1 Introduction
In this paper we consider the system of nonlinear Hadamard fractional differential equations involving coupled integral boundary conditions
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:
$$ \textstyle\begin{cases} D^{\beta }u(t)+ f_{1}(t,u(t),v(t))=0,\quad 1< t< e, \\ D^{\beta }v(t)+f_{2}(t,u(t),v(t))=0,\quad 1< t< e, \\ u(1)=v(1)=u'(1)=v'(1)=0,\\ u(e)=\int _{1}^{e} h(s) v(s)\frac{ds}{s},\\ v(e)= \int _{1}^{e} g(s) u(s)\frac{ds}{s}, \end{cases} $$
(1.1)
(H1)
\(f_{i}\) (\(i=1,2\)) are nonnegative continuous functions on \([1,e]\times \mathbb{R}^{+}\times \mathbb{R}^{+}\),
(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 [1‐48] and the references therein. In [1‐9], the authors used iterative techniques to study existence and uniqueness of solutions for fractional boundary value problems. In [1] the authors studied positive solutions for the p-Laplacian fractional Riemann–Stieltjes integral boundary value problem
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 [10‐32], the authors used fixed point methods to study the existence of (positive) solutions fractional order equations. In [10] Mawhin’s continuation theorem was used to study the following fractional order boundary value problem at resonance:
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 [23‐32]. In [23], 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
Hadamard fractional differential equations are also popular in the literature; see [33‐48] and the references therein. In [33], 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
where \({}^{C}D^{p_{i}}\), \({}^{H}D^{q_{i}}\) are respectively the Caputo and Hadamard fractional derivatives. In [34] the authors established positive solutions for the coupled Hadamard fractional integral boundary value problems
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:
or
$$ \textstyle\begin{cases} -D_{t}^{\beta }(\varphi _{p}(-D_{t}^{\alpha }z))(t)=f(t,z(t),D_{t}^{ \gamma }z(t)),\quad t\in (0,1), \\ D_{t}^{\alpha }z (0)=D_{t}^{\alpha +1} z(0)=D_{t}^{\gamma }z(0)=0, \\ D_{t}^{\alpha }z (1)=0,\qquad D_{t}^{\gamma }z(1)=\int _{0}^{1} D_{t}^{ \gamma }z(s)\,d A(s), \end{cases} $$
(1.2)
$$ \textstyle\begin{cases} {}^{c}D^{q} x(t)=f(t,x(t),x'(t)),\quad t\in [0,T], \\ x(0)=\alpha I_{\eta }^{\gamma ,\delta }x(\zeta ),\qquad x(T)=\beta {}^{ \rho }I^{p}x(\xi ),\quad 0< \zeta ,\xi \le T, \end{cases} $$
(1.3)
$$ \textstyle\begin{cases} ({}^{c}D^{\alpha }+\lambda {}^{c}D^{\alpha -1} )u(t)=f(t,u(t),v(t),{}^{c}D ^{p_{1}}v(t),I^{q_{1}}v(t)),\quad t\in (0,1), \\ ({}^{c}D^{\beta }+\mu {}^{c}D^{\beta -1} )v(t)=g(t,u(t),{}^{c}D^{p_{2}}u(t),I ^{q_{2}}u(t),v(t)),\quad t\in (0,1), \\ u(0)=u'(0)=u''(0)=0,\qquad u(1)=\int _{0}^{1}u(s)\,dH_{1}(s)+\int _{0}^{1} v(s)\,dH _{2}(s), \\ v(0)=v'(0)=v''(0)=0,\qquad v(1)=\int _{0}^{1}u(s)\,dK_{1}(s)+\int _{0}^{1} v(s)\,dK _{2}(s). \end{cases} $$
(1.4)
$$ \textstyle\begin{cases} {}^{C}D^{p_{1}} {}^{H}D^{q_{1}}x(t)=f(t,x(t),y(t)), \quad t\in [a,b], \\ {}^{H}D^{q_{2}} {}^{C}D^{p_{2}}x(t)=g(t,x(t),y(t)), \quad t\in [a,b], \\ \alpha _{1}x(a)+\alpha _{2} {}^{C}D^{p_{2}}y(a)=0,\qquad \beta _{1}x(b)+\beta _{2} {}^{C}D^{p_{2}}y(b)=0, \\ \alpha _{3}y(a)+\alpha _{4} {}^{H}D^{q_{1}}x(a)=0,\qquad \beta _{3}y(b)+\beta _{4} {}^{H}D^{q_{1}}x(b)=0, \end{cases} $$
(1.5)
$$ \textstyle\begin{cases} D^{\alpha }u(t)+\lambda f(t,u(t),v(t))=0,\quad t\in (1,e), \lambda >0, \\ D^{\beta }v(t)+\lambda g(t,u(t),v(t))=0,\quad t\in (1,e), \lambda >0, \\ u^{(j)}(1)=v^{(j)}(1)=0,\quad 0\le j\le n-2,\\ u(e)=\mu \int _{1}^{e} v(s) \frac{ds}{s},\\ v(e)=\nu \int _{1}^{e} u(s)\frac{ds}{s}, \end{cases} $$
(1.6)
\((\mathrm{H})_{\mathrm{Yang}1}\)
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 \(\liminf_{v\to +\infty }\min_{t\in [\theta _{1},\theta _{2}] }\frac{g(t,u,v)}{v}=+\infty \);
\((\mathrm{H})_{\mathrm{Yang}2}\)
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 \(\liminf_{u\to +\infty }\min_{t\in [\theta _{1},\theta _{2}] }\frac{g(t,u,v)}{u}=+\infty \).
Anzeige
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.
2 Preliminaries
In this section, we first provide some material for Hadamard fractional calculus; for details, see the book [49].
Definition 2.1
The Hadamard derivative of fractional order q for a function \(g: [1,\infty )\rightarrow \mathbb{R}\) is defined as
where \(n=[q]+1\); \([q]\) denotes the integer part of the real number q and \(\log (\cdot )=\log _{e}(\cdot )\).
$$ D^{q}g(t)=\frac{1}{\varGamma (n-q)} \biggl(t\frac{d}{dt} \biggr)^{n} \int _{1}^{t} (\log t-\log s )^{n-q-1}g(s)\frac{ds}{s},\quad n-1< q< n, $$
Definition 2.2
The Hadamard fractional integral of order q for a function \(g: [1,\infty )\rightarrow \mathbb{R}\) is defined as
provided the integral exists.
$$ I^{q}g(t)=\frac{1}{\varGamma (q)} \int _{1}^{t} (\log t-\log s ) ^{q-1} g(s)\frac{ds}{s},\quad q>0, $$
Anzeige
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
can be transformed into the following Hammerstein type integral equations:
where
here, \(d_{g,h}\), \(d_{g}\), \(d_{h}\)
are three positive constants defined in the proof.
$$ \textstyle\begin{cases} D^{\beta }u(t)+x(t)=0,\qquad D^{\beta }v(t)+y(t)=0,\quad t\in (1,e), \\ u(1)=v(1)=u'(1)=v'(1)=0,\\ u(e)=\int _{1}^{e} h(s) v(s)\frac{ds}{s},\\ v(e)= \int _{1}^{e} g(s) u(s)\frac{ds}{s}, \end{cases} $$
(2.1)
$$ \textstyle\begin{cases} u(t)=\int _{1}^{e} G_{1}(t,s) x(s)\frac{ds}{s}+\frac{d_{h} (\log t)^{ \beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} x(s)\frac{ds}{s}\\ \hphantom{u(t)=}{}+\frac{ (\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} y(s) \frac{ds}{s},\\ v(t)=\int _{1}^{e} G_{1}(t,s)y(s)\frac{ds}{s}+\frac{d _{g}(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1} ^{e} h(t)G_{1}(t,s)\frac{dt}{t} y(s)\frac{ds}{s}\\ \hphantom{v(t)=}{}+\frac{(\log t)^{ \beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} x(s)\frac{ds}{s}, \end{cases} $$
(2.2)
$$ G_{1}(t,s)=\frac{1}{\varGamma (\beta )} \textstyle\begin{cases} (\log t)^{\beta -1}(1-\log s)^{\beta -1}- (\log t-\log s ) ^{\beta -1}, & 1\le s\le t\le e, \\ (\log t)^{\beta -1}(1-\log s)^{\beta -1}, & 1\le t\le s \le e; \end{cases} $$
(2.3)
Proof
From Lemma 2.3 of [34] we have
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
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
This implies that
Let \(d_{g,h}=1-\int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \cdot \int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}\), \(d_{h}=\int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t}\), \(d_{g}= [4]\int _{1}^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}\). Then
Consequently, we have
Similarly, we also obtain that
This completes the proof. □
$$ \begin{aligned}[b] & u(t)=c_{11}(\log t)^{\beta -1}+c_{12}( \log t)^{\beta -2}+c_{13}( \log t)^{\beta -3}- \frac{1}{\varGamma (\beta )} \int _{1}^{t} (\log t- \log s )^{\beta -1}x(s)\frac{ds}{s}, \\ & v(t)=c_{21}(\log t)^{ \beta -1}+c_{22}(\log t)^{\beta -2}+c_{23}(\log t)^{\beta -3}- \frac{1}{ \varGamma (\beta )} \int _{1}^{t} (\log t-\log s )^{\beta -1}y(s) \frac{ds}{s}, \end{aligned} $$
$$ \begin{aligned}[b] & u(t)=c_{11}(\log t)^{\beta -1}- \frac{1}{\varGamma (\beta )} \int _{1} ^{t} (\log t-\log s )^{\beta -1}x(s)\frac{ds}{s}, \\ & v(t)=c _{21}(\log t)^{\beta -1}-\frac{1}{\varGamma (\beta )} \int _{1}^{t} (\log t- \log s )^{\beta -1}y(s)\frac{ds}{s}. \end{aligned} $$
$$\begin{aligned}& \begin{aligned}[b] & c_{11}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}\\ &\quad = c_{21} \int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1} ^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t}, \end{aligned} \\& \begin{aligned}[b] & c_{21}-\frac{1}{\varGamma (\beta )} \int _{1}^{e} (1- \log s )^{\beta -1}y(s) \frac{ds}{s}\\ &\quad =c_{11} \int _{1} ^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t}. \end{aligned} \end{aligned}$$
$$ \begin{aligned}[b] & \begin{bmatrix} 1 & -\int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \\ - \int _{1} ^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t} & 1 \end{bmatrix} \begin{bmatrix} c_{11} \\ c_{21} \end{bmatrix} \\ & \quad = \begin{bmatrix} \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1} ^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t} \\ \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1- \log s )^{\beta -1}y(s)\frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t} \end{bmatrix}. \end{aligned} $$
$$ \begin{aligned}[b] \begin{bmatrix} c_{11} \\ c_{21} \end{bmatrix} &= \frac{1}{d_{g,h}} \begin{bmatrix} 1 & \int _{1}^{e} h(t) (\log t)^{\beta -1} \frac{dt}{t} \\ \int _{1} ^{e} g(t) (\log t)^{\beta -1} \frac{dt}{t} & 1 \end{bmatrix} \\ & \quad{} \cdot \begin{bmatrix} \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1-\log s)^{\beta -1} x(s) \frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} h(t) \int _{1} ^{t} (\log t -\log s )^{\beta -1}y(s)\frac{ds}{s} \frac{dt}{t} \\ \frac{1}{\varGamma (\beta )}\int _{1}^{e} (1- \log s )^{\beta -1}y(s)\frac{ds}{s}- \frac{1}{\varGamma (\beta )} \int _{1}^{e} g(t) \int _{1}^{t} (\log t -\log s )^{\beta -1}x(s)\frac{ds}{s}\frac{dt}{t} \end{bmatrix}. \end{aligned} $$
$$\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}$$
$$\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}$$
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:
(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
\(t,s\in [1,e]\),
(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 [14]. Let \(\beta \in (n-1,n]\) with \(n\in \mathbb{N}\), \(n\ge 3\). Then the function
has the following properties:
$$ G(z,l)=\frac{1}{\varGamma (\beta )} \textstyle\begin{cases} z^{\beta -1}(1-l)^{\beta -1}-(z-l)^{\beta -1}, &0\le l\le z\le 1, \\ z^{\beta -1}(1-l)^{\beta -1}, & 0\le z\le l\le 1, \end{cases} $$
(R1)
\(G(z,l)=G(1-l,1-z)\) for \(z,l\in [0,1]\);
(R2)
\(\varGamma (\beta )k(z)q(l)\le G(z,l)\le (\beta -1)q(l)\) for \(z,l\in [0,1]\);
(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
Thus (I1), (I2) hold. This completes the proof. □
$$ \begin{aligned}[b] & \varGamma (\beta )k(\log t)q(\log s)\le G(\log t, \log s)\le (\beta -1)q( \log s), \\ & \varGamma (\beta )k(\log t)q(\log s)\le G(\log t,\log s) \le (\beta -1)k(\log t),\quad \text{for }t,s\in [1,e]. \end{aligned} $$
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:
and
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.
$$ \textstyle\begin{cases} A_{1}(u,v)(t) =\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{1}(u,v)(t) =}{}+\frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{1}(u,v)(t) =}{}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s},\\ A_{2}(u,v)(t)=\int _{1}^{e} G_{1}(t,s)f_{2}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{2}(u,v)(t)=}{}+\frac{d _{g}(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1} ^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s}\\ \hphantom{A_{2}(u,v)(t)=}{}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s}, \end{cases} $$
(2.6)
$$ A(u,v) (t)=\bigl(A_{1}(u,v),A_{2}(u,v)\bigr) (t)\quad \text{for }t\in [1,e]. $$
(2.7)
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
and
Note that \(\beta -1>1\), so we have
This completes the proof. □
$$ \begin{aligned}[b] A_{1}(u,v) (t) &= \int _{1}^{e} G_{1}(t,s) f_{1}\bigl(s,u(s),v(s)\bigr)\frac{ds}{s}\\ &\quad {}+ \frac{d _{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} f_{2}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \le \int _{1}^{e} \frac{\beta -1}{{\varGamma (\beta )}}\log s(1-\log s)^{ \beta -1} f_{1}\bigl(s,u(s),v(s)\bigr)\frac{ds}{s}\\ &\quad {}+ \frac{d_{h} }{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{1}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} f_{2}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s}, \end{aligned} $$
$$ \begin{aligned}[b] A_{1}(u,v) (t) &\ge \int _{1}^{e} \frac{(\log t)^{\beta -1}(1-\log t)}{ \varGamma (\beta )} \log s(1- \log s)^{\beta -1} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{d_{h} (\log t)^{\beta -1}(1-\log t)}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s) \frac{dt}{t} f_{1}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s} \\ & \quad {}+\frac{ (\log t)^{\beta -1}(1-\log t)}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s) \frac{dt}{t} f_{2}\bigl(s,u(s),v(s)\bigr) \frac{ds}{s}. \end{aligned} $$
$$ A_{1}(u,v) (t)\ge \frac{(\log t)^{\beta -1}(1-\log t)}{\beta -1} \bigl\Vert A _{1}(u,v) \bigr\Vert \quad \text{for }u,v\in P, t\in [1,e]. $$
Lemma 2.7
(see [50])
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
then
\(i(A,\varOmega \cap P,P)=0\), where
i
denotes the fixed point index on
P.
$$ \omega -A\omega \neq \lambda \omega _{0},\quad \forall \lambda \geq 0, \omega \in \partial \varOmega \cap P, $$
Lemma 2.8
(see [50])
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
then
\(i(A,\varOmega \cap P,P)=1\).
$$ \omega -\lambda A\omega \neq 0, \forall \lambda \in [0,1], \omega \in \partial \varOmega \cap P, $$
3 Main results
Let
Now we list our assumptions for the nonlinearities \(f_{i}\) (\(i=1,2\)).
$$\begin{aligned}& \kappa _{3}=\frac{\beta }{d_{g,h}\varGamma (2\beta +1)} \int _{1}^{e} g(t) ( \log t)^{\beta -1}(1- \log t)\frac{dt}{t},\\& \kappa _{4}=\frac{\beta }{d _{g,h}\varGamma (2\beta +1)} \int _{1}^{e} h(t) (\log t)^{\beta -1}(1- \log t)\frac{dt}{t}, \\& \kappa _{5}=\frac{\beta (\beta -1)}{d_{g,h}\varGamma (2\beta +1)} \int _{1} ^{e} g(t)\frac{dt}{t},\qquad \kappa _{6}=\frac{\beta (\beta -1)}{d_{g,h} \varGamma (2\beta +1)} \int _{1}^{e} h(t)\frac{dt}{t}. \end{aligned}$$
(H3)
There are \(a_{1i}, b_{1i}\ge 0\) (\(i=1,2\)) and \(l_{1},l_{2}>0\) such that
$$ \begin{aligned}[b] & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}< 1,\qquad b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}< 1, \\ &\det \begin{pmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a _{11}\kappa _{3} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\ge \begin{pmatrix} a_{11}x+b_{11}y-l_{1} \\ a_{12}x+b_{12}y-l_{2} \end{pmatrix}, \quad \forall (t,x,y)\in [1,e]\times \mathbb{R}^{+} \times \mathbb{R}^{+}. \end{aligned} $$
(H4)
There are \(a_{2i}, b_{2i}\ge 0\) (\(i=1,2\)) and \(r_{1}>0\) such that
$$ \begin{aligned}[b] & (\kappa _{2}+\kappa _{5}d_{h})a_{21}+\kappa _{6}a_{22}< 1,\qquad (\kappa _{2}+d _{g}\kappa _{6})b_{22}+\kappa _{5}b_{21}< 1, \\ & \det \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d_{g}\kappa _{6})b_{22}- \kappa _{5}b_{21} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\le \begin{pmatrix} a_{21}x+b_{21}y \\ a_{22}x+b_{22}y \end{pmatrix},\quad \forall (t,x,y)\in [1,e]\times [0,r_{1}] \times [0,r_{1}]. \end{aligned} $$
(H5)
There are \(a_{3i}, b_{3i}\ge 0\) (\(i=1,2\)) and \(r_{2}>0\) such that
$$ \begin{aligned}[b] & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}< 1,\qquad b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}< 1, \\ & \det \begin{pmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a _{31}\kappa _{3} \end{pmatrix}>0, \\ & \begin{pmatrix} f_{1}(t,x,y) \\ f_{2}(t,x,y) \end{pmatrix}\ge \begin{pmatrix} a_{31}x+b_{31}y \\ a_{32}x+b_{32}y \end{pmatrix},\quad \forall (t,x,y)\in [1,e]\times [0,r_{2}] \times [0,r_{2}]. \end{aligned} $$
(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
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
then this, together with Lemma 2.6, implies that
From (3.1) we have
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
Combining this with (H3), we have
and
Solving this matrix inequality, we have
Note (3.2), and we find
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
From Lemma 2.7 we have
$$ u(t)=A_{1}(u,v) (t)+\lambda \varphi _{1}(t), \qquad v(t)=A_{2}(u,v) (t)+ \lambda \varphi _{1}(t)\quad \text{for }t\in [1,e], $$
(3.1)
$$ u,v\in P_{0}. $$
(3.2)
$$ u(t)\ge A_{1}(u,v) (t), \qquad v(t)\ge A_{2}(u,v) (t)\quad \text{for }t\in [1,e]. $$
(3.3)
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \left ( \textstyle\begin{array}{l} \int _{1}^{e} \mu (t) (\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s)) \frac{ds}{s}\\ \quad {}+\frac{d_{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s)) \frac{ds}{s} \\ \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s} )\frac{dt}{t} \\ \int _{1}^{e} \mu (t) ( \int _{1}^{e} G_{1}(t,s)f _{2}(s,u(s),v(s))\frac{ds}{s}\\ \quad {}+\frac{d_{g}(\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f _{2}(s,u(s),v(s))\frac{ds}{s} \\ \quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s} ) \frac{dt}{t} \end{array}\displaystyle \right ) \\ &\quad \ge \begin{pmatrix} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t}+\kappa _{4} \int _{1}^{e} \mu (t)f_{2}(t,u(t),v(t)) \frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)f _{2}(t,u(t),v(t))\frac{dt}{t}+ \kappa _{3}\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t} \end{pmatrix}. \end{aligned} $$
(3.4)
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \left ( \textstyle\begin{array}{l} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)(a_{11}u(t)+b_{11}v(t)-l _{1})\frac{dt}{t}\\ \quad {}+\kappa _{4} \int _{1}^{e} \mu (t)(a_{12}u(t)+b_{12}v(t)-l _{2})\frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)(a_{12}u(t)+b_{12}v(t)-l_{2})\frac{dt}{t}\\ \quad {}+ \kappa _{3}\int _{1} ^{e} \mu (t)(a_{11}u(t)+b_{11}v(t)-l_{1})\frac{dt}{t} \end{array}\displaystyle \right ), \end{aligned} $$
(3.5)
$$ \begin{aligned}[b] & \begin{pmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g} \kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a _{11}\kappa _{3} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix} \\ &\quad \le \begin{pmatrix} ((\kappa _{1}+\kappa _{3}d_{h})l_{1}+\kappa _{4}l_{2})\int _{1}^{e} \mu (t)\frac{dt}{t} \\ ((\kappa _{1}+d_{g}\kappa _{4})l_{2}+\kappa _{3}l_{1})\int _{1}^{e} \mu (t)\frac{dt}{t} \end{pmatrix}= \begin{pmatrix} \frac{(\kappa _{1}+\kappa _{3}d_{h})l_{1}+\kappa _{4}l_{2}}{\varGamma ( \beta +2)} \\ \frac{(\kappa _{1}+d_{g}\kappa _{4})l_{2}+\kappa _{3}l _{1}}{\varGamma (\beta +2)} \end{pmatrix}. \end{aligned} $$
Hence, there exist \(M_{1}>0\), \(M_{2}>0\) such that
$$ \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix} \le \begin{pmatrix} M_{1} \\ M_{2} \end{pmatrix}. $$
$$ \begin{pmatrix} \Vert v \Vert \\ \Vert u \Vert \end{pmatrix}\le \begin{pmatrix} \frac{M_{1}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \\ \frac{M_{2}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \end{pmatrix}. $$
$$ (u,v)\neq A(u,v)+\lambda (\varphi _{1},\varphi _{1}), \quad \text{for } (u,v) \in \partial B_{R_{1}}\cap (P\times P), \forall \lambda \ge 0. $$
$$ i\bigl(A, B_{R_{1}}\cap (P \times P),P \times P\bigr) = 0. $$
(3.6)
Next we claim that
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
Multiplying by \(\mu (t)\) and integrating from 1 to e, Lemmas 2.4 and 2.5 enable us to obtain
Substituting (H4) into this matrix inequality, we obtain
Consequently, we get
Therefore, (H4) implies that
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
$$ (u,v)\neq \lambda A(u,v),\quad \text{for }(u,v)\in \partial B_{r_{1}} \cap (P\times P), \forall \lambda \in [0,1], $$
(3.7)
$$ u(t)\le A_{1}(u,v) (t),\qquad v(t)\le A_{2}(u,v) (t)\quad \text{for } t\in [1,e]. $$
(3.8)
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix} \\ &\quad \le \left ( \textstyle\begin{array}{l} \int _{1}^{e} \mu (t) (\int _{1}^{e} G_{1}(t,s) f_{1}(s,u(s),v(s)) \frac{ds}{s}\\ \quad {}+\frac{d_{h} (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )} \int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s)) \frac{ds}{s} \\ \quad {}+\frac{ (\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f_{2}(s,u(s),v(s))\frac{ds}{s} )\frac{dt}{t} \\ \int _{1}^{e} \mu (t) ( \int _{1}^{e} G_{1}(t,s)f _{2}(s,u(s),v(s))\frac{ds}{s}\\ \quad {}+\frac{d_{g}(\log t)^{\beta -1}}{d_{g,h} \varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} h(t)G_{1}(t,s)\frac{dt}{t} f _{2}(s,u(s),v(s))\frac{ds}{s} \\ \quad {}+\frac{(\log t)^{\beta -1}}{d_{g,h}\varGamma (\beta )}\int _{1}^{e} \int _{1}^{e} g(t)G_{1}(t,s)\frac{dt}{t} f_{1}(s,u(s),v(s))\frac{ds}{s} ) \frac{dt}{t} \end{array}\displaystyle \right ) \\ &\quad \le \begin{pmatrix} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t}+\kappa _{6} \int _{1}^{e} \mu (t)f_{2}(t,u(t),v(t)) \frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)f _{2}(t,u(t),v(t))\frac{dt}{t}+ \kappa _{5}\int _{1}^{e} \mu (t)f_{1}(t,u(t),v(t)) \frac{dt}{t} \end{pmatrix}. \end{aligned} $$
(3.9)
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \le \begin{pmatrix} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)(a_{21}u(t)+b_{21}v(t)) \frac{dt}{t}+\kappa _{6} \int _{1}^{e} \mu (t)(a_{22}u(t)+b_{22}v(t)) \frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)(a _{22}u(t)+b_{22}v(t))\frac{dt}{t}+ \kappa _{5}\int _{1}^{e} \mu (t)(a _{21}u(t)+b_{21}v(t))\frac{dt}{t} \end{pmatrix}. \end{aligned} $$
(3.10)
$$ \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{22}-\kappa _{5}b_{21} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$
Hence,
$$ \int _{1}^{e} u(t)\mu (t)\frac{dt}{t}=0, \qquad \int _{1}^{e} v(t)\mu (t) \frac{dt}{t}=0. $$
$$ i\bigl(A, B_{r_{1}}\cap (P \times P),P \times P\bigr) = 1. $$
(3.11)
From (3.6) and (3.11) we have
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. □
$$ \begin{aligned} &i\bigl(A,(B_{R_{1}}\backslash \overline{B}_{r_{1}})\cap (P \times P), P \times P\bigr)\\ &\quad =i\bigl(A, B_{R_{1}}\cap (P \times P),P \times P\bigr)-i\bigl(A, B_{r_{1}} \cap (P \times P),P \times P\bigr)=0-1=-1. \end{aligned} $$
Theorem 3.2
Proof
We use similar methods as in Theorem 3.1 to prove this theorem. We first claim that
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
Similar to (3.4), (3.5), from (H5) we obtain
and
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
$$ (u,v)\neq A(u,v)+\lambda (\varphi _{2},\varphi _{2}),\quad \text{for } (u,v) \in \partial B_{r_{2}}\cap (P \times P), \forall \lambda \ge 0, $$
(3.12)
$$ u(t)\ge A_{1}(u,v) (t),\qquad v(t)\ge A_{2}(u,v) (t) \quad \text{for } t\in [1,e]. $$
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \ge \begin{pmatrix} (\kappa _{1}+\kappa _{3}d_{h})\int _{1}^{e} \mu (t)(a_{31}u(t)+b_{31}v(t)) \frac{dt}{t}+\kappa _{4} \int _{1}^{e} \mu (t)(a_{32}u(t)+b_{32}v(t)) \frac{dt}{t} \\ (\kappa _{1}+d_{g}\kappa _{4}) \int _{1}^{e} \mu (t)(a _{32}u(t)+b_{32}v(t))\frac{dt}{t}+ \kappa _{3}\int _{1}^{e} \mu (t)(a _{31}u(t)+b_{31}v(t))\frac{dt}{t} \end{pmatrix}, \end{aligned} $$
$$ \begin{aligned}[b] & \begin{pmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g} \kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a _{31}\kappa _{3} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} v(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} u(t) \mu (t)\frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \end{aligned} $$
$$ i\bigl(A, B_{r_{2}}\cap (P \times P),P \times P\bigr) = 0. $$
(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
and similar to (3.9), (3.10), and by (H6) we have
Thus
Solving this matrix inequality, we have
Note that \((u,v)\in S_{2}\), and from Lemma 2.6, we find \(u,v\in P_{0}\). Thus, we obtain
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
From Lemma 2.8 we have
$$ u(t)\le A_{1}(u,v) (t),\qquad v(t)\le A_{2}(u,v) (t) \quad \text{for } t\in [1,e], $$
$$ \begin{aligned}[b] & \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix} \\ &\quad \le \left ( \textstyle\begin{array}{l} (\kappa _{2}+\kappa _{5}d_{h})\int _{1}^{e} \mu (t)(a_{41}u(t)+b_{41}v(t)+l _{3})\frac{dt}{t}\\ \quad {}+\kappa _{6} \int _{1}^{e} \mu (t)(a_{42}u(t)+b_{42}v(t)+l _{4})\frac{dt}{t} \\ (\kappa _{2}+d_{g}\kappa _{6}) \int _{1}^{e} \mu (t)(a_{42}u(t)+b_{42}v(t)+l_{4})\frac{dt}{t}\\ \quad {}+ \kappa _{5}\int _{1} ^{e} \mu (t)(a_{41}u(t)+b_{41}v(t)+l_{3}))\frac{dt}{t} \end{array}\displaystyle \right ). \end{aligned} $$
$$ \begin{aligned} & \begin{pmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{42}-\kappa _{5}b_{41} \end{pmatrix} \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t)\mu (t) \frac{dt}{t} \end{pmatrix}\\ &\quad \le \begin{pmatrix} \frac{(\kappa _{2}+\kappa _{5}d_{h})l_{3}+\kappa _{6}l_{4}}{\varGamma ( \beta +2)} \\ \frac{(\kappa _{2}+d_{g}\kappa _{6})l_{4}+\kappa _{5}l _{3}}{\varGamma (\beta +2)} \end{pmatrix}. \end{aligned} $$
Hence, there exist \(M_{3}>0\), \(M_{4}>0\) such that
$$ \begin{pmatrix} \int _{1}^{e} u(t)\mu (t)\frac{dt}{t} \\ \int _{1}^{e} v(t) \mu (t)\frac{dt}{t} \end{pmatrix}\le \begin{pmatrix} M_{3} \\ M_{4} \end{pmatrix}. $$
$$ \begin{pmatrix} \Vert u \Vert \\ \Vert v \Vert \end{pmatrix}\le \begin{pmatrix} \frac{M_{3}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \\ \frac{M_{4}(\beta -1)\varGamma (2\beta +2)}{\beta ^{2}\varGamma (\beta )} \end{pmatrix}. $$
$$ (u,v)\neq \lambda A(u,v),\quad \text{for }(u,v)\in \partial B_{R_{2}} \cap (P\times P), \forall \lambda \in [0,1]. $$
(3.14)
$$ i\bigl(A, B_{R_{2}}\cap (P \times P),P \times P\bigr) = 1. $$
(3.15)
From (3.13) and (3.15) we have
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. □
$$ \begin{aligned} &i\bigl(A,(B_{R_{2}}\backslash \overline{B}_{r_{2}})\cap (P \times P), P \times P\bigr)\\ &\quad =i\bigl(A, B_{R_{2}}\cap (P \times P),P \times P\bigr)-i\bigl(A, B_{r_{2}} \cap (P \times P),P \times P\bigr)=1-0=1. \end{aligned} $$
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
and
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
and
As a result, (H3)–(H4) hold.
$$\begin{aligned}& a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12} \kappa _{4}=10\times 0.012+600 \times 0.00144< 1,\\& b_{12}(\kappa _{1}+d_{g}\kappa _{4})+b_{11}\kappa _{3}= 7\times 0.012+63 0 \times 0.00144< 1, \\& (\kappa _{2}+\kappa _{5}d_{h})a_{21}+ \kappa _{6}a_{22}=0.134\times 3+0.017 \times 4 < 1,\\& ( \kappa _{2}+d_{g}\kappa _{6})b_{22}+ \kappa _{5}b_{21}=0.134 \times 2+0.017\times 3< 1, \\& \begin{vmatrix} b_{11}(\kappa _{1}+\kappa _{3}d_{h})+b_{12}\kappa _{4} & a_{11}(\kappa _{1}+\kappa _{3}d_{h})+a_{12}\kappa _{4}-1 \\ b_{12}(\kappa _{1}+d_{g}\kappa _{4})+b_{11}\kappa _{3}-1 & a_{12}(\kappa _{1}+d_{g}\kappa _{4})+a_{11}\kappa _{3} \end{vmatrix}= \begin{vmatrix} 7.57 & -0.016 \\ -0.009 & 7.21 \end{vmatrix}>0, \end{aligned}$$
$$ \begin{vmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{21}-\kappa _{6}a_{22} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{21}-\kappa _{6}b_{22} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{22}-\kappa _{5}a_{21} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{22}-\kappa _{5}b_{21} \end{vmatrix}= \begin{vmatrix} 0.53 & -0.436 \\ -0.587 & 0.681 \end{vmatrix}>0. $$
$$\begin{aligned}& \liminf_{a_{11}x+b_{11}y\to +\infty }\frac{f_{1}(t,x,y)}{a_{11}x+b _{11}y}=\liminf _{10x+630y\to +\infty } \frac{(10x+630 y)^{\gamma _{1}}}{10x+630y}=+\infty ,\\& \quad \text{uniformly on }t\in [1,e], \\& \liminf_{a_{12}x+b_{12}y\to +\infty }\frac{f_{2}(t,x,y)}{a_{12}x+b _{12}y}=\liminf _{600x+7 y\to +\infty }\frac{(600x+7 y)^{\gamma _{2}}}{600x+7 y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \limsup_{a_{21}x+b_{21}y\to 0^{+}} \frac{f_{1}(t,x,y)}{a_{21}x+b_{21}y}=\limsup _{3x+3y\to 0^{+}} \frac{(10x+630 y)^{\gamma _{1}}}{3x+3y}=0,\quad \text{uniformly on }t \in [1,e], \end{aligned}$$
$$ \limsup_{a_{22}x+b_{22}y\to 0^{+}} \frac{f_{2}(t,x,y)}{a_{22}x+b_{22}y}=\limsup _{4x+2y\to 0^{+}} \frac{(600x+7 y)^{\gamma _{2}}}{4x+2y}=0,\quad \text{uniformly on }t\in [1,e]. $$
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
and
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
and
As a result, (H5)–(H6) hold.
$$\begin{aligned}& a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32} \kappa _{4}=8\times 0.012+620 \times 0.00144< 1,\\& b_{32}(\kappa _{1}+d_{g}\kappa _{4})+b_{31}\kappa _{3}= 7\times 0.012+63 0 \times 0.00144< 1, \\& (\kappa _{2}+\kappa _{5}d_{h})a_{41}+ \kappa _{6}a_{42}=0.134\times 3+0.017 \times 4 < 1,\\& ( \kappa _{2}+d_{g}\kappa _{6})b_{42}+ \kappa _{5}b_{41}=0.134 \times 2+0.017\times 3< 1, \\& \begin{vmatrix} b_{31}(\kappa _{1}+\kappa _{3}d_{h})+b_{32}\kappa _{4} & a_{31}(\kappa _{1}+\kappa _{3}d_{h})+a_{32}\kappa _{4}-1 \\ b_{32}(\kappa _{1}+d_{g}\kappa _{4})+b_{31}\kappa _{3}-1 & a_{32}(\kappa _{1}+d_{g}\kappa _{4})+a_{31}\kappa _{3} \end{vmatrix}= \begin{vmatrix} 7.57 & -0.0112 \\ -0.009 & 7.45 \end{vmatrix}>0, \end{aligned}$$
$$ \begin{vmatrix} 1-(\kappa _{2}+\kappa _{5}d_{h})a_{41}-\kappa _{6} a_{42} & -(\kappa _{2}+ \kappa _{5}d_{h})b_{41}-\kappa _{6}b_{42} \\ -(\kappa _{2}+d_{g}\kappa _{6})a_{42}-\kappa _{5}a_{41} & 1-(\kappa _{2}+d _{g}\kappa _{6})b_{42}-\kappa _{5}b_{41} \end{vmatrix}= \begin{vmatrix} 0.53 & -0.436 \\ -0.587 & 0.681 \end{vmatrix}>0. $$
$$\begin{aligned}& \liminf_{a_{31}x+b_{31}y\to 0^{+}} \frac{f_{1}(t,x,y)}{a_{31}x+b_{31}y}=\liminf _{8x+630y\to 0^{+}}\frac{(8x+630 y)^{\gamma _{3}}}{8x+630y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \liminf_{a_{32}x+b_{32}y\to 0^{+}} \frac{f_{2}(t,x,y)}{a_{32}x+b_{32}y}=\liminf _{620x+7 y\to 0^{+}}\frac{(620x+7 y)^{\gamma _{4}}}{620x+7 y}=+\infty ,\quad \text{uniformly on }t\in [1,e], \\& \limsup_{a_{41}x+b_{41}y\to +\infty } \frac{f_{1}(t,x,y)}{a_{41}x+b _{41}y}=\limsup _{3x+3y\to +\infty } \frac{(8x+630 y)^{\gamma _{3}}}{3x+3y}=0,\quad \text{uniformly on }t\in [1,e], \end{aligned}$$
$$ \limsup_{a_{42}x+b_{42}y\to +\infty } \frac{f_{2}(t,x,y)}{a_{42}x+b _{42}y}=\limsup _{4x+2y\to +\infty } \frac{(620x+7 y)^{\gamma _{4}}}{4x+2y}=0,\quad \text{uniformly on }t\in [1,e]. $$
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