\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 135, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/135\hfil Positive solutions for a system] {Positive solutions for a system of second-order boundary-value problems involving first-order derivatives} \author[K. Wang, Z. Yang \hfil EJDE-2012/135\hfilneg] {Kun Wang, Zhilin Yang} % in alphabetical order \address{Kun Wang \newline Department of Mathematics \\ Qingdao Technological University\\ Qingdao, 266033, China} \email{wangkun880304@163.com} \address{Zhilin Yang \newline Department of Mathematics \\ Qingdao Technological University \\ Qingdao, 266033, China} \email{zhilinyang@sina.com} \thanks{Submitted May 31, 2012. Published August 17, 2012.} \subjclass[2000]{34B18, 45G15, 45M20, 47H07, 47H11} \keywords{System of second-order boundary-value problems; positive solution; \hfill\break\indent first-order derivative; fixed point index; $\mathbb{R}_+^2$-monotone matrix; concave function} \begin{abstract} In this article we study the existence and multiplicity of positive solutions for the system of second-order boundary value problems involving first order derivatives \begin{gather*} -u''=f(t, u, u', v, v'),\\ -v''=g(t, u, u', v, v'),\\ u(0)=u'(1)=0,\quad v(0)=v'(1)=0. \end{gather*} Here $f,g\in C([0,1]\times \mathbb{R}_+^{4},\mathbb{R}_+) (\mathbb{R}_+:=[0,\infty))$. We use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing Jensen's integral inequality for concave functions and $\mathbb{R}_+^2$-monotone matrices. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In this article we study the existence and multiplicity of positive solutions for the system of second-order boundary value problems involving first order derivatives \begin{equation}\label{second-order} \begin{gathered} -u''=f(t, u, u', v, v'),\\ -v''=g(t, u, u', v, v'),\\ u(0)=u'(1)=0,\quad v(0)=v'(1)=0, \end{gathered} \end{equation} where $f\in C([0,1]\times \mathbb{R}_+^{4},\mathbb{R}_+)$ and $g\in C([0,1]\times \mathbb{R}_+^{4},\mathbb{R}_+)$. By a positive solution of \eqref{second-order}, we mean a pair of functions $(u,v)\in C^{2}[0,1]\times C^{2}[0,1]$ that solve \eqref{second-order} and satisfy $u(t)\geq 0$, $v(t)\geq 0$ for all $t\in [0,1]$, with at least one of them positive on $(0,1]$. Boundary-value problems for systems of nonlinear second-order ordinary differential equations arise from physics, biology, chemistry, and other applied sciences, and, as a result, play an important role in both theory and application. Recently, there are many articles in this direction. We refer the reader to \cite{Caglar,Hu,Infante,Kong,PingKang,Lishan,Liuw,Lu,thompson,Zhongli,Xi,% Jinbao,Yang,Youming} and the references cited therein. It should remarked that in the works cited above, only a few of them involve first-order derivatives in their nonlinearities. In \cite{Wang}, the authors study the existence and multiplicity of positive solutions for the system \begin{gather*} \begin{aligned} &(-1)^mw^{(2m)}\\ &=f(t, w, w',-w''',\dots, (-1)^{m-1}w^{(2m-1)}, z, z',-z''',\dots, (-1)^{n-1}z^{(2n-1)}), \end{aligned}\\ \begin{aligned} &(-1)^nz^{(2n)}\\ &=g(t, w, w',-w''',\dots, (-1)^{m-1}w^{(2m-1)}, z, z',-z''',\dots, (-1)^{n-1}z^{(2n-1)}), \end{aligned}\\ w^{(2i)}(0)=w^{(2i+1)}(1)=0\quad (i=0,1,\dots, m-1),\\ z^{(2j)}(0)=z^{(2j+1)}(1)=0\quad (j=0,1,\dots, n-1). \end{gather*} The hypotheses imposed on the nonlinearities $f$ and $g$ are formulated in terms of two linear functions $h_1(x)$ and $h_2(y)$. The main results in \cite{Wang} are established by using fixed point index theory based on a priori estimates of positive solutions achieved by utilizing new integral inequalities and nonnegative matrices. In \cite{Yang6}, motivated by \cite{Yang3}, Yang and Kong studied the system of second-order boundary value problems involving first-order derivatives \begin{equation}\label{Yang} -u''_{i}=f_i(t,u_1,u_1',\dots,u_n,u_n'), u_i(0)=u'_i(1)=0,i=1,\dots,n. \end{equation} To obtain the a priori estimates of positive solutions, the authors develop some integral identities and inequalities so that the main conditions imposed on $f_i'$s in \cite{Yang6} can be formulated in terms of simple linear functions of the form $g_i(x_1,\dots,x_{2n}):=\sum^{n}_{i=1}a_i(x_{2i-1}+2x_{2i})$. More precisely, for example, (H2) in \cite{Yang6} states that there exist a nonnegative matrix $A = (a_{ij})_{n\times n}$ and a constant $c> 0$ such that the matrix $A- I$ is an $\mathbb{R}^n_ +$-monotone matrix and \[ f_i(t, x)\geq \sum_{j=1}^n a_{ij}(x_{2j-1} + 2x_{2j}) -c \] for all $(t, x) \in [0, 1] \times \mathbb{R}^{2n}_+$, $i = 1,\dots, n$. Motivated by \cite{Wang,Yang6,Yang3}, in this paper, we study the existence and multiplicity of positive solutions for \eqref{second-order}. We use fixed point index theory to establish our main results based on a priori estimates of positive solutions for some associated problems, generalizing the corresponding ones for the single boundary value problem \[ -u''=f(t,u,u'),\quad u(0)=u'(1)=0 \] in \cite{Yang3}. Our generalizations are not routine, as our conditions imposed on the nonlinearities $f$ and $g$, unlike these in \cite{Yang3}, involve both linear functions on $\mathbb{R}_+^4$ and concave functions on $\mathbb{R}_+ $; these functions describe how the nonlinearities $f,g$ grow and enable us to treat the three cases of them: one with both superlinear, one with both sublinear and the last with one superlinear and the other sublinear. Also, it is of interest to note that, for nonnegative constants $p,q,\xi,\eta$ and nonnegative concave functions $\varphi$, $\psi$, we have to prove the ratio \[ \frac{\int_0^1(pu(t)+2qu'(t))\varphi(t)dt} {\int_0^1(\xi u(t)+2\eta u'(t))\psi(t)dt} \] is bounded away from both 0 and $\infty$ (see Lemma \ref{lem2.2} below for more details). This is a great difference between this article and \cite{Wang,Yang6}. We use fixed point index theory to establish our main results based on a priori estimates of positive solutions achieved by utilizing Jensen's integral equality for concave functions and $\mathbb{R}_+^2$-monotone matrices. More precisely, Jensen's inequality is mainly applied to derive the boundedness of weighted integrals of positive solutions for some problems associated to \eqref{second-order}, whereas $\mathbb{R}_+^2$-monotone matrices are employed to solve systems of inequalities resulting from some weighted integrals and thereby achieve the boundedness of associated weighted integrals. This article is organized as follows. Section 2 contains some preliminary results, including two new integral inequalities and a new integral identity. Our main results, namely Theorems \ref{thm3.1}--\ref{thm3.3}, are stated and proved in Section 3. Finally, in Section 4, we presented four examples of nonlinearities to illustrate our main results. \section{Preliminaries} Let $E:=C^1([0,1],\mathbb{R})$ and \[ P:=\{u\in E: u(t)\geq 0,u'(t)\geq 0, \forall t\in [0,1]\}, \|u\|:=\max\{\|u\|_{0},\|u'\|_{0}\}, \] where $\|u\|_{0}:=\max\{|u(t)|:t\in[0,1]\}$. Clearly, $(E, \|\cdot\|)$ is a real Banach space and $P$ is a cone in $E$. For $(u,v)\in E^2$, let \[ \|(u,v)\|:=\max\{\|u \|, \|v\|\}. \] Then $E^2$ is also a real Banach space under the above norm and $P^2$ is a cone in $E^2$. Let $k(t,s):=\min\{t,s\}$ and \[ (Tu)(t):=\int_0^1k(t,s)u(s)ds. \] Then $T: E\to E$ is a completely continuous, positive, linear operator, with the spectral radius $r(T)=\frac{4}{\pi^2}$ and \begin{equation}\label{varphi} (T\varphi)(s)=\int_0^1k(t,s)\varphi(t)dt=r(T)\varphi(s) \end{equation} where $\varphi(t):= \sin\frac{\pi}{2}t$. In our setting, problem \eqref{second-order} is equivalent to the system of nonlinear integral equations \begin{equation}\label{integro-ordinary(2)} \begin{gathered} u(t)=\int_0^1k(t,s)f(s, u(s),u'(s),v(s),v'(s))ds,\\ v(t)=\int_0^1k(t,s)g(s, u(s),u'(s),v(s),v'(s))ds.\\ \end{gathered} \end{equation} Define the operators $A_{i}(i=1,2):P^2\to P$ and $A: P^2 \to P^2$ by \begin{gather*} A_1(u,v)(t):=\int_0^1k(t,s)f(s, u(s),u'(s),v(s),v'(s))ds,\\ A_2(u,v)(t):=\int_0^1k(t,s)g(s, u(s),u'(s),v(s),v'(s))ds,\\ A(u,v)(t):=(A_{1}(u,v),A_{2}(u,v)). \end{gather*} Now $f\in C([0,1]\times \mathbb{R}_+^{4},\mathbb{R}_+)$ and $g\in C([0,1]\times \mathbb{R}_+^{4},\mathbb{R}_+)$ imply that $A_{i}$ and $A$ are completely continuous operators. Clearly, the existence of positive solutions for \eqref{second-order} is equivalent to that of positive fixed points of $A: P^2\to P^2$. To establish the priori estimates of positive solutions for some problems associated with \eqref{second-order}, we need two transcendental equations(see \cite{Yang3}). For any $\xi>\eta>0$, let $\mu(\xi,\eta)\in(1/\xi,1/\eta)$ denote the minimal positive solution of the transcendental equation \begin{equation}\label{transcendental equation1} \eta\mu\sin\sqrt{\xi\mu-\eta^2\mu^2}-\sqrt{\xi\mu -\eta^2\mu^2}\cos\sqrt{\xi\mu-\eta^2\mu^2}=0. \end{equation} Also, for any $\eta>\xi>0$, let $\nu(\xi,\eta)\in(1/\eta,1/\xi)$ denote the unique solution of the transcendental equation \begin{equation}\label{transcendental equation2} \eta\nu\sinh\sqrt{\eta^2\nu^2-\xi\nu}-\sqrt{\eta^2\nu^2 -\xi\nu}\cosh\sqrt{\eta^2\nu^2-\xi\nu}=0. \end{equation} in $(1/\eta,\infty)$. Let \begin{equation}\label{e-function} \varphi_{\xi,\eta}(t):=\begin{cases} \frac{\pi}{2}\sin\frac{\pi t}{2},&\xi>0,\eta=0,\\ te^t ,&\xi=\eta>0,\\ \frac{\xi\mu(\xi,\eta)}{\sqrt{\xi\mu(\xi,\eta)-\eta^2\mu^2(\xi,\eta)}} \exp(\eta\mu(\xi,\eta)t)\\ \times \sin(\sqrt{\xi\mu(\xi,\eta)- \eta^2\mu^2(\xi,\eta)}t),&\xi>\eta>0,\\ \frac{\xi\nu(\xi,\eta)}{\sqrt{\eta^2\nu^2(\xi,\eta)-\xi\nu(\xi,\eta)}} \exp(\eta\nu(\xi,\eta)t)\\ \times \sinh(\sqrt{\eta^2\nu^2(\xi,\eta) -\xi\nu(\xi,\eta)}t), &\eta>\xi>0,\\ \end{cases} \end{equation} \begin{equation}\label{lambda} \lambda(\xi,\eta):=\begin{cases} \frac{\pi^2}{4\xi},& \xi>0,\eta=0\\ \frac{1}{\xi},& \xi=\eta>0,\\ \mu(\xi,\eta),& \xi>\eta>0,\\ \nu(\xi,\eta),& \eta>\xi>0.\\ \end{cases} \end{equation} for all $\xi>0,\eta\geq0$. Direct calculation shows \begin{equation}\label{transcendental equation3} \int_0^1\varphi_{\xi,\eta}(t)dt=1. \end{equation} \begin{lemma} \label{lem2.1} Suppose $\psi\in C([0,1],\mathbb{R}_+)$ is not identically vanishing on $[0,1]$, and $v\in C([0,1],\mathbb{R}_+)$ is a concave function. Let $\varrho(\psi):=\int_0^1t\psi(t)dt>0$. Then we have \begin{equation}\label{estimate1} \int_0^1\psi(t)v(t)dt\geq v(1)\varrho(\psi). \end{equation} \end{lemma} \begin{proof} By the concavity of $v$ and the nonnegativity of $\psi$, we have \[ \int_0^1\psi(t)v(t)dt=\int_0^1\psi(t)v(t\cdot1+(1-t)\cdot0)dt \geq v(1)\int_0^1t\psi(t)dt =v(1)\varrho(\psi). \] This completes the proof. \end{proof} Denote \[ P_0:=\{u\in P: u\text{ is concave on } [0,1],u(0)=u'(1)=0\}. \] \begin{lemma} \label{lem2.2} Let $\xi_i>0, \eta_i\geq0, \varphi_{(\xi_i,\eta_i)}(i=1,2,3)$ be defined by \eqref{e-function}. Define \[ \beta(\xi_1,\eta_1,\xi_2,\eta_2,\xi_3,\eta_3):=\sup_{u\in P_0\setminus\{0\}}\frac{\int_0^1(\xi_1u(t)+2\eta_1u'(t))\varphi_{\xi_2,\eta_2}(t)dt} {\int_0^1(\xi_3u(t)+2\eta_3u'(t))\varphi_{\xi_3,\eta_3}(t)dt}. \] Then $0<\beta(\xi_1,\eta_1,\xi_2,\eta_2,\xi_3,\eta_3)<\infty$. \end{lemma} \begin{proof} If $u\in P_0$, then by Lemma \ref{lem2.1}, we have \begin{equation}\label{integro-eqution2} \begin{aligned} &\int_0^1(\xi_3u(t)+2\eta_3u'(t))\varphi_{\xi_3,\eta_3}(t)dt\\ &\geq\int_0^1\xi_3u(t)\varphi_{\xi_3,\eta_3}(t)dt =\xi_3\int_0^1u(t)\varphi_{\xi_3,\eta_3}(t)dt\\ &\geq u(1)\xi_3\varrho(\varphi_{\xi_3,\eta_3}) =\xi_3\|u\|_{0}\varrho(\varphi_{\xi_3,\eta_3})\\ \end{aligned} \end{equation} and \begin{equation}\label{second} \begin{aligned} &\int_0^1(\xi_1u(t)+2\eta_1u'(t))\varphi_{\xi_2,\eta_2}(t)dt\\ &=\xi_1\int_0^1u(t)\varphi_{\xi_2,\eta_2}(t)dt+2\eta_1\int_0^1u'(t)\varphi_{\xi_2,\eta_2}(t)dt\\ &\leq \xi_1\|\varphi_{\xi_2,\eta_2}\|_{0}\int_0^1u(t)dt+2\eta_1\|\varphi_{\xi_2,\eta_2}\|_0\int_0^1u'(t)dt\\ &\leq \xi_1\|\varphi_{\xi_2,\eta_2}\|_{0}\|u\|_{0}+2\eta_1\|\varphi_{\xi_2,\eta_2}\|_{0}u(1)\\ &=(\xi_1+2\eta_1)\|\varphi_{\xi_2,\eta_2}\|_{0}\|u\|_{0}. \end{aligned} \end{equation} Combining \eqref{integro-eqution2} and \eqref{second}, we obtain \[ \beta(\xi_1,\eta_1,\xi_2,\eta_2,\xi_3,\eta_3) \leq\frac{(\xi_1+2\eta_1)\|\varphi_{\xi_2,\eta_2}\|_{0}} {\xi_3\varrho(\varphi_{\xi_3,\eta_3})}<\infty. \] This completes the proof. \end{proof} \begin{lemma} \label{lem2.3} If $ u\in C^{2}[0,1]$, $u(0)=u'(1)=0$, $\xi>0$, $\eta\geq 0$, then \begin{equation}\label{integro-eqution1} \int_0^1-u''(t)\varphi_{\xi,\eta}(t)dt=\lambda(\xi,\eta)\int_0^1(\xi u(t)+2\eta u'(t))\varphi_{\xi,\eta}(t)dt, \end{equation} where $\varphi_{\xi,\eta}$ and $\lambda(\xi,\eta)$ are defined by \eqref{e-function} and \eqref{lambda} respectively. \end{lemma} \begin{proof} We just prove \eqref{integro-eqution1} in the case $\xi>\eta>0$; the remaining cases can be proved in the same way. Let \[ a:=\eta\mu(\xi,\eta),\quad b:=\sqrt{\xi\mu(\xi,\eta)-\eta^2\mu^2(\xi,\eta)}. \] Then $\varphi_{\xi,\eta}(t)=\frac{\xi a}{\eta b}e^{at}\sin bt$, and \begin{equation}\label{lemma2.2.1} a^2+b^2=\xi\mu(\xi,\eta),a\sin b-b\cos b=0. \end{equation} Integrate by parts over $[0,1]$ and use \eqref{lemma2.2.1} to obtain \begin{equation}\label{lemma2.2.2} \begin{aligned} &\int_0^1-u''(t)\varphi_{\xi,\eta}(t)dt \\ &=\frac{\xi a}{\eta b}\int_0^1-u''(t)e^{at}\sin btdt\\ &=\frac{\xi a}{\eta b}\int_0^1u'(t)e^{at}(a\sin bt+b\cos bt)dt\\ &=\mu(\xi,\eta)\int_0^12\eta u'(t)\varphi_{\xi,\eta}(t)dt +\frac{\xi a}{\eta b}\int_0^1u'(t)e^{at}(b\cos bt-a\sin bt)dt\\ &=\mu(\xi,\eta)\int_0^1(\xi u(t)+2\eta u'(t))\varphi_{\xi,\eta}(t)dt. \end{aligned} \end{equation} This completes the proof. \end{proof} \begin{lemma}[\cite{Guo}] \label{lem2.4} Let $E$ be a real Banach space and $P$ a cone on $E$. Suppose that $\Omega\subset E$ is a bounded open set and that $T:\overline{\Omega}\bigcap P\to P$ is a completely continuous operator. If there exists $w_0\in P\backslash \{0\}$ such that \[ w-Tw\neq \lambda w_0, \forall \lambda\geq 0, \omega\in\partial\Omega\cap P, \] then $i(T,\Omega\bigcap P,P)=0$, where $i$ indicates the fixed point index on $P$. \end{lemma} \begin{lemma}[\cite{Guo}] \label{lem2.5} Let $E$ be a real Banach space and $P$ a cone on $E$. Suppose that $\Omega\subset E$ is a bounded open set with $0\in\Omega$ and that $T:\overline{\Omega}\cap P\to P$ is a completely continuous operator. If \[ w-\lambda Tw\neq0, \forall \lambda\in[0,1], \quad w\in\partial\Omega\cap P, \] then $i(T,\Omega\cap P,P)=1$. \end{lemma} \begin{lemma}[{\cite[Lemma 2.4]{Yang1}}] \label{lem2.6} If $p$ is concave on $[d,\infty)$, with $\lim_{y\to\infty}p(y)/y\geq 0$, then $p$ is increasing on $[d,\infty)$ and \begin{equation}\label{concave} p(y+z-d)\leq p(y)+p(z)-p(d) \end{equation} for all $y,z\in [d,\infty)$. \end{lemma} \section{Existence of positive solutions for \eqref{second-order}} \noindent\textbf{Definition} A real matrix $B$ is said to be nonnegative if all elements of $B$ are nonnegative. \noindent\textbf{Definition} (see \cite[p.112]{Berman}) A real square matrix $M=(m_{ij})_{2\times 2}$ is called $\mathbb{R}_+^2$-monotone, if for any column vector $x\in\mathbb{R}^2$, $Mx\in\mathbb{R}_+^2\Rightarrow x\in\mathbb{R}_+^2$. For simplicity, we denote by $x:=(x_1,x_2,x_3,x_4)\in\mathbb{R}_+^4$ and $I_\rho:=[0,\rho]$ for $\rho>0$. Now we list our hypotheses on $f$ and $g$. \begin{itemize} \item[(H1)] $f, g\in C([0,1]\times\mathbb{R}_+^{4},\mathbb{R}_+)$. \item[(H2)] There exist $p\in C(\mathbb{R}_+,\mathbb{R}_+)$ and $q\in C(\mathbb{R}_+,\mathbb{R}_+)$ such that: (1) $p$ is concave; (2) There are two constants $c>0$ and $\mu_1>1$ such that \[ f(t,x)\geq p(x_3)-c, g(t,x)\geq q(x_1)-c,\quad \forall(t,x)\in[0,1]\times\mathbb{R}_+^4, \] and \[ p(q(t))\geq \frac{\pi^4\mu_1}{16}t-c, \quad \forall t\in\mathbb{R}_+. \] \item[(H3)] For every $N>0$, there exist two functions $\Phi_N, \Psi_N\in C(\mathbb{R}_+,\mathbb{R}_+)$ such that \[ f(t,x)\leq\Phi_N(x_{2}+x_{4}),\quad g(t,x)\leq\Psi_N(x_{2}+x_{4}) \] for all $x\in I_N\times\mathbb{R}_+\times I_N\times\mathbb{R}_+$, $t\in [0,1]$, and \[ \int_0^{\infty}\frac{\tau d\tau}{\Phi_N(\tau)+\Psi_N(\tau)+\delta}=\infty \] for all $\delta>0$. \item[(H4)] There are constants $a_{i}>0$, $b_{i}\geq0$, $c_{i}>0$, $d_{i}\geq 0$ $(i=1,2)$ and $r>0$ such that \[ \begin{pmatrix} f(t,x) \\ g(t,x) \end{pmatrix} \leq \begin{pmatrix} a_1 & 2b_1 & c_1 &2d_1 \\ a_2 & 2b_2 & c_2 & 2d_2 \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \] for all $(t,x)\in[0,1]\times I_r^4$ and the matrix \[ B_1:=\begin{pmatrix}\lambda(a_1,b_1)-1,&-\beta(c_1,d_1,a_1,b_1,c_2,d_2) \\ -\beta(a_2,b_2,c_2,d_2,a_1,b_1),& \lambda(c_2,d_2)-1\end{pmatrix} \] is an $\mathbb{R}_+^2$-monotone matrix, where the entries $\beta(c_1,d_1,a_1,b_1,c_2,d_2)$ and $\beta(a_2,b_2,c_2,d_2,a_1,b_1)$ are defined as in Lemma \ref{lem2.2}. \item[(H5)] There exist $\widetilde{p}\in C(\mathbb{R}_+,\mathbb{R}_+)$ and $\widetilde{q}\in C(\mathbb{R}_+,\mathbb{R}_+)$ such that: (1) $\widetilde{p}$ is concave, $\widetilde{p}(0)=\widetilde{q}(0)=0$; (2) There are two constants $r_2>0$ and $\mu_2>1$ such that \begin{gather*} f(t,x)\geq \widetilde{p}(x_3), g(t,x)\geq \widetilde{q}(x_1), \forall(t,x)\in[0,1]\times I_{r_2}^4, \\ \widetilde{p}(\widetilde{q}(t))\geq \frac{\pi^4\mu_2}{16}t, \forall t\in[0,r_2] \end{gather*} \item[(H6)] There are nonnegative constants $a_{i}>0$, $b_{i}\geq0$, $c_{i}>0$, $d_{i}\geq 0$ $(i=3,4)$ and $c>0$ such that \[ \begin{pmatrix} f(t,x) \\ g(t,x) \end{pmatrix} \leq \begin{pmatrix} a_3 & 2b_3 & c_3 &2d_3 \\ a_4 & 2b_4 & c_4 & 2d_4 \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \\ x_3 \\ x_4 \end{pmatrix} + \begin{pmatrix} c \\ c \end{pmatrix} \] for all $(t,x)\in[0,1]\times\mathbb{R}_+^4$ and the matrix \[ B_2:=\begin{pmatrix} \lambda(a_3,b_3)-1,&-\beta(c_3,d_3,a_3,b_3,c_4,d_4) \\ -\beta(a_4,b_4,c_4,d_4,a_3,b_3),& \lambda(c_4,d_4)-1\end{pmatrix} \] is an $\mathbb{R}_+^2$-monotone matrix, where the entries $\beta(c_3,d_3,a_3,b_3,c_4,d_4)$ and $\beta(a_4,b_4,c_4,d_4,a_3,b_3)$ are defined as in Lemma \ref{lem2.2}. \item[(H7)] $f(t,x)$ and $g(t,x)$ are increasing in $x\in \mathbb{R}_+^4$, and there is a constant $\omega>0$ such that \[ \int_0^1f(s,\omega,\omega,\omega,\omega)ds<\omega, \quad \int_0^1g(s,\omega,\omega,\omega,\omega)ds<\omega. \] \end{itemize} \begin{remark}[{\cite[p.113]{Berman}}] \label{rmk1} \rm A real square matrix $M$ is $\mathbb{R}_+^2$-monotone if and only if $M$ is nonsingular and $M^{-1}$ is nonnegative. \end{remark} \begin{remark} \label{rmk2} \rm Let $l_{ij}(i,j=1,2)$ be four nonnegative constants. Then it is easy to see that the matrix $D:=\begin{pmatrix}l_{11}-1&-l_{12} \\ -l_{21}& l_{22}-1\end{pmatrix}$ is an $\mathbb{R}_+^2$-monotone matrix if and only if $l_{11}>1$, $l_{22}>1,\det D=(l_{11}-1)(l_{22}-1)-l_{12}l_{21}>0$. \end{remark} \begin{remark} \label{rmk3}\rm $f(t,x)$ is said to be increasing in $x$ if \[f(t,x)\leq f(t,y)\] holds for every pair $x,y\in \mathbb {R}^{4}_+$ with $x\leq y$ for all $t\in[0,1]$, where the partial ordering $\leq$ in $\mathbb{R}^{4}_+$ is understood componentwise. \end{remark} We adopt the convention in the sequel that $\widehat{c}_1,\widehat{c}_2,\dots$ stand for different positive constants and $\Omega_\rho:=\{v\in E:\|v\|<\rho\}$ for $\rho>0$. \begin{theorem} \label{thm3.1} If {\rm (H1)--(H4)} hold, then \eqref{second-order} has at least one positive solution. \end{theorem} \begin{proof} By (H2), we obtain \begin{equation}\label{Thm31} A_1(u,v)(t)\geq\int_0^1k(t,s)p(v(s))ds-\widehat{c}_1,\quad A_2(u,v)(t)\geq\int_0^1k(t,s)q(u(s))ds-\widehat{c}_1 \end{equation} for all $(u,v)\in P^2$, $t\in [0,1]$. We claim that the set \[ \mathcal{M}_1:=\{(u,v)\in P^2: (u,v)=A(u,v)+\lambda(\sigma,\sigma), \lambda\geq0\} \] is bounded, where $\sigma(t):=te^{-t}$. Indeed, if $(u_{0},v_{0})\in\mathcal{M}_{1}$, then there exist a constant $\lambda_{0}\geq0$ such that $(u_{0},v_{0})=A(u_{0},v_{0})+\lambda_{0}(\sigma,\sigma)$, which can be written in the form \begin{gather*} u_0(t)=\int_0^1k(t,s)f(s,u_0(s),u'_0(s),v_0(s),v_0'(s))ds+\lambda_0\sigma(t),\\ v_0(t)=\int_0^1k(t,s)g(s,u_0(s),u'_0(s),v_0(s),v_0'(s))ds+\lambda_0 \sigma(t). \end{gather*} By (H2) and \eqref{Thm31}, we have \begin{equation}\label{theorem1-1} u_{0}(t)\geq\int_0^1k(t,s)p(v_{0}(s))ds-\widehat{c}_1,v_{0}(t) \geq\int_0^1k(t,s)q(u_{0}(s))ds-\widehat{c}_1 \end{equation} for all $t\in [0,1]$. The nonnegativity and concavity of $p$ imply $\lim_{y\to\infty}p(y)/y\geq 0$. We also note $\max_{(t,s)\in[0,1]\times[0,1]}k(t,s)=1$. Now Lemma \ref{lem2.6} and Jensen's inequality imply \begin{equation}\label{theorem1-3} p(v_{0}(t))\geq p(v_{0}(t)+\widehat{c}_1)-p(\widehat{c}_1)\geq\int_0^1k(t,s)p(q(u_{0}(s)))ds -p(\widehat{c}_1). \end{equation} This, together with \eqref{theorem1-1} and (H2), implies \begin{equation}\label{theorem1-4} \begin{aligned} u_{0}(t)&\geq\int_0^1k(t,s) \Big[\int_0^1k(s,\tau)p(q(u_{0}(\tau)))d\tau-p(\widehat{c}_1)\Big]ds-\widehat{c}_1\\ &\geq\int_0^1\int_0^1k(t,s)k(s,\tau)p(q(u_{0}(\tau)))\,ds\,d\tau-\widehat{c}_2\\ &\geq\int_0^1\int_0^1k(t,s)k(s,\tau)\Big[\frac{\pi^4\mu_1}{16}u_{0}(\tau)-c\Big]ds\, d\tau-\widehat{c}_2\\ &\geq\frac{\pi^4\mu_1}{16}\int_0^1\int_0^1k(t,s)k(s,\tau)u_{0}(\tau)\,ds\,d\tau -\widehat{c}_3. \end{aligned} \end{equation} Multiply both sides of the last inequality by $\varphi(t):=\sin(\pi t/2)$ and integrate over $[0,1]$ and use \eqref{varphi} twice to obtain \begin{equation}\label{theorem1-5} \begin{aligned} \int_0^1\varphi(t)u_{0}(t)dt &\geq\frac{\pi^4\mu_1}{16}\int_0^1\int_0^1\int_0^1\varphi(t)k(t,s)k(s,\tau)u_{0}(\tau) dt\,ds\,d\tau-\frac{2\widehat{c}_3}{\pi}\\ &=\mu_1\int_0^1\varphi(t)u_{0}(t)dt-\frac{2\widehat{c}_3}{\pi}, \end{aligned} \end{equation} so that \begin{equation}\label{theorem1-6} \int_0^1\varphi(t)u_{0}(t)dt\leq\frac{2\widehat{c}_3}{\pi(\mu_1-1)}. \end{equation} By Lemma \ref{lem2.1}, we have \begin{equation}\label{theorem1-7} \| u_{0}\|_{0}=u_0(1)\leq\frac{2\widehat{c}_3}{\pi\varrho(\varphi)(\mu_1-1)} =\frac{\widehat{c}_3\pi}{2(\mu_1-1)}. \end{equation} Multiply the first inequality of \eqref{theorem1-1} by $\varphi(t)$, integrate over $[0,1]$ and use \eqref{varphi} to obtain \[ \int_0^1u_0(t)\varphi(t)dt\geq\frac{4}{\pi^2}\int_0^1p(v_{0}(t))\varphi(t)dt -\frac{2}{\pi}\widehat{c}_1. \] This, along with \eqref{theorem1-6}, implies \begin{equation}\label{theorem1-8} \int_0^1p(v_{0}(t))\varphi(t)dt\leq\frac{\pi^2}{4} \Big(\frac{2\widehat{c}_1}{\pi}+\int_0^1u_{0}(t)\varphi(t) dt\Big) \leq \frac{\widehat{c}_1\pi}{2}+\frac{\widehat{c}_3\pi}{2(\mu_1-1)}. \end{equation} By Lemma \ref{lem2.1}, we have \begin{equation}\label{theorem1-9} \begin{aligned} \|v_0\|_0=v_0(1) &\leq\frac{1}{\varrho(\varphi)} \int_0^1v_{0}(t)\varphi(t)dt\\ &=\frac{\|v_{0}\|_0}{\varrho(\varphi) p(\|v_0\|_0)} \int_0^1\varphi(t)\frac{v_{0}(t)}{\|v_0\|_0}p(\|v_{0}\|_0)dt\\ &\leq\frac{\|v_{0}\|_0}{\varrho(\varphi) p(\|v_{0}\|_0)} \int_0^1\varphi(t)p(v_{0}(t))dt, \end{aligned} \end{equation} so that \[ p(\|v_{0}\|_0)\leq\frac{1}{\varrho(\varphi) }\int_0^1\varphi(t)p(v_{0}(t))dt \leq\frac{\widehat{c}_1\pi^3}{8}+\frac{\widehat{c}_3\pi^4}{16(\mu_1-1)}. \] (H2) implies that $p$ is strictly increasing and $\lim_{x\to\infty}p(x)=\infty$ (see Lemma \ref{lem2.6}). Consequently, there exists $\widehat{c}_4>0$ such that \[ \|v_{0}\|_0\leq \widehat{c}_4. \] Let $N:=\max \{\frac{\widehat{c}_3\pi}{2(\mu_1-1)}, \widehat{c}_4\}$. Then \begin{equation}\label{theorem1-10} \|u\|_0\leq N,\quad \|v\|_0\leq N, \quad \forall (u,v)\in\mathcal{M}_1. \end{equation} This establishes the a priori bound of $\mathcal{M}_1$ for $\|(u,v)\|_0$. Now it remains to derive the a priori bound of $\mathcal{M}_1$ for $\|(u',v')\|_0$. To this end, we let \[ \Lambda:=\{\mu\geq 0: \text{there exists $(u,v)\in P^2$ \ such that } (u,v)=A(u,v)+\mu(\sigma,\sigma)\}. \] Now \eqref{theorem1-10} imply that $\mu_{0}:=\sup\Lambda<\infty$. By (H3), there are two functions $\Phi_N, \Psi_N\in C(\mathbb{R}_+,\mathbb{R}_+)$ such that \begin{gather*} f(t,u(t),u'(t),v(t),v'(t))\leq\Phi_N(u'(t)+v'(t)),\\ g(t,u(t),u'(t),v(t),v'(t))\leq\Psi_N(u'(t)+v'(t)) \end{gather*} for all $(u,v)\in\mathcal {M}_{1}$, $t\in [0,1]$. Hence, for all $(u,v)\in\mathcal{M}_1$ and for some $\mu\geq 0$, we have \begin{align*} -u''(t) &=f(t,u(t),u'(t),v(t),v'(t))+\mu(2-t)e^{-t}\\ &\leq\Phi_N(u'(t)+v'(t))+\mu(2-t)e^{-t}\\ &\leq\Phi_N(u'(t)+v'(t))+2\mu_0, \end{align*} \begin{align*} -v''(t)&=g(t,u(t),u'(t),v(t),v'(t))+\mu(2-t)e^{-t}\\ &\leq\Psi_N(u'(t)+v'(t))+\mu(2-t)e^{-t}\\ &\leq\Psi_N(u'(t)+v'(t))+2\mu_0, \end{align*} so that \begin{align*} -(u''(t)+v''(t))(u'(t)+v'(t)) &\leq(u'(t) +v'(t))(\Phi_N(u'(t)+v'(t))\\ &\quad +\Psi_N(u'(t)+v'(t))+4\mu_0). \end{align*} This implies \[ \int_0^{u'(0)+v'(0)}\frac{\tau d\tau}{\Phi_N(\tau)+\Psi_N(\tau)+4\mu_0} \leq\int_0^1(u'(t)+v'(t))dt=u(1)+v(1)\leq 2N \] for all $(u,v)\in\mathcal {M}_{1}$. By (H3) again, there exists a constant $N_1>0$ such that \[ \|u'+v'\|_0=u'(0)+v'(0)\leq N_1,\quad \forall (u,v)\in\mathcal {M}_{1}. \] This establishes the a priori bound of $\mathcal{M}_1$ for $\|(u',v')\|_0$ and, in turn, implies that $\mathcal{M}_1$ is bounded(notice that we have achieved the a priori bound of $\mathcal{M}_1$ for $\|(u,v)\|_0$ in \eqref{theorem1-10}). Taking $R>\max\{\sup\{\|(u,v)\|:(u,v)\in\mathcal{M}_1\},r\}$, we have \[ (u,v)\neq A(u,v)+\lambda(\sigma,\sigma), \quad \forall (u,v)\in\partial\Omega_R\cap P^2, \lambda\geq0. \] Now Lemma \ref{lem2.4} yields \begin{equation}\label{theorem2} i(A,\Omega_R\cap P^2,P^2)=0. \end{equation} Let \[ \mathcal{M}_2:=\{(u,v)\in\overline{\Omega}_r\cap P^2:(u,v)=\lambda A(u,v), \lambda\in[0,1]\}. \] Now we want to prove that $\mathcal{M}_2=\{0\}$. Indeed, if $(u,v)\in\mathcal{M}_2$, then $(u,v)\in P_0^2$ and $(u,v)=\lambda A(u,v)$ for some $\lambda\in[0,1]$, written componentwise as \begin{gather*} u(t)=\lambda\int_0^1k(t,s)f(s,u(s),u'(s),v(s),v'(s))ds,\\ v(t)=\lambda\int_0^1k(t,s)g(s,u(s),u'(s),v(s),v'(s))ds, \end{gather*} which are equivalent to \[ -u''(t)=\lambda f(t,u(t),u'(t),v(t),v'(t)), -v''(t)=\lambda g(t,u(t),u'(t),v(t),v'(t)). \] By (H4), we have \begin{gather*} -u''(t)\leq a_{1}u(t)+2b_{1}u'(t)+c_{1}v(t)+2d_{1}v'(t), \\ -v''(t)\leq a_{2}u(t)+2b_{2}u'(t)+c_{2}v(t)+2d_{2}v'(t). \end{gather*} Multiply the last two inequalities by $\varphi_{a_{1},b_{1}}(t)$ and $\varphi_{c_{2},d_{2}}(t)$ respectively and integrate over $[0,1]$ and use Lemmas \ref{lem2.2} and \ref{lem2.3} to obtain \begin{align*} &\lambda(a_{1},b_{1})\int_0^1(a_{1}u(t)+2b_{1}u'(t))\varphi_{a_{1},b_{1}}(t)dt\\ &\leq\int_0^1(a_{1}u(t)+2b_{1}u'(t))\varphi_{a_{1},b_{1}}(t)dt+\int_0^1(c_{1}v(t)+2d_{1}v'(t)) \varphi_{a_{1},b_{1}}(t)dt\\ &\leq\int_0^1(a_{1}u(t)+2b_{1}u'(t))\varphi_{a_{1},b_{1}}(t)dt\\ &\quad +\beta(c_{1},d_{1},a_{1},b_{1},c_{2},d_{2})\int_0^1(c_2v(t) +2d_2v'(t))\varphi_{c_2,d_2}(t)dt, \end{align*} \begin{align*} &\lambda(c_2,d_2)\int_0^1(c_2v(t)+2d_2v'(t))\varphi_{c_2,d_2}(t)dt\\ &\leq\int_0^1(a_{2}u(t)+2b_{2}u'(t))\varphi_{c_2,d_2}(t)dt +\int_0^1(c_{2}v(t)+2d_{2}v'(t)) \varphi_{c_2,d_2}(t)dt\\ &\leq\beta(a_2,b_2,c_2,d_2,a_1,b_1)\int_0^1(a_1u(t) +2b_1u'(t))\varphi_{a_1,b_1}(t)dt\\ &\quad +\int_0^1(c_2v(t)+2d_2v'(t))\varphi_{c_2,d_2}(t)dt, \end{align*} which can be written in the form \begin{align*} &\begin{pmatrix}\lambda(a_1,b_1)-1&-\beta(c_1,d_1,a_1,b_1,c_2,d_2) \\ -\beta(a_2,b_2,c_2,d_2,a_1,b_1)& \lambda(c_2,d_2)-1\end{pmatrix}\\ &\cdot \begin{pmatrix} \int_0^1(a_1u(t)+2b_1u'(t))\varphi_{a_1,b_1}(t)dt\\ \int_0^1(c_2v(t)+2d_2v'(t))\varphi_{c_2,d_2}(t)dt \end{pmatrix} \\ &=B_1 \begin{pmatrix} \int_0^1(a_1u(t)+2b_1u'(t))\varphi_{a_1,b_1}(t)dt \\ \int_0^1(c_2v(t)+2d_2v'(t))\varphi_{c_2,d_2}(t)dt \end{pmatrix}\leq\begin{pmatrix}0 \\ 0 \end{pmatrix}. \end{align*} (H4) again implies \[ \begin{pmatrix} \int_0^1(a_1u(t)+2b_1u'(t))\varphi_{a_1,b_1}(t)dt \\ \int_0^1(c_2v(t)+2d_2v'(t))\varphi_{c_2,d_2}(t)dt \end{pmatrix} \leq B_1^{-1}\begin{pmatrix}0 \\0 \end{pmatrix}=\begin{pmatrix}0 \\0 \end{pmatrix}. \] Consequently, \[ \int_0^1(a_1u(t)+2b_1u'(t))\varphi_{a_1,b_1}(t)dt =\int_0^1(c_2v(t)+2d_2v'(t))\varphi_{c_2,d_2}(t)dt=0 \] and $u=v=0$, whence $\mathcal{M}_2= \{0\}$, as required. As a result of this, we have \[(u,v)\neq\lambda A(u,v), \quad \forall (u,v)\in\partial\Omega_r\cap P^2, \lambda\in[0,1]. \] Now Lemma \ref{lem2.5} yields \[ i(A,\Omega_r\cap P^2,P^2)=1. \] This together with \eqref{theorem2} implies \[ i(A,(\Omega_R\backslash\overline{\Omega}_r)\cap P^2,P^2)=0-1=-1. \] Therefore, $A$ has at least one fixed point $(u,v)$ on $(\Omega_R\backslash\overline{\Omega}_r)\cap P^2$ and thus \eqref{second-order} has at least one positive solution. This completes the proof. \end{proof} \begin{theorem} \label{thm3.2} If {\rm (H1), (H5), (H6)} hold, then \eqref{second-order} has at least one positive solution. \end{theorem} \begin{proof} By (H5), for all $(u,v)\in\overline{\Omega}_{r_2}\cap P^2$, we have \[ A_1(u,v)(t)\geq\int_0^1k(t,s)\widetilde{p}(v(s))ds, \quad A_2(u,v)(t)\geq\int_0^1k(t,s)\widetilde{q}(u(s))ds. \] Let \[ \mathcal{M}_3:=\{(u,v)\in\overline{\Omega}_{r_2}\cap P^2 :(u,v) =A(u,v)+\lambda(\sigma,\sigma),\lambda\geq0\}. \] Now we want to prove that $\mathcal{M}_3\subset\{0\}$, where $\sigma(t):=te^{-t}$. If $(\widetilde{u},\widetilde{v})\in\mathcal{M}_3$, there exists $\widetilde{\lambda}\geq0$ such that $(\widetilde{u},\widetilde{v})=A(\widetilde{u},\widetilde{v}) +\widetilde{\lambda}(\sigma,\sigma)$, which implies \begin{equation}\label{theorem2-1} \widetilde{u}(t)\geq\int_0^1k(t,s)\widetilde{p}(\widetilde{v}(s))ds, \widetilde{v}(t)\geq\int_0^1k(t,s)\widetilde{q}(\widetilde{u}(s))ds \end{equation} Note $\max_{(t,s)\in[0,1]\times[0,1]}k(t,s)=1$. By (H5) and Jensen's inequality, we have \begin{align*} \widetilde{u}(t) &\geq\int_0^1k(t,s)\widetilde{p}(\widetilde{v}(s))ds\\ &\geq\int_0^1k(t,s)\widetilde{p}(\int_0^1k(s,\tau)\widetilde{q}(\widetilde{u}(\tau))d\tau)ds\\ &\geq\int_0^1\int_0^1k(t,s)k(s,\tau)\widetilde{p}(\widetilde{q}(\widetilde{u}(\tau)))d\tau ds\\ &\geq\int_0^1\int_0^1k(t,s)k(s,\tau)\widetilde{p}(\widetilde{q}(\widetilde{u}(\tau)))\,ds\,d\tau\\ &\geq\frac{\pi^4\mu_2}{16}\int_0^1\int_0^1k(t,s)k(s,\tau)\widetilde{u}(\tau)\,ds\,d\tau\,. \end{align*} Multiply both sides of the above inequality by $\varphi(t):=\sin\frac{\pi}{2}t$ and integrate over $[0,1]$ and use \eqref{varphi} to obtain \[ \int_0^1\widetilde{u}(t)\varphi(t)dt \geq\mu_2\int_0^1\widetilde{u}(t)\varphi(t)dt, \] so that $\int_0^1\widetilde{u}(t)\varphi(t)dt=0$, and whence $\widetilde{u}(t)\equiv 0$. This, together with \eqref{theorem2-1}, yields $\widetilde{p}(\widetilde{v}(t))\equiv 0$, and, in particular, \[ \widetilde{p}(\|\widetilde{v}\|_0)=\widetilde{p}(\widetilde{v}(1))= 0. \] Note that (H5) implies that $\widetilde{p}$ is strictly increasing on $[0, \varepsilon]$ for sufficiently small $\varepsilon>0$ (see Lemma \ref{lem2.6}) and thus $\widetilde{v}(1)=0$. Hence $\widetilde{u}=\widetilde{v}=0$, and $\mathcal{M}_3\subset\{0\}$, as required. As a result of this, we have \[ (u,v)\neq A(u,v)+\lambda(\varphi,\varphi), \forall (u,v)\in\partial\Omega_{r_2}\cap P^2, \lambda\geq0. \] Now Lemma \ref{lem2.4} yields \begin{equation}\label{theorem6} i(A,\Omega_{r_2}\cap P^2,P^2)=0. \end{equation} Let \[ \mathcal{M}_{4}:=\{(u,v)\in P^2:(u,v)=\lambda A(u,v), \lambda\in[0,1]\}. \] We now assert that $\mathcal{M}_4$ is bounded. Indeed, if $(u,v)\in\mathcal{M}_{4}$, then $(u,v)\in P_0^2$ and $(u,v)=\lambda A(u,v)$ for some $\lambda\in[0,1]$, which can be written componentwise as \begin{gather*} u(t)=\lambda\int_0^1k(t,s)f(s,u(s),u'(s),v(s),v'(s))d s,\\ v(t)=\lambda\int_0^1k(t,s)g(s,u(s),u'(s),v(s),v'(s))ds. \end{gather*} Differentiate the last equations twice to obtain \[ -u''(t)=\lambda f(t,u(t),u'(t),v(t),v'(t)), \quad -v''(t)=\lambda g(t,u(t),u'(t),v(t),v'(t)), \] for $t\in [0,1]$. By (H6), we have \begin{gather}\label{theorem3} -u''(t)\leq a_{3}u(t)+2b_{3}u'(t)+c_{3}v(t)+2d_{3}v'(t)+\widetilde{c},\\ \label{theorem4} -v''(t)\leq a_{4}u(t)+2b_{4}u'(t)+c_{4}v(t)+2d_{4}v'(t)+\widetilde{c} \end{gather} Multiply the last two inequalities by $\varphi_{a_3,b_3}(t)$ and $\varphi_{c_4,d_4}(t)$ respectively and integrate over $[0,1]$, and use Lemmas \ref{lem2.2} and \ref{lem2.3} to obtain \[\begin{aligned} \lambda(a_3,b_3)&\int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt\\ &\leq\int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt+ \int_0^1(c_3v(t)+2d_3v'(t))\varphi_{a_3,b_3}(t)dt+\widetilde{c}\\ &\leq\int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt\\ &+ \beta(c_3,d_3,a_3,b_3,c_4,d_4)\int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt+\widetilde{c}, \end{aligned}\] and \begin{align*} &\lambda(c_4,d_4)\int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt\\ &\leq\int_0^1(a_4u(t)+2b_4u'(t))\varphi_{c_4,d_4}(t)dt +\int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt+\widetilde{c}\\ &\leq\beta(a_4,b_4,c_4,d_4,a_3,b_3)\int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt\\ &\quad +\int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt+\widetilde{c}, \end{align*} which can be written in the form \begin{align*} &\begin{pmatrix} \lambda(a_3,b_3)-1 &-\beta(c_3,d_3,a_3,b_3,c_4,d_4) \\ -\beta(a_4,b_4,c_4,d_4,a_3,b_3)& \lambda(c_4,d_4)-1\end{pmatrix}\\ &\cdot \begin{pmatrix}\int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt \\ \int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt \end{pmatrix}\\ &=B_2 \begin{pmatrix}\int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt \\ \int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt \end{pmatrix} \leq\begin{pmatrix}\widetilde{c} \\ \widetilde{c} \end{pmatrix}. \end{align*} (H6) again implies \[ \begin{pmatrix} \int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt \\ \int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt \end{pmatrix} \leq B_2^{-1}\begin{pmatrix}\widetilde{c} \\ \widetilde{c} \end{pmatrix}:=\begin{pmatrix}\widetilde{c}_1 \\ \widetilde{c}_2 \end{pmatrix}. \] Let $\widetilde{c}_3:=\max\{\widetilde{c}_1,\widetilde{c}_2\}>0$. Then we have \[ \int_0^1(a_3u(t)+2b_3u'(t))\varphi_{a_3,b_3}(t)dt\leq \widetilde{c}_3, \int_0^1(c_4v(t)+2d_4v'(t))\varphi_{c_4,d_4}(t)dt\leq \widetilde{c}_3, \] for all $ (u,v)\in\mathcal{M}_4$. By Lemma \ref{lem2.1}, we obtain \[ \|u\|_0=u(1)\leq\frac{\widetilde{c}_3}{a_3\varrho(\varphi_{a_3,b_3})}, \quad \|v\|_0=v(1)\leq\frac{\widetilde{c}_3}{c_4\varrho(\varphi_{c_4,d_4})}, \] for all $(u,v)\in\mathcal{M}_4$. Let $\widetilde{N}=\max\{\frac{\widetilde{c}_3}{a_3\varrho(\varphi_{a_3,b_3})}, \frac{\widetilde{c}_3}{c_4\varrho(\varphi_{c_4,d_4})}\}>0$. Then by \eqref{theorem3} and \eqref{theorem4}, we have \begin{gather*} -u''(t)\leq (a_3+c_3)\widetilde{N}+2b_3u'(t)+2d_3v'(t)+\widetilde{c}, \\ -v''(t)\leq (a_4+c_4)\widetilde{N}+2b_4u'(t)+2d_4v'(t)+\widetilde{c}, \end{gather*} for all $(u,v)\in\mathcal{M}_4$. Adding the above inequalities yields \[ -u''(t)-v''(t)\leq (a_3+a_4+c_3+c_4)\widetilde{N} +2(b_3+b_4)u'(t)+2(d_3+d_4)v'(t)+2\widetilde{c}. \] Let \[ \widetilde{N}_2:=(a_3+a_4+c_3+c_4)\widetilde{N}+2\widetilde{c}, \widetilde{L}:=2(b_3+b_4+d_3+d_4)+1. \] Noticing $u'(1)=v'(1)=0$, we obtain \[ u'(t)+v'(t)\leq\frac{\widetilde{N}_2}{\widetilde{L}}(e^{\widetilde{L} -\widetilde{L}t}-1), \] so that \[ \|u'+v'\|_0=u'(0)+v'(0)\leq \frac{\widetilde{N}_2}{\widetilde{L}}(e^{\widetilde{L}}-1). \] This proves the boundedness of $\mathcal{M}_4$, as asserted. Taking $R>\max\{\sup\{\|(u,v)\|:(u,v)\in\mathcal{M}_4\},r_2\}$, we have \[ (u,v)\neq\lambda A(u,v), \quad \forall(u,v)\in\partial\Omega_R\cap(P\times P),\; \lambda\in[0,1]. \] Now Lemma \ref{lem2.5} yields \begin{equation}\label{theorem5} i(A,\Omega_R\cap P^2,P^2)=1. \end{equation} Combining \eqref{theorem6} and \eqref{theorem5} gives \[ i(A,(\Omega_R\backslash\overline{\Omega}_{r_2})\cap P^2,P^2)=1. \] Hence $A$ has at least one fixed point on $(\Omega_R\backslash\overline{\Omega}_{r_2})\cap P^2$. Thus \eqref{second-order} has at least one positive solution. This completes the proof. \end{proof} \begin{theorem} \label{thm3.3} If {\rm (H1)--(H3), (H5), (H7)} hold, then \eqref{second-order} has at least two positive solutions. \end{theorem} \begin{proof} By (H7), we have \[ f(t,x)\leq f(t,\omega,\omega,\omega,\omega),\quad g(t,x)\leq g(t,\omega,\omega,\omega,\omega), \] for all $t\in[0,1]$ and all $x\in I_\omega^4$. Consequently, we have for all $(u,v)\in \partial\Omega_{\omega}\cap P^2$, \begin{align*} \|A_1(u,v)\|_0=A_1(u,v)(1) &=\int_0^1sf(s,u(s),u'(s),v(s),v'(s))ds\\ &\leq\int_0^1f(s,u(s),u'(s),v(s),v'(s))ds\\ &\leq\int_0^1f(s,\omega,\omega,\omega,\omega)ds\\ &<\omega=\|(u,v)\|, \end{align*} \begin{align*} \|A_2(u,v)\|_0=A_2(u,v)(1) &=\int_0^1sg(s,u(s),u'(s),v(s),v'(s))ds\\ &\leq\int_0^1g(s,u(s),u'(s),v(s),v'(s))ds\\ &\leq\int_0^1g(s,\omega,\omega,\omega,\omega)ds\\ &<\omega=\|(u,v)\|, \end{align*} \begin{align*} \|(A_1(u,v))'\|_0 &=(A_1(u,v))'(0)=\int_0^1f(s,u(s),u'(s),v(s),v'(s))ds\\ &\leq\int_0^1f(s,\omega,\omega,\omega,\omega)ds\\ &<\omega=\|(u,v)\|, \end{align*} and \begin{align*} \|(A_2(u,v))'\|_0 &=(A_2(u,v))'(0)=\int_0^1g(s,u(s),u'(s),v(s),v'(s))ds\\ &\leq\int_0^1g(s,\omega,\omega,\omega,\omega)ds\\ &<\omega=\|(u,v)\|. \end{align*} The preceding inequalities imply \[ \|A(u,v)\|=\|(A_1(u,v),A_2(u,v))\|<\omega=\|(u,v)\|; \] thus \[ (u,v)\neq\lambda A(u,v), \quad \forall (u,v)\in\partial\Omega_{\omega}\cap P^2,\; 0\leq\lambda\leq1. \] Now Lemma \ref{lem2.5} yields \begin{equation}\label{theorem7} i(A,\Omega_{\omega}\cap P^2,P^2)=1. \end{equation} By (H2), (H3) and (H5), we find that \eqref{theorem2} and \eqref{theorem6} hold. Note that we can choose $R>\omega>r_2$ in \eqref{theorem2} and \eqref{theorem6} (see the proofs of Theorems \ref{thm3.1} and \ref{thm3.2}). Combining \eqref{theorem2}, \eqref{theorem6} and \eqref{theorem7}, we obtain \begin{gather*} i(A,(\Omega_{R}\backslash\overline{\Omega}_{\omega})\cap P^2,P^2)=0-1=-1, \\ i(A,(\Omega_{\omega}\backslash\overline{\Omega}_{r_2})\cap P^2,P^2)=1-0=1. \end{gather*} Therefore, $A$ has at least two fixed points, with one on $(\Omega_{R}\backslash\overline{\Omega}_{\omega})\cap P^2$ and the other on $(\Omega_{\omega}\backslash\overline{\Omega}_{r_2})\cap P^2$. Hence \eqref{second-order} has at least two positive solutions. \end{proof} \section{Examples} In this section we present four examples to illustrate our main results. \begin{example} \label{examp4.1} \rm Let $(a_{ij})_{2\times4}$ be a positive matrix with $1<\alpha_i\leq 2(i=1,2)$ and \[ f(t,x):=\Big(\sum_{j=1}^{4} a_{1j}x_j\Big)^{\alpha_1}, \quad g(t,x):=\Big(\sum_{j=1}^{4} a_{2j}x_j\Big)^{\alpha_2},\quad x\in\mathbb{R}_+^{4},\; t\in [0,1]. \] Then (H1)-(H4) holds with both $f$ and $g$ superlinear. By Theorem \ref{thm3.1}, Equaton \eqref{second-order} has at least one positive solution. It suffices to verify (H2)-(H4). (1) Let $p(y):=y$ and $q(y):=a_{21} y^{\alpha_2}$. Then $p$ is concave and $p(q(y))/y\to \infty(y\to \infty)$. It is easy to see that there exists $c>0$ such that \[ f(t,x)\geq p(x_3)-c, g(t,x)\geq q(x_1)-c \] for all $x\in\mathbb{R}_+^4$ and $t\in [0,1]$. This implies that (H2) holds true. (2) Assumption (H3) holds with \[ \Phi_N(t):=\big((a_{12}+a_{14})t+2N\big)^{\alpha_1}, \quad \Psi_N(t):=\big((a_{22}+a_{24})t+2N\big)^{\alpha_2} \] for every $N>0$. (3) Note that there is $r>0$ such that \[ \begin{pmatrix} f(t,x) \\ g(t,x) \end{pmatrix} \leq \frac{1}{3} \begin{pmatrix} 1& 2 & 1 & 2 \\ 1& 2 & 1 &2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \] for all $x\in I_r^4, t\in [0,1]$. Let $a:=\frac{1}{3}$. Then \[ \lambda (a,a)=\frac{1}{a}=3, \beta(a,a,a,a,a,a)=1. \] Obviously, the matrix \[ B_1: = \begin{pmatrix} \lambda (a,a)-1 & -\beta(a,a,a,a,a,a) \\ -\beta(a,a,a,a,a,a) & \lambda (a,a)-1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \] is $\mathbb{R}_2^+$-monotone. Therefore (H4) holds with $a_i=b_i=c_i=d_i=a(i=1,2)$ and $B_1$ as defined above. \end{example} \begin{example} \label{examp4.2}\rm Let $(b_{ij})_{2\times 4}$ be a positive matrix with $0<\alpha_i<1(i=3,4)$ and \[ f(t,x):=\Big(\sum_{j=1}^{4} b_{1j}x_j\Big)^{\alpha_3},\quad g(t,x):=\Big(\sum_{j=1}^{4}b_{2j}x_j\Big)^{\alpha_4}, \quad x\in\mathbb{R}_+^{4},\; t\in [0,1]. \] Now (H1), (H5) and (H6) hold with both $f$ and $g$ sublinear. By Theorem \ref{thm3.2}, \eqref{second-order} has at least one positive solution. \end{example} \begin{example} \label{examp4.3} \rm Let $(c_{ij})_{2\times 4}$ be a positive matrix and \[ f(t,x):=\frac{\sum_{j=1}^{4}c_{1j}x_j^{\frac{5}{3}}}{\sum_{j=1}^{4}x_j+1},\quad g(t,x):=\frac{\sum_{j=1}^{4}c_{2j}x_j^3}{\sum_{j=1}^{4}x_j+1},\quad x\in\mathbb{R}_+^{4}, \; t\in [0,1]. \] Now (H1)--(H4) hold with $f$ sublinear and $g$ superlinear at $\infty$. By Theorem \ref{thm3.1}, Equation \eqref{second-order} has at least one positive solution. \end{example} \begin{example} \label{examp4.4} \rm Let $(a_{ij})_{2\times 4}$, $(b_{ij})_{2\times 4}$ be two positive matrices, with $1<\beta_i\leq 2$, $0<\gamma_i<1(i=1,2)$ and \[ \Big(\sum_{j=1}^{4} a_{1j}\Big)^{\beta_1} +\Big(\sum_{j=1}^{4}b_{1j}\Big)^{\gamma_1}<1, \quad \Big(\sum_{j=1}^{4}a_{2j}\Big)^{\beta_2} +\Big(\sum_{j=1}^{4} b_{2j}\Big)^{\gamma_2}<1. \] Let \begin{gather*} f(t,x):=\Big(\sum_{j=1}^{4} a_{1j}x_j\Big)^{\beta_1} +\Big(\sum_{j=1}^{4} b_{1j}x_j\Big)^{\gamma_1}, \\ g(t,x):=\Big(\sum_{j=1}^{4}a_{2j}x_j\Big)^{\beta_2} +\Big(\sum_{j=1}^{4} b_{2j}x_j\Big)^{\gamma_2} \end{gather*} for $x\in\mathbb{R}_+^4$, $t\in [0,1]$. Now (H1)--(H3), (H5) and (H7) hold with both $f$ and $g$ superlinear at $\infty$ and sublinear at $0$. By Theorem \ref{thm3.3}, \eqref{second-order} has at least two positive solutions. \end{example} \begin{thebibliography}{99} \bibitem{Berman} A. Berman, J. 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