Existence and Asymptotic Estimates of the Maximal and Minimal Solutions for a Coupled Tempered Fractional Differential System with Different Orders

Abstract

In this paper, we focus on the existence and asymptotic estimates of the maximal and minimal solutions for a coupled tempered fractional differential system with different orders. By introducing an order reduction technique and some new growth conditions, we establish some new results on the existence of positive extremal solutions for the tempered fractional differential system, meanwhile, we also obtain the asymptotic estimate of the positive extreme solution by an iterative technique, which possesses a sharp asymptotic estimate. In particular, the iterative sequences converging to maximal and minimal solutions starting from two known initial values are easy to compute. Moreover, the weight function ${\hslash }_i$ is allowed to have an infinite number of singular points in $[0,1]$.

1. Introduction

In this paper, we focus on the existence and asymptotic estimates of the maximal and minimal solutions for the following tempered fractional coupled system with different orders:

$$\left\{\begin{array}{c}-{{}_0{}^{R}D_t}^{a,\lambda }u\left(t\right)={\hslash }_1\left(t\right)f_1\left(e^{\lambda t}v\left(t\right),{{}_0{}^{R}D_t}^{b,\lambda }v\left(t\right)\right),t\in (0,1),\\ -{{}_0{}^{R}D_t}^{c,\lambda }v\left(t\right)={\hslash }_2\left(t\right)f_2\left(e^{\lambda t}u\left(t\right),{{}_0{}^{R}D_t}^{d,\lambda }u\left(t\right)\right),t\in (0,1),\\ u\left(0\right)=0,v=0,\hfill \\ {{}_0{}^{R}D_t}^{d,\lambda }u\left(1\right)=\alpha {{}_0{}^{R}D_t}^{d,\lambda }u\left({\tau }_1\right),{{}_0{}^{R}D_t}^{b,\lambda }v\left(1\right)=\beta {{}_0{}^{R}D_t}^{b,\lambda }v\left({\tau }_2\right),\end{array}\right.$$

(1)

where $a,c\in (1,2],b,d\in (0,1]$ are real constants satisfying $a-d>1,c-b>1$, $0\le \alpha ,\beta \le 1,$ $\lambda$ is a positive constant, $0<{\tau }_1,{\tau }_2<1$, the functions $f_1,f_2\in C(R_{+}^2,(0,+\infty ))$ are increasing functions with respect to the first and second variables, the weight functions ${\hslash }_i\in (L^1(0,1),(0,+\infty ))$ can have an infinite number of singular points in $[0,1]$. In the system (1), ${{}_0{}^{R}D_t}^{a,\lambda }u\left(t\right)$ denote the tempered fractional derivative of u, which is a exponential optimization for the Riemann–Liouville fractional derivative ${{}_0{}^R\mathbb{D}_t}^{a}u\left(t\right)$, and has the following mathematical relationship:

$${{}_0{}^{R}D_t}^{a,\lambda }u\left(t\right)=e^{-\lambda t}{{}_0{}^R\mathbb{D}_t}^a\left(e^{\lambda t}u\left(t\right)\right).$$

For the definition of the Riemann–Liouville fractional derivative and integral, we refer the reader to [1,2,3,4,5].

In recent years, many researchers have been conducting continuous research on fractional equations due to their widespread applications in chemical engineering, automatic control, and thermoelasticity [6,7,8,9,10]. In particular, since the heredity of fractional derivatives can describe certain types of motion with nonlinear or non-stationary characteristics, such as fractional harmonic oscillators, fractional damped oscillators, and fractional Brownian motion [11,12], several classical definitions of fractional derivatives such as the Riemann–Liouville fractional derivative and the Caputo fractional derivative have been introduced to model and analyze more accurately complex systems in the real world such as the design of new materials, control systems, signal processing, physical phenomena [13,14,15,16,17], the fractal-fractional Sirs and coronavirus model [18,19], etc. In addition, in mathematical theory, some nonlinear analysis theories and methods, such as spaces theories and smooth theories [20,21,22,23,24,25,26,27,28,29,30,31], operator theories [32,33,34,35], monotone iterative techniques [36,37,38,39,40,41,42,43,44], spectral analysis [45,46], the variational method [47,48,49,50,51,52,53,54,55], the method of upper and lower solutions [56,57,58,59,60,61], and so on, were developed to solve various fractional problems [62,63,64,65,66,67,68,69].

The tempered fractional derivative was introduced to overcome the limitations of Brownian motion in modeling long-range dependent phenomena occurring in financial time series, Nile river data, and fractal analysis, etc. [70]. It is also helpful for describing counting processes when the inter-arrival times are heavy tailed or arrivals are delayed in the time-fractional Poisson process [71,72]. Recently, Zhang et al. [73] studied the following tempered fractional equation:

$$\left\{\begin{array}{c}-{{}_0{}^{R}D_t}^{\alpha ,\lambda }u\left(t\right)=p\left(t\right)h\left(e^{\lambda t}u\left(t\right),{{}_0{}^{R}D_t}^{\beta ,\lambda }u\left(t\right)\right),t\in (0,1),\\ {{}_0{}^{R}D_t}^{\beta ,\lambda }u\left(0\right)=0,{{}_0{}^{R}D_t}^{\beta ,\lambda }u\left(1\right)=0,\hfill \end{array}\right.$$

(2)

where $\alpha \in (0,1],\beta \in (1,2]$ with $\alpha -\beta >1$, $h\in C(R_{+}^2,R_{+})$, $p\in L^1([0,1],(0,+\infty ))$. By using the Guo–Krasnoselskii fixed point theorem, the authors established the existence of positive solutions for tempered fractional Equation (2). In [59], Quan and Liu studied a class of fractional differential systems with time delays

$$\left\{\begin{array}{c}{\mathbb{D}}_{0^{+}}^{\alpha }u\left(t\right)+f(t,v\left(t\right),v_t)=0,t\in (0,1),\hfill \\ {\mathbb{D}}_{0^{+}}^{\beta }v\left(t\right)+g(t,u\left(t\right),u_t)=0,t\in (0,1),\hfill \\ u\left(t\right)=\varphi \left(t\right),v\left(t\right)=\psi \left(t\right),t\in [-\tau ,0],\hfill \\ {\mathbb{D}}_{0^{+}}^{{\gamma }_1}u\left(1\right)=a{\mathbb{D}}_{0^{+}}^{{\gamma }_1}u\left(\xi \right),{\mathbb{D}}_{0^{+}}^{{\gamma }_2}v\left(1\right)=b{\mathbb{D}}_{0^{+}}^{{\gamma }_2}v\left(\eta \right),\hfill \end{array}\right.$$

(3)

where ${\gamma }_1,{\gamma }_2\in (0,1)$, $1+{\gamma }_1\le \alpha \le 2,1+{\gamma }_2\le \beta \le 2$, $\xi ,\eta \in (0,1)$ and $a,b\in R$, the functions $f,g\in C\left(\right[0,1]\times R\times C[-\tau ,0\left]\right)$. $u_t=u(t+\theta ),v_t=v(t+\theta )$, in which $\theta \in (-\tau ,0],\varphi ,\psi \in C[0,1]$ and $\varphi \left(0\right)=0,\psi \left(0\right)=0$. By employing the upper and lower solutions method, some new results for the existence of solutions for Equation (3) were established.

However compared to the above existing works, we notice that no results on extremal solutions for the system (1) have been reported. Thus motivated by the above works, in this paper, we first introduce a technique of the reduction of order and combine the properties of the Green function to transform the original system (1) into an equivalent integral system, then we construct a nonlinear equation and derive the property of the solution of this nonlinear equation, and then by using these properties, we further establish the existence of extremal solutions for the tempered fractional coupled system with different orders (1). In addition, differently from [59,73], in this paper, we adopt a new twin iterative technique, so that we not only obtain the maximal and minimal solutions of the system but we also construct iterative sequences converging to an extremal solution and obtain the asymptotic properties of an extremal solution.

2. Preliminaries and Lemmas

For the convenience of the reader, it is necessary to recall some properties about the Riemann–Liouville fractional derivative and integral.

Lemma 1

([2]). The Riemann–Liouville fractional derivative and integral have some properties as follows:

(a) Let $x\left(t\right)\in C[0,1]\cap L^1[0,1]$ and $\theta >0$, then

$$I^{\theta }{{}_0{}^R\mathbb{D}_t}^{\theta }x\left(t\right)=x\left(t\right)+\sum _{i=1}^n{{\rm Y}}_i{t}^{\theta -i},$$

(4)

where ${{\rm Y}}_i\in R,n=\left[\theta \right]+1$.

(b) Let $x\left(t\right)\in L^1(0,1)$, and $\theta >\mu >0$, so one has

$$I^{\theta }{I}^{\mu }x\left(t\right)=I^{\theta +\mu }x\left(t\right),{{}_0{}^R\mathbb{D}_t}^{\mu }{I}^{\theta }x\left(t\right)=I^{\theta -\mu }x\left(t\right),{{}_0{}^R\mathbb{D}_t}^{\mu }{I}^{\mu }x\left(t\right)=x\left(t\right).$$

(5)

(c) Let $\theta >0,\mu >0$ with $\mu -\theta >1$, so one gets

$${{}_0{}^R\mathbb{D}_t}^{\theta }{t}^{\mu -1}={\displaystyle \frac{\Gamma \left(\mu \right)}{\Gamma (\mu -\theta )}}{t}^{\mu -\theta -1}.$$

(6)

Lemma 2.

Let $p_i\left(t\right)\in L^1[0,1]$ be a positive function, and the following linear system

$$\left\{\begin{array}{c}-{{}_0{}^{R}D_t}^{a-d,\lambda }u\left(t\right)=p_1\left(t\right),t\in (0,1),\\ -{{}_0{}^{R}D_t}^{c-b,\lambda }v\left(t\right)=p_2\left(t\right),t\in (0,1),\\ u\left(0\right)=v\left(0\right)=0,\hfill \\ u\left(1\right)=\alpha u\left({\tau }_1\right),v\left(1\right)=\beta v\left({\tau }_2\right),\hfill \end{array}\right.$$

(7)

has a unique positive solution $\left(u\right(t),v(t\left)\right)$, which can be expressed as the following integral form

$$\left\{\begin{array}{c}u\left(t\right)={\int }_0^1{H}_1(t,s)p_1\left(s\right)ds,\\ v\left(t\right)={\int }_0^1{H}_2(t,s)p_2\left(s\right)ds,\end{array}\right.$$

(8)

where

$$H_1(t,s)=\left\{\begin{array}{c}{\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}-\alpha {({\tau }_1-s)}^{a-d-1}{e}^{-\lambda {\tau }_1}{t}^{a-d-1}-\left(e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}\right){(t-s)}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s},\hfill \\ t\ge s,{\tau }_1\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}-\left(e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}\right){(t-s)}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s},t\ge s,{\tau }_1\le s,\hfill \\ {\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}-\alpha {({\tau }_1-s)}^{a-d-1}{e}^{-\lambda {\tau }_1}{t}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s},t\le s,{\tau }_1\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s},t\le s,{\tau }_1\le s,\hfill \end{array}\right.$$

(9)

and

$$H_2(t,s)=\left\{\begin{array}{c}{\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}-\beta {({\tau }_2-s)}^{c-b-1}{e}^{-\lambda {\tau }_2}{t}^{c-b-1}-\left(e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}\right){(t-s)}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s},\hfill \\ t\ge s,{\tau }_2\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}-\left(e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}\right){(t-s)}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s},t\ge s,{\tau }_2\le s,\hfill \\ {\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}-\beta {({\tau }_2-s)}^{c-b-1}{e}^{-\lambda {\tau }_2}{t}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s},t\le s,{\tau }_2\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s},t\le s,{\tau }_2\le s,\hfill \end{array}\right.$$

(10)

where

$${\rho }_{\alpha }=(e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1})\Gamma (a-d),{\rho }_{\beta }=(e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1})\Gamma (c-b),$$

are the Green functions of the system (7).

Proof. 

It follows from Lemma 1 that (7) can be transformed as the following integral form

$$e^{\lambda t}u\left(t\right)=-{\displaystyle \frac{1}{\Gamma (a-d)}}{\int }_0^t{(t-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)ds+m_1{t}^{a-d-1}+m_2{t}^{a-d-2},t\in [0,1],$$

and

$$e^{\lambda t}v\left(t\right)=-{\displaystyle \frac{1}{\Gamma (c-b)}}{\int }_0^t{(t-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)ds+n_1{t}^{c-b-1}+n_2{t}^{c-b-2},t\in [0,1].$$

Since $a-d,c-b\in (1,2)$ and $u\left(0\right)=0,v\left(0\right)=0$, we have $m_2=n_2=0$. Thus, we get

$$\left\{\begin{array}{c}{e}^{\lambda t}u\left(t\right)=-{\displaystyle \frac{1}{\Gamma (a-d)}}{\int }_0^t{(t-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)ds+m_1{t}^{a-d-1},t\in [0,1],\hfill \\ e^{\lambda t}v\left(t\right)=-{\displaystyle \frac{1}{\Gamma (c-b)}}{\int }_0^t{(t-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)ds+n_1{t}^{c-b-1},t\in [0,1].\hfill \end{array}\right.$$

(11)

Taking $t=1$ and $t={\tau }_i$, respectively, we have

$$\left\{\begin{array}{c}{e}^{\lambda }u\left(1\right)=-{\displaystyle \frac{1}{\Gamma (a-d)}}{\int }_0^1{(1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)ds+m_1,\hfill \\ e^{\lambda }v\left(1\right)=-{\displaystyle \frac{1}{\Gamma (c-b)}}{\int }_0^1{(1-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)ds+n_1,\hfill \end{array}\right.$$

(12)

and

$$\left\{\begin{array}{c}{e}^{\lambda {\tau }_1}u\left({\tau }_1\right)=-{\displaystyle \frac{1}{\Gamma (a-d)}}{\int }_0^{{\tau }_1}{({\tau }_1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)ds+m_1{{\tau }_1}^{a-d-1},\hfill \\ e^{\lambda {\tau }_2}v\left({\tau }_2\right)=-{\displaystyle \frac{1}{\Gamma (c-b)}}{\int }_0^{{\tau }_2}{({\tau }_2-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)ds+n_1{{\tau }_2}^{c-b-1}.\hfill \end{array}\right.$$

(13)

It follows from $u\left(1\right)=\alpha u\left({\tau }_1\right),v\left(1\right)=\beta v\left({\tau }_2\right)$, (12) and (13) that

$$\begin{array}{c}-{\int }_0^1{\displaystyle \frac{e^{-\lambda }{(1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)}{\Gamma (a-d)}}ds+m_1{e}^{-\lambda }=-{\int }_0^{{\tau }_1}{\displaystyle \frac{\alpha e^{-\lambda {\tau }_1}{({\tau }_1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)}{\Gamma (a-d)}}ds+\alpha m_1{e}^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1},\\ -{\int }_0^1{\displaystyle \frac{e^{-\lambda }{(1-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)}{\Gamma (c-b)}}ds+n_1{e}^{-\lambda }=-{\int }_0^{{\tau }_2}{\displaystyle \frac{\beta e^{-\lambda {\tau }_2}{({\tau }_2-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)}{\Gamma (c-b)}}ds+\beta n_1{e}^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1},\end{array}$$

that is,

$$\left\{\begin{array}{c}{m}_1={\displaystyle \frac{1}{e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}}}\left({\int }_0^1{\displaystyle \frac{e^{-\lambda }{(1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)}{\Gamma (a-d)}}ds-{\int }_0^{{\tau }_1}{\displaystyle \frac{\alpha e^{-\lambda {\tau }_1}{({\tau }_1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)}{\Gamma (a-d)}}ds\right),\hfill \\ n_1={\displaystyle \frac{1}{e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}}}\left({\int }_0^1{\displaystyle \frac{e^{-\lambda }{(1-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)}{\Gamma (c-b)}}ds-{\int }_0^{{\tau }_2}{\displaystyle \frac{\beta e^{-\lambda {\tau }_2}{({\tau }_2-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)}{\Gamma (c-b)}}ds\right).\hfill \end{array}\right.$$

Substituting $m_1$ and $n_1$ into (11), one has

$$\begin{array}{c}u\left(t\right)=-{\int }_0^t{\displaystyle \frac{e^{-\lambda t}{(t-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)}{\Gamma (a-d)}}ds\hfill \\ +{\displaystyle \frac{e^{-\lambda t}{t}^{a-d-1}}{e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}}}\left({\int }_0^1{\displaystyle \frac{e^{-\lambda }{(1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)}{\Gamma (a-d)}}ds-{\int }_0^{{\tau }_1}{\displaystyle \frac{\alpha e^{-\lambda {\tau }_1}{({\tau }_1-s)}^{a-d-1}{e}^{\lambda s}{p}_1\left(s\right)}{\Gamma (a-d)}}ds\right)\hfill \\ ={\int }_0^1{\displaystyle \frac{e^{-\lambda t}{e}^{\lambda s}{e}^{-\lambda }{\left(t(1-s)\right)}^{a-d-1}{p}_1\left(s\right)}{{\rho }_{\alpha }}}ds\hfill \\ -{\int }_0^{{\tau }_1}{\displaystyle \frac{e^{-\lambda t}{e}^{\lambda s}\alpha e^{-\lambda {\tau }_1}{\left(t({\tau }_1-s)\right)}^{a-d-1}{p}_1\left(s\right)}{{\rho }_{\alpha }}}ds-{\int }_0^t{\displaystyle \frac{e^{-\lambda t}{e}^{\lambda s}(e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}){(t-s)}^{a-d-1}{p}_1\left(s\right)ds}{{\rho }_{\alpha }}}.\hfill \end{array}$$

When $t\ge {\tau }_1$, we can rewrite $u\left(t\right)$ as

$$\begin{array}{c}u\left(t\right)={\int }_0^{{\tau }_1}{\displaystyle \frac{{\left(t(1-s)\right)}^{a-d-1}{e}^{-\lambda }-\alpha {\left(t({\tau }_1-s)\right)}^{a-d-1}{e}^{-\lambda {\tau }_1}-\left(e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}\right){(t-s)}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_1\left(s\right)ds\hfill \\ +{\int }_{{\tau }_1}^t{\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}-\left(e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}\right){(t-s)}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_1\left(s\right)ds\hfill \\ +{\int }_t^1{\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_1\left(s\right)ds,\hfill \end{array}$$

When $t<{\tau }_1$, we can rewrite $u\left(t\right)$ as

$$\begin{array}{c}u\left(t\right)={\int }_0^t{\displaystyle \frac{{\left(t(1-s)\right)}^{a-d-1}{e}^{-\lambda }-\alpha {\left(t({\tau }_1-s)\right)}^{a-d-1}{e}^{-\lambda {\tau }_1}-\left(e^{-\lambda }-\alpha e^{-\lambda {\tau }_1}{{\tau }_1}^{a-d-1}\right){(t-s)}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_1\left(s\right)ds\hfill \\ +{\int }_t^{{\tau }_1}{\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}-\alpha {({\tau }_1-s)}^{a-d-1}{e}^{-\lambda {\tau }_1}{t}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_1\left(s\right)ds\hfill \\ +{\int }_{{\tau }_1}^1{\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{t}^{a-d-1}}{{\rho }_{\alpha }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_1\left(s\right)ds.\hfill \end{array}$$

Similarly, we also have

$$\begin{array}{c}v\left(t\right)=-{\int }_0^t{\displaystyle \frac{e^{-\lambda t}{(t-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)}{\Gamma (c-b)}}ds\hfill \\ +{\displaystyle \frac{e^{-\lambda t}{t}^{c-b-1}}{e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}}}\left({\int }_0^1{\displaystyle \frac{e^{-\lambda }{(1-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)}{\Gamma (c-b)}}ds-{\int }_0^{{\tau }_1}{\displaystyle \frac{\beta e^{-\lambda {\tau }_2}{({\tau }_2-s)}^{c-b-1}{e}^{\lambda s}{p}_2\left(s\right)}{\Gamma (c-b)}}ds\right)\hfill \\ ={\int }_0^1{\displaystyle \frac{e^{-\lambda t}{e}^{\lambda s}{e}^{-\lambda }{\left(t(1-s)\right)}^{c-b-1}{p}_2\left(s\right)}{{\rho }_{\beta }}}ds\hfill \\ -{\int }_0^{{\tau }_2}{\displaystyle \frac{e^{-\lambda t}{e}^{\lambda s}\beta e^{-\lambda {\tau }_2}{\left(t({\tau }_2-s)\right)}^{c-b-1}{p}_2\left(s\right)}{{\rho }_{\beta }}}ds-{\int }_0^t{\displaystyle \frac{e^{-\lambda t}{e}^{\lambda s}(e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}){(t-s)}^{c-b-1}{p}_2\left(s\right)ds}{{\rho }_{\beta }}},\hfill \end{array}$$

and if $t\ge {\tau }_2$,

$$\begin{array}{c}v\left(t\right)={\int }_0^{{\tau }_2}{\displaystyle \frac{{\left(t(1-s)\right)}^{c-b-1}{e}^{-\lambda }-\beta {\left(t({\tau }_2-s)\right)}^{c-b-1}{e}^{-\lambda {\tau }_2}-\left(e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}\right){(t-s)}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_2\left(s\right)ds\hfill \\ +{\int }_{{\tau }_2}^t{\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}-\left(e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}\right){(t-s)}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_2\left(s\right)ds\hfill \\ +{\int }_t^1{\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_2\left(s\right)ds,\hfill \end{array}$$

if $t\le {\tau }_2$,

$$\begin{array}{c}v\left(t\right)={\int }_0^t{\displaystyle \frac{{\left(t(1-s)\right)}^{c-b-1}{e}^{-\lambda }-\beta {\left(t({\tau }_2-s)\right)}^{c-b-1}{e}^{-\lambda {\tau }_2}-\left(e^{-\lambda }-\beta e^{-\lambda {\tau }_2}{{\tau }_2}^{c-b-1}\right){(t-s)}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_2\left(s\right)ds\hfill \\ +{\int }_t^{{\tau }_2}{\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}-\beta {({\tau }_2-s)}^{c-b-1}{e}^{-\lambda {\tau }_2}{t}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_2\left(s\right)ds\hfill \\ +{\int }_{{\tau }_2}^1{\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{t}^{c-b-1}}{{\rho }_{\beta }}}{e}^{-\lambda t}{e}^{\lambda s}{p}_2\left(s\right)ds.\hfill \end{array}$$

Consequently, the Green function $H_1(t,s),H_2(t,s)$ of (7) can be expressed by (9) and (10). The proof is completed. □

Obviously, the Green functions $H_1(t,s)$ and $H_2(t,s)$ have the following estimate.

Lemma 3.

The Green functions $H_1(t,s)$ and $H_2(t,s)$ are continuous functions and have the following properties:

$\left(\mathbf{1}\right)$ For any $t,s\in [0,1]$, $H_i(t,s)\ge 0,i=1,2.$

$\left(\mathbf{2}\right)$ For any $t,s\in [0,1]$, $H_i(t,s)$ satisfies

$$0\le H_1(t,s)\le G_1\left(s\right)t^{a-d-1}{e}^{-\lambda t},$$

$$0\le H_2(t,s)\le G_2\left(s\right)t^{c-b-1}{e}^{-\lambda t},$$

where

$$G_1\left(s\right)={\displaystyle \frac{{(1-s)}^{a-d-1}{e}^{-\lambda }{e}^{\lambda s}}{{\rho }_{\alpha }}},G_2\left(s\right)={\displaystyle \frac{{(1-s)}^{c-b-1}{e}^{-\lambda }{e}^{\lambda s}}{{\rho }_{\beta }}}.$$

Lemma 4.

Suppose that $a-d,c-b\in (1,2)$; then, the system (1) is equivalent to the following system

$$\left\{\begin{array}{c}-{{}_0{}^{R}D_t}^{a-d,\lambda }x\left(t\right)={\hslash }_1\left(t\right)f_1\left(I^b\left(e^{\lambda t}y\left(t\right)\right),y\left(t\right)\right),\\ -{{}_0{}^{R}D_t}^{c-b,\lambda }y\left(t\right)={\hslash }_2\left(t\right)f_2\left(I^d\left(e^{\lambda t}x\left(t\right)\right),x\left(t\right)\right),\hfill \\ x\left(0\right)=0x\left(1\right)=\alpha x\left({\tau }_1\right),\hfill \\ y\left(0\right)=0,y\left(1\right)=\beta y\left({\tau }_2\right).\hfill \end{array}\right.$$

(14)

Proof. 

Suppose that $(u,v)$ is a positive solution of the system (1); make the following integral transformation

$$v\left(t\right)=e^{-\lambda t}{I}^b\left(e^{\lambda t}y\left(t\right)\right),u\left(t\right)=e^{-\lambda t}{I}^d\left(e^{\lambda t}x\left(t\right)\right),t\in [0,1].$$

(15)

Since $b,d\in (0,1]$, by the part (b) of Lemma 1, we have

$$y\left(t\right)={{}_0{}^{R}D_t}^{b,\lambda }v\left(t\right),x\left(t\right)={{}_0{}^{R}D_t}^{d,\lambda }u\left(t\right),$$

and by taking $t=1$, we have

$$y\left(1\right)={{}_0{}^{R}D_t}^{b,\lambda }v\left(1\right),x\left(1\right)={{}_0{}^{R}D_t}^{d,\lambda }u\left(1\right).$$

It follows from the boundary conditions of the system (1) that

$${{}_0{}^{R}D_t}^{d,\lambda }u\left(1\right)=\alpha {{}_0{}^{R}D_t}^{d,\lambda }u\left({\tau }_1\right)\mathrm{and}{{}_0{}^{R}D_t}^{b,\lambda }v\left(1\right)=\beta {{}_0{}^{R}D_t}^{b,\lambda }v\left({\tau }_2\right)$$

and $u\left(0\right)=0,v\left(0\right)=0$ that

$$x\left(0\right)=0,x\left(1\right)=\alpha x\left({\tau }_1\right),\mathrm{and}y\left(0\right)=0,y\left(1\right)=\beta y\left({\tau }_2\right).$$

In addition,

$$\begin{array}{c}{{}_0{}^{R}D_t}^{a,\lambda }u\left(t\right)=e^{-\lambda t}{{}_0{}^R\mathbb{D}_t}^a\left(e^{\lambda t}u\left(t\right)\right)\hfill \\ =e^{-\lambda t}{\displaystyle \frac{d^2}{d{t}^2}}{I}^{2-a}\left(I^d\left(e^{\lambda t}u\left(t\right)\right)\right)\hfill \\ =e^{-\lambda t}{\displaystyle \frac{d^2}{d{t}^2}}{I}^{2-(a-d)}\left(e^{\lambda t}u\left(t\right)\right)\hfill \\ ={{}_0{}^{R}D_t}^{a-d,\lambda }x\left(t\right)\hfill \\ =-{\hslash }_1\left(t\right)f_1\left(I^b\left(e^{\lambda t}y\left(t\right)\right),y\left(t\right)\right)\hfill \\ =-{\hslash }_1\left(t\right)f_1\left(e^{\lambda t}v\left(t\right),{{}_0{}^{R}D_t}^{b,\lambda }v\left(t\right)\right),\hfill \end{array}$$

(16)

and

$${{}_0{}^{R}D_t}^{c,\lambda }v\left(t\right)={{}_0{}^{R}D_t}^{c-b,\lambda }y\left(t\right)=-{\hslash }_2\left(t\right)f_2\left(I^d\left(e^{\lambda t}x\left(t\right)\right),x\left(t\right)\right)=-{\hslash }_2\left(t\right)f_2\left(e^{\lambda t}u\left(t\right),{{}_0{}^{R}D_t}^{d,\lambda }u\left(t\right)\right),$$

(17)

which implies that the problem (1) can be converted into the lower order problem (14).

Conversely, assume that $(x,y)$ is a positive solution of the system (14). Make the same integral transformation as (15), then, by Lemma 1, (16), (17) and simple computations, we get that $(u,v)$ is a positive solution of the system (1), which implies that the problem (14) can be transformed into (1), and the details of the proof can be found in [37].

Thus, the problem (1) is equivalent to the lower order problem (14). □

Now, we introduce some conditions and assumptions, which will be used in the rest of paper.

$\left(\mathbf{H}\right)$ There exists some positive constants $\aleph >0$ and ${\epsilon }_i>2$ such that

$$0<\aleph =max\left\{\underset{s+t>0}{sup}{\displaystyle \frac{f_1(s,t)}{{(s+t+1)}^{{\epsilon }_1}}},\underset{s+t>0}{sup}{\displaystyle \frac{f_2(s,t)}{{(s+t+1)}^{{\epsilon }_2}}}\right\}<+\infty .$$

For convenience, we denote four constants and give another assumption.

$$M_1={\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)ds,M_2={\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)ds,{\kappa }_1={\displaystyle \frac{e^{\lambda }}{\lambda \Gamma \left(b\right)}},{\kappa }_2={\displaystyle \frac{e^{\lambda }}{\lambda \Gamma \left(d\right)}}.$$

$\left(\mathbf{G}\right)$

$${\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)ds{\left({\kappa }_1+2\right)}^{{\epsilon }_1}+{\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)ds{\left({\kappa }_2+2\right)}^{{\epsilon }_2}<1.$$

Now denote

$$L_1={\left[{\displaystyle \frac{1}{3^{{\epsilon }_1(1+{\epsilon }_2)M_2^{{\epsilon }_1}{M}_1}}}\right]}^{{\displaystyle \frac{1}{{\epsilon }_1{\epsilon }_2-1}}},L_2={\left[{\displaystyle \frac{1}{3^{{\epsilon }_2(1+{\epsilon }_1)M_1^{{\epsilon }_2}{M}_2}}}\right]}^{{\displaystyle \frac{1}{{\epsilon }_1{\epsilon }_2-1}}},$$

and let $X=C[0,1]$ with the maximum norm

$$\left|\right|x\left|\right|=\underset{t\in [0,1]}{max}\left|x\left(t\right)\right|,$$

then, X is a Banach space. Define a Banach space $E=X\times X=\left\{(x,y):x\left(t\right),y\left(t\right)\in C[0,1]\right\}$ with the norm $\left|\right|(x,y)\left|\right|=\left|\right|x\left|\right|+\left|\right|y\left|\right|,$ and a cone K in Banach space E

$$K=\left\{(x,y)\in E:0\le x\left(t\right)\le L_1{e}^{-\lambda t}{t}^{a-d-1},0\le y\left(t\right)\le L_2{e}^{-\lambda t}{t}^{c-b-1},t\in [0,1]\right\}.$$

From Lemmas 2 and 4, define a nonlinear operator

$$T(x,y)=\left(T_1(x,y),T_2(x,y)\right).$$

where

$$\left\{\begin{array}{c}{T}_1(x,y)\left(t\right)={\int }_0^1{H}_1(t,s){\hslash }_1\left(s\right)f_1\left(I^b\left(e^{\lambda s}y\left(s\right)\right),y\left(s\right)\right)ds,t\in [0,1],\hfill \\ T_2(x,y)\left(t\right)={\int }_0^1{H}_2(t,s){\hslash }_2\left(s\right)f_2\left(I^d\left(e^{\lambda s}x\left(s\right)\right),x\left(s\right)\right)ds,t\in [0,1],\hfill \end{array}\right.$$

(18)

then, the existence of positive solutions to the integral system (14) is equivalent to finding a fixed point of the operator T.

Lemma 5.

Suppose that $\left(\mathbf{H}\right)$ holds, then the operator $T:E\to E$ is completely continuous.

Proof. 

Firstly, we perform the following simple operations based on the relevant properties of Gamma and Beta functions

$$\begin{array}{c}{I}^b\left(t^{c-b-1}\right)={\displaystyle \frac{1}{\Gamma \left(b\right)}}{\int }_0^t{(t-s)}^{b-1}{s}^{c-b-1}ds={\displaystyle \frac{1}{\Gamma \left(b\right)}}{\int }_0^1{(t-tx)}^{b-1}{\left(tx\right)}^{c-b-1}tdx\hfill \\ ={\displaystyle \frac{t^{c-1}}{\Gamma \left(b\right)}}{\int }_0^1{(1-x)}^{b-1}{\left(x\right)}^{c-b-1}dx={\displaystyle \frac{B(b,c-b)}{\Gamma \left(b\right)}}{t}^{c-1}={\displaystyle \frac{\Gamma (c-b)}{\Gamma \left(c\right)}}{t}^{c-1},\hfill \end{array}$$

(19)

and

$$\begin{array}{c}{I}^d\left(t^{a-d-1}\right)={\displaystyle \frac{1}{\Gamma \left(d\right)}}{\int }_0^t{(t-s)}^{d-1}{s}^{a-d-1}ds={\displaystyle \frac{1}{\Gamma \left(d\right)}}{\int }_0^1{(t-tx)}^{d-1}{\left(tx\right)}^{a-d-1}tdx\hfill \\ ={\displaystyle \frac{t^{a-1}}{\Gamma \left(d\right)}}{\int }_0^1{(1-x)}^{d-1}{\left(x\right)}^{a-d-1}dx={\displaystyle \frac{B(d,a-d)}{\Gamma \left(d\right)}}{t}^{a-1}={\displaystyle \frac{\Gamma (a-d)}{\Gamma \left(a\right)}}{t}^{a-1}.\hfill \end{array}$$

(20)

Since $c-b>1,c\in (1,2]$ and $a-d>1,a\in (1,2]$, noting that the $\Gamma \left(z\right)$ is an increasing function for $z>1$, we have

$$\begin{array}{c}0\le I^b\left(e^{\lambda s}y\left(s\right)\right)\le L_2^{-1}{\displaystyle \frac{\Gamma (c-b)}{\Gamma \left(c\right)}}{s}^{c-1}\le L_2^{-1}{s}^{c-1},\hfill \\ 0\le I^d\left(e^{\lambda s}x\left(s\right)\right)\le L_1^{-1}{\displaystyle \frac{\Gamma (a-d)}{\Gamma \left(a\right)}}{s}^{a-1}\le L_1^{-1}{s}^{a-1}.\hfill \end{array}$$

(21)

By the condition (H), we have

$$\aleph {(s+t+1)}^{{\epsilon }_i}\ge f_i(s,t),\mathrm{for}s+t>0.$$

It follows from the definitions K that for any $(x,y)\in K$,

$$0\le x\left(t\right)\le L_1{e}^{-\lambda t}{t}^{a-d-1},0\le y\left(t\right)\le L_2{e}^{-\lambda t}{t}^{c-b-1}.$$

Thus, for any $(x,y)\in K$, by (21), one has

$$\begin{array}{c}{T}_1(x,y)\left(t\right)={\int }_0^1{H}_1(t,s){\hslash }_1\left(s\right)f_1\left(I^b\left(e^{\lambda s}y\left(s\right)\right),y\left(s\right)\right)ds\hfill \\ \le e^{-\lambda t}{t}^{a-d-1}{\int }_0^1{G}_1\left(s\right){\hslash }_1\left(s\right)f_1\left(I^b\left(e^{\lambda s}y\left(s\right)\right),y\left(s\right)\right)ds\hfill \\ \le e^{-\lambda t}{t}^{a-d-1}{\int }_0^1{G}_1\left(s\right){\hslash }_1\left(s\right)f_1\left(L_2{I}^b{s}^{c-b-1},L_2{s}^{c-b-1}{e}^{-\lambda s}\right)ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{a-d-1}}{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)f_1\left(L_2{s}^{c-1},L_2{s}^{c-b-1}{e}^{-\lambda s}\right)ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{a-d-1}\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right){\left(L_2{s}^{c-1}+L_2{s}^{c-b-1}{e}^{-\lambda s}+1\right)}^{{\epsilon }_1}ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{a-d-1}\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right){\left(L_2{s}^{c-1}+L_2{s}^{c-b-1}{e}^{-\lambda s}+L_2\right)}^{{\epsilon }_1}ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{a-d-1}\times \aleph L_2^{{\epsilon }_1}}{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right){\left(s^{c-1}+s^{c-b-1}{e}^{-\lambda s}+1\right)}^{{\epsilon }_1}ds\hfill \\ \le e^{-\lambda t}{t}^{a-d-1}\times 3^{{\epsilon }_1}{L}_2^{{\epsilon }_1}{M}_1\hfill \\ \le L_1{e}^{-\lambda t}{t}^{a-d-1}<+\infty ,\hfill \end{array}$$

(22)

and

$$\begin{array}{c}{T}_2(x,y)\left(t\right)={\int }_0^1{H}_2(t,s){\hslash }_2\left(s\right)f_2\left(I^d\left(e^{\lambda s}x\left(s\right)\right),x\left(s\right)\right)ds\hfill \\ \le e^{-\lambda t}{t}^{c-b-1}{\int }_0^1{G}_2\left(s\right){\hslash }_2\left(s\right)f_2\left(I^d\left(e^{\lambda s}x\left(s\right)\right),x\left(s\right)\right)ds\hfill \\ \le e^{-\lambda t}{t}^{c-b-1}{\int }_0^1{G}_2\left(s\right){\hslash }_2\left(s\right)f_2\left(L_1{I}^d{s}^{a-d-1},L_1{s}^{a-d-1}{e}^{-\lambda s}\right)ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{c-b-1}}{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)f_2\left(L_1{s}^{a-1},L_1{s}^{a-d-1}{e}^{-\lambda s}\right)ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{c-b-1}\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right){\left(L_1{s}^{a-1}+L_1{s}^{a-d-1}{e}^{-\lambda s}+1\right)}^{{\epsilon }_2}ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{c-b-1}\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right){\left(L_1{s}^{a-1}+L_1{s}^{a-d-1}{e}^{-\lambda s}+L_1\right)}^{{\epsilon }_2}ds\hfill \\ \le {\displaystyle \frac{e^{-\lambda t}{t}^{c-b-1}\times \aleph L_1^{{\epsilon }_2}}{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right){\left(s^{a-1}+s^{a-d-1}{e}^{-\lambda s}+1\right)}^{{\epsilon }_2}ds\hfill \\ \le e^{-\lambda t}{t}^{c-b-1}\times 3^{{\epsilon }_2}{L}_1^{{\epsilon }_2}{M}_2\hfill \\ \le L_2{e}^{-\lambda t}{t}^{c-b-1}<+\infty .\hfill \end{array}$$

(23)

Hence, we obtain

$$0\le T_1(x,y)\le L_1{e}^{-\lambda t}{t}^{a-d-1},0\le T_2(x,y)\le L_2{e}^{-\lambda t}{t}^{c-b-1},$$

which implies that $T\left(K\right)\subset K$ is well defined and uniformly bounded.

Since $H_i(t,s)$ is also a uniformly bounded continuous function in $[0,1]\times [0,1]$, for any $ϵ>0$ and $s,t_1,t_2\in [0,1]$, there exists a constant $\sigma >0$ such that for $|t_1-t_2|<\sigma$, one has

$$\begin{array}{c}|H_1(t_1,s)-H_1(t_2,s)|<{\displaystyle \frac{1}{2}}{\left(3^{{\epsilon }_1}\aleph L_2^{{\epsilon }_1}{\int }_0^1{\hslash }_1\left(s\right)ds\right)}^{-1}ϵ,\hfill \\ |H_2(t_1,s)-H_2(t_2,s)|<{\displaystyle \frac{1}{2}}{\left(3^{{\epsilon }_2}\aleph L_1^{{\epsilon }_2}{\int }_0^1{\hslash }_2\left(s\right)ds\right)}^{-1}ϵ,\hfill \end{array}$$

Hence, we can obtain

$$\begin{array}{c}|T_1(x,y)\left(t_1\right)-T_1(x,y)\left(t_2\right)|={\int }_0^1|H_1(t_1,s)-H_1(t_2,s)|{\hslash }_1\left(s\right)f_1\left(I^b\left(e^{\lambda s}y\left(s\right)\right),y\left(s\right)\right)ds\hfill \\ \le {\int }_0^1|H_1(t_1,s)-H_1(t_2,s)|\times 3^{{\epsilon }_1}\aleph L_2^{{\epsilon }_1}{\hslash }_1\left(s\right)ds<{\displaystyle \frac{1}{2}}ϵ,\hfill \end{array}$$

(24)

and

$$\begin{array}{c}|T_2(x,y)\left(t_1\right)-T_2(x,y)\left(t_2\right)|={\int }_0^1|H_2(t,s)-H_2(t_2,s)|{\hslash }_2\left(s\right)f_2\left(I^d\left(e^{\lambda s}x\left(s\right)\right),x\left(s\right)\right)ds\\ \le {\int }_0^1|H_2(t,s)-H_2(t_2,s)|\times 3^{{\epsilon }_2}\aleph L_1^{{\epsilon }_2}{\hslash }_2\left(s\right)ds<{\displaystyle \frac{1}{2}}ϵ,\hfill \end{array}$$

(25)

which implies that T is equicontinuous, and then by the Arezela–Ascoli theorem, T is completely continuous. The proof is completed. □

The following Lemma is important for proving our main result.

Lemma 6.

Suppose that $\left(\mathbf{G}\right)$ holds, then the following equation

$$\left({\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)ds{\left({\kappa }_1(x+1)+x+2\right)}^{{\epsilon }_1}+{\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)ds{\left({\kappa }_2(x+1)+x+2\right)}^{{\epsilon }_2}\right){(x+1)}^{-1}=1$$

(26)

has a unique solution η in $(0,+\infty )$.

Proof. 

Let

$$F\left(x\right)=1-\left({\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right){\left({\kappa }_1(x+1)+x+2\right)}^{{\epsilon }_1}+{\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right){\left({\kappa }_2(x+1)+x+2\right)}^{{\epsilon }_2}\right){(x+1)}^{-1},$$

(27)

it follows from $\left(\mathbf{G}\right)$ and ${\epsilon }_i>2$ that

$$F\left(0\right)=1-\left({\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)ds{\left({\kappa }_1+2\right)}^{{\epsilon }_1}+{\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)ds{\left({\kappa }_2+2\right)}^{{\epsilon }_2}\right)>0,$$

(28)

and

$$F(+\infty )=-\infty .$$

(29)

On the other hand, differentiating (27) with respect to x, and noting that $1-{\epsilon }_i<0$ and ${\epsilon }_i>2$, one has

$$\begin{array}{c}{F}^{\prime }\left(x\right)=-M_1{(x+1)}^{-2}{\left[({\kappa }_1+1)x+{\kappa }_1+2\right]}^{{\epsilon }_1-1}\left[({\epsilon }_1-1)({\kappa }_1+1)x+{\kappa }_1({\epsilon }_1-1)+{\epsilon }_1-2\right]\hfill \\ -M_2{(x+1)}^{-2}{\left(({\kappa }_2+1)x+{\kappa }_2+2\right)}^{{\epsilon }_2-1}\left[({\epsilon }_2-1)({\kappa }_2+1)x+{\kappa }_2({\epsilon }_2-1)+{\epsilon }_2-2\right]<0.\hfill \end{array}$$

(30)

Thus, (28)–(30) guarantee that Equation (26) has a unique positive solution $\eta$ in $(0,+\infty )$. □

3. Main Results

In this section, we give our main result and proof.

Theorem 1.

Assume that $\left(\mathbf{H}\right)$ and $\left(\mathbf{G}\right)$ hold, then the following conclusions are valid.

$\left({\mathbf{C}}_{\mathbf{1}}\right)\mathbf{Existence}$: The tempered fractional system (1) has a positive maximal solution ${\omega }^{*}=({\overline{u}}^{*},{\overline{v}}^{*})$ and a positive minimal solution ${\varpi }^{*}=({\underline{u}}^{*},{\underline{v}}^{*})$.

$\left({\mathbf{C}}_{\mathbf{2}}\right)\mathbf{Asymptotic}\mathbf{properties}$: For the above solutions $({\overline{u}}^{*},{\overline{v}}^{*})$, $({\underline{u}}^{*},{\underline{v}}^{*})$ of the tempered fractional system (1), the following properties are satisfied:

$$0\le {\displaystyle \frac{{\overline{u}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{a-d-1}}}\le L_1,0\le {\displaystyle \frac{{\overline{v}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{c-b-1}}}\le L_2,$$

$$0\le {\displaystyle \frac{{\underline{u}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{a-d-1}}}\le L_1,0\le {\displaystyle \frac{{\underline{v}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{c-b-1}}}\le L_2.$$

$\left({\mathbf{C}}_{\mathbf{3}}\right)\mathbf{Iterative}\mathbf{sequences}$: For the known initial values ${\varpi }_0=(0,0),{\omega }_0=(\eta +1,\eta +1)$, construct iterative sequences as

$$\begin{array}{c}{\omega }_n\left(t\right)={\int }_0^1{H}_i(t,s){\hslash }_i\left(s\right)f_i(I^b\left(e^{\lambda s}{\omega }_{n-1}\left(s\right)\right),{\omega }_{n-1}\left(s\right))ds,\hfill \\ {\varpi }_n\left(t\right)={\int }_0^1{H}_i(t,s){\hslash }_i\left(s\right)f_i(I^b\left(e^{\lambda s}{\varpi }_{n-1}\left(s\right)\right),{\varpi }_{n-1}\left(s\right))ds.\hfill \end{array}$$

Then,

$$\underset{n\to +\infty }{lim}{\omega }_n\left(t\right)={\omega }^{*}\left(t\right),\underset{n\to +\infty }{lim}{\varpi }_n\left(t\right)={\varpi }^{*}\left(t\right)$$

uniformly hold for $t\in [0,1]$.

Proof. 

Firstly, take

$$P_{*}=\left\{(x,y)\in K:0\le \left|\right|x\left|\right|\le \eta +1,0\le \left|\right|y\left|\right|\le \eta +1\right\}.$$

In the following, we prove that $T\left(P_{*}\right)\subset P_{*}$ is a compact operator. In fact, for any $(x,y)\in P_{*}$, we have

$$0\le x\left(t\right)+y\left(t\right)\le \underset{t\in [0,1]}{max}x\left(t\right)+\underset{t\in [0,1]}{max}y\left(t\right)\le 2\eta +2.$$

Thus, it follows from $\left(\mathbf{H}\right)$ and Lemma 6 that

$$\begin{array}{c}\left|\right|T_1(x,y)\left|\right|=\underset{t\in [0,1]}{max}{\int }_0^1{H}_1(t,s){\hslash }_1\left(s\right)f_1(I^b\left(e^{\lambda s}y\left(s\right)\right),y\left(s\right))ds\hfill \\ \le {\int }_0^1{G}_1\left(s\right){\hslash }_1\left(s\right)f_1(I^b\left(e^{\lambda s}y\left(s\right)\right),y\left(s\right))ds\hfill \\ \le {\int }_0^1{G}_1\left(s\right){\hslash }_1\left(s\right){\displaystyle \frac{f_1(I^b\left(e^{\lambda s}y\left(s\right)\right),y\left(s\right))}{{\left(I^b\left(e^{\lambda s}y\left(s\right)\right)+y\left(s\right)+1\right)}^{{\epsilon }_1}}}{\left(I^b\left(e^{\lambda s}y\left(s\right)\right)+y\left(s\right)+1\right)}^{{\epsilon }_1}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right){\left[I^b\left(e^{\lambda s}y\left(s\right)\right)+y\left(s\right)+1\right]}^{{\epsilon }_1}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right){\left[I^b\left(e^{\lambda s}(\eta +1)\right)+\eta +1+1\right]}^{{\epsilon }_1}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right){\left({\kappa }_1(\eta +1)+\eta +1+1\right)}^{{\epsilon }_1}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)ds{\left({\kappa }_1(\eta +1)+\eta +1+1\right)}^{{\epsilon }_1}+{\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)ds{\left({\kappa }_2(\eta +1)+\eta +1+1\right)}^{{\epsilon }_2}\hfill \\ =\eta +1,\hfill \end{array}$$

(31)

and

$$\begin{array}{c}\left|\right|T_2(x,y)\left|\right|={\int }_0^1{H}_2(t,s){\hslash }_2\left(s\right)f_2\left(I^d\left(e^{\lambda s}x\left(s\right)\right),x\left(s\right)\right)ds\hfill \\ \le {\int }_0^1{G}_2\left(s\right){\hslash }_2\left(s\right)f_2(I^d\left(e^{\lambda s}x\left(s\right)\right),x\left(s\right))ds\hfill \\ \le {\int }_0^1{G}_2\left(s\right){\hslash }_2\left(s\right){\displaystyle \frac{f_2(I^d\left(e^{\lambda s}x\left(s\right)\right),x\left(s\right))}{{\left(I^d\left(e^{\lambda s}x\left(s\right)\right)+x\left(s\right)+1\right)}^{{\epsilon }_2}}}{\left(I^d\left(e^{\lambda s}x\left(s\right)\right)+x\left(s\right)+1\right)}^{{\epsilon }_2}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right){\left[I^d\left(e^{\lambda s}x\left(s\right)\right)+x\left(s\right)+1\right]}^{{\epsilon }_2}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right){\left[I^d\left(e^{\lambda s}(\eta +1)\right)+\eta +1+1\right]}^{{\epsilon }_2}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right){\left({\kappa }_2(\eta +1)+\eta +1+1\right)}^{{\epsilon }_2}ds\hfill \\ \le {\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)ds{\left({\kappa }_1(\eta +1)+\eta +1+1\right)}^{{\epsilon }_1}+{\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)ds{\left({\kappa }_2(\eta +1)+\eta +1+1\right)}^{{\epsilon }_2}\hfill \\ =\eta +1,\hfill \end{array}$$

(32)

which implies that $T\left(P_{*}\right)\subset P_{*}$. From Lemma 5, we know that $T\left(P_{*}\right)\subset P_{*}$ is a completely continuous operator.

In the following, we construct an iterative sequence starting from a known initial value. Let

$${\varpi }_0\left(t\right)=({\underline{u}}_0\left(t\right),{\underline{v}}_0\left(t\right))=(0,0),$$

and

$${\varpi }_1\left(t\right)=({\underline{u}}_1\left(t\right),{\underline{v}}_1\left(t\right))=\left(T_1({\underline{u}}_0,{\underline{v}}_0)\left(t\right),T_2({\underline{u}}_0,{\underline{v}}_0)\left(t\right)\right)=\left(T_1(0,0)\left(t\right),T_2(0,0)\left(t\right)\right),t\in [0,1].$$

It is clear that ${\varpi }_0\left(t\right)=(0,0)\in P_{*}$, then we have ${\varpi }_1\left(t\right)\in T\left(P_{*}\right)$. Keep this iterative process, and we write

$${\varpi }_{n+1}\left(t\right)=T{\varpi }_n\left(t\right)=T^{n+1}{\varpi }_0\left(t\right),n=1,2,····,$$

Since $T\left(P_{*}\right)\subset P_{*}$, we get ${\varpi }_n\left(t\right)\in P_{*}$ for $n\ge 1$. Moreover, it follows from the fact that T is a compact operator that ${\varpi }_n$ is a sequentially compact set.

On the other hand, since ${\varpi }_1\left(t\right)\ge 0={\varpi }_0\left(t\right)$ and T is a increasing operator, we have

$${\varpi }_2\left(t\right)=T{\varpi }_1\left(t\right)\ge T{\varpi }_0\left(t\right)={\varpi }_1\left(t\right),t\in [0,1].$$

Thus, by induction, without the loss of generality, we have

$$(0,0)\le {\varpi }_n\le {\varpi }_{n+1}\le (\eta +1,\eta +1),n=1,2,···,$$

which implies that $\left\{{\varpi }_n\left(t\right)\right\}$ is an increasing compact sequence with upper bounds. Consequently, there exists ${\varpi }^{*}=({\underline{u}}^{*},{\underline{v}}^{*})\in P_{*}$ such that

$${\varpi }_n\to {\varpi }^{*},n\to +\infty .$$

Since $T{\varpi }_n={\varpi }_{n-1}$ and T is a continuous operator, take the limit on both sides, we have $T{\varpi }^{*}={\varpi }^{*}$, which implies that ${\varpi }^{*}=({\underline{u}}^{*},{\underline{v}}^{*})$ is a non-negative solution of the system (1). Furthermore, $f(0,0)>0$, and thus ${\varpi }^{*}$ is a positive solution of (1). In addition, since ${\varpi }^{*}\in P_{*}\subset K$, for $({\underline{u}}^{*}\left(t\right),{\underline{v}}^{*}\left(t\right))\in P_{*}$, one has

$$0\le {\displaystyle \frac{{\underline{u}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{a-d-1}}}\le L_1,0\le {\displaystyle \frac{{\underline{v}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{c-b-1}}}\le L_2,t\in (0,1].$$

Next, we choose

$${\omega }_0\left(t\right)=({\overline{u}}_0\left(t\right),{\overline{v}}_0\left(t\right))=(\eta +1,\eta +1)$$

as another initial value, and construct the iterative sequence

$$\begin{array}{c}{\omega }_1\left(t\right)=({\overline{u}}_1\left(t\right),{\overline{v}}_1\left(t\right))=\left(T_1({\overline{u}}_0,{\overline{v}}_0)\left(t\right),T_2({\overline{u}}_0,{\overline{v}}_0)\left(t\right)\right)T{\omega }_0\left(t\right),\hfill \\ {\omega }_2\left(t\right)=({\overline{u}}_2\left(t\right),{\overline{v}}_2\left(t\right))=\left(T_1({\overline{u}}_1,{\overline{v}}_1)\left(t\right),T_2({\overline{u}}_1,{\overline{v}}_1)\left(t\right)\right)=T{\omega }_1\left(t\right),\hfill \\ \cdots \cdots \cdots \hfill \\ {\omega }_{n+1}\left(t\right)=T{\omega }_n\left(t\right)=T^{n+1}{\omega }_0\left(t\right),t\in [0,1],n=1,2,···.\hfill \end{array}$$

Note that ${\omega }_0\left(t\right)=(\eta +1,\eta +1)\in P_{*}$, then, ${\omega }_1\left(t\right)\in P_{*}$ and ${\omega }_{n+1}\left(t\right)\in P_{*}$, and thus we have

$$0\le {\overline{u}}_1\left(t\right)\le \left|\right|{\overline{u}}_1\left(t\right)\left|\right|\le \eta +1={\overline{u}}_0\left(t\right),0\le {\overline{v}}_1\left(t\right)\le \left|\right|{\overline{v}}_1\left(t\right)\left|\right|\le \eta +1={\overline{v}}_0\left(t\right),$$

i.e.,

$${\omega }_1\left(t\right)\le {\omega }_0\left(t\right).$$

Thus, from the monotonicity of T, we have ${\omega }_2\left(t\right)=T{\omega }_1\left(t\right)\le T{\omega }_0\left(t\right)={\omega }_1\left(t\right)$, and then by induction, we also have

$$0\le \cdots \le {\omega }_{n+1}\left(t\right)\le {\omega }_n\left(t\right)\le \cdots \le {\omega }_0\left(t\right),n=1,2,···.$$

It follows from Lemma 5 that $\left\{{\omega }_n\right\}$ is a sequentially compact set with upper bound ${\omega }_0\left(t\right)$. Consequently, there exists ${\omega }_{*}\in P_{*}$ such that ${\omega }_n\left(t\right)\to {\omega }_{*}=({\overline{u}}^{*},{\overline{v}}^{*})$. Since $T{\omega }_n\left(t\right)={\omega }_{n-1}\left(t\right)$, letting $n\to +\infty$, according to the continuity of T, we have $T{\omega }_{*}={\omega }_{*}$. Thus, $f(0,0)>0$ guarantees that ${\omega }_{*}$ is also a positive solution of the system (1) with asymptotic properties

$$0\le {\displaystyle \frac{{\underline{u}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{a-d-1}}}\le L_1,0\le {\displaystyle \frac{{\underline{v}}^{*}\left(t\right)}{e^{-\lambda t}{t}^{c-b-1}}}\le L_2.$$

In the end, we prove that ${\omega }_{*}=({\overline{u}}^{*},{\overline{v}}^{*})$ and ${\varpi }_{*}=({\underline{u}}^{*},{\underline{v}}^{*})$ are a pair of extremal solutions of the system (1). In fact, let $\varkappa$ be any positive solution of (1) in $P_{*}$, then we have

$${\varpi }_0\left(t\right)=(0,0)\le \varkappa \le (\eta +1,\eta +1)={\omega }_0\left(t\right),$$

and

$${\varpi }_1\left(t\right)=T{\varpi }_0\left(t\right)\le T\varkappa =\varkappa \le T{\omega }_0\left(t\right)={\omega }_1\left(t\right).$$

By induction, we have

$${\varpi }_n\left(t\right)\le \varkappa \le {\omega }_n\left(t\right),n=1,2,···,$$

(33)

then, by taking the limit on both sides of (33), we have

$${\varpi }_{*}\le \varkappa \le {\omega }_{*},$$

which indicates that ${\omega }_{*}$ and ${\varpi }_{*}$ are a pair of maximal and minimal positive solutions of the system (1), respectively. The proof is completed. □

4. Example

Example 1.

Let

$$\begin{array}{c}{f}_1\left(s,t\right)={\left(s+t\right)}^{5}ln\left(1+{\displaystyle \frac{1}{s+t+1}}\right)+{\left(s+t+1\right)}^5+1,\hfill \\ f_2\left(s,t\right)={\left(s+t\right)}^{6}ln\left(1+{\displaystyle \frac{1}{s+t+1}}\right)+{\left(s+t+1\right)}^6+2,\hfill \end{array}$$

and $a={\displaystyle \frac{9}{5}},c={\displaystyle \frac{7}{5}},b={\displaystyle \frac{1}{5}},d={\displaystyle \frac{2}{5}},\lambda =2,\alpha ={\displaystyle \frac{1}{6}},\beta ={\displaystyle \frac{1}{7}},{\tau }_1={\displaystyle \frac{3}{4}},{\tau }_2={\displaystyle \frac{4}{5}}$, and so we consider the following singular tempered fractional system (1)

$$\left\{\begin{array}{c}-{{}_0{}^{R}D_t}^{{\displaystyle \frac{9}{5}},2}u\left(t\right)={\hslash }_1\left(t\right)f_1(e^{2t}v\left(t\right),{{}_0{}^{R}D_t}^{{\displaystyle \frac{1}{5}},2}v\left(t\right)),t\in (0,1),\hfill \\ -{{}_0{}^{R}D_t}^{{\displaystyle \frac{7}{5}},2}v\left(t\right)={\hslash }_2\left(t\right)f_2(e^{2t}u\left(t\right),{{}_0{}^{R}D_t}^{{\displaystyle \frac{1}{5}},2}u\left(t\right)),t\in (0,1),\hfill \\ u\left(0\right)=0,v\left(0\right)=0,\hfill \\ {{}_0{}^{R}D_t}^{{\displaystyle \frac{2}{5}},2}u\left(1\right)={\displaystyle \frac{1}{3}}\times {{}_0{}^{R}D_t}^{{\displaystyle \frac{2}{5}},2}u\left({\displaystyle \frac{1}{5}}\right),{{}_0{}^{R}D_t}^{{\displaystyle \frac{1}{5}},2}v\left(1\right)={\displaystyle \frac{2}{3}}\times {{}_0{}^{R}D_t}^{{\displaystyle \frac{1}{5}},2}v\left({\displaystyle \frac{2}{5}}\right),\hfill \end{array}\right.$$

(34)

where the weight functions

$${\hslash }_1\left(t\right)={\displaystyle \frac{\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right)\Gamma \left(\frac{7}{5}\right)}{18\times {10}^8{t}^{\frac{1}{3}}{(1-t)}^{\frac{1}{4}}{\left(\frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}+2\right)}^5}},{\hslash }_2\left(t\right)={\displaystyle \frac{\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right)\Gamma \left(\frac{6}{5}\right)}{18\times {10}^7{t}^{\frac{1}{2}}{(1-t)}^{\frac{1}{3}}{\left(\frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}+2\right)}^6}}.$$

may have an infinite number of singular points in $[0,1]$. Then, the system (34) has a positive maximal solution ${\omega }^{*}=(\overline{u},\overline{v})$ and a positive minimal solution ${\varpi }^{*}=(\underline{u},\underline{v})$, which possess the following asymptotic properties:

$$0\le {\displaystyle \frac{\overline{u}\left(t\right)}{e^{-2t}{t}^{\frac{2}{5}}}}\le 45.0743,0\le {\displaystyle \frac{\overline{v}\left(t\right)}{e^{-2t}{t}^{\frac{1}{5}}}}\le 100.6319,$$

$$0\le {\displaystyle \frac{\underline{u}\left(t\right)}{e^{-2t}{t}^{\frac{2}{5}}}}\le 45.0743,0\le {\displaystyle \frac{\underline{v}\left(t\right)}{e^{-2t}{t}^{\frac{1}{5}}}}\le 100.6319.$$

Proof. 

By simple calculations, we have

$$H_1(t,s)=\left\{\begin{array}{c}{\displaystyle \frac{{(1-s)}^{\frac{2}{5}}{e}^{-2}{t}^{\frac{2}{5}}-\frac{1}{6}{(\frac{3}{4}-s)}^{\frac{2}{5}}{e}^{-\frac{3}{2}}{t}^{\frac{2}{5}}-\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right){(t-s)}^{\frac{2}{5}}}{(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}})\Gamma \left(\frac{7}{5}\right)}}{e}^{-2t}{e}^{2s},t\ge s,{\displaystyle \frac{3}{4}}\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{\frac{2}{5}}{e}^{-2}{t}^{\frac{2}{5}}-\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right){(t-s)}^{\frac{2}{5}}}{(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{4}{3}\right)}^{\frac{2}{5}})\Gamma \left(\frac{7}{5}\right)}}{e}^{-2t}{e}^{2s},t\ge s,{\displaystyle \frac{3}{4}}\le s,\hfill \\ {\displaystyle \frac{{(1-s)}^{\frac{2}{5}}{e}^{-2}{t}^{\frac{2}{5}}-\frac{1}{6}{(\frac{3}{4}-s)}^{\frac{2}{5}}{e}^{-\frac{3}{2}}{t}^{\frac{2}{5}}}{(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}})\Gamma \left(\frac{7}{5}\right)}}{e}^{-2t}{e}^{2s},t\le s,{\displaystyle \frac{3}{4}}\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{\frac{2}{5}}{e}^{-2}{t}^{\frac{2}{5}}}{(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}})\Gamma \left(\frac{7}{5}\right)}}{e}^{-2t}{e}^{2s},t\le s,{\displaystyle \frac{3}{4}}\le s,\hfill \end{array}\right.$$

and

$$H_2(t,s)=\left\{\begin{array}{c}{\displaystyle \frac{{(1-s)}^{\frac{1}{5}}{e}^{-2}{t}^{\frac{1}{5}}-\frac{1}{7}{(\frac{4}{5}-s)}^{\frac{1}{5}}{e}^{-\frac{8}{5}}{t}^{\frac{1}{5}}-\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right){(t-s)}^{\frac{1}{5}}}{(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}})\Gamma \left(\frac{6}{5}\right)}}{e}^{-2t}{e}^{2s},t\ge s,{\displaystyle \frac{4}{5}}\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{\frac{1}{5}}{e}^{-2}{t}^{\frac{1}{5}}-\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right){(t-s)}^{\frac{1}{5}}}{(e^{-2}-\beta e^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}})\Gamma \left(\frac{6}{5}\right)}}{e}^{-2t}{e}^{2s},t\ge s,{\displaystyle \frac{4}{5}}\le s,\hfill \\ {\displaystyle \frac{{(1-s)}^{\frac{1}{5}}{e}^{-2}{t}^{\frac{1}{5}}-\frac{1}{7}{(\frac{4}{5}-s)}^{\frac{1}{5}}{e}^{-\frac{8}{5}}{t}^{\frac{1}{5}}}{(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}})\Gamma \left(\frac{6}{5}\right)}}{e}^{-2t}{e}^{2s},t\le s,{\displaystyle \frac{4}{5}}\ge s,\hfill \\ {\displaystyle \frac{{(1-s)}^{\frac{1}{5}}{e}^{-2}{t}^{\frac{1}{5}}}{(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}})\Gamma \left(\frac{6}{5}\right)}}{e}^{-2t}{e}^{2s},t\le s,{\displaystyle \frac{4}{5}}\le s.\hfill \end{array}\right.$$

Let ${\epsilon }_1=5,{\epsilon }_2=6$, then we have

$$0<\underset{s+t>0}{sup}{\displaystyle \frac{f_1(s,t)}{{\left(s+t+1\right)}^5}}=\underset{s+t>0}{sup}{\displaystyle \frac{{\left(s+t\right)}^{5}ln\left(1+\frac{1}{s+t+1}\right)+{\left(s+t+1\right)}^5+1}{{\left(s+t+1\right)}^5}}=2<+\infty ,$$

and

$$0<\underset{s+t>0}{sup}{\displaystyle \frac{f_2\left(s,t\right)}{{\left(s+t+1\right)}^6}}=\underset{s+t>0}{sup}{\displaystyle \frac{{\left(s+t\right)}^{6}ln\left(1+\frac{1}{s+t+1}\right)+2{\left(s+t+1\right)}^6+2}{{\left(s+t+1\right)}^6}}=3<+\infty ,$$

thus, we have $0<\aleph =3<\infty$, and the condition $\left(\mathbf{H}\right)$ holds.

Next, we verify the condition $\left(\mathbf{G}\right)$. Noting that

$${\kappa }_1={\displaystyle \frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}},{\kappa }_2={\displaystyle \frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}},$$

we have

$$\begin{array}{c}0<{\int }_0^1{\hslash }_1\left(s\right)ds={\displaystyle \frac{\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right)\Gamma \left(\frac{7}{5}\right)}{18\times {10}^8{\left(\frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}+2\right)}^5}}{\int }_0^1{s}^{-{\displaystyle \frac{1}{3}}}{(1-s)}^{-{\displaystyle \frac{1}{4}}}ds={\displaystyle \frac{\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right)\Gamma \left(\frac{7}{5}\right)}{18\times {10}^8{\left(\frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}+2\right)}^5}}B\left({\displaystyle \frac{2}{3}},{\displaystyle \frac{3}{4}}\right)\hfill \\ =5.4322\times {10}^{-13}<+\infty ,\hfill \\ 0<{\int }_0^1{\hslash }_2\left(s\right)ds={\displaystyle \frac{\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right)\Gamma \left(\frac{6}{5}\right)}{18\times {10}^7{\left(\frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}+2\right)}^6}}{\int }_0^1{s}^{-{\displaystyle \frac{1}{2}}}{(1-s)}^{-{\displaystyle \frac{1}{3}}}ds={\displaystyle \frac{\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right)\Gamma \left(\frac{6}{5}\right)}{18\times {10}^7{\left(\frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}+2\right)}^6}}B\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{2}{3}}\right)\hfill \\ =5.8618\times {10}^{-13}<+\infty .\hfill \end{array}$$

Consequently,

$$\begin{array}{c}{M}_1={\displaystyle \frac{3}{\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right)\Gamma \left(\frac{7}{5}\right)}}\times {\displaystyle \frac{\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right)\Gamma \left(\frac{7}{5}\right)}{18\times {10}^8{\left(\frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}+2\right)}^5}}B\left({\displaystyle \frac{2}{3}},{\displaystyle \frac{3}{4}}\right)\hfill \\ ={\displaystyle \frac{1}{6\times {10}^8{\left(\frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}+2\right)}^5}}B\left({\displaystyle \frac{2}{3}},{\displaystyle \frac{3}{4}}\right)=1.7974\times {10}^{-11},\hfill \end{array}$$

$$\begin{array}{c}{M}_2={\displaystyle \frac{3}{\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right)\Gamma \left(\frac{6}{5}\right)}}\times {\displaystyle \frac{\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right)\Gamma \left(\frac{6}{5}\right)}{18\times {10}^7{\left(\frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}+2\right)}^6}}B\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{4}}\right)\hfill \\ ={\displaystyle \frac{1}{6\times {10}^7{\left(\frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}+2\right)}^6}}B\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{4}}\right)=1.646\times {10}^{-11}.\hfill \end{array}$$

Thus, we have

$$\begin{array}{c}{\displaystyle \frac{\aleph }{{\rho }_{\alpha }}}{\int }_0^1{\hslash }_1\left(s\right)ds{({\kappa }_1+2)}^{{\epsilon }_1}+{\displaystyle \frac{\aleph }{{\rho }_{\beta }}}{\int }_0^1{\hslash }_2\left(s\right)ds{({\kappa }_2+2)}^{{\epsilon }_2}\hfill \\ =M_1{({\kappa }_1+2)}^{{\epsilon }_1}+M_2{({\kappa }_2+2)}^{{\epsilon }_2}\hfill \\ ={\displaystyle \frac{3\times {\left(\frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}+2\right)}^5}{\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right)\Gamma \left(\frac{7}{5}\right)}}{\displaystyle \frac{\left(e^{-2}-\frac{1}{6}{e}^{-\frac{3}{2}}{\left(\frac{3}{4}\right)}^{\frac{2}{5}}\right)\Gamma \left(\frac{7}{5}\right)}{18\times {10}^8{\left(\frac{e^2}{2\Gamma \left(\frac{1}{5}\right)}+2\right)}^5}}{\int }_0^1{s}^{-{\displaystyle \frac{1}{3}}}{(1-s)}^{-{\displaystyle \frac{1}{4}}}ds\hfill \\ +{\displaystyle \frac{3\times {\left(\frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}+2\right)}^6}{\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right)\Gamma \left(\frac{6}{5}\right)}}{\displaystyle \frac{\left(e^{-2}-\frac{1}{7}{e}^{-\frac{8}{5}}{\left(\frac{4}{5}\right)}^{\frac{1}{5}}\right)\Gamma \left(\frac{6}{5}\right)}{18\times {10}^7{\left(\frac{e^2}{2\Gamma \left(\frac{2}{5}\right)}+2\right)}^6}}{\int }_0^1{s}^{-{\displaystyle \frac{1}{2}}}{(1-s)}^{-{\displaystyle \frac{1}{3}}}ds\hfill \\ ={\displaystyle \frac{1}{6}}\left({\int }_0^1{s}^{-{\displaystyle \frac{1}{3}}}{(1-s)}^{-{\displaystyle \frac{1}{4}}}ds\times {10}^{-8}+{\int }_0^1{s}^{-{\displaystyle \frac{1}{2}}}{(1-s)}^{-{\displaystyle \frac{1}{3}}}ds\times {10}^{-7}\right)\hfill \\ ={\displaystyle \frac{1}{6}}\left[B\left({\displaystyle \frac{2}{3}},{\displaystyle \frac{3}{4}}\right)\times {10}^{-8}+B\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{2}{3}}\right)\times {10}^{-7}\right]=4.6238\times {10}^{-8}<1,\hfill \end{array}$$

which implies that the condition $\left(\mathbf{G}\right)$ holds.

In the end, by simple calculations, we have

$$L_1={\left({\displaystyle \frac{1}{3^{35}{(1.646\times {10}^{-11})}^{5}1.7974\times {10}^{-11}}}\right)}^{{\displaystyle \frac{1}{29}}}=45.0743,$$

$$L_2={\left({\displaystyle \frac{1}{3^{36}{(1.7974\times {10}^{-11})}^{6}1.646\times {10}^{-11}}}\right)}^{{\displaystyle \frac{1}{29}}}=100.6319.$$

So, Theorem 3.1 guarantees that the system (34) has a positive maximal solution ${\omega }^{*}=(\overline{u},\overline{v})$ and a positive minimal solution ${\varpi }^{*}=(\underline{u},\underline{v})$ satisfying the following asymptotic properties

$$0\le {\displaystyle \frac{\overline{u}\left(t\right)}{e^{-2t}{t}^{\frac{2}{5}}}}\le 45.0743,0\le {\displaystyle \frac{\overline{v}\left(t\right)}{e^{-2t}{t}^{\frac{1}{5}}}}\le 100.6319,$$

$$0\le {\displaystyle \frac{\underline{u}\left(t\right)}{e^{-2t}{t}^{\frac{2}{5}}}}\le 45.0743,0\le {\displaystyle \frac{\underline{v}\left(t\right)}{e^{-2t}{t}^{\frac{1}{5}}}}\le 100.6319.$$

□

5. Conclusions

Fractional derivatives and integrals can be used to describe tissue growth diffusion with long-term memory effects, or to simulate the dynamic behavior of viscoelasticity, electrochemistry control, and porous media, which are more effective than traditional integer order operators. In particular, the tempered fractional derivative was introduced to overcome the limitations of Brownian motion in modeling long-range dependent phenomena occurring in financial time series, Nile river data, and fractal analysis. In this paper, we focus on the existence and asymptotic estimates of the maximal and minimal solutions for a class of a tempered fractional coupled system with different orders. Through the technique of the reduction order, it is transformed into an equivalent integral system with order in $(1,2]$, and then we construct a nonlinear equation and derive the property of the solution of the nonlinear equation, which is important for getting the extremal solution of the tempered fractional coupled system. In addition, differently from [59,73], we introduce a new twin iterative technique and not only obtain the maximal and minimal solutions of the system but also construct iterative sequences converging to extremal solutions and obtain asymptotic properties of extremal solutions. Specifically, the weight function is allowed to have an infinite number of singular points in $[0,1]$. In future research, we will continue to focus on the study of more complex low-order singular tempered fractional equations and systems.