\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 52, pp. 1--15.\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-2012/52\hfil Positive solutions for a system] {Positive solutions for a system of higher order boundary-value problems involving all derivatives of odd orders} \author[K. Wang, Z. Yang\hfil EJDE-2012/52\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 October 24, 2011. Published March 30, 2012.} \subjclass[2000]{34B18, 45G15, 45M20, 47H07, 47H11} \keywords{Systenm of higher order boundary value problem; positive solution; \hfill\break\indent nonnegative matrix; fixed point index} \begin{abstract} In this article we study the existence of positive solutions for the system of higher order boundary-value problems involving all derivatives of odd orders \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*} Here $f,g\in C([0,1]\times\mathbb{R}_+^{m+n+2},\mathbb{R}_+)$ $(\mathbb{R}_+:=[0,+\infty))$. Our hypotheses imposed on the nonlinearities $f$ and $g$ are formulated in terms of two linear functions $h_1(x)$ and $h_2(y)$. We use fixed point index theory to establish our main results based on a priori estimates of positive solutions achieved by utilizing nonnegative 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 higher order boundary-value problems: $$\label{higher-order} \begin{gathered} (-1)^mw^{(2m)}=f(t, w, w',\dots, (-1)^{m-1}w^{(2m-1)}, z, z',\dots, (-1)^{n-1}z^{(2n-1)}),\\ (-1)^nz^{(2n)}=g(t, w, w',\dots, (-1)^{m-1}w^{(2m-1)}, z, z',\dots, (-1)^{n-1}z^{(2n-1)}),\\ w^{(2i)}(0)=w^{(2i+1)}(1)=0\quad (i=0,1,\dots, m-1),\\ z^{(2j)}(0)=z^{(2j+1)}(1)=0\quad (j=0,1,\dots, n-1), \end{gathered}$$ where $m,n\geq 2, f\in C([0,1]\times \mathbb{R}_+^{m+n+2},\mathbb{R}_+)$ and $g\in C([0,1]\times \mathbb{R}_+^{m+n+2},\mathbb{R}_+)$ $(\mathbb{R}_+:=[0,+\infty))$. By a positive solution of \eqref{higher-order}, we mean a pair of functions $(w,z)\in C^{2m}[0,1]\times C^{2n}[0,1]$ that solve \eqref{higher-order} and satisfy $w(t)\geq 0$, $z(t)\geq 0$ for all $t\in[0,1]$, with at least one of them positive on $(0,1]$. The so-called Lidstone problem $$\label{Lidstone} \begin{gathered} (-1)^nu^{(2n)}=f(t, u, -u'',\dots, (-1)^{n-1}u^{(2n-2)}),\\ u^{(2i)}(0)=u^{(2i)}(1)=0, \quad (i=0,1,\dots, n-1), \end{gathered}$$ has been extensively studied in recent years; see \cite{Yao,YuhongMa,YuanmingWang1,YuanmingWang2,Wei1,Wei2} and the references cited therein. The existence of positive solutions for systems of nonlinear differential equations have been studied by many authors; see for instance, \cite{Dunninger1,Henderson1,Liu1,Cao1}, to cite a few. In \cite{Yang2}, the author studied the existence of positive solutions of the system \begin{gather*} (-1)^mu^{(2m)}=f_1(t, u,-u'',\dots, (-1)^{m-1}u^{(2m-2)}, v,-v'',\dots, (-1)^{n-1}v^{(2n-2)}),\\ (-1)^nv^{(2n)}=f_2(t, u,-u'',\dots, (-1)^{m-1}u^{(2m-2)}, v,-v'',\dots, (-1)^{n-1}v^{(2n-2)}),\\ \alpha_0u^{(2i)}(0)-\beta_0u^{(2i+1)}(0) =\alpha_1u^{(2i)}(1)+\beta_1u^{(2i+1)}(1)=0\quad (i=0,1,\dots, m-1),\\ \alpha_0v^{(2j)}(0)-\beta_0v^{(2j+1)}(0) =\alpha_1v^{(2j)}(1)+\beta_1v^{(2j+1)}(1)=0\quad (j=0,1,\dots,n-1), \end{gather*} where $m,n\geq 1$ and $f_1,f_2\in C([0,1]\times \mathbb{R}_+^{m+n},\mathbb{R}_+)$. The main results obtained in \cite{Yang2} are presented in terms of nonnegative matrices and the author used the method of order reduction to overcome the difficulty arising from high order derivatives. Furthermore, in \cite{Kang}, Kang et al., using the fixed point theorem of cone expansion and compression type due to Krasnosel'skill, established some simple criteria for the existence, multiplicity and nonexistence of positive solutions of the following systems of singular boundary value problems with integral boundary value conditions: \begin{gather*} (-1)^pu^{(2p)}=\lambda a_1(t)f(t,u,-u'',\dots,(-1)^{p-1}u^{2p-2},v,-v'', \dots,(-1)^{q-1}v^{2q-2}),\\ (-1)^qv^{(2q)}=\mu a_2(t)g(t,u,-u'',\dots,(-1)^{p-1}u^{2p-2},v,-v'', \dots,(-1)^{q-1}v^{2q-2}),\\ a_iu^{(2i)}(0)-b_iu^{(2i+1)}(0)=\int_0^1m_i(s)u^{(2i)}(s)ds,\quad 0\leq i\leq p-1,\\ c_iu^{(2i)}(1)-d_iu^{(2i+1)}(1)=\int_0^1n_i(s)u^{(2i)}(s)ds, \quad 0\leq i\leq p-1,\\ \alpha_jv^{(2j)}(0)-\beta_jv^{(2j+1)}(0)=\int_0^1\varphi_j(s)v^{(2j)}(s)ds, \quad 0\leq j\leq q-1,\\ \gamma_jv^{(2j)}(1)-\delta_jv^{(2j+1)}(1)=\int_0^1\psi_j(s)v^{(2j)}(s)ds, \quad 0\leq j\leq q-1, \end{gather*} where $00$ such that $f(t,x,y)\geq a_1h_1(x)+b_1h_2(y)-c, g(t,x,y)\geq a_2h_1(x)+b_2h_2(y)-c,$ for all $(t,x,y)\in[0,1]\times \mathbb{R}_+^{m+n+2}$ and the matrix $B_1:=\begin{pmatrix} a_1-1&b_1 \\ a_2& b_2-1\end{pmatrix}$ is invertible with $B_1^{-1}$ nonnegative. \item[(F3)] For every $N>0$, there exist two functions $\Phi_N, \Psi_N\in C(\mathbb{R}_+,\mathbb{R}_+)$ such that $f(t,x,y)\leq\Phi_N(x_{m+1}+y_{n+1}), g(t,x,y)\leq\Psi_N(x_{m+1}+y_{n+1})$ for all $(x_1,\dots,x_m)\in\underbrace{[0,N]\times\dots\times[0,N]}_m$, $(y_1,\dots,y_n)\in\underbrace{[0,N]\times\dots\times [0,N]}_n$ and $x_{m+1},y_{n+1}\geq0$, and $\int_0^{\infty}\frac{\tau d\tau}{\Phi_N(\tau)+\Psi_N(\tau)+\delta}=\infty$ for all $\delta>0$. \item[(F4)] There are four nonnegative constants $c_1, c_2, d_1, d_2$ and a positive constant $r$ such that $f(t,x,y)\leq c_1 h_1(x)+d_1 h_2(y), \quad g(t,x,y)\leq c_2 h_1(x)+d_2 h_2(y)$ for all $(t,x,y)\in[0,1]\times([0,r])^{m+n+2}$ and $B_2:=\begin{pmatrix} 1-c_1&-d_1 \\ -c_2& 1-d_2\end{pmatrix}$ is invertible with $B_2^{-1}$ nonnegative. \item[(F5)] There are four nonnegative constants $l_1,l_2,m_1,m_2$ and a postive constant $c$ such that $f(t,x,y)\leq l_1 h_1(x)+m_1 h_2(y)+c, g(t,x,y)\leq l_2 h_1(x)+m_2 h_2(y)+c,$ for all $(t,x,y)\in[0,1]\times\mathbb {R}_+^{m+n+2}$ and $B_3:=\begin{pmatrix} 1-l_1&-m_1 \\ -l_2& 1-m_2\end{pmatrix}$ is invertible with $B_3^{-1}$ nonnegative. \item[(F6)] There are four nonnegative constants $p_1,p_2,q_1,q_2$ and a positive constant $r$ such that $f(t,x,y)\geq p_1h_1(x)+q_1h_2(y), g(t,x,y)\geq p_2h_1(x)+q_2h_2(y),$ for all $(t,x,y)\in[0,1]\times([0,r])^{m+n+2}$ and $B_4:=\begin{pmatrix} p_1-1&q_1 \\ p_2& q_2-1\end{pmatrix}$ is invertible with $B_4^{-1}$ nonnegative. \item[(F7)] $f(t,x,y)$ and $g(t,x,y)$ are increasing in $x$ and $y$,and there is a constant $\Lambda>0$ such that $\int_0^1f(s,\underbrace{\Lambda,\dots,\Lambda}_{m+n+2})ds<\Lambda, \int_0^1g(s,\underbrace{\Lambda,\dots,\Lambda}_{m+n+2})ds<\Lambda.$ \end{itemize} \begin{remark}[{\cite[Remark 2]{Yang2}}] \label{rmk1} \rm Let $l_{ij}(i,j=1,2)$ be four nonnegative constants. Then it is easy to see that the matrix $B:=\begin{pmatrix}l_{11}-1&l_{12} \\ l_{21}& l_{22}-1\end{pmatrix}$ is invertible with $B^{-1}$ nonnegative if and only if one of the following two conditions is satisfied: \begin{itemize} \item[(1)] $l_{11}> 1$, $l_{22}>1$, $l_{12}=l_{21}=0$. \item[(2)] $l_{11}\leq1$, $l_{22}\leq1$, $\det B=(1-l_{11})(1-l_{22})-l_{12}l_{21}<0$. \end{itemize} \end{remark} \begin{remark}[{\cite[Remark 3]{Yang2}}] \label{rmk2} \rm Let $l_{ij}(i,j=1,2)$ be four nonnegative constants. Then it is easy to see that the matrix $D:=\begin{pmatrix} 1-l_{11}&-l_{12} \\ -l_{21}& 1-l_{22}\end{pmatrix}$ is invertible with $D^{-1}$ nonnegative if and only if $l_{11}<1,l_{22}<1,\det D=(1-l_{11})(1-l_{22})-l_{12}l_{21}>0$. \end{remark} \begin{remark} \label{rmk3}\rm $f(t,x,y)$ is said to be increasing in $x$ and $y$ if $f(t,x,y)\leq f(t,x',y')$ holds for every pair $(x,y),(x',y')\in \mathbb {R}^{m+n+2}_+$ with $(x,y)\leq(x',y')$, where the partial ordering $\leq$ in $\mathbb{R}_+^{m+n+2}$ is understood componentwise. \end{remark} We have the following comments about the functions $f$ and $g$. \begin{itemize} \item[(1)] Condition (F3) is of Berstein-Nagumo type; \item[(2)] $f$ and $g$ grow superlinearly both at $+\infty$ and at $0$ if (F2) and (F4) hold; \item[(3)] $f$ and $g$ grow sublinearly both at $+\infty$ and at $0$ if (F5) and (F6) hold. \end{itemize} We adopt the convention in the sequel that $n_1,n_2,\dots$ stand for different positive constants. $G_1$ and $G_2$ are defined by \eqref{convention1} and \eqref{convention2}. \begin{theorem} \label{thm3.1} If {\rm (F1)--(F4)} hold, then \eqref{higher-order} has at least one positive solution. \end{theorem} \begin{proof} It suffices to prove that \eqref{integro-or2} has at least one positive solution. We claim that the set $\mathcal {M}_{1}:=\{(u,v)\in P\times P: (u,v)=A(u,v)+\lambda(\varphi,\varphi), \lambda\geq0\}$ is bounded, where $\varphi(t):=te^{-t}$. Indeed, if $(u_{0},v_{0})\in\mathcal{M}_{1}$, then there exist a constant $\lambda_{0}\geq0$ such that $(u_{0},v_{0})=A(u_{0},v_{0})+\lambda_{0}(\varphi,\varphi)$, which can be written in the form $-u_0''(t)=G_1(u_0,v_0)(t)+\lambda_0(2-t)e^{-t}, -v_0''(t)=G_2(u_0,v_0)(t)+\lambda_0(2-t)e^{-t}.$ By (F2), we have \begin{align*} -u_0''(t)&\geq a_1((T^{m-1}u_0)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u_0)'(t))+b_1((T^{n-1}v_0)(t)\\ &\quad +2\sum_{i=0}^{n-1}(T^{n-i-1}v_0)'(t))-c, \end{align*} \begin{align*} -v_0''(t)&\geq a_2((T^{m-1}u_0)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u_0)'(t))+b_2((T^{n-1}v_0)(t)\\ &\quad +2\sum_{i=0}^{n-1}(T^{n-i-1}v_0)'(t))-c. \end{align*} Multiply by $\psi(t):=te^{t}$ on both sides of the last two inequalities and integrate over $[0,1]$, and use Lemmas \ref{lem2.1} and \ref{lem2.2} to obtain $\int_0^1(u_0(t)+2u_0'(t))te^tdt\geq a_1\int_0^1(u_0(t)+2u_0'(t))te^tdt+b_1\int_0^1(v_0(t)+2v_0'(t))te^tdt-c$ and $\int_0^1(v_0(t)+2v_0'(t))te^tdt\geq a_2\int_0^1(u_0(t)+2u_0'(t))te^tdt+b_2\int_0^1(v_0(t)+2v_0'(t))te^tdt-c.$ The above two inequalities can be written as \begin{align*} \begin{pmatrix} a_1-1&b_1 \\a_2&b_2-1 \end{pmatrix} \begin{pmatrix}\int_0^1(u_0(t)+2u_0'(t))te^tdt \\ \int_0^1(v_0(t)+2v_0'(t))te^tdt \end{pmatrix} &=B_1 \begin{pmatrix}\int_0^1(u_0(t)+2u_0'(t))te^tdt \\\int_0^1(v_0(t)+2v_0'(t))te^tdt \end{pmatrix} \\ &\leq\begin{pmatrix}c \\c \end{pmatrix} . \end{align*} Now (F2) implies $\begin{pmatrix}\int_0^1(u_0(t)+2u_0'(t))te^tdt \\\int_0^1(v_0(t)+2v_0'(t))te^tdt \end{pmatrix} \leq B_1^{-1}\begin{pmatrix}c \\c \end{pmatrix} :=\begin{pmatrix}N_1 \\ N_2 \end{pmatrix} .$ Let $N:=\max\{N_1, N_2\}>0$. Then we have $\int_0^1(u_0(t)+2u_0'(t))te^tdt\leq N, \quad \int_0^1(v_0(t)+2v_0'(t))te^tdt\leq N, \quad \forall (u_0,v_0)\in\mathcal {M}_{1}.$ Now Lemma \ref{lem2.3} implies $$\label{ethm1} \|u_0\|_{0}=u_0(1)\leq N, \|v_0\|_{0}=v_0(1)\leq N, \quad \forall (u_0,v_0)\in\mathcal {M}_{1}.$$ Furthermore, this estimate leads to $\|T^{m-1}u_0\|_0=(T^{m-1}u_0)(1)\leq N, \|T^{n-1}v_0\|_0=(T^{n-1}v_0)(1)\leq N,$ for all $(u_0,v_0)\in\mathcal {M}_{1}$ and \begin{gather*} \|(T^{m-i-1}u_0)'\|_0=(T^{m-i-1}u_0)'(0)=\int_0^1(T^{m-i-2}u_0)(t)dt\leq N, \\ \|(T^{n-j-1}v_0)'\|_0=(T^{n-j-1}v_0)'(0)=\int_0^1(T^{n-j-2}v_0)(t)dt\leq N, \forall (u_0,v_0)\in\mathcal {M}_{1}, \end{gather*} $i=0,\dots,m-2$, $j=0,\dots,n-2$. Let $$\mathbb{H}:=\{\mu\geq0:\text{ there exists }(u,v)\in P\times P, \text{ such that }(u,v)=A(u,v)+\mu(\varphi,\varphi)\}.$$ Now \eqref{ethm1} implies that $\mu_{0}:=\sup\mathbb{H}<+\infty$. By (F3), there are two functions $\Phi_N, \Psi_N\in C(\mathbb{R}_+,\mathbb{R}_+)$ such that $G_1(u,v)(t)\leq\Phi_N(u'(t)+v'(t)), \quad G_2(u,v)(t)\leq\Psi_N(u'(t)+v'(t)),$ for all $(u,v)\in\mathcal {M}_{1}$. Hence we obtain \begin{align*} -u''(t)&=G_1(u,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))+\mu(2-t)e^{-t}\\ &\leq\Phi_N(u'(t)+v'(t))+2\mu_0, \end{align*} \begin{align*} -v''(t)&=G_2(u,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))+\mu(2-t)e^{-t}\\ &\leq\Psi_N(u'(t)+v'(t))+2\mu_0, \end{align*} so that $-u''(t)-v''(t)\leq\Phi_N(u'(t)+v'(t))+\Psi_N(u'(t)+v'(t))+4\mu_0$ for all $(u,v)\in\mathcal {M}_{1}$, $\mu\in\mathbb{H}$, and $\int_0^{u'(0)+v'(0)}\frac{\tau d\tau}{\Phi_N(\tau)+\Psi_N(\tau)+4\mu_0} \leq\int_0^1u'(t)+v'(t)dt=u(1)+v(1)\leq 2N$ for all $(u,v)\in\mathcal {M}_{1}$. By (F3) 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 means that $\mathcal{M}_1$ is bounded. Taking $R>\sup\{\|(u,v)\|:(u,v)\in\mathcal{M}_1\}$, we have $(u,v)\neq A(u,v)+\lambda(\varphi,\varphi), \quad \forall (u,v)\in\partial\Omega_R\cap(P\times P), \lambda\geq0.$ Now Lemma \ref{lem2.4} yields $$\label{ethm2} i(A,\Omega_R\cap(P\times P),P\times P)=0.$$ Let $\mathcal{M}_2:=\{(u,v)\in\overline{\Omega}_r\cap(P\times P):(u,v)=\lambda A(u,v), \lambda\in[0,1]\}.$ Now we want to prove that $\mathcal{M}_2=\{0\}$. Indeed, if $(u,v)\in\mathcal{M}_2$, then there is $\lambda\in[0,1]$ such that $(u(t),v(t))=(\lambda\int_0^1k(t,s)G_1(u,v)(s)ds,\lambda\int_0^1k(t,s)G_2(u,v)(s)ds)$ which can be written in the form $-u''(t)=\lambda G_1(u,v)(t), \quad -v''(t)=\lambda G_2(u,v)(t).$ By (F4), we have \begin{gather*} \begin{aligned} -u''(t)&\leq c_1((T^{m-1}u)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u)'(t))+d_1((T^{n-1}v)(t)\\ &\quad +2\sum_{i=0}^{n-1}(T^{n-i-1}v)'(t)), \end{aligned} \\ \begin{aligned} -v''(t)&\leq c_2((T^{m-1}u)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u)'(t))+d_2((T^{n-1}v)(t)\\ &\quad +2\sum_{i=0}^{n-1}(T^{n-i-1}v)'(t)). \end{aligned} \end{gather*} Multiply by $\psi(t):=te^{t}$ on both sides of the above and integrate over $[0,1]$, and use Lemmas \ref{lem2.1} and \ref{lem2.2} to obtain \begin{align*} \begin{pmatrix} 1-c_1&-d_1 \\-c_2&1-d_2 \end{pmatrix} \begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\ \int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} & =B_2 \begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\ \int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} \\ & \leq\begin{pmatrix}0 \\0 \end{pmatrix} . \end{align*} (F4) again implies $\begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\ \int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} \leq B_2^{-1}\begin{pmatrix}0 \\0 \end{pmatrix} =\begin{pmatrix}0 \\0 \end{pmatrix} .$ Consequently, $$\int_0^1(u(t)+2u'(t))te^tdt=\int_0^1(v(t)+2v'(t))te^tdt=0$$ and whence $u\equiv0$, $v\equiv0$, as required. Thus 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 $i(A,\Omega_r\cap(P\times P),P\times P)=1.$ This together with \eqref{ethm2} implies $i(A,(\Omega_R\backslash\overline{\Omega}_r)\cap(P\times P),P\times P)=0-1=-1.$ Therefore, $A$ has at least one fixed point on $(\Omega_R\backslash\overline{\Omega}_r)\cap(P\times P)$ and \eqref{integro-or2} has at least one positive solution $(u,v)$, and thus \eqref{higher-order} has at least one positive solution $(w,z)=(T^{m-1}u,T^{n-1}v)$. This completes the proof. \end{proof} \begin{theorem} \label{thm3.2} If {\rm (F1), (F5), (F6)} hold, then \eqref{higher-order} has at least one positive solution. \end{theorem} \begin{proof} Let $\mathcal{M}_{3}:=\{(u,v)\in P\times P:(u,v)=\lambda A(u,v), \lambda\in[0,1]\}.$ We now assert that $\mathcal{M}_3$ is bounded. Indeed, if $(u,v)\in\mathcal{M}_{3}$, then there is $\lambda\in[0,1]$ such that $u(t)=\lambda\int_0^1k(t,s)G_1(u,v)(s)ds, v(t)=\lambda\int_0^1k(t,s)G_2(u,v)(s)ds,$ which can be written in the form $-u''(t)=\lambda G_1(u,v)(t), \quad -v''(t)=\lambda G_2(u,v)(t).$ By (F5), we have \label{ethm3} \begin{aligned} -u''(t)&\leq l_1((T^{m-1}u)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u)'(t))+m_1((T^{n-1}v)(t)\\ &\quad +2\sum_{i=0}^{n-1}(T^{n-i-1}v)'(t))+c \end{aligned} and \label{ethm4} \begin{aligned} -v''(t)&\leq l_2((T^{m-1}u)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u)'(t))+m_2((T^{n-1}v)(t)\\ &\quad +2\sum_{i=0}^{n-1}(T^{n-i-1}v)'(t))+c. \end{aligned} Multiply by $\psi(t):=te^{t}$ on both sides of the above two inequalities and integrate over $[0,1]$, and then use Lemmas \ref{lem2.1} and \ref{lem2.2} to obtain $\int_0^1(u(t)+2u'(t))te^tdt\leq l_1\int_0^1(u(t)+2u'(t))te^tdt+m_1\int_0^1(v(t)+2v'(t))te^tdt+c$ and $\int_0^1(v(t)+2v'(t))te^tdt\leq l_2\int_0^1(u(t)+2u'(t))te^tdt+m_2\int_0^1(v(t)+2v'(t))te^tdt+c,$ which can be written in the form \begin{align*} \begin{pmatrix} 1-l_1&-m_1 \\-l_2&1-m_2 \end{pmatrix} \begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\\int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} &=B_3 \begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\\int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} \\ &\leq\begin{pmatrix}c \\c \end{pmatrix} . \end{align*} (F5) again implies $\begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\\int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} \leq B_3^{-1}\begin{pmatrix}c \\c \end{pmatrix} :=\begin{pmatrix}n_3 \\n_4 \end{pmatrix} .$ Let $N=\max\{n_3,n_4\}>0$. Then we have $\int_0^1(u(t)+2u'(t))te^tdt\leq N, \int_0^1(v(t)+2v'(t))te^tdt\leq N, \forall (u,v)\in\mathcal{M}_3.$ By Lemma \ref{lem2.3}, we obtain \begin{gather*} \|u\|_0=u(1)\leq\int_0^1(u(t)+2u'(t))te^tdt\leq N,\\ \|v\|_0=v(1)\leq\int_0^1(v(t)+2u'(t))te^tdt\leq N, \end{gather*} for all $(u,v)\in\mathcal{M}_3$. Furthermore, those estimates lead to $\|T^{m-1}u\|_0=(T^{m-1})(1)\leq N, \quad \|T^{n-1}u\|_0=(T^{n-1})(1)\leq N,$ for all $(u,v)\in\mathcal{M}_3$ and \begin{gather*} \|(T^{m-i-1}u)'\|_0=(T^{m-i-1}u)'(0)=\int_0^1(T^{m-i-2}u)(t)dt\leq N, \\ \|(T^{n-j-1}v)'\|_0=(T^{n-j-1}v)'(0)=\int_0^1(T^{n-j-2}v)(t)dt\leq N \end{gather*} for all $(u,v)\in\mathcal{M}_3$ and $i=0,\dots,m-2$, $j=0,\dots,n-2$. By \eqref{ethm3} and \eqref{ethm4}, we have \begin{gather*} -u''(t)\leq (l_1(2m-1)+m_1(2n-1))N+2l_1u'(t)+2m_1v'(t)+c, \\ -v''(t)\leq (l_2(2m-1)+m_2(2n-1))N+2l_2u'(t)+2m_2v'(t)+c, \end{gather*} for all $(u,v)\in\mathcal{M}_3$. So we have \begin{align*} -(u''(t)+v''(t))&\leq (l_1(2m-1)+m_1(2n-1)+l_2(2m-1)+m_2(2n-1))N\\ &\quad +2(l_1+l_2)u'(t)+2(m_1+m_2)v'(t)+2c. \end{align*} Noticing $u'(1)=v'(1)=0$ and letting $N_2:=(l_1(2m-1)+m_1(2n-1)+l_2(2m-1)+m_2(2n-1))N+2c$ and $L:=2(l_1+l_2+m_1+m_2)+1$, we obtain $u'(t)+v'(t)\leq\frac{N_2}{L}(e^{L-Lt}-1),$ so that $\|u'+v'\|_0=u'(0)+v'(0)\leq \frac{N_2}{L}(e^L-1).$ This proves the boundedness of $\mathcal{M}_3$. Taking $R>\sup\{\|(u,v)\|:(u,v)\in\mathcal{M}_3\}$, 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 $$\label{ethm5} i(A,\Omega_R\cap(P\times P),P\times P)=1.$$ Let $\mathcal{M}_4:=\{(u,v)\in \overline{\Omega}_r\cap (P\times P): (u,v)=A(u,v)+\lambda(\varphi,\varphi),\lambda\geq 0\}$ where $\varphi(t):=te^{-t}$. We want to prove that $\mathcal{M}_4\subset\{0\}$. Indeed, if $(u,v)\in\mathcal{M}_4$, then there is $\lambda\geq0$ such that $u(t)=\int_0^1k(t,s)G_1(u,v)(s)ds+\lambda\varphi(t), \quad v(t)=\int_0^1k(t,s)G_2(u,v)(s)ds+\lambda\varphi(t),$ which can be written in the form $-u''(t)=G_1(u,v)(t)+\lambda(2-t)e^{-t}, \quad -v''(t)=G_2(u,v)(t)+\lambda(2-t)e^{-t}.$ By (F6), we have \begin{align*} &-u''(t)\\ &\geq p_1\big((T^{m-1}u)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u)'(t)\big) +q_1\big((T^{n-1}v)(t)+2\sum_{i=0}^{n-1}(T^{n-i-1}v)'(t)\big), \\ &-v''(t)\\ &\geq p_2\big((T^{m-1}u)(t)+2\sum_{i=0}^{m-1}(T^{m-i-1}u)'(t)\big) +q_2\big((T^{n-1}v)(t)+2\sum_{i=0}^{n-1}(T^{n-i-1}v)'(t)\big). \end{align*} Multiply by $\psi(t):=te^{t}$ on both sides of the above two inequalities and integrate over $[0,1]$ and use Lemmas \ref{lem2.1} and \ref{lem2.2} to obtain $\int_0^1(u(t)+2u'(t))te^tdt\geq p_1\int_0^1(u(t)+2u'(t))te^tdt+q_1\int_0^1(v(t)+2v'(t))te^tdt$ and $\int_0^1(v(t)+2v'(t))te^tdt\geq p_2\int_0^1(u(t)+2u'(t))te^tdt+q_2\int_0^1(v(t)+2v'(t))te^tdt,$ which can be written in the form \begin{align*} \begin{pmatrix} p_1-1&q_1 \\p_2&q_2-1 \end{pmatrix} \begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\ \int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} &=B_4 \begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\\int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} \\ &\leq\begin{pmatrix}0 \\0 \end{pmatrix} . \end{align*} Now (F6) implies $\begin{pmatrix}\int_0^1(u(t)+2u'(t))te^tdt \\\int_0^1(v(t)+2v'(t))te^tdt \end{pmatrix} \leq B_4^{-1}\begin{pmatrix}0 \\0 \end{pmatrix} =\begin{pmatrix}0 \\0 \end{pmatrix} .$ Therefore, $\int_0^1(u(t)+2u'(t))te^tdt=\int_0^1(v(t)+2v'(t))te^tdt=0$ and $u\equiv0$, $v\equiv0$. This implies $\mathcal{M}_4\subset\{0\}$, as required. As a result of this, we obtain $(u,v)\neq A(u,v)+\lambda(\varphi,\varphi), \quad \forall (u,v)\in\partial\Omega_r\cap(P\times P),\; \lambda\geq0.$ Now Lemma \ref{lem2.4} yields $$\label{ethm6} i(A,\Omega_r\cap(P\times P),P\times P)=0.$$ Combining \eqref{ethm5} and \eqref{ethm6} gives $i(A,(\Omega_R\backslash\overline{\Omega}_r)\cap(P\times P),P\times P)=1.$ Hence $A$ has at least one fixed point on $(\Omega_R\backslash\overline{\Omega}_r)\cap(P\times P)$. Thus \eqref{higher-order} has at least one positive solution $(w,z)=(T^{m-1}u,T^{n-1}v)$. This complete the proof. \end{proof} \begin{theorem} \label{thm3.3} If {\rm (F1)--(F3), (F6), (F7)} hold, then \eqref{higher-order} has at least two positive solutions. \end{theorem} \begin{proof} By (F7), the following inequalities $f(t,x,y)\leq f(t,\underbrace{\Lambda,\dots,\Lambda}_{m+n+2})<\Lambda,\quad g(t,x,y)\leq g(t,\underbrace{\Lambda,\dots,\Lambda}_{m+n+2})<\Lambda,$ hold for all $t\in[0,1]$ and all $(x,y)\in\underbrace{[0,\Lambda]\times\dots\times [0,\Lambda]}_{m+n+2}$. Consequently, we have for all $(u,v)\in \partial\Omega_{\Lambda}\cap(P\times P)$, \begin{align*} \|A_1(u,v)\|_0&=A_1(u,v)(1)=\int_0^1sG_1(u,v)(s)ds\leq\int_0^1G_1(u,v)(s)ds\\ &\leq\int_0^1f(s,\Lambda,\dots,\Lambda)ds<\Lambda=\|(u,v)\|, \end{align*} \begin{align*} \|A_2(u,v)\|_0&=A_2(u,v)(1)=\int_0^1sG_2(u,v)(s)ds\leq\int_0^1G_2(u,v)(s)ds\\ & \leq\int_0^1g(s,\Lambda,\dots,\Lambda)ds<\Lambda=\|(u,v)\|, \end{align*} \begin{align*} \|(A_1(u,v))'\|_0&=(A_1(u,v))'(0)=\int_0^1G_1(u,v)(s)ds\\ &\leq\int_0^1f(s,\Lambda,\dots,\Lambda)ds<\Lambda=\|(u,v)\|, \end{align*} \begin{align*} \|(A_2(u,v))'\|_0&=(A_2(u,v))'(0)=\int_0^1G_2(u,v)(s)ds\\ &\leq\int_0^1g(s,\Lambda,\dots,\Lambda)ds<\Lambda=\|(u,v)\|. \end{align*} The preceding inequalities imply $\|A(u,v)\|=\|(A_1(u,v),A_2(u,v))\|<\Lambda=\|(u,v)\|$, and thus $(u,v)\neq\lambda A(u,v), \quad \forall (u,v)\in\partial\Omega_{\Lambda}\cap(P\times P),\; 0\leq\lambda\leq1.$ Now Lemma \ref{lem2.5} yields $$\label{ethm7} i(A,\Omega_{\Lambda}\cap(P\times P),P\times P)=1.$$ By (F2), (F3) and (F6), we know that \eqref{ethm2} and \eqref{ethm6} hold. Note we can choose $R>\Lambda>r$ in \eqref{ethm2} and \eqref{ethm6} (see the proofs of Theorems \ref{thm3.1} and \ref{thm3.2}). Combining \eqref{ethm2}, \eqref{ethm6} and \eqref{ethm7}, we obtain $i(A,(\Omega_{R}\backslash\overline{\Omega}_{\Lambda})\cap(P\times P),P\times P)=0-1=-1,$ and $i(A,(\Omega_{\Lambda}\backslash\overline{\Omega}_{r})\cap(P\times P),P\times P)=1-0=1.$ Therefore, $A$ has at least two fixed points, with one on $(\Omega_{R}\backslash\overline{\Omega}_{\Lambda})\cap(P\times P)$ and the other on $(\Omega_{\Lambda}\backslash\overline{\Omega}_{r})\cap(P\times P)$. Hence \eqref{higher-order} has at least two positive solutions. \end{proof} \section{Examples} In this section we present three examples that illustrate our main results. \begin{example} \label{examp4.1} \rm Suppose $(\xi_{ij})_{2\times (m+1)}$ and $(\eta_{ij})_{2\times (n+1)}$ be two positive matrices and $1<\alpha_i\leq 2(i=1,2.)$. Let \begin{gather*} f(t,x,y):=\Big(\sum_{j=1}^{m+1} \xi_{1j}x_j+\sum_{j=1}^{n+1} \eta_{1j}y_j\Big)^{\alpha_1} \quad t\in [0,1],\;x\in\mathbb{R}_+^{m+1},\;y\in\mathbb{R}_+^{n+1}, \\ g(t,x,y):=\Big(\sum_{j=1}^{m+1} \xi_{2j}x_j+\sum_{j=1}^{n+1} \eta_{2j}y_j\Big)^{\alpha_2} \quad t\in [0,1],\;x\in\mathbb{R}_+^{m+1},\;y\in\mathbb{R}_+^{n+1}. \end{gather*} Now (F1)--(F4) hold. By Theorem \ref{thm3.1}, \eqref{higher-order} has at least one positive solution. \end{example} \begin{example} \label{examp4.2} \rm Suppose $(\xi'_{ij})_{2\times (m+1)}$ and $(\eta'_{ij})_{2\times (n+1)}$ be two positive matrices and $0<\alpha_i<1(i=3,4.)$. Let \begin{gather*} f(t,x,y):=\Big(\sum_{j=1}^{m+1} \xi_{1j}'x_j +\sum_{j=1}^{n+1} \eta_{1j}'y_j\Big)^{\alpha_3} \quad t\in [0,1],\;x\in\mathbb{R}_+^{m+1},\; y\in\mathbb{R}_+^{n+1}, \\ g(t,x,y):=\Big(\sum_{j=1}^{m+1} \xi_{2j}'x_j+\sum_{j=1}^{n+1} \eta_{2j}'y_j\Big)^{\alpha_4} \quad t\in [0,1],\;x\in\mathbb{R}_+^{m+1},\; y\in\mathbb{R}_+^{n+1}. \end{gather*} Now (F1), (F5) and (F6) are satisfied. By Theorem \ref{thm3.2}, \eqref{higher-order} has at least one positive solution. \end{example} \begin{example} \label{examp4.3} \rm Suppose $(\xi_{ij})_{2\times (m+1)}$, $(\xi'_{ij})_{2\times (m+1)}$, $(\eta_{ij})_{2\times (n+1)}$ and $(\eta'_{ij})_{2\times (n+1)}$ be four positive matrices, $1<\beta_i\leq2(i=1,2)$, $0<\gamma_i<1$ $(i=1,2)$. Let \begin{gather*} f(t,x,y):=\Big(\sum_{j=1}^{m+1} \xi_{1j}x_j+\sum_{j=1}^{n+1} \eta_{1j}y_j\Big)^{\beta_1} +\Big(\sum_{j=1}^{m+1} \xi_{1j}'x_j+\sum_{j=1}^{n+1} \eta_{1j}'y_j\Big)^{\gamma_1} \\ t\in [0,1],x\in\mathbb{R}_+^{m+1},y\in\mathbb{R}_+^{n+1}, \\ g(t,x,y):=\Big(\sum_{j=1}^{m+1} \xi_{2j}x_j+\sum_{j=1}^{n+1} \eta_{2j}y_j\Big)^{\beta_2} +\Big(\sum_{j=1}^{m+1} \xi_{2j}'x_j+\sum_{j=1}^{n+1} \eta_{2j}'y_j\Big)^{\gamma_2} \\ t\in [0,1],x\in\mathbb{R}_+^{m+1},y\in\mathbb{R}_+^{n+1}. \end{gather*} Now (F1)-(F3), (F6) and (F7) are satisfied. By Theorem \ref{thm3.3}, \eqref{higher-order} has at least two positive solutions. \end{example} \begin{thebibliography}{99} \bibitem{Cao1} Z. Cao, D. Jiang; \emph{Periodic solutions of second order sigular coupled systems}, Nonlinear Anal. 71 (2009), 3661-3667. \bibitem{Dancer} E. N. Dancer; \emph{Global solution branches for positive mappings}, Arch. Ration. Mech. Anal. 52 (1973), 181-192. \bibitem{Dunninger1} D. R. Dunninger, H. 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