcal{L}}_1+{\mathcal{L}}_2)e\right]\left(t\right)+{\sigma }_1\left(t\right)+{\sigma }_2\left(t\right),\hfill \\ N_{1}e\left(0\right)=(N_2-N_3)e\left(Y\right)+C_1+C_2,\hfill \end{array}\right.$$

(10)

and:

$$\left\{\begin{array}{c}{l}^{\prime }\left(t\right)=-[A_1\left(t\right)-A_2\left(t\right)]l\left(t\right)-\left[({\mathcal{L}}_1-{\mathcal{L}}_2)l\right]\left(t\right)+{\sigma }_1\left(t\right)-{\sigma }_2\left(t\right),\hfill \\ N_{1}l\left(0\right)=(N_2+N_3)l\left(Y\right)+C_1-C_2.\hfill \end{array}\right.$$

(11)

By Lemma 4, both (10) and (11) have a unique solution.

Note that $k={\displaystyle \frac{e+l}{2}},h={\displaystyle \frac{e-l}{2}}$, and the pair $(k,h)$ is the only solution of Problem (9). This ends the proof. ☐

4. Main Results

Definition 2.

A pair $(S,W)$ is seen as a coupled lower and upper approximative solution to (1) if:

$$\left\{\begin{array}{c}{S}^{\prime }\left(t\right)\le Q_1(S,W)\left(t\right),t\in I,\hfill \\ W^{\prime }\left(t\right)\ge Q_2(W,S)\left(t\right),t\in I,\hfill \\ g_1(S\left(0\right),S\left(Y\right),W\left(0\right),W\left(Y\right))\le 0,\hfill \\ g_2(W\left(0\right),W\left(Y\right),S\left(0\right),S\left(Y\right))\ge 0.\hfill \end{array}\right.$$

Theorem 2.

Suppose that (5), (8) hold, $Q_1,Q_2\in C(E,E)$ and $\left(H_1\right)$ the function $(S,W)\in \mathsf{\Omega }$ is the coupled lower and upper approximative solution to Problem (1) satisfying $S\left(t\right)\le W\left(t\right),t\in I$; $\left(H_2\right)$ there exist $A_1,A_2\in C(I,[0,+\infty ))$ with $A_1\left(t\right)\ge A_2\left(t\right)\ge 0$ for $t\in I$, and the positive linear operators ${\mathcal{L}}_1-{\mathcal{L}}_2,{\mathcal{L}}_1+{\mathcal{L}}_2\in C(E,E)$ satisfy:

$$\begin{array}{cc}\hfill & Q_i(x,y)\left(t\right)-Q_j(v,w)\left(t\right)\hfill \\ \hfill \le & A_1\left(t\right)[v\left(t\right)-x\left(t\right)]+A_2\left(t\right)[w\left(t\right)-y\left(t\right)]+\left[{\mathcal{L}}_1(v-x)\right]\left(t\right)+\left[{\mathcal{L}}_2(w-y)\right]\left(t\right)\hfill \end{array}$$

for $i\ge j$, i, $j=1,2$ for $t\in I$, $S\le x\le v\le W$, $S\le w\le y\le W$; $\left(H_3\right)$ the functions $g_1,g_2\in C({\mathbb{R}}^4,\mathbb{R})$ are non-decreasing of the third variable and satisfy:

$$g_i({\overline{x}}_1,{\overline{x}}_2,y,y_1)-g_j(x_1,x_2,y,{\overline{y}}_1)\le N_1({\overline{x}}_1-x_1)-N_2({\overline{x}}_2-x_2)+N_3(y_1-{\overline{y}}_1)$$

for $i\ge j$, i, $j=1,2$, $S\left(0\right)\le x_1\le {\overline{x}}_1\le W\left(0\right),S\left(0\right)\le x_2\le {\overline{x}}_2\le W\left(0\right)$, $S\left(Y\right)\le y\le \overline{y}\le W\left(Y\right)$, $S\left(Y\right)\le y_1\le {\overline{y}}_1\le W\left(Y\right)$, where $N_1\ge N_2+N_3$, $N_2>0$, $N_3>0$ and ${ϵ}_1=\frac{N_2}{N_1}$, ${ϵ}_2=\frac{N_3}{N_1}$.

Then, Problem (1) has solution $(x,y)\in [S,W]\times [S,W]$, where $[S,W]=\{\xi \in \mathsf{\Omega }:S(t)\le \xi (t)\le W(t),t\in I\}$.

Proof. 

Fist, using the following equations, one constructs two sequences $\left\{S_n\left(t\right)\right\}$ and $\left\{W_n\left(t\right)\right\}$:

$$\left\{\begin{array}{c}{S}_n^{\prime }\left(t\right)=-B(S_n,W_n)\left(t\right)+F_1(S_{n-1},W_{n-1})\left(t\right),t\in I,\hfill \\ W_n^{\prime }\left(t\right)=-B(W_n,S_n)\left(t\right)+F_2(W_{n-1},S_{n-1})\left(t\right),t\in I,\hfill \\ N_1{S}_n\left(0\right)=N_2{S}_n\left(Y\right)-N_3{W}_n\left(Y\right)+{\phi }_1(S_{n-1},W_{n-1}),\hfill \\ N_1{W}_n\left(0\right)=N_2{W}_n\left(Y\right)-N_3{S}_n\left(Y\right)+{\phi }_2(W_{n-1},S_{n-1}),\hfill \end{array}\right.$$

(12)

with $n=1,2,\dots$, where $S_0=S$, $W_0=W$, $F_i(x,y)\left(t\right)=Q_i(x,y)\left(t\right)+A_1\left(t\right)x\left(t\right)+A_2\left(t\right)y\left(t\right)+\left({\mathcal{L}}_{1}x\right)\left(t\right)+\left({\mathcal{L}}_{1}y\right)\left(t\right)$ and ${\phi }_i(u,v)=N_{1}u\left(0\right)-N_{2}u\left(Y\right)+N_{3}v\left(Y\right)-g_i(u\left(0\right),u\left(Y\right),v\left(0\right),v\left(Y\right)),i=1,2.$

Apparently, from Theorem 1, Problem (12) has only one solution. Now, we are going to divide this proof into five parts.

Step 1. One shows that $S\le S_1$ and $W_1\le W$.

Set $r\left(t\right)=S\left(t\right)-S_1\left(t\right),s\left(t\right)=W_1\left(t\right)-W\left(t\right)$. From $\left(H_1\right)$, we get:

$$\begin{array}{cc}\hfill r^{\prime }\left(t\right)& =S^{\prime }\left(t\right)-S_1^{\prime }\left(t\right)\hfill \\ \hfill & \le Q_1(S,W)\left(t\right)+B(S_1,W_1)\left(t\right)-F_1(S,W)\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)r\left(t\right)+A_2\left(t\right)s\left(t\right)-\left({\mathcal{L}}_{1}r\right)\left(t\right)+\left({\mathcal{L}}_{2}s\right)\left(t\right),t\in I,\hfill \\ \hfill s^{\prime }\left(t\right)& =W_1^{\prime }\left(t\right)-W^{\prime }\left(t\right)\hfill \\ \hfill & \le B(W_1,S_1)\left(t\right)+F_2(W,S)\left(t\right)-Q_2(W,S)\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)s\left(t\right)+A_2\left(t\right)r\left(t\right)-\left({\mathcal{L}}_{1}s\right)\left(t\right)+\left({\mathcal{L}}_{2}r\right)\left(t\right),t\in I,\hfill \\ \hfill r\left(0\right)& =S\left(0\right)-S_1\left(0\right)\hfill \\ \hfill & {\displaystyle \le \frac{1}{N_1}{g}_1(S\left(0\right),S\left(Y\right),W\left(0\right),W\left(Y\right))+\frac{N_2}{N_1}r\left(Y\right)+\frac{N_3}{N_1}s\left(Y\right)}\hfill \\ \hfill & {\displaystyle \le \frac{N_2}{N_1}r\left(Y\right)+\frac{N_3}{N_1}s\left(Y\right),}\hfill \end{array}$$

and:

$$\begin{array}{cc}\hfill s\left(0\right)& =W_1\left(0\right)-W\left(0\right)\hfill \\ \hfill & {\displaystyle \le -\frac{1}{N_1}{g}_2(W\left(0\right),W\left(Y\right),S\left(0\right),W\left(Y\right))+\frac{N_2}{N_1}s\left(Y\right)+\frac{N_3}{N_1}r\left(Y\right)}\hfill \\ \hfill & {\displaystyle \le \frac{N_2}{N_1}s\left(Y\right)+\frac{N_3}{N_1}r\left(Y\right).}\hfill \end{array}$$

From Lemma 2 and $N_1\ge N_2>0,N_1\ge N_3>0$, we have $r\left(t\right)\le 0$ and $s\left(t\right)\le 0$, so $S\le S_1$ and $W_1\le W$.

Step 2. We show that $W_n\le W_{n-1}$ and $S_{n-1}\le S_n$ if $W_{n-1}\le W_{n-2},S_{n-2}\le S_{n-1}$, $n=2,3,\dots$.

Set $e\left(t\right)=S_{n-1}\left(t\right)-S_n\left(t\right),l\left(t\right)=W_n\left(t\right)-W_{n-1}\left(t\right)$. From $\left(H_1\right),\left(H_2\right)$, we get:

$$\begin{array}{cc}\hfill e^{\prime }\left(t\right)& =S_{n-1}^{\prime }\left(t\right)-S_n^{\prime }\left(t\right)\hfill \\ \hfill & =-B(S_{n-1},W_{n-1})\left(t\right)+F_1(S_{n-2},W_{n-2})\left(t\right)+B(S_n,W_n)\left(t\right)-F_1(S_{n-1},W_{n-1})\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)e\left(t\right)+A_2\left(t\right)l\left(t\right)-\left({\mathcal{L}}_{1}e\right)\left(t\right)+\left({\mathcal{L}}_{2}l\right)\left(t\right)+Q_1(S_{n-2},W_{n-2})\hfill \\ \hfill & -Q_1(S_{n-1},W_{n-1})+A_1\left(t\right)[S_{n-2}\left(t\right)-S_{n-1}\left(t\right)]+A_2\left(t\right)[W_{n-2}\left(t\right)-W_{n-1}\left(t\right)]\hfill \\ \hfill & +\left[{\mathcal{L}}_1(S_{n-2}-S_{n-1})\right]\left(t\right)+\left[{\mathcal{L}}_2(W_{n-2}-W_{n-1})\right]\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)e\left(t\right)+A_2\left(t\right)l\left(t\right)-\left({\mathcal{L}}_{1}e\right)\left(t\right)+\left({\mathcal{L}}_{2}l\right)\left(t\right),t\in I,\hfill \end{array}$$

and:

$$\begin{array}{cc}\hfill l^{\prime }\left(t\right)& =W_n^{\prime }\left(t\right)-W_{n-1}^{\prime }\left(t\right)\hfill \\ \hfill & =-B(W_n,S_n)\left(t\right)+F_2(W_{n-1},S_{n-1})\left(t\right)+B(W_{n-1},S_{n-1})\left(t\right)-F_2(W_{n-2},S_{n-2})\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)l\left(t\right)+A_2\left(t\right)e\left(t\right)-\left({\mathcal{L}}_{1}l\right)\left(t\right)+\left({\mathcal{L}}_{2}e\right)\left(t\right)+Q_2(W_{n-2},S_{n-2})\hfill \\ \hfill & -Q_2(W_{n-1},S_{n-1})+A_1\left(t\right)[W_{n-1}\left(t\right)-W_{n-2}\left(t\right)]+A_2\left(t\right)[S_{n-1}\left(t\right)-S_{n-2}\left(t\right)]\hfill \\ \hfill & +\left[{\mathcal{L}}_1(W_{n-1}-W_{n-2})\right]\left(t\right)+\left[{\mathcal{L}}_2(S_{n-1}-S_{n-2})\right]\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)l\left(t\right)+A_2\left(t\right)e\left(t\right)-\left({\mathcal{L}}_{1}l\right)\left(t\right)+\left({\mathcal{L}}_{2}e\right)\left(t\right),t\in I.\hfill \end{array}$$

By $\left(H_1\right)$ and $\left(H_3\right)$, we have:

$$\begin{array}{cc}\hfill N_{1}e\left(0\right)& =N_1[S_{n-1}\left(0\right)-S_n\left(0\right)]\hfill \\ \hfill & =N_1{S}_{n-1}\left(0\right)-N_2{S}_n\left(Y\right)+N_3{W}_n\left(Y\right)-{\phi }_1(S_{n-1},W_{n-1})\hfill \\ \hfill & =N_{2}e\left(Y\right)+N_{3}l\left(Y\right)+g_1(S_{n-1}\left(0\right),S_{n-1}\left(Y\right),W_{n-1}\left(0\right),W_{n-1}\left(Y\right))\hfill \\ \hfill & -g_1(S_{n-2}\left(0\right),S_{n-2}\left(Y\right),W_{n-2}\left(0\right),W_{n-2}\left(Y\right))+N_1[S_{n-2}\left(0\right)-S_{n-1}\left(0\right)]\hfill \\ \hfill & +N_2[S_{n-1}\left(Y\right)-S_{n-2}\left(Y\right)]+N_3[W_{n-2}\left(Y\right)-W_{n-1}\left(Y\right)]\hfill \\ \hfill & \le N_{2}e\left(Y\right)+N_{3}l\left(Y\right),\hfill \end{array}$$

and:

$$\begin{array}{cc}\hfill N_{1}l\left(0\right)& =N_1[W_n\left(0\right)-W_{n-1}\left(0\right)]\hfill \\ \hfill & =N_2{W}_n\left(Y\right)-N_3{S}_n\left(Y\right)+{\phi }_2(W_{n-1},S_{n-1})-N_1{W}_{n-1}\left(0\right)\hfill \\ \hfill & =N_{2}l\left(Y\right)+N_{3}e\left(Y\right)-g_2(W_{n-1}\left(0\right),W_{n-1}\left(Y\right),S_{n-1}\left(0\right),S_{n-1}\left(Y\right))\hfill \\ \hfill & +g_2(W_{n-2}\left(0\right),W_{n-2}\left(Y\right),S_{n-2}\left(0\right),S_{n-2}\left(Y\right))+N_1[W_{n-1}\left(0\right)-W_{n-2}\left(0\right)]\hfill \\ \hfill & -N_2[W_{n-1}\left(Y\right)-W_{n-2}\left(Y\right)]+N_3[S_{n-1}\left(Y\right)-S_{n-2}\left(Y\right)]\hfill \\ \hfill & \le N_{2}l\left(Y\right)+N_{3}e\left(Y\right).\hfill \end{array}$$

Using Lemma 2 and $N_1\ge N_2>0,N_1\ge N_3>0$, we have $e\le 0,l\le 0$, so $W_n\le W_{n-1}$, $S_{n-1}\le S_n$.

From mathematical induction, one gets that $\left\{S_n\right\}$ is a nondecreasing sequence, and $\left\{W_n\right\}$ is a nonincreasing sequence.

Step 3. We claim that $S_1\le W_1$ if $S\le W$.

Let $u\left(t\right)=S_1\left(t\right)-W_1\left(t\right)$. Using $\left(H_2\right)$ and $\left(H_3\right)$, one receives:

$$\begin{array}{cc}\hfill u^{\prime }\left(t\right)& =S_1^{\prime }\left(t\right)-W_1^{\prime }\left(t\right)\hfill \\ \hfill & =-B(S_1,W_1)\left(t\right)+F_1(S,W)\left(t\right)+B(W_1,S_1)\left(t\right)-F_2(W,S)\left(t\right)\hfill \\ \hfill & =-[A_1\left(t\right)-A_2\left(t\right)]u\left(t\right)-\left[({\mathcal{L}}_1-{\mathcal{L}}_2)u\right]\left(t\right)+Q_1(S,W)\left(t\right)+A_1\left(t\right)S\left(t\right)\hfill \\ \hfill & +A_2\left(t\right)W\left(t\right)+\left({\mathcal{L}}_{1}S\right)\left(t\right)+\left({\mathcal{L}}_{2}W\right)\left(t\right)-Q_2(W,S)\left(t\right)\hfill \\ \hfill & -A_1\left(t\right)W\left(t\right)-A_2\left(t\right)S\left(t\right)-\left({\mathcal{L}}_{1}W\right)\left(t\right)-\left({\mathcal{L}}_{2}S\right)\left(t\right)\hfill \\ \hfill & \le -[A_1\left(t\right)-A_2\left(t\right)]u\left(t\right)-\left[({\mathcal{L}}_1-{\mathcal{L}}_2)u\right]\left(t\right),t\in I,\hfill \end{array}$$

and:

$$\begin{array}{cc}\hfill N_{1}u\left(0\right)& =N_1[S_1\left(0\right)-W_1\left(0\right)]\hfill \\ \hfill & =(N_2+N_3)u\left(Y\right)+{\phi }_1(S,W)-{\phi }_2(W,S)\hfill \\ \hfill & =(N_2+N_3)u\left(Y\right)+N_1[S\left(0\right)-W\left(0\right)]+(N_2+N_3)[W\left(Y\right)-S\left(Y\right)]\hfill \\ \hfill & -g_1(S\left(0\right),S\left(Y\right),W\left(0\right),W\left(Y\right))+g_2(W\left(0\right),W\left(Y\right),S\left(0\right),S\left(Y\right))\hfill \\ \hfill & \le (N_2+N_3)u\left(Y\right).\hfill \end{array}$$

From Lemma 1, one arrives at $u\le 0$, and $S_1\le W_1$. Utilizing the mathematical induction, we gain $W_n\le S_n$, $n=1,2,\dots$.

Step 4. By the above three steps, we obtain:

$$S_0\le S_1\le \cdots \le S_n\le \cdots \le W_n\le \cdots \le W_1\le W_0,$$

and every $S_n,W_n$ satisfies (12) for all n. We can get that the sequences $\left\{S_n\left(t\right)\right\},\left\{W_n\left(t\right)\right\}$ are monotonically and bounded. Then, there are $\rho ,r$ satisfying that ${\displaystyle \underset{n\to \infty }{\lim }{S}_n\left(t\right)=\rho \left(t\right)}$, ${\displaystyle \underset{n\to \infty }{\lim }{W}_n\left(t\right)=r\left(t\right)}$ and satisfying the equations:

$$\left\{\begin{array}{c}{\rho }^{\prime }\left(t\right)=Q_1(\rho ,r)\left(t\right),t\in I,\hfill \\ r^{\prime }\left(t\right)=Q_2(r,\rho )\left(t\right),t\in I,\hfill \\ g\left(\rho \right(0),\rho (Y),r(0),r(Y\left)\right)=0,\hfill \\ g\left(r\right(0),r(Y),\rho (0),\rho (Y\left)\right)=0.\hfill \end{array}\right.$$

It proves that a pair $(\rho ,r)$ is the solution of Problem (1).

Step 5. One verifies that $(\rho ,r)$ is an extremal solution of (1) in $[S,W]$.

Let $(x,y)$ be any pair of solutions for (1) satisfying $S\left(t\right)\le x\left(t\right),y\left(t\right)\le W\left(t\right)$. We assume that there is a positive integer n satisfying $S_{n-1}\left(t\right)\le x\left(t\right),y\left(t\right)\le W_{n-1}\left(t\right)$. Set $u\left(t\right)=S_n\left(t\right)-x\left(t\right)$ and $v\left(t\right)=y\left(t\right)-W_n\left(t\right)$. By using $\left(H_2\right)$ and $\left(H_3\right)$, one gets:

$$\begin{array}{cc}\hfill u^{\prime }\left(t\right)& =S_n^{\prime }\left(t\right)-x^{\prime }\left(t\right)\hfill \\ \hfill & =-B(S_n,W_n)\left(t\right)+F_1(S_{n-1},W_{n-1})-Q_1(x,y)\left(t\right)\hfill \\ \hfill & =-A_1\left(t\right)S_n\left(t\right)-A_2\left(t\right)W_n\left(t\right)-\left({\mathcal{L}}_1{S}_n\right)\left(t\right)-\left({\mathcal{L}}_2{W}_n\right)\left(t\right)+Q_1(S_{n-1},W_{n-1})\left(t\right)\hfill \\ \hfill & +A_1\left(t\right)S_{n-1}\left(t\right)+A_2\left(t\right)W_{n-1}\left(t\right)+\left({\mathcal{L}}_1{S}_{n-1}\right)\left(t\right)+\left({\mathcal{L}}_2{W}_{n-1}\right)\left(t\right)-Q_1(x,y)\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)u\left(t\right)+A_2\left(t\right)v\left(t\right)-\left({\mathcal{L}}_{1}u\right)\left(t\right)+\left({\mathcal{L}}_{2}v\right)\left(t\right),t\in I,\hfill \\ \hfill v^{\prime }\left(t\right)& =y^{\prime }\left(t\right)-W_n^{\prime }\left(t\right)\hfill \\ \hfill & =Q_2(x,y)\left(t\right)+B(W_n,S_n)\left(t\right)-F_2(W_{n-1},S_{n-1})\hfill \\ \hfill & =Q_2(y,x)\left(t\right)+A_1\left(t\right)W_n\left(t\right)+A_2\left(t\right)S_n\left(t\right)+\left({\mathcal{L}}_1{W}_n\right)\left(t\right)+\left({\mathcal{L}}_2{S}_n\right)\left(t\right)\hfill \\ \hfill & -Q_2(W_{n-1},S_{n-1})\left(t\right)-A_1\left(t\right)W_{n-1}\left(t\right)-A_2\left(t\right)S_{n-1}\left(t\right)\hfill \\ \hfill & -\left({\mathcal{L}}_1{W}_{n-1}\right)\left(t\right)-\left({\mathcal{L}}_2{S}_{n-1}\right)\left(t\right)\hfill \\ \hfill & \le -A_1\left(t\right)v\left(t\right)+A_2\left(t\right)u\left(t\right)-\left({\mathcal{L}}_{1}v\right)\left(t\right)+\left({\mathcal{L}}_{2}u\right)\left(t\right),t\in I,\hfill \\ \hfill N_{1}u\left(0\right)& =N_1[S_n\left(0\right)-x\left(0\right)]\hfill \\ \hfill & =N_2{S}_n\left(Y\right)-N_3{W}_n\left(Y\right)+{\phi }_1(S_{n-1},W_{n-1})-N_{1}x\left(0\right)\hfill \\ \hfill & =N_2{S}_n\left(Y\right)-N_3{W}_n\left(Y\right)+N_1{S}_{n-1}\left(0\right)-N_2{S}_{n-1}\left(Y\right)+N_3{W}_{n-1}\left(Y\right)-N_{1}x\left(0\right)\hfill \\ \hfill & -g_1(S_{n-1}\left(0\right),S_{n-1}\left(Y\right),W_{n-1}\left(0\right),W_{n-1}\left(Y\right))+g_1(x\left(0\right),x\left(Y\right),y\left(0\right),y\left(Y\right))\hfill \\ \hfill & \le N_2{S}_n\left(Y\right)-N_{2}x\left(Y\right)-N_3{W}_n\left(Y\right)+N_{3}y\left(Y\right)\hfill \\ \hfill & \le N_{2}u\left(Y\right)+N_{3}v\left(Y\right),\hfill \end{array}$$

and:

$$\begin{array}{cc}\hfill N_{1}v\left(0\right)& =N_1[y\left(0\right)-W_n\left(0\right)]\hfill \\ \hfill & =N_{1}y\left(0\right)-N_2{W}_n\left(Y\right)+N_3{S}_n\left(Y\right)+{\phi }_1(W_{n-1},S_{n-1})\hfill \\ \hfill & =N_{1}y\left(0\right)-N_2{W}_n\left(Y\right)+N_3{S}_n\left(Y\right)-N_1{W}_{n-1}\left(0\right)+N_2{W}_{n-1}\left(Y\right)-N_3{S}_{n-1}\left(Y\right)\hfill \\ \hfill & +g_1(W_{n-1}\left(0\right),W_{n-1}\left(Y\right),S_{n-1}\left(0\right),S_{n-1}\left(Y\right))-g_1(y\left(0\right),y\left(Y\right),x\left(0\right),x\left(Y\right))\hfill \\ \hfill & \le -N_2{W}_n\left(Y\right)+N_{2}y\left(Y\right)+N_3{S}_n\left(Y\right)-N_{3}x\left(Y\right)\hfill \\ \hfill & \le N_{2}v\left(Y\right)+N_{3}u\left(Y\right).\hfill \end{array}$$

From Lemma 2, one see that $S_n\left(t\right)\le x\left(t\right)$, $y\left(t\right)\le W_n\left(t\right)$ on I. Since $S\left(t\right)\le x\left(t\right)$, $y\left(t\right)\le W\left(t\right)$, by induction, one gets $S_n\left(t\right)\le x\left(t\right)$, $y\left(t\right)\le W_n\left(t\right)$ for every $n\in \mathbb{N}$. Now, take the limit as $n\to \infty$, then $\rho \left(t\right)\le x\left(t\right)$, $y\left(t\right)\le r\left(t\right)$. The proof is over. ☐

5. Example

Example 1.

Assume that:

$$\left\{\begin{array}{c}{z}^{\prime }\left(t\right)=Q_1(z,w)\left(t\right),t\in [0,Y],\hfill \\ w^{\prime }\left(t\right)=Q_2(w,z)\left(t\right),t\in [0,Y],\hfill \\ g_1(z\left(0\right),z\left(Y\right),w\left(0\right),w\left(Y\right))=0,\hfill \\ g_2(w\left(0\right),w\left(Y\right),z\left(0\right),z\left(Y\right))=0,\hfill \end{array}\right.$$

(13)

with:

$$\begin{array}{c}{\displaystyle Q_1(z,w)\left(t\right)=-tz\left(t\right)-\frac{Y}{5t}{\int }_0^{t}z\left(s\right)ds+\frac{3Y}{10t^3}{\int }_0^t{s}^{2}w\left(s\right)ds,}\hfill \\ {\displaystyle Q_2(w,z)\left(t\right)=-tw\left(t\right)-\frac{Y}{5t}{\int }_0^{t}w\left(s\right)ds+\frac{3Y}{10t^3}{\int }_0^t{s}^{2}z\left(s\right)ds,}\hfill \\ g\left(z\right(0),z(Y),w(0),w(Y\left)\right)=4z\left(0\right)-2z\left(Y\right)+w\left(Y\right)-2,\hfill \\ g\left(w\right(0),w(Y),z(0),z(Y\left)\right)=4w\left(0\right)-2w\left(Y\right)+z\left(Y\right)-2.\hfill \end{array}$$

Set $S=0$, $W=2$. It is easily verified that W is an upper solution, S is a lower solution and $S\le W$. Take ${\displaystyle \left({\mathcal{L}}_{1}z\right)\left(t\right)=\frac{Y}{5t}{\int }_0^{t}z\left(s\right)ds}$, ${\displaystyle \left({\mathcal{L}}_{2}w\right)\left(t\right)=\frac{3Y}{10t^3}{\int }_0^t{s}^{2}w\left(s\right)ds}$, $A_1\left(t\right)=t$, $A_2\left(t\right)=0$, $N_1=4$, $N_2=2$, $N_3=1$ and $Y=\frac{1}{2}$. It is easy to see that $\left(H_1\right)$, $\left(H_2\right)$ and $\left(H_3\right)$ are satisfied. By simple computation, we obtain ${\mathcal{L}}_1\left(t\right)=0.2Y$, ${\mathcal{L}}_2\left(t\right)=0.1Y$, and:

$$\begin{array}{c}{\displaystyle {\int }_0^Y[A_1\left(t\right)+A_2\left(t\right)+{\mathcal{L}}_1\left(t\right)+{\mathcal{L}}_2\left(t\right)]dt=0.2<\frac{N_2-N_3}{N_1}=0.25,}\hfill \\ {\displaystyle {\int }_0^Y[A_1\left(t\right)-A_2\left(t\right)+{\mathcal{L}}_1\left(t\right)-{\mathcal{L}}_2\left(t\right)]dt=0.15<\frac{N_2+N_3}{N_1}=0.75,}\hfill \\ {\displaystyle \frac{N_1}{N_1-(N_2-N_3)e^{-{\int }_0^Y[A_1\left(r\right)+A_2\left(r\right)]dr}}\parallel {\mathcal{L}}_1+{\mathcal{L}}_2\parallel Y<1,}\hfill \\ {\displaystyle \frac{N_1}{N_1-(N_2+N_3)e^{-{\int }_0^Y[A_1\left(r\right)+A_2\left(r\right)]dr}}\parallel {\mathcal{L}}_1-{\mathcal{L}}_2\parallel Y<1.}\hfill \end{array}$$

Hence, utilizing Theorem 2, (13) has an extreme solution in $[S,W]\times [S,W]$.

This equation can be used as a simple model of the physical field, and this example shows that the conclusion in this paper applies to Problem (13).

6. Conclusions

In this paper, we obtain the existence results of coupled causal differential equations by means of the monotone iterative technique. In practice, after modeling in our field, if such coupled differential equations are obtained, we do not have to calculate the solution; we only need to use the sufficient conditions in this paper to judge the existence of the solution.