Uniqueness Methods and Stability Analysis for Coupled Fractional Integro-Differential Equations via Fixed Point Theorems on Product Space

Abstract

In this paper, we obtain unique solution and stability results for coupled fractional differential equations with p-Laplacian operator and Riemann–Stieltjes integral conditions that expand and improve the works of some of the literature. In order to obtain the existence and uniqueness of solutions for coupled systems, several fixed point theorems for operators in ordered product spaces are given without requiring the existence conditions of upper–lower solutions or the compactness and continuity of operators. By applying the conclusions of the operator theorem studied, sufficient conditions for the unique solution of coupled fractional integro-differential equations and approximate iterative sequences for uniformly approximating unique solutions were obtained. In addition, the Hyers–Ulam stability of the coupled system is discussed. As applications, the corresponding results obtained are well demonstrated through some concrete examples.

1. Introduction

The idea of fractional calculus originated from a letter of Leibnitz and Hospital dated 30 September 1695, which opened up a new research avenue. Fractional calculus, as a generalization of ordinary calculus of positive integers, attracted more attention, due the contribution of Riemann, Liouville, Gronwall, and Letnikov [1,2,3,4,5]. Compared to the models of integro-differential equations, fractional differential equations (FDEs) have been confirmed to be reasonably good tools for the description of the memory and global properties of various processes and phenomena, due to the fact that the existence of a “memory” term in FDEs guarantees the history and its influence on the present and future. In the last two decades, there have been great achievements in applying FDEs to practical problems in many disciplines: image processing and signal detection [6,7], biological systems [8,9], and many others have been mined. In the wake of these developments, the forms of FDEs have become more abundant and diverse, including a variety of boundary value conditions, such as initial value [10,11], m-point (two-point, three-point, multipoint) boundary value [12,13,14], integral boundary value [15,16], and integro-differential boundary value [17], and include nonlinear terms with different properties, such as monotonic incrementality [18], and concavity and convexity [19,20], etc.

In a large number of nonlinear fractional differential models in dynamic systems, the existence-uniqueness of solutions and stability results have been considered as hot research topics; we highlight some recent results. The existence-uniqueness analysis of a kind of FDE with p-Laplacian operator was executed in [21], using the Guo–Krasnoselskii fixed point theorem. The authors in [22] analyzed a unique solution to a fractional integro-differential equation using the contraction mapping principle. Furthermore, Ulam–Hyers stability was discussed. Ref. [23] studied the existence of a unique solution and its stability in the sense of Hyers and Ulam stability for a class of $\psi$-Hilfer FDEs. The sufficient conditions of unique solution and stability provided to FDEs by the topological degree method was studied in [24]. For more recent cases, we suggest that readers review the works in [25,26,27].

In the last few decades, one of the most rapidly growing developments has been coupled differential equations. The objective fact is that many complex problems—such as automatic control theory, condensed matter physical fractal and solute convection, and dispersion in porous media—are described by coupled differential equations, which is one of the most important reasons for this growth.

Recently, within the framework of coupled systems, fractional derivatives have been introduced that perfectly combine the memory effects and nonlocality of fractional-order operators with the advantages of coupled systems in describing coupling and nonlinearity systems. For a small sample of works on the study of solutions in cases of coupled FDEs, we refer the reader to works [28,29,30,31,32,33].

In [28], the authors discussed a coupled system with three-point boundary conditions:

$$\left\{\begin{array}{c}{}^{c}D_{0^{+}}^{{\eta }_1}x(t)={\lambda }_1{a}_1(t)f_1(t,x(t),y(t)),\hfill \\ {}^{c}D_{0^{+}}^{{\eta }_2}y(t)={\lambda }_2{a}_2(t)f_2(t,x(t),y(t)),\hfill \\ x^{(i)}(0)=0,y^{(i)}(0)=0,i=0,1,···,n-1,\hfill \\ x(1)={\gamma }_{1}x({\xi }_1),y(1)={\gamma }_{2}y({\xi }_2),\hfill \end{array}\right.$$

where ${}^{c}D_{0^{+}}^{{\eta }_1}, {}^{c}D_{0^{+}}^{{\eta }_2}$ signify Caputo fractional derivatives. By means of fixed point theorem, the existence and multiplicity of positive solutions are established.

The coupled equation below with uncoupled integral conditions was introduced in [29]:

$$\left\{\begin{array}{c}{D}_{0^{+}}^{\gamma }y(t)+a_1(t)f_1(t,z(t))=0,\hfill \\ D_{0^{+}}^{\eta }z(t)+a_2(t)f_2(t,y(t))=0,\hfill \\ y(0)=0,y(1)={\int }_0^1{b}_1(\iota )y(\iota )\mathrm{d}\iota ,\hfill \\ z(0)=0,z(1)={\int }_0^1{b}_2(\iota )z(\iota )\mathrm{d}\iota ,\hfill \end{array}\right.$$

where $D_{0^{+}}^{\gamma }$, $D_{0^{+}}^{\eta }$ signify Riemann–Liouville (R–L) fractional derivatives, $1<\gamma ,\eta \le 2,$ $a_j\in C((0,1),$ $[0,+\infty )),$ $b_j\in L^1[0,1]$ is non-negative, $f_j\in C([0,1]\times [0,+\infty ),[0,+\infty )),j=1,2.$ The conclusions of existence–nonexistence to positive solutions are gained by the contraction mapping principle, the Leray–Schauder alternative theorem, and cone fixed point theorems.

In [30], a class of coupled FDEs with three-point coupled conditions are discussed:

$$\left\{\begin{array}{c}{a}_2{}^{c}D^{q+2}+a_1{}^{c}D^{q+1}+a_0{}^{c}D^{q}y(s)=f_1(s,y(s),z(s)),\hfill \\ b_2{}^{c}D^{p+2}+b_1{}^{c}D^{p+1}+b_0{}^{c}D^{p}z(s)=f_2(s,y(s),z(s)),\hfill \\ y(0)=0,y^{\prime }(0)=0,y(1)={\lambda }_{1}z(\eta ),\hfill \\ z(0)=0,z^{\prime }(0)=0,z(1)={\lambda }_{3}y(\eta ),\hfill \end{array}\right.$$

where ${}^{c}D^q$, ${}^{c}D^p$ signify Caputo fractional derivatives, $0<q,p<1$. The existence-uniqueness of solutions was derived by Banach’s contraction principle and Leray–Schauder’s alternative.

Based on the above description, and to the best of our knowledge, only a few studies in the literature have considered unique positive solutions and the stability of coupled FDEs under p-Laplacian operator and Riemann–Stieltjes (R–S) integral conditions. Hence, for the sake of enriching the aforementioned studies, coupled-type FDEs under a p-Laplacian operator and an R–S integral condition are considered:

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_{11}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]+f_1(t,u(t),v(t))+g_1(t,u(t),v(t))=0,t\in [0,1],\hfill \\ D_{0^{+}}^{{\eta }_{21}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]+f_2(t,u(t),v(t))+g_2(t,u(t),v(t))=0,t\in [0,1],\hfill \\ u(0)=0,D_{0^{+}}^{{\alpha }_{1i}}u(0)=0,i-1<{\eta }_{12}-{\alpha }_{1i}\le i(i=2,···,m_1-1),\hfill \\ v(0)=0,D_{0^{+}}^{{\alpha }_{2i}}v(0)=0,i-1<{\eta }_{22}-{\alpha }_{2i}\le i(i=2,···,m_2-1),\hfill \\ D_{0^{+}}^{{\beta }_1}u(1)={\tau }_1{\int }_0^1{k}_1(s)u(s)\mathrm{d}{A}_{12}(s),{\eta }_{12}-{\beta }_1>1,\hfill \\ D_{0^{+}}^{{\beta }_2}v(1)={\tau }_2{\int }_0^1{k}_2(s)v(s)\mathrm{d}{A}_{22}(s),{\eta }_{22}-{\beta }_2>1,\hfill \\ [{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right){]}_{t=0}=0,D_{0^{+}}^{{\beta }_{1i}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]{|}_{t=0}=0,\hfill \\ n_1-i-1<{\eta }_{11}-{\beta }_{1i}<n_1-i,i=1,···,n_1-2,\hfill \\ [{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)]{{|}_{t=0}=0,D_{0^{+}}^{{\beta }_{2i}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]|}_{t=0}=0,\hfill \\ n_2-i-1<{\eta }_{21}-{\beta }_{2i}<n_2-i,i=1,···,n_2-2,\hfill \\ D_{0^{+}}^{{\gamma }_{11}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]{{|}_{t=1}={\int }_0^1{a}_1(s)D_{0^{+}}^{{\gamma }_{12}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]|}_{t=\xi }\mathrm{d}s\hfill \\ +{\int }_0^1{b}_1(s)D_{0^{+}}^{{\gamma }_{13}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(s)\right)\right]\mathrm{d}{A}_{11}(s),\xi \in (0,1),\hfill \\ D_{0^{+}}^{{\gamma }_{21}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]{{|}_{t=1}={\int }_0^1{a}_2(s)D_{0^{+}}^{{\gamma }_{22}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]|}_{t=\xi }\mathrm{d}s\hfill \\ +{\int }_0^1{b}_2(s)D_{0^{+}}^{{\gamma }_{23}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(s)\right)\right]\mathrm{d}{A}_{21}(s),\xi \in (0,1),\hfill \end{array}\right.$$

(1)

where $D_{0^{+}}^{{\eta }_{11}},D_{0^{+}}^{{\eta }_{12}},D_{0^{+}}^{{\eta }_{21}},D_{0^{+}}^{{\eta }_{22}}$ indicate R–L fractional derivatives, ${\int }_0^1{k}_i(s)\mathrm{d}{A}_{ij}(s)$ denotes an R–S integral, $n_j-1<{\eta }_{j1}\le n_j,m_j-1<{\eta }_{j2}\le m_j(n_j,m_j>3),{\gamma }_{j1}-{\gamma }_{j2}>0,$ ${\gamma }_{j1}-{\gamma }_{j3}>0,{\eta }_{j1}-{\gamma }_{j1}-1>0,{\eta }_{j1}-{\gamma }_{j2}-1>0,{\eta }_{j1}-{\gamma }_{j3}-1>0,j=1,2.$ ${\phi }_p(s)={|s|}^{p-2}s,p>1,{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. Let $f_1,f_2:[0,1]\times [0,+\infty )\times [0,+\infty )\to [0,+\infty ),g_1,g_2:[0,1]\times [0,+\infty )\to [0,+\infty ),a_j,b_j,k_j:[0,1]\to [0,+\infty )$ are continuous, and ${\tau }_1,{\tau }_2$ are positive constants. Actually, through different values of the parameters $a_j,b_j,k_j$ and ${\tau }_j(j=1,2)$, our equation can change into the one in [29,34,35,36].

Although the achievements of fixed point theory for nonlinear operators continue to accumulate and develop rapidly, providing theoretical support for solving various nonlinear differential, integral, and difference equations, the theory still has limitations, especially when solving coupled fractional differential equations. In fact, these methods are only applicable to the coupled equations of the following nonlinear terms:

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_1}{\psi }_1(t)=f(t,{\psi }_2(t))\mathrm{or}f(t,{\psi }_2(t),D_{0^{+}}^{{\beta }_1}{\psi }_2(t)),{\beta }_1<{\eta }_1,\hfill \\ D_{0^{+}}^{{\eta }_2}{\psi }_2(t)=g(t,{\psi }_1(t))\mathrm{or}f(t,{\psi }_1(t),D_{0^{+}}^{{\beta }_2}{\psi }_1(t)),{\beta }_2<{\eta }_2,\hfill \end{array}\right.$$

However, in cases of fractional coupled equations with the following mixed nonlinear terms,

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_1}{\psi }_1(t)=f(t,{\psi }_1(t),{\psi }_2(t)),\hfill \\ D_{0^{+}}^{{\eta }_2}{\psi }_2(t)=g(t,{\psi }_1(t),{\psi }_2(t)),\hfill \end{array}\right.$$

(2)

the above fixed point theorems for noncompact and discontinuous operators are no longer applicable and have limitations. However, coupled differential equations with mixed nonlinear terms have wide applications in many models, such as infectious disease models, chronic disease models, and spring systems, making new operator fixed point theorems particularly urgent [37,38,39]. To break these limitations, we propose some fixed point theorems of $\left(T_3,T_4\right)$ and $\left(A_1+B_1,A_2+B_2\right)$ in ordered Banach product space to solve problems with the form (1), where the operators do not require compactness and continuity.

The main contributions of this paper are as follows: (a) This paper utilizes a new operator fixed point theorem to solve differential equations with the forms (1) and (2), which extends the research space and operator properties, breaking the limitations of the original non-compact and non-continuous operator fixed point theorem in solving coupled equations with mixed nonlinear terms, such as [28,29,30]. This is an extension of the research results in this field. (b) The coupled equations (1) that we studied contain a p-Laplacian operator and an R–S integral, an R–L fractional derivative of the boundary value condition, which includes various characteristics in the expansive existing literature. Moreover, by setting the value of the parameters $a_j,b_j,k_j$, and ${\tau }_j(j=1,2)$ in problems (1), the coupled equations can reduce to the problems in [29,34,35,36], which is an extension and improvement of the existing works, and which also demonstrates the effectiveness and universality of operator fixed point theorems’ application. (c) The Hyers–Ulam stability of solutions for nonlinear p-Laplacian equations is established.

2. Preliminaries

For the reader’s convenience, we present some necessary lemmas and basic results:

Definition 1

([1]). The R–L fractional derivative of the function $\delta :[0,+\infty )\to R$ is

$$D_{0^{+}}^{\eta }\delta (s)={\displaystyle \frac{1}{\Gamma (m-\eta )}}{\displaystyle \frac{{\mathrm{d}}^{\mathrm{m}}}{\mathrm{d}{s}^m}}{\int }_0^s{(s-\iota )}^{m-\eta -1}\delta (\iota )\mathrm{d}\iota ,\eta >0.$$

The R–L fractional integral of δ is

$$I_{0^{+}}^{\eta }\delta (s)={\displaystyle \frac{1}{\Gamma (\eta )}}{\int }_0^s{(s-\iota )}^{\eta -1}\delta (\iota )\mathrm{d}\iota ,\eta >0,$$

where $m=[\eta ]+1$, $[\eta ]$ represents the integer part of η.

Lemma 1

([2]). If $\delta \in C(0,1)\bigcap L^1(0,+\infty ),D_{0^{+}}^{\eta }\delta \in C(0,1)\bigcap L^1(0,+\infty )$, $\eta >0$ then

$$I_{0^{+}}^{\eta }{D}_{0^{+}}^{\eta }\delta (t)=\delta (t)+a_1{t}^{\eta -1}+a_2{t}^{\eta -2}+···+a_m{t}^{\eta -m},$$

where $a_i\in R,i=1,2,···,m(m=[\eta ]+1)$.

Lemma 2.

If $\sigma \in C[0,1],j=1,2$ then the equation

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_{j1}}\delta (t)+\sigma (t)=0,n_j-1<{\eta }_{j1}\le n_j,n_j>3,\hfill \\ \delta (0)=0,D_{0^{+}}^{{\beta }_{ji}}\delta (0)=0,n_j-i-1<{\eta }_{j1}-{\beta }_{ji}\le n_j-i,i=1,···,n_j-2,\hfill \\ D_{0^{+}}^{{\gamma }_{j1}}\delta (1)={\int }_0^1{a}_j(s)D_{0^{+}}^{{\gamma }_{j2}}\delta (\xi )\mathrm{d}s+{\int }_0^1{b}_j(s)D_{0^{+}}^{{\gamma }_{j3}}\delta (s)\mathrm{d}{A}_{j1}(s)\hfill \end{array}\right.$$

(3)

has a unique solution,

$${\displaystyle \delta (t)={\int }_0^1{H}_{j3}(t,s)\sigma (s)\mathrm{d}s,}$$

where

$$H_{j3}(t,s)=H_{j0}(t,s)+t^{{\eta }_{j1}-1}{H}_{j1}(\xi ,s)+t^{{\eta }_{j1}-1}{\int }_0^1{H}_{j2}(\tau ,s)b_j(\tau )\mathrm{d}{A}_{j1}(\tau ),$$

with

$$H_{j0}(t,s)={\displaystyle \frac{1}{\Gamma ({\eta }_{j1})}}\left\{\begin{array}{c}\begin{array}{cc}{t}^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}-{(t-s)}^{{\eta }_{j1}-1},\hfill & 0\le s\le t\le 1;\hfill \\ t^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1},\hfill & 0\le t\le s\le 1.\hfill \end{array}\hfill \end{array}\right.$$

The size and form of $H_{j1}(t,s),H_{j2}(t,s),$ have been adjusted. The formulas below that exceed the template have also been modified accordingly. This sentence needs to be deleted.

$$H_{j1}(t,s)=\left\{\begin{array}{c}\begin{array}{cc}{\displaystyle \frac{{\int }_0^1{a}_j(\tau )\mathrm{d}\tau t^{{\eta }_{j1}-{\gamma }_{j2}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}-{(t-s)}^{{\eta }_{j1}-{\gamma }_{j2}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}},\hfill & 0\le s\le t\le 1;\hfill \\ {\displaystyle \frac{{\int }_0^1{a}_j(\tau )\mathrm{d}\tau t^{{\eta }_{j1}-{\gamma }_{j2}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}},\hfill & 0\le t\le s\le 1.\hfill \end{array}\hfill \end{array}\right.$$

$$H_{j2}(t,s)=\left\{\begin{array}{c}\begin{array}{cc}{\displaystyle \frac{t^{{\eta }_{j1}-{\gamma }_{j3}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}-{(t-s)}^{{\eta }_{j1}-{\gamma }_{j3}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}},\hfill & 0\le s\le t\le 1;\hfill \\ {\displaystyle \frac{t^{{\eta }_{j1}-{\gamma }_{j3}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}},\hfill & 0\le t\le s\le 1.\hfill \end{array}\hfill \end{array}\right.$$

$$M_j={\displaystyle \frac{1}{\Gamma ({\eta }_{j1}-{\gamma }_{j1})}}-{\displaystyle \frac{{\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{\int }_0^1{a}_j(s)\mathrm{d}s-{\int }_0^1{\displaystyle \frac{s^{{\eta }_{j1}-{\gamma }_{j3}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{b}_j(s)\mathrm{d}{A}_{j1}(s).$$

Proof. 

From the first formula of (3) and Lemma 1, we can obtain

$$\begin{array}{cc}\hfill \delta (t)& =-I_{0^{+}}^{{\eta }_{j1}}\sigma (t)+c_1{t}^{{\eta }_{j1}-1}+c_2{t}^{{\eta }_{j1}-2}+···+c_{n_j}{t}^{{\eta }_{j1}-n_j}.\hfill \end{array}$$

From $\delta (0)=0$, we obtain $c_{n_j}=0$. It follows from $D_{0^{+}}^{{\beta }_{j1}}{t}^{{\eta }_{j1}}={\displaystyle \frac{\Gamma ({\eta }_{j1}+1)}{\Gamma ({\eta }_{j1}+1-{\beta }_{j1})}}{t}^{{\eta }_{j1}-{\beta }_{j1}}$ that

$$\begin{array}{cc}\hfill D_{0^{+}}^{{\beta }_{j1}}\delta (t)=& -I_{0^{+}}^{{\eta }_{j1}-{\beta }_{j1}}\sigma (t)+c_1{\displaystyle \frac{\Gamma ({\eta }_{j1})t^{{\eta }_{j1}-{\beta }_{j1}-1}}{\Gamma ({\eta }_{j1}-{\beta }_{j1})}}+c_2{\displaystyle \frac{\Gamma ({\eta }_{j1}-1)t^{{\eta }_{j1}-{\beta }_{j1}-2}}{\Gamma ({\eta }_{j1}-1-{\beta }_{j1})}}+···\hfill \\ \hfill & +c_{n_j-2}{\displaystyle \frac{\Gamma ({\eta }_{j1}-n_j+3)t^{{\eta }_{j1}-n_j+2-{\beta }_{j1}}}{\Gamma ({\eta }_{j1}-n_j+3-{\beta }_{j1})}}+c_{n_j-1}{\displaystyle \frac{\Gamma ({\eta }_{j1}-n_j+2)t^{{\eta }_{j1}-n_j+1-{\beta }_{j1}}}{\Gamma ({\eta }_{j1}-n_j+2-{\beta }_{j1})}}.\hfill \end{array}$$

Then, by $n_j-2<{\eta }_{j1}-{\beta }_{j1}\le n_j-1$ and $D_{0^{+}}^{{\beta }_{j1}}\delta (0)=0$ we deduce $c_{n_j-1}=0.$ Similarly, from $D_{0^{+}}^{{\beta }_{ji}}\delta (0)=0,n_j-i-1<{\eta }_{j1}-{\beta }_{ji}\le n_j-i,i=2,···n_j-2$ it is easy to verify $c_{n_j-2}=···=c_2=0.$ Thus, one obtains

$$\delta (t)=-I_{0^{+}}^{{\eta }_{j1}}\sigma (t)+c_1{t}^{{\eta }_{j1}-1}=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\sigma (s)\mathrm{d}s+c_1{t}^{{\eta }_{j1}-1}.$$

(4)

Then, from $D_{0^{+}}^{{\gamma }_{j1}}\delta (1)={\int }_0^1{a}_j(s)D_{0^{+}}^{{\gamma }_{j2}}\delta (\xi )\mathrm{d}s+{\int }_0^1{b}_j(s)D_{0^{+}}^{{\gamma }_{j3}}\delta (s)\mathrm{d}{A}_{j1}(s)$ we conclude that

$$\begin{array}{cc}\hfill c_1=& {\displaystyle \frac{1}{M_j\Gamma ({\eta }_{j1})}}[{\int }_0^1{\displaystyle \frac{{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j1})}}\sigma (s)\mathrm{d}s-{\int }_0^1{\int }_0^{\xi }{\displaystyle \frac{{(\xi -\tau )}^{{\eta }_{j1}-{\gamma }_{j2}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{a}_j(s)\sigma (\tau )\mathrm{d}\tau \mathrm{d}s\hfill \\ \hfill & -{\int }_0^1{\int }_0^s{\displaystyle \frac{{(s-\tau )}^{{\eta }_{j1}-{\gamma }_{j3}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{b}_j(s)\sigma (\tau )\mathrm{d}\tau \mathrm{d}{A}_{j1}(s)].\hfill \end{array}$$

(5)

Substituting (5) in (4), one observes

$$\begin{array}{cc}\hfill \delta (t)=& -{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\sigma (s)\mathrm{d}s+{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})}}{\int }_0^1{\displaystyle \frac{{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j1})}}\sigma (s)\mathrm{d}s\hfill \\ \hfill & -{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})}}{\int }_0^1{\int }_0^{\xi }{\displaystyle \frac{{(\xi -\tau )}^{{\eta }_{j1}-{\gamma }_{j2}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{a}_j(s)\sigma (\tau )\mathrm{d}\tau \mathrm{d}s\hfill \\ \hfill & -{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})}}{\int }_0^1{\int }_0^s{\displaystyle \frac{{(s-\tau )}^{{\eta }_{j1}-{\gamma }_{j3}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{b}_j(s)\sigma (\tau )\mathrm{d}\tau \mathrm{d}{A}_{j1}(s).\hfill \end{array}$$

(6)

From the definitions of $M_j$, the following holds:

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j1})}}=& {\displaystyle \frac{1}{\Gamma ({\eta }_{j1})}}+{\int }_0^1{\displaystyle \frac{{\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{a}_j(s)\mathrm{d}s\hfill \\ \hfill & +{\int }_0^1{\displaystyle \frac{s^{{\eta }_{j1}-{\gamma }_{j3}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{b}_j(s)\mathrm{d}{A}_{j1}(s).\hfill \end{array}$$

(7)

Plugging (7) into (6), one obtains

$$\begin{array}{cc}\hfill \delta (t)=& -{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\sigma (s)\mathrm{d}s+[{\displaystyle \frac{1}{\Gamma ({\eta }_{j1})}}+{\int }_0^1{\displaystyle \frac{{\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{a}_j(s)\mathrm{d}s\hfill \\ \hfill & +{\int }_0^1{\displaystyle \frac{s^{{\eta }_{j1}-{\gamma }_{j3}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{b}_j(s)\mathrm{d}{A}_{j1}(s)]t^{{\eta }_{j1}-1}{\int }_0^1{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}\sigma (s)\mathrm{d}s\hfill \\ \hfill & -{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})}}{\int }_0^1{\int }_0^{\xi }{\displaystyle \frac{{(\xi -\tau )}^{{\eta }_{j1}-{\gamma }_{j2}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{a}_j(s)\sigma (\tau )\mathrm{d}\tau \mathrm{d}s\hfill \\ \hfill & -{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})}}{\int }_0^1{\int }_0^s{\displaystyle \frac{{(s-\tau )}^{{\eta }_{j1}-{\gamma }_{j3}-1}}{\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{b}_j(s)\sigma (\tau )\mathrm{d}\tau \mathrm{d}{A}_{j1}(s)\hfill \\ \hfill =& -{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\sigma (s)\mathrm{d}s+{\int }_0^1{\displaystyle \frac{t^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\sigma (s)\mathrm{d}s\hfill \\ \hfill & +{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{\int }_0^1{a}_j(\tau )\mathrm{d}\tau {\int }_0^1{\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}\sigma (s)\mathrm{d}s\hfill \\ \hfill & -{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{\int }_0^1{a}_j(\tau )\mathrm{d}\tau {\int }_0^{\xi }{(\xi -s)}^{{\eta }_{j1}-{\gamma }_{j2}-1}\sigma (s)\mathrm{d}s\hfill \\ \hfill & +{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{\int }_0^1{s}^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(s)\mathrm{d}{A}_{j1}(s){\int }_0^1{(1-\tau )}^{{\eta }_{j1}-{\gamma }_{j1}-1}\sigma (\tau )\mathrm{d}\tau \hfill \\ \hfill & -{\displaystyle \frac{t^{{\eta }_{j1}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{\int }_0^1{\int }_0^s{(s-\tau )}^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(s)\sigma (\tau )\mathrm{d}\tau \mathrm{d}{A}_{j1}(s)\hfill \\ \hfill =& {\int }_0^1{H}_{j0}(t,s)\sigma (s)\mathrm{d}s+{\int }_0^1{t}^{{\eta }_{j1}-1}{H}_{j1}(\xi ,s)\sigma (s)\mathrm{d}s\hfill \\ \hfill & +{\int }_0^1{\int }_0^1{t}^{{\eta }_{j1}-1}{H}_{j2}(\tau ,s)b_j(\tau )\mathrm{d}{A}_{j1}(\tau )\sigma (s)\mathrm{d}s\hfill \\ \hfill =& {\int }_0^1{H}_{j3}(t,s)\sigma (s)\mathrm{d}s.\hfill \end{array}$$

□

Lemma 3.

If $M_j>0,j=1,2$ then

(i) for any $t,s\in [0,1],H_{j0}(t,s),H_{j1}(t,s),H_{j2}(t,s),H_{j3}(t,s)$ are continuous.

(ii) for any $t,s\in [0,1],$ we conclude

$$0\le a_{j3}(s)t^{{\eta }_{j1}-1}\le H_{j3}(t,s)\le b_{j3}{t}^{{\eta }_{j1}-1},$$

where $a_{j3}(s)=a_{j0}(s)+a_{j1}(s){\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}+{\int }_0^1{a}_{j2}(s){\tau }^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(\tau )\mathrm{d}{A}_{j1}(\tau ),b_{j3}={\displaystyle \frac{1}{\Gamma ({\eta }_{j1})}}+b_{j1}{\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}+{\int }_0^1{b}_{j2}{\tau }^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(\tau )\mathrm{d}{A}_{j1}(\tau ).$

Proof. 

From the definition of $H_{ji}(t,s)(i=0,1,2,3,j=1,2)$, we can know that (i) holds. Then, as $0\le s\le t\le 1$ we observe that

$$H_{j0}(t,s)={\displaystyle \frac{t^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}-{(t-s)}^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\le {\displaystyle \frac{t^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}},$$

and

$$\begin{array}{cc}\hfill H_{j0}(t,s)& ={\displaystyle \frac{t^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}-t^{{\eta }_{j1}-1}{(1-\frac{s}{t})}^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\hfill \\ \hfill & \ge {\displaystyle \frac{t^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}-t^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}\hfill \\ \hfill & ={\displaystyle \frac{t^{{\eta }_{j1}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}\left(1-{(1-s)}^{{\gamma }_{j1}}\right)}{\Gamma ({\eta }_{j1})}}=a_{j0}(s)t^{{\eta }_{j1}-1}\ge 0.\hfill \end{array}$$

When $0\le t\le s\le 1$, $0\le a_{j0}(s)t^{{\eta }_{j1}-1}\le H_{j0}(t,s)\le {\displaystyle \frac{t^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}$ clearly holds. That is, for any $t,s\in [0,1]\le 1,$ it gives

$$0\le a_{j0}(s)t^{{\eta }_{j1}-1}\le H_{j0}(t,s)\le {\displaystyle \frac{t^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}.$$

Due to $-{(t-s)}^{{\eta }_{j1}-{\gamma }_{j2}-1}>-t^{{\eta }_{j1}-{\gamma }_{j2}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j2}-1}$, $-{(t-s)}^{{\eta }_{j1}-{\gamma }_{j3}-1}>-t^{{\eta }_{j1}-{\gamma }_{j3}-1}{(1-s)}^{{\eta }_{j1}-{\gamma }_{j3}-1}$, similarly to the proof process, the following inequalities hold:

$$\begin{array}{cc}\hfill 0& \le {\displaystyle \frac{{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}[1-{(1-s)}^{{\gamma }_{j1}-{\gamma }_{j2}}]{\int }_0^1{a}_j(\tau )\mathrm{d}\tau }{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}}{t}^{{\eta }_{j1}-{\gamma }_{j2}-1}\hfill \\ \hfill & \le H_{j1}(t,s)\le {\displaystyle \frac{{\int }_0^1{a}_j(\tau )\mathrm{d}\tau ·t^{{\eta }_{j1}-{\gamma }_{j2}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j2})}},\hfill \\ \hfill 0& \le {\displaystyle \frac{{(1-s)}^{{\eta }_{j1}-{\gamma }_{j1}-1}[1-{(1-s)}^{{\gamma }_{j1}-{\gamma }_{j3}}]}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}{t}^{{\eta }_{j1}-{\gamma }_{j3}-1}\hfill \\ \hfill & \le H_{j2}(t,s)\le {\displaystyle \frac{t^{{\eta }_{j1}-{\gamma }_{j3}-1}}{M_j\Gamma ({\eta }_{j1})\Gamma ({\eta }_{j1}-{\gamma }_{j3})}}.\hfill \end{array}$$

Hence, we conclude that

$$\begin{array}{cc}\hfill H_{j3}(t,s)=& H_{j0}(t,s)+t^{{\eta }_{j1}-1}{H}_{j1}(\xi ,s)+t^{{\eta }_{j1}-1}{\int }_0^1{H}_{j2}(\tau ,s)b_j(\tau )\mathrm{d}{A}_{j1}(\tau )\hfill \\ \hfill \le & {\displaystyle \frac{t^{{\eta }_{j1}-1}}{\Gamma ({\eta }_{j1})}}+t^{{\eta }_{j1}-1}{b}_{j1}{\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}+t^{{\eta }_{j1}-1}{\int }_0^1{b}_{j2}{\tau }^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(\tau )\mathrm{d}{A}_{j1}(\tau )\hfill \\ \hfill =& [{\displaystyle \frac{1}{\Gamma ({\eta }_{j1})}}+b_{j1}{\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}+{\int }_0^1{b}_{j2}{\tau }^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(\tau )\mathrm{d}{A}_{j1}(\tau )]t^{{\eta }_{j1}-1}\hfill \end{array}$$

$$\begin{array}{cc}\hfill H_{j3}(t,s)\ge & a_{j0}(s)t^{{\eta }_{j1}-1}+t^{{\eta }_{j1}-1}{a}_{j1}(s){\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}+t^{{\eta }_{j1}-1}{\int }_0^1{a}_{j2}(s){\tau }^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(\tau )\mathrm{d}{A}_{j1}(\tau )\hfill \\ \hfill =& [a_{j0}(s)+a_{j1}(s){\xi }^{{\eta }_{j1}-{\gamma }_{j2}-1}+{\int }_0^1{a}_{j2}(s){\tau }^{{\eta }_{j1}-{\gamma }_{j3}-1}{b}_j(\tau )\mathrm{d}{A}_{j1}(\tau )]t^{{\eta }_{j1}-1}\ge 0,\hfill \end{array}$$

i.e., (ii) holds.  □

Lemma 4.

Let $\sigma \in C[0,1],j=1,2$; then,

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_{j2}}\delta (t)+\sigma (t)=0,m_j-1<{\eta }_{j2}\le m_j,\hfill \\ \delta (0)=0,D_{0^{+}}^{{\alpha }_{ji}}\delta (0)=0,i-1<{\eta }_{j2}-{\alpha }_{ji}\le i,i=2,···,m_j-1,\hfill \\ D_{0^{+}}^{{\beta }_j}\delta (1)={\tau }_j{\int }_0^1{k}_j(s)\delta (s)\mathrm{d}{A}_{j2}(s),\hfill \end{array}\right.$$

(8)

has a unique solution:

$${\displaystyle \delta (t)={\int }_0^1{H}_{j5}(t,s)\sigma (s)\mathrm{d}s,}$$

where

$$H_{j5}(t,s)=H_{j4}(t,s)+{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j){\psi }_j(s)t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})(1-Q_j)}},$$

with

$$H_{j4}(t,s)=\left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{t^{{\eta }_{j2}-1}{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}-{(t-s)}^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})}},0\le s\le t\le 1;\hfill \\ {\displaystyle \frac{t^{{\eta }_{j2}-1}{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{\Gamma ({\eta }_{j2})}},0\le t\le s\le 1.\hfill \end{array}\hfill \end{array}\right.$$

$$Q_j={\int }_0^1{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)}{\Gamma ({\eta }_{j2})}}{s}^{{\eta }_{j2}-1}{k}_j(s)\mathrm{d}{A}_{j2}(s),{\psi }_j(s)={\int }_0^1{H}_{j4}(\tau ,s)k_j(\tau )\mathrm{d}{A}_{j2}(\tau ).$$

Proof. 

From Lemma 4 and the first formula of (8), one observes that

$$\begin{array}{cc}\hfill \delta (t)& =-I_{0^{+}}^{{\eta }_{j2}}\sigma (t)+c_1{t}^{{\eta }_{j2}-1}+c_2{t}^{{\eta }_{j2}-2}+···+c_{m_j}{t}^{{\eta }_{j2}-m_j}.\hfill \end{array}$$

It follows from $\delta (0)=0$ that $c_{m_j}=0$. By $D_{0^{+}}^{{\alpha }_{j(m_j-1)}}\delta (0)=0$, we have

$$\begin{array}{cc}\hfill D_{0^{+}}^{{\alpha }_{j(m_j-1)}}\delta (t)=& -I_{0^{+}}^{{\eta }_{j2}-{\alpha }_{j(m_j-1)}}\sigma (t)+c_1{\displaystyle \frac{\Gamma ({\eta }_{j2})t^{{\eta }_{j2}-{\alpha }_{j(m_j-1)}-1}}{\Gamma ({\eta }_{j2}-{\alpha }_{j(m_j-1)})}}\hfill \\ \hfill & +c_2{\displaystyle \frac{\Gamma ({\eta }_{j2}-1)t^{{\eta }_{j2}-{\alpha }_{j(m_j-1)}-2}}{\Gamma ({\eta }_{j2}-1-{\alpha }_{j(m_j-1)})}}\hfill \\ \hfill & +···+c_{m_j-2}{\displaystyle \frac{\Gamma ({\eta }_{j2}-m_j+3)t^{{\eta }_{j2}-m_j+2-{\alpha }_{j(m_j-1)}}}{\Gamma ({\eta }_{j2}-m_j+3-{\alpha }_{j(m_j-1)})}}\hfill \\ \hfill & +c_{m_j-1}{\displaystyle \frac{\Gamma ({\eta }_{j2}-m_j+2)t^{{\eta }_{j2}-m_j+1-{\alpha }_{j(m_j-1)}}}{\Gamma ({\eta }_{j2}-m_j+2-{\alpha }_{j(m_j-1)})}}.\hfill \end{array}$$

Then, from $m_j-2<{\eta }_{j2}-{\alpha }_{j(m_j-1)}\le m_j-1$ one obtains $c_{m_j-1}=0.$ Then, from $D_{0^{+}}^{{\alpha }_{ji}}\delta (0)=0,i-1<{\eta }_{j2}-{\alpha }_{ji}\le i,i=2,···,m_j-2$ we conclude that

$$c_{m_j-2}=···=c_2=0.$$

Thus, one obtains

$$\delta (t)=-I_{0^{+}}^{{\eta }_{j2}}\sigma (t)+c_1{t}^{{\eta }_{j2}-1}=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})}}\sigma (s)\mathrm{d}s+c_1{t}^{{\eta }_{j2}-1}.$$

(9)

It follows from $D_{0^{+}}^{{\beta }_j}{t}^{{\eta }_{j2}-1}={\displaystyle \frac{\Gamma ({\eta }_{j2})}{\Gamma ({\eta }_{j2}-{\beta }_j)}}{t}^{{\eta }_{j2}-{\beta }_j-1}$ and $D_{0^{+}}^{{\beta }_j}\delta (1)={\tau }_j{\int }_0^1{k}_j(s)\delta (s)\mathrm{d}{A}_{j2}(s)$ that

$$c_1={\int }_0^1{\displaystyle \frac{{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{\Gamma ({\eta }_{j2})}}\sigma (s)\mathrm{d}s+{\int }_0^1{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)}{\Gamma ({\eta }_{j2})}}{k}_j(s)\delta (s)\mathrm{d}{A}_{j2}(s).$$

(10)

Substituting (10) in (9), we conclude that

$$\begin{array}{cc}\hfill \delta (t)=& -{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})}}\sigma (s)\mathrm{d}s+{\int }_0^1{\displaystyle \frac{t^{{\eta }_{j2}-1}{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{\Gamma ({\eta }_{j2})}}\sigma (s)\mathrm{d}s\hfill \\ \hfill & +{\int }_0^1{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})}}{k}_j(s)\delta (s)\mathrm{d}{A}_{j2}(s)\hfill \\ \hfill =& {\int }_0^1{H}_{j4}(t,s)\sigma (s)\mathrm{d}s+{\int }_0^1{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})}}{k}_j(s)\delta (s)\mathrm{d}{A}_{j2}(s),\hfill \end{array}$$

(11)

Integrating (11), we have

$${\int }_0^1{k}_j(s)\delta (s)\mathrm{d}{A}_{j2}(s)={\displaystyle \frac{1}{1-Q_j}}{\int }_0^1\left[{\int }_0^1{H}_{j4}(\tau ,s)k_j(\tau )\mathrm{d}{A}_{j2}(\tau )\right]\sigma (s)\mathrm{d}s.$$

(12)

Combining (11) and (12), we obtain

$$\begin{array}{cc}\hfill \delta (t)& ={\int }_0^1{H}_{j4}(t,s)\sigma (s)\mathrm{d}s+{\int }_0^1\left[{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})(1-Q_j)}}{\int }_0^1{H}_{j4}(\tau ,s)k_j(\tau )\mathrm{d}{A}_{j2}(\tau )\right]\sigma (s)\mathrm{d}s\hfill \\ \hfill & ={\int }_0^1{H}_{j5}(t,s)\sigma (s)\mathrm{d}s.\hfill \end{array}$$

□

Lemma 5.

If $0\le Q_j<1,j=1,2$ then

(i) $H_{j4}(t,s),H_{j5}(t,s)$ are continuous for any $t,s\in [0,1]$;

(ii) for any $t,s\in [0,1],$ we deduce

$$0\le a_{j5}(s)t^{{\eta }_{j2}-1}\le H_{j5}(t,s)\le b_{j5}{t}^{{\eta }_{j2}-1},$$

where $a_{j5}(s)={\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j){\psi }_j(s)}{\Gamma ({\eta }_{j2})(1-Q_j)}},b_{j5}={\displaystyle \frac{1}{\Gamma ({\eta }_{j2})(1-Q_j)}}.$

Proof. 

From $H_{j4}(t,s)$ and $H_{j5}(t,s)$, we can ascertain that (i) is established. Then, using method similar to Lemma 4, we easily observe that

$$0\le H_{j4}(t,s)\le {\displaystyle \frac{t^{{\eta }_{j2}-1}{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{\Gamma ({\eta }_{j2})}},\forall t,s\in [0,1].$$

From ${\psi }_j(s)={\int }_0^1{H}_{j4}(\tau ,s)k_j(\tau )\mathrm{d}{A}_{j2}(\tau )$ and $Q_j={\int }_0^1{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)}{\Gamma ({\eta }_{j2})}}{s}^{{\eta }_{j2}-1}{k}_j(s)\mathrm{d}{A}_{j2}(s),$ one observes that

$$0\le {\psi }_j(s)\le {\int }_0^1{\displaystyle \frac{t^{{\eta }_{j2}-1}{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{\Gamma ({\eta }_{j2})}}{k}_j(t)\mathrm{d}{A}_{j2}(t)={\displaystyle \frac{Q_j{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)}}.$$

Then, for any $t,s\in [0,1]$ we obtain

$$\begin{array}{cc}\hfill H_{j5}(t,s)& =H_{j4}(t,s)+{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j){\psi }_j(s)t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})(1-Q_j)}}\ge {\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j){\psi }_j(s)t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})(1-Q_j)}}\ge 0,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill H_{j5}(t,s)& \le {\displaystyle \frac{t^{{\eta }_{j2}-1}{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{\Gamma ({\eta }_{j2})}}+{\displaystyle \frac{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})(1-Q_j)}}{\displaystyle \frac{Q_j{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{{\tau }_j\Gamma ({\eta }_{j2}-{\beta }_j)}}\hfill \\ \hfill & \le {\displaystyle \frac{t^{{\eta }_{j2}-1}{(1-s)}^{{\eta }_{j2}-{\beta }_j-1}}{\Gamma ({\eta }_{j2})(1-Q_j)}}\le {\displaystyle \frac{t^{{\eta }_{j2}-1}}{\Gamma ({\eta }_{j2})(1-Q_j)}},\hfill \end{array}$$

i.e., (ii) holds.  □

Assuming that E is a real Banach space, $\theta$ is a zero element of E, and $P\subset E$ is a non-empty closed convex set, if P is a cone then (1) $z\in P,\mu \ge 0⟹\mu z\in P;(2)z\in P,-z\in P\Rightarrow z=\theta$. P is normal if $\forall x_1,y_1\in E,$ $\theta \le x_1\le y_1,$ and if there exists $L>0$, such that $\parallel x_1\parallel \le L\parallel y_1\parallel$, where L is a normality constant. Let $h>\theta$, $P_h=\{z\in E\mid {\lambda }_{1}h\le z\le {\lambda }_{2}h,{\lambda }_1,{\lambda }_2>0\},inf(P)=\{z\in P:z\mathrm{is}\mathrm{an}\mathrm{interior}\mathrm{point}\mathrm{of}P\}$.

For $(z_1,z_2)\in E\times E$, the norm is $||(z_1,z_2)||=max\{||z_1||,||z_2||\}$ or $||(z_1,z_2)||=||z_1||+||z_2||$. According to [5], $(E\times E,||(z_1,z_2)||)$ is a Banach space. There is a partial order in $E\times E$:

$$({\tau }_1,{\upsilon }_1)\le ({\tau }_2,{\upsilon }_2)\iff {\tau }_1\le {\tau }_2,{\upsilon }_1\le {\upsilon }_2.$$

Let $\tilde{P}=P\times P=\{(z_1,z_2)\in E\times E|z_1\ge 0,z_2\ge 0\};$ then, the cone $\tilde{P}\subset E\times E$, and $\tilde{P}$ is normal. Given $h=(h_1,h_2)\in \tilde{P},$ we define

$$\widetilde{P_h}=P_{h_1}\times P_{h_2}=\{(x_1,x_2)|x_1\in P_{h_1},x_2\in P_{h_2}\};$$

then, $\widetilde{P_h}\subset \tilde{P}$.

Lemma 6

([5]). Let P be a normal cone in Banach space E, $h=(h_1,h_2)\in \tilde{P},h_1,h_2\ne \theta$. $T_1,T_2:\tilde{P}\to P$ are increasing operators. We assume that

($A_1^{\prime }$) there exists $h_0=(h_0^{(1)},h_0^{(2)})\in {\tilde{P}}_h$, such that $T_1{h}_0\in P_{h_1},T_2{h}_0\in P_{h_2};$

($A_2^{\prime }$) for any $s\in (0,1),y,z\in P,$ there exist ${\phi }_1,{\phi }_2:(0,1)\to (0,1)$, such that

$$T_1(sy,sz)\ge {\phi }_1(s)T_1(y,z),T_2(sy,sz)\ge {\phi }_2(s)T_2(y,z),$$

where ${\phi }_1(s),{\phi }_2(s)>s.$

Then,

$(L_1^{\prime })$ $\left(T_1(y,z),T_2(y,z)\right)=(y,z)$ has a unique solution $(y^{\ast },z^{\ast })\in {\tilde{P}}_h$;

$(L_2^{\prime })$ for any initial values $y_0,z_0\in {\tilde{P}}_h$, we construct the iterative sequences as follows:

$$(y_n,z_n)=\left(T_1(y_{n-1},z_{n-1}),T_2(y_{n-1},z_{n-1})\right),n=1,2,···,$$

where $y_n\to y^{\ast },z_n\to z^{\ast },$ as $n\to \infty$.

3. Operators Fixed Point Theorems on Ordered Product Spaces

In this section, some conclusions of the operators $(T_3,T_4)$ and $(A_1+B_1,A_2+B_2)$ in ordered product space are given.

Theorem 1.

Let $h=(h_1,h_2)\in P\times P,h_1,h_2\ne \theta$. $T_3,T_4:\tilde{P}\to P$ are decreasing operators. Assume that

($l_1$) there exists $h_0=(h_0^{(1)},h_0^{(2)})\in {\tilde{P}}_h$, such that $T_3(h_0^{(1)},h_0^{(2)})\in P_{h_1},T_4(h_0^{(1)},h_0^{(2)})\in P_{h_2};$

($l_2$) for any $s\in (0,1),y,z\in P$, there exist two increasing functions ${\varphi }_1(s),{\varphi }_2(s)\in (s,1]$, such that

$$T_3(sy,sz)\le {\varphi }_1^{-1}(s)T_3(y,z),T_4(sy,sz)\le {\varphi }_2^{-1}(s)T_4(y,z);$$

Then, $\left(T_3(y,z),T_4(y,z)\right)=(y,z)$ has a unique solution, $(y^{\ast },z^{\ast })\in {\tilde{P}}_h$.

Proof. 

According to Lemma 6, we define an operator $\tilde{A}:\tilde{P}\to \tilde{P}$

$$\tilde{A}(y,z)=(T_1(y,z),T_2(y,z)),$$

where $T_1,T_2$ satisfies the condition of Lemma 6. We can easily ascertain that $\tilde{A}:\tilde{P}\to \tilde{P}$ is an increase from the definition of $T_1,T_2$. Using the condition ($H_2^{\prime }$) of Lemma 6, there exists ${\phi }_i(s)>s$, such that

$$\begin{array}{cc}\hfill \tilde{A}(sy,sz)& =(T_1(sy,sz),T_2(sy,sz))\ge ({\phi }_1(s)T_1(y,z),{\phi }_2(s)T_2(y,z))\hfill \\ \hfill & \ge (\phi (s)T_1(y,z),\phi (s)T_2(y,z))\ge \phi (s)\tilde{A}(y,z),\hfill \end{array}$$

(13)

where $\phi (s)=min\{{\phi }_1(s),{\phi }_2(s)\},\phi (s)>s.$ According to ($H_1^{\prime }$), for any $(h_0^{(1)},h_0^{(2)})\in {\tilde{P}}_h$ one obtains

$$\tilde{A}(h_0^{(1)},h_0^{(2)})=(T_1(h_0^{(1)},h_0^{(2)}),T_2(h_0^{(1)},h_0^{(2)}))\subset P_{h_1}\times P_{h_2}=\widetilde{P_h},i.e.,\tilde{A}:{\tilde{P}}_h\to {\tilde{P}}_h.$$

(14)

Define an operator $B:\tilde{P}\to \tilde{P}$ as

$$B(y,z)=(T_3(y,z),T_4(y,z)),\forall y,z\in P.$$

In the following context, by the conditions ($l_1$)–($l_2$), we prove that the operator $B^2$ is an increasing operator and satisfies (13) and (14).

Step 1: Since $T_3,T_4$ are decreasing operators, for any ${\tau }_1,{\upsilon }_1,{\tau }_2,{\upsilon }_2\in P$, let ${\tau }_1\le {\tau }_2,$ ${\upsilon }_1\le {\upsilon }_2;$ we conclude that

$$B({\tau }_1,{\upsilon }_1)=(T_3({\tau }_1,{\upsilon }_1),T_4({\tau }_1,{\upsilon }_1))\ge (T_3({\tau }_2,{\upsilon }_2),T_4({\tau }_2,{\upsilon }_2))=B({\tau }_2,{\upsilon }_2),$$

i.e., B is a decreasing operator. Then, one observes that

$$B^2({\tau }_1,{\upsilon }_1)=B(T_3({\tau }_1,{\upsilon }_1),T_4({\tau }_1,{\upsilon }_1))\le B(T_3({\tau }_2,{\upsilon }_2),T_4({\tau }_2,{\upsilon }_2))=B^2({\tau }_2,{\upsilon }_2);$$

hence, $B^2$ is an increasing operator.

Step 2: From $T_3(h_0^{(1)},h_0^{(2)})\in P_{h_1}$ in the condition ($l_1$) there exists $a_1\in (0,1)$, such that

$$\begin{array}{c}\hfill a_1{h}_1\le T_3(h_0^{(1)},h_0^{(2)})\le {\displaystyle \frac{1}{a_1}}{h}_1,\end{array}$$

(15)

For any $y\in P_{h_1},z\in P_{h_2},$ due to $(h_0^{(1)},h_0^{(2)})\in P_{h_1}\times P_{h_2}$ in the condition ($l_1$), there exist $m_1,m_2>0,$ such that

$$\begin{array}{c}\hfill m_1{h}_0^{(1)}\le y\le m_2{h}_0^{(1)},m_1{h}_0^{(2)}\le z\le m_2{h}_0^{(2)}.\end{array}$$

(16)

Then, according to the condition ($l_2$) and (15) and (16), it holds that

$$\begin{array}{c}\hfill T_3(y,z)\le T_3(m_1{h}_0^{(1)},m_1{h}_0^{(2)})\le {\varphi }_1^{-1}(m_1)T_3(h_0^{(1)},h_0^{(2)})\le {\varphi }_1^{-1}(m_1){\displaystyle \frac{1}{a_1}}{h}_1.\end{array}$$

(17)

Using ($l_2$) again, we obtain

$$T_3(s^{-1}y,s^{-1}z)\ge {\varphi }_1(s)T_3(y,z).$$

Then, one can see that

$$\begin{array}{c}\hfill T_3(y,z)\ge T_3(m_2{h}_0^{(1)},m_2{h}_0^{(2)})\ge {\varphi }_1(m_2)T_3(h_0^{(1)},h_0^{(2)})\ge {\varphi }_1(m_2)a_1{h}_1.\end{array}$$

(18)

In view of (17) and (18), we conclude that $T_3:\widetilde{P_h}\to P_{h_1}.$ In the same way, it is easy to obtain $T_4:\widetilde{P_h}\to P_{h_2}.$ Then, from the definition of B, one observes $B^2:\widetilde{P_h}\to \widetilde{P_h}.$

Step 3: For every $s\in (0,1)$, $x,y\in P$, from the condition ($l_2$) we deduce that

$$\begin{array}{cc}\hfill B(sy,sz)& =(T_3(sy,sz),T_4(sy,sz))\hfill \\ \hfill & \le ({\varphi }_1^{-1}(s)T_3(y,z),{\varphi }_2^{-1}(s)T_4(y,z))\hfill \\ \hfill & \le {\varphi }^{-1}(s)B(y,z),\hfill \end{array}$$

(19)

where $\varphi (s)=min\{{\varphi }_1(s),{\varphi }_2(s)\},$ we can know that $\varphi (s)\in (s,1],$ and $\varphi$ is increasing about s. Then, by (19) it holds that

$$\begin{array}{cc}\hfill B(y,z)=B(s{s}^{-1}y,s{s}^{-1}z)\le {\varphi }^{-1}(s)B(s^{-1}y,s^{-1}z),\mathrm{i}.\mathrm{e}.,B(s^{-1}y,s^{-1}z)& \ge \varphi (s)B(y,z).\hfill \end{array}$$

(20)

Hence, according to the monotonicity of B and (19) and (20) one can see that

$$\begin{array}{cc}\hfill B^2(sy,sz)& =B(T_3(sy,sz),T_4(sy,sz))\hfill \\ \hfill & \ge B({\varphi }^{-1}(s)T_3(y,z),{\varphi }^{-1}(s)T_4(y,z))\hfill \\ \hfill & \ge \varphi (\varphi (s))B(T_3(y,z),T_4(y,z))\ge \varphi (s)B^2(y,z).\hfill \end{array}$$

So far, it is proved that the increasing operator $B^2$ satisfies (13) and (14). Thus, by Lemma 6, $B^2$ has a unique solution, $(x^{\ast },y^{\ast })\in {\tilde{P}}_h$, and there exist $y_0,z_0\in {\tilde{P}}_h$ and the iterative sequence

$$(y_n,z_n)=B^2(y_{n-1},z_{n-1}),n=1,2,···,$$

When $n\to \infty$, we have $y_n\to x^{\ast },z_n\to z^{\ast }$. Then, we can obtain

$$B^2(B(y^{\ast },z^{\ast }))=B(B^2(y^{\ast },z^{\ast }))=B(y^{\ast },z^{\ast }).$$

From the fact that $B^2$ has a unique fixed point, $B(y^{\ast },z^{\ast })=(y^{\ast },z^{\ast })$ holds. That is, B has a fixed point $(y^{\ast },z^{\ast })$.

Next, we expound that B has a unique fixed point. Let $(\tilde{y},\tilde{z})$ be another fixed point of B. Then,

$$B^2(\tilde{y},\tilde{z})=B(B(\tilde{y},\tilde{z}))=B(\tilde{y},\tilde{z})=(\tilde{y},\tilde{z}),$$

is in contradiction with the unique fixed point of $B^2$; thus, B has a unique fixed point $(y^{\ast },z^{\ast })$.  □

Theorem 2.

Assume that ($l_1$)–($l_2$) hold; then, for any given ${\kappa }_1,{\kappa }_2>0$ the operator equation $(y,z)=({\kappa }_1{T}_3(y,z),{\kappa }_2{T}_4(y,z))$ has a unique solution $(y_{{\kappa }_1,{\kappa }_2}^{\ast },z_{{\kappa }_1,{\kappa }_2}^{\ast })\in {\tilde{P}}_h$.

Proof. 

Let $T_{{\kappa }_1}={\kappa }_1{T}_3,T_{{\kappa }_2}={\kappa }_2{T}_4$. Since $T_{{\kappa }_1},T_{{\kappa }_2}$ satisfy ($l_1$)–($l_2$), from Theorem 1, the equation $(y,z)=(T_{{\kappa }_1}(y,z),T_{{\kappa }_2}(y,z))$ has a unique solution $(y_{{\kappa }_1,{\kappa }_2}^{\ast },z_{{\kappa }_1,{\kappa }_2}^{\ast })\in {\tilde{P}}_h$; that is, $(y,z)=({\kappa }_1{T}_3(y,z),{\kappa }_2{T}_4(y,z))$ has a unique solution $(y_{{\kappa }_1,{\kappa }_2}^{\ast },z_{{\kappa }_1,{\kappa }_2}^{\ast })\in {\tilde{P}}_h$.  □

Theorem 3.

Let $h=(h_1,h_2)\in P\times P,h_1,h_2\ne \theta$. $A_1,A_2,B_1,B_2:\tilde{P}\to P$ are increasing operators. Assume that

($l_3$) there exists $h_0=(h_0^{(1)},h_0^{(2)})\in {\tilde{P}}_h$, such that $A_1{h}_0,B_2{h}_0\in P_{h_1},A_2{h}_0,B_2{h}_0\in P_{h_2};$

($l_4$) for any $s\in (0,1),y,z\in P$, there exist ${\alpha }_1\in (0,1),\phi (s)\in (s,1]$, such that

$$\begin{array}{cc}\hfill & A_1(sy,sz)\ge s^{{\alpha }_1}{A}_1(y,z),A_2(sy,sz)\ge s{A}_2(y,z),\hfill \\ \hfill & B_1(sy,sz)\ge s{B}_1(y,z),B_2(sy,sz)\ge \phi (s)B_2(y,z);\hfill \end{array}$$

($l_5$) for any $y,z\in P,$ there exist some constants $p_1>0,0<p_2<1$, such that

$$A_1(y,z)\ge p_1{B}_1(y,z),B_2(y,z)\ge {\displaystyle \frac{p_2}{1-p_2}}{A}_2(y,z).$$

Then,

$(L_1)$ $\left(A_1(y,z)+B_2(y,z),A_2(y,z)+B_2(y,z)\right)=(y,z)$ has a unique solution $(y^{\ast },z^{\ast })\in {\tilde{P}}_h$;

$(L_2)$ For any initial value $(y_0,z_0)\in {\tilde{P}}_h$, construction iteration sequences

$$(y_n,z_n)=\left(A_1(y_{n-1},z_{n-1})+B_1(y_{n-1},z_{n-1}),A_2(y_{n-1},z_{n-1})+B_2(y_{n-1},z_{n-1})\right),$$

where $n=1,2,···,$ when $n\to \infty$ $||y_n-y^{\ast }||\to 0,||z_n-z^{\ast }||\to 0$.

Proof. 

Define the operators $T_1=A_1+B_1,T_2=A_2+B_2$ as

$$T_1(y,z)=A_1(y,z)+B_1(y,z),T_2(y,z)=A_2(y,z)+B_2(y,z),\forall y,z\in P.$$

Step 1: Prove that $T_1,T_2:\tilde{P}\to P$ are increasing operators. Take any ${\tau }_1,{\tau }_2,{\upsilon }_1,{\upsilon }_2\in P,$ let ${\tau }_1\le {\tau }_2,$ and ${\upsilon }_1\le {\upsilon }_2.$ Since $A_1,A_2,B_1,B_2:\tilde{P}\to P$ are increasing operators, it holds that

$$\begin{array}{c}\hfill T_1({\tau }_1,{\upsilon }_1)=A_1({\tau }_1,{\upsilon }_1)+B_1({\tau }_1,{\upsilon }_1)\le A_1({\upsilon }_2,{\upsilon }_2)+B_1({\tau }_2,{\upsilon }_2)=T_1({\tau }_2,{\upsilon }_2).\end{array}$$

Similarly, we compute $T_2({\tau }_1,{\upsilon }_1)\le T_2({\tau }_2,{\upsilon }_2).$ That is, $T_1,T_2:\tilde{P}\to P$ are increasing operators.

Step 2: For $h_0=(h_0^{(1)},h_0^{(2)})\in {\tilde{P}}_h$, using $A_1{h}_0,B_1{h}_0\in P_{h_1}$ and $A_2{h}_0,B_2{h}_0\in P_{h_2}$ in condition ($l_3$), $\exists c_i\in (0,1)(i=1,2,3,4)$, such that

$$\begin{array}{cc}\hfill & c_1{h}_1\le A_1{h}_0\le {\displaystyle \frac{1}{c_1}}{h}_1,c_2{h}_1\le B_1{h}_0\le {\displaystyle \frac{1}{c_2}}{h}_1,\hfill \\ \hfill & c_3{h}_2\le A_2{h}_0\le {\displaystyle \frac{1}{c_3}}{h}_2,c_4{h}_2\le B_2{h}_0\le {\displaystyle \frac{1}{c_4}}{h}_2,\hfill \end{array}$$

(21)

it is easy to obtain

$$\begin{array}{cc}\hfill & \left(c_1+c_2\right)h_1\le T_1(h_0^{(1)},h_0^{(2)})\le \left({\displaystyle \frac{1}{c_1}}+{\displaystyle \frac{1}{c_2}}\right)h_1,\hfill \\ \hfill & \left(c_3+c_4\right)h_2\le T_2(h_0^{(1)},h_0^{(2)})\le \left({\displaystyle \frac{1}{c_3}}+{\displaystyle \frac{1}{c_4}}\right)h_2,\hfill \end{array}$$

i.e., $T_1{h}_0\in P_{h_1},T_2{h}_0\in P_{h_2}.$

Step 3: from the condition ($l_5$), one observes

$$\begin{array}{c}\hfill A_1(y,z)\ge {\displaystyle \frac{A_1(y,z)+B_1(y,z)}{1+p_1^{-1}}}={\displaystyle \frac{T_1(y,z)}{1+p_1^{-1}}},(1-p_2)B_2(y,z)\ge p_2{A}_2(y,z).\end{array}$$

(22)

Then, according to condition ($l_4$), one can see that

$$\begin{array}{cc}\hfill T_1(sy,sz)-t{T}_1(y,z)& =A_1(sy,sz)+B_1(sy,sz)-t{A}_1(y,z)-t{B}_1(y,z)\hfill \\ \hfill & \ge (s^{{\alpha }_1}-s)A_1(y,z)\ge {\displaystyle \frac{s^{{\alpha }_1}-s}{1+p_1^{-1}}}{T}_1(y,z),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill T_2(sy,sz)\ge & s{A}_2(y,z)+\phi (s)B_2(y,z)=s{A}_2(y,z)+[p_2\phi (s)+(1-p_2)s]B_2(y,z)\hfill \\ \hfill & +(1-p_2)(\phi (s)-s)B_2(y,z)\ge s{A}_2(y,z)+[p_2\phi (s)+(1-p_2)s]B_2(y,z)\hfill \\ \hfill & +p_2(\phi (s)-s)A_2(y,z)=[p_2\phi (s)+(1-p_2)s](A_2(y,z)+B_2(y,z))\hfill \\ \hfill \ge & [p_2\phi (s)+(1-p_2)s]T_2(y,z).\hfill \end{array}$$

Let ${\phi }_1(s)={\displaystyle \frac{s^{{\alpha }_1}-s}{1+p_1^{-1}}}+s,{\phi }_2(s)=p_2\phi (s)+(1-p_2)s;$ then, there exists $s<{\phi }_1(s),{\phi }_2(s)<1$, such that

$$\begin{array}{cc}\hfill & T_1(sy,sz)\ge {\phi }_1(s)T_1(y,z),T_2(sy,sz)\ge {\phi }_2(s)T_2(y,z).\hfill \end{array}$$

Therefore, according to Lemma 6 the conclusion $(L1)$–$(L2)$ holds.  □

Theorem 4.

Let $h=(h_1,h_2)\in P\times P,h_1,h_2\ne \theta$. $A_1,A_2,B_1,B_2:\tilde{P}\to P$ are decreasing operators. Assume that ($l_3$), ($l_5$) holds and

($l_6$) for any $s\in (0,1),y,z\in P$ there exists ${\alpha }_1\in (0,1)$ and increasing function $\phi (s)\in (s,1]$, such that

$$\begin{array}{cc}\hfill & A_1(sy,sz)\le s^{-{\alpha }_1}{A}_1(y,z),A_2(sy,sz)\le s^{-1}{A}_2(y,z),\hfill \\ \hfill & B_1(sy,sz)\le s^{-1}{B}_1(y,z),B_2(sy,sz)\le {\phi }^{-1}(s)B_2(y,z).\hfill \end{array}$$

Then, $(L1)$ holds.

Proof. 

Define the operators $T_3=A_1+B_1,T_4=A_2+B_2$ by

$$T_3(y,z)=A_1(y,z)+B_1(y,z),T_4(y,z)=A_2(y,z)+B_2(v),\forall v\in P.$$

Step 1: Since $A_1,A_2,B_1,B_2:\tilde{P}\to P$ are decreasing operators, we can easily draw a conclusion that $T_3,T_4:\tilde{P}\to P$ are decreasing operators. By the condition ($l_3$), one observes that (21) holds, and it is easy to obtain $T_3(h_0^{(1)},h_0^{(2)})\in P_{h_1},T_4(h_0^{(1)},h_0^{(2)})\in P_{h_2}.$

Step 2: We verify the condition ($l_2$) in Theorem 1. From the condition ($l_6$), one can see that

$$\begin{array}{cc}\hfill & A_1(y,z)=A_1(s{s}^{-1}x,s{s}^{-1}y)\le s^{-{\alpha }_1}{A}_1(s^{-1}y,s^{-1}z),\hfill \end{array}$$

Hence, it is easy to obtain

$$A_1(s^{-1}y,s^{-1}z)\ge s^{{\alpha }_1}{A}_1(y,z).$$

(23)

Again, applying the same approach, it gives

$$\begin{array}{cc}\hfill A_2(s^{-1}y,s^{-1}z)& \ge s{A}_2(y,z),B_1(s^{-1}y,s^{-1}z)\ge s{B}_1(y,z),\hfill \\ \hfill & B_2(s^{-1}y,s^{-1}z)\ge \phi (s)B_2(y,z).\hfill \end{array}$$

(24)

According to the condition ($l_5$) and the corresponding Theorem 2 proof, (22) holds. Then, by (23) and (24), we deduce

$$\begin{array}{cc}\hfill & T_3(s^{-1}y,s^{-1}z)-s{T}_3(y,z)\ge (s^{{\alpha }_1}-s)A_1(y,z)\ge {\displaystyle \frac{s^{{\alpha }_1}-s}{1+p_1^{-1}}}{T}_3(y,z),\hfill \\ \hfill & T_4(s^{-1}y,s^{-1}z)\ge s{A}_2(y,z)+\phi (s)B_2(y,z)\ge [p_2\phi (s)+(1-p_2)s]T_4(y,z).\hfill \end{array}$$

Let ${\varphi }_1(s)={\displaystyle \frac{s^{{\alpha }_1}-s}{1+p_1^{-1}}}+s,{\varphi }_2(s)=p_2\phi (s)+(1-p_2)s;$ then, ${\varphi }_1(s),{\varphi }_2(s)\in (s,1]$, ${\varphi }_1,{\varphi }_2$ are increasing about s, and

$$\begin{array}{cc}\hfill & T_3(s^{-1}y,s^{-1}z)\ge {\varphi }_1(s)T_3(y,z),T_4(s^{-1}y,s^{-1}z)\ge {\varphi }_2(s)T_4(y,z).\hfill \end{array}$$

Thus, we deduce

$$\begin{array}{cc}\hfill & T_3(sy,sz)\le {\varphi }_1^{-1}(s)T_3(y,z),T_4(sy,sz)\le {\varphi }_2^{-1}(s)T_4(y,z).\hfill \end{array}$$

From Theorem 1, we can obtain conclusion $(L1)$.  □

Remark 1.

In Theorems 3 and 4, the condition ($l_4$) turns into

($l_4^{\prime }$) for any $s\in (0,1),y,z\in P$ there exists ${\phi }_i(s)\in (s,1](i=1,2,3,4)$, such that

$$\begin{array}{cc}\hfill & A_1(sy,sz)\ge {\phi }_1(s)A_1(y,z),A_2(sy,sz)\ge {\phi }_2(s)A_2(y,z),\hfill \\ \hfill & B_1(sy,sz)\ge {\phi }_3(s)B_1(y,z),B_2(sy,sz)\ge {\phi }_4(s)B_2(y,z);\hfill \end{array}$$

or the condition ($l_6$) changes into

($l_6^{\prime }$) for any $s\in (0,1),y,z\in P$ there exists some increasing functions ${\varphi }_i(s)\in (s,1]$ $(i=1,2,3,4)$, such that

$$\begin{array}{cc}\hfill & A_1(sy,sz)\le {\varphi }_1^{-1}(s)A_1(y,z),A_2(sy,sz)\le {\varphi }_2^{-1}(s)A_2(y,z),\hfill \\ \hfill & B_1(sy,sz)\le {\varphi }_3^{-1}(s)B_1(y,z),B_2(sy,sz)\le {\varphi }_4^{-1}(s)B_2(y,z).\hfill \end{array}$$

At this time, the condition ($l_5$) is not required; then, the conclusions of Theorems 3 and 4 are still valid.

Remark 2.

In Theorems 1–4, when $A_1,B_1,A_2,B_2:inf(P\times P)\to inf(P)$ are increasing operators and the corresponding conditions hold for any $y,z\in inf(P)$, without conditions ($l_1$) or ($l_3$), the corresponding conclusions are still valid.

4. Unique Solution to Coupled FDEs

In this chapter, the above conclusions of operator fixed point theorems are used to discuss coupled system (1).

Let $E=\{z|z\in C[0,1]\}$, the norm $||z||={max}_{t\in [0,1]}|z|$, $P=\{z\in C[0,1]|z(t)\ge 0,t\in [0,1]\}$; then, E is Banach space, P is a normal cone on E, and the normal constant is 1.

Lemma 7.

Let $f_1,f_2,g_1,g_2\in C([0,1]\times [0,+\infty )\times [0,+\infty ),[0,+\infty )),$ and $F_i(\tau ,u(\tau ),v(\tau ))=$ $f_i(\tau ,u(\tau ),v(\tau ))+g_i(\tau ,u(\tau ),v(\tau )),i=1,2$. The problem (1) has the following integral forms:

$$\left\{\begin{array}{c}\begin{array}{c}u(t)={\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,u(\tau ),v(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,\hfill \\ v(t)={\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )F_2(\tau ,u(\tau ),v(\tau ))\mathrm{d}\tau \right)\mathrm{d}s.\hfill \end{array}\hfill \end{array}\right.$$

(25)

Proof. 

Let ${\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)=x_1(t),$ ${\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)=x_2(t);$ then, some parts of Equation (1),

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_{11}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]+f_1(t,u(t),v(t))+g_1(t,u(t),v(t))=0,t\in [0,1],\hfill \\ D_{0^{+}}^{{\eta }_{21}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]+f_2(t,u(t),v(t))+g_2(t,u(t),v(t))=0,t\in [0,1],\hfill \\ [{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right){]}_{t=0}=0,D_{0^{+}}^{{\beta }_{1i}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]{|}_{t=0}=0,\hfill \\ n_1-i-1<{\eta }_{11}-{\beta }_{1i}<n_1-i,i=1,···,n_1-2,\hfill \\ [{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)]{{|}_{t=0}=0,D_{0^{+}}^{{\beta }_{2i}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]|}_{t=0}=0,\hfill \\ n_2-i-1<{\eta }_{21}-{\beta }_{2i}<n_2-i,i=1,···,n_2-2,\hfill \\ D_{0^{+}}^{{\gamma }_{11}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]{{|}_{t=1}={\int }_0^1{a}_1(s)D_{0^{+}}^{{\gamma }_{12}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(t)\right)\right]|}_{t=\xi }\mathrm{d}s\hfill \\ +{\int }_0^1{b}_1(s)D_{0^{+}}^{{\gamma }_{13}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{12}}u(s)\right)\right]\mathrm{d}{A}_{11}(s),\xi \in (0,1),\hfill \\ D_{0^{+}}^{{\gamma }_{21}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]{{|}_{t=1}={\int }_0^1{a}_2(s)D_{0^{+}}^{{\gamma }_{22}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(t)\right)\right]|}_{t=\xi }\mathrm{d}s\hfill \\ +{\int }_0^1{b}_2(s)D_{0^{+}}^{{\gamma }_{23}}\left[{\phi }_p\left(D_{0^{+}}^{{\eta }_{22}}v(s)\right)\right]\mathrm{d}{A}_{21}(s),\xi \in (0,1),\hfill \end{array}\right.$$

change into

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_{11}}{x}_1(t)+f_1(t,u(t),v(t))+g_1(t,u(t),v(t))=0,\hfill \\ x_1(0)=0,D_{0^{+}}^{{\beta }_{1i}}{x}_1(0)=0,\hfill \\ D_{0^{+}}^{{\gamma }_{11}}{x}_1(1)={\int }_0^1{a}_1(s)D_{0^{+}}^{{\gamma }_{12}}{x}_1(\xi )\mathrm{d}s+{\int }_0^1{b}_1(s)D_{0^{+}}^{{\gamma }_{13}}{x}_1(s)\mathrm{d}{A}_{11}(s),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\eta }_{21}}{x}_2(t)+f_2(t,u(t),v(t))+g_2(t,u(t),v(t))=0,\hfill \\ x_2(0)=0,D_{0^{+}}^{{\beta }_{2i}}{x}_2(0)=0,\hfill \\ D_{0^{+}}^{{\gamma }_{21}}{x}_2(1)={\int }_0^1{a}_2(s)D_{0^{+}}^{{\gamma }_{22}}{x}_2(\xi )\mathrm{d}s+{\int }_0^1{b}_2(s)D_{0^{+}}^{{\gamma }_{23}}{x}_2(s)\mathrm{d}{A}_{21}(s).\hfill \end{array}\right.$$

By Lemma 2, we have

$$\left\{\begin{array}{c}\begin{array}{c}{x}_1(t)={\int }_0^1{H}_{13}(t,s)[f_1(s,u(s),v(s))+g_1(s,u(s),v(s))]\mathrm{d}s,\hfill \\ x_2(t)={\int }_0^1{H}_{23}(t,s)[f_2(s,u(s),v(s))+g_2(s,u(s),v(s))]\mathrm{d}s.\hfill \end{array}\hfill \end{array}\right.$$

Then,

$$\left\{\begin{array}{c}\begin{array}{c}{D}_{0^{+}}^{{\eta }_{12}}u(t)={\phi }_q\left({\int }_0^1{H}_{13}(t,s)[f_1(s,u(s),v(s))+g_1(s,u(s),v(t))]\mathrm{d}s\right),\hfill \\ D_{0^{+}}^{{\eta }_{22}}v(t)={\phi }_q\left({\int }_0^1{H}_{23}(t,s)[f_2(s,u(s),v(s))+g_2(s,u(s),v(t))]\mathrm{d}s\right),\hfill \end{array}\hfill \end{array}\right.$$

where ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. With the rest of the conditions and by Lemma 4 we obtain conclusion (25).  □

Theorem 5.

Let $f_1,f_2,g_1,g_2:[0,1]\times [0,+\infty )\times [0,+\infty )\to [0,+\infty )$ be continuous, $f_1(t,1,1),f_2(t,1,1),$ $g_1(t,1,1),g_2(t,1,1)\equiv / 0,t\in [0,1]$. Assume that

($H_1$) for $t\in [0,1],y,z\in [0,+\infty )$, $f_1(t,y,z),f_2(t,y,z),g_1(t,y,z),g_2(t,y,z)$ are decreasing in y and z;

($H_2$) there exist two increasing functions ${\varphi }_1(t),{\varphi }_2(t)\in (t,1]$, $\forall \lambda \in (0,1),x,y\in [0,+\infty )$, such that

$$f_1(t,\lambda y,\lambda z)+g_1(t,\lambda y,\lambda z)\le {\varphi }_1^{-{\displaystyle \frac{1}{q-1}}}(t)[f_1(t,y,z)+g_1(t,y,z)],$$

$$f_2(t,\lambda y,\lambda z)+g_2(t,\lambda y,\lambda z)\le {\varphi }_2^{-{\displaystyle \frac{1}{q-1}}}(t)[f_2(t,y,z)+g_2(t,y,z)].$$

Then, problem (1) has a unique positive solution $(y^{\ast },z^{\ast })$ in ${\tilde{P}}_h$.

Proof. 

Define some operators $T_3,T_4:\tilde{P}\to E$ by

$$T_3(y,z)(t)={\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )[f_1(\tau ,y(\tau ),z(\tau ))+g_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,$$

$$T_4(y,z)(t)={\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )[f_2(\tau ,y(\tau ),z(\tau ))+g_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s.$$

Evidently, by Lemma 7 $(y,z)$ is the solution to (1) $\iff (y,z)=\left(T_3(y,z),T_4(y,z)\right)$.

Step 1: By $H_{13}(t,s),H_{23}(t,s),H_{15}(t,s),H_{25}(t,s)\ge 0$ of Lemmas 3 and 5, and the definitions of $f_1,f_2,g_1,g_2$, we deduce $T_3,T_4:\tilde{P}\to P$. From the monotonicity of $f_1,f_2,g_1,g_2$ of condition ($H_1$), we can know that $T_3,T_4$ are decreasing operators.

Step 2: Let $h_0^{(1)}(t)=h_1(t)=t^{{\eta }_{12}-1},h_0^{(2)}(t)=h_2(t)=t^{{\eta }_{22}-1},t\in [0,1],$ and $F_i(\tau ,u(\tau ),v(\tau ))=f_i(\tau ,u(\tau ),v(\tau ))+g_i(\tau ,u(\tau ),v(\tau )),i=1,2$. For $h_0=(h_0^{(1)},h_0^{(2)})=(h_1,h_2)\in {\tilde{P}}_h$, by (ii) of Lemma 3 and (ii) of Lemma 5 we conclude that

$$\begin{array}{cc}\hfill T_3(h_0^{(1)},h_0^{(2)})(t)=& {\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,{\tau }^{{\eta }_{12}-1},{\tau }^{{\eta }_{22}-1})\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill \ge & {\int }_0^1{a}_{15}(s)t^{{\eta }_{12}-1}{\phi }_q\left({\int }_0^1{a}_{13}(\tau )s^{{\eta }_{11}-1}{F}_1(\tau ,1,1)]\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \left\{{\int }_0^1{a}_{15}(s)s^{({\eta }_{11}-1)(q-1)}{\phi }_q\left({\int }_0^1{a}_{13}(\tau )F_1(\tau ,1,1)\mathrm{d}\tau \right)\mathrm{d}s\right\}·t^{{\eta }_{12}-1},\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill T_3(h_0^{(1)},h_0^{(2)})(t)\le & {\int }_0^1{b}_{15}{t}^{{\eta }_{12}-1}{\phi }_q\left({\int }_0^1{b}_{13}{s}^{{\eta }_{11}-1}{F}_1(\tau ,0,0)\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \left\{b_{15}{b}_{13}^{q-1}{\int }_0^1{s}^{({\eta }_{11}-1)(q-1)}{\phi }_q\left({\int }_0^1{F}_1(\tau ,0,0)\mathrm{d}\tau \right)\mathrm{d}s\right\}·t^{{\eta }_{12}-1},\hfill \end{array}$$

(27)

From $f_1(t,1,1),g_1(t,1,1)\equiv / 0$ and (26) and (27), one obtains $T_3{h}_0\in P_{h_1}.$ In a similar way, we obtain

$$\begin{array}{cc}\hfill \{{\int }_0^1& a_{25}(s)s^{({\eta }_{21}-1)(q-1)}{\phi }_q\left({\int }_0^1{a}_{23}(\tau )F_2(\tau ,0,0)\mathrm{d}\tau \right)\mathrm{d}s\}·t^{{\eta }_{22}-1}\hfill \\ \hfill & \le T_4(h_1,h_2)(t)\le \left\{b_{25}{b}_{23}^{q-1}{\int }_0^1{s}^{({\eta }_{21}-1)(q-1)}{\phi }_q\left({\int }_0^1{F}_2(\tau ,1,1)\mathrm{d}\tau \right)\mathrm{d}s\right\}·t^{{\eta }_{22}-1},\hfill \end{array}$$

implying that $T_4{h}_0\in P_{h_2}.$

Step 3: In view of ($H_2$), and invoking Lemmas 3 and 5, there exist ${\varphi }_1(t),{\varphi }_2(t)\in (t,1]$, such that

$$\begin{array}{cc}\hfill T_3(\lambda y,\lambda z)(t)=& {\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,\lambda y(\tau ),\lambda z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill \le & {\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau ){\varphi }_1^{-{\displaystyle \frac{1}{q-1}}}(\lambda )F_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& {\varphi }_1^{-1}(\lambda ){\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& {\varphi }_1^{-1}(\lambda )T_3(y,z)(t),\hfill \end{array}$$

$$\begin{array}{cc}\hfill T_4(\lambda y,\lambda z)(t)=& {\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )F_2(\tau ,\lambda y(\tau ),\lambda z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill \le & {\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau ){\varphi }_2^{-{\displaystyle \frac{1}{q-1}}}(\lambda )F_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& {\varphi }_2^{-1}(\lambda ){\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )F_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& {\varphi }_2^{-1}(\lambda )T_4(y,z)(t).\hfill \end{array}$$

Hence, by Theorem 1 Equation (1) has a unique positive solution $(y^{\ast },z^{\ast })$ in ${\tilde{P}}_h$.  □

When the properties of the nonlinear terms $f_1,f_2,g_1,g_2$ change, the sufficient conditions of the unique solution to problem (1) are as follows, which is deduced according to Theorem 2:

Theorem 6.

Let $f_1,f_2,g_1,g_2:[0,1]\times [0,+\infty )\times [0,+\infty )\to [0,+\infty )$ be continuous, $f_1(t,0,0),f_2(t,0,0),$ $g_1(t,0,0),g_2(t,0,0)\equiv / 0$. Assume that

($H_3$) for $s\in [0,1],y,z\in [0,+\infty )$, $f_1(t,y,z),f_2(t,y,z),g_1(t,y,z),g_2(t,y,z)$ are increasing in y and z.

($H_4$) there exist ${\alpha }_1\in (0,1),\phi (\lambda )\in (\lambda ,1]$, such that

$$f_1(t,\lambda y,\lambda z)\ge {\lambda }^{{\displaystyle \frac{{\alpha }_1}{q-1}}}{f}_1(t,y,z),f_2(t,\lambda y,\lambda z)\ge {\lambda }^{{\displaystyle \frac{1}{q-1}}}{f}_2(t,y,z),$$

$$g_1(t,\lambda y,\lambda z)\ge {\lambda }^{{\displaystyle \frac{1}{q-1}}}{g}_1(t,y,z),g_2(t,\lambda y,\lambda z)\ge {\phi }^{{\displaystyle \frac{1}{q-1}}}(\lambda )g_2(t,y,z).$$

($H_5$) there exist $p_1>0,0<p_2<1$, such that $f_1(t,y,z)\ge p_1^{{\displaystyle \frac{1}{q-1}}}{g}_1(t,y,z),g_2(t,y,z)\ge {({\displaystyle \frac{p_2}{1-p_2}})}^{{\displaystyle \frac{1}{q-1}}}{f}_2(t,y,z),$ $\forall t\in [0,1],y,z\in [0,+\infty )$.

Then, we obtain

(a1) the problem (1) has a unique positive solution $(y^{\ast },z^{\ast })$ in ${\tilde{P}}_h$;

(b1) for any initial values $y_0,x_0\in P_h$, constructing successively the iterative sequences

$$\left\{\begin{array}{c}\begin{array}{c}{y}_n(t)={\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,y_{n-1}(\tau ),z_{n-1}(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,\hfill \\ z_n(t)={\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )F_2(\tau ,y_{n-1}(\tau ),z_{n-1}(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,\hfill \end{array}\hfill \end{array}\right.$$

where $n=1,2,···,$ we have $y_n\to y^{\ast },z_n\to z^{\ast },$ as $n\to \infty .$ $F_i(\tau ,y(\tau ),z(\tau ))=[f_i(\tau ,y(\tau ),z(\tau ))$ $+g_i(\tau ,y(\tau ),z(\tau ))],i=1,2.$

Proof. 

Define four operators $A_1,A_2,B_1,B_2:\tilde{P}\to E$ by

$$A_1(y,z)(t)={\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )f_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,$$

$$A_2(y,z)(t)={\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )f_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,$$

$$B_1(y,z)(t)={\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )g_1(\tau ,y(\tau ),z(\tau )\mathrm{d}\tau \right)\mathrm{d}s,$$

$$B_2(y,z)(t)={\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )g_2(\tau ,y(\tau ),z(\tau )\mathrm{d}\tau \right)\mathrm{d}s.$$

Evidently by Lemma 7, $(x,y)$ is the solution to (1) $\iff (y,z)=(A_1(y,z)+B_1(y,z),A_2(y,z)$ $+B_2(y,z))$. The proof process may be divided into four steps:

Step 1: From Lemmas 3 and 5 and the definitions of $f_1,f_2,g_1,g_2$, we shall obtain $A_1(\tilde{P}),A_2(\tilde{P}),$ $B_1(\tilde{P}),$ $B_2(\tilde{P})\subset P$. By the condition ($H_3$), one observes that $A_1,A_2,B_1,B_2$ are increasing operators.

Step 2: For $h_0=(h_0^{(1)},h_0^{(2)})=(t^{{\eta }_{12}-1},t^{{\eta }_{22}-1})=(h_1,h_2)\in {\tilde{P}}_h$, by the properties of Green’s function in Lemmas 3 and 5 we deduce

$$\begin{array}{cc}\hfill \{{\int }_0^1{a}_{15}(s)& s^{({\eta }_{11}-1)(q-1)}{\phi }_q\left({\int }_0^1{a}_{13}(\tau )f_1(\tau ,0,0)\mathrm{d}\tau \right)\mathrm{d}s\}·h_1(t)\hfill \\ \hfill & \le A_1{h}_0(t)\le \left\{b_{15}{b}_{13}^{q-1}{\int }_0^1{s}^{({\eta }_{11}-1)(q-1)}{\phi }_q\left({\int }_0^1{f}_1(\tau ,1,1)\mathrm{d}\tau \right)\mathrm{d}s\right\}·h_1(t),\hfill \end{array}$$

$$\begin{array}{cc}\hfill \{{\int }_0^1{a}_{25}(s)& s^{({\eta }_{21}-1)(q-1)}{\phi }_q\left({\int }_0^1{a}_{23}(\tau )f_2(\tau ,0,0)\mathrm{d}\tau \right)\mathrm{d}s\}·h_2(t)\hfill \\ \hfill & \le A_2{h}_0(t)\le \left\{b_{25}{b}_{23}^{q-1}{\int }_0^1{s}^{({\eta }_{21}-1)(q-1)}{\phi }_q\left({\int }_0^1{f}_2(\tau ,1,1)\mathrm{d}\tau \right)\mathrm{d}s\right\}·h_2(t),\hfill \end{array}$$

$$\begin{array}{cc}\hfill \{{\int }_0^1{a}_{15}(s)& s^{({\eta }_{11}-1)(q-1)}{\phi }_q\left({\int }_0^1{a}_{13}(\tau )g_1(\tau ,0,0)\mathrm{d}\tau \right)\mathrm{d}s\}·h_1(t)\hfill \\ \hfill & \le B_1{h}_0(t)\le \left\{b_{15}{b}_{13}^{q-1}{\int }_0^1{s}^{({\eta }_{11}-1)(q-1)}{\phi }_q\left({\int }_0^1{g}_1(\tau ,1,1)\mathrm{d}\tau \right)\mathrm{d}s\right\}·h_1(t),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \{{\int }_0^1{a}_{25}(s)& s^{({\eta }_{21}-1)(q-1)}{\phi }_q\left({\int }_0^1{a}_{23}(\tau )g_2(\tau ,0,0)\mathrm{d}\tau \right)\mathrm{d}s\}·h_2(t)\hfill \\ \hfill & \le B_2{h}_0(t)\le \left\{b_{25}{b}_{23}^{q-1}{\int }_0^1{s}^{({\eta }_{21}-1)(q-1)}{\phi }_q\left({\int }_0^1{g}_2(\tau ,1,1)\mathrm{d}\tau \right)\mathrm{d}s\right\}·h_2(t),\hfill \end{array}$$

which imply $A_1{h}_0,B_1{h}_0\in P_{h_1}$ and $A_2{h}_0,B_2{h}_0\in P_{h_2}$ hold, i.e., the condition $(l_3)$ of Theorem 2 holds.

Step 3: Based on ($H_4$), for any $\lambda \in (0,1)$ there exist ${\alpha }_1\in (0,1),\phi (\lambda )\in (\lambda ,1]$, such that

$$\begin{array}{cc}\hfill A_1(\lambda y,\lambda z)(t)\ge & {\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau ){\lambda }^{{\displaystyle \frac{{\alpha }_1}{q-1}}}{f}_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& {\lambda }^{{\alpha }_1}{\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )f_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& {\lambda }^{{\alpha }_1}{A}_1(y,z)(t),\hfill \end{array}$$

$$\begin{array}{cc}\hfill A_2(\lambda y,\lambda z)(t)\ge & {\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau ){\lambda }^{{\displaystyle \frac{1}{q-1}}}{f}_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \lambda {\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )f_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \lambda A_2(y,z)(t),\hfill \end{array}$$

$$\begin{array}{cc}\hfill B_1(\lambda y,\lambda z)(t)\ge & {\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau ){\lambda }^{{\displaystyle \frac{1}{q-1}}}{g}_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \lambda {\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )g_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \lambda B_1(y,z)(t),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill B_2(\lambda y,\lambda z)(t)\ge & {\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau ){\phi }^{{\displaystyle \frac{1}{q-1}}}(\lambda )g_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \phi (\lambda ){\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )g_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill =& \phi (\lambda )B_2(y,z)(t);\hfill \end{array}$$

hence, the condition $(l_4)$ of Theorem 2 holds.

Step 4: Considering the condition ($H_5$), there exist $p_1>0,0<p_2<1$, such that

$$\begin{array}{cc}\hfill A_1(y,z)(t)& ={\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )f_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill & \ge {\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )p_1^{{\displaystyle \frac{1}{q-1}}}{g}_1(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill & =p_1{B}_1(y,z)(t),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill B_2(y,z)(t)& ={\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )g_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill & \ge {\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau ){({\displaystyle \frac{p_2}{1-p_2}})}^{{\displaystyle \frac{1}{q-1}}}{f}_2(\tau ,y(\tau ),z(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill & ={\displaystyle \frac{p_2}{1-p_2}}{A}_2(y,z)(t),\hfill \end{array}$$

which verify the condition ($l_5$) of Theorem 2. Thus, according to Theorem 2, the conclusions (a1)–(b1) hold.  □

Theorem 7.

Let $f_1,f_2,g_1,g_2:[0,1]\times [0,+\infty )\times [0,+\infty )\to [0,+\infty )$ be continuous, $f_1(t,1,1),f_2(t,1,1),$ $g_1(t,1,1),g_2(t,1,1)\equiv / 0$. Assume that ($H_1$), ($H_5$) hold and

($H_6$) there exists ${\alpha }_1\in (0,1)$ and increasing function $\phi (\lambda )\in (\lambda ,1]$, such that

$$f_1(t,\lambda y,\lambda z)\le {\lambda }^{-{\displaystyle \frac{{\alpha }_1}{q-1}}}{f}_1(t,y,z),f_2(t,\lambda y,\lambda z)\le {\lambda }^{-{\displaystyle \frac{1}{q-1}}}{f}_2(t,y,z),$$

$$g_1(t,\lambda y,\lambda z)\le {\lambda }^{-{\displaystyle \frac{1}{q-1}}}{g}_1(t,y,z),g_2(t,\lambda y,\lambda z)\le {\phi }^{-{\displaystyle \frac{1}{q-1}}}(\lambda )g_2(t,y,z).$$

Then, problem (1) has a unique positive solution $(y^{\ast },z^{\ast })$ in ${\tilde{P}}_h$.

Proof. 

The proof process is similar to Theorem 6, so it is omitted.  □

5. Hyers–Ulam Stability

In this segment, the Hyers–Ulam stability of the systems (1) was studied.

Definition 2

(Hyers–Ulam stability).

($H_7$) Let $F_i(\tau ,u(\tau ),v(\tau ))=[f_i(\tau ,u(\tau ),v(\tau ))+g_i(\tau ,u(\tau ),v(\tau ))],i=1,2.$ There exist $K_1$ and $K_2$, satisfying the following:

for any ${\delta }_1,{\delta }_2>0,$ if

$$\left\{\begin{array}{c}\begin{array}{c}|u(t)-{\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,u(\tau ),v(\tau ))\mathrm{d}\tau \right)\mathrm{d}s|\le {\delta }_1,\hfill \\ |v(t)-{\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )F_2(\tau ,u(\tau ),v(\tau ))\mathrm{d}\tau \right)\mathrm{d}s|\le {\delta }_2,\hfill \end{array}\hfill \end{array}\right.$$

holds and there exists $(x,y)$, satisfying

$$\left\{\begin{array}{c}\begin{array}{c}x(t)={\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,x(\tau ),y(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,\hfill \\ y(t)={\int }_0^1{H}_{25}(t,s){\phi }_q\left({\int }_0^1{H}_{23}(s,\tau )F_2(\tau ,x(\tau ),y(\tau ))\mathrm{d}\tau \right)\mathrm{d}s,\hfill \end{array}\hfill \end{array}\right.$$

(28)

such that

$$||(u,v)-(x,y)||\le K_1{\delta }_1+K_2{\delta }_2,$$

then the problem (1) is Hyers–Ulam stable.

Lemma 8

([25]). Let ${\phi }_p$ be a p-Laplacian operator; then,

(i) if $1<p\le 2,z_1{z}_2>0$ and $|z_1|,|z_2|\ge \omega >0$ then

$$|{\phi }_p(z_1)-{\phi }_p(z_2)|\le (p-1){\omega }^{p-2}|z_1-z_2|;$$

(ii) if $p\ge 2$ and $|z_1|,|z_2|\le \omega$ then

$$|{\phi }_p(z_1)-{\phi }_p(z_2)|\le (p-1){\omega }^{p-2}|z_1-z_2|.$$

Theorem 8.

Let $(u,v)$ be the unique solution of (1), and $(x,y)$ satisfies (28). Assume that

($H_7$) There exist some real valued constant ${\kappa }_i(i=1,2,···,4)$, for $x,y,u,v\in [0,+\infty ),$ there is

$$|f_1(t,u,v)-f_1(t,x,y)|+|g_1(t,u,v)-g_1(t,x,y)|\le {\kappa }_1|u-x|+{\kappa }_2|v-y|,$$

$$|f_2(t,u,v)-f_2(t,x,y)|+|g_2(t,u,v)-g_2(t,x,y)|\le {\kappa }_3|u-x|+{\kappa }_4|v-y|;$$

then, (1) is Hyers–Ulam stable.

Proof. 

From (ii) of Lemmas 3 and 6, we obtain

$$\begin{array}{cc}\hfill |u(t)-x(t)|=& |{\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,u(\tau ),v(\tau ))\mathrm{d}\tau \right)\mathrm{d}s\hfill \\ \hfill & -{\int }_0^1{H}_{15}(t,s){\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,x(\tau ),y(\tau ))\mathrm{d}\tau \right)\mathrm{d}s|\hfill \\ \hfill \le & {\int }_0^1{H}_{15}(t,s)|{\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )F_1(\tau ,u(\tau ),v(\tau ))\mathrm{d}\tau \right)\hfill \\ \hfill & -{\phi }_q\left({\int }_0^1{H}_{13}(s,\tau )L_1(\tau ,x(\tau ),y(\tau ))\mathrm{d}\tau \right)|\mathrm{d}s\hfill \\ \hfill \le & (q-1){\omega }^{q-2}{\int }_0^1{H}_{15}(t,s){\int }_0^1{H}_{13}(s,\tau )\left(\right|f_1(\tau ,u(\tau ),v(\tau ))-f_1(\tau ,x(\tau ),y(\tau ))|\hfill \\ \hfill & +|g_1(\tau ,u(\tau ),v(\tau ))-g_1(\tau ,x(\tau ),y(\tau ))|)\mathrm{d}\tau \mathrm{d}s\hfill \\ \hfill \le & (q-1){\omega }^{q-2}{b}_{15}{b}_{13}[{\kappa }_1||u-x||+{\kappa }_2||v-y||]=K_{11}{\delta }_1+K_{12}{\delta }_2,\hfill \end{array}$$

where $K_{11}=(q-1){\omega }^{q-2}{b}_{15}{b}_{13}{\kappa }_1$, $K_{12}=(q-1){\omega }^{q-2}{b}_{15}{b}_{13}{\kappa }_2,$ $L_I(\tau ,u(\tau ),v(\tau ))=f_I(\tau ,u(\tau ),v(\tau ))+g_I(\tau ,u(\tau ),v(\tau )),i=1,2$. By the same method, we can obtain

$$\begin{array}{cc}\hfill |v(t)-y(t)|\le & (q-1){\omega }^{q-2}{b}_{25}{b}_{23}[{\kappa }_3||u-x||+{\kappa }_4||v-y||]=K_{21}{\delta }_1+K_{22}{\delta }_2,\hfill \end{array}$$

where $K_{21}=(q-1){\omega }^{q-2}{b}_{25}{b}_{23}{\kappa }_3$, $K_{22}=(q-1){\omega }^{q-2}{b}_{25}{b}_{23}{\kappa }_4$. Letting $K_1=max\{K_{11},K_{21}\}$, $K_2=max\{K_{12},K_{22}\}$, or $K_1=K_{11}+K_{21}$, $K_2=K_{12}+K_{22}$, we have

$$||(u,v)-(x,y)||\le K_1{\delta }_1+K_2{\delta }_2,$$

Then, from the ($H_7$) of Definition 2, (1) is Hyers–Ulam stable.  □

6. Applications

Example 1.

Consider the following equation:

$$\left\{\begin{array}{c}{D}_{0^{+}}^{{\displaystyle \frac{13}{4}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{21}{5}}}u(t))]+F_1(t,u(t),v(t))=0,t\in [0,1],\hfill \\ D_{0^{+}}^{{\displaystyle \frac{9}{2}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(t))]+F_2(t,u(t),v(t))=0,t\in [0,1],\hfill \\ u(0)=0,D_{0^{+}}^{{\displaystyle \frac{9}{4}}}u(0)=0,D_{0^{+}}^{{\displaystyle \frac{5}{4}}}u(0)=0,\hfill \\ v(0)=0,D_{0^{+}}^{{\displaystyle \frac{18}{5}}}v(0)=0,D_{0^{+}}^{{\displaystyle \frac{13}{5}}}v(0)=0,D_{0^{+}}^{{\displaystyle \frac{8}{5}}}v(0)=0,D_{0^{+}}^{{\displaystyle \frac{3}{5}}}v(0)=0,\hfill \\ D_{0^{+}}^{{\displaystyle \frac{13}{6}}}u(1)=2{\int }_0^1{s}^{{\displaystyle \frac{1}{2}}}u(s)\mathrm{d}{s}^2,D_{0^{+}}^{{\displaystyle \frac{15}{4}}}v(1)=3{\int }_0^1{s}^{{\displaystyle \frac{1}{3}}}v(s)\mathrm{d}{s}^3,\hfill \\ [{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{21}{5}}}u(t)){]}_{t=0}=0,D_{0^{+}}^{{\displaystyle \frac{4}{3}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{21}{5}}}u(t)\}]{|}_{t=0}=0,\hfill \\ D_{0^{+}}^{{\displaystyle \frac{1}{3}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{21}{5}}}u(t))]{|}_{t=0}=0,{[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(t))]|}_{t=0}=0,\hfill \\ D_{0^{+}}^{{\displaystyle \frac{13}{5}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(t))]{{|}_{t=0}=0,D_{0^{+}}^{{\displaystyle \frac{8}{5}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(t))]|}_{t=0}=0,\hfill \\ D_{0^{+}}^{{\displaystyle \frac{3}{5}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(t))]{|}_{t=0}=0,\hfill \\ D_{0^{+}}^{{\displaystyle \frac{13}{6}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{21}{5}}}u(t))]{{|}_{t=1}={\int }_0^1{s}^{{\displaystyle \frac{3}{4}}}{D}_{0^{+}}^{{\displaystyle \frac{11}{6}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{21}{5}}}u(t))]|}_{t={\displaystyle \frac{1}{6}}}\mathrm{d}s\hfill \\ +{\int }_0^1{s}^{{\displaystyle \frac{1}{5}}}{D}_{0^{+}}^{{\displaystyle \frac{4}{3}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{21}{5}}}u(s))]\mathrm{d}{\displaystyle \frac{s}{3}},\hfill \\ D_{0^{+}}^{{\displaystyle \frac{17}{5}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(t))]{{|}_{t=1}={\int }_0^1{s}^{{\displaystyle \frac{4}{7}}}{D}_{0^{+}}^{{\displaystyle \frac{16}{5}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(t))]|}_{t={\displaystyle \frac{1}{6}}}\mathrm{d}s\hfill \\ +{\int }_0^1{s}^{{\displaystyle \frac{1}{4}}}{D}_{0^{+}}^{{\displaystyle \frac{14}{5}}}[{\phi }_4(D_{0^{+}}^{{\displaystyle \frac{16}{3}}}v(s)])\mathrm{d}{\displaystyle \frac{s}{4}},\hfill \end{array}\right.$$

(29)

where

$$\begin{array}{cc}\hfill & {\eta }_{11}={\displaystyle \frac{13}{4}}\in (3,4),{\eta }_{12}={\displaystyle \frac{21}{5}}\in (4,5),{\eta }_{21}={\displaystyle \frac{9}{2}}\in (4,5),{\eta }_{22}={\displaystyle \frac{16}{3}}\in (5,6),\hfill \\ \hfill & {\alpha }_{12}={\displaystyle \frac{9}{4}},{\alpha }_{13}={\displaystyle \frac{5}{4}},{\alpha }_{22}={\displaystyle \frac{18}{5}},{\alpha }_{23}={\displaystyle \frac{13}{5}},{\alpha }_{24}={\displaystyle \frac{8}{5}},{\alpha }_{25}={\displaystyle \frac{3}{5}},\hfill \\ \hfill & {\beta }_{11}={\displaystyle \frac{1}{3}},{\beta }_{12}={\displaystyle \frac{4}{3}},{\beta }_{21}={\displaystyle \frac{3}{5}},{\beta }_{22}={\displaystyle \frac{8}{5}},{\beta }_{23}={\displaystyle \frac{13}{5}},\hfill \\ \hfill & {\gamma }_{11}={\displaystyle \frac{13}{6}},{\gamma }_{12}={\displaystyle \frac{11}{6}},{\gamma }_{13}={\displaystyle \frac{4}{3}},{\gamma }_{21}={\displaystyle \frac{17}{5}},{\gamma }_{22}={\displaystyle \frac{16}{5}},{\gamma }_{23}={\displaystyle \frac{14}{5}},\hfill \\ \hfill & A_{11}(s)={\displaystyle \frac{s}{3}},A_{21}(s)={\displaystyle \frac{s}{4}},A_{12}(s)=s^2,A_{22}(s)=s^3,\hfill \\ \hfill & {\beta }_1={\displaystyle \frac{13}{6}},{\beta }_2={\displaystyle \frac{15}{4}},{\tau }_1=2,{\tau }_2=3,\xi ={\displaystyle \frac{1}{6}},p=4,\hfill \\ \hfill & k_1(s)=s^{{\displaystyle \frac{1}{2}}},k_2(s)=s^{{\displaystyle \frac{1}{3}}},a_1(s)=s^{{\displaystyle \frac{3}{4}}},a_2(s)=s^{{\displaystyle \frac{4}{7}}},b_1(s)=s^{{\displaystyle \frac{1}{5}}},b_2(s)=s^{{\displaystyle \frac{1}{4}}}.\hfill \end{array}$$

Case 1: When $f_1(t,u,v)+g_1(t,u,v)=F_1(t,u,v)=t^{{\displaystyle \frac{2}{3}}}+{(u+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}+{(v+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}},$ $f_2(t,u,v)+g_2(t,u,v)=F_2(t,u,v)=t^{{\displaystyle \frac{7}{5}}}+{(u+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}+{(v+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}$, let the Equation (29) be expressed as (29)_a.

Conclusion: We see that (29)_a has a unique positive solution, $(x^{\ast },y^{\ast })\in P_{h_1}\times P_{h_2},h_1(t)=t^{{\displaystyle \frac{17}{5}}},$ $h_2(t)=t^{{\displaystyle \frac{13}{3}}}$. In addition, (2.9)_a is Hyers–Ulam stable.

Proof. 

(i) From the above parameters, we can obtain ${\gamma }_{j1}-{\gamma }_{j2}>0,$ ${\gamma }_{j1}-{\gamma }_{j3}>0,{\eta }_{j1}-{\gamma }_{j1}-1>0,$ ${\eta }_{j1}-{\gamma }_{j2}-1>0,{\eta }_{j1}-{\gamma }_{j3}-1>0,j=1,2,$

$$M_1={\displaystyle \frac{1}{\Gamma (\frac{13}{4}-\frac{13}{6})}}-{\displaystyle \frac{{\frac{1}{6}}^{\frac{13}{4}-\frac{11}{6}-1}}{\Gamma (\frac{13}{4}-\frac{11}{6})}}{\int }_0^1{s}^{{\displaystyle \frac{3}{4}}}\mathrm{d}s-{\int }_0^1{\displaystyle \frac{s^{\frac{13}{4}-\frac{4}{3}-1}}{\Gamma (\frac{13}{4}-\frac{4}{3})}}{s}^{{\displaystyle \frac{1}{5}}}\mathrm{d}{\displaystyle \frac{s}{3}}\approx 0.5752>0,$$

$$M_2={\displaystyle \frac{1}{\Gamma (\frac{9}{2}-\frac{17}{5})}}-{\displaystyle \frac{{\frac{1}{6}}^{\frac{9}{2}-\frac{16}{5}-1}}{\frac{9}{2}-\frac{16}{5}}}{\int }_0^1{s}^{{\displaystyle \frac{4}{7}}}\mathrm{d}s-{\int }_0^1{\displaystyle \frac{s^{\frac{9}{2}-\frac{14}{5}-1}}{\Gamma (\frac{9}{2}-\frac{14}{5})}}{s}^{{\displaystyle \frac{1}{4}}}\mathrm{d}{\displaystyle \frac{s}{4}}\approx 0.3805>0,$$

$$Q_1={\int }_0^1{\displaystyle \frac{2\Gamma (\frac{21}{5}-\frac{13}{6})}{\Gamma (\frac{21}{5})}}{s}^{{\displaystyle \frac{21}{5}}-1}{s}^{{\displaystyle \frac{1}{2}}}(s)\mathrm{d}{s}^2\approx 0.0918<1,$$

$$Q_2={\int }_0^1{\displaystyle \frac{3\Gamma (\frac{16}{3}-\frac{15}{4})}{\Gamma (\frac{16}{3})}}{s}^{{\displaystyle \frac{16}{3}}-1}{s}^{{\displaystyle \frac{1}{3}}}(s)\mathrm{d}{s}^3\approx 0.0231<1.$$

(ii) $F_1(t,x,y),F_2(t,x,y)$ are decreasing in x and y.

(iii) For $q={\displaystyle \frac{4}{3}},$ there exist ${\varphi }_1(\lambda )={\lambda }^{{\displaystyle \frac{1}{18}}}\in (\lambda ,1],{\varphi }_2(\lambda )={\lambda }^{{\displaystyle \frac{1}{6}}},$ such that

$$\begin{array}{cc}\hfill F_1(t,\lambda x,\lambda y)& =t^{{\displaystyle \frac{2}{3}}}+{(\lambda x+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}+{(\lambda y+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}}\le t^{{\displaystyle \frac{2}{3}}}+{\lambda }^{-{\displaystyle \frac{1}{5}}}{(x+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}+{\lambda }^{-{\displaystyle \frac{1}{6}}}{(y+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}}\hfill \\ \hfill & \le {\lambda }^{-{\displaystyle \frac{1}{6}}}\left[t^{{\displaystyle \frac{2}{3}}}+{(x+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}+{(y+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}}\right]={\varphi }_1^{-{\displaystyle \frac{1}{q-1}}}(\lambda )F_1(t,x,y),\hfill \\ \hfill F_2(t,\lambda x,\lambda y)& =t^{{\displaystyle \frac{7}{5}}}+{(\lambda x+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}+{(\lambda y+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}\le t^{{\displaystyle \frac{7}{5}}}+{\lambda }^{-{\displaystyle \frac{1}{2}}}{(x+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}+{\lambda }^{-{\displaystyle \frac{1}{3}}}{(y+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}\hfill \\ \hfill & \le {\lambda }^{-{\displaystyle \frac{1}{2}}}\left[t^{{\displaystyle \frac{7}{5}}}+{(x+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}+{(y+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}\right]={\varphi }_2^{-{\displaystyle \frac{1}{q-1}}}(\lambda )F_2(t,x,y).\hfill \end{array}$$

Hence, by (i)–(iii) and Theorem 5, (29)_a has a unique positive solution $(x^{\ast },y^{\ast })$.

(iv) There exist ${\kappa }_1=1,{\kappa }_2=1,{\kappa }_3=1,{\kappa }_4=1$, such that

$$\begin{array}{cc}\hfill |F_1(t,u,v)-F_1(t,x,y)|=& |\left(t^{{\displaystyle \frac{2}{3}}}+{(u+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}+{(v+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}}\right)\hfill \\ \hfill & -\left(t^{{\displaystyle \frac{2}{3}}}+{(x+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}+{(y+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}}\right)|\hfill \\ \hfill \le & |{(u+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}-{(x+{\displaystyle \frac{3}{2}})}^{-{\displaystyle \frac{1}{5}}}|+|{(v+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}}-{(y+{\displaystyle \frac{7}{3}})}^{-{\displaystyle \frac{1}{6}}}|\hfill \\ \hfill \le & |x-u|+|v-y|,\hfill \\ \hfill |F_2(t,u,v)-F_2(t,x,y)|=& |\left(t^{{\displaystyle \frac{7}{5}}}+{(u+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}+{(v+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}\right)\hfill \\ & -\left(t^{{\displaystyle \frac{7}{5}}}+{(x+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}+{(y+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}\right)|\hfill \\ \hfill \le & |{(u+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}-{(x+{\displaystyle \frac{7}{4}})}^{-{\displaystyle \frac{1}{2}}}|+|{(v+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}-{(y+{\displaystyle \frac{5}{2}})}^{-{\displaystyle \frac{1}{3}}}|\hfill \\ \hfill \le & |u-x|+|v-y|,\hfill \end{array}$$

Thus, (29)_a is Hyers–Ulam stable.  □

Case 2: When $f_1(t,x,y)=t^{{\displaystyle \frac{2}{5}}}+{(x+2)}^{{\displaystyle \frac{1}{5}}}+{(y+1)}^{{\displaystyle \frac{1}{6}}},f_2(t,x,y)=t+{(x+1)}^{{\displaystyle \frac{1}{3}}}+{(y+1)}^{{\displaystyle \frac{1}{4}}},g_1(t,x,y)={\displaystyle \frac{1}{3}}t+{\displaystyle \frac{1}{4}}{(x+1)}^{{\displaystyle \frac{1}{5}}}+{\displaystyle \frac{1}{5}}{(y+1)}^{{\displaystyle \frac{1}{6}}},g_2(t,x,y)={\displaystyle \frac{1}{4}}{t}^{{\displaystyle \frac{3}{7}}}+{\displaystyle \frac{1}{5}}{(x+1)}^{{\displaystyle \frac{1}{3}}}+{\displaystyle \frac{1}{6}}{(y+1)}^{{\displaystyle \frac{1}{4}}}$, let the Equation (29) be expressed as (29)_b.

Conclusion: We see that (29)_b has a unique positive solution, $(x^{\ast },y^{\ast })\in P_{h_1}\times P_{h_2},h_1=t^{{\displaystyle \frac{17}{5}}},$ $h_2=t^{{\displaystyle \frac{13}{3}}}$. In addition, (29)_b is Hyers–Ulam stable.

Proof. 

(v) The following, $f_1(t,x,y),f_2(t,x,y),g_1(t,x,y),g_2(t,x,y)$, are increasing in x and y.

(vi) For $q={\displaystyle \frac{4}{3}},$ there exist ${\alpha }_1={\displaystyle \frac{1}{15}},\phi (\lambda )={\lambda }^{{\displaystyle \frac{1}{9}}}$, such that

$$\begin{array}{cc}\hfill f_1(t,\lambda x,\lambda y)& =t^{{\displaystyle \frac{2}{5}}}+{(\lambda x+2)}^{{\displaystyle \frac{1}{5}}}+{(\lambda y+1)}^{{\displaystyle \frac{1}{6}}}\ge t^{{\displaystyle \frac{2}{5}}}+{\lambda }^{{\displaystyle \frac{1}{5}}}{(x+2)}^{{\displaystyle \frac{1}{5}}}+{\lambda }^{{\displaystyle \frac{1}{6}}}{(y+1)}^{{\displaystyle \frac{1}{6}}}\hfill \\ \hfill & \ge {\lambda }^{{\displaystyle \frac{1}{5}}}\left[t^{{\displaystyle \frac{2}{5}}}+{(x+2)}^{{\displaystyle \frac{1}{5}}}+{(y+1)}^{{\displaystyle \frac{1}{6}}}\right]={\lambda }^{{\displaystyle \frac{{\alpha }_1}{q-1}}}{f}_1(t,x,y),\hfill \\ \hfill f_2(t,\lambda x,\lambda y)& =t+{(\lambda x+1)}^{{\displaystyle \frac{1}{3}}}+{(\lambda y+1)}^{{\displaystyle \frac{1}{4}}}\ge t+{\lambda }^{{\displaystyle \frac{1}{3}}}{(x+1)}^{{\displaystyle \frac{1}{3}}}+{\lambda }^{{\displaystyle \frac{1}{4}}}{(y+1)}^{{\displaystyle \frac{1}{4}}}\hfill \\ \hfill & \ge \lambda \left[t+{(x+1)}^{{\displaystyle \frac{1}{3}}}+{(y+1)}^{{\displaystyle \frac{1}{4}}}\right]\ge {\lambda }^{{\displaystyle \frac{1}{q-1}}}{f}_2(t,x,y),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill g_1(t,\lambda x,\lambda y)& ={\displaystyle \frac{1}{3}}t+{\displaystyle \frac{1}{4}}{(\lambda x+1)}^{{\displaystyle \frac{1}{5}}}+{\displaystyle \frac{1}{5}}{(\lambda y+1)}^{{\displaystyle \frac{1}{6}}}\hfill \\ \hfill & \ge \lambda [{\displaystyle \frac{1}{3}}t+{\displaystyle \frac{1}{4}}{(x+1)}^{{\displaystyle \frac{1}{5}}}+{\displaystyle \frac{1}{5}}{(y+1)}^{{\displaystyle \frac{1}{6}}}]\ge {\lambda }^{{\displaystyle \frac{1}{q-1}}}{g}_1(t,x,y),\hfill \\ \hfill g_2(t,\lambda x,\lambda y)=& {\displaystyle \frac{1}{4}}{t}^{{\displaystyle \frac{3}{7}}}+{\displaystyle \frac{1}{5}}{(\lambda x+1)}^{{\displaystyle \frac{1}{3}}}+{\displaystyle \frac{1}{6}}{(\lambda y+1)}^{{\displaystyle \frac{1}{4}}}\hfill \\ \hfill & \ge {\lambda }^{{\displaystyle \frac{1}{3}}}[{\displaystyle \frac{1}{4}}{t}^{{\displaystyle \frac{3}{7}}}+{\displaystyle \frac{1}{5}}{(x+1)}^{{\displaystyle \frac{1}{3}}}+{\displaystyle \frac{1}{6}}{(y+1)}^{{\displaystyle \frac{1}{4}}}]={\phi }^{{\displaystyle \frac{1}{q-1}}}(\lambda )g_2(t,x).\hfill \end{array}$$

(vii) Letting $p_1=3^{{\displaystyle \frac{1}{3}}},p_2={\displaystyle \frac{{\frac{1}{6}}^{\frac{1}{3}}}{1+{\frac{1}{6}}^{\frac{1}{3}}}}$, one obtains

$$\begin{array}{cc}\hfill f_1(t,x,y)& =t^{{\displaystyle \frac{2}{5}}}+{(x+2)}^{{\displaystyle \frac{1}{5}}}+{(y+1)}^{{\displaystyle \frac{1}{6}}}\hfill \\ \hfill & \ge 3\left[{\displaystyle \frac{1}{3}}t+{\displaystyle \frac{1}{4}}{(x+1)}^{{\displaystyle \frac{1}{5}}}+{\displaystyle \frac{1}{5}}{(y+1)}^{{\displaystyle \frac{1}{6}}}\right]=p_1^{{\displaystyle \frac{1}{q-1}}}{g}_1(t,x,y),\hfill \\ \hfill g_2(t,x,y)& ={\displaystyle \frac{1}{4}}{t}^{{\displaystyle \frac{3}{7}}}+{\displaystyle \frac{1}{5}}{(x+1)}^{{\displaystyle \frac{1}{3}}}+{\displaystyle \frac{1}{6}}{(y+1)}^{{\displaystyle \frac{1}{4}}}\hfill \\ \hfill & \ge {\displaystyle \frac{1}{6}}\left[t+{(x+1)}^{{\displaystyle \frac{1}{3}}}+{(y+1)}^{{\displaystyle \frac{1}{4}}}\right]={({\displaystyle \frac{p_2}{1-p_2}})}^{{\displaystyle \frac{1}{q-1}}}{f}_2(t,x,y).\hfill \end{array}$$

Therefore, from (i), (vi)–(vii), by Theorem 7, (29)_b has a unique positive solution $(x^{\ast },y^{\ast })$.

(viii) There exist ${\kappa }_1={\displaystyle \frac{1}{5}}(2^{-{\displaystyle \frac{4}{5}}}+{\displaystyle \frac{1}{4}}),{\kappa }_2={\displaystyle \frac{1}{5}},{\kappa }_3={\displaystyle \frac{2}{5}},{\kappa }_4={\displaystyle \frac{7}{24}}$, such that

$$\begin{array}{cc}\hfill |f_1(t,u,v)-f_1(t,x,y)|+& |g_1(t,u,v)-g_1(t,x,y)|=|\left(t^{{\displaystyle \frac{2}{5}}}+{(u+2)}^{{\displaystyle \frac{1}{5}}}+{(v+1)}^{{\displaystyle \frac{1}{6}}}\right)\hfill \\ \hfill & -\left(t^{{\displaystyle \frac{2}{5}}}+{(x+2)}^{{\displaystyle \frac{1}{5}}}+{(y+1)}^{{\displaystyle \frac{1}{6}}}\right)|+|\left({\displaystyle \frac{1}{3}}t+{\displaystyle \frac{1}{4}}{(u+1)}^{{\displaystyle \frac{1}{5}}}+{\displaystyle \frac{1}{5}}{(v+1)}^{{\displaystyle \frac{1}{6}}}\right)\hfill \\ \hfill & -\left({\displaystyle \frac{1}{3}}t+{\displaystyle \frac{1}{4}}{(x+1)}^{{\displaystyle \frac{1}{5}}}+{\displaystyle \frac{1}{5}}{(y+1)}^{{\displaystyle \frac{1}{6}}}\right)|\hfill \\ \hfill \le & |{(u+2)}^{{\displaystyle \frac{1}{5}}}-{(x+2)}^{{\displaystyle \frac{1}{5}}}|+{\displaystyle \frac{1}{4}}|{(u+1)}^{{\displaystyle \frac{1}{5}}}-{(x+1)}^{{\displaystyle \frac{1}{5}}}|\hfill \\ \hfill & +{\displaystyle \frac{6}{5}}|{(v+1)}^{{\displaystyle \frac{1}{6}}}-{(y+1)}^{{\displaystyle \frac{1}{6}}}|\hfill \\ \hfill \le & {\displaystyle \frac{1}{5}}(2^{-{\displaystyle \frac{4}{5}}}+{\displaystyle \frac{1}{4}})|x-u|+{\displaystyle \frac{1}{5}}|v-y|,\hfill \\ \hfill |f_2(t,u,v)-f_2(t,x,y)|& +|g_2(t,u,v)-g_2(t,x,y)|=|\left(t+{(u+1)}^{{\displaystyle \frac{1}{3}}}+{(v+1)}^{{\displaystyle \frac{1}{4}}}\right)\hfill \\ \hfill & -\left(t+{(x+1)}^{{\displaystyle \frac{1}{3}}}+{(y+1)}^{{\displaystyle \frac{1}{4}}}\right)|+|\left({\displaystyle \frac{1}{4}}{t}^{{\displaystyle \frac{3}{7}}}+{\displaystyle \frac{1}{5}}{(u+1)}^{{\displaystyle \frac{1}{3}}}+{\displaystyle \frac{1}{6}}{(v+1)}^{{\displaystyle \frac{1}{4}}}\right)\hfill \\ \hfill & -\left({\displaystyle \frac{1}{4}}{t}^{{\displaystyle \frac{3}{7}}}+{\displaystyle \frac{1}{5}}{(x+1)}^{{\displaystyle \frac{1}{3}}}+{\displaystyle \frac{1}{6}}{(y+1)}^{{\displaystyle \frac{1}{4}}}\right)|\hfill \\ \hfill \le & {\displaystyle \frac{6}{5}}|{(u+1)}^{{\displaystyle \frac{1}{3}}}-{(x+1)}^{{\displaystyle \frac{1}{3}}}|+{\displaystyle \frac{7}{6}}|{(v+1)}^{{\displaystyle \frac{1}{4}}}-{(y+1)}^{{\displaystyle \frac{1}{4}}}|\hfill \\ \hfill \le & {\displaystyle \frac{2}{5}}|u-x|+{\displaystyle \frac{7}{24}}|v-y|,\hfill \end{array}$$

which implies that ($H_8$) of Theorem 8 holds. Thus, (29)_b is Hyers–Ulam stable.  □

7. Conclusions

This paper mainly studies several fixed point theorems for non-compact and non-continuous operators on product spaces to discuss a class of coupled fractional differential equations with mixed nonlinear terms. Firstly, considering the difficulty in practical applications caused by the requirement of operator compactness in classical operator theory and the limitations of non-compact and non-continuous operator theory in handling coupled differential equations with mixed nonlinear terms, several types of fixed point theorems for operators on Banach product spaces have been obtained, as shown in Theorems 1–4. The obtained operator fixed point conclusions expand and enrich the relevant research results on operator research space, providing methods and ideas for studying iterative schemes for unique and approximate solutions of coupled differential equations with mixed nonlinear terms. Secondly, a class of coupled fractional differential equations with p-Laplace operator and R–S integral boundary conditions were discussed, using the operator conclusions on the product space studied. Sufficient conditions for the existence and uniqueness of solutions to the equations were obtained, as detailed in Theorems 5–7. In addition, based on the setting of parameter values and changes in R–S integration in the studied equation, the equation can be transformed into some equations from the existing literature ([29,34,35,36]), which indicates that the equation studied is more general and, to some extent, proves the universality of the operator theorem obtained. Finally, the Hyers–Ulam stability of coupled fractional differential equations was discussed, as shown in Theorem 8. The operator theory studied in this paper is designed for coupled differential equations with mixed nonlinear terms, so the limitation of the study is obvious: that is, it is invalid for equations with non-mixed nonlinear terms. In future research, it may be possible to improve the research sets and operator properties in Banach product spaces, thereby expanding the categories of problems that can be studied.