\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 13, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/13\hfil Positive solutions] {Positive solutions for a second-order system with integral boundary conditions} \author[W. Song, W. Gao\hfil EJDE-2011/13\hfilneg] {Wenjing Song, Wenjie Gao} \address{Wenjing Song \newline Institute of Mathematics, Jilin University, Changchun 130012, China. Institute of Applied Mathematics, Jilin University of Finance and Economics, Changchun 130017, China} \email{swj-78@163.com} \address{Wenjie Gao \newline Institute of Mathematics, Jilin University, Changchun 130012, China} \email{wjgao@jlu.edu.cn} \thanks{Submitted November 29, 2010. Published January 26, 2011.} \thanks{Supported by grant 10771085 from NSFC, by Key Lab of Symbolic Computation and \hfill\break\indent Knowledge Engineering of Ministry of Education, and by the 985 program of Jilin University} \subjclass[2000]{34B15, 34B27} \keywords{Positive solution; integral boundary condition; fixed point theorem} \begin{abstract} This article concerns the existence of positive solutions to a second-order system with integral boundary conditions. By applying Krasnoselskii fixed point theorem, we show the existence of solutions under certain conditions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} In this article, we investigate the existence of positive solutions to the following system of second order ordinary differential equations with integral boundary conditions: $$\label{e1.1} \begin{gathered} x''(t)=-f(t,x(t),y(t)),\quad (t,x,y)\in(0,1)\times[0,+\infty)\times[0,+\infty),\\ y''(t)=-g(t,x(t),y(t)),\quad (t,x,y)\in(0,1)\times[0,+\infty)\times[0,+\infty),\\ x(0)-ax'(0)=\int_0^1 \varphi_0(s)y(s)ds,\quad x(1)+bx'(1)=\int_0^1 \varphi_1(s)y(s)ds,\\ y(0)-ay'(0)=\int_0^1 \psi_0(s)x(s)ds,\quad y(1)+by'(1)=\int_0^1 \psi_1(s)x(s)ds, \end{gathered}$$ where $f,g\in C([0,1]\times[0,+\infty)\times[0,+\infty),[0,+\infty))$, $\varphi_0,\varphi_1,\psi_0,\psi_1\in C([0,1],[0,+\infty))$, $a$ and $b$ are positive real parameters. Boundary value problems with positive solutions describe many phenomena in the applied sciences found in the theory of nonlinear diffusion generated by nonlinear sources, thermal ignition of gases, and concentration in chemical or biological problems. Readers may refer to \cite{b2,c1,g1} for details. In the past few years, much effort has been devoted to the study of the existence of positive solutions to ordinary differential equations or systems with different kinds of boundary conditions, see \cite{a1,l1,l2,l3,m1}. On the other hand, problems with integral boundary conditions arise naturally in thermal conduction problems \cite{c2}, semiconductor problems \cite{i1} and hydrodynamic problems \cite{c3}. Many authors have investigated scalar problems with integral boundary conditions; see for instance \cite{b1,k1,y2,z1}. Particularly, in \cite{b1}, Boucherif discussed the following boundary value problem with integral boundary condition: \label{e1.2} \begin{gathered} y''(t)=f(t,y(t)),\quad 00$and$k_2(t)>0$for all$t\in[0,1]$, and$G(t,s)>0$for all$(t,s)\in[0,1]\times[0,1]$. Moreover, we have the following propositions: \begin{proposition} \label{prop2.1} There exists a positive continuous function$\gamma:[0,1]\to\mathbb{R}$such that$G(t,s)\geq \gamma(t)G(s,s)$for all$t,s\in[0,1]$. Moreover,$\gamma_0:=\min\{\gamma(t):t\in[0,1]\}>0$. \end{proposition} The proof of the above proposition is similar to \cite[Lemma 2]{b1}, and we omit it here. \begin{proposition} \label{prop2.2} Under assumption {\rm (H0)}, for all$t,s\in[0,1]$, we have$G(t,s)\leq G(s,s)$. \end{proposition} The of the above proposition follows standard argument, it is omitted here. Let us denote two operators$A, B:=C[0,1]\times C[0,1]\to C[0,1]$as follows: \begin{gather*} A(x,y)(t)=\int_0^1 G(t,s)f(s,x(s),y(s))ds+\int_0^1 \Phi(t,s)y(s)ds,\\ B(x,y)(t)=\int_0^1 G(t,s)g(s,x(s),y(s))ds+\int_0^1 \Psi(t,s)x(s)ds.\\ \end{gather*} Then we define an operator$T:C[0,1]\times C[0,1]\to C[0,1]\times C[0,1]$as $$\label{e2.2} Tz(t)=\int_0^1 H(t,s)F(s,x(s),y(s))ds +\int_0^1 K(t,s)z(s)ds =\begin{pmatrix} A(x,y)(t)\\ B(x,y)(t) \end{pmatrix},$$ where \begin{gather*} z(t)=\begin{pmatrix} x(t)\\ y(t) \end{pmatrix}, \; (x,y)\in C[0,1]\times C[0,1],\quad H(t,s)=\begin{pmatrix} G(t,s)&0\\ 0&G(t,s) \end{pmatrix}, \\ F(s,x(s),y(s))=\begin{pmatrix} f(s,x(s),y(s))\\ g(s,x(s),y(s)) \end{pmatrix} \quad K(t,s)=\begin{pmatrix} 0&\Phi(t,s)\\ \Psi(t,s)&0 \end{pmatrix}. \end{gather*} It is clear that the existence of a positive solution for \eqref{e2.1} is equivalent to the existence of a nontrivial fixed point of$T$in$C[0,1]\times C[0,1]$. To obtain a positive solution of \eqref{e1.1}, we need the following lemma. \begin{lemma} \label{lem2.1} Assume {\rm (H0)--(H2)} hold. Then$T:C[0,1]\times C[0,1]\to C[0,1]\times C[0,1]$is a completely continuous operator. \end{lemma} \begin{proof} Firstly, we prove that$T$is a compact operator. That is, for any bounded subset$D\subset C[0,1]\times C[0,1]$, we show that$T(D)$is relatively compact in$C[0,1]\times C[0,1]$. Since$D\subset C[0,1]\times C[0,1]$is a bounded subset, there exists a constant$\overline{M}>0$such that$\| z\|=\| x\|+\| y\|\leq \overline{M}$for any$z\in D. By applying (H0), (H1) and Proposition 2.2, we obtain \begin{align*} \| A(x,y)\| &=\max_{0\leq t\leq 1}\big|\int_0^1 G(t,s)f(s,x(s),y(s))ds +\int_0^1 \Phi(t,s)y(s)ds\big|\\ &\leq L\int_0^1 | G(s,s)| ds+M_{\Phi}\overline{M}<+\infty. \end{align*} HereL=\max\{f(t,x,y):0\leq t\leq1,| x|\leq \overline{M}, | y|\leq \overline{M}\}+\max\{g(t,x,y):0\leq t\leq1,| x|\leq \overline{M},| y|\leq \overline{M}\}$. Similarly, we can obtain $$\| B(x,y)\|\leq L\int_0^1 | G(s,s)| ds+M_{\Psi}\overline{M}<+\infty.$$ Then from the definition of the norm of the product space$C[0,1]\times C[0,1], we have \begin{align*} \| T(z)\|&=\| A(x,y)\|+\|B(x,y)\| \\ &\leq 2L\int_0^1 | G(s,s)| ds+(M_{\Phi}+M_{\Psi})\overline{M} <+\infty. \end{align*} Therefore,T(D)$is uniformly bounded with the norm of$C[0,1]\times C[0,1]$. Moreover, for any$t\in(0,1), we have \begin{align*} &|\frac{d}{dt} A(x,y)(t)|\\ &=\Big| \Big(\int_0^{t}G(t,s)f(s,x(s),y(s))ds+ \int_{t}^1 G(t,s)f(s,x(s),y(s))ds\Big)' \\ &\quad -\frac{1}{1+a+b}\int_0^1 \varphi_0(s)y(s)ds +\frac{1}{1+a+b}\int_0^1 \varphi_1(s)y(s)ds\Big|\\ & =\Big| \frac{1}{1+a+b}\Big[-\int_0^1 sf(s,x(s),y(s))ds -a\int_0^{t}f(s,x(s),y(s))ds \\ &\quad+(b+1)\int_{t}^1 f(s,x(s),y(s))ds -\int_0^1 \varphi_0(s)y(s)ds+\int_0^1 \varphi_1(s)y(s)ds\Big]\Big|\\ &\leq\frac{1}{1+a+b}[(2+a+b)L+2K\overline{M}]<+\infty , \end{align*} where \begin{align*} K&=\max\{\varphi_0(t):0\leq t\leq1\}+\max\{\varphi_1(t):0\leq t\leq1\}\\ &\quad + \max\{\psi_0(t):0\leq t\leq1\} +\max\{\psi_1(t):0\leq t\leq1\}. \end{align*} Thus, it is easy to prove thatA(D)$is equicontinuous. This together with the Arzel\'a-Ascoli theorem guarantees that$A(D)$is relatively compact in$C[0,1]$. Similarly, we can prove that$B(D)$is relatively compact in$C[0,1]$. Therefore,$T(D)$is relatively compact in$C[0,1]\times C[0,1]$. On the other hand, according to the definition of$T$, it is easily seen that$T$is continuous. We obtain that$T$is completely continuous. The proof is complete. \end{proof} We shall discuss the existence of a positive solution of \eqref{e1.1} by using the following fixed point theorem of cone expansion and compression. \begin{lemma}[{\cite[Theorem 4]{b1}}] \label{lem2.2} Let$E$be a Banach space and$K\subset E$be a cone. Suppose$\Omega_1$and$\Omega_2$are two bounded open sets in Banach space$E$such that$\theta\in\Omega_1$,$\overline{\Omega_1}\subset \Omega_2$and suppose that the operator$T:K\cap(\overline{\Omega_2}\setminus\Omega_1)\to K$is completely continuous such that \begin{itemize} \item[(A1)]$\| Tx\|\leq \| x\|$for all$x\in K\cap\partial\Omega_1$and$\| Tx\|\geq \| x\|$for all$x\in K\cap\partial\Omega_2$or \item[(A2)]$ \| Tx\|\geq \| x\|$for all$x\in K\cap\partial\Omega_1$and$\| Tx\|\leq \| x\|$for all$x\in K\cap\partial\Omega_2$. \end{itemize} Then$T$has a fixed point in$K\cap(\overline{\Omega_2}\setminus\Omega_1)$. \end{lemma} To use Lemma \ref{lem2.2}, let$E=C[0,1]\times C[0,1]$, $$P=\{u\in C[0,1],\ u(t)\geq 0,\ t\in[0,1]\},$$ and $$P_0=\{(u,v)\in P\times P,\min_{0\leq t\leq 1}((u,v))= \min_{0\leq t\leq 1}(u(t)+v(t))\geq\frac{1-M}{1-m}\gamma_0\| (u,v)\|\},$$ where $$M=\max\{M_{\Phi},M_{\Psi}\},\quad m=\min\{m_{\Phi},m_{\Psi}\}.$$ It is easy to see that$P_0$is a cone in$E$. \begin{lemma} \label{lem2.3} Under Assumptions {\rm (H0)--(H2)}, the operator$T:P_0\to P_0$is a completely continuous. \end{lemma} \begin{proof} By Lemma \ref{lem2.1}, we only need to prove that$T(P_0)\subset P_0$. Define an operator$N:C[0,1]\times C[0,1]\to C[0,1]\times C[0,1]$by$N(z)(t)=\int_0^1 K(t,s)z(s)ds$. Then$N(P\times P)\subset P\times P. Noting that \begin{align*} \| Nz(t)\| &=\max_{0\leq t\leq 1}|\int_0^1 \Phi(t,s)y(s)ds|+ \max_{0\leq t\leq 1}|\int_0^1 \Psi(t,s)x(s)ds|\\ &\leq \max\{M_{\Phi},M_{\Psi}\}\| z\|, \end{align*} one has\| N \|\leq \max\{M_{\Phi},M_{\Psi}\}<1$. Then$I-N$is invertible. Similarly to \cite[Lemma 3]{b1}, we have $$Tz(t)=\int_0^1 H(t,s)F(s,x(s),y(s))ds +\int_0^1 R(t,s)\int_0^1 H(s,\tau)F(\tau,x(\tau),y(\tau))d\tau ds,$$ where$R(t,s)$is the resolvent kernel by$R(t,s)=\sum_{j=1}^{\infty}K_{j}(t,s)$and$K_{j}(t,s)=\int_0^1 K (t,\tau)K _{j-1}(\tau,s)d\tau$,$j=2,3,\dots$, and$K _1(t,s)=K(t,s). Let $R(t,s)=\begin{pmatrix} R_1(t,s) & R_2(t,s)\\ R_3(t,s) & R_4(t,s) \end{pmatrix}.$ It can be easily verified that \begin{gather*} \frac{m^2}{1-m^2}\leq R_1(t,s),\; R_4(t,s)\leq\frac{M^2}{1-M^2},\\ \frac{m}{1-m^2}\leq R_2(t,s),\; R_3(t,s)\leq\frac{M}{1-M^2}\,. \end{gather*} From Propositions 2.1, 2.2 and (H1), we find that \label{e2.3} \begin{aligned} \|Tz(t)\| &\leq \frac{1}{1-M}[\int_0^1 G(\tau,\tau)f(\tau,x(\tau),y(\tau))d\tau\\ &\quad + \int_0^1 G(\tau,\tau)g(\tau,x(\tau),y(\tau))d\tau], \end{aligned} \label{e2.4} \begin{aligned} \min_{0\leq t\leq 1}Tz(t) &\geq \frac{\gamma_0}{1-m}[\int_0^1 G(\tau,\tau)f(\tau,x(\tau),y(\tau)) d\tau\\ &\quad +\int_0^1 G(\tau,\tau)g(\tau,x(\tau),y(\tau))d\tau]. \end{aligned} By \eqref{e2.3} and \eqref{e2.4}, we have $$\min_{0\leq t\leq1}Tz(t)\geq \frac{1-M}{1-m}\gamma_0\| Tz(t)\|.$$ Therefore,T(P_0)\subset P_0$. \end{proof} \section{Main results} In this section, we show the existence of positive solutions to \eqref{e1.1}. Firstly, we introduce some notation. \begin{gather*} f_{\beta}=\liminf_{| x| +| y|\to\beta}\min_{0\leq t\leq 1}\frac{f(t,x,y)}{| x| +| y|},\quad f^{\beta}=\limsup_{| x| +| y|\to\beta}\max_{0\leq t\leq 1}\frac{f(t,x,y)}{| x| +| y|},\\ g_{\beta}=\liminf_{| x| +| y|\to\beta}\min_{0\leq t\leq 1}\frac{g(t,x,y)}{| x| +| y|},\quad g^{\beta}=\limsup_{| x| +|y|\to\beta}\max_{0\leq t\leq 1}\frac{g(t,x,y)}{| x| +| y|}, \end{gather*} where$\beta=0$or$\infty$. \begin{theorem} \label{thm3.1} Assume that {\rm (H0)--(H2)} hold. If $$f^{0},g^{0}<\frac{1-M}{2\int_0^1 G(s,s)ds}\quad\text{and}\quad f_{\infty},g_{\infty}>\frac{(1-m)^2}{2\gamma_0^2(1-M) \int_0^1 G(s,s)ds},$$ then Problem \eqref{e1.1} has at least one positive solution. \end{theorem} \begin{proof} Since$f^{0},g^{0}<\frac{1-M}{2\int_0^1 G(s,s)ds}$, there exists an$r>0$, such that$f(t,x,y)\leq (f^{0}+\varepsilon_1)(| x|+| y |)$, and$g(t,x,y)\leq (g^{0}+\varepsilon_1)(|x|+| y |)$for$t\in[0,1]$,$| x|+| y |\leq r$, where$\varepsilon_1$satisfies$f^{0}+\varepsilon_1\leq\frac{1-M}{2\int_0^1 G(s,s)ds}$and$g^{0}+\varepsilon_1\leq\frac{1-M}{2\int_0^1 G(s,s)ds}$. Let$\Omega_1=\{z=(x,y)\in P\times P,\ \| z\|\frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds} $, there exists an$R>r>0$, such that$f(t,x,y)\geq (f_{\infty}-\varepsilon _2)(| x|+| y|)$and$g(t,x,y)\geq (g_{\infty}-\varepsilon _2)(| x|+| y|)$for$t\in [0,1],\ | x|+| y|\geq R$, where$\varepsilon_2$satisfies$f_{\infty}-\varepsilon _2\geq \frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$and$g_{\infty}-\varepsilon _2\geq \frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$. Let$\Omega_2=\{z=(x,y)\in P\times P,\ \| z\|\frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds} $, then \eqref{e1.1} has at least one positive solution. \end{theorem} Next we discuss the multiplicity of positive solutions for Problem \eqref{e1.1}. We obtain the following results. \begin{theorem} \label{thm3.3} Assume that {\rm (H0)--(H2)} hold, and \begin{itemize} \item[(i)]$f_0>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$and$g_{\infty}>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$; \item[(ii)] There exists an$l>0$such that$\max_{0\leq t\leq1,(x,y)\in\partial\Omega_1}f(t,x,y)<\frac{1-M}{2\int_0^1 G(s,s)ds} l$and$\max_{0\leq t\leq1,(x,y)\in\partial\Omega_1}g(t,x,y) <\frac{1-M}{2\int_0^1 G(s,s)ds}l$, where$\Omega_1:=\{z=(x,y)\in P\times P,\|z\|\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$, we can choose$\varepsilon_3>0$such that$f_0-\varepsilon_3\geq \frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$, and also there exists an$00$, such that$g_{\infty}-\varepsilon_4\geq \frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$, and also there exists an$l_2>l$, such that$g(t,x,y)\geq (g_{\infty}-\varepsilon_4)(| x|+| y|)$for$t\in[0,1], | x|+| y|\geq l_2$. Let$\Omega_{\widetilde{l}_2}:=\{z=(x,y)\in P\times P, \| z\|<\widetilde{l}_2\}$, where$\widetilde{l}_2=\frac{1-m}{(1-M)\gamma_0}l_2$. For any$z=(x,y)\in\partial \Omega_{\widetilde{l}_2}\cap P_0$, we have $$\label{e3.4} \begin{split} \| Tz\|&\geq \frac{\gamma_0}{1-m}\int_0 ^1 G(s,s)ds\cdot(g_{\infty}-\varepsilon_4)(\frac{1-M}{1-m}\gamma_0\| z\|) \\ &\geq\| z\|. \end{split}$$ By (ii), for$z=(x,y)\in\partial\Omega_1\cap P_0$, we have $$\label{e3.5} \begin{split} \|Tz(t)\|&\leq\frac{1}{1-M}[\int_0^1 G(\tau,\tau) f(\tau,x(\tau),y(\tau))d\tau+ \int_0^1 G(\tau,\tau)g(\tau,x(\tau),y(\tau))d\tau] \\ &< \frac{1}{1-M} \int_0^1 G(s,s)ds\cdot\frac{1-M}{\int _0^1 G(s,s)ds}l=l. \end{split}$$ Therefore, from \eqref{e3.3}, \eqref{e3.5} and Lemma \ref{lem2.2}, it follows that \eqref{e1.1} has at least one positive solution$z_1\in P_0$with$l_1\leq \| z_1\|\frac{(1-m)^2}{\gamma_0^2(1-M) \int_0^1 G(s,s)ds}$or$g_0,f_{\infty}>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$or$g_0,g_{\infty}>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$; \item[(ii)] There exists an$l>0$such that$\max_{0\leq t\leq1,(x,y)\in\partial\Omega_1}f(t,x,y) <\frac{1-M}{2\int_0^1 G(s,s)ds} l$and$\max_{0\leq t\leq1,(x,y)\in\partial\Omega_1}g(t,x,y) <\frac{1-M}{2\int_0^1 G(s,s)ds} l$, where$\Omega_1:=\{z=(x,y)\in P\times P,\|z\|