\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 22, pp. 1--20.\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-2010/22\hfil Monotone positive solutions] {Monotone positive solutions for $p$-Laplacian equations with sign changing coefficients and multi-point boundary conditions} \author[J. Xia, Y. Liu \hfil EJDE-2010/22\hfilneg] {Jianye Xia, Yuji Liu} \address{Jianye Xia \newline Department of Mathematics, Guangdong University of Finance \\ Guangzhou 510320, China} \email{JianyeXia@sohu.com} \address{Yuji Liu \newline Department of Mathematics, Hunan Institute of Science and Technology\\ Yueyang 414000, China} \email{liuyuji888@sohu.com} \thanks{Submitted October 26, 2009. Published February 4, 2010.} \thanks{Supported by grants 06JJ5008 and 7004569 from the Natural Science Foundation of \hfill\break\indent Hunan and Guangdong provinces, China} \subjclass[2000]{34B10, 34B15, 35B10} \keywords{Second order differential equation; positive solution \hfill\break\indent multi-point boundary value problem} \begin{abstract} We prove the existence of three monotone positive solutions for the second-order multi-point boundary value problem, with sign changing coefficients, \begin{gather*} [p(t)\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x'(0)=-\sum_{i=1}^la _ix'(\xi_i)+\sum_{i=l+1}^ma_ix'(\xi_i),\\ x(1)+\beta x'(1)=\sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^mb_ix(\xi_i) -\sum_{i=1}^mc_ix'(\xi_i). \end{gather*} To obtain these results, we use a fixed point theorem for cones in Banach spaces. Also we present an example that illustrates the main results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} As is well known, a differential equation defined on the interval $a\leq t\leq b$ having the form \begin{gather*} [p(t)x'(t)]'+(q(t)+\lambda r(t))x(t)=0,\\ a_1x(a)+a_2x'(a)=a_3x(b)+a_4x'(b)=0, \end{gather*} is called a Sturm-Liouville boundary-value problem or Sturm-Liouville system. Here $p(t)>0$, $q(t)$, the weighting function $r(t)$, the constants $a_1, a_2, a_3, a_4$ are given, and the eigenvalue $\lambda$ is an unspecified parameter. Sturm-Liouville boundary-value problems for nonlinear second-order $p$-Laplacian differential equations have been studied extensively; see for example \cite{a1,a2,b1,d1,g1,g2,h1,l1,l2,l3,l4,m1,m2}. The study of existence of positive solutions for such problems is complicated since there is no Green's function for the p-Laplacian $(p\neq 2)$. Some authors have extend Sturm-Liouville boundary conditions to nonlinear cases. For example, He, Ge and Peng \cite{h1}, by means of the Leggett-Williams fixed-point theorem, established criteria for the existence of at least three positive solutions to the one-dimensional p-Laplacian boundary-value problem $$\label{e1} \begin{gathered} (\phi(y'))' + g(t)f(t,y) = 0,\quad t\in (0,1),\\ y(0) - B_0(y'(0)) = 0,\\ y(1) + B_1(y'(1)) = 0, \end{gathered}$$ where $\phi(v)=|v|^{p-2}v$ with $v>1$, under the assumptions $xB_0(x)\geq 0$, $xB_1(x)\geq 0$ and that there exist constants $M_i>0$ such that $$\label{e2} |B_i(x)|\leq M_i|x|,\quad x\in \mathbb{R}.$$ In \cite{l2,l3,l4,h1}, the authors extended \eqref{e1} to a more general case. They established some existence results of at least one positive solution of Sturm-Liouville boundary value problems of higher order differential equations. Liu \cite{l5} studied the boundary-value problem \label{e3} \begin{gathered}{} [\phi(x^{(n-1)}(t))]'=f(t,x(t),x'(t),\dots,x^{(n-2)}(t)),\quad 00$. By using Green's functions (which is complicate for studying \eqref{e1}) and Guo-Krasnoselskii fixed point theorem \cite{d1,g2}, the existence and multiplicity of positive solutions for \eqref{e4} were given. There is no paper discussing the existence of multiple positive solutions of \eqref{e4} by using Leggett-Williams fixed point theorem. Liu in \cite{l3,l4} also studied some Sturm-Liouville type multi-point boundary value problems. In recent papers \cite{l1,p1}, the authors studied the four-point boundary-value problem $$\label{e5} \begin{gathered} (\phi(x'))' + f(t,x(t),x'(t)) = 0,\quad t\in (0,1),\\ \alpha x(0) -\beta x'(\xi) = 0,\\ \gamma x(1) + \delta x'(\eta) = 0, \end{gathered}$$ where$\phi(x)=|x|^{p-2}x$,$p > 1$,$\phi^{-1}(x)=|x|^{q-2}x$with$1/p + 1/q = 1$,$\alpha>0$,$\beta\geq 0$,$\gamma>0$,$\delta\geq 0$,$0<\xi<\eta<1$,$f$is continuous and nonnegative. When$\xi\to 0$and$\eta\to 1$, \eqref{e5} converges to a Sturm-Liouville boundary-value problem. So \eqref{e5} can be seen as a generalized Sturm-Liouville boundary-value problem. Xu \cite{x1} proved the existence of at least one or two positive solutions of the differential equation $$\label{e6} \begin{gathered}{} [\phi(x'(t))]'+a(t)f(x(t))=0,\quad t\in (0,1),\\ x'(0)=\sum_{i=1}^ma _ix'(\xi_i),\\ x(1)=\sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^sb_ix(\xi_i) -\sum_{i=s+1}^mc_ix'(\xi_i), \end{gathered}$$ where$0<\xi_1<\dots<\xi_m<1$,$a_i,b_i,c_i\geq 0$,$a$and$f$are continuous functions,$\phi(x)=|x|^{p-2}x$with$p>1$. Motivated by above mentioned papers, we investigate the boundary-value problem $$\label{e7} \begin{gathered}{} [p(t)\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x'(0)=-\sum_{i=1}^la _ix'(\xi_i)+\sum_{i=l+1}^ma_ix'(\xi_i),\\ x(1)+\beta x'(1)=\sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^mb_ix(\xi_i) -\sum_{i=1}^mc_ix'(\xi_i), \end{gathered}$$ where \begin{itemize} \item$0<\xi_1<\dots<\xi_m<1$,$\beta\geq 0$,$1\leq k,l\leq m$and$ a_i\geq 0,b_i\ge0,c_i\geq 0$for all$i=1,\dots,m$; \item$f$is defined on$[0,1]\times \mathbb{R}\times \mathbb{R}$, continuous, nonnegative with$f(t,0,0)\not\equiv 0$on each subinterval of$[0,1]$; \item$p$is defined on$[0,1]$, continuous and positive; \item$\phi$is called$p$-Laplacian,$\phi(x)=|x|^{p-2}x$with$p>1$, its inverse function is denoted by$\phi^{-1}(x)$with$\phi^{-1}(x)=|x|^{q-2}x$with$1/p+1/q=1$. \end{itemize} Sufficient conditions for the existence of at least three monotone positive solutions of \eqref{e7} are established by using a fixed point theorem for cones in Banach spaces. We remark that the fixed point theorem used here is different form the one in \cite{x1}. Our results improve the the results in \cite{x1} since: Three positive solutions are obtained, while one or two positive solutions were obtained in \cite{x1}; the nonlinearity in \eqref{e7} depends on$t,x, x'$, while the nonlinearity in \cite{x1} depends only on$t,x$. The main result for this article is presented in Section 2, and an example is given in Section 3. \section{Main Results} In this section, we first present some background definitions and state an important fixed point theorem. Then the main results are given and proved. \begin{definition} \label{def2.1} \rm Let$X$be a semi-ordered real Banach space. The nonempty convex closed subset$P$of$X$is called a cone in$X$if$x+y\in P$and$ax\in P$for all$x,y\in P$and$a\geq 0$, and$x\in X$and$-x\in X$imply$x=0$. \end{definition} \begin{definition} \label{def2.2} \rm Let$X$be a semi-ordered real Banach space and$P$a cone in$X$. A map$ \psi :P\to [0,+\infty)$is a nonnegative continuous concave (or convex) functional map provided$\psi $is nonnegative, continuous and satisfies $$\psi (tx+(1-t)y)\geq \text{ (or \leq) } t\psi (x)+(1-t)\psi(y) \quad\text{for all }x,y\in P,t\in [0,1].$$ \end{definition} \begin{definition} \label{def2.3} \rm Let$X$be a semi-ordered real Banach space. An operator$T;X\to X$is completely continuous if it is continuous and maps bounded sets into pre-compact sets. \end{definition} Let$a_1,a_2,a_3,a_4>0$be positive constants,$\psi$be a nonnegative continuous functional on the cone$P$. Define the sets as follows: \begin{gather*} P(\beta_1;a_4)=\{x\in P:\beta_1(x)0$ such that $\|x\|\leq M\beta_1(x)$ for all $x\in P$; \item[(3)] there exist positive numbers $a_1a_2$ for all $x\in P(\beta_1,\alpha_1;a_2,a_4)$ with $\beta_2(Tx)>a_3$; \item[(E3)] $\{x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4): \alpha_1(x)>a_2\}\neq \emptyset$ and $\alpha_1(Tx)>b$ for all $x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4)$; \item[(E4)] $0\not\in R(\beta_1,\psi;a_1,a_4)$ and \psi(Tx)a_2,\quad \psi(x_2)>a_1,\quad \alpha_1(x_2)0, $$\tau(t_0)=0 and \tau(1)=1. Thus$$ \frac{dx}{dt}=\frac{dx}{d\tau}\frac{d\tau}{dt} =\frac{dx}{d\tau}\frac{\phi^{-1}\big(\frac{1}{p(t)}\big)} {\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds} $$which implies$$ p(t)\phi(x'(t))=\phi\Big(\frac{dx}{d\tau}\Big) \frac{1}{\phi\Big(\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds\Big)}. $$Hence$$ \phi'\Big(\frac{dx}{d\tau}\Big)\frac{\phi^{-1}\big(\frac{1}{p(t)}\big)} {\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds}\frac{d^2x}{d\tau^2} =\phi\Big(\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds\Big) \big[\frac{}{}p(t)\phi(x'(t))\big]'\leq 0 for all t\in [t_0,1]. Note that \phi'(x)\geq 0. It follows that \frac{d^2x}{d\tau^2}\leq 0. Together with x''(\tau)\leq 0(\tau\in [0,1]). Then x' is decreasing on [0,1]. We get that there exist t_0\leq \eta\leq t\leq \xi\leq 1 such that \begin{align*} \frac{x(t_0)-x(1)}{t_0-1}-\frac{x(t)-x(1)}{t-1} &= -\frac{(t-1)[x(t_0)-x(t)]+(t_0-t)[x(1)-x(t)]} {(t-1)(t_0-1)}\\ &= -\frac{(t-1)(t_0-t)x'(\eta)+(t_0-t)(1-t)tx'(\xi)}{(t-1)(t_0-1)}\\ &\leq -\frac{(t-1)(t_0-t)x'(\xi)+(t_0-t)(1-t)tx'(\xi)}{(t-1)(t_0-1)}=0. \end{align*} It follows that for t\in (t_0,1), x(t)\ge x(1)+(t-1)\frac{x(t_0)-x(1)}{t_0-1} =x(1)\Big(1-\frac{1-t}{1-t_0}\Big)+\frac{1-t}{1-t_0}x(t_0)\ge (1-t)x(t_0). $$If t_0>0, for t\in (0,t_0), similarly to above discussion, we have$$ x(t)\geq tx(t_0),\;t\in (0,t_0). Then one gets that x(t)\geq \min\{t,1-t\}\max_{t\in [0,1]}x(t) for all t\in [0,1]. The proof is complete. \end{proof} Consider the boundary-value problem $$\label{e9} \begin{gathered}{} [p(t)\phi(x'(t))]'+\theta(t)=0,\quad t\in (0,1),\\ x'(0)=-\sum_{i=1}^la _ix'(\xi_i)+\sum_{i=l+1}^ma_ix'(\xi_i),\\ x(1)+\beta x'(1)=\sum_{i=1}^kb_ix(\xi_i) -\sum_{i=k+1}^sb_ix(\xi_i)-\sum_{i=s+1}^mc_ix'(\xi_i). \end{gathered}$$ \begin{lemma} \label{lem2.2} Suppose that {\rm (A1)--(A5)} hold. If x is a solution of \eqref{e9}, then x is concave, decreasing and positive on (0,1). \end{lemma} \begin{proof} Suppose x satisfies \eqref{e9}. It follows from the assumptions that px' is decreasing on [0,1]. Lemma \ref{lem2.1} implies that x' is decreasing on [0,1]. Then x is concave on [0,1]. First, we prove that x'(0)\leq 0. One sees from (A2) and the deceasing property of p(t)\phi(x'(t)) that \begin{align*} x'(1)&= -\sum_{i=1}^la _i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)\phi^{-1}(p(\xi_i)\phi(x'(\xi_i))) +\sum_{i=l+1}^ma_i\big(\frac{1}{p(\xi_i)}\big) \phi^{-1}(p(\xi_i)\phi(x'(\xi_i)))\\ &\leq -\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big) \phi^{-1}(p(\xi_l)\phi(x'(\xi_l)))\\ &\quad +\sum_{i=l+1}^ma_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big) \phi^{-1}(p(\xi_{l+1})\phi(x'(\xi_{l+1})))\\ &= \Big(\sum_{i=l+1}^ma_i\phi^{-1} \big(\frac{1}{p(\xi_i)}\big)-\sum_{i=1}^la_i\phi^{-1} \big(\frac{1}{p(\xi_i)}\big) \Big) \phi^{-1}(p(\xi_{l+1})\phi(x'(\xi_{l+1})))\\ &\quad +\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big) [\phi^{-1}(p(\xi_{l+1})\phi(x'(\xi_{l+1})))-\phi^{-1}(p(\xi_l) \phi(x'(\xi_l)))]\\ &\leq \Big(\sum_{i=l+1}^ma_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big) -\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big) \Big)\phi^{-1}(p(\xi_{l+1}\phi(x'(\xi_{l+1})))\\ &\leq \Big(\sum_{i=l+1}^ma_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big) -\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big) \Big)\phi^{-1}(p(0)\phi(x'(0))). \end{align*} It follows that \Big(1-\sum_{i=l+1}^ma_i\Big(\frac{p(0)}{p(\xi_i)}\Big) +\sum_{i=1}^la_i\frac{p(0)}{p(\xi_i)}\Big)x'(0)\leq 0. It follows that x'(0)\leq 0. Then x'(t)\leq 0 for all t\in [0,1]. Second, we prove that x(1)\geq 0. It follows from the boundary conditions in \eqref{e9} that \begin{align*} x(1)+\beta x'(1) &= \sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^sb_ix(\xi_i) -\sum_{i=s+1}^mc_ix'(\xi_i)\\ &\geq \sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^sb_ix(\xi_i)\\ &\geq \sum_{i=1}^kb_ix(\xi_k)-\sum_{i=k+1}^sb_ix(\xi_{k+1})\\ &= \Big(\sum_{i=1}^kb_i-\sum_{i=k+1}^sb_i\Big)x(\xi_k) +\sum_{i=k+1}^sb_i[x(\xi_k)-x(\xi_{k+1})]\\ &\geq \Big(\sum_{i=1}^kb_i-\sum_{i=k+1}^sb_i\Big)x(\xi_k)\\ &\geq \Big(\sum_{i=1}^kb_i-\sum_{i=k+1}^sb_i\Big)x(1). \end{align*} Then \Big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^sb_i\Big)x(1)+\beta x'(1) \geq 0. $$We get x(1)\geq 0 since x'(1)\leq 0 and \beta\geq 0. Then x(t)> x(1)\geq 0 for all t\in [0,1). The proof is complete. \end{proof} \begin{lemma} \label{lem2.3} Suppose that {\rm (A1),(A2'), (A3)--(A5)} hold. Let $\mu =\phi\Big(\frac{1}{\sum_{i=l+1}^ma_i\phi^{-1} \big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1.$ If y is a solution of \eqref{e9}, then$$ y(t)=B_\theta-\int_t^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_\theta)-\frac{1}{p(s)} \int_0^s\theta(u)du\Big)ds,\quad t\in [0,1], $$where$$ A_\theta\in \Big[-\phi^{-1} \Big(\frac{\int_0^1\theta(u)du}{\mu p(0)}\Big),0\Big], \begin{align*} A_\theta &= -\sum_{i=1}^la_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_\theta)-\frac{1}{p(\xi_i)} \int_0^{\xi_i}\theta(u)du\Big)\\ &\quad +\sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_\theta)-\frac{1}{p(\xi_i)} \int_0^{\xi_i}\theta(u)du\Big) \end{align*} and \begin{align*} B_\theta &= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i} \Big[-\beta\phi^{-1} \Big(\frac{1}{p(1)}p(0)\phi(A_\theta)-\frac{1}{p(1)} \int_0^1\theta(u)du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_\theta)-\frac{1}{p(s)} \int_0^s\theta(u)du\Big)ds\\ &\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_\theta)-\frac{1}{p(s)} \int_0^s\theta(u)du\Big)ds \\ &\quad -\sum_{i=1}^mc_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_\theta)-\frac{1}{p(\xi_i)} \int_0^{\xi_i}\theta(u)du\Big)ds\Big]. \end{align*} \end{lemma} \begin{proof} Since y is solution of \eqref{e9}, \begin{gather*} y'(t)=\phi^{-1}\Big(\frac{1}{p(t)}p(0)\phi(y'(0)) -\frac{1}{p(t)}\int_0^t\theta(u)du\Big),\\ y(t)=y(1)-\int_t^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(y'(0)) -\frac{1}{p(s)}\int_0^s\theta(u)du\Big)ds. \end{gather*} The boundary conditions in \eqref{e9} imply \begin{align*} y'(0) &= -\sum_{i=1}^la_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0)) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\ &\quad +\sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0)) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big) \end{align*} and \begin{align*} &y(1)+\beta\phi^{-1} \Big(\frac{1}{p(1)}p(0)\phi(y'(0)) -\frac{1}{p(1)}\int_0^1\theta(u)du\Big)\\ &= \sum_{i=1}^kb_i\Big(y(1)-\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(y'(0))-\frac{1}{p(s)}\int_0^s\theta(u)du \Big)ds\Big)\\ &\quad - \sum_{i=k+1}^mb_i \Big(y(1)-\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(y'(0))-\frac{1}{p(s)}\int_0^s\theta(u)du \Big)ds\Big)\\ &\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0)) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big). \end{align*} It follows that \begin{align*} y(1)&= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i} \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(y'(0)) -\frac{1}{p(1)}\int_0^1\theta(u)du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(y'(0)) -\frac{1}{p(s)}\int_0^s\theta(u)du\Big)ds\\ &\quad+\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(y'(0)) -\frac{1}{p(s)}\int_0^s\theta(u)du\Big)ds \\ &\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0)) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)ds\Big]. \end{align*} Lemma \ref{lem2.2} implies that y'(0)\leq 0. One finds that \begin{align*} y'(0)&= -\sum_{i=1}^la_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0)) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\ &\quad +\sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0)) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\ &\geq \sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0)) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big). \end{align*} If y'(0)\neq 0, one gets $$\label{e10} 1-\sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0) -\frac{1}{\phi(y'(0))}\frac{1}{p(\xi_i)} \int_0^{\xi_i}\theta(u)du\Big)\leq 0.$$ Let G(c)=c-\sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(c) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big). $$Then$$ \frac{G(c)}{c}=1-\sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)-\frac{1}{\phi(c)}\frac{1}{p(\xi_i)} \int_0^{\xi_i}\theta(u)du\Big). $$Note that$$ \mu=\phi\Big(\frac{1}{\sum_{i=l+1}^ma_i\phi^{-1} \big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1. $$It is easy to see that \frac{G(c)}{c} is decreasing on (-\infty,0) and on (0,+\infty). First, since \lim_{t\to 0^+} G(c)/c =+\infty and$$ \lim_{c\to +\infty}\frac{G(c)}{c} =1-\sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\Big)>0, we get that \frac{G(c)}{c}>0 for all c>0. Second, we have \lim_{c\to 0^-} G(c)/c=-\infty and \begin{align*} \frac{G\Big(-\phi^{-1}\Big(\frac{\int_0^1\theta(u)du}{\mu p(0)}\Big) \Big)}{-\phi^{-1}\Big(\frac{\int_0^1\theta(u)du}{\mu p(0)}\Big)} &= 1-\!\sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)+\frac{p(0)\mu} {\int_0^1\theta(u)du}\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\ &= 1-\sum_{i=t+1}^ma_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)+\mu\frac{\int_0^{\xi_i}\theta(u)du} {\int_0^1\theta(u)du}\frac{1}{p(\xi_i)}p(0)\Big)\\ &\geq 1-\phi^{-1}(1+\mu)\sum_{i=l+1}^ma_i\phi^{-1} \Big(\frac{p(0)}{p(\xi_i)}\Big) = 0. \end{align*} It follows from \eqref{e10} that \frac{G(y'(0))}{y'(0)}\leq 0. We get 0\geq y'(0)\ge-\phi^{-1}\Big(\frac{\int_0^1\theta(u)du}{\mu p(0)}\Big). $$The proof is complete. \end{proof} Define the nonlinear operator T:X\to X by $(Tx)(t)= B_x-\int_t^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x) -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds,$ for t\in [0,1], where$$ A_\sigma\in \Big[-\phi^{-1}\Big(\frac{\int_0^1f(u,x(u),x'(u))du} {\Big(\phi\Big(\frac{1}{\sum_{i=t+1}^ma_i\phi^{-1} \big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1 \Big)p(0)}\Big),0\Big], \begin{align*} A_x&= -\sum_{i=1}^la_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)-\frac{1}{p(\xi_i)} \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)\\ &\quad +\sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}f(u,x(u),x'(u))du\Big), \end{align*} \begin{align*} B_x &= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x) -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds\\ &\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds \\ &\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x) -\frac{1}{p(\xi_i)} \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\Big]. \end{align*} \begin{lemma} \label{lem2.4} Suppose that {\rm (A1),(A2'), (A3), (A4)} hold. Then \begin{itemize} \item[(i)] the following equalities hold: \begin{gather*} [p(t)\phi((Ty)'(t))]'+f(t,y(t),y'(t))=0,\quad t\in (0,1),\\ (Ty)'(0)=-\sum_{i=1}^la_i(Ty)'(\xi_i)+\sum_{i=l+1}^ma _i(Ty)'(\xi_i),\\ (Ty)(1)+\beta (Ty)'(1)=\sum_{i=1}^kb_i(Ty)(\xi_i)-\sum_{i=k+1}^mb_i(Ty)(\xi_i) -\sum_{i=1}^mc_i(Ty)'(\xi_i); \end{gather*} \item[(ii)] Ty\in P for each y\in P; \item[(iii)] x is a positive solution of \eqref{e7} if and only if x is a solution of the operator equation y=Ty in P; \item[(iv)] T:P\to P is completely continuous. \end{itemize} \end{lemma} \begin{proof} The proofs of (i), (ii) and (iii) are simple. To prove (iv), it suffices to prove that T is continuous on P and T is relative compact. We divide the proof into two steps: \textbf{Step 1.} For each bounded subset D\subset P, and each x_0\in D, since f(t,u,v) is continuous, we can prove that T is continuous at y(t). \textbf{Step 2.} For each bounded subset D\subset P, prove that T is relative compact on D. It is similar to that of the proof of Lemmas in \cite{l6} and are omitted. \end{proof} \begin{lemma} \label{lem2.5} Suppose that {\rm (A1), (A2'), (A3), (A4)} hold. Then there exists a constant M>0 such that \max_{t\in [0,1]}(Tx)(t)\leq M\max_{t\in [0,1]}|(Tx)'(t)|\quad \text{for each }x\in P. $$\end{lemma} \begin{proof} For x\in P, Lemma \ref{lem2.2} implies that (Tx)(t)\geq 0 and (Tx)'(t)\leq 0 for all t\in [0,1]. Lemma \ref{lem2.4} implies$$ (Tx)(1)+\beta (Tx)'(1)=\sum_{i=1}^kb_i(Tx)(\xi_i)-\sum_{i=k+1}^mb_i(Tx)(\xi_i) -\sum_{i=1}^mc_i(Tx)'(\xi_i). Then there exist numbers \eta_i\in [\xi_i,1] such that \begin{align*} (Tx)(1) &= \frac{(Tx)(1)-\sum_{i=1}^kb_i(Tx)(1) +\sum_{i=k+1}^mb_i(Tx)(1)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &= \frac{-\beta (Tx)'(1)-\sum_{i=1}^mc_i(Tx)'(\xi_i)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad +\frac{\sum_{i=1}^kb_i[(Tx)(\xi_i)-(Tx)(1)] -\sum_{i=k+1}^mb_i[(Tx)(\xi_i)-(Tx)(1)]} {1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &= \frac{-\beta (Tx)'(1)-\sum_{i=1}^mc_i(Tx)'(\xi_i)} {1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad +\frac{\sum_{i=1}^kb_i(\xi_i-1)(Tx)'(\eta_i) -\sum_{i=k+1}^mb_i(\xi_i-1)(Tx)'(\eta_i)} {1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}. \end{align*} It follows that \begin{align*} &|(Tx)(t)|\\ &= |(Tx)(t)-(Tx)(1)+(Tx)(1)|\\ &\leq |(Tx)(1)|+(1-t)|(Tx)'(\xi)|\quad\text{where }\xi\in [t,1]\\ &\leq \Big(1+\frac{\beta p(1)+\sum_{i=1}^kb_i(1-\xi_i)+\sum_{i=k+1}^mb_i(1-\xi_i) +\sum_{i=1}^mc_i}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\Big) \max_{t\in [0,1]}|(Tx)'(t)|. \end{align*} Then \begin{align*} &\max_{t\in [0,1]}|(Tx)(t)|\\ &\le\Big(1+\frac{\beta p(1)+\sum_{i=1}^kb_i(1-\xi_i)+\sum_{i=k+1}^mb_i(1-\xi_i) +\sum_{i=1}^mc_i}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\Big)\max_{t\in [0,1]}|(Tx)'(t)|. \end{align*} It follows that there exists a constant M>0 such that for all x\in P, $\max_{t\in [0,1]}(Tx)(t)\leq M\beta_1((Tx))\,.$ \end{proof} Denote \begin{gather*} \mu = \phi\Big(\frac{1}{\sum_{i=t+1}^ma_i\phi^{-1} \big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1,\\ M = 1+\frac{\beta p(1)+\sum_{i=1}^kb_i(1-\xi_i)+\sum_{i=k+1}^mb_i(1-\xi_i) +\sum_{i=1}^mc_i}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}, \\ \begin{aligned} L_1 &= \beta\phi^{-1}\Big(\frac{1+\mu}{\mu p(1)}\Big) +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds+\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds \\ &\quad +\sum_{i=1}^mc_i\phi^{-1}\left(\frac{1+\mu \xi_i}{\mu p(\xi_i)}\right)ds+\Big(1-\sum_{i=1}^kb_i\Big) \int_0^1\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds, \end{aligned}\\ \begin{aligned} L_2&= \beta\phi^{-1}\Big(\frac{1}{ p(1)}\Big) +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds+\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\big(\frac{s}{ p(s)}\big)ds \\ &\quad +\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{\xi_i}{p(\xi_i)}\Big)ds +\Big(1-\sum_{i=1}^kb_i\Big) \int_0^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds. \end{aligned} \end{gather*} \begin{theorem} \label{thm2.6} Choose k\in (0,1). Suppose that {\rm (A1), (A2'), (A3), (A4)} hold. Let e_1,e_2,c be positive numbers and \begin{gather*} Q= \min\Big\{\phi\Big(\frac {c\big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\big)}{L_1}\Big),\quad \frac{\mu p(1)\phi(c)}{1+\mu} \Big\};\\ W= \phi\Big(\frac {e_2\big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\big)}{L_2}\Big);\\ E= \phi\Big(\frac{e_1\big(1-\sum_{i=1}^kb_i +\sum_{i=k+1}^mb_i\big)}{L_1}\Big). \end{gather*} If Mc>e_2>\frac{e_1}{k}>e_1>0 and \begin{itemize} \item[(A6)] f(t,u,v)\leq Q for all t\in [0,1],u\in [0,Mc],v\in [-c,c]; \item[(A7)] f(t,u,v)\geq W for all t\in [0,k],u\in [e_2,e_2/k],v\in [-c,c]; \item[(A8)] f(t,u,v)\leq E for all t\in [0,1],u\in [0,e_1],v\in [-c,c]; \end{itemize} then \eqref{e7} has at least three solutions x_1,x_2,x_3 such that x_1(0)e_2,\quad x_3(0)>e_1,\quad x_3(k)0 such that $\|x\|\le M\beta_1(x)$ for all $x\in P$. Then (1) and (2) of Theorem \ref{thm2.1} hold. Corresponding to Theorem \ref{thm2.1}, we have $a_4=c$, $a_3=\frac{e_2}{k}$, $a_2=e_2$, $a_1=e_1$. Now, we check that \eqref{e3} of Theorem \ref{thm2.1} holds. One sees that $00$, $a_4>0$. The rest is divided into four steps. \textbf{Step 1.} Prove that $T(\overline{P_1(\beta_1;a_4)})\subseteq\overline{P_1(\beta_1;a_4)}$; For $x\in \overline{P_1(\beta_1;a_4)}$, we have $\beta_1(x)\le a_4=c$. Then \begin{gather*} 0\leq x(t)\leq \max_{t\in [0,1]}x(t)\leq Mc\quad\text{for } t\in [0,1],\\ -c\leq x'(t)\leq c\quad\text{for all }t\in [0,1]. \end{gather*} So (A6) implies that $f(t,x(t),x'(t))< Q$ for $t\in [0,1]$. Then Lemma \ref{lem2.3} implies $$0\geq A_x\ge-\phi^{-1}\Big(\frac{\int_0^1f(u,x(u),x'(u))du}{\mu p(0)}\Big)\geq -\phi^{-1}\Big(\frac{Q}{\mu p(0)}\Big).$$ Since $$c\ge\max\big\{\frac{e_2}{\sigma_0},\;La,\;\phi^{-1} \big(\frac{2}{p(0)}\big)a,\;\phi^{-1}\big(\frac{1}{p(1)}\big)a\big\},$$ we obtain $$\phi(a)\leq \min\big\{\phi\big(\frac{c}{L}\big),\; \frac{\phi(c)p(0)}{2},\;\phi(c)p(1)\big\}=Q.$$ Then $\max_{t\in [0,1]}(Ty)(t)=(Ty)(0)$ implies \begin{align*} &(Tx)(0)\\ &= B_x-\int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x) -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\\ &= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x) -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds\\ &\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds\\ &\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x) -\frac{1}{p(\xi_i)} \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\Big]\\ &\quad -\int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x) -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\\ &= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x) -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds\\ &\quad -\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\Big(\frac{1}{p(s)}p(0) \phi(A_x)-\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds \\ &\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x) -\frac{1}{p(\xi_i)}\int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\\ &\quad -\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\ &\leq \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i} \Big[\beta\phi^{-1}\Big(\frac{Q}{\mu p(1)}+\frac{Q}{p(1)}\Big) \\ &\quad +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \big(\frac{Q}{\mu p(s)}+\frac{Qs}{p(s)}\big)ds +\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\big(\frac{Q}{\mu p(s)}+\frac{Qs}{p(s)}\big)ds \\ &\quad +\sum_{i=1}^mc_i\phi^{-1}\big(\frac{Q}{\mu p(\xi_i)}+\frac{Q\xi_i}{p(\xi_i)}\big)ds +\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1} \big(\frac{Q}{\mu p(s)}+\frac{Qs}{p(s)}\big)ds\Big]\\ &= \frac{\phi^{-1}(Q)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[\beta\phi^{-1}\Big(\frac{1+\mu}{\mu p(1)}\Big) +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1+\mu s}{\mu p(s)}\Big)ds\\ &\quad +\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1} \Big(\frac{1+\mu s}{\mu p(s)}\Big)ds +\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1+\mu \xi_i}{\mu p(\xi_i)}\Big)ds \\ &\quad +\Big(1-\sum_{i=1}^kb_i\Big) \int_0^1\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds\Big] \leq c. \end{align*} On the other hand, similarly to above discussion, since $(Ty)'(t)$ is decreasing and $(Ty)'(0)\leq 0$, from Lemma \ref{lem2.3} we have $(Tx)'(1)= \phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x) -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big),$ \begin{align*} \max_{t\in [0,1]}|(Tx)'(t)|&= -\phi^{-1} \Big(\frac{1}{p(1)}p(0)\phi(A_x) -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big)\\ &\leq \phi^{-1}\Big(\frac{Q}{\mu p(1)}+\frac{Q}{p(1)}\Big) \leq c. \end{align*} It follows that $\|Tx\|=\max\big\{\max_{t\in [0,1]}|(Tx)(t)|,\max_{t\in [0,1]}|(Tx)'(t)|\big\}\leq c$. Then $T(\overline{P(\beta_1;a_4)})\subseteq\overline{P(\beta_1;a_4)}$. This completes the proof of (E1) in Theorem \ref{thm2.1}. \textbf{Step 2.} Prove that $\alpha_1(Tx)>a_2$ for all $x\in P(\beta_1,\alpha_1;a_2,a_4)$ with $\beta_2(Tx)>a_3$; For $y\in P(\beta_1,\alpha_1;a_2,a_4)=P(\beta_1,\alpha_1;e_2,c)$ with $\beta_2(Ty)>a_3=\frac{e_2}{k}$, we have $\alpha_1(y)=\min_{t\in [0,k]}y(t)\geq e_2, \quad \beta_1 (y)=\max_{t\in [0,1]}|y'(t)|\leq c, \quad \max_{t\in [0,k]}(Ty)(t)>\frac{e_2}{k}.$ Then $$\alpha_1(Ty)=\min_{t\in [0,k]}(Ty)(t)\ge k\beta_2(Ty)>k\frac{e_2}{k}=e_2=a_2.$$ This completes the proof of (E2) in Theorem \ref{thm2.1}. \textbf{Step 3.} Prove that $\{x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4):\alpha_1(x)>a_2\}\neq \emptyset$ and $\alpha_1(Tx)>b$ for all $x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4)$; It is easy to check that $\{x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4):\alpha_1(x)>a_2\}\neq \emptyset$. For $y\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4)$, one has that $$\alpha_1(y)=\min_{t\in [0,k]}y(t)\geq e_2,\quad \beta_2(y)=\max_{t\in [0,k]}y(t)\leq \frac{e_2}{k},\quad \beta_1(y)=\max_{t\in [0,1]}| y'(t)|\leq c.$$ Then $$e_2\leq y(t)\leq \frac{e_2}{k},\quad t\in [0,k],\;| y'(t)|\leq c.$$ Thus (A7) implies $$f(t,y(t), y'(t))\geq W,\quad t\in [0,k].$$ Since $$\alpha_1(Ty)=\min_{t\in [0,k]}(Ty)(t)\geq k\max_{t\in [0,1]}(Ty)(t),$$ we obtain $\alpha_1(Ty)\geq k\max_{t\in [0,1]}(Ty)(t)$. Then $$\alpha_1(Ty)\geq k\max_{t\in [0,1]}(Ty)(t)\geq k(Ty)(0).$$ From $A_x\leq 0$, we obtain \begin{align*} \alpha_1(T_1y) &\geq k\Big[\frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big(-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)-\frac{1}{p(1)} \int_0^1f(u,x(u),x'(u))du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du \Big)ds\\ &\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds \\ &\quad -\sum_{i=1}^mc_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)-\frac{1}{p(\xi_i)} \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\Big)\\ &\quad -\int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x) -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\ &= \frac{k}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x) -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds\\ &\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds \\ &\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x) -\frac{1}{p(\xi_i)} \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\\ &\quad -\Big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\Big) \int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x)\\ &\quad -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\ &= \frac{k}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x) -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\ &\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds\\ &\quad -\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)} \int_0^sf(u,x(u),x'(u))du\Big)ds \\ &\quad -\sum_{i=1}^mc_i\phi^{-1} \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)-\frac{1}{p(\xi_i)} \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\\ &\quad-\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1} \Big(\frac{1}{p(s)}p(0)\phi(A_x)\\ &\quad -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\ &\geq \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[\beta\phi^{-1}\big(\frac{W}{p(1)}\big) +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}\big( \frac{Ws}{p(s)}\big)ds+\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1} \big(\frac{Ws}{p(s)}\big)ds \\ &\quad +\sum_{i=1}^mc_i\phi^{-1}\big(\frac{W\xi_i}{p(\xi_i)}\big)ds +\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1} \big(\frac{Ws}{p(s)}\big)ds\Big]\\ &= \frac{\phi^{-1}(W)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\ &\quad\times \Big[\beta\phi^{-1}\big(\frac{1}{ p(1)}\big) +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1} \big(\frac{s}{ p(s)}\big)ds+\sum_{i=k+1}^mb_i \int_0^{\xi_i}\phi^{-1}\big(\frac{s}{ p(s)}\big)ds \\ &\quad +\sum_{i=1}^mc_i\phi^{-1}\big(\frac{\xi_i}{ p(\xi_i)}\big)ds +\Big(1-\sum_{i=1}^kb_i\Big) \int_0^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds\Big] = e_2. \end{align*} This completes the proof of (E3) in Theorem \ref{thm2.1}. \textbf{Step 4.} Prove that $0\not\in R(\beta_1,\psi;a_1,a_4)$ and that $\psi(Tx)e_2,\quad \beta(x_3)>e_1,\quad \alpha (x_3)e_2>\frac{e_1}{k}>e_1>0$ and \begin{gather*} f(t,u,v)\leq Q\quad\text{for all }t\in [0,1],u\in [0,Mc],v\in [-20000,20000];\\ f(t,u,v)\geq W\quad\text{for all }t\in [0,1/2],u\in [100,200],v\in [-20000,20000];\\ f(t,u,v)\leq E\quad\text{for all }t\in [0,1],u\in [0,2], v\in [-20000,20000]; \end{gather*} Then (A6), (A7) and (A8) hold. 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