\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{ Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 96, pp. 1--52.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{7mm}} \begin{document} \title[\hfilneg EJDE-2008/96\hfil Solutions to second order BVPs] {Solutions to second order non-homogeneous multi-point BVPs using a fixed-point theorem} \author[Y. Liu\hfil EJDE-2008/96\hfilneg] {Yuji Liu} \address{Yuji Liu \newline Department of Mathematics, Guangdong University of Business Studies, Guangzhou 510320, China} \email{liuyuji888@sohu.com} \thanks{Submitted March 3, 2008. Published July 25, 2008.} \thanks{Supported by grant 06JJ50008 from the Natural Science Foundation of Hunan \hfill\break\indent province and grant 7004569 from the Natural Science Foundation of Guangdong \hfill\break\indent province , China} \subjclass[2000]{34B10, 34B15, 35B10} \keywords{Second order; $p$-Laplacian; positive solution;\hfill\break\indent mixed type multi-point boundary-value problem} \begin{abstract} In this article, we study five non-homogeneous multi-point bound\-ary-value problems (BVPs) of second order differential equations with the one-dimensional $p$-Laplacian. These problems have a common equation (in different function domains) and different boundary conditions. We find conditions that guarantee the existence of at least three positive solutions. The results obtained generalize several known ones and are illustrated by examples. It is also shown that the approach for getting three positive solutions by using multi-fixed-point theorems can be extended to nonhomogeneous BVPs. The emphasis is on the nonhomogeneous boundary conditions and the nonlinear term involving first order derivative of the unknown. Some open problems are also proposed. \end{abstract} \maketitle \numberwithin{equation}{section} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Multi-point boundary-value problems (BVPs) for second order differential equations without $p$-Laplacian have received a wide attention because of their potential applications and BVPs are fascinating and challenging fields of study, one may see the textbook by Ge \cite{g}. There are four classes of such BVPs (including their special cases) studied in known papers: $$\label{e1} \begin{gathered} x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad x(1)=\sum_{i=1}^m\beta_ix(\xi_i), \end{gathered}$$ $$\label{e2} \begin{gathered} x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i),\quad x(1)=\sum_{i=1}^m\beta_ix(\xi_i), \end{gathered}$$ $$\label{e3} \begin{gathered} x''(t)+f(t,x(t),x'(t))=0,\;\;t\in (0,1),\\ x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i), \end{gathered}$$ and $$\label{e4} \begin{gathered} x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i),\quad x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i), \end{gathered}$$ where $0<\xi_1<\dots<\xi_m<1$, $\alpha_i,\beta_i\in \mathbb{R}$ and $f$ is a continuous function. The main methods to get solutions or multiple positive solutions of these BVPs are as follows: (i) fixed point index theory \cite{iw1,iw2,zf,zs}, or fixed point theorems in cones in Banach spaces \cite{d,fg2,gl,l3,m1,m2,mc}, such as the Guo-Krasnoselskii's fixed-point theorem \cite{chc,m4,mt,s,sczw,sz}; Leggett-Williams theorems \cite{l1,l4,oz}, the five-functional fixed point theorem \cite{aah}; the fixed point theorem of Avery and Peterson \cite{ko1}, etc. (ii) Mawhin's continuation theorem \cite{flg,iz,ko2,ko3,l1,l2,ly1,ly2,lz,m2}; (iii) the shooting methods \cite{kw,p}; (iv) upper and lower solution methods and monotone iterative techniques \cite{aop,cr,zm}; upper and lower solution methods and Leray-Schauder degree theory \cite{kk1,kk2,kk3,mo}, or the approach of a combination of nonlinear alternative of Leray-Schauder with the properties of the associated vector field at the $(x,x')$ plane \cite{gp}; (v) the critical point theory and variational methods \cite{la}; (vi) topological degree theory \cite{gmyz,g}; the Schauder's fixed point theorem in suitable Banach space \cite{l4,m1,m2,m3,nm}. In all the above papers, the boundary conditions (BCs) are homogeneous. However, in many applications, BVPs consist of differential equations coupled with nonhomogeneous BCs, for example \begin{gather*} y''=\frac{1}{\lambda}(1+y^2)^{\frac{1}{2}},\quad t\in (a,b),\\ y(a)=a\alpha,\quad y(b)=\beta \end{gather*} and \begin{gather*} y''=-\frac{(1+y'(t))^2}{2(y(t)-\alpha)},\quad t\in (a,b),\\ y(a)=\alpha,\quad y(b)=\beta \end{gather*} which are very well known and were proposed in 1690 and 1696, respectively. In 1964, the BVPs studied by Zhidkov and Shirikov \cite{zs2} and by Lee \cite{lee} are also nonhomogeneous. There are also several papers concerned with the existence of positive solutions of BVPs for differential equations with non-homogeneous BCs. Ma Ruyun \cite{m5} studied existence of positive solutions of the following BVP consisting of second order differential equations and three-point BC $$\label{e5} \begin{gathered} x''(t))+a(t)f(x(t))=0,\quad t\in (0,1),\\ x(0)=0,\quad x(1)-\alpha x(\eta)=b, \end{gathered}$$ Kwong and Kong \cite{kw} studied the BVP \label{e6} \begin{gathered} y''(t)=-f(t,y(t)),\quad 00$such that: \eqref{e6} has at least two positive solutions if$b\in (0,b^*)$; \eqref{e6} has at least one solution if$b=0$or$b=b^*$; \eqref{e6} has no positive solution if$b>b^*$. Palamides [\cite{p}], under superlinear and/or sublinear growth rate in$f$, proved the existence of positive solutions (and monotone in some cases) of the boundary-value problem $$\label{e7} \begin{gathered} y''(t)=-f(t,y(t),y'(t)),\quad 0\le t\leq1, \\ \alpha y(0)-\beta y'(0)=0,\quad y(1)-\sum_{i=1}^{m-2}\alpha_iy(\xi_i)=b\ge0, \end{gathered}$$ where$\alpha>0$,$\beta>0$, the function$f$is continuous, and$f(t,y,y')\ge 0$, for all$t\in[0,1]$,$y\ge 0$,$y'\in \mathbb{R}$. The approach is based on an analysis of the corresponding vector field on the face-plane and on Kneser's property for the solution's funnel. Sun, Chen, Zhang and Wang \cite{sczw} studied the existence of positive solutions for the three-point boundary-value problem $$\label{e8} \begin{gathered} u''(t)+a(t)f(u(t))=0,\quad 0\le t\le 1,\\ u'(0)=0,\quad u(1)-\sum_{i=1}^{m-2}\alpha_iu(\xi_i)=b\ge0, \end{gathered}$$ where$\xi_i\in(0,1)$,$\alpha_i\ge 0$are given. It was proved that there exists$b^*>0$such that \eqref{e8} has at least one positive solution if$b\in (0,b^*)$and no positive solution if$b>b^*$. To study the existence of positive solutions of BVPs \eqref{e5}, \eqref{e6}, \eqref{e7}, \eqref{e8}, the Green's functions of the corresponding problems are established and play an important role in the proofs of the main results. In recent papers, using lower and upper solutions methods, Kong and Kong \cite{kk1,kk2,kk3} established results for solutions and positive solutions of the following two problems $$\label{e9} \begin{gathered} x''(t))+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x'(0)-\sum_{i=1}^m\alpha_i x'(\xi_i)=\lambda_1,\quad x(1)-\sum_{i=1}^m\beta_i x(\xi_i)=\lambda_2, \end{gathered}$$ and $$\label{e10} \begin{gathered} x''(t))+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)-\sum_{i=1}^m\alpha_i x(\xi_i)=\lambda_1,\quad x(1)-\sum_{i=1}^m\beta_i x(\xi_i)=\lambda_2, \end{gathered}$$ respectively. We note that the boundary conditions in \cite{kk1,kk2} are two-parameter non-homogeneous BCs. There, the existence of lower and upper solutions with certain relations are assumed. In recent years, there have been many exciting results concerning the existence of positive solutions of BVPs of second order differential equations with$p$-Laplacian subjected to different multi-point boundary conditions: $$\label{e11} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad x(1)=\sum_{i=1}^m\beta_ix(\xi_i), \end{gathered}$$ $$\label{e12} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i),\quad x(1)=\sum_{i=1}^m\beta_ix(\xi_i), \end{gathered}$$ $$\label{e13} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i). \end{gathered}$$ These results, can be found in \cite{fg1,fg3,fgj,flg,fpg,gmyz,jfg,jg,jgz,l5,sg,sgz,sqg,wg1,wg2,wg3,wg4,x,zwg}. In above mentioned papers, to obtain positive solutions, two kinds of assumptions are supposed. The first one is imposed on$\alpha_i,\beta_i$, the other one called growth conditions is imposed on the nonlinearity$f$. To define a cone$P$in Banach spaces and to define the nonlinear operator$T:P\to P$are important steps in the the proofs of the results. It is easy to see that \begin{gather*} x''(t)=-2,\quad t\in (0,1),\\ x(0)=x(1)=0 \end{gather*} has unique positive solution$x(t)=-t^2+t$, but the BVP \begin{gather*} x''(t)=-2,\quad t\in (0,1),\\ x(0)=A, \quad x(1)=B \end{gather*} has positive solution$x(t)=-t^2+(B-A+1)t+A$if and only if$A\ge 0$and$B\ge 0$. It shows us that the presence of nonhomogeneous BCs can induce nonexistence of positive solutions of a BVP. Motivated by the facts mentioned above, this paper is concerned with the more generalized BVPs for second order differential equation with$p$-Laplacian coupled with nonhomogeneous multi-point BCs; i.e., $$\label{e14} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i)+A,\quad x(1)=\sum_{i=1}^m\beta_ix(\xi_i)+B, \end{gathered}$$ $$\label{e15} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\sum_{i=1}^m\alpha _ix(\xi_i)+A,\quad x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i)+B, \end{gathered}$$ $$\label{e16} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\alpha x'(0)+A,\quad x(1)=\sum_{i=1}^m\beta_ix(\xi_i)+B, \end{gathered}$$ $$\label{e17} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\sum_{i=1}^m\alpha _ix'(\xi_i)+A,\quad x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i)+B, \end{gathered}$$ and $$\label{e18} \begin{gathered}{} [\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\sum_{i=1}^m\alpha _ix(\xi_i)+A,\quad x(1)=\sum_{i=1}^m\beta_ix(\xi_i)+B, \end{gathered}$$ where$0<\xi_1<\dots<\xi_m<1$,$A,B\in \mathbb{R}$,$\alpha_i\ge0,\alpha\ge 0$,$\beta_i\ge0$for all$i=1,\dots,m$,$f$is continuous and nonnegative,$\phi$is called$p$-Laplacian,$\phi(x)=|x|^{p-2}x$for$x\neq 0$and$\phi(0)=0$with$p>1$, its inverse function is denoted by$\phi^{-1}(x)$with$\phi^{-1}(x)=|x|^{q-2}x$for$x\neq 0$and$\phi^{-1}(0)=0 $with$1/p+1/q=1$. The purpose is to establish sufficient conditions for the existence of at least three positive solutions for \eqref{e14}--\eqref{e18}. The results in this paper are new since there exists no paper concerned with the existence of at least three positive solutions of these nonhomogeneous multi-point BVPs even when$\phi(x)=x$. Maybe it is the first time to use the multi-fixed-point theorem to solve these kinds of BVPs. The remainder of this paper is organized as follows: The main results are presented in Section 2 (Theorems \ref{thm210}, \ref{thm214}, \ref{thm218}, \ref{thm222}, \ref{thm226}). Some examples to show the main results are given in Section 3. \section{Main Results} In this section, we first present some background definitions in Banach spaces and state an important three fixed point theorem. Then the main results are given and proved. \begin{definition} \label{def21}\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$ax\in P$for all$x\in P$and$a\ge 0$and$x\in X$and$-x\in X$imply$x=0$. \end{definition} \begin{definition} \label{def22}\rm A map$ \psi :P\to [0,+\infty)$is a nonnegative continuous concave or convex functional map provided$\psi $is nonnegative, continuous and satisfies $$\psi (tx+(1-t)y)\ge t\psi (x)+(1-t)\psi(y),$$ or $$\psi (tx+(1-t)y)\le t\psi (x)+(1-t)\psi(y),$$ for all$x,y\in P$and$t\in [0,1]$. \end{definition} \begin{definition} \label{def23} \rm An operator$T;X\to X$is completely continuous if it is continuous and maps bounded sets into relative compact sets. \end{definition} \begin{definition} \label{def24} \rm Let$a,b,c,d,h>0$be positive constants,$\alpha,\psi $be two nonnegative continuous concave functionals on the cone$P$,$\gamma ,\beta,\theta$be three nonnegative continuous convex functionals on the cone$P$. Define the convex sets: \begin{gather*} P_c=\{x\in P:\|x\|0$ such that $$\alpha(x)\le \beta(x),\quad \|x\|\le M\gamma (x)\quad\text{for all }x\in P.$$ Furthermore, Suppose that $h,d,a,b,c>0$ are constants with $da\}\neq \emptyset$ and $\alpha(Tx)>a$ for every $x$ in $P(\gamma,\theta,\alpha;a,b,c)$; \item[(C2)] $\{y\in Q(\gamma,\theta,\psi;h,d,c)|\beta(x)a$ for $y\in P(\gamma,\alpha;a,c)$ with $\theta(Ty)>b$; \item[(C4)] $\beta(Tx)a,\quad \beta(y_3)>d,\quad \alpha (y_3)y(1)\ge 0$ for $t\in (0,1)$ since $y'(t)\le0$ on $[0,1]$. The proof is complete. \end{proof} \begin{lemma} \label{lem28} Suppose that {\rm (H2)--(H4)} hold. If $y$ is a solution of \eqref{e20}, then $$y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds,$$ with $$\label{e21} \phi^{-1}(A_h)=\sum_{i=1}^m\alpha_i\phi^{-1} \Big(A_h-\int_0^{\xi_i}h(s)ds\Big)+A,$$ and \begin{align*} B_h&=\frac{1}{1-\sum_{i=1}^m\beta_i}\Big[-\int_0^1\phi^{-1} \Big(A_h-\int_0^sh(u)du\Big)ds\\ &\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i} \phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds\Big], \end{align*} where $a= \phi\left(\frac{A}{1-\sum_{i=1}^m\alpha_i}\right)$ and $$b= \phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big) -\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\int_0^1h(s)ds.$$ \end{lemma} \begin{proof} It follows from \eqref{e20} that $$y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds,$$ and BCs in \eqref{e20} imply that $$\label{e22} y'(0)=\sum_{i=1}^m\alpha_i\phi^{-1}\Big(\phi(y'(0)) -\int_0^{\xi_i}h(s)ds\Big)+A,$$ and \begin{align*} y(0)&= \frac{1}{1-\sum_{i=1}^m\beta_i}\Big[-\int_0^1\phi^{-1} \Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds \\ &\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i} \phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds\Big]. \end{align*} Let $$G(c)=\phi^{-1}(c)-\sum_{i=1}^m\alpha_i\phi^{-1} \Big(c-\int_0^{\xi_i}h(s)ds\Big)-A.$$ It is easy to see that \begin{align*} G(a)&= G\Big(\phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big)\\ &\geq \frac{A}{1-\sum_{i=1}^m\alpha_i} -\sum_{i=1}^m\alpha_i\phi^{-1} \Big(\phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big) -A = 0. \end{align*} On the other hand, one sees that \begin{align*} \frac{G(b)}{\phi^{-1}(b)} &=1-\sum_{i=1}^m\alpha_i\phi^{-1} \Big(1-\frac{\int_0^{\xi_i}h(s)ds}{b}\Big)-\frac{A}{\phi^{-1}(b)} \\ &= 1-\sum_{i=1}^m\alpha_i\phi^{-1} \Big(1-\frac{\int_0^{\xi_i}h(s)ds}{ \phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big)-\frac{\phi \big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}{1-\phi \big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\int_0^1h(s)ds}\Big) \\ &\quad-\frac{A}{\phi^{-1} \Big(\phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big)-\frac{\phi \big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}{1-\phi \big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\int_0^1h(s)ds\Big)} \\ &\ge 1-\sum_{i=1}^m\alpha_i\phi^{-1}\Big( 1+\frac{1-\phi \big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} {\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} \Big)-\frac{1-\sum_{i=1}^m\alpha_i}{1+\sum_{i=1}^m\alpha_i}\\ &= 1-\sum_{i=1}^m\alpha_i\frac{2}{1+\sum_{i=1}^m\alpha_i} -\frac{1-\sum_{i=1}^m\alpha_i}{1+\sum_{i=1}^m\alpha_i} = 0. \end{align*} Hence $G(b)\le 0$. It is easy to know that $\frac{G(c)}{\phi^{-1}(c)}$ is continuous and decreasing on $(-\infty,0)$ and continuous and increasing on $(0,+\infty)$, Hence $G(a)\ge 0$ and $G(b)\le 0$ and $\lim_{c\to 0^+}\frac{G(c)}{\phi^{-1}(c)}= +\infty,\quad \lim_{c\to +\infty}\frac{G(c)}{\phi^{-1}(c)} = 1-\sum_{i=1}^m\alpha_i>0,$ we get that there exists unique constant $A_h=\phi(y'(0))\in [b,a]$ such that \eqref{e21} holds. The proof is completed. \end{proof} Note $h=B/(1-\sum_{i=1}^m\beta_i)$, and let $x(t)-h=y(t)$. Then \eqref{e14} is transformed into the boundary-value problem $$\label{e23} \begin{gathered}{} [\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ y'(0)-\sum_{i=1}^m\alpha_i y'(\xi_i)=A,\quad y(1)-\sum_{i=1}^m\beta_i y(\xi_i)=0, \end{gathered}$$ Let \begin{align*} P_1=\Big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1],\; y'(t)\le 0 \text{ is decreasing on }[0,1],\\ &\ \ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in [0,1]}y(t) \text{ for all }t\in [0,1] \Big\}. \end{align*} Then $P_1$ is a cone in $X$. Since \begin{align*} |y(t)|&=\big|\frac{\sum_{i=1}^m\beta_i y(\xi_i)-\sum_{i=1}^m\beta_i y(1)}{1-\sum_{i=1}^m\beta_i }\big|\\ &\leq \frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i }\max_{t\in [0,1]}|y'(t)|\\ &=\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i }\gamma(y), \end{align*} we obtain $$\max_{t\in [0,1]}|y(t)|\leq \frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i }\gamma(y).$$ It is easy to see that there exists a constant $M>0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_1$. Define the nonlinear operator $T_1:P_1\to X$ by \begin{align*} (T_1y)(t)= B_y+\int_0^t\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_1, \end{align*} where $A_y$ satisfies $$\label{e24} \phi^{-1}(A_y)=\sum_{i=1}^m\alpha_i\phi^{-1} \Big(A_y-\int_0^{\xi_i}f(s,y(s)+h,y'(s))ds\Big)+A,$$ and $B_y$ satisfies \begin{align*} B_y&= \frac{1}{1-\sum_{i=1}^m\beta_i} \Big(-\int_0^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &+\sum_{i=1}^m\beta_i\int_0^{\xi_i} \phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big). \end{align*} Then for $y\in P_1$, \begin{align*} (T_1y)(t) &= -\int_t^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds. \end{align*} \begin{lemma} \label{lem29} Suppose that {\rm (H1)--(H3)} hold. It is easy to show that \begin{itemize} \item[(i)] the following equalities hold: \begin{gather*} [\phi((T_1y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ (T_1y)'(0)-\sum_{i=1}^m\alpha_i (T_1y)'(\xi_i)=A,\quad (T_1y)(1)-\sum_{i=1}^m\beta_i (T_1y)(\xi_i)=0; \end{gather*} \item[(ii)] $T_1y\in P_1$ for each $y\in P_1$; \item[(iii)] $x$ is a solution of \eqref{e14} if and only if $x=y+h$ and $y$ is a solution of the operator equation $y=T_1y$ in $P_1$; \item[(iv)] $T_1:P_1\to P_1$ 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 $T_1$ is continuous and $T_1$ is compact. We divide the proof into two steps: \noindent\textbf{Step 1.} Prove that $T_1$ is continuous about $y$. Suppose $y_n\in X$ and $y_n\to y_0\in X$. Let $A_{y_n}$ be decided by \eqref{e21} corresponding to $y_n$ for $n=0,1,2,\dots$. We will prove that $A_{y_n}\to A_{y_0}$ as $n$ tends to infinity. Since $y_n\to y_0$ uniformly on [0,1] and $f$ is continuous, we have that for $\epsilon =1$, there exists $N$, when $n>N,$ for each $t\in [0,1]$, such that $$0\le f(t,y_n(t)+h,y_n'(t))\le 1+f(t,y_0(t)+h,y_0'(t))\le 1+\max_{t\in [0,1]}f(t,y_0(t)+h,y_0'(t)).$$ Hence Lemma \ref{lem28} implies that $A_{y_n}$ is an element in the interval \begin{align*} &\Big[\phi\Big(\frac{A(1+\sum_{i=1}^m\alpha_i)} {1-\sum_{i=1}^m\alpha_i}\Big) -\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} \int_0^1f(s,y_n(s)+h,y_n'(s))ds,\\ &\phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big]\\ &\subseteq \Big[\phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big) -\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} \Big(1+\max_{t\in [0,1]}f(t,y_0(t)+h,y_0'(t))\Big),\\ &\quad \phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big]. \end{align*} It follows that $\{A_{y_n}\}$ is bounded. If $\{A_{y_n}\}$ does not converge to $A_{y_0}$, then there exist two subsequences $\{A_{y_{n_k}}^1\}$ and $\{A_{y_{n_k}}^2\}$ of $\{A_{y_n}\}$ with $$A_{y_{n_k}}^1\to C_1,\;\;A_{y_{n_k}}^2\to C_2,\quad k\to +\infty,\quad C_1\neq C_2.$$ By the construction of $A_{y_n}$, $$\phi^{-1}(A_{y_{n_k}}^1)=\sum_{i=1}^m\alpha_i\phi^{-1} \Big(A_{y_{n_k}}^1-\int_0^{\xi_i}f(s,y_{n_k}(s)+h,y_{n_k}'(s))ds\Big)+A.$$ Since $f(t,y_n(t)+h,y_n'(t))$ is uniformly bounded, by Lebesgue's dominated convergence theorem, letting $k\to +\infty$, we get $$\phi^{-1}(C_1)=\sum_{i=1}^m\alpha_i\phi^{-1} \Big(C_1-\int_0^{\xi_i}f(s,y_0(s)+h,y_0'(s))ds\Big)+A.$$ Since $A_{y_0}$ satisfies $$\phi^{-1}(A_{y_0})=\sum_{i=1}^m\alpha_i\phi^{-1} \Big(A_{y_0}-\int_0^{\xi_i}f(s,y_0(s)+h,y_0'(s))ds\Big)+A,$$ Lemma \ref{lem28} implies that $A_{y_0}=C_1$. Similarly, we can prove that $A_{y_0}=C_2$. This contradicts to $C_1\neq C_2$. Therefore for each $y_n\to y_0$, we have $A_{y_n}\to A_{y_0}$. It follows that $A_y$ is continuous about $y$. So the continuity of $T_1$ is obvious. \noindent\textbf{Step 2.} Prove that $T_1$ is compact. Let $\Omega\subseteq P_1$ bet a bounded set. Suppose that $\Omega\subseteq \{y\in P_1:\|y\|\le M\}$. For $y\in \Omega$, we have $$0\le \int_0^1f(s,y(s)+h,y'(s))ds\le \max_{t\in [0,1],u\in [h,M+h],v\in [-M,M]}f(t,u,v)=:D.$$ It follows from the definition of $T_1$ and Lemma \ref{lem28} that \begin{align*} &|(T_1y)(t)|\\ &= \Big|-\int_t^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big|\\ &\le \int_0^1\phi^{-1} \Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big) +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} \int_0^1f(u,y(u)+h,y'(u))du \\ &\quad +\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1} \Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big)\\ &\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} \int_0^1f(u,y(u)+h,y'(u))du+\int_0^sf(u,y(u)+h,y'(u))du\Big)ds \\ &\le \phi^{-1} \Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big)+\frac{D}{1-\phi \big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\Big) \\ &\quad +\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i} \phi^{-1} \Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big) +\frac{D}{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\Big)\\ &= \phi^{-1} \left(E\right)+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i} \phi^{-1} \left(E\right), \end{align*} \begin{align*} & |(T_1y)'(t)|\\ &= \big|\phi^{-1} \left(\phi(A_y)-\int_0^tf(u,y(u)+h,y'(u))du\right)ds\big|\\ &\le \phi^{-1} \Big(\phi\left(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\right) \\ &\quad+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)} \int_0^1f(u,y(u)+h,y'(u))du +\int_0^1f(u,y(u)+h,y'(u))du\Big)ds\\ &\le \phi^{-1} \Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big) +\frac{D}{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\Big)\\ &= \phi^{-1}(E), \end{align*} where $$E=\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big) +\frac{D}{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}.$$ For the uniform continuity of $\phi$ on the interval $[-E,E]$, for each $\epsilon>0$, there exists a $\rho>0$ such that $$|\phi^{-1}(Y_1)-\phi^{-1}(Y_2)|<\epsilon,\quad Y_1,Y_2\in [-E,E],\quad |Y_1-Y_2|<\rho.$$ Put $$Y_1=\phi(A_y)-\int_0^{t_1}f(u,y(u)+h,y'(u))du,\quad Y_2=\phi(A_y)-\int_0^{t_2}f(u,y(u)+h,y'(u))du.$$ Since $|Y_1-Y_2|=|\int_{t_1}^{t_2}f(u,y(u)+h,y'(u))du|\le D|t_1-t_2|$, it is easy to see that there exists $\delta>0$ (independent of $\epsilon$ ) such that $|Y_1-Y_2|<\rho$ for all $t_1,t_2$ with $|t_1-t_2|<\delta$. Hence there is $\delta>0$ (independent of $\epsilon$) such that $$|(Ty)'(t_1)-(Ty)'(t_2)|=\phi^{-1}(Y_1)-\phi^{-1}(Y_2)|<\epsilon,$$ whenever $t_1,t_2\in [0,1]$ and $|t_1-t_2|<\delta$. This shows that $(Ty)(t)$ is equi-continuous on [0,1]. The Arzela-Askoli theorem guarantees that $T(\Omega)$ is relative compact, which means that $T$ is compact. Hence the continuity and the compactness of $T$ imply that $T$ is completely continuous. \end{proof} \begin{theorem} \label{thm210} Suppose that {\rm (H1)--(H3)} hold and there exist positive constants $e_1,e_2,c$, \begin{gather*} \begin{aligned} L&= \int_0^1\phi^{-1} \left(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left( 1+\sum_{i=1}^m\alpha_i\right)}+s\right)ds\\ &\quad+\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1} \left(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left( 1+\sum_{i=1}^m\alpha_i\right)}+s\right)ds, \end{aligned}\\ Q= \min\big\{\phi\big(\frac{c}{L}\big),\;\frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2)-\phi\left( 1+\sum_{i=1}^m\alpha_i\right)}} \big\};\\ W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} \left(s-k\right)ds}\Big);\quad E= \phi\big(\frac{e_1}{L}\big). \end{gather*} such that $$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0,\quad Q\ge \phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big),\quad Q>W.$$ If \begin{itemize} \item[(A1)] $f(t,u,v)W$ for all $t\in [k,1-k],u\in [e_2+h,e_2/\sigma_0^2+h],v\in [-c,c]$; \item[(A3)] $f(t,u,v)< E$ for all $t\in [0,1],u\in [h,e_1/\sigma_0+h],v\in [-c,c]$; \end{itemize} then \eqref{e14} has at least three decreasing positive solutions $x_1,x_2,x_3$ such that $$x_1(0)e_2+h,\quad x_3(0)>e_1+h,\quad x_3(1-k)0,B\le 0 and A>0,B<0, the author could not get the sufficient conditions guaranteeing the existence of multiple positive solutions of \eqref{e14}. \end{remark} \begin{proof}[Proof of Theorem \ref{thm210}] To apply Lemma \ref{lem25}, we prove that its hypotheses are satisfied. By the definitions, it is easy to see that \alpha,\psi  are two nonnegative continuous concave functionals on the cone P_1, \gamma ,\beta,\theta are three nonnegative continuous convex functionals on the cone P_1 and \alpha(y)\le \beta(y) for all y\in P_1, there exist constants M>0 such that \|y\|\le M\gamma (y)\quad\text{for all }y\in P_1. Lemma \ref{lem28} implies that x=x(t) is a positive solution of \eqref{e14} if and only if x(t)=y(t)+h and y(t) is a solution of the operator equation y=T_1y and T_1:P_1\to P_1 is completely continuous. Corresponding to Lemma \ref{lem25},$$ c=c,\quad h=\sigma_0e_1,\quad d=e_1,\quad a=e_2,\quad b=\frac{e_2}{\sigma_0}. $$Now, we prove that all conditions of Lemma \ref{lem25} hold. One sees that 0a\}=\{y\in P_1\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big) |\alpha(y)>e_2\}\neq \emptyset$$ and $\alpha(T_1y)>e_2\quad\text{for every }y\in P_1\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big)$. Choose $y(t)=\frac{e_2}{2\sigma_0}$ for all $t\in [0,1]$. Then $y\in P_1$ and $$\alpha(y)=\frac{e_2}{2\sigma_0}>e_2,\;\theta(y)=\frac{e_2}{2\sigma_0}\le \frac{e_2}{\sigma_0},\;\gamma(y)=0< c.$$ It follows that $\{y\in P_1(\gamma,\theta,\alpha;a,b,c):\alpha(y)>a\}\neq \emptyset$. For $y\in P_1(\gamma,\theta,\alpha;a,b,c)$, one has $$\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2,\quad \theta(y)=\max_{t\in [k,1-k]}y(t)\le \frac{e_2}{\sigma_0},\quad \gamma(y)=\max_{t\in [0,1]}| y'(t)|\le c.$$ Then $$e_2\le y(t)\le \frac{e_2}{\sigma_0^2},\quad t\in [k,1-k],\; |y'(t)|\le c.$$ Thus (A2) implies $$f(t,y(t)+h,| y'(t)|)\ge W,\quad n\in [k,1-k].$$ Since $$\alpha(T_1y)=\min_{t\in [k,1-k]}(T_1y)(t)\ge \sigma_0\max_{t\in [0,1]}(T_1y)(t),$$ we get \begin{align*} \alpha(T_1y) &\geq \sigma_0\Big[-\int_0^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big]\\ &\ge \sigma_0\Big[-\int_0^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big]\\ &\ge \sigma_0\Big[\int_0^1\phi^{-1} \left(\phi\left(\frac{-A}{1-\sum_{i=1}^m\alpha_i}\right) +\int_0^sf(u,y(u)+h,y'(u))du\right)ds\Big]\\ &\ge \sigma_0\Big[\int_k^{1-k}\phi^{-1} \left(\phi\left(\frac{-A}{1-\sum_{i=1}^m\alpha_i}\right) +\int_k^sf(u,y(u)+h,y'(u))du\right)ds\Big]\\ &\ge \sigma_0\int_k^{1-k}\phi^{-1} \big(W(s-k)\big)ds\\ &= e_2. \end{align*} This completes Step 2. \noindent\textbf{Step 3.} Prove that $\{y\in Q(\gamma,\theta,\psi;h,d,c):\beta(y)a$ for $y\in P_1(\gamma,\alpha;a,c)$ with $\theta(T_1y)>b$; For $y\in P_1(\gamma,\alpha;a,c)=P_1(\gamma,\alpha;e_2,c)$ with $\theta(T_1y)>b=\frac{e_2}{\sigma_0}$, we have that $\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2$ and $\gamma (y)=\max_{t\in [0,1]}|y(t)|\le c$ and $\max_{t\in [k,1-k]}(T_1y)(t)>\frac{e_2}{\sigma_0}$. Then $$\alpha(T_1y)=\min_{t\in [k,1-k]}(T_1y)(t)\ge \sigma_0\beta(T_1y)>\sigma_0\frac{e_2}{\sigma_0}=e_2=a.$$ This completes Step 4. \noindent\textbf{Step 5.} Prove that $\beta(T_1y)e_2,\quad \beta(y_3)>e_1,\quad \alpha (y_3)e_2+h,\\ \max_{t\in [0,1]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]}x_3(t)e_2+h$, $x_3(0)>e_1+h$, $x_3(1-k)y(0)\ge 0$ for $t\in (0,1)$ since $y'(t)\ge 0$ for all $t\in [0,1]$. The proof is complete. \end{proof} \begin{lemma} \label{lem212} Suppose that {\rm (H4), (H6), (H7)} hold. If $y$ is a solution of \eqref{e25}, then $$y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds,$$ with $$\label{e26} \phi^{-1}(A_h)=\sum_{i=1}^m\beta_i\phi^{-1} \Big(A_h+\int_{\xi_i}^1h(s)ds\Big)+B,$$ and $$B_h=\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1} \Big(A_h+\int_s^1h(u)du\Big)ds,$$ where \begin{gather*} a= \phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big),\\ b=\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)} {1-\sum_{i=1}^m\beta_i}\Big) +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}\int_0^1h(s)ds. \end{gather*} \end{lemma} \begin{proof} It follows from \eqref{e25} that $$y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds,$$ and the BCs in \eqref{e25} imply \begin{gather*} y'(1)=\sum_{i=1}^m\beta_i\phi^{-1}\left(\phi(y'(1)) +\int_{\xi_i}^1h(s)ds\right)+B, \\ y(0)= \frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1} \Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds. \end{gather*} Let $$G(c)=\phi^{-1}(c)-\sum_{i=1}^m\beta_i\phi^{-1} \Big(c+\int_{\xi_i}^1h(s)ds\Big)-B.$$ It is easy to see that \begin{align*} G(a) &= G\Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)\Big)\\ &\le \frac{B}{1-\sum_{i=1}^m\beta_i} -\sum_{i=1}^m\beta_i\phi^{-1} \Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)\Big)-B\\ &= 0. \end{align*} On the other hand, one sees that \begin{align*} \frac{G(b)}{\phi^{-1}(b)} &= 1-\sum_{i=1}^m\beta_i\phi^{-1} \Big(1+\frac{\int_{\xi_i}^1h(s)ds}{b}\Big) -\frac{B}{\phi^{-1}(b)} \\ &= 1-\sum_{i=1}^m\beta_i\phi^{-1} \Big(1+\frac{\int_{\xi_i}^1h(s)ds}{ \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)} {1-\sum_{i=1}^m\beta_i}\Big) +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} \int_0^1h(s)ds}\Big) \\ &\quad -\frac{B}{\phi^{-1} \Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)} {1-\sum_{i=1}^m\beta_i}\Big) +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} \int_0^1h(s)ds\Big)} \\ &\ge 1-\sum_{i=1}^m\beta_i\phi^{-1} \Big( 1+\frac{1-\phi \Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)} {\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} \Big)-\frac{1-\sum_{i=1}^m\beta_i}{1+\sum_{i=1}^m\beta_i} = 0. \end{align*} Hence $G(b)\ge 0$. It is easy to know that $\frac{G(c)}{\phi^{-1}(c)}$ is continuous and decreasing on $(-\infty,0)$ and continuous and increasing on $(0,+\infty)$, Hence $G(a)\le 0$ and $G(b)\ge 0$ and $\lim_{c\to 0^-}\frac{G(c)}{\phi^{-1}(c)}= +\infty,\quad \lim_{c\to -\infty}\frac{G(c)}{\phi^{-1}(c)}= 1-\sum_{i=1}^m\beta_i>0.$ Then there exists unique constant $A_h=\phi(y'(1))\in [a,b]$ such that \eqref{e26} holds. The proof is complete. \end{proof} Note $h=\frac{A}{1-\sum_{i=1}^m\alpha_i}$, and $x(t)-h=y(t)$. Then \eqref{e15} is transformed into the boundary-value problem $$\label{e27} \begin{gathered}{} [\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ y(0)-\sum_{i=1}^m\alpha_i y(\xi_i)=0,\quad y'(1)-\sum_{i=1}^m\beta_i y'(\xi_i)=B, \end{gathered}$$ Let \begin{align*} P_2=\big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1], y'(t)\text{ is decreasing on }[0,1],\\ &\ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in [0,1]}y(t)\text{ for all }t\in [0,1]\big\}. \end{align*} Then $P_2$ is a cone in $X$. Since \begin{align*} |y(t)|&=\big|\frac{\sum_{i=1}^m\alpha_i y(\xi_i)-\sum_{i=1}^m\alpha_i y(0)}{1-\sum_{i=1}^m\alpha_i}\big|\\ &\le\frac{\sum_{i=1}^m\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i }\max_{t\in [0,1]}|y'(t)| =\frac{\sum_{i=1}^m\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i}\gamma(y), \end{align*} we have $$\max_{t\in [0,1]}|y(t)|\le\frac{\sum_{i=1}^m \alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i }\gamma(y).$$ It is easy to see that there exists a constant $M>0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_1$. Define the nonlinear operator $T_2:P_2\to X$ by $(T_2y)(t)= B_y+\int_0^t\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_2,$ where $A_y$ satisfies $$\label{e28} \phi^{-1}(A_y)=\sum_{i=1}^m\beta_i\phi^{-1}\left(A_y +\int_{\xi_i}^1f(s,y(s)+h,y'(s))ds\right)+B,$$ and $B_y$ satisfies \begin{align*} B_y&= \frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1} \left(A_y+\int_s^1f(u,y(u)+h,y'(u))ds\right)ds. \end{align*} Then for $y\in P_2$, \begin{align*} (T_2y)(t)&= \int_0^t\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1} \left(A_y+\int_s^1f(u,y(u)+h,y'(u))ds\right)ds\,. \end{align*} \begin{lemma} \label{lem213} Suppose that {\rm (H5), (H6), (H7)} hold. Then \begin{itemize} \item[(i)] the following equalities hold: \begin{gather*}{} [\phi((T_2y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ (T_2y)(0)-\sum_{i=1}^m\alpha_i (T_2y)(\xi_i)=0,\quad (T_2y)'(1)-\sum_{i=1}^m\beta_i (T_2y)'(\xi_i)=B; \end{gather*} \item[(ii)] $T_2y\in P_2$ for each $y\in P_2$; \item[(iii)] $x$ is a solution of \eqref{e15} if and only if $x=y+h$ and $y$ is a solution of the operator equation $y=T_2y$ in cone $P_2$; \item[(iv)] $T_2:P_2\to P_2$ is completely continuous. \end{itemize} \end{lemma} The proofs of the above lemma is similar to that of Lemma \ref{lem29} and it is omitted. \begin{theorem} \label{thm214} Suppose that {\rm (H5)--(H7)} hold and there exist positive constants $e_1,e_2,c$, \begin{gather*} \begin{aligned} L&= \int_0^1\phi^{-1} \Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1} \Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds, \end{aligned}\\ Q= \min\big\{\phi\left(\frac {c}{L}\right),\;\frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2)-\phi\left( 1+\sum_{i=1}^m\beta_i\right)}} \big\};\quad W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} (1-k-s)ds}\Big);\\ E= \phi(\frac{e_1}{L}). \end{gather*} such that $$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0,\quad Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.$$ If (A1), (A2) and (A3) from in Theorem \ref{thm210} hold, then \eqref{e15} has at least three increasing positive solutions $x_1,x_2,x_3$ such that $$x_1(1)e_2+h,\quad x_3(1)>e_1+h,\quad x_3(k)0 such that \|y\|\le M\gamma (y)\quad\text{for all }y\in P_2. Lemma \ref{lem213} implies that x=x(t) is a positive solution of \eqref{e15} if and only if x(t)=y(t)+h and y(t) is a solution of the operator equation y=T_2y and T_2:P_2\to P_2 is completely continuous. Corresponding to Lemma \ref{lem25},$$ c=c,\quad h=\sigma_0e_1,\quad d=e_1,\quad a=e_2,\quad b=\frac{e_2}{\sigma_0}. $$Now, we prove that all conditions of Lemma \ref{lem25} hold. One sees that 0a\} =\{y\in P_2\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big): \alpha(y)>e_2\}\neq \emptyset$$ and $\alpha(T_2y)>e_2\quad\text{for every }y\in P_2\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big)$; Choose $y(t)=\frac{e_2}{2\sigma_0}$ for all $t\in [0,1]$. Then $y\in P_2$ and $$\alpha(y)=\frac{e_2}{2\sigma_0}>e_2,\quad \theta(y)=\frac{e_2}{2\sigma_0}\le \frac{e_2}{\sigma_0},\quad \gamma(y)=0< c.$$ It follows that $\{y\in P_2(\gamma,\theta,\alpha;a,b,c)|\alpha(y)>a\}\neq \emptyset$. For $y\in P_2(\gamma,\theta,\alpha;a,b,c)$, one has $$\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2,\quad \theta(y)=\max_{t\in [k,1-k]}y(t)\le \frac{e_2}{\sigma_0},\quad \gamma(y)=\max_{t\in [0,1]}|y'(t)|\le c.$$ Then $$e_2\le y(t)\le \frac{e_2}{\sigma_0^2},\quad t\in [k,1-k],\quad |y'(t)|\le c.$$ Thus (A2) implies $f(t,y(t)+h,| y'(t)|)\ge W,\;\;t\in [k,1-k]$. Since $$\alpha(T_2y)=\min_{t\in [k,1-k]}(T_2y)(t)\ge \sigma_0\max_{t\in [0,1]}(T_2y)(t),$$ from Lemma \ref{lem212}, we have \begin{align*} \alpha(T_2y) &\geq \sigma_0\max_{t\in [0,1]}(T_2y)(t)\\ &= \sigma_0\Big[\int_0^1\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds \\ &\quad+\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1} \Big(A_h+\int_s^1f(u,y(u)+h,y'(u))ds\Big)ds\Big]\\ &\ge \sigma_0\Big[\int_0^1\phi^{-1} \Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big) +\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1} \Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)\\ &\quad +\int_s^1f(u,y(u)+h,y'(u))ds\Big)ds\Big] \\ &\ge \sigma_0\int_0^1\phi^{-1} \left(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)+\int_s^1f(u,y(u)+h,y'(u))du\right)ds\\ &\ge \sigma_0\int_0^1\phi^{-1} \Big(\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\ &\ge \sigma_0\int_k^{1-k}\phi^{-1} \Big(\int_s^{1-k}f(u,y(u)+h,y'(u))du\Big)ds\\ &\ge \sigma_0\int_k^{1-k}\phi^{-1} \left(W(1-k-s)\right)ds = e_2. \end{align*} This completes Step 2. \noindent\textbf{Step 3.} Prove that the set $\{y\in Q(\gamma,\theta,\psi;h,d,c)|\beta(y)a$ for $y\in P_2(\gamma,\alpha;a,c)$ with $\theta(T_2y)>b$; For $y\in P_2(\gamma,\alpha;a,c)=P_2(\gamma,\alpha;e_2,c)$ with $\theta(T_2y)>b=\frac{e_2}{\sigma_0}$, we have that $\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2$ and $\gamma (y)=\max_{t\in [0,1]}|y(t)|\le c$ and $\max_{t\in [k,1-k]}(T_2y)(t)>\frac{e_2}{\sigma_0}$. Then $$\alpha(T_2y)=\min_{t\in [k,1-k]}(T_2y)(t)\ge \sigma_0\beta(T_2y)>\sigma_0\frac{e_2}{\sigma_0}=e_2=a.$$ This completes Step 4. \noindent\textbf{Step 5.} Prove that $\beta(T_2y)e_2,\quad \beta(y_3)>e_1,\quad \alpha(y_3)e_2+h,\\ \max_{t\in [0,1]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]}x_3(t)e_2+h,\quad x_3(1)>e_1+h,\quad x_3(k)0$, the BCs in \eqref{e29} and (H4) imply $y'(t)>0$ for all $t\in [0,1]$. Then $$y(1)=\sum_{i=1}^m\beta_iy(\xi_i)\le \sum_{i=1}^m\beta_iy(1).$$ Then $y(1)\le 0$. On the other hand, $y(0)=\alpha y'(0)+D\ge 0$. This is a contradiction since $y'(t)>0$. If $y'(1)\le 0$ and $y'(0)\le 0$, then we get that $y'(t)\le 0$ for all $t\in [0,1]$. The BCs in \eqref{e29} and (H4) imply $$y(1)=\sum_{i=1}^m\beta_iy(\xi_i)\ge \sum_{i=1}^m\beta_iy(1).$$ It follows that $y(1)\ge 0$. Then $y(t)\ge y(1)\ge 0$ for all $t\in [0,1]$. (H4) implies $y(t)>0$ for all $t\in (0,1)$. If $y'(1)\le 0$ and $y'(0)>0$, then $y(0)=\alpha y'(0)+D\ge 0$. It follows from \eqref{e29} and (H4) that $$y(1)=\sum_{i=1}^m\beta_iy(\xi_i)\ge \sum_{i=1}^m\beta_i\min\{y(0),y(1)\}.$$ If $y(1)\le y(0)$, then $y(1)\ge \sum_{i=1}^m\beta_i y(1)$ implies that $y(1)\ge 0$. If $y(1)>y(0)$, then $y(1)>0$ since $y(0)\ge 0$. It follows from (H4) that $y(t)\ge \min\{y(0),y(1)\}\ge 0$. Then $y$ is positive on $(0,1)$. The proof is complete. \end{proof} \begin{lemma} \label{lem216} Assume {\rm (H4), (H10)} hold and $D\ge 0$. If $y$ is a solution of \eqref{e29}, then $$y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big),\quad t\in [0,1],$$ where $A_h\in [b,0]$, \begin{align*} &\Big[\alpha\phi^{-1}\Big(A_h+\int_{\xi}^1h(u)du\Big)+D\Big] \Big(1-\sum_{i=1}^m\beta_i\Big)\\ &= -\Big(1-\sum_{i=1}^m\beta_i\Big) \int_0^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\ &\quad -\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds, \end{align*} \begin{gather*} B_h=\alpha\phi^{-1}\Big(A_h+\int_{\xi}^1h(u)du\Big)+D,\\ b=-\int_0^1h(u)du-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big),\\ a= 1+\alpha\Big(1-\sum_{i=1}^m\beta_i\Big)-\sum_{i=1}^m\beta_i\xi_i. \end{gather*} \end{lemma} \begin{proof} From \eqref{e29} it follows that \begin{gather*} y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(1))+\int_s^1(u)du\Big)ds,\\ \begin{aligned} y(0)\Big(1-\sum_{i=1}^m\beta_i\Big) &= -\int_0^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds\\ &\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i} \phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds, \end{aligned}\\ y(0)=\alpha\phi^{-1}\Big(\phi(y'(1))+\int_{0}^1h(u)du\Big)+D. \end{gather*} Thus \begin{align*} &\Big[\alpha\phi^{-1}\Big(\phi(y'(1))+\int_{0}^1h(u)du\Big)+D\Big] \Big(1-\sum_{i=1}^m\beta_i\Big)\\ &= -\int_0^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds\\ &\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i} \phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds. \end{align*} Let \begin{align*} G(c) &= \Big[\alpha\phi^{-1}\Big(c+\int_{0}^1h(u)du\Big)+D\Big] \Big(1-\sum_{i=1}^m\beta_i\Big)\\ &\quad +\int_0^1\phi^{-1}\Big(c+\int_s^1h(u)du\Big)ds\\ &\quad -\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1} \Big(c+\int_s^1h(u)du\Big)ds\\ &= \Big[\alpha\phi^{-1}\Big(c+\int_{0}^1h(u)du\Big)+D\Big] \Big(1-\sum_{i=1}^m\beta_i\Big)\\ &\quad +\Big(1-\sum_{i=1}^m\beta_i\Big) \int_0^1\phi^{-1}\Big(c+\int_s^1h(u)du\Big)ds\\ &\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1} \Big(c+\int_s^1h(u)du\Big)ds\\. \end{align*} It is easy to see that $G(c)$ is increasing on $(-\infty,+\infty)$, $G(0)\ge 0$. Since $$a=1+\alpha\Big(1-\sum_{i=1}^m\beta_i\Big)-\sum_{i=1}^m\beta_i\xi_i,$$ we get \begin{align*} G(b)&=G\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big) -\int_0^1h(u)du\Big) \\ &= \Big[\alpha\phi^{-1} \Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big)+D\Big] \Big(1-\sum_{i=1}^m\beta_i\Big) \\ &\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1} \Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big) -\int_0^sh(u)du\Big)ds \\ &\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big) -\int_0^sh(u)du\Big)ds \\ &\le \Big[\alpha\phi^{-1} \Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big) +D\Big] \Big(1-\sum_{i=1}^m\beta_i\Big)\\ &\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1} \Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big)ds\\ &\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big)ds = 0, \end{align*} we get $\phi(y'(1))\ge -\int_0^1h(u)du-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)=b$. The proof is complete. \end{proof} Let \begin{align*} P_3=\big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1], y'(t)\text{ is decreasing on }[0,1],\\ &\ \ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in [0,1]}y(t)\text{ for all }t\in [0,1] \big\}. \end{align*} Then $P_3$ is a cone in $X$. Since, for $y\in P_3$, we have \begin{align*} |y(t)| &= |y(t)-y(1)+y(1)|\\ &\le |y'(\theta)|(1-t)+|y(1)|\\ &\le \max_{t\in [0,1]}|y'(t)|+\Big|\frac{\sum_{i=1}^m\beta_i y(\xi_i)-\sum_{i=1}^m\beta_i y(1)}{1-\sum_{i=1}^m\beta_i }\Big|\\ &\le \Big(1+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i }\Big)\max_{t\in [0,1]}|y'(t)|\\ &= \Big(1+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i }\Big)\gamma(y), \end{align*} we get $$\max_{t\in [0,1]}|y(t)| \le\Big(1+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i }\Big)\gamma(y).$$ It is easy to see that there exists a constant $M>0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_3$. Suppose that (H10) holds. Let $x(t)-h=y(t)$. Then \eqref{e16} is transformed into \begin{gather*} [\phi(y'(t))]'+f(t,y(t)+h,y'(t))=0,\quad t\in (0,1),\\ y(0)-\alpha y'(0)=A-\frac{B}{1-\sum_{i=1}^m\beta_i}\ge 0,\\ y(1)- \sum_{i=1}^m\beta_iy(\xi_i)= 0, \end{gather*} Define the nonlinear operator $T_3:P_3\to X$ by \begin{align*} (T_3y)(t)&= B_y+\int_0^t\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_3, \end{align*} where \begin{align*} &\Big[\alpha\phi^{-1}\Big(A_y+\int_{0}^1f(u,y(u)+h,y'(u))du\Big) +A-\frac{B}{1-\sum_{i=1}^m\beta_i}\Big]\Big(1-\sum_{i=1}^m\beta_i\Big)\\ &= -\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1 \phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\ &\quad -\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds, \end{align*} and $$B_h=\alpha\phi^{-1}\Big(A_y+\int_{0}^1f(u,y(u)+h,y'(u))duu\Big) +A-\frac{B}{1-\sum_{i=1}^m\beta_i}.$$ For $y\in P_3$, the definition of $A_y$ implies \begin{align*} (T_3y)(t) &= \alpha\phi^{-1}\Big(A_y+\int_{0}^1f(u,y(u)+h,y'(u))du\Big) +A-\frac{B}{1-\sum_{i=1}^m\beta_i}\\ &\quad +\int_0^t\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\ &= -\int_t^1\phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\ &\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds. \end{align*} \begin{lemma} \label{lem217} Suppose that {\rm (H8)--(H10)} hold. Then \begin{itemize} \item[(i)] $T_3y\in P_3$ for each $y\in P_3$; \item[(ii)] $x$ is a solution of \eqref{e17} if and only if $x=y+h$ and $y$ is a solution of the operator equation $T_3y=y$ in $P_3$; \item[(iii)] $T_3:P_3\to P_3$ is completely continuous; \item[(iv)] the following equalities hold: \begin{gather*}{} [\phi((T_3y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ (T_3y)(0)-\alpha (T_3y)'(\xi)=A-\frac{B}{1-\sum_{i=1}^m\beta_i},\\ (T_3y)(1)-\sum_{i=1}^m\beta_i (T_3y)(\xi_i)=0; \end{gather*} \end{itemize} \end{lemma} The proof of the above lemma is similar to that of Lemma \ref{lem29}, so it is omitted. \begin{theorem} \label{thm218} Suppose that {\rm (H8)--(H10)} hold and that there exist positive constants $e_1,e_2,c$, \begin{gather*} L= \phi^{-1}(2)+\frac{\phi^{-1}(2)}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i(1-\xi_i),\\ Q= \min\big\{\phi\big(\frac {c}{L}\big), \frac{\phi(c)}{2} \big\};\\ W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} (s-k)ds}\Big);\quad E= \phi\big(\frac{e_1}{L}\big). \end{gather*} such that $c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{\big(A-\frac{B}{1-\sum_{i=1}^m\beta_i}\big) \left(1-\sum_{i=1}^m\beta_i \right)} {1+\alpha-\sum_{i=1}^m\beta_i\xi_i}\Big),\quad Q>W.$$ If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold, then \eqref{e16} has at least three positive solutions $x_1,x_2,x_3$ such that \begin{gather*} \max_{t\in [0,1]}x_1(t)e_2+h,\\\ \min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]}x_3(t)0$, we get that$y'(t)>0$for all$t\in [0,1]$. The BCs in \eqref{e30} and (H4) imply that$ y(0)=\alpha y'(0)\ge 0$. Then$y(t)>y(0)\ge 0$for all$t\in [0,1]$. If$y'(0)\ge 0$and$y'(1)\le0$, then$y(0)=\alpha y'(0)\ge 0$. It follows from \eqref{e30} that $$y(1)=\sum_{i=1}^m\beta_iy(\xi_i)+D\ge \sum_{i=1}^m\beta_i\min\{y(0),y(1)\}.$$ If$y(1)\le y(0)$, then$y(1)\ge \sum_{i=1}^m\beta_i y(1)$implies that$y(1)\ge 0$. If$y(1)>y(0)$, then$y(1)>0$since$y(0)\ge 0$. It follows that$y(t)\ge \min\{y(0),y(1)\}\ge 0$. Then$y$is positive on$(0,1)$. The proof is complete. \end{proof} \begin{lemma} \label{lem220} Assume {\rm (H4), (H13)},$D\ge 0$. If$y$is a solution of \eqref{e30}, then $$y(t)=B_h-\int_t^1\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big),\quad t\in [0,1],$$ where$A_h\in [a,b], \begin{align*} &\alpha\phi^{-1}\Big(A_h-\int_0^{\xi}h(u)du\Big) \Big(1-\sum_{i=1}^m\beta_i\Big)\\ &\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1} \Big(A_h-\int_0^sh(u)du\Big)ds\\ &\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds-D=0, \end{align*} and \begin{gather*} B_h=\alpha\phi^{-1}\Big(\phi(y'(0))-\int_0^{\xi}h(u)du\Big),\\ a=\phi\big(\frac{D}{c}\big),\quad b=\phi\big(\frac{D}{c}\big)+\int_0^1h(u)du,\quad c=\alpha\big(1-\sum_{i=1}^m\beta_i\big) +1-\sum_{i=1}^m\beta_i\xi_i. \end{gather*} \end{lemma} \begin{proof} It follows from \eqref{e30} that \begin{gather*} y(t)=y(1)-\int_t^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds,\\ y(1)\Big(1-\sum_{i=1}^m\beta_i\Big) = -\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds+D,\\ y(1)-\int_0^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds=\alpha y'(0). \end{gather*} Thus \begin{align*} &\Big(1-\sum_{i=1}^m\beta_i\Big) \Big(\int_0^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds+\alpha y'(0)\Big)\\ &= -\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds+D. \end{align*} It follows from the proof of Lemma \ref{lem219} thaty'(0)\ge 0. Let \begin{align*} G(c)&= \Big(1-\sum_{i=1}^m\beta_i\Big) \int_0^1\phi^{-1}\Big(c-\int_0^sh(u)du\Big)ds +\alpha\Big(1-\sum_{i=1}^m\beta_i\Big) \phi^{-1}(c)\\ &\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(c-\int_0^sh(u)du\Big)ds-D. \end{align*} It is easy to see thatG(c)$is increasing on$(-\infty,+\infty)$and$G(a)=G\left(\phi(\frac{D}{a})\right)\le0. Since \begin{align*} G(b)&=G\Big(\phi\big(\frac{D}{c}\big)+\int_0^1h(u)du\Big)\\ &\geq \Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1} \Big(\phi\big(\frac{D}{c}\big)+\int_s^1h(u)du\Big)ds\\ &\quad +\alpha\Big(1-\sum_{i=1}^m\beta_i\Big) \phi^{-1}\Big(\phi\big(\frac{D}{c}\big)+\int_0^1h(u)du\Big)\\ &\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(\phi\big(\frac{D}{c}\big)+\int_s^1h(u)du\Big)ds-D\\ &\ge \Big(1-\sum_{i=1}^m\beta_i\Big)\frac{D}{c} +\alpha\Big(1-\sum_{i=1}^m\beta_i\Big) \frac{D}{c}+\sum_{i=1}^m\beta_i(1-\xi_i)\frac{D}{c}-D\\ &= 0, \end{align*} we geta\le \phi(y'(0))\le \phi\big(\frac{D}{c}\big)+\int_0^1h(u)du$. The proof is complete. \end{proof} Suppose that (H13) holds. Let$x(t)-h=y(t). Then \eqref{e16} is transformed into \begin{gather*}{} [\phi(y'(t))]'+f(t,y(t)+h,y'(t))=0,\quad t\in (0,1),\\ y(0)-\alpha y'(0)=0,\\ y(1)-\sum_{i=1}^m\beta_iy(\xi_i)=B-\frac{A}{1-\sum_{i=1}^m\beta_i}\ge 0.\end{gather*} Let \begin{align*} P_3'=\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1], y'(t)\text{ is decreasing on }[0,1],\\ &\ \ \ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in [0,1]}y(t)\text{ for all }t\in [0,1] \}. \end{align*} ThenP_3'$is a cone in$X$. Since, for$y\in P_3'$, we have $$|y(t)|\le|y(t)-y(0)|+|y(0)|\le \max_{t\in [0,1]}|y'(t)|+\alpha\max_{t\in [0,1]}|y'(t)| =(1+\alpha)\gamma(y),$$ we get$\max_{t\in [0,1]}|y(t)|\le(1+\alpha)\gamma(y)$. It is easy to see that there exists a constant$M>0$such that$\|y\|\le M\gamma (y)$for all$y\in P_3'$. Define the nonlinear operator$T_3':P_3'\to Xby $(T_3'y)(t)= B_y-\int_t^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_3',$ where \begin{align*} &\Big(1-\sum_{i=1}^m\beta_i\Big)\Big(\int_0^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds+\alpha \phi^{-1}(A_y)\Big)\\ &= -\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds +B-\frac{A}{1-\sum_{i=1}^m\beta_i}. \end{align*} and $B_y= \int_0^1\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds +\alpha\phi^{-1}(A_y).$ Then \begin{align*} (T_3'y)(t) &= \int_0^1\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds +\alpha\phi^{-1}(A_y)\\ &\quad -\int_t^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &= \int_0^t\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds +\alpha\phi^{-1}(A_y),\quad y\in P_3'. \end{align*} \begin{lemma} \label{lem221} Assume {\rm (H11)--(H13)}. Then the following holds: \begin{itemize} \item[(i)]T_3'y\in P_3'$for each$y\in P_3'$; \item[(ii)]$x$is a solution of \eqref{e16}' if and only if$x=y+h$and$y$is a solution of the operator equation$T_3'y=y$in$P_3'$; \item[(iii)]$T_3':P_3'\to P_3'$is completely continuous; \item[(iv)] the following equalities hold: \begin{gather*}{} [\phi((T_3'y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ (T_3'y)(0)-\alpha (T_3'y)(\xi)=0,\\ (T_3'y)(1)-\sum_{i=1}^m\beta_i (T_3'y)(\xi_i)=B-\Big(1-\sum_{i=1}^m\beta_i\Big)A; \end{gather*} \end{itemize} \end{lemma} The proof of the above lemma is similar to that of Lemma \ref{lem29}, so it is omitted. \begin{theorem} \label{thm222} Suppose that {\rm H11)--(H13)} hold and there exist positive constants$e_1,e_2,c$, \begin{gather*} Q= \min\big\{\phi\Big(\frac{c}{\int_0^1\phi^{-1}(2-s)ds+\alpha \phi^{-1}(2)}\Big),\;\frac{\phi(c)}{2} \big\};\\ W= \phi\Big(\frac{e_2}{\sigma_0\min\big\{\int_k^{1-k}\phi^{-1} \left(1-k-s\right)ds,\;\int_k^{1-k}\phi^{-1}(s-k)ds\big\}}\Big);\\ E= \phi\Big(\frac{e_1}{\int_0^1\phi^{-1}(2-s)ds+\alpha\phi^{-1}(2)}\Big). \end{gather*} such that $$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>\frac{\alpha}{ \alpha\Big(1-\sum_{i=1}^m\beta_i\Big)+1-\sum_{i=1}^m\beta_i\xi_i} \Big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\Big),$$ and $$Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i \big)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.$$ If {\rm (A1)--(A3)} defined in Theorem \ref{thm210} hold, then \eqref{e16} has at least three positive solutions$x_1,x_2,x_3$such that \begin{gather*} \max_{t\in [0,1]}x_1(t)e_2+h,\\ \min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]}x_3(t)y(0)\ge 0$ for $t\in (0,1)$ since $y'(t)\ge 0$ for all $t\in [0,1]$. The proof is complete. \end{proof} \begin{lemma} \label{lem224} Assume {\rm (H4)--(H16)}. If $y$ is a solution of \eqref{e31}, then $$y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds$$ and there exists unique $A_h\in [a,b]$ such that $$\label{e32} \phi^{-1}(A_h)=\sum_{i=1}^m\beta_i\phi^{-1}\left(A_h+\int_{\xi_i}^1h(s)ds\right)+D,$$ and \begin{gather*} B_h=\sum_{i=1}^m\alpha_i\phi^{-1}\Big(A_h+\int_{\xi_i}^1h(u)du\Big),\\ a=\phi\Big(\frac{D}{1-\sum_{i=1}^m\beta_i}\Big), \\ b=\phi\Big(\frac{D\big(1+\sum_{i=1}^m\beta_i\big)}{1-\sum_{i=1}^m\beta_i}\Big)+ \frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi\big( \frac{1+\sum_{i=1}^m\beta_i}{2}\big)}\int_0^1h(s)ds. \end{gather*} \end{lemma} \begin{proof} It follows from \eqref{e31} that $$y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds$$ The BCs in \eqref{e31} imply \begin{gather*} y'(1)=\sum_{i=1}^m\beta_i\phi^{-1}\Big(\phi(y'(1))+\int_{\xi_i}^1h(s)ds\Big) +D,\\ y(0)=\sum_{i=1}^m\alpha_i\phi^{-1}\Big(\phi(y'(1))+\int_{\xi_i}^1h(u)du\Big). \end{gather*} Let $G(c)= \phi^{-1}(c)-\sum_{i=1}^m\beta_i\phi^{-1} \Big(c+\int_{\xi_i}^1h(s)ds\Big)-D.$ It is easy to see that \begin{align*} G(a)&= \frac{D}{1-\sum_{i=1}^m\beta_i}-\sum_{i=1}^m\beta_i\phi^{-1} \Big(\phi\Big(\frac{D}{1-\sum_{i=1}^m\beta_i}\Big) +\int_{\xi_i}^1h(s)ds\Big)-D\\ &\le \frac{D}{1-\sum_{i=1}^m\beta_i}-\sum_{i=1}^m\beta_i \frac{D}{1-\sum_{i=1}^m\beta_i}-D=0. \end{align*} On the other hand, \begin{align*} \frac{G(b)}{\phi^{-1}(b)} &=1-\sum_{i=1}^m\beta_i\phi^{-1} \Big(1+\frac{\int_{\xi_i}^1h(s)ds}{b}\Big)-\frac{D}{\phi^{-1}(b)}\\ &= 1-\sum_{i=1}^m\beta_i\phi^{-1} \bigg(1+\frac{\int_{\xi_i}^1h(s)ds}{ \phi\Big(\frac{D\big(1+\sum_{i=1}^m\beta_i\big)} {1-\sum_{i=1}^m\beta_i}\Big) +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} \int_0^1h(s)ds}\bigg) \\ &\quad -\frac{D}{\phi^{-1} \Big(\phi\Big(\frac{D\big(1+\sum_{i=1}^m\beta_i\big)} {1-\sum_{i=1}^m\beta_i}\Big) +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi \big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}\int_0^1h(s)ds\Big)}\\ &\le 1-\sum_{i=1}^m\beta_i\phi^{-1} \Big( 1+\frac{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} \Big)-\frac{1-\sum_{i=1}^m\beta_i}{1+\sum_{i=1}^m\beta_i} = 0. \end{align*} Hence $G(b)\ge 0$. It is easy to know that $\frac{G(c)}{\phi^{-1}(c)}$ is continuous and decreasing on $(-\infty,0)$ and continuous and increasing on $(0,+\infty)$, Hence $G(a)\le 0$ and $G(b)\ge 0$ and $\lim_{c\to 0^-}\frac{G(c)}{\phi^{-1}(c)}= +\infty,\quad \lim_{c\to -\infty}\frac{G(c)}{\phi^{-1}(c)} = 1-\sum_{i=1}^m\beta_i>0.$ Then there exists unique constant $A_h=\phi(y'(1))\in [a,b]$ such that \eqref{e32} holds. The proof is complete. \end{proof} Suppose (H16) holds. Let $x(t)-A=y(t)$. Then \eqref{e17} is transformed into $$\label{e33} \begin{gathered}{} [\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ y(0)-\sum_{i=1}^m\alpha_i y'(\xi_i)=0,\\ y'(1)-\sum_{i=1}^m\beta_i y'(\xi_i)=B-\frac{A}{1-\sum_{i=1}^m\beta_i}\ge 0, \end{gathered}$$ Let \begin{align*} P_4=\{&y\in X:y(t)\ge 0\text{ for all }t\in [0,1], y'(t)\text{ is decreasing on }[0,1],\\ &\ \ \ \ \ \ \ \ y(t)\ge ty(1)\text{ for all }t\in [0,1]\}. \end{align*} Then $P_4$ is a cone in $X$. For $y\in P_4$, since $$|y(t)|=|y(t)-y(0)|+|y(0)|\le \Big(1+\sum_{i=1}^m\alpha_i \Big)\max_{t\in [0,1]}|y'(t)|=\Big(1+\sum_{i=1}^m\alpha_i \Big)\gamma(y),$$ we get $$\max_{t\in [0,1]}|y(t)|\le\Big(1+\sum_{i=1}^m\alpha_i \Big)\gamma(y).$$ It is easy to see that there exists a constant $M>0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_1$. Define the operator $T_4:P_4\to X$ by $(T_4y)(t)= B_y+\int_0^t\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_4,$ where $$\label{e34} \phi^{-1}(A_h)=\sum_{i=1}^m\beta_i\phi^{-1} \Big(A_h+\int_{\xi_i}^1f(u,y(u)+h,y'(u))du\Big) +B-\frac{A}{1-\sum_{i=1}^m\beta_i},$$ and $B_y= \sum_{i=1}^m\alpha_i\phi^{-1}\left(A_y+\int_{\xi_i}^1f(u,y(u)+h,y'(u))du\right).$ Then \begin{align*} (T_4y)(t)&= \sum_{i=1}^m\alpha_i\phi^{-1} \Big(A_y+\int_{\xi_i}^1f(u,y(u)+h,y'(u))du\Big)\\ &\quad +\int_0^t\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds. \end{align*} It follows from Lemma \ref{lem224} that \begin{align*} \phi\Big(\frac{\Big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\Big)} {1-\sum_{i=1}^m\beta_i}\Big) & \le A_y\\ &\le \phi\Big(\frac{\Big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\Big) \Big(1+\sum_{i=1}^m\beta_i\Big)} {1-\sum_{i=1}^m\beta_i}\Big)\\ &\quad + \frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} \int_0^1f(u,y(u)+h,y'(u))du. \end{align*} \begin{lemma} \label{lem225} Suppose that {\rm (H14)--(H16)}. Then \begin{itemize} \item[(i)] the following equalities hold: \begin{gather*}{} [\phi((T_4y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ (T_4y)(0)-\sum_{i=1}^m\alpha_i (T_4y)'(\xi_i)=0,\\ (T_4y)'(1)-\sum_{i=1}^m\beta_i (T_4y)'(\xi_i)=B-\frac{A}{1-\sum_{i=1}^m\beta_i}; \end{gather*} \item[(ii)] $T_4y\in P_4$ for each $y\in P_4$; \item[(iii)] $x$ is a solution of \eqref{e17} if and only if $x=y+h$ and $y$ is a solution of the operator equation $y=T_4y$ in $P_4$; \item[(iv)] $T_4:P_4\to P_4$ is completely continuous. \end{itemize} \end{lemma} The proof of the above lemma is similar to that of Lemma \ref{lem29}; so we omit it. \begin{theorem} \label{thm226} Suppose that {\rm (H14)--(H16)} hold, and there exist positive constants $e_1,e_2,c$, and \begin{gather*} \begin{aligned} L&= \int_0^1\phi^{-1} \Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+1-s\Big)ds\\ &\quad +\sum_{i=1}^m\alpha_i\phi^{-1} \Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+1-\xi_i\Big), \end{aligned}\\ Q= \min\big\{\phi(\frac{c}{L}),\; \frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2) -\phi\left(1+\sum_{i=1}^m\beta_i\right)}} \big\};\\ W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} \left(1-k-s\right)ds}\Big);\quad E= \phi\big(\frac{e_1}{L}\big). \end{gather*} such that $c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{\big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\big) \left(1+\sum_{i=1}^m\beta_i \right)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.$$ If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold, then \eqref{e17} has at least three increasing positive solutions $x_1,x_2,x_3$ such that \begin{gather*} \max_{t\in [0,1]}x_1(t)e_2+h,\\ \min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]}x_3(t)0$such that$\|y\|\le M\gamma (y)$for all$y\in P_5$. Define the operator$T_5:P_5\to Xby $(T_5y)(t)= B_y+\int_0^t\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_5,$ where \begin{align*} &\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i} \sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &+\int_0^1\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\ &-\sum_{i=1}^m\beta_i\int_0^{\xi_i}\phi^{-1} \left(A_y-\int_0^sf(u,y(u)+h,y'(u))du \right)ds-B=0. \end{align*} and $B_y= \frac{1} {1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i} \phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds.$ \begin{lemma} \label{lem229} Assume {\rm (H17), (H18), (H19)}. Then \begin{itemize} \item[(i)] the following equalities hold: \begin{gather*}{} [\phi((T_5y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\ (T_5y)(0)-\sum_{i=1}^m\alpha_i (T_5y)(\xi_i)=0,\\ (T_5y)(1)-\sum_{i=1}^m\beta_i (T_5y)(\xi_i)=B-\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}A\ge 0; \end{gather*} \item[(ii)]T_5y\in P_5$for each$y\in P_5$; \item[(iii)]$x$is a solution of \eqref{e18} if and only if$x=y+h$and$y$is a solution of the operator equation$y=T_5y$; \item[(iv)]$T_5:P_5\to P_5$is completely continuous. \end{itemize} \end{lemma} The proof is similar to that of Lemma \ref{lem29}; therefore, it is omitted. \begin{theorem} \label{thm230} Suppose that {\rm (H17), (H18), (H19)} hold, and that there exist positive constants$e_1,e_2,c, \begin{gather*} \begin{aligned} L&= \int_0^1\phi^{-1} \Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\big( 1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1 \phi^{-1} \Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\big( 1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds, \end{aligned} \\ Q= \min\big\{\phi\big(\frac{c}{L}\big),\; \frac{\phi(c)}{1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2) -\phi\big(1+\sum_{i=1}^m\beta_i\big)}}\big\}; \\ W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} \left(1-k-s\right)ds}\Big);\quad E= \phi\big(\frac{e_1}{L}\big). \end{gather*} such thatc\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.$$ If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold, then \eqref{e18} has at least three positive solutions$x_1,x_2,x_3$such that \begin{gather*} \max_{t\in [0,1]}x_1(t)e_2+h,\\ \min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]}x_3(t)0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_6$. Define the operator $T_6:P_6\to X$ by \begin{align*} (T_6y)(t)&= B_y-\int_t^1\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h',y'(u))du\Big)ds,\quad y\in P_6, \end{align*} where $A_h\in [b,0]$ satisfies \begin{align*} &-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i} \sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\ &\quad -\int_0^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\ &\quad +\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\phi^{-1} \Big(A_h+\int_s^1h(u)du\Big)ds-A=0, \end{align*} and $B_h$ satisfies $$B_h=-\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i \int_{\xi_i}^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds.$$ \end{proof} \begin{lemma} \label{lem233} Suppose that {\rm (H19)--(H21)} hold. Then \begin{itemize} \item[(i)] the following equalities hold: \begin{gather*}{} [\phi((T_6y)'(t))]'+f\left(t,y(t)+h',y'(t)\right)=0,\quad t\in (0,1),\\ (T_6y)(0)-\sum_{i=1}^m\alpha_i (T_6y)(\xi_i)=A-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}B\ge 0,\\ (T_6y)(1)-\sum_{i=1}^m\beta_i (T_6y)(\xi_i)=0; \end{gather*} \item[(ii)] $T_6y\in P_6$ for each $y\in P_6$; \item[(iii)] $x$ is a solution of \eqref{e18} if and only if $x=y+h$ and $y$ is a solution of the operator equation $y=T_6y$; \item[(iv)] $T_6:P_6\to P_6$ is completely continuous. \end{itemize} \end{lemma} The proof of the above lemma is similar to that of Lemma \ref{lem29}; we omit it. \begin{theorem} \label{thm234} Suppose that {\rm (H19)--(H21)} hold and that there exist positive constants $e_1,e_2,c$, \begin{gather*} \begin{aligned} L&= \int_0^1\phi^{-1} \Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2) -\phi\big(1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds\\ &\quad+\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1 \phi^{-1} \Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\big( 1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds, \end{aligned}\\ Q= \min\big\{\phi\big(\frac{c}{L}\big),\; \frac{\phi(c)}{1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\left( 1+\sum_{i=1}^m\beta_i\right)}}\big\}; \\ W = \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} \left(1-k-s\right)ds}\Big);\quad E = \phi\big(\frac{e_1}{L}\big). \end{gather*} such that $c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i \big)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.$$ If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold, then \eqref{e18} has at least three positive solutions $x_1,x_2,x_3$ such that \begin{gather*} \max_{t\in [0,1]}x_1(t)e_2+h,\\ \min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]} x_3(t)e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big),\quad Q>W.$$ If $$f_0(u)=\begin{cases} \frac{150}{299}x,& x\in [0,4],\\ \frac{600}{299}, & x\in [4,44],\\ (x-44)\frac{4000-\frac{600}{299}}{54-44}+\frac{600}{299},& x\in [44,54],\\ 4000, & x\in [54,20004],\\ x-16004, &x\ge 20004, \end{cases}$$ and $$f(t,u,v)=f_0(u)+\frac{1+\sin t}{10000}+\frac{u^2+v^2}{2\times 10^12},$$ it is easy to see that$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)} {1-\sum_{i=1}^m\alpha_i}\Big),\quad Q>W$$ and \begin{itemize} \item[(A1)]$f(t,u,v)<\frac{120\times 20000}{299}$for all$t\in [0,1],u\in [4,20004],v\in [-20000,20000]$; \item[(A2)]$f(t,u,v)>1600$for all$t\in [1/4,3/4],u\in [54,804],v\in [-20000,20000]$; \item[(A3)]$f(t,u,v)\le \frac{1200}{299}$for all$t\in [0,1],u\in [4,44],v\in [-20000,20000]$; \end{itemize} then Theorem \ref{thm210} implies that \eqref{e41} has at least three decreasing and positive solutions$x_1,x_2,x_3$such that$x_1(0)<14$,$x_2(3/4)>54$,$x_3(0)>14$,$x_3(3/4)<54$. \subsection*{Example 3.2} Consider the boundary-value problem $$\label{e42} \begin{gathered} x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\frac{1}{2}x'(1/4)+2,\quad x'(1)=\frac{1}{4}x'(1/4)+\frac{1}{4}x'(1/2)+5. \end{gathered}$$ Corresponding to \eqref{e17}, one sees that$\phi(x)=x=\phi^{-1}(x)$,$\alpha=2$,$\xi_1=1/4$,$\xi_2=1/2$,$\alpha_1=1/2$,$\alpha_2=0$,$\beta_1=1/4=\beta_2$,$A=2$,$B=5$,$h=2$. Choose$k=1/3$, then$\sigma_0=1/3$,$e_1=20$,$e_2=80,c=30000and \begin{gather*} \begin{aligned} L&= \int_0^1\phi^{-1} \Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}+1-s\Big)ds\\ &\quad +\sum_{i=1}^m\alpha_i\phi^{-1} \Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)} {1-\phi\Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}+1-\xi_i\Big) =\frac{55}{8}, \end{aligned}\\ Q= \min\big\{\phi\left(\frac {c}{L}\right),\; \frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2) -\phi\big( 1+\sum_{i=1}^m\beta_i\big)}} \big\}=\frac{48000}{11};\\ W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} \left(1-k-s\right)ds}\Big)=4320;\quad E= \phi\big(\frac{e_1}{L}\big)=\frac{32}{11}. \end{gather*} such thatc\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{\big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\big) \left(1+\sum_{i=1}^m\beta_i\right)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.$$ If \begin{itemize} \item[(A1)]$f(t,u,v)<\frac{48000}{11}$for all$t\in [0,1],u\in [2,30002],v\in [-30000,30000]$; \item[(A2)]$f(t,u,v)>4320$for all$t\in [1/3,2/3],u\in [82,722],v\in [-30000,30000]$; \item[(A3)]$f(t,u,v)\le \frac{32}{11}$for all$t\in [0,1],u\in [2,62],v\in [-30000,30000]$; \end{itemize} then Theorem \ref{thm226} implies that \eqref{e42} has at least three positive solutions$x_1,x_2,x_3$such that \begin{gather*} \max_{t\in [0,1]}x_1(t)<22,\quad \min_{t\in [1/3,2/3]}x_2(t)>82,\\ \max_{t\in [0,1]}x_3(t)>22,\quad \min_{t\in [k,1-k]}x_3(t)<82. \end{gather*} \subsection*{Example 3.3} Consider the boundary-value problem $$\label{e43} \begin{gathered} x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\ x(0)=\frac{1}{2}x(1/4)+\frac{1}{3}x(1/2)+2,\\ x(1)=\frac{1}{4}x(1/4)+\frac{1}{4}x(1/2)+8. \end{gathered}$$ Corresponding to \eqref{e18}, one sees that$\phi(x)=x=\phi^{-1}(x)$,$\xi_1=1/4$,$\xi_2=1/2$,$\alpha_1=1/2$,$\alpha_2=1/3$,$\beta_1=1/4=\beta_2$,$A=2,B=8$,$h\frac{2}{1-\sum_{I=1}^m\beta_i}=16$. Choose$k=1/4$, then$\sigma_0=1/4$. Choose$e_1=50$,$e_2=250$,$c=400000and \begin{gather*} \begin{aligned} L&= \int_0^1\phi^{-1} \Big(1+\frac{\phi\Big(1+\sum_{i=1}^m\beta_i\Big)}{\phi(2)-\phi\big( 1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds\\ &\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1 \phi^{-1} \Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2) -\phi\big(1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds=\frac{481}{64}, \end{aligned}\\ Q= \min\big\{\phi\left(\frac {c}{L}\right),\;\;\;\frac{\phi(c)}{1+\frac{\phi\Big(1+\sum_{i=1}^m\beta_i\Big)}{\phi(2)-\phi\left( 1+\sum_{i=1}^m\beta_i\right)}} \big\}=\frac{25600000}{481};\\ W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1} \left(1-k-s\right)ds}\Big)=8000;\quad E= \phi\big(\frac{e_1}{L}\big)=\frac{3200}{481}. \end{gather*} such thatc\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$, $$Q\ge \phi\Big(\frac{B\left(1+\sum_{i=1}^m\beta_i\right)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.$$ If \begin{itemize} \item[(A1)]$f(t,u,v)<\frac{25600000}{481}$for all$t\in [0,1],u\in [4,400004],v\in [-400000,400000]$; \item[(A2)]$f(t,u,v)>8000$for all$t\in [1/4,3/4],u\in [254,4004],v\in [-400000,400000]$; \item[(A3)]$f(t,u,v)\le \frac{3200}{481}$for all$t\in [0,1],u\in [4,204],v\in [-400000,400000]$; \end{itemize} then Theorem \ref{thm230} implies that \eqref{e43} has at least three positive solutions$x_1,x_2,x_3$such that \begin{gather*} \max_{t\in [0,1]}x_1(t)<54,\quad \min_{t\in [k,1-k]}x_2(t)>254, \\ \max_{t\in [0,1]}x_3(t)>54,\quad \min_{t\in [k,1-k]}x_3(t)<254. \end{gather*} \subsection*{Acknowledgements} The author is grateful to the anonymous referee for his/her detailed reading and constructive comments which improve the presentation of this article. 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