# Positive solutions for a system of nonlinear Hadamard fractional differential equations involving coupled integral boundary conditions

## Abstract

In this paper we use the fixed point index to study the existence of positive solutions for a system of nonlinear Hadamard fractional differential equations involving coupled integral boundary conditions. Here we use appropriate nonnegative matrices to depict the coupling behavior for our nonlinearities.

## Introduction

In this paper we consider the system of nonlinear Hadamard fractional differential equations involving coupled integral boundary conditions

$$\textstyle\begin{cases} D^{\beta }u(t)+ f_{1}(t,u(t),v(t))=0,\quad 1< t< e, \\ D^{\beta }v(t)+f_{2}(t,u(t),v(t))=0,\quad 1< t< e, \\ u(1)=v(1)=u'(1)=v'(1)=0,\\ u(e)=\int _{1}^{e} h(s) v(s)\frac{ds}{s},\\ v(e)= \int _{1}^{e} g(s) u(s)\frac{ds}{s}, \end{cases}$$
(1.1)

where $$\beta \in (2,3]$$, $$D^{\beta }$$ is the Hadamard fractional derivative of fractional order β, and $$f_{i}$$ ($$i=1,2$$), h, g satisfy the following conditions:

1. (H1)

$$f_{i}$$ ($$i=1,2$$) are nonnegative continuous functions on $$[1,e]\times \mathbb{R}^{+}\times \mathbb{R}^{+}$$,

2. (H2)

$$h,g\ge 0$$ (0) on $$[1,e]$$ with $$\int _{1}^{e} h(t) ( \log t)^{\beta -1} \frac{dt}{t}\cdot \int _{1}^{e} g(t) (\log t)^{ \beta -1} \frac{dt}{t} \in (0,1)$$.

Fractional-order differential equations is a rapidly developing area of research; we refer the reader to [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48] and the references therein. In [1,2,3,4,5,6,7,8,9], the authors used iterative techniques to study existence and uniqueness of solutions for fractional boundary value problems. In  the authors studied positive solutions for the p-Laplacian fractional Riemann–Stieltjes integral boundary value problem

$$\textstyle\begin{cases} -D_{t}^{\beta }(\varphi _{p}(-D_{t}^{\alpha }z))(t)=f(t,z(t),D_{t}^{ \gamma }z(t)),\quad t\in (0,1), \\ D_{t}^{\alpha }z (0)=D_{t}^{\alpha +1} z(0)=D_{t}^{\gamma }z(0)=0, \\ D_{t}^{\alpha }z (1)=0,\qquad D_{t}^{\gamma }z(1)=\int _{0}^{1} D_{t}^{ \gamma }z(s)\,d A(s), \end{cases}$$
(1.2)

where $$D_{t}^{\alpha }$$, $$D_{t}^{\beta }$$, $$D_{t}^{\gamma }$$ are the Riemann–Liouville fractional derivatives, and they not only obtained existence and uniqueness of positive solutions for (1.2), but also constructed an iteration sequence for the unique positive solution. In [10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32], the authors used fixed point methods to study the existence of (positive) solutions fractional order equations. In  Mawhin’s continuation theorem was used to study the following fractional order boundary value problem at resonance:

$$\textstyle\begin{cases} {}^{c}D^{q} x(t)=f(t,x(t),x'(t)),\quad t\in [0,T], \\ x(0)=\alpha I_{\eta }^{\gamma ,\delta }x(\zeta ),\qquad x(T)=\beta {}^{ \rho }I^{p}x(\xi ),\quad 0< \zeta ,\xi \le T, \end{cases}$$
(1.3)

where $${}^{c}D^{q}$$ is the Caputo fractional derivative, $$I_{\eta }^{ \gamma ,\delta }$$ is a Erdélyi–Kober type integral, and $${}^{\rho }I^{p}$$ denotes the generalized Riemann–Liouville type integral boundary conditions. For fractional differential systems, see [23,24,25,26,27,28,29,30,31,32]. In , using the Leray–Schauder alternative and the Banach contraction principle, the authors studied existence and uniqueness of solutions for the system of nonlinear Caputo type sequential fractional integro-differential equations

$$\textstyle\begin{cases} ({}^{c}D^{\alpha }+\lambda {}^{c}D^{\alpha -1} )u(t)=f(t,u(t),v(t),{}^{c}D ^{p_{1}}v(t),I^{q_{1}}v(t)),\quad t\in (0,1), \\ ({}^{c}D^{\beta }+\mu {}^{c}D^{\beta -1} )v(t)=g(t,u(t),{}^{c}D^{p_{2}}u(t),I ^{q_{2}}u(t),v(t)),\quad t\in (0,1), \\ u(0)=u'(0)=u''(0)=0,\qquad u(1)=\int _{0}^{1}u(s)\,dH_{1}(s)+\int _{0}^{1} v(s)\,dH _{2}(s), \\ v(0)=v'(0)=v''(0)=0,\qquad v(1)=\int _{0}^{1}u(s)\,dK_{1}(s)+\int _{0}^{1} v(s)\,dK _{2}(s). \end{cases}$$
(1.4)

Hadamard fractional differential equations are also popular in the literature; see [33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48] and the references therein. In , the authors used the Banach contraction principle, the Leray–Schauder’s alternative, and Krasnoselskii’s fixed-point theorem to study the existence and uniqueness of solutions for the coupled system of nonlinear sequential Caputo and Hadamard fractional differential equations with coupled separated boundary conditions

$$\textstyle\begin{cases} {}^{C}D^{p_{1}} {}^{H}D^{q_{1}}x(t)=f(t,x(t),y(t)), \quad t\in [a,b], \\ {}^{H}D^{q_{2}} {}^{C}D^{p_{2}}x(t)=g(t,x(t),y(t)), \quad t\in [a,b], \\ \alpha _{1}x(a)+\alpha _{2} {}^{C}D^{p_{2}}y(a)=0,\qquad \beta _{1}x(b)+\beta _{2} {}^{C}D^{p_{2}}y(b)=0, \\ \alpha _{3}y(a)+\alpha _{4} {}^{H}D^{q_{1}}x(a)=0,\qquad \beta _{3}y(b)+\beta _{4} {}^{H}D^{q_{1}}x(b)=0, \end{cases}$$
(1.5)

where $${}^{C}D^{p_{i}}$$, $${}^{H}D^{q_{i}}$$ are respectively the Caputo and Hadamard fractional derivatives. In  the authors established positive solutions for the coupled Hadamard fractional integral boundary value problems

$$\textstyle\begin{cases} D^{\alpha }u(t)+\lambda f(t,u(t),v(t))=0,\quad t\in (1,e), \lambda >0, \\ D^{\beta }v(t)+\lambda g(t,u(t),v(t))=0,\quad t\in (1,e), \lambda >0, \\ u^{(j)}(1)=v^{(j)}(1)=0,\quad 0\le j\le n-2,\\ u(e)=\mu \int _{1}^{e} v(s) \frac{ds}{s},\\ v(e)=\nu \int _{1}^{e} u(s)\frac{ds}{s}, \end{cases}$$
(1.6)

where $$\alpha ,\beta \in (n-1,n]$$ and $$n\ge 3$$, $$D^{\alpha }$$, $$D^{ \beta }$$ are the Hadamard fractional derivatives and their nonlinearities f, g satisfy the following conditions:

$$(\mathrm{H})_{\mathrm{Yang}1}$$ :

There exists $$[\theta _{1},\theta _{2}]\subset (1,e)$$ such that $$\liminf_{u\to +\infty }\min_{t\in [\theta _{1},\theta _{2}] }\frac{f(t,u,v)}{u}=+\infty$$ and $$\liminf_{v\to +\infty }\min_{t\in [\theta _{1},\theta _{2}] }\frac{g(t,u,v)}{v}=+\infty$$;

or
$$(\mathrm{H})_{\mathrm{Yang}2}$$ :

There exists $$[\theta _{1},\theta _{2}]\subset (1,e)$$ such that $$\liminf_{v\to +\infty }\min_{t\in [\theta _{1},\theta _{2}] }\frac{f(t,u,v)}{v}=+\infty$$ and $$\liminf_{u\to +\infty }\min_{t\in [\theta _{1},\theta _{2}] }\frac{g(t,u,v)}{u}=+\infty$$.

Motivated by the above, in this paper we study the existence of positive solutions for the system of nonlinear Hadamard fractional differential equations (1.1) involving coupled integral boundary conditions. We use appropriate nonnegative matrices to depict the coupling behavior for our nonlinearities, and note that they can grow both superlinearly and sublinearly. We remark here that our conditions for nonlinear terms are not as restrictive as those in $$(\mathrm{H})_{\mathrm{Yang}1}$$ and $$(\mathrm{H})_{\mathrm{Yang}2}$$; see (H3)–(H6) in Sect. 3.

## Preliminaries

In this section, we first provide some material for Hadamard fractional calculus; for details, see the book .

### Definition 2.1

The Hadamard derivative of fractional order q for a function $$g: [1,\infty )\rightarrow \mathbb{R}$$ is defined as

$$D^{q}g(t)=\frac{1}{\varGamma (n-q)} \biggl(t\frac{d}{dt} \biggr)^{n} \int _{1}^{t} (\log t-\log s )^{n-q-1}g(s)\frac{ds}{s},\quad n-1< q< n,$$

where $$n=[q]+1$$; $$[q]$$ denotes the integer part of the real number q and $$\log (\cdot )=\log _{e}(\cdot )$$.

### Definition 2.2

The Hadamard fractional integral of order q for a function $$g: [1,\infty )\rightarrow \mathbb{R}$$ is defined as

$$I^{q}g(t)=\frac{1}{\varGamma (q)} \int _{1}^{t} (\log t-\log s ) ^{q-1} g(s)\frac{ds}{s},\quad q>0,$$

provided the integral exists.

In what follows, we calculate the Green’s functions associated with (1.1) and study some properties of these Green’s functions.

### Lemma 2.3

(see [34, Lemma 2.3])

Let $$x,y\in C[1,e]$$. Then the integral boundary value problem

$$\textstyle\begin{cases} D^{\beta }u(t)+x(t)=0,\qquad D^{\beta }v(t)+y(t)=0,\quad t\in (1,e), \\ u(1)=v(1)=u'(1)=v'(1)=0,\\ u(e)=\int _{1}^{e} h(s) v(s)\frac{ds}{s},\\ v(e)= \int _{1}^{e} g(s) u(s)\frac{ds}{s}, \end{cases}$$
(2.1)

can be transformed into the following Hammerstein type integral equations:

$$\textstyle\begin{cases} u(t)=\int _{1}^{e} G_{1}(t,s) x(s)\frac{ds}{s}+\frac{d_{h} (\log t)^{ \beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} x(s)\frac{ds}{s}\\ \hphantom{u(t)=}{}+\frac{ (\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} y(s) \frac{ds}{s},\\ v(t)=\int _{1}^{e} G_{1}(t,s)y(s)\frac{ds}{s}+\frac{d _{g}(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1} ^{e} h(t)G_{1}(t,s)\frac{dt}{t} y(s)\frac{ds}{s}\\ \hphantom{v(t)=}{}+\frac{(\log t)^{ \beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} x(s)\frac{ds}{s}, \end{cases}$$
(2.2)

where

$$G_{1}(t,s)=\frac{1}{\varGamma (\beta )} \textstyle\begin{cases} (\log t)^{\beta -1}(1-\log s)^{\beta -1}- (\log t-\log s ) ^{\beta -1}, & 1\le s\le t\le e, \\ (\log t)^{\beta -1}(1-\log s)^{\beta -1}, & 1\le t\le s \le e; \end{cases}$$
(2.3)

here, $$d_{g,h}$$, $$d_{g}$$, $$d_{h}$$ are three positive constants defined in the proof.

### Proof

From Lemma 2.3 of  we have

\begin{aligned}[b] & u(t)=c_{11}(\log t)^{\beta -1}+c_{12}( \log t)^{\beta -2}+c_{13}( \log t)^{\beta -3}- \frac{1}{\varGamma (\beta )} \int _{1}^{t} (\log t- \log s )^{\beta -1}x(s)\frac{ds}{s}, \\ & v(t)=c_{21}(\log t)^{ \beta -1}+c_{22}(\log t)^{\beta -2}+c_{23}(\log t)^{\beta -3}- \frac{1}{ \varGamma (\beta )} \int _{1}^{t} (\log t-\log s )^{\beta -1}y(s) \frac{ds}{s}, \end{aligned}

where $$c_{1i},c_{2i}\in \mathbb{R}$$, $$i=1,2,3$$. Note that $$u(1)=v(1)=u'(1)=v'(1)=0$$ implies $$c_{12},c_{13},c_{22},c_{23}=0$$. Then we have

\begin{aligned}[b] & u(t)=c_{11}(\log t)^{\beta -1}- \frac{1}{\varGamma (\beta )} \int _{1} ^{t} (\log t-\log s )^{\beta -1}x(s)\frac{ds}{s}, \\ & v(t)=c _{21}(\log t)^{\beta -1}-\frac{1}{\varGamma (\beta )} \int _{1}^{t} (\log t- \log s )^{\beta -1}y(s)\frac{ds}{s}. \end{aligned}

Using the conditions $$u(e)=\int _{1}^{e} h(s) v(s)\frac{ds}{s}$$, $$v(e)= \int _{1}^{e} g(s) u(s)\frac{ds}{s}$$, we obtain

\begin{aligned}& \begin{aligned}[b] & c_{11}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}\\ &\quad = c_{21} \int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1} ^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t}, \end{aligned} \\& \begin{aligned}[b] & c_{21}-\frac{1}{\varGamma (\beta )} \int _{1}^{e} (1- \log s )^{\beta -1}y(s) \frac{ds}{s}\\ &\quad =c_{11} \int _{1} ^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t}. \end{aligned} \end{aligned}

This implies that

\begin{aligned}[b] & \begin{bmatrix} 1 & -\int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \\ - \int _{1} ^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t} & 1 \end{bmatrix} \begin{bmatrix} c_{11} \\ c_{21} \end{bmatrix} \\ & \quad = \begin{bmatrix} \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1} ^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t} \\ \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1- \log s )^{\beta -1}y(s)\frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t} \end{bmatrix}. \end{aligned}

Let $$d_{g,h}=1-\int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \cdot \int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}$$, $$d_{h}=\int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t}$$, $$d_{g}= \int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}$$. Then

\begin{aligned}[b] \begin{bmatrix} c_{11} \\ c_{21} \end{bmatrix} &= \frac{1}{d_{g,h}} \begin{bmatrix} 1 & \int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \\ \int _{1} ^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t} & 1 \end{bmatrix} \\ & \quad{} \cdot \begin{bmatrix} \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1} ^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t} \\ \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1- \log s )^{\beta -1}y(s)\frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t} \end{bmatrix}. \end{aligned}

Consequently, we have

\begin{aligned} u(t) &= \frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}\\ &\quad {}- \frac{(\log t)^{\beta -1}}{d _{g,h}\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1}^{t} (\log t - \log s )^{\beta -1}y(s)\frac{ds}{s}\frac{dt}{t} \\ & \quad {}+ \frac{d_{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1} ^{e} (1- \log s )^{\beta -1}y(s) \frac{ds}{s}\\ &\quad {}- \frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s) \frac{ds}{s}\frac{dt}{t} \\ & \quad {}-\frac{1}{\varGamma (\beta )} \int _{1}^{t} (\log t-\log s ) ^{\beta -1}x(s)\frac{ds}{s} \\ & = \frac{ (\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )} \int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \int _{1}^{e} (1- \log s )^{\beta -1}y(s) \frac{ds}{s}\\ &\quad {} - \frac{( \log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t} \\ & \quad {}+ \frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} (1- \log s)^{\beta -1} x(s) \frac{ds}{s} \\ &\quad {}- \frac{d_{h} (\log t)^{\beta -1}}{d _{g,h}\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t - \log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t} \\ & \quad {}-\frac{1}{\varGamma (\beta )} \int _{1}^{t} (\log t-\log s ) ^{\beta -1}x(s)\frac{ds}{s}+ \frac{(\log t)^{\beta -1}}{\varGamma ( \beta )} \int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s} \\ &\quad {}-\frac{( \log t)^{\beta -1}}{\varGamma (\beta )} \int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s} \\ & = \int _{1}^{e} G_{1}(t,s) x(s) \frac{ds}{s}+\frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \biggl[ \int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t} \int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}\\ &\quad {}- \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t} \biggr] \\ & \quad {}+ \frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \biggl[ \int _{1} ^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \int _{1}^{e} (1- \log s )^{\beta -1}y(s) \frac{ds}{s} \\ &\quad {}- \int _{1}^{e} h(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t} \biggr] \\ & = \int _{1}^{e} G_{1}(t,s) x(s) \frac{ds}{s}+\frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \biggl[ \int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t} \int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}\\ &\quad {}- \int _{1}^{e} g(t) \int _{s}^{e} (\log t -\log s )^{\beta -1}x(s)\frac{dt}{t}\frac{ds}{s} \biggr] \\ & \quad {}+ \frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \biggl[ \int _{1} ^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \int _{1}^{e} (1- \log s )^{\beta -1}y(s) \frac{ds}{s}\\ &\quad {} - \int _{1}^{e} h(t) \int _{s}^{e} (\log t -\log s )^{\beta -1}y(s)\frac{dt}{t} \frac{ds}{s} \biggr] \\ & = \int _{1}^{e} G_{1}(t,s) x(s) \frac{ds}{s}+\frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} x(s)\frac{ds}{s}\\ &\quad {}+\frac{ (\log t)^{ \beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} y(s)\frac{ds}{s}. \end{aligned}

Similarly, we also obtain that

\begin{aligned} v(t) &=c_{21}(\log t)^{\beta -1}- \frac{1}{\varGamma (\beta )} \int _{1}^{t} (\log t-\log s )^{\beta -1}y(s)\frac{ds}{s}\\ &\quad {}+ \frac{1}{ \varGamma (\beta )} \int _{1}^{e} (\log t)^{\beta -1} (1-\log s ) ^{\beta -1}y(s)\frac{ds}{s} \\ & \quad {}-\frac{1}{\varGamma (\beta )} \int _{1}^{e} (\log t)^{\beta -1} (1- \log s )^{\beta -1}y(s)\frac{ds}{s} \\ & = \int _{1}^{e} G_{1}(t,s)y(s) \frac{ds}{s}-\frac{1}{\varGamma (\beta )} \int _{1}^{e} (\log t)^{\beta -1} (1-\log s )^{\beta -1}y(s)\frac{ds}{s}\\ &\quad {}+ \frac{d_{g}( \log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} (1-\log s)^{ \beta -1} x(s) \frac{ds}{s} \\ & \quad {}- \frac{d_{g}(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1} ^{e} h(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}y(s) \frac{ds}{s}\frac{dt}{t}\\ &\quad {}+ \frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma ( \beta )} \int _{1}^{e} (1- \log s )^{\beta -1}y(s) \frac{ds}{s} \\ & \quad {}- \frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s) \frac{ds}{s}\frac{dt}{t} \\ & = \int _{1}^{e} G_{1}(t,s)y(s) \frac{ds}{s}+\frac{d_{g}(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \biggl[ \int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \int _{1}^{e} (1-\log s)^{\beta -1} y(s) \frac{ds}{s}\\ &\quad {}- \int _{1}^{e} h(t) \int _{1} ^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t} \biggr] \\ & \quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \biggl[ \int _{1} ^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t} \int _{1}^{e} (1-\log s)^{ \beta -1} x(s) \frac{ds}{s}\\ &\quad {}- \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t} \biggr] \\ & = \int _{1}^{e} G_{1}(t,s)y(s) \frac{ds}{s}+\frac{d_{g}(\log t)^{\beta -1}}{d _{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} y(s)\frac{ds}{s}\\ &\quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} x(s) \frac{ds}{s}. \end{aligned}

This completes the proof. □

From Lemma 2.3, we note (1.1) is equivalent to the Hammerstein type integral equations

$$\textstyle\begin{cases} u(t)=\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{u(t)=}{}+\frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{u(t)=}{}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s},\\ v(t)=\int _{1}^{e} G_{1}(t,s)f_{2}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{v(t)=}{}+\frac{d _{g}(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1} ^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{v(t)=}{}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s}. \end{cases}$$
(2.4)

### Lemma 2.4

The function $$G_{1}(t,s)$$ satisfies the following inequalities:

1. (I1)

$$\frac{1}{\varGamma (\beta )} (\log t)^{\beta -1}(1-\log t)\log s(1- \log s)^{\beta -1}\le G_{1}(t,s)\le \frac{\beta -1}{{\varGamma (\beta )}}\log s(1-\log s)^{\beta -1}$$ for $$t,s\in [1,e]$$,

2. (I2)

$$\frac{1}{\varGamma (\beta )} (\log t)^{\beta -1}(1-\log t)\log s(1- \log s)^{\beta -1}\le G_{1}(t,s)\le \frac{\beta -1}{{\varGamma (\beta )}}(\log t)^{\beta -1}(1-\log t)$$ for $$t,s\in [1,e]$$.

### Proof

We note a result from . Let $$\beta \in (n-1,n]$$ with $$n\in \mathbb{N}$$, $$n\ge 3$$. Then the function

$$G(z,l)=\frac{1}{\varGamma (\beta )} \textstyle\begin{cases} z^{\beta -1}(1-l)^{\beta -1}-(z-l)^{\beta -1}, &0\le l\le z\le 1, \\ z^{\beta -1}(1-l)^{\beta -1}, & 0\le z\le l\le 1, \end{cases}$$

has the following properties:

1. (R1)

$$G(z,l)=G(1-l,1-z)$$ for $$z,l\in [0,1]$$;

2. (R2)

$$\varGamma (\beta )k(z)q(l)\le G(z,l)\le (\beta -1)q(l)$$ for $$z,l\in [0,1]$$;

3. (R3)

$$\varGamma (\beta )k(z)q(l)\le G(z,l)\le (\beta -1)k(z)$$ for $$z,l\in [0,1]$$, where $$k(z)=\frac{z^{\beta -1}(1-z)}{\varGamma (\beta )}$$, $$q(l)=\frac{l(1-l)^{ \beta -1}}{\varGamma (\beta )}$$.

Now, we turn our attention to $$G_{1}$$. If logt, logs are regarded as z, l, then from (R2), (R3) we have

\begin{aligned}[b] & \varGamma (\beta )k(\log t)q(\log s)\le G(\log t, \log s)\le (\beta -1)q( \log s), \\ & \varGamma (\beta )k(\log t)q(\log s)\le G(\log t,\log s) \le (\beta -1)k(\log t),\quad \text{for }t,s\in [1,e]. \end{aligned}

Thus (I1), (I2) hold. This completes the proof. □

Let $$\mu (t)=\frac{1}{\varGamma (\beta )}\log t(1-\log t)^{\beta -1}$$ for $$t\in [1,e]$$.

### Lemma 2.5

Let $$\kappa _{1}=\frac{\beta ^{2}\varGamma (\beta )}{ \varGamma (2\beta +2)}$$, $$\kappa _{2}=\frac{\beta -1}{\varGamma (\beta +2)}$$. Then, for any $$s\in [1,e]$$, the following inequalities hold:

$$\kappa _{1} \mu (s)\le \int _{1}^{e} G_{1}(t,s)\mu (t) \frac{dt}{t} \le \kappa _{2} \mu (s).$$
(2.5)

This is a direct result from Lemma 2.4(I1), so we omit its proof.

Let $$E:=C[1,e]$$, $$\|u\|:=\max_{t\in [1,e]}|u(t)|$$, $$P:=\{u\in E:u(t) \geq 0, \forall t\in [1,e]\}$$. Then $$(E,\|\cdot \|)$$ is a real Banach space and P is a cone on E. From Lemma 2.3 and (2.4), we define operators $$A_{i}:P\times P \to P$$ as follows:

$$\textstyle\begin{cases} A_{1}(u,v)(t) =\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{1}(u,v)(t) =}{}+\frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{1}(u,v)(t) =}{}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s},\\ A_{2}(u,v)(t)=\int _{1}^{e} G_{1}(t,s)f_{2}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{2}(u,v)(t)=}{}+\frac{d _{g}(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1} ^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{2}(u,v)(t)=}{}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s}, \end{cases}$$
(2.6)

and

$$A(u,v) (t)=\bigl(A_{1}(u,v),A_{2}(u,v)\bigr) (t)\quad \text{for }t\in [1,e].$$
(2.7)

Note $$A_{i}:P\times P\to P$$, $$A:P\times P\to P\times P$$ are completely continuous operators and $$(u,v)$$ solves (1.1) if and only if $$(u,v)$$ is a fixed point of the operator A.

### Lemma 2.6

Let $$P_{0}=\{z\in P: z(t)\ge \frac{(\log t)^{ \beta -1}(1-\log t)}{\beta -1}\|z\|, \forall t\in [1,e] \}$$. Then $$P_{0}$$ is also a cone on E, and $$A_{i}(P\times P)\subset P_{0}$$, $$i=1,2$$.

### Proof

We only prove $$A_{1}(P\times P)\subset P_{0}$$. From Lemma 2.4(I1), for $$t\in [1,e]$$, we have

\begin{aligned}[b] A_{1}(u,v) (t) &= \int _{1}^{e} G_{1}(t,s) f_{1}\bigl(s,u(s),v(s)\bigr)\frac{ds}{s}\\ &\quad {}+ \frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} f_{2}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \le \int _{1}^{e} \frac{\beta -1}{{\varGamma (\beta )}}\log s(1-\log s)^{ \beta -1} f_{1}\bigl(s,u(s),v(s)\bigr)\frac{ds}{s}\\ &\quad {}+ \frac{d_{h} }{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{1}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} f_{2}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s}, \end{aligned}

and

\begin{aligned}[b] A_{1}(u,v) (t) &\ge \int _{1}^{e} \frac{(\log t)^{\beta -1}(1-\log t)}{ \varGamma (\beta )} \log s(1- \log s)^{\beta -1} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{d_{h} (\log t)^{\beta -1}(1-\log t)}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{ (\log t)^{\beta -1}(1-\log t)}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} f_{2}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s}. \end{aligned}

Note that $$\beta -1>1$$, so we have

$$A_{1}(u,v) (t)\ge \frac{(\log t)^{\beta -1}(1-\log t)}{\beta -1} \bigl\Vert A _{1}(u,v) \bigr\Vert \quad \text{for }u,v\in P, t\in [1,e].$$

This completes the proof. □

### Lemma 2.7

(see )

Let E be a real Banach space and P be a cone on E. Suppose that $$\varOmega \subset E$$ is a bounded open set and that $$A:\overline{\varOmega }\cap P\to P$$ is a continuous compact operator. If there exists $$\omega _{0}\in P\backslash \{0\}$$ such that

$$\omega -A\omega \neq \lambda \omega _{0},\quad \forall \lambda \geq 0, \omega \in \partial \varOmega \cap P,$$

then $$i(A,\varOmega \cap P,P)=0$$, where i denotes the fixed point index on P.

### Lemma 2.8

(see )

Let E be a real Banach space and P be a cone on E. Suppose that $$\varOmega \subset E$$ is a bounded open set with $$0\in \varOmega$$ and that $$A:\overline{\varOmega }\cap P \to P$$ is a continuous compact operator. If

$$\omega -\lambda A\omega \neq 0, \forall \lambda \in [0,1], \omega \in \partial \varOmega \cap P,$$

then $$i(A,\varOmega \cap P,P)=1$$.

## Main results

Let

\begin{aligned}& \kappa _{3}=\frac{\beta }{d_{g,h}\varGamma (2\beta +1)} \int _{1}^{e} g(t) ( \log t)^{\beta -1}(1- \log t)\frac{dt}{t},\\& \kappa _{4}=\frac{\beta }{d _{g,h}\varGamma (2\beta +1)} \int _{1}^{e} h(t) (\log t)^{\beta -1}(1- \log t)\frac{dt}{t}, \\& \kappa _{5}=\frac{\beta (\beta -1)}{d_{g,h}\varGamma (2\beta +1)} \int _{1} ^{e} g(t)\frac{dt}{t},\qquad \kappa _{6}=\frac{\beta (\beta -1)}{d_{g,h} \varGamma (2\beta +1)} \int _{1}^{e} h(t)\frac{dt}{t}. \end{aligned}

Now we list our assumptions for the nonlinearities $$f_{i}$$ ($$i=1,2$$).

1. (H3)

There are $$a_{1i}, b_{1i}\ge 0$$ ($$i=1,2$$) and $$l_{1},l_{2}>0$$ such that

\begin{aligned}[b] & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}< 1,\qquad b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}< 1, \\ &\det \begin{pmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a _{11}\kappa _{3} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\ge \begin{pmatrix} a_{11}x+b_{11}y-l_{1} \\ a_{12}x+b_{12}y-l_{2} \end{pmatrix}, \quad \forall (t,x,y)\in [1,e]\times \mathbb{R}^{+} \times \mathbb{R}^{+}. \end{aligned}
2. (H4)

There are $$a_{2i}, b_{2i}\ge 0$$ ($$i=1,2$$) and $$r_{1}>0$$ such that

\begin{aligned}[b] & (\kappa _{2}+\kappa _{5}d_{h})a_{21}+\kappa _{6}a_{22}< 1,\qquad (\kappa _{2}+d _{g}\kappa _{6})b_{22}+\kappa _{5}b_{21}< 1, \\ & \det \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d_{g}\kappa _{6})b_{22}- \kappa _{5}b_{21} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\le \begin{pmatrix} a_{21}x+b_{21}y \\ a_{22}x+b_{22}y \end{pmatrix},\quad \forall (t,x,y)\in [1,e]\times [0,r_{1}] \times [0,r_{1}]. \end{aligned}
3. (H5)

There are $$a_{3i}, b_{3i}\ge 0$$ ($$i=1,2$$) and $$r_{2}>0$$ such that

\begin{aligned}[b] & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}< 1,\qquad b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}< 1, \\ & \det \begin{pmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a _{31}\kappa _{3} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\ge \begin{pmatrix} a_{31}x+b_{31}y \\ a_{32}x+b_{32}y \end{pmatrix},\quad \forall (t,x,y)\in [1,e]\times [0,r_{2}] \times [0,r_{2}]. \end{aligned}
4. (H6)

There are $$a_{4i}, b_{4i}\ge 0$$ ($$i=1,2$$) and $$l_{3},l_{4}>0$$ such that

\begin{aligned}[b] & (\kappa _{2}+\kappa _{5}d_{h})a_{41}+\kappa _{6} a_{42}< 1,\qquad (\kappa _{2}+d _{g}\kappa _{6})b_{42}+\kappa _{5}b_{41}< 1, \\ & \det \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d_{g}\kappa _{6})b_{42}- \kappa _{5}b_{41} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\le \begin{pmatrix} a_{41}x+b_{41}y+l_{3} \\ a_{42}x+b_{42}y+l_{4} \end{pmatrix}, \quad \forall (t,x,y)\in [1,e]\times \mathbb{R}^{+}\times \mathbb{R}^{+}. \end{aligned}

Let $$B_{\rho }:=\{u\in E:\|u\|<\rho \}$$ for $$\rho >0$$ in the sequel.

### Theorem 3.1

Suppose that (H1)–(H4) hold. Then (1.1) has a positive solution.

### Proof

Let $$S_{1}=\{(u,v)\in P\times P: (u,v)=A(u,v)+\lambda (\varphi _{1},\varphi _{1}), \forall \lambda \ge 0\}$$, where $$\varphi _{1}$$ is a fixed element in $$P_{0}$$. We claim that $$S_{1}$$ is a bounded set in $$P\times P$$. Note if there exists $$(u,v)\in S_{1}$$ such that

$$u(t)=A_{1}(u,v) (t)+\lambda \varphi _{1}(t), \qquad v(t)=A_{2}(u,v) (t)+ \lambda \varphi _{1}(t)\quad \text{for }t\in [1,e],$$
(3.1)

then this, together with Lemma 2.6, implies that

$$u,v\in P_{0}.$$
(3.2)

From (3.1) we have

$$u(t)\ge A_{1}(u,v) (t), \qquad v(t)\ge A_{2}(u,v) (t)\quad \text{for }t\in [1,e].$$
(3.3)

From the definitions of $$A_{i}$$ ($$i=1,2$$), multiplying by $$\mu (t)$$ and integrating from 1 to e, Lemmas 2.4 and 2.5 enable us to obtain

\begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \left ( \textstyle\begin{array}{l} \int _{1}^{e} \mu (t) (\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s)) \frac{ds}{s}\\ \quad {}+\frac{d_{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s)) \frac{ds}{s} \\ \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s} )\frac{dt}{t} \\ \int _{1}^{e} \mu (t) ( \int _{1}^{e} G_{1}(t,s)f _{2}(s,u(s),v(s))\frac{ds}{s}\\ \quad {}+\frac{d_{g}(\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f _{2}(s,u(s),v(s))\frac{ds}{s} \\ \quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s} ) \frac{dt}{t} \end{array}\displaystyle \right ) \\ &\quad \ge \begin{pmatrix} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t}+\kappa _{4} \int _{1}^{e} \mu (t)f_{2}(t,u(t),v(t)) \frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)f _{2}(t,u(t),v(t))\frac{dt}{t}+ \kappa _{3}\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t} \end{pmatrix}. \end{aligned}
(3.4)

Combining this with (H3), we have

\begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \left ( \textstyle\begin{array}{l} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)(a_{11}u(t)+b_{11}v(t)-l _{1})\frac{dt}{t}\\ \quad {}+\kappa _{4} \int _{1}^{e} \mu (t)(a_{12}u(t)+b_{12}v(t)-l _{2})\frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)(a_{12}u(t)+b_{12}v(t)-l_{2})\frac{dt}{t}\\ \quad {}+ \kappa _{3}\int _{1} ^{e} \mu (t)(a_{11}u(t)+b_{11}v(t)-l_{1})\frac{dt}{t} \end{array}\displaystyle \right ), \end{aligned}
(3.5)

and

\begin{aligned}[b] & \begin{pmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a _{11}\kappa _{3} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix} \\ &\quad \le \begin{pmatrix} ((\kappa _{1}+\kappa _{3}d_{h})l_{1}+\kappa _{4}l_{2})\int _{1}^{e} \mu (t)\frac{dt}{t} \\ ((\kappa _{1}+d_{g}\kappa _{4})l_{2}+\kappa _{3}l_{1})\int _{1}^{e} \mu (t)\frac{dt}{t} \end{pmatrix}= \begin{pmatrix} \frac{(\kappa _{1}+\kappa _{3}d_{h})l_{1}+\kappa _{4}l_{2}}{\varGamma ( \beta +2)} \\ \frac{(\kappa _{1}+d_{g}\kappa _{4})l_{2}+\kappa _{3}l _{1}}{\varGamma (\beta +2)} \end{pmatrix}. \end{aligned}

Solving this matrix inequality, we have

$( ∫ 1 e v ( t ) μ ( t ) d t t ∫ 1 e u ( t ) μ ( t ) d t t ) ≤ ( a 12 ( κ 1 + d g κ 4 ) + a 11 κ 3 1 − a 11 ( κ 1 + κ 3 d h ) + a 12 κ 4 1 − b 12 ( κ 1 + d g κ 4 ) + b 11 κ 3 b 11 ( κ 1 + κ 3 d h ) + b 12 κ 4 ) det ( b 11 ( κ 1 + κ 3 d h ) + b 12 κ 4 a 11 ( κ 1 + κ 3 d h ) + a 12 κ 4 − 1 b 12 ( κ 1 + d g κ 4 ) + b 11 κ 3 − 1 a 12 ( κ 1 + d g κ 4 ) + a 11 κ 3 ) ( ( κ 1 + κ 3 d h ) l 1 + κ 4 l 2 Γ ( β + 2 ) ( κ 1 + d g κ 4 ) l 2 + κ 3 l 1 Γ ( β + 2 ) ) .$

Hence, there exist $$M_{1}>0$$, $$M_{2}>0$$ such that

$$\begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix} \le \begin{pmatrix} M_{1} \\ M_{2} \end{pmatrix}.$$

Note (3.2), and we find

$$\begin{pmatrix} \Vert v \Vert \\ \Vert u \Vert \end{pmatrix}\le \begin{pmatrix} \frac{M_{1}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \\ \frac{M_{2}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \end{pmatrix}.$$

This proves that $$S_{1}$$ is bounded in $$P\times P$$. As a result, if we choose $$R_{1}> \{r_{1},\frac{M_{1}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )} , \frac{M_{2}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )} \}$$ ($$r_{1}$$ is defined by (H4)), then we have

$$(u,v)\neq A(u,v)+\lambda (\varphi _{1},\varphi _{1}), \quad \text{for } (u,v) \in \partial B_{R_{1}}\cap (P\times P), \forall \lambda \ge 0.$$

From Lemma 2.7 we have

$$i\bigl(A, B_{R_{1}}\cap (P \times P),P \times P\bigr) = 0.$$
(3.6)

Next we claim that

$$(u,v)\neq \lambda A(u,v),\quad \text{for }(u,v)\in \partial B_{r_{1}} \cap (P\times P), \forall \lambda \in [0,1],$$
(3.7)

where $$r_{1}$$ is defined by (H4). Suppose (3.7) is not true. Then there exist $$(u,v)\in \partial B_{r_{1}}\cap (P\times P)$$ and $$\lambda \in [0,1]$$ such that $$(u,v) = \lambda A(u,v)$$, which implies that

$$u(t)\le A_{1}(u,v) (t),\qquad v(t)\le A_{2}(u,v) (t)\quad \text{for } t\in [1,e].$$
(3.8)

Multiplying by $$\mu (t)$$ and integrating from 1 to e, Lemmas 2.4 and 2.5 enable us to obtain

\begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix} \\ &\quad \le \left ( \textstyle\begin{array}{l} \int _{1}^{e} \mu (t) (\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s)) \frac{ds}{s}\\ \quad {}+\frac{d_{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s)) \frac{ds}{s} \\ \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s} )\frac{dt}{t} \\ \int _{1}^{e} \mu (t) ( \int _{1}^{e} G_{1}(t,s)f _{2}(s,u(s),v(s))\frac{ds}{s}\\ \quad {}+\frac{d_{g}(\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f _{2}(s,u(s),v(s))\frac{ds}{s} \\ \quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s} ) \frac{dt}{t} \end{array}\displaystyle \right ) \\ &\quad \le \begin{pmatrix} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t}+\kappa _{6} \int _{1}^{e} \mu (t)f_{2}(t,u(t),v(t)) \frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)f _{2}(t,u(t),v(t))\frac{dt}{t}+ \kappa _{5}\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t} \end{pmatrix}. \end{aligned}
(3.9)

Substituting (H4) into this matrix inequality, we obtain

\begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \le \begin{pmatrix} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)(a_{21}u(t)+b_{21}v(t)) \frac{dt}{t}+\kappa _{6} \int _{1}^{e} \mu (t)(a_{22}u(t)+b_{22}v(t)) \frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)(a _{22}u(t)+b_{22}v(t))\frac{dt}{t}+ \kappa _{5}\int _{1}^{e} \mu (t)(a _{21}u(t)+b_{21}v(t))\frac{dt}{t} \end{pmatrix}. \end{aligned}
(3.10)

Consequently, we get

$$\begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{22}-\kappa _{5}b_{21} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$$

Therefore, (H4) implies that

$( ∫ 1 e u ( t ) μ ( t ) d t t ∫ 1 e v ( t ) μ ( t ) d t t ) ≤ ( 1 − ( κ 2 + d g κ 6 ) b 22 − κ 5 b 21 ( κ 2 + κ 5 d h ) b 21 + κ 6 b 22 ( κ 2 + d g κ 6 ) a 22 + κ 5 a 21 1 − ( κ 2 + κ 5 d h ) a 21 − κ 6 a 22 ) ( 0 0 ) det ( 1 − ( κ 2 + κ 5 d h ) a 21 − κ 6 a 22 − ( κ 2 + κ 5 d h ) b 21 − κ 6 b 22 − ( κ 2 + d g κ 6 ) a 22 − κ 5 a 21 1 − ( κ 2 + d g κ 6 ) b 22 − κ 5 b 21 ) = ( 0 0 ) .$

Hence,

$$\int _{1}^{e} u(t)\mu (t)\frac{dt}{t}=0, \qquad \int _{1}^{e} v(t)\mu (t) \frac{dt}{t}=0.$$

Note that $$\mu (t)\not \equiv 0$$ for $$t\in [1,e]$$, so $$u(t)=v(t) \equiv 0$$, $$t\in [1,e]$$, which implies that $$\|u\|=\|v\|=0$$, contradicting $$(u,v)\in \partial B_{r_{1}}\cap (P\times P)$$. As a result, (3.7) holds. From Lemma 2.8 we have

$$i\bigl(A, B_{r_{1}}\cap (P \times P),P \times P\bigr) = 1.$$
(3.11)

From (3.6) and (3.11) we have

\begin{aligned} &i\bigl(A,(B_{R_{1}}\backslash \overline{B}_{r_{1}})\cap (P \times P), P \times P\bigr)\\ &\quad =i\bigl(A, B_{R_{1}}\cap (P \times P),P \times P\bigr)-i\bigl(A, B_{r_{1}} \cap (P \times P),P \times P\bigr)=0-1=-1. \end{aligned}

Therefore the operator A has at least one fixed point on $$(B_{R_{1}} \backslash \overline{B}_{r_{1}})\cap (P\times P)$$. Equivalently, (1.1) has at least one positive solution. This completes the proof. □

### Theorem 3.2

Suppose that (H1)–(H2), (H5)–(H6) hold. Then (1.1) has a positive solution.

### Proof

We use similar methods as in Theorem 3.1 to prove this theorem. We first claim that

$$(u,v)\neq A(u,v)+\lambda (\varphi _{2},\varphi _{2}),\quad \text{for } (u,v) \in \partial B_{r_{2}}\cap (P \times P), \forall \lambda \ge 0,$$
(3.12)

where $$\varphi _{2}\in P$$ is a given element. Suppose the claim is not true. Then there exist $$(u,v)\in \partial B_{r_{2}}\cap (P\times P)$$ and $$\lambda \ge 0$$ such that $$(u,v) =A(u,v)+\lambda (\varphi _{2},\varphi _{2})$$, which implies that

$$u(t)\ge A_{1}(u,v) (t),\qquad v(t)\ge A_{2}(u,v) (t) \quad \text{for } t\in [1,e].$$

Similar to (3.4), (3.5), from (H5) we obtain

\begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \begin{pmatrix} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)(a_{31}u(t)+b_{31}v(t)) \frac{dt}{t}+\kappa _{4} \int _{1}^{e} \mu (t)(a_{32}u(t)+b_{32}v(t)) \frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)(a _{32}u(t)+b_{32}v(t))\frac{dt}{t}+ \kappa _{3}\int _{1}^{e} \mu (t)(a _{31}u(t)+b_{31}v(t))\frac{dt}{t} \end{pmatrix}, \end{aligned}

and

\begin{aligned}[b] & \begin{pmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a _{31}\kappa _{3} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \end{aligned}

Thus $$u(t)=v(t)\equiv 0$$ for $$t\in [1,e]$$, and $$\|u\|=\|v\|=0$$, which contradicts $$(u,v)\in \partial B_{r_{2}}\cap {(P\times P)}$$. Consequently, (3.12) holds, and from Lemma 2.7 we have

$$i\bigl(A, B_{r_{2}}\cap (P \times P),P \times P\bigr) = 0.$$
(3.13)

Let $$S_{2}=\{(u,v)\in P\times P: (u,v)=\lambda A(u,v), \forall \lambda \in [0,1] \}$$. Now we prove that $$S_{2}$$ is bounded in $$P\times P$$. Note if there exists $$(u,v)\in S_{2}$$, then

$$u(t)\le A_{1}(u,v) (t),\qquad v(t)\le A_{2}(u,v) (t) \quad \text{for } t\in [1,e],$$

and similar to (3.9), (3.10), and by (H6) we have

\begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix} \\ &\quad \le \left ( \textstyle\begin{array}{l} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)(a_{41}u(t)+b_{41}v(t)+l _{3})\frac{dt}{t}\\ \quad {}+\kappa _{6} \int _{1}^{e} \mu (t)(a_{42}u(t)+b_{42}v(t)+l _{4})\frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)(a_{42}u(t)+b_{42}v(t)+l_{4})\frac{dt}{t}\\ \quad {}+ \kappa _{5}\int _{1} ^{e} \mu (t)(a_{41}u(t)+b_{41}v(t)+l_{3}))\frac{dt}{t} \end{array}\displaystyle \right ). \end{aligned}

Thus

\begin{aligned} & \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{42}-\kappa _{5}b_{41} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \le \begin{pmatrix} \frac{(\kappa _{2}+\kappa _{5}d_{h})l_{3}+\kappa _{6}l_{4}}{\varGamma ( \beta +2)} \\ \frac{(\kappa _{2}+d_{g}\kappa _{6})l_{4}+\kappa _{5}l _{3}}{\varGamma (\beta +2)} \end{pmatrix}. \end{aligned}

Solving this matrix inequality, we have

$( ∫ 1 e u ( t ) μ ( t ) d t t ∫ 1 e v ( t ) μ ( t ) d t t ) ≤ ( 1 − ( κ 2 + d g κ 6 ) b 42 − κ 5 b 41 ( κ 2 + κ 5 d h ) b 41 + κ 6 b 42 ( κ 2 + d g κ 6 ) a 42 + κ 5 a 41 1 − ( κ 2 + κ 5 d h ) a 41 − κ 6 a 42 ) det ( 1 − ( κ 2 + κ 5 d h ) a 41 − κ 6 a 42 − ( κ 2 + κ 5 d h ) b 41 − κ 6 b 42 − ( κ 2 + d g κ 6 ) a 42 − κ 5 a 41 1 − ( κ 2 + d g κ 6 ) b 42 − κ 5 b 41 ) ( ( κ 2 + κ 5 d h ) l 3 + κ 6 l 4 Γ ( β + 2 ) ( κ 2 + d g κ 6 ) l 4 + κ 5 l 3 Γ ( β + 2 ) ) .$

Hence, there exist $$M_{3}>0$$, $$M_{4}>0$$ such that

$$\begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t) \mu (t)\frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} M_{3} \\ M_{4} \end{pmatrix}.$$

Note that $$(u,v)\in S_{2}$$, and from Lemma 2.6, we find $$u,v\in P_{0}$$. Thus, we obtain

$$\begin{pmatrix} \Vert u \Vert \\ \Vert v \Vert \end{pmatrix}\le \begin{pmatrix} \frac{M_{3}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \\ \frac{M_{4}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \end{pmatrix}.$$

This proves that $$S_{2}$$ is bounded in $$P\times P$$. As a result, if we take $$R_{2}> \{r_{2},\frac{M_{3}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )}, \frac{M_{4}(\beta -1)\varGamma (2\beta +2)}{ \beta ^{2}\varGamma (\beta )} \}$$ ($$r_{2}$$ is defined by (H5)), we conclude that

$$(u,v)\neq \lambda A(u,v),\quad \text{for }(u,v)\in \partial B_{R_{2}} \cap (P\times P), \forall \lambda \in [0,1].$$
(3.14)

From Lemma 2.8 we have

$$i\bigl(A, B_{R_{2}}\cap (P \times P),P \times P\bigr) = 1.$$
(3.15)

From (3.13) and (3.15) we have

\begin{aligned} &i\bigl(A,(B_{R_{2}}\backslash \overline{B}_{r_{2}})\cap (P \times P), P \times P\bigr)\\ &\quad =i\bigl(A, B_{R_{2}}\cap (P \times P),P \times P\bigr)-i\bigl(A, B_{r_{2}} \cap (P \times P),P \times P\bigr)=1-0=1. \end{aligned}

Therefore the operator A has at least one fixed point on $$(B_{R_{2}} \backslash \overline{B}_{r_{2}})\cap (P\times P)$$. Equivalently, (1.1) has at least one positive solution. This completes the proof. □

### Example 3.3

Let $$\beta =2.5$$, $$h(t)=g(t)=\log t$$ for $$t\in [1,e]$$. Then $$d_{h}=d_{g}=\int _{1}^{e} (\log t)^{\beta } \frac{dt}{t}=\frac{2}{7}$$, $$d_{g,h}=1-\int _{1}^{e} h(t) (\log t)^{ \beta -1} \frac{dt}{t}\cdot \int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}=1-\frac{4}{49}=\frac{45}{49}$$. This implies that (H2) holds. Next, we calculate $$\kappa _{i}$$ ($$i=1,2,3,4,5,6$$) as follows:

\begin{aligned}& \kappa _{1}=\frac{\beta ^{2}\varGamma (\beta )}{\varGamma (2\beta +2)}=\frac{2.5^{2} \varGamma (2.5)}{\varGamma (7)}\approx 0.01154,\\& \kappa _{2}=\frac{\beta -1}{ \varGamma (\beta +2)}=\frac{1.5}{\varGamma (4.5)} \approx 0.129, \\& \begin{aligned} \kappa _{3}&=\kappa _{4}=\frac{\beta }{d_{g,h}\varGamma (2\beta +1)} \int _{1}^{e} (\log t) (\log t)^{\beta -1}(1-\log t)\frac{dt}{t}\\ &=\frac{2.5}{ \frac{45}{49}\varGamma (6)} \int _{1}^{e} (\log t)^{2.5}(1-\log t) \frac{dt}{t}\approx 0.00144, \end{aligned} \\& \kappa _{5}=\kappa _{6}= \frac{\beta (\beta -1)}{d_{g,h}\varGamma (2\beta +1)} \int _{1}^{e} ( \log t)\frac{dt}{t}= \frac{2.5\times 1.5}{\frac{45}{49}\varGamma (6)} \int _{1}^{e} (\log t)\frac{dt}{t} \approx 0.017. \end{aligned}

Case 1. Let $$a_{11}=10$$, $$a_{12}=600$$, $$b_{11}=630$$, $$b_{12}=7$$, $$a_{21}=3$$, $$a_{22}=4$$, $$b_{21}=3$$, $$b_{22}=2$$. Then we have

\begin{aligned}& a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12} \kappa _{4}=10\times 0.012+600 \times 0.00144< 1,\\& b_{12}(\kappa _{1}+d_{g}\kappa _{4})+b_{11}\kappa _{3}= 7\times 0.012+63 0 \times 0.00144< 1, \\& (\kappa _{2}+\kappa _{5}d_{h})a_{21}+ \kappa _{6}a_{22}=0.134\times 3+0.017 \times 4 < 1,\\& ( \kappa _{2}+d_{g}\kappa _{6})b_{22}+ \kappa _{5}b_{21}=0.134 \times 2+0.017\times 3< 1, \\& \begin{vmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g}\kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a_{11}\kappa _{3} \end{vmatrix}= \begin{vmatrix} 7.57 & -0.016 \\ -0.009 & 7.21 \end{vmatrix}>0, \end{aligned}

and

$$\begin{vmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{22}-\kappa _{5}b_{21} \end{vmatrix}= \begin{vmatrix} 0.53 & -0.436 \\ -0.587 & 0.681 \end{vmatrix}>0.$$

Let $$f_{1}(t,x,y)= (10x+630 y)^{\gamma _{1}}$$, $$f_{2}(t,x,y)=(600x+7 y)^{ \gamma _{2}}$$ for $$t\in [1,e]$$, $$x,y\in \mathbb{R}^{+}$$, $$\gamma _{1}, \gamma _{2}>1$$. Then we have

\begin{aligned}& \liminf_{a_{11}x+b_{11}y\to +\infty }\frac{f_{1}(t,x,y)}{a_{11}x+b _{11}y}=\liminf _{10x+630y\to +\infty } \frac{(10x+630 y)^{\gamma _{1}}}{10x+630y}=+\infty ,\\& \quad \text{uniformly on }t\in [1,e], \\& \liminf_{a_{12}x+b_{12}y\to +\infty }\frac{f_{2}(t,x,y)}{a_{12}x+b _{12}y}=\liminf _{600x+7 y\to +\infty }\frac{(600x+7 y)^{\gamma _{2}}}{600x+7 y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \limsup_{a_{21}x+b_{21}y\to 0^{+}} \frac{f_{1}(t,x,y)}{a_{21}x+b_{21}y}=\limsup _{3x+3y\to 0^{+}} \frac{(10x+630 y)^{\gamma _{1}}}{3x+3y}=0,\quad \text{uniformly on }t \in [1,e], \end{aligned}

and

$$\limsup_{a_{22}x+b_{22}y\to 0^{+}} \frac{f_{2}(t,x,y)}{a_{22}x+b_{22}y}=\limsup _{4x+2y\to 0^{+}} \frac{(600x+7 y)^{\gamma _{2}}}{4x+2y}=0,\quad \text{uniformly on }t\in [1,e].$$

As a result, (H3)–(H4) hold.

Case 2. Let $$a_{31}=8$$, $$a_{32}=620$$, $$b_{31}=630$$, $$b_{32}=7$$, $$a_{41}=3$$, $$a_{42}=4$$, $$b_{41}=3$$, $$b_{42}=2$$. Then we have

\begin{aligned}& a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32} \kappa _{4}=8\times 0.012+620 \times 0.00144< 1,\\& b_{32}(\kappa _{1}+d_{g}\kappa _{4})+b_{31}\kappa _{3}= 7\times 0.012+63 0 \times 0.00144< 1, \\& (\kappa _{2}+\kappa _{5}d_{h})a_{41}+ \kappa _{6}a_{42}=0.134\times 3+0.017 \times 4 < 1,\\& ( \kappa _{2}+d_{g}\kappa _{6})b_{42}+ \kappa _{5}b_{41}=0.134 \times 2+0.017\times 3< 1, \\& \begin{vmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g}\kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a_{31}\kappa _{3} \end{vmatrix}= \begin{vmatrix} 7.57 & -0.0112 \\ -0.009 & 7.45 \end{vmatrix}>0, \end{aligned}

and

$$\begin{vmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{42}-\kappa _{5}b_{41} \end{vmatrix}= \begin{vmatrix} 0.53 & -0.436 \\ -0.587 & 0.681 \end{vmatrix}>0.$$

Let $$f_{1}(t,x,y)= (8x+630 y)^{\gamma _{3}}$$, $$f_{2}(t,x,y)=(620x+7 y)^{ \gamma _{4}}$$ for $$t\in [1,e]$$, $$x,y\in \mathbb{R}^{+}$$, $$\gamma _{3}, \gamma _{4}\in (0,1)$$. Then we have

\begin{aligned}& \liminf_{a_{31}x+b_{31}y\to 0^{+}} \frac{f_{1}(t,x,y)}{a_{31}x+b_{31}y}=\liminf _{8x+630y\to 0^{+}}\frac{(8x+630 y)^{\gamma _{3}}}{8x+630y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \liminf_{a_{32}x+b_{32}y\to 0^{+}} \frac{f_{2}(t,x,y)}{a_{32}x+b_{32}y}=\liminf _{620x+7 y\to 0^{+}}\frac{(620x+7 y)^{\gamma _{4}}}{620x+7 y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \limsup_{a_{41}x+b_{41}y\to +\infty } \frac{f_{1}(t,x,y)}{a_{41}x+b _{41}y}=\limsup _{3x+3y\to +\infty } \frac{(8x+630 y)^{\gamma _{3}}}{3x+3y}=0,\quad \text{uniformly on }t\in [1,e], \end{aligned}

and

$$\limsup_{a_{42}x+b_{42}y\to +\infty } \frac{f_{2}(t,x,y)}{a_{42}x+b _{42}y}=\limsup _{4x+2y\to +\infty } \frac{(620x+7 y)^{\gamma _{4}}}{4x+2y}=0,\quad \text{uniformly on }t\in [1,e].$$

As a result, (H5)–(H6) hold.

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## Funding

This work was supported financially by the National Natural Science Foundation of China (11871302, 11601048), Natural Science Foundation of Chongqing (cstc2016jcyjA0181), Doctoral Scientific Research Foundation of Qufu Normal University and Youth Foundation of Qufu Normal University (BSQD20130140), and Natural Science Foundation of Chongqing Normal University (16XYY24).

## Author information

Authors

### Contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Jiqiang Jiang.

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### Competing interests

The authors declare that there is no conflict of interests regarding the publication of this paper. 