Lyapunov Inequalities for Systems of Tempered Fractional Differential Equations with Multi-Point Coupled Boundary Conditions via a Fix Point Approach

Abstract

In this paper, we study a system of nonlinear tempered fractional differential equations with multi-point coupled boundary conditions. By applying the properties of Green’s function and the operator and combining the method of matrix analysis, we obtain the corresponding Lyapunov inequalities under two Banach spaces. And, we have compared two Lyapunov inequalities under certain conditions. An example is given to verify our results.

1. Introduction

In this paper, we study the following system of nonlinear tempered fractional differential equations with multi-point coupled boundary conditions

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{}_0{}^R\mathbb{D}_t^{{\alpha }_1,\lambda }x\left(t\right)+f(t,x\left(t\right),y\left(t\right))=0,t\in (0,1),\hfill \\ {}_0{}^R\mathbb{D}_t^{{\alpha }_2,\lambda }y\left(t\right)+g(t,x\left(t\right),y\left(t\right))=0,t\in (0,1),\hfill \\ {\displaystyle x\left(0\right)=0,x\left(1\right)=\sum _{i=1}^n{a}_{1i}x\left({\xi }_i\right)+\sum _{j=1}^n{a}_{2j}y\left({\eta }_j\right),}\hfill \\ {\displaystyle y\left(0\right)=0,y\left(1\right)=\sum _{i=1}^n{a}_{3i}x\left({\xi }_i\right)+\sum _{j=1}^n{a}_{4j}y\left({\eta }_j\right),}\hfill \end{array}\right.\end{array}\end{array}$$

(1)

where $f,g:[0,1]\times R^2\to R$; $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_n<1$; $0<{\eta }_1<{\eta }_2<\cdots <{\eta }_n<1$; $a_{ij}\ge 0(i=1,2,3,4;j=1,2,\dots ,n)$; $1<{\alpha }_i\le 2(i=1,2)$; $\lambda \ge 0$; ${}_0{}^R\mathbb{D}_t^{{\alpha }_i,\lambda }(i=1,2)$ are the tempered fractional derivatives, which are defined by

$${}_0{}^R\mathbb{D}_t^{{\alpha }_i,\lambda }x\left(t\right)=e^{-\lambda t}{}_0{}^R\mathcal{D}_t^{{\alpha }_i}\left(e^{\lambda t}x\left(t\right)\right),$$

(2)

where ${}_0{}^R\mathcal{D}_t^{{\alpha }_i}(i=1,2)$ is the standard Riemann–Liouville derivative.

Fractional calculus has drawn extensive attention since it plays a significant role in numerous fields such as physics and engineering. For example, it can describe abnormal phenomena emerging in engineering and applied sciences, including fluid mechanics and physics [1,2]. Fractional calculus is a generalization of integer-order calculus, extending the derivatives and integrals of functions to non-integer orders. There are multiple ways to define fractional calculus. Among them, the most common are Riemann–Liouville-type fractional integrals and derivatives as well as Caputo-type fractional integrals and derivatives. Tempered fractional calculus is a kind of calculus that can be used to deal with stability problems and model turbulent and Brownian motion processes [3,4]. Under certain specific conditions, the tempered fractional derivative can be expressed as a combination form of Riemann–Liouville fractional derivatives through mathematical transformations. When $\lambda >0$, the tempered fractional derivative is an extension of Riemann–Liouville derivatives. In particular, when $\lambda =0$, the tempered fractional derivative (2) can be regarded as a Riemann–Liouville derivative [5].

Stability is a fundamental property of differential equations. There are many methods to study the stability of solutions. One of these methods is to establish a Lyapunov inequality and analyze the aforementioned stability by utilizing the properties of tempered fractional derivatives, through which the conclusion of system stability can be obtained [6,7,8]. The Lyapunov inequality has been applied and generalized in eigenvalue problems and oscillation theory [9,10,11]. Originally, the Lyapunov inequality was proposed by the mathematician Lyapunov [12]. Tiryaki [13], Ntouyas, and Ahmad [14] provided a review of Lyapunov inequalities and outlined the progress of related research. Li [15] provided a review of the stability of fractional calculus; put forward many of the latest research results on the stability of differential equations and corresponding analysis methods; and listed some examples of stability, such as linear fractional differential equations, differential equations with time-delay, and Lyapunov equations. Subsequently, many scholars have studied Lyapunov inequalities for fractional differential equations under different boundary conditions. Wang [16] and Çakmak [17] studied Lyapunov inequalities under anti-periodic boundary conditions. The Lyapunov-type inequality and tempered differential equations can also be used to handle the existence, uniqueness, and stability of solutions of differential equations with boundary value conditions [10,18,19,20,21,22,23,24,25,26,27,28,29].

In [30], Ferreira studied the following Riemann–Liouville fractional differential equation

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{(}_a^R{\mathcal{D}}^{\alpha }y)\left(t\right)+q\left(t\right)y\left(t\right)=0,a<t<b,\hfill \\ y\left(a\right)=0,y\left(b\right)=0,\hfill \end{array}\right.\end{array}\end{array}$$

(3)

where $q:[a,b]\to R$ is a real and continuous function, ${}_a{}^R\mathcal{D}^{\alpha }$ is the standard Riemann–Liouville fractional derivative, and $1<\alpha \le 2$. By transforming System (3) into the form of an integral equation and applying some properties of Green’s function, the author obtained the following Lyapunov-type inequality:

$$\begin{array}{c}\hfill {\int }_a^b\left|q\left(s\right)\right|ds>\Gamma \left(\alpha \right){\left({\displaystyle \frac{4}{b-a}}\right)}^{\alpha -1}.\end{array}$$

In [31], based on the previous work, Ferreira studied the following Caputo fractional differential equation:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{(}_a^C{D}^{\alpha }y)\left(t\right)+q\left(t\right)y\left(t\right)=0,a<t<b,\hfill \\ y\left(a\right)=0,y\left(b\right)=0,\hfill \end{array}\right.\end{array}\end{array}$$

(4)

where $q:[a,b]\to R$ is a real and continuous function, ${}_a{}^{C}D^{\alpha }$ is Caputo fractional derivative, and $1<\alpha \le 2$. By defining a norm on the Banach space, transforming the differential equation into an integral equation and applying the properties of Green’s function, the author obtained the following Lyapunov-type inequality:

$$\begin{array}{c}\hfill {\int }_a^b\left|q\left(s\right)\right|ds>{\displaystyle \frac{\Gamma \left(\alpha \right){\alpha }^{\alpha }}{{\left[(\alpha -1)(b-a)\right]}^{\alpha -1}}}.\end{array}$$

In [24], Jleli et al. studied the system of the Caputo fractional differential equation as follows:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{(}^C{D}_a^{{\alpha }_1}x)\left(t\right)=f(t,x\left(t\right),y\left(t\right)),a<t<b,\hfill \\ {(}^C{D}_a^{{\alpha }_2}y)\left(t\right)=g(t,x\left(t\right),y\left(t\right)),a<t<b,\hfill \\ x\left(a\right)=x\left(b\right)=y\left(a\right)=y\left(b\right)=0,\hfill \end{array}\right.\end{array}\end{array}$$

(5)

where $f,g:[a,b]\times C[a,b]\times C[a,b]\to R$; $1<{\alpha }_i<2(i=1,2)$; $(a,b)\in R\times R(a<b)$; ${}^{C}D_a^{{\alpha }_i}(i=1,2)$ is Caputo fractional derivative. By applying Perov’s fixed-point theorem on the norm and the properties of Green’s function, the authors obtained the following Lyapunov-type inequality:

$$\begin{array}{c}\hfill I_a^b({\alpha }_1,p_{11})+I_a^b({\alpha }_2,p_{22})+{\left[{\left(I_a^b({\alpha }_1,p_{11})-I_a^b({\alpha }_2,p_{22})\right)}^2+4I_a^b({\alpha }_1,p_{12})I_a^b({\alpha }_2,p_{21})\right]}^{{\displaystyle \frac{1}{2}}}\ge 2\end{array}$$

if System (5) has a nontrivial solution and positive functions $p_{ij}\in C[a,b](i,j=1,2)$ exist such that

$$\left|f(t,x,y)-f(t,u,v)\right|\le p_{11}\left(t\right)\left|x-u\right|+p_{12}\left(t\right)\left|y-v\right|,t\in [a,b],x,y,u,v\in R,$$

and

$$\left|g(t,x,y)-g(t,u,v)\right|\le p_{21}\left(t\right)\left|x-u\right|+p_{22}\left(t\right)\left|y-v\right|,t\in [a,b],x,y,u,v\in R.$$

In [9], Zou and Cui studied the system of Riemann–Liouville fractional differential equations with multi-point coupled boundary conditions

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{\mathcal{D}}_{0+}^{{\alpha }_1}x\left(t\right)+f(t,x\left(t\right),y\left(t\right))=0,a<t<b,\hfill \\ {\mathcal{D}}_{0+}^{{\alpha }_2}y\left(t\right)+g(t,x\left(t\right),y\left(t\right))=0,a<t<b,\hfill \\ {\displaystyle x\left(a\right)=0,x\left(b\right)=\sum _{i=1}^n{a}_{1i}x\left({\xi }_i\right)+\sum _{j=1}^n{a}_{2j}y\left({\eta }_j\right),}\hfill \\ {\displaystyle y\left(a\right)=0,y\left(b\right)=\sum _{i=1}^n{a}_{3i}x\left({\xi }_i\right)+\sum _{j=1}^n{a}_{4j}y\left({\eta }_j\right),}\hfill \end{array}\right.\end{array}\end{array}$$

(6)

where $f,g:[0,1]\times R^2\to R$; $a,b\in R$; $0<a<b$; $a<{\xi }_1<{\xi }_2<\cdots <{\xi }_n<b$; $a<{\eta }_1<{\eta }_2<\cdots <{\eta }_n<b$; $a_{ij}\ge 0(i=1,2,3,4;j=1,2,\dots ,n)$, $1<{\alpha }_i\le 2(i=1,2)$. By using the method of matrix analysis, the properties of Green’s function, and the operator T, they proved two Lyapunov-type inequalities under certain conditions as follows:

$$\begin{array}{c}\hfill \begin{array}{cc}& J_{11}(p_{11},p_{21})+J_{22}(p_{12},p_{22})\hfill \\ & +\sqrt{{[J_{11}(p_{11},p_{21})-J_{22}(p_{12},p_{22})]}^2+4J_{12}(p_{12},p_{22})J_{21}(p_{11},p_{21})}\ge 2,\hfill \end{array}\end{array}$$

(7)

and

$$\begin{array}{c}\hfill \begin{array}{cc}& {\xi }_{11}(p_{11},p_{21})+{\xi }_{22}(p_{12},p_{22})\hfill \\ & +\sqrt{{[{\xi }_{11}(p_{11},p_{21})-{\xi }_{22}(p_{12},p_{22})]}^2+4{\xi }_{12}(p_{12},p_{22}){\xi }_{21}(p_{11},p_{21})}\ge 2,\hfill \end{array}\end{array}$$

(8)

if System (6) has a nontrivial solution and positive functions $p_{11}$,$p_{12}$, $p_{21}$,$p_{22}\in C[a,b]$ exist such that

$$\left|f(t,x,y)\right|\le p_{11}\left(t\right)\left|x\right|+p_{12}\left(t\right)\left|y\right|,x,y\in R.$$

and

$$\left|g(t,x,y)\right|\le p_{21}\left(t\right)\left|x\right|+p_{22}\left(t\right)\left|y\right|,x,y\in R.$$

After transforming the differential equation into a matrix equation, we can solve the equation with the help of various operational properties and methods of matrices. Based on matrix analysis methods, various controllers can be designed to achieve control of the systems described by differential equations. In [32], Ji and Liu constructed a Lyapunov-like function, and in combination with matrix inequalities, proposed a design method for the feedback controller under specific conditions. When studying elliptic differential equations, Kuznetsov et al. also introduced a large number of matrices, which provided convenience for the research [33].

Moreover, many scholars have conducted in-depth research on the properties such as the existence, uniqueness, and stability of solutions to fractional-order differential equations under multi-point coupled boundary value conditions by using tools like the fixed-point theorem and have achieved a series of important results. In practical applications, coupled boundary conditions may involve uncertainties [34]. When uncertainties exist in boundary conditions, the analysis of coupled systems will become more complicated. In [35], Samadi et al. studied the system of nonlinear Caputo fractional-order differential equations with the P-Laplacian operator under coupled nonlocal boundary value conditions and obtained the existence results of solutions for the differential equations system by applying the fixed-point theorem. In [36], Salem and Alghamdi studied the nonlinear fractional-order Langevin equation under multi-point coupled integral boundary value conditions and obtained the uniqueness results of the solution by applying the fixed-point theorem.

Inspired by the above papers [9,24,35,36], we study the system of tempered fractional differential equations with multi-point boundary value conditions (1). This paper makes improvements and generalizations on the basis of references [9,24]. First, what this paper studies is the tempered fractional derivative, which can be regarded as a generalization of the Riemann–Liouville fractional derivative, and they can be transformed into each other under certain conditions. For example, when $\lambda =0$, we can find that ${}_0{}^R\mathbb{D}_t^{{\alpha }_i,\lambda }$ is equivalent to ${}_0{}^R\mathcal{D}_t^{{\alpha }_i}$. Secondly, this paper obtains the corresponding Green’s function and gives the proof of the properties of Green’s function. Next, by applying matrix analysis methods, using the properties of Green’s function, and the properties of the integral-operator T, we obtain the corresponding Lyapunov-type inequalities in two Banach spaces $E\times E$ and $E_1\times E_2$. Finally, by introducing a new function $g_{ij}\left(s\right)$, we generalize the second Lyapunov inequality and obtain the third Lyapunov inequality. And under certain conditions, we compare the first Lyapunov inequality with the third Lyapunov inequality; the relative stability results of the matrices of these two Lyapunov inequalities are obtained.

The organization of this paper as follows. In Section 2, we give some definitions and lemmas of Riemann–Liouville fractional differential equations. We obtain Green’s function and prove some properties of Green’s function. In Section 3, we introduce the method of matrix analysis, give three kinds of Lyapunov inequalities, and prove the relative superiority of two of them under certain conditions. In Section 4, we provide an example to verify our results. In Section 5, the conclusions about this paper are given.

2. Preliminaries

In this section, we first present some definitions, lemmas, and theorems that we need in this article.

Definition 1

([2,4]). Assume that $f\left(t\right)\in L[0,1]$ is piece-wise continuous on $(0,1)$, $\alpha >0$, $\lambda \ge 0$. Then the Riemann–Liouville tempered fractional derivative of order $\alpha >0$ for f is defined as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {}_0{}^R\mathbb{D}_t^{\alpha ,\lambda }f\left(t\right)& =e^{-\lambda t}{\mathcal{D}}_{0^{+}}^{\alpha }\left(e^{\lambda t}f\left(t\right)\right)\hfill \\ & ={\displaystyle \frac{e^{-\lambda t}}{\Gamma (n-\alpha )}}{\left({\displaystyle \frac{d}{dt}}\right)}^{\left(n\right)}{\int }_0^t{(t-s)}^{n-\alpha -1}f\left(s\right)e^{\lambda s}ds,t\in [0,1],n=\left[\alpha \right]+1,\hfill \end{array}\end{array}$$

where ${\mathcal{D}}_{0^{+}}^{\alpha }$ is the Riemann–Liouville fractional derivative

$${\mathcal{D}}_{0^{+}}^{\alpha }f\left(t\right)={\displaystyle \frac{1}{\Gamma (n-\alpha )}}{\left({\displaystyle \frac{d}{dt}}\right)}^{\left(n\right)}{\int }_0^t{(t-s)}^{n-\alpha -1}f\left(s\right)ds,t\in [0,1],n=\left[\alpha \right]+1,$$

provided that the right-hand side is point-wise defined on $(0,+\infty )$ and denotes the integral part of number α, where $\Gamma (n-\alpha )$ is Euler gamma function.

Definition 2

([2,4]). Assume that $f\left(t\right)\in L[0,1]$ is piece-wise continuous on $(0,1)$, $\alpha >0$, $\lambda \ge 0$. Then, the Riemann–Liouville tempered fractional integral of order $\alpha >0$ for f is defined as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {}_0{}^R\mathbb{I}_t^{\alpha ,\lambda }f\left(t\right)& =e^{-\lambda t}{I}_{0^{+}}^{\alpha }\left(e^{\lambda t}f\left(t\right)\right)\hfill \\ & ={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{(t-s)}^{\alpha -1}{e}^{-\lambda (t-s)}f\left(s\right)ds,t\in [0,1],\hfill \end{array}\end{array}$$

where $I_{0^{+}}^{\alpha }$ is the Riemann–Liouville fractional integral

$$I_{0^{+}}^{\alpha }f\left(t\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t{(t-s)}^{\alpha -1}f\left(s\right)ds,t\in [0,1],$$

provided that the right-hand side is point-wise defined on $(0,+\infty )$, where $\Gamma \left(\alpha \right)$ is Euler gamma function.

Lemma 1

([20,26]). Let $g\left(t\right)\in C[0,1]\cap L^1[0,1]$, $n-1<\alpha <n$, $n\in N^{+}$, then

$${}_0{}^R\mathbb{I}_t^{\alpha ,\lambda }{}_0{}^R\mathbb{D}_t^{\alpha ,\lambda }g\left(t\right)=g\left(t\right)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_n{t}^{\alpha -n},$$

where $c_i\in R$, $i=1,2,\dots ,n$.

For convenience, we list the following assumptions that we need:

$\left({\mathrm{H}}_1\right)$

$a_{ij}\ge 0(i=1,2,3,4;j=1,2,\dots ,n)$, ${\xi }_i\in (0,1)(i=1,2,\dots ,n)$, ${\eta }_j\in (0,1)(j=1,2,\dots ,n)$, ${\gamma }_{ij}\ge 0(i,j=1,2)$ and $\gamma \stackrel{△}{=}{\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}{\gamma }_{21}>0$, where

$$\begin{array}{c}\hfill \begin{array}{c}\hfill {\gamma }_{11}=1-\sum _{i=1}^n{a}_{1i}{\xi }_i^{{\alpha }_1-1}{e}^{\lambda (1-{\xi }_i)},{\gamma }_{12}=\sum _{j=1}^n{a}_{2j}{\eta }_j^{{\alpha }_2-1}{e}^{\lambda (1-{\eta }_j)},\\ \hfill {\gamma }_{21}=\sum _{i=1}^n{a}_{3i}{\xi }_i^{{\alpha }_1-1}{e}^{\lambda (1-{\xi }_i)},{\gamma }_{22}=1-\sum _{j=1}^n{a}_{4j}{\eta }_j^{{\alpha }_2-1}{e}^{\lambda (1-{\eta }_j)}.\end{array}\end{array}$$

$\left({\mathrm{H}}_2\right)$

$f,g:[0,1]\times R\times R\to R$ are continuous functions.

$\left({\mathrm{H}}_3\right)$

$0\le \lambda \le {\displaystyle \frac{\alpha -1}{1-{\xi }_1}}$, where $\alpha =min\left\{{\alpha }_1,{\alpha }_2\right\}$.

We firstly study the system of tempered fractional differential equation as follows:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{}_0{}^R\mathbb{D}_t^{{\alpha }_1,\lambda }x\left(t\right)+{\Phi }_1\left(t\right)=0,t\in (0,1),\hfill \\ {\displaystyle x\left(0\right)=0,x\left(1\right)=\sum _{i=1}^n{a}_{1i}x\left({\xi }_i\right)+\sum _{j=1}^n{a}_{2j}y\left({\eta }_j\right),}\hfill \\ {}_0{}^R\mathbb{D}_t^{{\alpha }_2,\lambda }y\left(t\right)+{\Phi }_2\left(t\right)=0,t\in (0,1),\hfill \\ {\displaystyle y\left(0\right)=0,y\left(1\right)=\sum _{i=1}^n{a}_{3i}x\left({\xi }_i\right)+\sum _{j=1}^n{a}_{4j}y\left({\eta }_j\right).}\hfill \end{array}\right.\end{array}\end{array}$$

(9)

Lemma 2.

Let ${\Phi }_1,{\Phi }_2\in C[0,1]$, if $(x,y)$ is a solution of the integral equation

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle x\left(t\right)={\int }_0^1{G}_{11}(t,s){\Phi }_1\left(s\right)ds+{\int }_0^1{G}_{12}(t,s){\Phi }_2\left(s\right)ds,}\\ {\displaystyle y\left(t\right)={\int }_0^1{G}_{21}(t,s){\Phi }_1\left(s\right)ds+{\int }_0^1{G}_{22}(t,s){\Phi }_2\left(s\right)ds,}\end{array}\right.\end{array}\end{array}$$

(10)

then $(x,y)$ is a solution of System (9), where

$$\begin{array}{c}\hfill \begin{array}{cc}& G_{11}(t,s)=G_{{\alpha }_1}(t,s)+{\displaystyle \frac{t^{{\alpha }_1-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right)G_{{\alpha }_1}\left({\xi }_i,s\right),\hfill \\ & G_{12}(t,s)={\displaystyle \frac{t^{{\alpha }_1-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{j=1}^n\left({\gamma }_{22}{a}_{2j}+{\gamma }_{12}{a}_{4j}\right)G_{{\alpha }_2}\left({\eta }_j,s\right),\hfill \\ & G_{21}(t,s)={\displaystyle \frac{t^{{\alpha }_2-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{i=1}^n\left({\gamma }_{21}{a}_{1i}+{\gamma }_{11}{a}_{3i}\right)G_{{\alpha }_1}\left({\xi }_i,s\right),\hfill \\ & G_{22}(t,s)=G_{{\alpha }_2}(t,s)+{\displaystyle \frac{t^{{\alpha }_2-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{j=1}^n\left({\gamma }_{21}{a}_{2j}+{\gamma }_{11}{a}_{4j}\right)G_{{\alpha }_2}\left({\eta }_j,s\right).\hfill \end{array}\end{array}$$

(11)

and

$$\begin{array}{c}\hfill \begin{array}{c}\hfill G_{{\alpha }_i}(t,s)={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}\left\{\begin{array}{c}{\displaystyle [t^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}-{(t-s)}^{{\alpha }_i-1}]e^{\lambda (s-t)},0\le s\le t\le 1,i=1,2,}\\ {\displaystyle \left[t^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-t)},0\le t\le s\le 1,i=1,2.}\end{array}\right.\end{array}\end{array}$$

(12)

Proof. 

For system (9), by (2) and Lemma 1, we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill e^{\lambda t}x\left(t\right)=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda s}{\Phi }_1\left(s\right)ds+c_1{t}^{{\alpha }_1-1}+c_2{t}^{{\alpha }_1-2},\\ \hfill e^{\lambda t}y\left(t\right)=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda s}{\Phi }_2\left(s\right)ds+d_1{t}^{{\alpha }_2-1}+d_2{t}^{{\alpha }_2-2}.\end{array}\end{array}$$

(13)

From the boundary conditions $x\left(0\right)=0$ and $y\left(0\right)=0$, we can infer that $c_2=0$ and $d_2=0$. Therefore, we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill e^{\lambda t}x\left(t\right)=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda s}{\Phi }_1\left(s\right)ds+c_1{t}^{{\alpha }_1-1},\\ \hfill e^{\lambda t}y\left(t\right)=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda s}{\Phi }_2\left(s\right)ds+d_1{t}^{{\alpha }_2-1}.\end{array}\end{array}$$

(14)

Let $x\left(1\right)=c$, and $y\left(1\right)=d$, so we conclude that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill e^{\lambda }x\left(1\right)=-{\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda s}{\Phi }_1\left(s\right)ds+c_1,\\ \hfill e^{\lambda }y\left(1\right)=-{\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda s}{\Phi }_2\left(s\right)ds+d_1,\end{array}\end{array}$$

then we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill c=x\left(1\right)=-{\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda (s-1)}{\Phi }_1\left(s\right)ds+c_1{e}^{-\lambda },\\ \hfill d=y\left(1\right)=-{\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda (s-1)}{\Phi }_2\left(s\right)ds+d_1{e}^{-\lambda }.\end{array}\end{array}$$

(15)

By (15), we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill c_1={\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda s}{\Phi }_1\left(s\right)ds+c{e}^{\lambda },\\ \hfill d_1={\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda s}{\Phi }_2\left(s\right)ds+d{e}^{\lambda }.\end{array}\end{array}$$

By (14), we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill e^{\lambda t}x\left(t\right)=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda s}{\Phi }_1\left(s\right)ds+({\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda s}{\Phi }_1\left(s\right)ds+c{e}^{\lambda })t^{{\alpha }_1-1},\\ \hfill e^{\lambda t}y\left(t\right)=-{\int }_0^t{\displaystyle \frac{{(t-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda s}{\Phi }_2\left(s\right)ds+({\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda s}{\Phi }_2\left(s\right)ds+d{e}^{\lambda })t^{{\alpha }_2-1},\end{array}\end{array}$$

namely,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill x\left(t\right)={\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_1-1}{t}^{{\alpha }_1-1}-{(t-s)}^{{\alpha }_1-1}}{\Gamma \left({\alpha }_1\right)}}{e}^{\lambda (s-t)}{\Phi }_1\left(s\right)ds+c{t}^{{\alpha }_1-1}{e}^{\lambda (1-t)},\\ \hfill y\left(t\right)={\int }_0^1{\displaystyle \frac{{(1-s)}^{{\alpha }_2-1}{t}^{{\alpha }_2-1}-{(t-s)}^{{\alpha }_2-1}}{\Gamma \left({\alpha }_2\right)}}{e}^{\lambda (s-t)}{\Phi }_2\left(s\right)ds+d{t}^{{\alpha }_2-1}{e}^{\lambda (1-t)}.\end{array}\end{array}$$

Hence, (13) implies

$$\begin{array}{c}\hfill \begin{array}{c}\hfill x\left(t\right)={\int }_0^1{G}_{{\alpha }_1}(t,s){\Phi }_1\left(s\right)ds+c{t}^{{\alpha }_1-1}{e}^{\lambda (1-t)},\\ \hfill y\left(t\right)={\int }_0^1{G}_{{\alpha }_2}(t,s){\Phi }_2\left(s\right)ds+d{t}^{{\alpha }_2-1}{e}^{\lambda (1-t)},\end{array}\end{array}$$

(16)

where $G_{{\alpha }_i}(i=1,2)$ are defined by (12).

It order to determine c and d, by (9), (15), and (16) we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill c=x\left(1\right)=\sum _{i=1}^n{a}_{1i}{\int }_0^1{G}_{{\alpha }_1}({\xi }_i,s){\Phi }_1\left(s\right)ds+\sum _{j=1}^n{a}_{2j}{\int }_0^1{G}_{{\alpha }_2}({\eta }_j,s){\Phi }_2\left(s\right)ds\\ \hfill +c\sum _{i=1}^n{a}_{1i}{\left({\xi }_i\right)}^{{\alpha }_1-1}{e}^{\lambda (1-{\xi }_i)}+d\sum _{j=1}^n{a}_{2j}{\left({\eta }_j\right)}^{{\alpha }_2-1}{e}^{\lambda (1-{\eta }_j)},\\ \hfill d=y\left(1\right)=\sum _{i=1}^n{a}_{3i}{\int }_0^1{G}_{{\alpha }_1}({\xi }_i,s){\Phi }_1\left(s\right)ds+\sum _{j=1}^n{a}_{4j}{\int }_0^1{G}_{{\alpha }_2}({\eta }_j,s){\Phi }_2\left(s\right)ds\\ \hfill +c\sum _{i=1}^n{a}_{3i}{\left({\xi }_i\right)}^{{\alpha }_1-1}{e}^{\lambda (1-{\xi }_i)}+d\sum _{j=1}^n{a}_{4j}{\left({\eta }_j\right)}^{{\alpha }_2-1}{e}^{\lambda (1-{\eta }_j)}.\end{array}\end{array}$$

Through simple simplification, by assumption $\left({\mathrm{H}}_1\right)$, we can obtain

$$\begin{array}{c}\hfill \begin{array}{c}\hfill c=\sum _{i=1}^n{a}_{1i}{\int }_0^1{G}_{{\alpha }_1}({\xi }_i,s){\Phi }_1\left(s\right)ds+\sum _{j=1}^n{a}_{2j}{\int }_0^1{G}_{{\alpha }_2}({\eta }_j,s){\Phi }_2\left(s\right)ds+c(1-{\gamma }_{11})+d{\gamma }_{12},\\ \hfill d=\sum _{i=1}^n{a}_{3i}{\int }_0^1{G}_{{\alpha }_1}({\xi }_i,s){\Phi }_1\left(s\right)ds+\sum _{j=1}^n{a}_{4j}{\int }_0^1{G}_{{\alpha }_2}({\eta }_j,s){\Phi }_2\left(s\right)ds+c{\gamma }_{21}+d(1-{\gamma }_{22}).\end{array}\end{array}$$

Writing the above two equations in matrix form, we can obtain

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left(\begin{array}{cc}{\gamma }_{11}& -{\gamma }_{12}\\ -{\gamma }_{21}& {\gamma }_{22}\end{array}\right)\left(\begin{array}{c}c\\ d\end{array}\right)\end{array}=\left(\begin{array}{c}{\displaystyle \sum _{i=1}^n{a}_{1i}{\int }_0^1{G}_{{\alpha }_1}({\xi }_i,s){\Phi }_1\left(s\right)ds+\sum _{j=1}^n{a}_{2j}{\int }_0^1{G}_{{\alpha }_2}({\eta }_j,s){\Phi }_2\left(s\right)ds}\\ {\displaystyle \sum _{i=1}^n{a}_{3i}{\int }_0^1{G}_{{\alpha }_1}({\xi }_i,s){\Phi }_1\left(s\right)ds+\sum _{j=1}^n{a}_{4j}{\int }_0^1{G}_{{\alpha }_2}({\eta }_j,s){\Phi }_2\left(s\right)ds}\end{array}\right).\end{array}$$

By $\left({\mathrm{H}}_1\right)$, since $\gamma ={\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}{\gamma }_{21}\ne 0$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill c=& {\displaystyle \frac{{\gamma }_{22}}{\gamma }}\sum _{i=1}^n{a}_{1i}{\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds+{\displaystyle \frac{{\gamma }_{12}}{\gamma }}\sum _{i=1}^n{a}_{3i}{\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds\hfill \\ & +{\displaystyle \frac{{\gamma }_{22}}{\gamma }}\sum _{j=1}^n{a}_{2j}{\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds+{\displaystyle \frac{{\gamma }_{12}}{\gamma }}\sum _{j=1}^n{a}_{4j}{\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds\hfill \\ \hfill =& {\displaystyle \frac{1}{\gamma }}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right){\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds\hfill \\ & +{\displaystyle \frac{1}{\gamma }}\sum _{j=1}^n\left({\gamma }_{22}{a}_{2j}+{\gamma }_{12}{a}_{4j}\right){\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds,\hfill \end{array}\end{array}$$

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill d=& {\displaystyle \frac{{\gamma }_{21}}{\gamma }}\sum _{i=1}^n{a}_{1i}{\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds+{\displaystyle \frac{{\gamma }_{11}}{\gamma }}\sum _{i=1}^n{a}_{3i}{\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds\hfill \\ & +{\displaystyle \frac{{\gamma }_{21}}{\gamma }}\sum _{j=1}^n{a}_{2j}{\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds+{\displaystyle \frac{{\gamma }_{11}}{\gamma }}\sum _{j=1}^n{a}_{4j}{\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds\hfill \\ \hfill =& {\displaystyle \frac{1}{\gamma }}\sum _{i=1}^n\left({\gamma }_{21}{a}_{1i}+{\gamma }_{11}{a}_{3i}\right){\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds\hfill \\ & +{\displaystyle \frac{1}{\gamma }}\sum _{j=1}^n\left({\gamma }_{21}{a}_{2j}+{\gamma }_{11}{a}_{4j}\right){\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds.\hfill \end{array}\end{array}$$

Hence, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill x\left(t\right)& ={\int }_0^1{G}_{{\alpha }_1}(t,s){\Phi }_1\left(s\right)ds+{\displaystyle \frac{t^{{\alpha }_1-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right){\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds\hfill \\ & +{\displaystyle \frac{t^{{\alpha }_1-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{j=1}^n\left({\gamma }_{22}{a}_{2j}+{\gamma }_{12}{a}_{4j}\right){\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds,\hfill \\ & ={\int }_0^1{G}_{11}(t,s){\Phi }_1\left(s\right)ds+{\int }_0^1{G}_{12}(t,s){\Phi }_2\left(s\right)ds,\hfill \\ \hfill y\left(t\right)& ={\int }_0^1{G}_{{\alpha }_2}(t,s){\Phi }_2\left(s\right)ds+{\displaystyle \frac{t^{{\alpha }_2-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{i=1}^n\left({\gamma }_{21}{a}_{1i}+{\gamma }_{11}{a}_{3i}\right){\int }_0^1{G}_{{\alpha }_1}\left({\xi }_i,s\right){\Phi }_1\left(s\right)ds\hfill \\ & +{\displaystyle \frac{t^{{\alpha }_2-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{j=1}^n\left({\gamma }_{21}{a}_{2j}+{\gamma }_{11}{a}_{4j}\right){\int }_0^1{G}_{{\alpha }_2}\left({\eta }_j,s\right){\Phi }_2\left(s\right)ds,\hfill \\ & ={\int }_0^1{G}_{21}(t,s){\Phi }_1\left(s\right)ds+{\int }_0^1{G}_{22}(t,s){\Phi }_2\left(s\right)ds.\hfill \end{array}\end{array}$$

Then, the proof is complete. □

Lemma 3.

Green’s function $G_{{\alpha }_i}(t,s)(i=1,2)$ has the following properties:

$\left(1\right)$

For any $t,s\in [0,1]$, $G_{{\alpha }_i}(t,s)\ge 0(i=1,2);$

$\left(2\right)$

${\displaystyle \underset{t\in [0,1]}{max}{G}_{{\alpha }_i}(t,s)=G_{{\alpha }_i}(s,s),s\in [0,1];}$

$\left(3\right)$

$G_{{\alpha }_i}(s,s)$ has a unique maximum, which is given by

$$\underset{s\in [0,1]}{max}{G}_{{\alpha }_i}(s,s)=G_{{\alpha }_i}({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}})={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_i-1},i=1,2;$$

$\left(4\right)$

For any $t,s\in [0,1]$, ${\displaystyle G_{{\alpha }_i}(t,s)\le \frac{1}{\Gamma \left({\alpha }_i\right)}\left[t^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-t)}(i=1,2)}$.

Proof. 

(1)

If $0\le t\le s\le 1$, we have

$$G_{{\alpha }_i}(t,s)={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}\left[t^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-t)}.$$

It is obvious that $G_{{\alpha }_i}(t,s)\ge 0$.

And, if $0\le s\le t\le 1$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill G_{{\alpha }_i}(t,s)& ={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}[t^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}-{(t-s)}^{{\alpha }_i-1}]e^{\lambda (s-t)},\hfill \\ & ={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}[{(t-ts)}^{{\alpha }_i-1}-{(t-s)}^{{\alpha }_i-1}]e^{\lambda (s-t)}.\hfill \end{array}\end{array}$$

(17)

Since $t-s\ge 0$ and $0<{\alpha }_i-1\le 1$, and $ts\le s$, we have $[{(t-ts)}^{{\alpha }_i-1}-{(t-s)}^{{\alpha }_i-1}]\ge 0$. Hence, by (17), we have $G_{{\alpha }_i}(t,s)\ge 0$. Then, we conclude the proof of $\left(1\right)$.

$\left(2\right)$

$G_{{\alpha }_i}(t,s)$ is an increasing function when $0\le t\le s\le 1$.

On the other hand, we only need to determine the monotone property of $G_{{\alpha }_i}(t,s)$ when $0\le s\le t\le 1$. For $t\in [s,1]$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\displaystyle \frac{d{G}_{{\alpha }_i}(t,s)}{dt}}& ={\displaystyle \frac{({\alpha }_i-1)}{\Gamma \left({\alpha }_i\right)}}\left[t^{{\alpha }_i-2}{(1-s)}^{{\alpha }_i-1}-{(t-s)}^{{\alpha }_i-2}\right]e^{\lambda (s-t)}\hfill \\ & -{\displaystyle \frac{\lambda }{\Gamma \left({\alpha }_i\right)}}\left[t^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}-{(t-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-t)},\hfill \end{array}\end{array}$$

(18)

since

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left[t^{{\alpha }_i-2}{(1-s)}^{{\alpha }_i-1}-{(t-s)}^{{\alpha }_i-2}\right]=t^{{\alpha }_i-2}\left[{(1-s)}^{{\alpha }_i-1}-{(1-{\displaystyle \frac{s}{t}})}^{{\alpha }_i-2}\right],\end{array}\end{array}$$

$-1<{\alpha }_i-2\le 0$, $1-s\ge 0$, $-{(1-{\displaystyle \frac{s}{t}})}^{{\alpha }_i-2}$ is an increasing function on $t\in [s,1]$, so we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}& \left[{(1-s)}^{{\alpha }_i-1}-{(1-{\displaystyle \frac{s}{t}})}^{{\alpha }_i-2}\right]\hfill \\ & \le \left[{(1-s)}^{{\alpha }_i-1}-{(1-s)}^{{\alpha }_i-2}\right]\hfill \\ & ={(1-s)}^{{\alpha }_i-2}(1-s-1)\le 0.\hfill \end{array}\end{array}$$

(19)

Hence, by (17) and (19), we also have ${\displaystyle \frac{d{G}_{{\alpha }_i}(t,s)}{dt}\le 0}$. Then, $G_{{\alpha }_i}(t,s)$ is a decreasing function and

$$G_{{\alpha }_i}(t,s)\le G_{{\alpha }_i}(s,s).$$

Then, we conclude the proof of $\left(2\right)$.

$\left(3\right)$

${\displaystyle G_{{\alpha }_i}(s,s)=\frac{1}{\Gamma \left({\alpha }_i\right)}\left[s^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}\right](i=1,2)}$, $s\in [0,1]$. Then, $G_{{\alpha }_i}(0,0)=G_{{\alpha }_i}(1,1)=0$, for $s\in (0,1)$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\displaystyle \frac{d{G}_{{\alpha }_i}(s,s)}{ds}}& ={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}\left[({\alpha }_i-1)s^{{\alpha }_i-2}{(1-s)}^{{\alpha }_i-1}-({\alpha }_i-1)s^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-2}\right]\hfill \\ & ={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}({\alpha }_i-1)s^{{\alpha }_i-2}{(1-s)}^{{\alpha }_i-2}(1-2s).\hfill \end{array}\end{array}$$

We have ${\displaystyle \frac{d{G}_{{\alpha }_i}(s,s)}{ds}=0}$ at ${\displaystyle s=\frac{1}{2}}$, and ${\displaystyle \frac{d{G}_{{\alpha }_i}(s,s)}{ds}>0}$ for ${\displaystyle s<\frac{1}{2}}$ and ${\displaystyle \frac{d{G}_{{\alpha }_i}(s,s)}{ds}<0}$ for ${\displaystyle s>\frac{1}{2}}$. Then, we conclude the proof of $\left(3\right)$.

$\left(4\right)$

By (12), it is obvious that property $\left(4\right)$ holds.

□

Lemma 4.

For $0<{\xi }_i<1$, and we have

$$\underset{s\in [0,1]}{max}{G}_{{\alpha }_i}({\xi }_i,s)=G_{{\alpha }_i}({\xi }_i,{\xi }_i)={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}{\xi }_i^{{\alpha }_i-1}{(1-{\xi }_i)}^{{\alpha }_i-1}.$$

Proof. 

Firstly, we define the two functions $g_1\left(s\right),g_2\left(s\right)$ as follows:

$$g_1\left(s\right)={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}\left[{\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}-{({\xi }_i-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-{\xi }_i)},0\le s\le {\xi }_i<1,$$

and

$$g_2\left(s\right)={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}\left[{\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-{\xi }_i)},0<{\xi }_i\le s\le 1.$$

For $s\in [0,{\xi }_i]$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\displaystyle \frac{d{g}_2\left(s\right)}{ds}}& ={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}\left[-{\xi }_i^{{\alpha }_i-1}({\alpha }_i-1){(1-s)}^{{\alpha }_i-2}\right]e^{\lambda (s-{\xi }_i)}+{\displaystyle \frac{\lambda }{\Gamma \left({\alpha }_i\right)}}\left[{\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-{\xi }_i)}\hfill \\ & ={\displaystyle \frac{e^{\lambda (s-{\xi }_i)}}{\Gamma \left({\alpha }_i\right)}}{\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-2}[-({\alpha }_i-1)+\lambda (1-s)].\hfill \end{array}\end{array}$$

Since $0\le \lambda \le {\displaystyle \frac{\alpha -1}{1-{\xi }_1}}$, it is obvious that for $s\in [{\xi }_i,1]$, $g_2\left(s\right)$ is monotonically decreasing, then we have

$$g_2\left(s\right)\le g_2\left({\xi }_i\right)=G_{{\alpha }_i}({\xi }_i,{\xi }_i)={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}{\xi }_i^{{\alpha }_i-1}{(1-{\xi }_i)}^{{\alpha }_i-1},s\in [{\xi }_i,1].$$

Next, we consider the monotonically of $g_1\left(s\right)$. If $0\le s\le {\xi }_i<1$, $1<{\alpha }_i\le 2(i=1,2)$, then we have $0\le ({\xi }_i-s)\le ({\xi }_i-{\xi }_{i}s)={\xi }_i(1-s)$, and thus ${({\xi }_i-s)}^{{\alpha }_i-1}\le {\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}$. For $s\in [0,{\xi }_i]$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\displaystyle \frac{d{g}_1\left(s\right)}{ds}}& ={\displaystyle \frac{1}{\Gamma \left({\alpha }_i\right)}}\left[-{\xi }_i^{{\alpha }_i-1}({\alpha }_i-1){(1-s)}^{{\alpha }_i-2}+({\alpha }_i-1){({\xi }_i-s)}^{{\alpha }_i-2}\right]e^{\lambda (s-{\xi }_i)}\hfill \\ & +{\displaystyle \frac{\lambda }{\Gamma \left({\alpha }_i\right)}}\left[{\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}-{({\xi }_i-s)}^{{\alpha }_i-1}\right]e^{\lambda (s-{\xi }_i)}.\hfill \end{array}\end{array}$$

Then, we have ${\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-1}-{({\xi }_i-s)}^{{\alpha }_i-1}\ge 0$, and since ${({\xi }_i-s)}^{{\alpha }_i-2}>{(1-s)}^{{\alpha }_i-2}>{\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-2}$, then ${({\xi }_i-s)}^{{\alpha }_i-2}-{\xi }_i^{{\alpha }_i-1}{(1-s)}^{{\alpha }_i-2}>0$. Thus, $g_1\left(s\right)$ is an increasing function on $s\in [0,{\xi }_i]$; then, we have $g_1\left(s\right)\le g_1\left({\xi }_i\right)=G_{{\alpha }_i}({\xi }_i,{\xi }_i)$. □

Lemma 5.

For $G_{ij}(t,s)(i,j=1,2)$, which, defined in Lemma 2, satisfy the following inequalities:

$\left(1\right)$

for any $t,s\in [0,1]$, $G_{ij}(t,s)\le {\lambda }_{ij}$;

$\left(2\right)$

for any $t,s\in [0,1]$, $G_{ij}(t,s)\le {\mu }_{ij}{t}^{{\alpha }_i-1}{(1-s)}^{{\alpha }_j-1}{e}^{\lambda (s-t)}$, where ${\lambda }_{ij},{\mu }_{ij}(i,j=1,2)$ are defined by

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\lambda }_{11}& ={\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}+{\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_1\right)}}\sum _{i=1}^n({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}){\xi }_i^{{\alpha }_1-1}{(1-{\xi }_i)}^{{\alpha }_1-1},\hfill \\ \hfill {\lambda }_{12}& ={\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_2\right)}}\sum _{j=1}^n({\gamma }_{22}{a}_{2j}+{\gamma }_{12}{a}_{4j}){\eta }_j^{{\alpha }_2-1}{(1-{\eta }_j)}^{{\alpha }_2-1},\hfill \\ \hfill {\lambda }_{21}& ={\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_2-1}+{\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_1\right)}}\sum _{i=1}^n({\gamma }_{21}{a}_{1i}+{\gamma }_{11}{a}_{3i}){\xi }_i^{{\alpha }_1-1}{(1-{\xi }_i)}^{{\alpha }_1-1},\hfill \\ \hfill {\lambda }_{22}& ={\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_2\right)}}\sum _{j=1}^n({\gamma }_{21}{a}_{2j}+{\gamma }_{11}{a}_{4j}){\eta }_j^{{\alpha }_2-1}{(1-{\eta }_j)}^{{\alpha }_2-1},\hfill \end{array}\end{array}$$

and

$$\begin{array}{c}\hfill \begin{array}{c}\hfill {\mu }_{11}={\displaystyle \frac{{\gamma }_{22}}{\gamma \Gamma \left({\alpha }_1\right)}},{\mu }_{12}={\displaystyle \frac{{\gamma }_{12}}{\gamma \Gamma \left({\alpha }_2\right)}},{\mu }_{21}={\displaystyle \frac{{\gamma }_{21}}{\gamma \Gamma \left({\alpha }_1\right)}},{\mu }_{22}={\displaystyle \frac{{\gamma }_{11}}{\gamma \Gamma \left({\alpha }_2\right)}}.\end{array}\end{array}$$

Proof. 

For simplicity, we only prove $G_{11}(t,s)$. For the proof of $G_{ij}(t,s)(i,j=1,2)$, we can use in the same method.

By Lemmas 2–4, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill G_{11}(t,s)& =G_{{\alpha }_1}(t,s)+{\displaystyle \frac{t^{{\alpha }_1-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right)G_{{\alpha }_1}\left({\xi }_i,s\right)\hfill \\ & \le G_{{\alpha }_i}(s,s)+{\displaystyle \frac{1}{\gamma }}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right)G_{{\alpha }_1}\left({\xi }_i,{\xi }_i\right)\hfill \\ & \le {\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}+{\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_1\right)}}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right){\xi }_i^{{\alpha }_1-1}{(1-{\xi }_1)}^{{\alpha }_1-1}\hfill \\ & ={\lambda }_{11},\hfill \end{array}\end{array}$$

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill G_{11}(t,s)& =G_{{\alpha }_1}(t,s)+{\displaystyle \frac{t^{{\alpha }_1-1}{e}^{\lambda (1-t)}}{\gamma }}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right)G_{{\alpha }_1}\left({\xi }_i,s\right)\hfill \\ & \le {\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}\left[t^{{\alpha }_1-1}{(1-s)}^{{\alpha }_1-1}\right]e^{\lambda (s-t)}+{\displaystyle \frac{t^{{\alpha }_1-1}{e}^{\lambda (1-t)}}{\gamma \Gamma \left({\alpha }_1\right)}}\sum _{i=1}^n\left({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}\right){\xi }^{{\alpha }_1-1}{(1-s)}^{{\alpha }_1-1}{e}^{\lambda (s-{\xi }_i)}\hfill \\ & ={\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}\left[t^{{\alpha }_1-1}{(1-s)}^{{\alpha }_1-1}\right]e^{\lambda (s-t)}\left[1+{\displaystyle \frac{{\gamma }_{22}{\sum }_{i=1}^n{a}_{1i}+{\gamma }_{12}{\sum }_{i=1}^n{a}_{3i}}{\gamma }}{\xi }^{{\alpha }_1-1}{e}^{\lambda (1-{\xi }_i)}\right]\hfill \\ & ={\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}\left[t^{{\alpha }_1-1}{(1-s)}^{{\alpha }_1-1}\right]e^{\lambda (s-t)}\left[1+{\displaystyle \frac{{\gamma }_{22}(1-{\gamma }_{11})+{\gamma }_{12}{\gamma }_{21}}{\gamma }}\right]\hfill \\ & ={\displaystyle \frac{{\gamma }_{22}}{\gamma \Gamma \left({\alpha }_1\right)}}\left[t^{{\alpha }_1-1}{(1-s)}^{{\alpha }_1-1}\right]e^{\lambda (s-t)}\hfill \\ & ={\mu }_{11}{t}^{{\alpha }_1-1}{(1-s)}^{{\alpha }_1-1}{e}^{\lambda (s-t)}.\hfill \end{array}\end{array}$$

Then, the proof is complete. □

3. Main Results

In this section, we will apply non-negative matrices to represent the Lyapunov-type inequalities of System (1). In the subsequent proof, we will utilize the definition of the nonnegative matrix. A nonnegative matrix converges to zero if

$$M^k\to 0,k\to +\infty .$$

We define a square matrix $M={\left(a_{ij}\right)}_{2\times 2}$ as nonnegative and write $M\ge 0$ if $a_{ij}\ge 0$ for $i,j=1,2$. Some other properties of the real square matrices can be found in paper [37]. For example, if $M_1$ and $M_2$ are two real square matrices and if $M_1-M_2\ge 0$, then we say $M_1\ge M_2$. We define $M_2^{+}$ as the set of non-negative matrices; let Trace$\left(M\right)$, det$\left(M\right)$, and $\rho \left(M\right)$ be the trace, determinant, and spectral radius of the matrix M, respectively [24]. For Trace$\left(M\right)$, det$\left(M\right)$, and $\rho \left(M\right)$, we have the following lemmas.

Lemma 6

([37]). Let $C\in M_2^{+}$, if $\rho \left(C\right)<1$, then

$$\underset{n\to +\infty }{lim}{C}^n=0.$$

Lemma 7

([24]). Let $C\in M_2^{+}$. Then,

$$\rho \left(C\right)={\displaystyle \frac{Trace\left(C\right)+\sqrt{{\left[Trace\left(C\right)\right]}^2-4det\left(C\right)}}{2}}.$$

Lemma 8

([38]). If $M_1\ge M_2\ge 0$, then

$$\rho \left(M_2\right)\le \rho \left(M_1\right).$$

The spectral radius is a function of the eigenvalues of a matrix. It is the supremum of the absolute values of the eigenvalues. The spectral radius of a matrix is smaller; the matrix is more stable in a relative sense. Conversely, the matrix has complex dynamic characteristics. Next, by using some properties of Green’s functions to define multiple expressions of solutions (10) and combining some properties of the spectral radius, the stability results of matrices are obtained.

Next, we list the following assumptions used in this paper:

$\left({\mathrm{H}}_4\right)$

There are two positive functions $m_{11}\left(t\right)$ and $m_{12}\left(t\right)$; for $t\in [0,1]$, we have

$$\left|f(t,x,y)\right|\le m_{11}\left(t\right)\left|x\right|+m_{12}\left(t\right)\left|y\right|,x,y\in R.$$

$\left({\mathrm{H}}_5\right)$

There are two positive functions $m_{21}\left(t\right)$ and $m_{22}\left(t\right)$; for $t\in [0,1]$, we have

$$\left|g(t,x,y)\right|\le m_{21}\left(t\right)\left|x\right|+m_{22}\left(t\right)\left|y\right|,x,y\in R.$$

Let $E=C[0,1]$ with the norm ${\displaystyle \parallel x\parallel =\underset{t\in [0,1]}{max}\left|x\left(t\right)\right|,x\in E}$. Then, E is a Banach space. We define the operator $T_1,T_2:E\times E\to E$ and $T:E\times E\to E\times E$ as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle T_1(x,y)\left(t\right)={\int }_0^1{G}_{11}(t,s)f(s,x\left(s\right),y\left(s\right))ds+{\int }_0^1{G}_{12}(t,s)g(s,x\left(s\right),y\left(s\right))ds,}\hfill \\ {\displaystyle T_2(x,y)\left(t\right)={\int }_0^1{G}_{21}(t,s)f(s,x\left(s\right),y\left(s\right))ds+{\int }_0^1{G}_{22}(t,s)g(s,x\left(s\right),y\left(s\right))ds,}\hfill \end{array}\right.\end{array}$$

and $T(x,y)=\left(T_1(x,y),T_2(x,y)\right)$, $(x,y)\in E\times E$.

For $m_{ij}\in C[0,1](i,j=1,2)$, we denote

$$J_{ij}(m_{1j},m_{2j})={\lambda }_{i1}{\int }_0^1{m}_{1j}\left(s\right)ds+{\lambda }_{i2}{\int }_0^1{m}_{2j}\left(s\right)ds,i,j=1,2.$$

(20)

Theorem 1.

Assume that the conditions $\left({\mathrm{H}}_1\right)$–$\left({\mathrm{H}}_5\right)$ are satisfied. If System (1) has nontrivial solutions, we have the following inequality:

$$\begin{array}{c}\hfill \begin{array}{cc}& J_{11}(m_{11},m_{21})+J_{22}(m_{12},m_{22})\hfill \\ & +\sqrt{{[J_{11}(m_{11},m_{21})-J_{22}(m_{12},m_{22})]}^2+4J_{12}(m_{12},m_{22})J_{21}(m_{11},m_{21})}\ge 2.\hfill \end{array}\end{array}$$

(21)

Proof. 

We mainly apply the proof by contradiction to prove Theorem 1. Let $(x^{\sigma },y^{\sigma })\in E\times E$ be a nontrivial solution of System (1). Assume that

$$\begin{array}{c}\hfill \begin{array}{cc}& J_{11}(m_{11},m_{21})+J_{22}(m_{12},m_{22})\hfill \\ & +\sqrt{{[J_{11}(m_{11},m_{21})-J_{22}(m_{12},m_{22})]}^2+4J_{12}(m_{12},m_{22})J_{21}(m_{11},m_{21})}<2.\hfill \end{array}\end{array}$$

By Lemma 2, $(x,y)\in C^2[0,1]\times C^2[0,1]$ is a solution to System (9) only if $(x,y)\in E\times E$ is a fixed point of T. Then, we have that $(x^{\sigma },y^{\sigma })$ is a nontrivial fixed point of T. By $\left({\mathrm{H}}_4\right)$–$\left({\mathrm{H}}_5\right)$ and Lemma 5, for $t\in [0,1]$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \left|x^{\sigma }\left(t\right)\right|& =\left|T_1(x^{\sigma },y^{\sigma })\left(t\right)\right|\hfill \\ & \le {\int }_0^1{G}_{11}(t,s)\left|f(s,x^{\sigma }\left(s\right),y^{\sigma }\left(s\right))\right|ds+{\int }_0^1{G}_{12}(t,s)\left|g(s,x^{\sigma }\left(s\right),y^{\sigma }\left(s\right))\right|ds\hfill \\ & \le {\int }_0^1{G}_{11}(t,s)(m_{11}\left(s\right)\left|x^{\sigma }\left(s\right)\right|+m_{12}\left(s\right)\left|y^{\sigma }\left(s\right)\right|)ds\hfill \\ & +{\int }_0^1{G}_{12}(t,s)(m_{21}\left(s\right)\left|x^{\sigma }\left(s\right)\right|+m_{22}\left(s\right)\left|y^{\sigma }\left(s\right)\right|)ds\hfill \\ & \le ({\lambda }_{11}{\int }_0^1{m}_{11}\left(s\right)ds+{\lambda }_{12}{\int }_0^1{m}_{21}\left(s\right)ds)\parallel x^{\sigma }\parallel \hfill \\ & +({\lambda }_{11}{\int }_0^1{m}_{12}\left(s\right)ds+{\lambda }_{12}{\int }_0^1{m}_{22}\left(s\right)ds)\parallel y^{\sigma }\parallel .\hfill \end{array}\end{array}$$

Therefore, by (20), we have

$$\parallel x^{\sigma }\parallel \le J_{11}(m_{11},m_{21})\parallel x^{\sigma }\parallel +J_{12}(m_{12},m_{22})\parallel y^{\sigma }\parallel .$$

(22)

Similarly, by $\left({\mathrm{H}}_5\right)$ and Lemma 5, we have

$$\parallel y^{\sigma }\parallel \le J_{21}(m_{11},m_{21})\parallel x^{\sigma }\parallel +J_{22}(m_{12},m_{22})\parallel y^{\sigma }\parallel .$$

(23)

Combining (22) and (23), we can write it in the following matrix form:

$$\left(\begin{array}{c}\parallel x^{\sigma }\parallel \\ \parallel y^{\sigma }\parallel \end{array}\right)\le \left(\begin{array}{cc}{J}_{11}(m_{11},m_{21})& J_{12}(m_{12},m_{22})\\ J_{21}(m_{11},m_{21})& J_{22}(m_{12},m_{22})\end{array}\right)\left(\begin{array}{c}\parallel x^{\sigma }\parallel \\ \parallel y^{\sigma }\parallel \end{array}\right).$$

For $n\in N$, through iteration we obtain

$$\left(\begin{array}{c}\parallel x^{\sigma }\parallel \\ \parallel y^{\sigma }\parallel \end{array}\right)\le {\left(\begin{array}{cc}{J}_{11}(m_{11},m_{21})& J_{12}(m_{12},m_{22})\\ J_{21}(m_{11},m_{21})& J_{22}(m_{12},m_{22})\end{array}\right)}^n\left(\begin{array}{c}\parallel x^{\sigma }\parallel \\ \parallel y^{\sigma }\parallel \end{array}\right).$$

Next, by Lemmas 6, 7 and (21), we infer that

$$\parallel x^{\sigma }\parallel =\parallel y^{\sigma }\parallel =0.$$

This implies that $(x^{\sigma },y^{\sigma })$ is a trivial solution, which contradicts the assumption. This proves (21). □

For $m_{ij}\in C[0,1](i,j=1,2)$, let

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill I_{ij}(m_{1j},m_{2j})& ={\mu }_{i1}{t}^{{\alpha }_i-1}{e}^{-\lambda t}{\int }_0^1{m}_{1j}\left(s\right){(1-s)}^{{\alpha }_1-1}{s}^{{\alpha }_j-1}ds\hfill \\ & +{\mu }_{i2}{t}^{{\alpha }_i-1}{e}^{-\lambda t}{\int }_0^1{m}_{2j}\left(s\right){(1-s)}^{{\alpha }_2-1}{s}^{{\alpha }_j-1}ds.\hfill \end{array}\end{array}$$

(24)

Theorem 2.

Assume that the conditions $\left({\mathrm{H}}_1\right)$–$\left({\mathrm{H}}_5\right)$ are satisfied. If System (1) has a nontrivial solutions, we have the following inequality:

$$\begin{array}{c}\hfill \begin{array}{cc}& I_{11}(m_{11},m_{21})+I_{22}(m_{12},m_{22})\hfill \\ & +\sqrt{{[I_{11}(m_{11},m_{21})-I_{22}(m_{12},m_{22})]}^2+4I_{12}(m_{12},m_{22})I_{21}(m_{11},m_{21})}\ge 2.\hfill \end{array}\end{array}$$

(25)

Proof. 

We mainly apply the proof by contradiction to prove Theorem 2. Let $(x^{\sigma },y^{\sigma })\in E\times E$ be a nontrivial solution of System (1). Assume that

$$\begin{array}{c}\hfill \begin{array}{cc}& I_{11}(m_{11},m_{21})+I_{22}(m_{12},m_{22})\hfill \\ & +\sqrt{{[I_{11}(m_{11},m_{21})-I_{22}(m_{12},m_{22})]}^2+4I_{12}(m_{12},m_{22})I_{21}(m_{11},m_{21})}<2.\hfill \end{array}\end{array}$$

(26)

By $\left({\mathrm{H}}_2\right)$, f and g are continuous functions. Therefore, f and g are bounded. $K>0$ exists such that

$$\left|f\right(s,x\left(s\right),y\left(s\right)\left)\right|\le K,\left|g\right(s,x\left(s\right),y\left(s\right)\left)\right|\le K,s\in [0,1].$$

Next, we define the subspaces $E_i$ of the vector space E as follows:

$$E_i=\left\{x\in E:\exists M>0,\left|x\left(t\right)\right|\le M{t}^{{\alpha }_i-1}{e}^{-\lambda t}(i=1,2),t\in [0,1]\right\}.$$

The norm of $E_i$ in the Banach space is

$${\parallel x\parallel }_i=max\left\{M>0,\left|x\left(t\right)\right|\le M{t}^{{\alpha }_i-1}{e}^{-\lambda t}(i=1,2),t\in [0,1]\right\}.$$

By Lemma 5, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \left|T_1(x,y)\left(t\right)\right|& \le {\int }_0^1{G}_{11}(t,s)\left|f(s,x\left(s\right),y\left(s\right))\right|ds+{\int }_0^1{G}_{12}(t,s)\left|g(s,x\left(s\right),y\left(s\right))\right|ds\hfill \\ & \le K{\mu }_{11}{t}^{{\alpha }_1-1}{\int }_0^1{(1-s)}^{{\alpha }_1-1}{e}^{\lambda (s-t)}ds+K{\mu }_{12}{t}^{{\alpha }_1-1}{\int }_0^1{(1-s)}^{{\alpha }_2-1}{e}^{\lambda (s-t)}ds.\hfill \end{array}\end{array}$$

(27)

Let $u=\lambda s$, then $ds={\displaystyle \frac{1}{\lambda }}du$, and then we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\int }_0^1{(1-s)}^{{\alpha }_1-1}{e}^{\lambda s}ds& ={\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_1}{\int }_0^{\lambda }{(\lambda -u)}^{{\alpha }_1-1}{e}^{u}du.\hfill \end{array}\end{array}$$

(28)

Further substitution in (28); let $v=\lambda -u$, then $du=-dv$,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_1}{\int }_0^{\lambda }{(\lambda -u)}^{{\alpha }_1-1}{e}^{u}du& ={\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_1}{\int }_{\lambda }^0{v}^{{\alpha }_1-1}{e}^{\lambda -v}(-dv)\hfill \\ & =e^{\lambda }{\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_1}{\int }_0^{\lambda }{v}^{{\alpha }_1-1}{e}^{-v}dv\hfill \\ & =e^{\lambda }{\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_1}\gamma ({\alpha }_1,\lambda ).\hfill \end{array}\end{array}$$

(29)

Similarly, we have

$${\int }_0^1{(1-s)}^{{\alpha }_2-1}{e}^{\lambda s}ds=e^{\lambda }{\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_2}\gamma ({\alpha }_2,\lambda ),$$

(30)

where $\gamma ({\alpha }_1,\lambda )={\int }_0^{\lambda }{v}^{{\alpha }_1-1}{e}^{-v}dv$, $\gamma ({\alpha }_2,\lambda )={\int }_0^{\lambda }{v}^{{\alpha }_2-1}{e}^{-v}dv$ are the lower incomplete gamma functions. By (29) and (30), (27) can then be simplified as

$$\left|T_1(x,y)\left(t\right)\right|\le \left(K{\mu }_{11}{\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_1}{e}^{\lambda }\gamma ({\alpha }_1,\lambda )+K{\mu }_{12}{\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_2}{e}^{\lambda }\gamma ({\alpha }_2,\lambda )\right)t^{{\alpha }_1-1}{e}^{-\lambda t}.$$

By the same way, we have

$$\left|T_2(x,y)\left(t\right)\right|\le \left(K{\mu }_{21}{\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_1}{e}^{\lambda }\gamma ({\alpha }_1,\lambda )+K{\mu }_{22}{\left({\displaystyle \frac{1}{\lambda }}\right)}^{{\alpha }_2}{e}^{\lambda }\gamma ({\alpha }_2,\lambda )\right)t^{{\alpha }_2-1}{e}^{-\lambda t}.$$

Evidently, the operator $T_i$ maps the space $E\times E$ into the subspace $E_i$ of the space E. Therefore, $(x^{\sigma },y^{\sigma })$ is a nontrivial fixed point of T in $E_1\times E_2$.

By $\left({\mathrm{H}}_4\right)$ and Lemma 5, for $t\in [0,1]$, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \left|x\right(t\left)\right|& =\left|T_1(x,y)\left(t\right)\right|\hfill \\ & \le {\int }_0^1{G}_{11}(t,s)\left|f(s,x\left(s\right),y\left(s\right))\right|ds+{\int }_0^1{G}_{12}(t,s)\left|g(s,x\left(s\right),y\left(s\right))\right|ds\hfill \\ & \le {\int }_0^1{G}_{11}(t,s)(m_{11}\left(s\right)\left|x\left(s\right)\right|+m_{12}\left(s\right)\left|y\left(s\right)\right|)ds\hfill \\ & +{\int }_0^1{G}_{12}(t,s)(m_{21}\left(s\right)\left|x\left(s\right)\right|+m_{22}\left(s\right)\left|y\left(s\right)\right|)ds\hfill \\ & \le {\mu }_{11}{t}^{{\alpha }_1-1}{\int }_0^1{(1-s)}^{{\alpha }_1-1}{e}^{\lambda (s-t)}(m_{11}\left(s\right){\parallel x\parallel }_1{s}^{{\alpha }_1-1}{e}^{-\lambda s}+m_{12}\left(s\right){\parallel y\parallel }_2{s}^{{\alpha }_2-1}{e}^{-\lambda s})\hfill \\ & +{\mu }_{12}{t}^{{\alpha }_1-1}{\int }_0^1{(1-s)}^{{\alpha }_2-1}{e}^{\lambda (s-t)}(m_{21}\left(s\right){\parallel x\parallel }_1{s}^{{\alpha }_1-1}{e}^{-\lambda s}+m_{22}\left(s\right){\parallel y\parallel }_2{s}^{{\alpha }_2-1}{e}^{-\lambda s})\hfill \\ & \le ({\mu }_{11}{\int }_0^1{m}_{11}\left(s\right){(1-s)}^{{\alpha }_1-1}{s}^{{\alpha }_1-1}ds+{\mu }_{12}{\int }_0^1{m}_{21}\left(s\right){(1-s)}^{{\alpha }_2-1}{s}^{{\alpha }_1-1}ds){\parallel x\parallel }_1{t}^{{\alpha }_1-1}{e}^{-\lambda t}\hfill \\ & +({\mu }_{11}{\int }_0^1{m}_{12}\left(s\right){(1-s)}^{{\alpha }_1-1}{s}^{{\alpha }_2-1}ds+{\mu }_{12}{\int }_0^1{m}_{22}\left(s\right){(1-s)}^{{\alpha }_2-1}{s}^{{\alpha }_2-1}ds){\parallel y\parallel }_2{t}^{{\alpha }_1-1}{e}^{-\lambda t}.\hfill \end{array}\end{array}$$

Therefore, by (24), we have

$${\parallel x\parallel }_1\le I_{11}(m_{11},m_{21}){\parallel x\parallel }_1+I_{12}(m_{12},m_{22}){\parallel y\parallel }_2.$$

(31)

Similarly, by $\left({\mathrm{H}}_5\right)$ and Lemma 5, we have

$${\parallel y\parallel }_1\le I_{21}(m_{11},m_{21}){\parallel x\parallel }_1+I_{22}(m_{12},m_{22}){\parallel y\parallel }_2.$$

(32)

Combining (31) and (32), we can write it in the following matrix form:

$$\left(\begin{array}{c}{\parallel x\parallel }_1\\ {\parallel y\parallel }_2\end{array}\right)\le \left(\begin{array}{cc}{I}_{11}(m_{11},m_{21})& I_{12}(m_{12},m_{22})\\ I_{21}(m_{11},m_{21})& I_{22}(m_{12},m_{22})\end{array}\right)\left(\begin{array}{c}{\parallel x\parallel }_1\\ {\parallel y\parallel }_2\end{array}\right).$$

Next, by Lemmas 6, 7 and (26), we infer that

$${\parallel x\parallel }_1={\parallel y\parallel }_2=0.$$

This implies that $(x,y)$ is a trivial solution, which contradicts the assumption. This proves (25). □

We define

$$g_{ij}\left(s\right)={(1-s)}^{{\alpha }_i-1}{s}^{{\alpha }_j-1}(i,j=1,2),s\in [0,1].$$

Next, differentiating $g_{ij}\left(s\right)$ on $(0,1)$, we have

$$g_{ij}^{\prime }\left(s\right)={(1-s)}^{{\alpha }_i-2}{s}^{{\alpha }_j-2}\left[{\alpha }_j-1-({\alpha }_i+{\alpha }_j-2)s\right].$$

This means that $g_{ij}^{\prime }\left(s\right)=0$ if and only if ${\displaystyle s_{ij}^{\sigma }=\frac{{\alpha }_j-1}{{\alpha }_i+{\alpha }_j-2}}$. When $0<s\le {\displaystyle \frac{{\alpha }_j-1}{{\alpha }_i+{\alpha }_j-2}}$, $g_{ij}^{\prime }\left(s\right)\ge 0$; when ${\displaystyle \frac{{\alpha }_j-1}{{\alpha }_i+{\alpha }_j-2}}\le s<1$, $g_{ij}^{\prime }\left(s\right)\le 0$, and $g_{ij}\left(0\right)=g_{ij}\left(1\right)=0$, by the continuity of $g_{ij}$, we have

$$g_{ij}\left(s\right)\le g_{ij}\left(s_{ij}^{\sigma }\right)={\displaystyle \frac{{({\alpha }_i-1)}^{{\alpha }_i-1}{({\alpha }_j-1)}^{{\alpha }_j-1}}{{({\alpha }_i+{\alpha }_j-2)}^{{\alpha }_i+{\alpha }_j-2}}}.$$

(33)

For $m_{ij}\in C[0,1](i,j=1,2)$, let

$$\tau (m_{1j},m_{2j})={\mu }_{i1}{g}_{1j}\left(s_{1j}^{\sigma }\right){\int }_0^1{m}_{1j}\left(s\right)ds+{\mu }_{i2}{g}_{2j}\left(s_{2j}^{\sigma }\right){\int }_0^1{m}_{2j}\left(s\right)ds,i=1,2.$$

(34)

The following theorems are immediate.

Theorem 3.

Assume that the conditions $\left({\mathrm{H}}_1\right)$–$\left({\mathrm{H}}_5\right)$ are satisfied. If System (1) has nontrivial solutions, we have the following inequality:

$$\begin{array}{c}\hfill \begin{array}{cc}& {\tau }_{11}(m_{11},m_{21})+{\tau }_{22}(m_{12},m_{22})\hfill \\ & +\sqrt{{[{\tau }_{11}(m_{11},m_{21})-{\tau }_{22}(m_{12},m_{22})]}^2+4{\tau }_{12}(m_{12},m_{22}){\tau }_{21}(m_{11},m_{21})}\ge 2.\hfill \end{array}\end{array}$$

(35)

In the following, we compare the cases of inequality (21) and inequality (35) when ${\alpha }_1={\alpha }_2$.

By (33), we have

$$g\left(s^{\sigma }\right)\triangleq g_{ij}\left(s_{ij}^{\sigma }\right)={\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1},i=1,2,$$

(36)

and the square non-negative matrices $M_1={\left(I_{ij}(m_{1j},m_{2j})\right)}_{2\times 2}$ and $M_2={\left({\tau }_{ij}(m_{1j},m_{2j})\right)}_{2\times 2}$ become

$$M_1=\left(\begin{array}{cc}{\lambda }_{11}& {\lambda }_{12}\\ {\lambda }_{21}& {\lambda }_{22}\end{array}\right)\left(\begin{array}{cc}{\displaystyle {\int }_0^1{m}_{11}\left(s\right)ds}& {\displaystyle {\int }_0^1{m}_{12}\left(s\right)ds}\\ {\displaystyle {\int }_0^1{m}_{21}\left(s\right)ds}& {\displaystyle {\int }_0^1{m}_{22}\left(s\right)ds}\end{array}\right),$$

and

$$M_2=g\left(s^{\sigma }\right)\left(\begin{array}{cc}{\mu }_{11}& {\mu }_{12}\\ {\mu }_{21}& {\mu }_{22}\end{array}\right)\left(\begin{array}{cc}{\displaystyle {\int }_0^1{m}_{11}\left(s\right)ds}& {\displaystyle {\int }_0^1{m}_{12}\left(s\right)ds}\\ {\displaystyle {\int }_0^1{m}_{21}\left(s\right)ds}& {\displaystyle {\int }_0^1{m}_{22}\left(s\right)ds}\end{array}\right).$$

Combining matrix $M_1$ and matrix $M_2$, as well as Lemma 5, we obtain

$${\lambda }_{11}={\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}+{\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_1\right)}}\sum _{i=1}^n({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}){\xi }_i^{{\alpha }_1-1}{(1-{\xi }_i)}^{{\alpha }_1-1},$$

(37)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\mu }_{11}g\left(s^{\sigma }\right)& ={\displaystyle \frac{{\gamma }_{22}}{\gamma \Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}\hfill \\ & ={\displaystyle \frac{{\gamma }_{22}-\gamma }{\gamma \Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}+{\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}\hfill \\ & ={\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_1\right)}}({\gamma }_{22}(1-{\gamma }_{11})+{\gamma }_{12}{\gamma }_{21}){\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}+{\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}\hfill \\ & ={\displaystyle \frac{1}{\Gamma \left({\alpha }_1\right)}}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}+{\displaystyle \frac{1}{\gamma \Gamma \left({\alpha }_1\right)}}\sum _{i=1}^n({\gamma }_{22}{a}_{1i}+{\gamma }_{12}{a}_{3i}){\xi }_i^{{\alpha }_1-1}{\left({\displaystyle \frac{1}{4}}\right)}^{{\alpha }_1-1}.\hfill \end{array}\end{array}$$

(38)

Compare (37) and (38); obviously, we have ${\lambda }_{11}\le {\mu }_{11}g\left(s^{\sigma }\right)$ if ${\displaystyle \frac{3}{4}}\le {\xi }_i<1(i=1,2,\dots ,n)$, and ${\lambda }_{11}\ge {\mu }_{11}g\left(s^{\sigma }\right)$ if $0<{\xi }_i\le {\displaystyle \frac{3}{4}}(i=1,2,\dots ,n)$. In the same way, for $i,j=1,2$, we can prove

(1)

${\lambda }_{ij}\le {\mu }_{ij}g\left(s^{\sigma }\right)$ when ${\displaystyle \frac{3}{4}}\le {\xi }_k,{\eta }_k<1(k=1,2,\dots ,n)$,

(2)

${\lambda }_{ij}\ge {\mu }_{ij}g\left(s^{\sigma }\right)$ when $0<{\xi }_k,{\eta }_k\le {\displaystyle \frac{3}{4}}(k=1,2,\dots ,n)$.

Therefore, if ${\displaystyle \frac{3}{4}}\le {\xi }_k,{\eta }_k<1(k=1,2,\dots ,n)$, we can obtain $0\le M_1\le M_2$. By Lemma 8, we have $\rho \left(M_1\right)<\rho \left(M_2\right)$; we can find that inequality (21) is an improvement on inequality (35). Similarly, if $0<{\xi }_k,{\eta }_k\le {\displaystyle \frac{3}{4}}(k=1,2,\dots ,n)$, we can obtain $M_1\ge M_2\ge 0$, $\rho \left(M_2\right)<\rho \left(M_1\right)$; we can find that inequality (35) is an improvement on inequality (21).

4. Example

In this section, we have provided an example to verify Theorem 1. The same method can be used to verify Theorems 2 and 3.

Example 1.

We consider the nonlinear tempered fractional differential equations with coupled boundary conditions as follows:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{-}_0^R{\mathbb{D}}_t^{{\alpha }_1,\lambda }x\left(t\right)={\displaystyle \frac{{\alpha }_1^{{\alpha }_1}\Gamma \left({\alpha }_1\right)}{{({\alpha }_1-1)}^{({\alpha }_1-1)}}}\left[t^{2}arctan\left(x\left(t\right)\right)+{\displaystyle \frac{e^{-t}}{2}}sin\left(y\left(t\right)\right)\right],0<t<1,\hfill \\ {-}_0^R{\mathbb{D}}_t^{{\alpha }_2,\lambda }y\left(t\right)={\displaystyle \frac{{\alpha }_2^{{\alpha }_2}\Gamma \left({\alpha }_2\right)}{{({\alpha }_2-1)}^{({\alpha }_2-1)}}}\left[{\displaystyle \frac{e^{-t}}{2}}x\left(t\right)+t^{2}ln(1+\left|y\left(t\right)\right|)\right],0<t<1,\hfill \\ {\displaystyle x\left(0\right)=0,x\left(1\right)=\frac{1}{4}x\left(\frac{1}{3}\right)+\frac{1}{2}x\left(\frac{1}{9}\right)+\frac{1}{6}y\left(\frac{1}{4}\right)+\frac{1}{3}y\left(\frac{1}{16}\right),}\hfill \\ {\displaystyle y\left(0\right)=0,y\left(1\right)=\frac{1}{2}x\left(\frac{1}{3}\right)+x\left(\frac{1}{9}\right)+\frac{1}{8}y\left(\frac{1}{4}\right)+\frac{1}{4}y\left(\frac{1}{16}\right).}\hfill \end{array}\right.\end{array}\end{array}$$

(39)

We choose $a_{1i}={\displaystyle \frac{i}{4}}$, $a_{2j}={\displaystyle \frac{j}{6}}$, $a_{3i}={\displaystyle \frac{i}{2}}$, $a_{4j}={\displaystyle \frac{j}{8}}$, ${\xi }_i={\displaystyle \frac{1}{3^i}}$, ${\eta }_j={\displaystyle \frac{1}{4^j}}(i,j=1,2)$, ${\alpha }_1={\alpha }_2=2$, and $\lambda ={\displaystyle \frac{3}{2}}$. It is easy to find that

$$m_{11}\left(t\right)=m_{22}\left(t\right)={\displaystyle \frac{{\alpha }_1^{{\alpha }_1}\Gamma \left({\alpha }_1\right)}{{({\alpha }_1-1)}^{({\alpha }_1-1)}}}{t}^2,m_{12}\left(t\right)=m_{22}\left(t\right)={\displaystyle \frac{{\alpha }_1^{{\alpha }_1}\Gamma \left({\alpha }_1\right)}{{({\alpha }_1-1)}^{({\alpha }_1-1)}}}{\displaystyle \frac{e^{-t}}{2}}.$$

Through simple calculations we obtain

$${\gamma }_{11}\approx 0.563,{\gamma }_{12}\approx 0.213,{\gamma }_{21}\approx 0.875,{\gamma }_{22}\approx 0.84,$$

and

$$\gamma ={\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}{\gamma }_{21}\approx 0.287>0.$$

By the definitions of ${\lambda }_{ij}(i=1,2)$ in Lemma 5, we have

$${\lambda }_{11}\approx 0.713,{\lambda }_{12}\approx 0.079,{\lambda }_{21}\approx 0.982,{\lambda }_{22}\approx 0.23.$$

It can be seen that assumptions $\left({\mathrm{H}}_1\right)$–$\left({\mathrm{H}}_5\right)$ are satisfied. Then, through calculation we can obtain

$$\begin{array}{c}\hfill \begin{array}{cc}& J_{11}(m_{11},m_{21})+J_{22}(m_{12},m_{22})\hfill \\ & +\sqrt{{[J_{11}(m_{11},m_{21})-J_{22}(m_{12},m_{22})]}^2+4J_{12}(m_{12},m_{22})J_{21}(m_{11},m_{21})}\hfill \\ & =0.0071+0.0344\hfill \\ & +\sqrt{{(0.00231+0.0209)}^2+4(0.0178+0.00178)(0.00729+0.0265)}\hfill \\ & =0.0415+\sqrt{0.00319}\hfill \\ & =0.098<2.\hfill \end{array}\end{array}$$

(40)

Hence, $(x,y)=(0,0)$ is the only solution to (39). Then, the Lyapunov inequality in Theorem 1 has been verified.

5. Conclusions

In this paper, we study the system of tempered fractional differential equations with multi-point coupled boundary conditions. We have proved some properties of Green’s function. The first Lyapunov inequality (21) is obtained by applying the properties of Green’s function and the method of matrix analysis in Banach space $E\times E$. The second Lyapunov inequality (25) is obtained by using the properties of Green’s function and operator T in the Banach space $E_1\times E_2$. The third Lyapunov inequality (35) is a generalization of the second Lyapunov inequality. By defining $g_{ij}\left(s\right)$, the first Lyapunov inequality (21) and the third Lyapunov inequality (35) are connected by the method of matrix analysis. Under certain conditions, we have compared the first Lyapunov inequality (21) and the third Lyapunov inequality (35). Finally, we provided an example to verify Theorem 1, and the expected results were obtained. The same method can be adopted to verify Theorems 2 and 3. In this paper, the case of $0\le \lambda \le {\displaystyle \frac{\alpha -1}{1-{\xi }_1}}$ is considered $(\alpha =min\left\{{\alpha }_1,{\alpha }_2\right\})$. In future research, we can consider the monotonicity of Green’s function of System (1) and the existence of the corresponding Lyapunov inequality when $\lambda >{\displaystyle \frac{\alpha -1}{1-{\xi }_1}}$.