\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 163, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/163\hfil Existence of positive solutions] {Existence of positive solutions for boundary-value problems with integral boundary conditions and sign changing nonlinearities} \author[S. Liu, M. Jia, Y. Tian \hfil EJDE-2010/163\hfilneg] {Shu Liu, Mei Jia, Yingqiang Tian} \address{College of Science, Shanghai University for Science and Technology, Shanghai 200093, China} \email[Shu Liu]{liu2008shu@126.com} \email[Mei Jia]{jiamei-usst@163.com } \email[Yingqiang Tian]{tian0013@163.com} \thanks{Submitted June 15, 2010. Published November 12, 2010.} \thanks{Supported by grant 10ZZ93 from the Innovation Program of Shanghai Municipal \hfill\break\indent Education Commission} \subjclass[2000]{34B15, 34B18} \keywords{Boundary value problems; $p$-Laplacian; integral boundary conditions; \hfill\break\indent fixed point theorem in double cones; positive solutions} \begin{abstract} In this article, we show the existence of positive solutions for boundary value problems with integral boundary conditions and sign changing nonlinearities. By using a fixed point theorem in double cones, we obtain sufficient conditions for the existence of two positive solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} The theory of boundary value problems with integral boundary conditions for ordinary differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics can all be reduced to nonlocal problems with integral boundary conditions. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to \cite{g1,k1,l1} and the references therein. For more information about the general theory of integral equations and their relation with boundary value problems we refer to \cite{a1,c2}. Moreover, boundary value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multi-point and nonlocal boundary value problems as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention. To identify a few, we refer the reader to \cite{b1,g3,m1,w1,w2,x1} and references therein. But the corresponding theory for boundary-value problem with integral boundary conditions and sign changing nonlinearities of one-dimensional $p$-Laplacian is not investigated till now. By using the fixed point theorem in double cones, Guo in \cite{g2} discussed the existence of positive solutions for second-order three-point boundary-value problem \begin{gather*} x''+f(t,x)=0, \quad 0\leq t\leq 1,\\ x(0)-\beta x'(0)=0,\quad x(1)=\alpha x(\eta), \end{gather*} where $f$ is allowed to change sign. Sufficient conditions are obtained by imposing growth conditions on $f$ which ensure the existence of at least two positive solutions for the above boundary value problems. Meanwhile, they proved a fixed point theorem in double cones which generalize the fixed point theorem in a cones to some degree. By using a theorem similar to the one in \cite{g2}, Cheung \cite{c1} proved the existence of two positive solutions for the problem $$(\Phi_p(u'))'+h(t)f(t,u)=0, \quad 01, \varphi_q=(\varphi_p)^{-1}, and \frac{1}{q}+\frac{1}{p}=1. We will assume the following conditions: \begin{itemize} \item[(H1)] f:[0,1] \times [0,+\infty)\to \mathbb{R} is continuous; \item[(H2)] f(t,0)\geq 0 (f\not\equiv 0) for t\in [0,1]; \item[(H3)] a>\sum_{i=1}^{m-2}a_i; g is non-negative, integrable, and \sigma:=\int_0^1 g(t) dt \in [0,1). \end{itemize} \section{Preliminaries} For a cone K in a Banach space X with norm \| \cdot \| and a constant r>0, let K_r=\{ x\in K:\| x \| a>0 such that \begin{itemize} \item[(C1)] \| Tx\| b for x\in \partial K'(b); \item[(C3)] Tx=T^*x, for x\in \partial K'_a(b)\cap\{u:T^*u=u\}, \end{itemize} then T has at least two fixed points y_1 and y_2 in K such that$$ 0\leq \| y_1\| 0$, so we can prove the lemma by the continuity of$H_h$. Namely, there is a unique$A_h\in (0,\int_0^1h(\tau)d\tau)$satisfying \eqref{e2.2} and \eqref{e2.3}. Meanwhile, we could have that there is unique$\sigma_h\in(0,1)$such that$A_h=\int_0^{\sigma_h}h(\tau)d\tau$by the continuity of$\int_0^th(\tau)d\tau$on$[0,1]$. \end{proof} \begin{lemma} \label{lem2.4} Assume that {\rm (H3)} holds. If$h\in C^+[0,1]$, then the solution of boundary value problem \eqref{e2.1} has the following properties: \begin{itemize} \item[(i)]$u(t)$is a concave function; \item[(ii)]$u(t)\geq 0$,$t\in [0,1]$; \item[(iii)]$u(\sigma_h)=\max_{0\leq t\leq 1}u(t)$, where$\sigma_h$is defined in lemma \ref{lem2.3}. \end{itemize} \end{lemma} \begin{proof} Suppose that$u(t)$is the solution of \eqref{e2.1}. (i) From the fact that$(\varphi_p(u'))'(t)=-h(t)\leq 0$we know that$\varphi_p(u')$is non-increasing. It follows that$u'(t)$is also non-increasing. Thus,$u(t)$is concave down on$[0, 1]$. (ii) From (i), we know that the minimum of$u(t)$is obtained at 0 or 1. So we only have to prove that$u(0)\geq 0$and also$u(1)\geq 0$. Firstly, we prove that$u(0)\geq 0$. If$u(0)<0$and$u'(0)\geq 0$, by the condition$au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i)$, we have $$au(0)-\sum_{i=1}^{m-2}a_iu(\xi_i)=bu'(0)\geq 0,$$ so $$au(0)\geq \sum_{i=1}^{m-2}a_iu(\xi_i). \label{e2.7}$$ From$a>\sum_{i=1}^{m-2}a_i>0$, we have $$au(0)< \sum_{i=1}^{m-2}a_iu(0). \label{e2.8}$$ From \eqref{e2.7} and \eqref{e2.8}, we obtain $$\sum_{i=1}^{m-2}a_iu(0)> \sum_{i=1}^{m-2}a_iu(\xi_i).$$ Hence, there is a$j$,$1\leq j\leq {m-2}$such that$0>u(0)>u(\xi_j)$. Then we obtain that$u(1)=\min_{t\in[0,1]}u(t)<0$and that there is$\eta_1\leq \eta_2\in [0,1]$such that$u(t)\leq 0$when$t\in [0,\eta_1]\cup[\eta_2,1]$and$u(t)\geq 0$when$t\in [\eta_1,\eta_2]$. From the boundary condition$u(1)=\int_0^1 g(s)u(s) ds, we have \label{e2.9} \begin{aligned} |u(1)| & =-u(1)=-\int_0^{\eta_1}g(s)u(s)ds -\int_{\eta_1}^{\eta_2}g(s)u(s) ds-\int_{\eta_2}^1g(s)u(s) ds\\ & \leq \int_0^{\eta_1}g(s)|u(s)| ds +\int_{\eta_1}^{\eta_2}g(s)|u(s)|ds +\int_{\eta_2}^1g(s)|u(s)|ds \\ &=|u(1)|\int_0^1g(s) ds. \end{aligned} We have\int_0^1g(s) ds \geq 1$which is contradiction to with (H3), so$u(0)\geq 0$. If$u(0)<0$and$u'(0)<0$, we can prove that$u(0)\geq 0$by using the same way as in \eqref{e2.9}. Similarly, we can prove$u(1)\geq 0$. Therefore, we obtain that$u(t)\geq 0$for$t\in [0,1]$. (iii) From the boundary conditions, we can easily show that the maximum of$u(t)$can not be at 0 or 1. If$u(0)$is the maximum, then$u'(0)\leq 0$which is contradicted with boundary condition$au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i)$. If$u(1)$is the maximum, it is contradicts the boundary condition$u(1)=\int_0^1 g(s)u(s) ds$. \end{proof} From Lemmas \ref{lem2.2} and \ref{lem2.3}, we have$u'(t)=\varphi_q(\int_t^{\sigma_h}h(\tau)d\tau)$. By the concavity of$u(t)$, we know$u(\sigma_h)=\max_{0\leq t\leq 1}u(t)$. \begin{lemma} \label{lem2.5} If$u\in C[0,1]$and$u(t)\geq 0$is a concave function, then for any$\delta\in(0,\frac{1}{2})$, we have $$\min_{\delta\leq t\leq {1-\delta}}u(t)\geq \delta \| u\|.$$ \end{lemma} \begin{proof} Let$u(\sigma)=\|u\|$. We discuss three cases: (i) If$\sigma<\delta$, then$\min_{\delta\leq t\leq {1-\delta}}u(t)=u(1-\delta)$. By the concavity of$u(t)$, we have $$\frac{u(\sigma)-u(1)}{1-\sigma}\leq \frac{u(1-\delta)-u(1)}{\delta};$$ i.e., $$\frac{u(1-\delta)}{\delta}\geq \frac{u(\sigma)}{1-\sigma}+(\frac{1}{\delta}-\frac{1}{1-\sigma})u(1).$$ For$\sigma< \delta<{1-\delta}$, we have${1-\sigma}>\delta$. So $$u(1-\delta)\geq \frac{\delta}{1-\sigma}u(\sigma)\geq {\delta u(\sigma)}.$$ (ii) If$\sigma>{1-\delta}$, then$\min_{\delta\leq t\leq {1-\delta}}u(t)=u(\delta)$and$\sigma >\delta$. By the concavity of$u(t)$, we have $$\frac{u(\sigma)-u(0)}{\sigma}\leq \frac{u(\delta)-u(0)}{\delta};$$ i.e., $$\frac{u(\delta)}{\delta}\geq \frac{u(\sigma)}{\sigma}+(\frac{1}{\delta}-\frac{1}{\sigma})u(0).$$ So $$u(\delta)\geq \frac{\delta}{\sigma}u(\sigma)\geq {\delta u(\sigma)}.$$ (iii) For$\delta\leq\sigma\leq{1-\delta}$, if$\min_{\delta\leq t\leq {1-\delta}}u(t)=u(1-\delta)$, the proof is the same as (i); if$\min_{\delta\leq t\leq {1-\delta}}u(t)=u(\delta)$, the proof is the same as (ii). From the three cases above, the proof is complete. \end{proof} Let$X=C[0,1]$and$\|u\|=\max_{0\leq t\leq 1}|u(t)|$, denote$K=\{u\in X:u(t)\geq 0, t\in [0,1]\}$and$K'=\{u\in K: u(t)$is a concave function on$[0,1]\}$. Obviously,$K,K'\subset X$are two cones of$X$with$K'\subset K$. For$u\in K$, from Lemma \ref{lem2.5} we know$\min_{t\in[\frac{1}{k},1-\frac{1}{k}]}u(t)\geq {\frac{1}{k}\|u\|}$with$k>\max\{\frac{1}{\xi_1},\frac{2}{1-\xi_{m-2}}\}$, and we can show$1-\frac{1}{k}>\xi_1$. Denote $$\alpha(u)=\min_{t\in[\frac{1}{k},1-\frac{1}{k}]}u(t),\quad \text{for } u\in K'.$$ We have$\alpha(u)\leq \|u\| \leq {k\alpha(u)}$. Define$T:K\to K$by $$Tu(t)=\begin{cases} \Big[\frac{b\varphi_q(\int_0^{\sigma_u}f(\tau,u(\tau))d\tau) +\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q(\int_s^{\sigma_u} f(\tau,u(\tau))d\tau)ds} {a-\sum_{i=1}^{m-2}a_{i}} \\ +\int_0^t\varphi_q(\int_s^{\sigma_u}f(\tau,u(\tau))d\tau)ds\Big]^+, & 0\leq t\leq \sigma_u, \\[4pt] \Big[\frac{\int_0^1g(s)(\int_s^1\varphi_q (\int_{\sigma_u}^rf(\tau,u(\tau))d\tau)dr)ds}{1-\int_0^1g(s)ds}\\ +\int_t^1\varphi_q(\int_{\sigma_u}^sf(\tau,u(\tau))d\tau)ds\Big]^+, & \sigma_u\leq t\leq 1, \end{cases}$$ where$B^+=\max\{B,0\}$and$\sigma_u$is defined in Lemma \ref{lem2.3}. Define$A:K\to X$by $$Au(t)=\begin{cases} \frac{b\varphi_q(\int_0^{\sigma_u}f(\tau,u(\tau))d\tau) +\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q (\int_s^{\sigma_u}f(\tau,u(\tau))d\tau)ds} {a-\sum_{i=1}^{m-2}a_{i}}\\ +\int_0^t\varphi_q(\int_s^{\sigma_u}f(\tau,u(\tau))d\tau)ds, & 0\leq t\leq \sigma_u, \\[4pt] \frac{\int_0^1g(s)(\int_s^1\varphi_q(\int_{\sigma_u}^r f(\tau,u(\tau))d\tau)dr)ds} {1-\int_0^1g(s)ds}\\ +\int_t^1\varphi_q(\int_{\sigma_u}^s f(\tau,u(\tau))d\tau)ds,& \sigma_u\leq t\leq 1. \end{cases}$$ For$u\in X$, denote$\theta : X\to K$while$(\theta u)(t)=\max \{u(t),0\}$, then$T=\theta \circ A$. For$u\in K'$, define$T^*:K'\to K'$by $$T^*u(t)=\begin{cases} \frac{b\varphi_q(\int_0^{\sigma_u}f^+(\tau,u(\tau))d\tau) +\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q(\int_s^{\sigma_u} f^+(\tau,u(\tau))d\tau)ds} {a-\sum_{i=1}^{m-2}a_{i}} \\ +\int_0^t\varphi_q(\int_s^{\sigma_u}f^+(\tau,u(\tau))d\tau)ds, & 0\leq t\leq \sigma_u, \\[4pt] \frac{\int_0^1g(s)(\int_s^1\varphi_q(\int_{\sigma_u}^r f^+(\tau,u(\tau))d\tau)dr)ds} {1-\int_0^1g(s)ds}\\ +\int_t^1\varphi_q(\int_{\sigma_u}^s f^+(\tau,u(\tau))d\tau)ds,& \sigma_u\leq t\leq 1. \end{cases}$$ \begin{lemma} \label{lem2.6}$T^*:K'\to K'$is completely continuous. \end{lemma} \begin{proof} Obviously, we know$T^*(u)\geq 0$. By$[\varphi_p((T^*u)'(t))]'=-f^+(t,u(t))\leq 0$, it is easy to show$\varphi_p((T^*u)'(t))$is non-increasing. Then$(T^*u)'(t)$is also monotone non-increasing, i.e.,$(T^*u)(t)$is concave function. Thus$T^*:K'\to K'$. We can have$T^*$is completely continuous from Ascoli-Arzela theorem and concavity of$f^+$. \end{proof} From Lemma \ref{lem2.2}, we have the following lemma. \begin{lemma} \label{lem2.7} A function$u(t)$is a solution of boundary value problem \eqref{e1.1} if and only if$u(t)$is a fixed point of the operator$A$. \end{lemma} \begin{lemma}[\cite{g2}] \label{lem2.8} If$A:K\to X$is completely continuous, then$T=\theta \circ A :K\to K$is also completely continuous. \end{lemma} \begin{lemma} \label{lem2.9} If$u$is a fixed point of operator$T$, then$u$is also a fixed point of operator$A$. \end{lemma} \begin{proof} Let$u$be a fixed point of operator of$T$. Obviously, if$Au(t)\geq 0$,$t\in [0,1]$, then$u(t)$is a fixed point of operator$A$. We prove that if$Tu(t)=u(t)$, then$Au(t)\geq 0$for$t\in[0,1]$. Suppose this is not true, then there is a$t_0\in (0,1)$such that$Au(t_0)<0=u(t_0)$. Let$(t_1,t_2)$be the maximal interval which contain$t_0$and such that$Au(t)<0$,$t\in (t_1,t_2)$. It follows$[t_1,t_2]\neq [0,1]$from (H2). If$t_2<1$, we have$u(t)=0$for$t\in [t_1,t_2]$,$Au(t)<0$for$t\in (t_1,t_2)$and$Au(t_2)=0$. Thus$(Au)'(t_2)\geq 0$. From (H2), we know$[\varphi_p((Au)'(t))]'=-f(t,0)\leq 0$, so$t_1=0$. On the other hand, from the fact that$Au(t)<0$for$t\in [0,t_2)$and$Au(t_2)=0, (Au)'(t_2)\geq 0$and$(Au)''(t)\leq 0$, we can obtain$(Au)'(0)>0$. Without loss of generality, we suppose that there exists$i=k$such that$u(\xi_i)\leq 0$for$1\leq i\leq k$and$u(\xi_i)>0$for$k0$,$a>\sum_{i=1}^{m-2}a_i$and$aAu(0)-bAu'(0)=\sum_{i=1}^{m-2}a_iAu(\xi_i)<0, we have \begin{align*} |aAu(0)-bAu'(0)| &>a|Au(0)|>\sum_{i=1}^ka_i|Au(0)|+\sum_{i=k+1}^{m-2}a_i|Au(0)|\\ &>\sum_{i=1}^ka_i|Au(\xi_i)|-\sum_{i=k+1}^{m-2}a_iAu(\xi_i)\\ & =-(\sum_{i=1}^{m-2}a_iAu(\xi_i))\\ &=|\sum_{i=1}^{m-2}a_iAu(\xi_i)|, \end{align*} a contradiction. Sot_2=1$. If$t_1>0$, then we have$Au(t)=0$for$t\in [t_1,t_2]$,$Au(t)<0$for$t\in (t_1,1)$and$Au(t_1)=0$. Thus$(Au)'(t_1)\leq 0$. We have$[\varphi_p((Au)'(t))]'=-f(t,0)\leq 0$by (H2). This implies$Au(t)<0$for$t\in (t_1,1]$and$Au(1)=\min_{t\in [t_1,1]}Au(t)$. We can prove that$Au(t)\geq 0$for$t\in [0,t_1]$. If there exists a$t_3\not\in [t_1,1]$such that$Au(t_3)<0$and there is a maximal interval$[t_4,t_5]$which contains$t_3$such that$Au(t)<0$for$t\in (t_4,t_5)$. Obviously$[t_4,t_5)\cap [t_1,1]=\emptyset$, so$1\not\in(t_4,t_5)$; i.e.,$t_5<1$, this is a contradiction with the above discussion. Thus we can show$Au(t)\geq 0$for$t\in [0,t_1]$. For$Au(1)<0, we have $$Au(1)=\int_0^1g(s)Au(s)ds.$$ Then \begin{align*} |Au(1)| &=-Au(1)=-\int_0^{t_1}g(s)Au(s)ds-\int_{t_1}^1g(s)Au(s)ds\\ &\leq\int_0^{t_1}g(s)|Au(1)|ds+\int_{t_1}^1g(s)|Au(1)|ds\\ &=|Au(1)|\int_0^1g(s)ds. \end{align*} So|Au(1)|\leq |Au(1)|\int_0^1g(s)ds$which is a contradiction with$(H_3)$. Thus$t_1=0$. The above also contradicts$[t_1,t_2]\neq [0,1]. Thus the proof is complete. \end{proof} \section{Main result} Denote $$t^*=\frac{\xi_{m-2}+1}{2}, \quad M=\frac{1}{q}(\frac{1-\xi_{m-2}}{2}-\frac{1}{k})^q ,\quad N=\frac{1}{\max\big\{\frac{a+b}{a-\sum_{i=1}^{m-2}a_i}, \frac{1}{1-\int_0^1g(s)ds}\big\}},$$ \begin{align*} N_0=\min\Big\{&\frac{b+a_1\xi_1}{(b+a\xi_1)q}(\xi_1-\frac{1}{k})^q,\; \frac{1}{q}\int_{\frac{1}{k}}^{\xi_1}g(s)((\xi_1-\frac{1}{k})^q -(\xi_1-s)^q)ds,\\ &\frac{b+a_1\xi_1}{(b+a\xi_1)q}(1-\frac{1}{k}-\xi_1)^q,\; \frac{1}{q}\int_{\xi_1}^{1-\frac{1}{k}}g(s)((1-\xi_1-\frac{1}{k})^q -(s-\xi_1)^q)ds\Big\}. \end{align*} It is easy to see thatN_0>0$and$\frac{1}{k}\sigma _u, we have \begin{align*} Tu(\overline{t}) & =\Big[\frac{\int_0^1g(s)(\int_s^1\varphi_q(\int_{\sigma_u}^rf(\tau,u(\tau))d\tau)dr)ds} {1-\int_0^1g(s)ds}+\int_{\overline{t}}^1\varphi_q(\int_{\sigma_u}^sf(\tau,u(\tau))d\tau)ds\Big]^+ \\ & <\frac{\int_0^1g(s)(\int_0^1\varphi_q(\int_0^1\varphi_p(\frac{c_2}{N})d\tau)dr)ds}{1-\int_0^1g(s)ds}+\int_0^1\varphi_q(\int_0^1\varphi_p(\frac{c_2}{N})d\tau)ds \\ & =\frac{c_2}{N}\frac{1}{1-\int_0^1g(s)ds} \leq c_2. \end{align*} From the above, we have \| Tu \| < c_2$, So (C1) of Lemma \ref{lem2.1} is satisfied. For$u\in \partial K'_{c_2}$; i.e.,$u\in K'$and$\|u\|=c_2$. From Lemma \ref{lem2.4} and (H5), we can prove that$\| T^*u \| c_2$and$\alpha(u)\leq \frac{1}{k}c_3$. We know$\|u\|\leq k\alpha(u)\leq c_3$and$\alpha(u)\geq \frac{1}{k}\|u\|\ >\frac{1}{k}c_2$by Lemma \ref{lem2.5}. By$T^*:K'\to K'$and the definition of$T^*$, we can get $$\min_{0\leq t \leq 1}u(t)=\min\{u(0),u(1)\}\geq 0.$$ From the boundary condition$au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i), we have \begin{align*} au(0)&=bu'(0)+\sum_{i=1}^{m-2}a_iu(\xi_i)\\ &\geq bu'(0)+a_1u(\xi_1)\\ &\geq bu'(\xi)+a_1u(\xi_1)\\ &=\frac{b}{\xi_1}(u(\xi_1)-u(0))+a_1u(\xi_1), \end{align*} where\xi \in (0,\xi_1)$. So we can obtain $$u(0)\geq \frac{b+a_1\xi_1}{b+a\xi_1}u(\xi_1).$$ From the definition$T^*and the use of condition (H7), we have \begin{align*} u(\xi_1) &>\int_0^{\xi_1}\varphi_q(\int_s^{\sigma_u}f^+(\tau,u(\tau))d\tau)ds > \int_{\frac{1}{k}}^{\xi_1}\varphi_q(\int_s^{\xi_1}f^+(\tau,u(\tau))d\tau)ds\\ &>\frac{c_1}{qN_0}(\xi_1-\frac{1}{k})^q, \quad\text{if } \sigma_u>\xi_1, \end{align*} and \begin{align*} u(\xi_1) &>\int_{\xi_1}^{1-\frac{1}{k}}\varphi_q(\int_{\sigma_u}^sf^+(\tau,u(\tau))d\tau)ds >\int_{\xi_1}^{1-\frac{1}{k}}\varphi_q(\int_{\xi_1}^sf^+(\tau,u(\tau))d\tau)ds\\ &>\frac{c_1}{qN_0}(1-\frac{1}{k}-\xi_1)^q, \quad\text{if } \sigma_u\leq \xi_1. \end{align*} Hence, $$u(0)\geq\frac{b+a_1\xi_1}{b+a_1\xi}\cdot\frac{c_1}{qN_0}\min\{(\xi_1-\frac{1}{k})^q,(1-\frac{1}{k}-\xi_1)^q\}\geq c_1.$$ From the boundary conditionu(1)=\int_0^1 g(s)u(s) ds, we have \begin{align*} u(1) &>\int_0^{\xi_1}g(s)u(s)ds>\int_{\frac{1}{k}}^{\xi_1}g(s)(\int_{\frac{1}{k}}^{\xi_1}\varphi_q(\int_r^{\xi_1}f^+(\tau,u(\tau))d\tau)dr)ds\\ &>\frac{c_1}{qN_0}(\int_{\frac{1}{k}}^{\xi_1}g(s)((\xi_1-\frac{1}{k})^q-(\xi_1-s)^q)ds), \quad\text{if } \sigma_u>\xi_1, \end{align*} and \begin{align*} u(1) &>\int_{\xi_1}^{1-\frac{1}{k}}g(s)u(s)ds>\int_{\xi_1}^{1-\frac{1}{k}}g(s)(\int_s^{1-\frac{1}{k}}\varphi_q(\int_{\xi_1}^rf^+(\tau,u(\tau))d\tau)dr)ds\\ &>\frac{c_1}{qN_0}(\int_{\xi_1}^{1-\frac{1}{k}}g(s)((1-\xi_1-\frac{1}{k})^q-(s-\xi_1)^q)ds), \quad \text{if }\sigma_u\leq \xi_1. \end{align*} So,u(1)\geq c_1$. Therefore, for$u\in \partial K'_{c_2}(\frac{1}{k}c_3)\cap\{u:u=T^*u\}$, we have $$c_1\leq u(t)\leq \|u\| \leq k\alpha(u)=c_3.$$ It follows$f(t,u(t))\geq 0$,$t\in [0,1]$from (H4). Thus,$T^*u=Tu$. So the condition of Lemma \ref{lem2.1} is satisfied. Then by Lemma \ref{lem2.7}, we know that operator$T$has two fixed points$u_1$and$u_2$satisfying$\$ 0<\|u_1 \|