\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 171, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/171\hfil Global properties and multiple solutions] {Global properties and multiple solutions for boundary-value problems of impulsive differential equations} \author[J. Wang, B. Yan \hfil EJDE-2013/171\hfilneg] {Jing Wang, Baoqiang Yan} % in alphabetical order \address{Jing Wang \newline School of Mathematical Sciences, Shandong Normal University, Jinan, 250014, China} \email{wnself36@gmail.com} \address{Baoqiang Yan (corresponding author)\newline School of Mathematical Sciences, Shandong Normal University, Jinan, 250014, China} \email{yanbaoqiang666@gmail.com} \thanks{Submitted June 4, 2013. Published July 26, 2013.} \subjclass[2000]{34B09, 34B15, 34B37} \keywords{Eigenvalues; bifurcation technique; global behavior; \hfill\break\indent multiple solutions} \begin{abstract} This article presents global properties and existence of multiple solutions for a class of boundary value problems of impulsive differential equations. We first show that the spectral properties of the linearization of these problems are similar to the well-know properties of the standard Sturm-Liouville problems. These spectral properties are then used to prove two Rabinowitz-type global bifurcation theorems. Finally, we use the global bifurcation theorems to obtain multiple solutions for the above problems having specified nodal properties. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} In this article, we study the global properties for the boundary-value problem $$\begin{gathered} -x''(t)=\lambda f(t,x),\quad t\in(0,1),\quad t\neq\frac{1}{2},\\ \Delta x|_{t=\frac{1}{2}}=\lambda\beta x(\frac{1}{2}),\\ \Delta x'|_{t=\frac{1}{2}}=-\lambda\beta x'(\frac{1}{2}-0),\\ x(0)=x(1)=0, \end{gathered}\label{e1.1}$$ where $\lambda\neq 0$, $\beta\neq0$, $\Delta x|_{t=\frac{1}{2}}=x(\frac{1}{2}+0)-x(\frac{1}{2})$, $\Delta x'|_{t=\frac{1}{2}}=x'(\frac{1}{2}+0)-x'(\frac{1}{2}-0)$, and $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ is continuous. The theory of impulsive differential equations has been a significant development in recent years and played a very important role in modern applied mathematical models of real processes rising in phenomena studied in physics, population dynamics, chemical technology, biotechnology and economics (see \cite{c2,d1,g1,l3,t1,t2}). There have appeared numerous papers on impulsive differential equations and many of them are committed to study the existence of solutions for boundary value problems of second order impulsive differential equations. The general methods to solve these problems include topological degree theory (see \cite{l4,s1,y2}), the method of upper and lower solutions (see \cite{l1,l2,l5}) and variational method (see \cite{c1,r1,r2,s1,t2,x1,z1}). In natural sciences, there are various concrete problems involving bifurcation phenomena, for example, Taylor vortices \cite{b1} and catastrophic shifts in ecosystems \cite{s3}. Rabinowitz \cite{r2} established the global bifurcation theorem from the trivial solution and in \cite{r3} the bifurcation from infinity was studied. In recent years, this theory has been successfully applied to Sturm-Liouville problems for ordinary differential equations, integral equations and partial differential equations (see \cite{m1,m2,m3,r1}). In \cite{l6}, using bifurcation techniques, Liu and O'Regan studied the existence of multiple solutions for the second order impulsive differential equation $$\begin{gathered} x''(t)+ra(t)f(t,x(t))=0, \quad t\in(0, 1),\; t\neq t_{i},\\ \Delta x|_{t=t_{i}}=\alpha_{i}x(t_{i}-0), \quad i=1,2,\dots,k,\\ x(0)=x(1)=0, \end{gathered}\label{e1.2}$$ in which they convert \eqref{e1.2} into \begin{gathered} y''(t)+\frac{r}{\prod_{00(\alpha<0)$and$|{\lambda-\mu}|=o(1), \| w\|=o(|\alpha|)$, at$\alpha=0$. Moreover, each of$ C_{\mu}^{+}, C_{\mu}^{-}$either \begin{itemize} \item[(1)] meets infinity in$\Sigma$, or \item[(2)] meets$(\hat{\mu},0)$where$\mu\neq\hat{\mu}\in\sigma(L)$, or \item[(3)] contains a pair of points$(\lambda,u), (\lambda,-\mu), u\neq 0$. \end{itemize} \end{lemma} \begin{lemma}[\cite{r3}] \label{lem1.2} Assume that$L$is compact and linear,$H(\lambda,u)$is continuous on$\mathbb{R}\times E$,$H(\lambda,u)=o(\| u\|)$at$u=\infty$uniformly on bounded$\lambda$intervals, and$\| u\|^{2}H(\lambda,\frac{u}{\| u\|^{2}})$is compact. If$\mu\in\sigma(L)$is of odd multiplicity, then$\Sigma$possesses an unbounded component$ D_{\mu}$which meets$(\mu,\infty)$. Moreover if$\Lambda\subset\mathbb{R}$is an interval such that$\Lambda\cap\sigma(L)={\mu}$and$\mathcal{M}$is a neighborhood of$(\mu,\infty)$whose projection on$\mathbb{R}$lies in$\Lambda$and whose projection on$E$is bounded away from$0$, then either \begin{itemize} \item[(1)]$ D_{\mu}\setminus\mathcal{M}$is bounded in$\mathbb{R}\times E$in which case$ D_{\mu}\setminus\mathcal{M}$meets$\Theta=\{(\lambda,0)|\lambda\in\mathbb{R}\}$or \item[(2)]$ D_{\mu}\setminus\mathcal{M}$is unbounded. \end{itemize} If$(2)$occurs and$ D_{\mu}\setminus\mathcal{M}$has a bounded projection on$\mathbb{R}$, then$ D_{\mu}\setminus\mathcal{M}$meets$(\mu,\infty)$where$\mu\neq\hat{\mu}\in\sigma(L)$. \end{lemma} \begin{lemma}[\cite{r3}] \label{lem1.3} Suppose the assumptions of Lemma \ref{lem1.2} hold. If$\mu\in\sigma(L)$is simple, then$ D_{\mu}$can be decomposed into two subcontinua$ D_{\mu}^{+}$,$ D_{\mu}^{-}$and there exists a neighborhood$\mathcal{O}\subset\mathcal{M}$of$(\mu,\infty)$such that$(\lambda,u)\in D_{\mu}^{+}( D_{\mu}^{-})\cap\mathcal{O}$, and$(\lambda,u)\neq(\mu,\infty)$implies$(\lambda,u)=(\lambda,\alpha v+w)$where$\alpha>0(\alpha<0)$and$|{\lambda-\mu}|=o(1)$,$\| w\|=o(|\alpha|)$, at$|\alpha|=\infty$. \end{lemma} \section{Preliminaries} Let$PC[0,1]=\{x:[0,1]\to\mathbb{R}:x(t)$is continuous at$t\neq\frac{1}{2}$, and left continuous at$t=\frac{1}{2}$, and$x(\frac{1}{2}+0)=\lim_{t\to\frac{1}{2}^{+}}x(t)$exists$\}$with the norm $$\| x\|=\sup_{t\in[0,1]}| x(t)|.$$ Let$PC'[0,1]=\{x\in PC[0,1]:x'(t)$is continuous at$t\neq\frac{1}{2}$, and$x'(\frac{1}{2}-0)=\lim_{t\to\frac{1}{2}^{-}}x'(t)$, and$x'(\frac{1}{2}+0)=\lim_{t\to\frac{1}{2}^{+}}x'(t)$exist$\}$with the norm $$\| x\|_1=\max\{\sup_{t\in[0,1]}| x(t)|,\sup_{t\in[0,1]}| x'(t)|\}.$$ Let$E=\{x\in PC'[0,1]:x(0)=x(1)=0\}$. It is well known that$E$is a Banach space with the norm$\|\cdot\|_1$. Let$S_{k}^{+}$denote the set of functions in$E$which have exactly$k-1$simple nodal zeros in$(0,1)$and are positive near$t=0$. (By a nodal zero we mean the function changes sign at the zeros and at at a simple nodal zero, the derivative of the function is nonzero.) And set$S_{k}^{-}=-S_{k}^{+}$,$S_{k}=S_{k}^{+}\cup S_{k}^{-}$. They are disjoint in$E$. Finally, let$\Phi_{k}^{\pm}=\mathbb{R}\times S_{k}^{\pm}$and$\Phi_{k}=\mathbb{R}\times S_{k}$. For the rest of the paper, we always assume the initial-value problem \begin{gather*} -x''(t)=\lambda f(t,x),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{t=\frac12}=\lambda\beta x(\frac{1}{2}),\\ \Delta x'|_{t=\frac12}=-\lambda\beta x'(\frac{1}{2}-0),\\ x(t_0)=x'(t_0)=0, \end{gather*} has the unique trivial solution$x\equiv0$on$[0,1]$for any$t_0\in[0,1]$. \begin{lemma}[\cite{g2}] \label{lem2.1}$x(t)\in PC[J,\mathbb{R}]\cap C^{2}[J',\mathbb{R}]$is the solution of \eqref{e1.1} equivalent to$x(t)\in PC'[J,\mathbb{R}]$is the solution of the integral equation $$x(t)=\begin{cases} \lambda\int_{0}^{1}G(t,s)f(s,x(s))ds -\lambda\beta t[x(\frac{1}{2})-\frac{1}{2}x'(\frac{1}{2}-0)], &t\in[0, \frac{1}{2}],\\ \lambda\int_{0}^{1}G(t,s)f(s,x(s))ds+\lambda\beta (1-t)[x(\frac{1}{2}) +\frac{1}{2}x'(\frac{1}{2}-0)], &t\in(\frac{1}{2}, 1], \end{cases}$$ where$J=[0,1], J'=J\setminus{\{\frac{1}{2}\}}$, $$G(t,s)=\begin{cases} s(1-t), & 0\leq s\leq t\leq 1,\\ t(1-s), & 0\leq t\leq s\leq 1. \end{cases}$$ \end{lemma} \begin{lemma} \label{lem2.2} If$a>0$, then the linear boundary-value problem $$\begin{gathered} -u''(t)=\lambda au(t),\quad t\in(0, 1),\; t\neq\frac{1}{2},\\ \Delta u|_{t=\frac{1}{2}}=\lambda\beta u(\frac{1}{2}),\\ \Delta u'|_{t=\frac{1}{2}}=-\lambda\beta u'(\frac{1}{2}-0),\\ u(0)=u(1)=0, \end{gathered}\label{e2.1}$$ possess an increasing sequence of eigenvalues $$0<\lambda_1<\lambda_2<\dots<\lambda_{k}<\dots, \quad \lim_{k\to+\infty}\lambda_k=+\infty.$$ And eigenfunction$u_k$corresponding to$\lambda_k$has exactly$k-1$nodal zeros on$(0,1)$. \end{lemma} \begin{proof} Let$u(t)$be the solution of \eqref{e2.1}. We consider the three following cases: \noindent\textbf{Case (i)}$\lambda=0$. Then$u(t)$can be written as $$u(t)= \begin{cases} c_1t+c_2,& t\in[0, \frac{1}{2}],\\ c_{3}t+c_{4},& t\in(\frac{1}{2}, 1]. \end{cases}$$ From \eqref{e2.1}, we have \begin{gather*} c_2=0,\\ -\frac{1}{2}c_1-c_2+\frac{1}{2}c_{3}+c_{4}=0,\\ -c_1+c_{3}=0,\\ c_{3}+c_{4}=0, \end{gather*} which implies$c_1=c_2=c_{3}=c_{4}=0$. Then$u(t)\equiv 0$. Thus$\lambda=0$is not the eigenvalue of \eqref{e2.1}. \noindent\textbf{Case (ii)}$\lambda<0$. Then$u(t)$can be written as $$u(t)= \begin{cases} c_1e^{\sqrt{-\lambda a}t}+c_2e^{-\sqrt{-\lambda a}t}, & t\in[0, \frac{1}{2}],\\ c_{3}e^{\sqrt{-\lambda a}t}+c_{4}e^{-\sqrt{-\lambda a}t},\quad &t\in(\frac{1}{2}, 1]. \end{cases}$$ From \eqref{e2.1}, we have \begin{gather*} c_1+c_2=0,\\ (1+\lambda\beta)e^{\frac{\sqrt{-\lambda a}}{2}}c_1 +(1+\lambda\beta)e^{-\frac{\sqrt{-\lambda a}}{2}}c_2 -e^{\frac{\sqrt{-\lambda a}}{2}}c_{3}-e^{-\frac{\sqrt{-\lambda a}}{2}}c_{4}=0,\\ (1-\lambda\beta)e^{\frac{\sqrt{-\lambda a}}{2}}c_1 -(1-\lambda\beta)e^{-\frac{\sqrt{-\lambda a}}{2}}c_2 -e^{\frac{\sqrt{-\lambda a}}{2}}c_{3}+e^{-\frac{\sqrt{-\lambda a}}{2}}c_{4}=0,\\ e^{\sqrt{-\lambda a}}c_{3}+e^{-\sqrt{-\lambda a}}c_{4}=0, \end{gather*} which implies$c_1=c_2=c_{3}=c_{4}=0$. Then$u(t)\equiv 0$. Thus$\lambda<0$is not the eigenvalue of \eqref{e2.1}. \noindent\textbf{Case (iii)}$\lambda>0$. Then$u(t)$can be written as $$u(t)= \begin{cases} c_1\cos(\sqrt{\lambda a}t)+c_2\sin(\sqrt{\lambda a}t), & t\in[0, \frac{1}{2}],\\ c_{3}\cos(\sqrt{\lambda a}t)+c_{4}\sin(\sqrt{\lambda a}t), & t\in(\frac{1}{2}, 1]. \end{cases}$$ From \eqref{e2.1} we know$c_1=0$and \begin{gather*} \cos(\frac{\sqrt{\lambda a}}{2})c_{3} +\sin(\frac{\sqrt{\lambda a}}{2})c_{4}-(1+\lambda\beta) \sin(\frac{\sqrt{\lambda a}}{2})c_2=0,\\ -\sin(\frac{\sqrt{\lambda a}}{2})c_{3} +\cos(\frac{\sqrt{\lambda a}}{2})c_{4}-(1-\lambda\beta) \cos(\frac{\sqrt{\lambda a}}{2})c_2=0,\\ \cos(\sqrt{\lambda a})c_{3}+\sin(\sqrt{\lambda a})c_{4}=0. \end{gather*} The determinant of the coefficient matrix$\det=\sin\sqrt{\lambda a}$. Letting$\det=0$, we have$\lambda_{k}=\frac{k^{2}\pi^{2}}{a}$,$k=1,2,\dots$. Then $$c_{3}=0,\quad c_{4}=(1+\beta(-1)^{k+1}\frac{k^{2}\pi^{2}}{a})c_2,$$ which implies$\lambda_{k}=\frac{k^{2}\pi^{2}}{a}$is the eigenvalues of \eqref{e2.1}, and $$0<\lambda_1<\lambda_2<\dots<\lambda_{k}<\dots,\quad \lim_{k\to+\infty}\lambda_k=+\infty.$$ The eigenfunction corresponding to$\lambda_{k}$is $$u_{k}(t)=\begin{cases} \sin(k\pi t), & t\in[0, \frac{1}{2}],\\ (1+\beta(-1)^{k+1}\frac{k^{2}\pi^{2}}{a})\sin(k\pi t), & t\in(\frac{1}{2}, 1]. \end{cases}$$ Now we prove that$u_{k}$has exactly$k-1$nodal zeros on$(0,1)$. Actually, if$k$is odd, then$u_{k}(t)$has$\frac{k-1}{2}$zeros on$(0,\frac{1}{2}]$,$u_{k}(t)$has$\frac{k-1}{2}$zeros on$(\frac{1}{2},1)$, and if$k$is even, then$u_{k}(t)$has$\frac{k}{2}$zeros on$(0,\frac{1}{2}]$,$u_{k}(t)$has$\frac{k-2}{2}$zeros on$(\frac{1}{2},1)$. Thus$u_{k}(t)=0$has$k-1$zeros on$(0,1)$. Using the fact $$u_{k}'(t)=\begin{cases} k\pi\cos(k\pi t), & t\in[0, \frac{1}{2}),\\ (1+\beta(-1)^{k+1}\frac{k^{2}\pi^{2}}{a})k\pi\cos(k\pi t), & t\in(\frac{1}{2}, 1], \end{cases}$$ we see that$u_{k}(t)$has exactly$k-1$nodal zeros on$(0,1)$. The proof is complete. \end{proof} \begin{lemma} \label{lem2.3} For each$k\geq 1$the algebraic multiplicity of eigenvalue$\lambda_k$is equal to$1$. \end{lemma} \begin{proof} Define the operator$K: PC'[0,1]\to PC'[0,1]$as follows: $$(Kx)(t)=\begin{cases} \int_{0}^{1}G(t,s)ax(s)ds-\beta t[x(\frac{1}{2})-\frac{1}{2}x'(\frac{1}{2}-0)] , & t\in[0, \frac{1}{2}],\\ \int_{0}^{1}G(t,s)ax(s)ds+\beta (1-t)[x(\frac{1}{2}) +\frac{1}{2}x'(\frac{1}{2}-0)] ,& t\in(\frac{1}{2}, 1]. \end{cases}$$ We need to prove only that$\ker(I-\lambda_{k}K)^2\subset \ker(I-\lambda_{k}K)$. For any$y\in ker(I-\lambda_{k}K)^2$, we have$(I-\lambda_{k}K)^2 y=0$, so$(I-\lambda_{k}K)y\in ker(I-\lambda_{k}K)$. Let$\bar{\beta}_k=1+\beta(-1)^{k+1}\lambda_k$. From Lemma \ref{lem2.2}, there exists a$\gamma$satisfying \begin{gather*} (I-\lambda_{k}K)y=\gamma\sin(\sqrt{\lambda_ka}t),\quad t\in[0, \frac{1}{2}],\\ (I-\lambda_{k}K)y=\gamma\bar{\beta}_k\sin(\sqrt{\lambda_ka}t),\quad t\in(\frac{1}{2}, 1]; \end{gather*} that is,$y$satisfies \begin{gather*} y''+\lambda_kay=-\gamma\lambda_ka\sin(\sqrt{\lambda_ka}t),\quad t\in[0, \frac{1}{2}),\\ y''+\lambda_kay=-\gamma\lambda_ka\bar{\beta}_k\sin(\sqrt{\lambda_ka}t),\quad t\in(\frac{1}{2}, 1], \end{gather*} and$y(0)=y(1)=0$. Now we prove$\gamma=0. Actually, the general solution of the above differential equation is of the form \begin{gather*} \begin{aligned} y(t)&=c_1 \cos(\sqrt{\lambda_ka}t)+c_2 \sin(\sqrt{\lambda_ka}t) -\frac{\gamma}{2}\sin(\sqrt{\lambda_ka}t)\\ &\quad +\frac{\gamma\sqrt{\lambda_ka} t}{2}\cos(\sqrt{\lambda_ka}t),\quad t\in[0, \frac{1}{2}], \end{aligned} \\ \begin{aligned} y(t)&= c_1\cos(\sqrt{\lambda_ka}t)+c_2 \sin(\sqrt{\lambda_ka}t) -\frac{\gamma\bar{\beta}_k}{4}\sin(\sqrt{\lambda_ka}t) +\frac{\gamma\bar{\beta}_k}{4}\sin(\sqrt{\lambda_ka}\\ &\quad -\sqrt{\lambda_ka}t) +\frac{\gamma\sqrt{\lambda_ka}\bar{\beta}_kt}{2} \cos(\sqrt{\lambda_ka}t)-\frac{\gamma\sqrt{\lambda_ka}\bar{\beta}_k}{4} \cos(\sqrt{\lambda_ka}t),\quad t\in(\frac{1}{2}, 1]. \end{aligned} \end{gather*} Fromy(0)=y(1)=0$, we have$c_1=0$and $$c_2 \sin(\sqrt{\lambda_ka})-\frac{\gamma\bar{\beta}_k}{4} \sin(\sqrt{\lambda_ka})+\frac{\gamma\sqrt{\lambda_ka} \bar{\beta}_k}{2}\cos(\sqrt{\lambda_ka}) -\frac{\gamma\sqrt{\lambda_ka}\bar{\beta}_k}{4}\cos(\sqrt{\lambda_ka})=0.$$ By Lemma \ref{lem2.2},$\lambda_k=k^2\pi^2/a$satisfies$\sin(\sqrt{\lambda_ka})=0, \cos(\sqrt{\lambda_ka})\neq 0$which implies$\gamma=0$. Thus$y\in ker(I-\lambda_{k}K)$. Therefore$\ker(I-\lambda_{k}K)^2\subset\ker(I-\lambda_{k}K)$. The proof is complete. \end{proof} \begin{lemma} \label{lem2.4} For any positive integer$k$,$S_{k}, S_{k}^{+}$and$S_{k}^{-}$are open in$E$. \end{lemma} \begin{proof} We prove only that$S_{k}$is open in$E$. Let$t_1, t_2, \dots, t_{k-1}\in(0,1)$are$k-1$simple nodal zeros on$(0,1)$of$u(t)\in S_{k}$. Suppose$u'(t_{j})=c_{j}>0$. Then there exists$\delta_{j}>0$which satisfies for any$t\in [t_{j}-\delta_{j}, t_{j}+\delta_{j}], u'(t)$is continuous,$u'(t)>\frac{c_{j}}{2}$,$u(t_j-\delta_j)<0$and$u(t_j+\delta_j)>0$. Letting$\varphi\in E$and$\|\varphi-u\|_1\leq\sigma, where \begin{align*} \sigma&=\min\Big\{\frac{c_{j}}{2}, \frac{u(t_{j}+\delta_{j})}{2}, -\frac{u(t_{j}-\delta_{j})}{2}, -\frac{1}{2}\max_{t\in[t_{j-1} +\delta_{j-1},t_{j}-\delta_{j}]}u(t), \\ &\quad \frac{1}{2}\max_{t\in[t_{j} +\delta_{j},t_{j+1}-\delta_{j+1}]}u(t)\Big\}, \end{align*} we have \begin{gather*} \varphi'(t)=\varphi'(t)-u'(t)+u'(t)\geq -\sigma +\frac{c_{j}}{2}>0,t\in[t_{j}-\delta_{j}, t_{j}+\delta_{j}],\\ \varphi(t_{j}+\delta_{j})>u(t_{j}+\delta_{j})-\sigma>0,\\ \varphi(t_{j}-\delta_{j})0. Since for any $t\in[t_{j-1}+\delta_{j-1}, t_{j}-\delta_{j}], u(t)<0$, and for any $t\in[t_{j}+\delta_{j}, t_{j+1}-\delta_{j+1}], u(t)>0$, one has $\varphi(t)u(t)+\sigma>0$ for any $t\in[t_{j}+\delta_{j}, t_{j+1}-\delta_{j+1}]$. Therefore for any $t\in[t_{j-1}+\delta_{j-1}, t_{j+1}-\delta_{j+1}], \varphi(t)$ has the unique simple zero $t_{j}^{*}$. If $u'(t_j)=c_j<0$, we can get for any $t\in[t_{j-1}+\delta_{j-1}, t_{j+1}-\delta_{j+1}], \varphi(t)$ has the unique simple zero $t_j^*$ by the same method as above also. Because $u(t)$ has exactly $k-1$ simple zeros on $(0,1)$, $\varphi(t)$ has exactly $k-1$ simple zeros on $(0,1)$. Therefore $\varphi(t)\in S_{k}$, which implies $S_{k}$ are open in $E$. By the same argument, we can prove $S_{k}^{+}$ and $S_{k}^{-}$ are open in $E$. The proof is complete. \end{proof} \section{Main results} In this section, we assume that \begin{itemize} \item[(H1)] There exist two positive numbers $f_{0}$ and $f_{\infty}$ such that $f_{0}=\lim_{|x|\to 0}\frac{f(t,x)}{x}$ and $f_{\infty}=\lim_{|x|\to\infty}\frac{f(t,x)}{x}$ both uniformly with respect to $t\in[0,1]$. \end{itemize} Let $\zeta,\xi\in C([0,1]\times\mathbb{R},\mathbb{R})$ be such that $$f(t,x)=f_{0}x+\zeta(t,x),\quad f(t,x)=f_{\infty}x+\xi(t,x).\label{e3.1}$$ Clearly, if (H1) holds, we have $$\lim_{|x|\to 0}\frac{\zeta(t,x)}{x}=0,\quad \lim_{|x|\to\infty}\frac{\xi(t,x)}{x}=0\label{e3.2}$$ both uniformly with respect to $t\in[0,1]$. Now we define three operators $L_{0}, L_{\infty}$, and $A$ as follows: $$(L_{0}x)(t)=\begin{cases} \int_{0}^{1}G(t,s)f_{0}x(s)ds-\beta t[x(\frac{1}{2}) -\frac{1}{2}x'(\frac{1}{2}-0)], & t\in[0, \frac{1}{2}],\\ \int_{0}^{1}G(t,s)f_{0}x(s)ds+\beta(1-t)[x(\frac{1}{2}) +\frac{1}{2}x'(\frac{1}{2}-0)],& t\in(\frac{1}{2}, 1], \end{cases} \label{e3.3}$$ $$(L_{\infty}x)(t)=\begin{cases} \int_{0}^{1}G(t,s)f_{\infty}x(s)ds-\beta t[x(\frac{1}{2}) -\frac{1}{2}x'(\frac{1}{2}-0)],& t\in[0, \frac{1}{2}],\\ \int_{0}^{1}G(t,s)f_{\infty}x(s)ds+\beta(1-t)[x(\frac{1}{2}) +\frac{1}{2}x'(\frac{1}{2}-0)], &t\in(\frac{1}{2},1], \end{cases} \label{e3.4}$$ and $$(Ax)(t)=\begin{cases} \int_{0}^{1}G(t,s)f(s,x(s))ds-\beta t[x(\frac{1}{2}) -\frac{1}{2}x'(\frac{1}{2}-0)],&t\in[0, \frac{1}{2}],\\ \int_{0}^{1}G(t,s)f(s,x(s))ds+\beta(1-t)[x(\frac{1}{2}) +\frac{1}{2}x'(\frac{1}{2}-0)], &t\in(\frac{1}{2}, 1]. \end{cases}\label{e3.5}$$ Obviously, $L_{0}$ and $L_{\infty}$ are compact and linear. From Lemma \ref{lem2.1}, we know that $x(t)$ is the solution of \eqref{e1.1} if and only if $x(t)$ is the solution of $$x=\lambda Ax.\label{e3.6}$$ Let $\Gamma=\mathbb{R}\times E$. A solution of \eqref{e3.6} is a pair $(\lambda,x)\in\Gamma$. The known curve of solutions $\{(\lambda,0)|\lambda\in\mathbb{R}\}$ will henceforth be referred to as the trivial solutions. The closure of the set on nontrivial solutions of \eqref{e3.6} will be denoted by $\Sigma$ as in Lemma \ref{lem1.1}. Now we are ready to give our main results. \begin{theorem} \label{thm3.1} Let {\rm (H1)} hold. In addition assume that for some $k\in\mathbb{N}$, either $$\frac{k^2\pi^2}{f_0}<\lambda<\frac{k^2\pi^2}{f_\infty}, \quad\text{or}\quad \frac{k^2\pi^2}{f_\infty}<\lambda<\frac{k^2\pi^2}{f_0}.$$ Then problem \eqref{e1.1} has at least two solutions $x_k^+$ and $x_k^-$, $x_k^+$ has exactly $k-1$ zeros in (0,1) and is positive near $t=0$, and $x_k^-$ has exactly $k-1$ zeros in $(0,1)$ and is negative near $t=0$. \end{theorem} \begin{theorem} \label{thm3.2} Let {\rm (H1)} hold. Suppose that there exist two integer $k>0$ and $j\geq0$ such that either \begin{itemize} \item[(i)] $\frac{(k+j)^2\pi^2}{f_0}<\lambda<\frac{k^2\pi^2}{f_\infty}$; \item[(ii)] $\frac{(k+j)^2\pi^2}{f_\infty}<\lambda<\frac{k^2\pi^2}{f_0}$ \end{itemize} Then problem \eqref{e1.1} has at least $2(j+1)$ solutions $x_{k+i}^+, x_{k+i}^-$, $i=0,1,\dots,j, x_{k+i}^+$ has exactly $k+i-1$ zeros in $(0,1)$ and is positive near $t=0$, $x_{k+i}^-$ has exactly $k+i-1$ zeros in $(0,1)$ and are negative near $t=0$. \end{theorem} To use the terminology of Rabinowitz \cite{r1,r2}, we first give the following lemmas. \begin{lemma} \label{lem3.3} Suppose that {\rm (H1)} is satisfied. Then the operator $A$ given in \eqref{e3.5} is Fr\'echet differentiable at $x=\theta$, and $A'(\theta)=L_{0}$. \end{lemma} \begin{proof} From \eqref{e3.1}, we obtain \begin{align*} &\| Ax-L_{0}x\|_1\\ &= \max\Big\{\sup_{t\in[0,1]}|\int_{0}^{1}G(t,s)f(s,x(s))ds -\int_{0}^{1}G(t,s)f_{0}x(s)ds|,\\ &\quad \sup_{t\in[0,1]}|\int_{0}^{1}G_t(t,s)f(s,x(s))ds -\int_{0}^{1}G_t(t,s)f_{0}x(s)ds|\Big\}\\ &= \max\Big\{\sup_{t\in[0,1]}|\int_{0}^{1}G(t,s)\zeta(s,x(s))ds|, \sup_{t\in[0,1]}|\int_{0}^{1}G_t(t,s)\zeta(s,x(s))ds|\Big\}\\ &\leq C\int_{0}^{1}|\zeta(s,x(s))| ds, \end{align*} where $C$ depends on bounds for $G$ and $G_{t}$. Consequently, from \eqref{e3.2}, $$\lim_{\| x\|_1\to 0}\frac{\| Ax-L_{0}x\|_1}{\| x\|_1} \leq\lim_{\| x\|_1\to 0}\frac{\int_{0}^{1}|\zeta(s,x(s))| ds}{\| x\|_1}=0, \quad \forall x\in E.$$ This means that the operator $A$ given in \eqref{e3.5} is Fr\'echet differentiable at $x=\theta$, and $A'(\theta)=L_{0}$. The proof is complete. \end{proof} \begin{lemma} \label{lem3.4} Suppose that {\rm (H1)} is satisfied. Then the operator $A$ given in \eqref{e3.5} is Fr\'echet differentiable at $x=\infty$, and $A'(\infty)=L_{\infty}$. \end{lemma} \begin{proof} For each $\varepsilon>0$, by \eqref{e3.2}, there exists $R>0$ such that $$|\xi(t,x)|\leq\varepsilon|x|,\quad \text{for } |x|>R, t\in[0,1].$$ Let $M=\max_{| x|\leq R}|\xi(t,x)|$. Then we have $$|\xi(t,x)|\leq\varepsilon| x|+M,\quad\forall x\in\mathbb{R}, t\in[0,1].$$ Therefore, for any $x\in E$, $t\in[0,1]$, \begin{align*} &\|Ax-L_{\infty}x\|_1\\ &= \max\Big\{\sup_{t\in[0,1]}|\int_{0}^{1}G(t,s)\xi(s,x(s))ds|, \sup_{t\in[0,1]}|\int_{0}^{1}G_{t}(t,s)\xi(s,x(s))ds|\Big\}\\ &\leq C_1\int_{0}^{1}|\xi(s,x(s))| ds\\ &\leq C_1(M+\varepsilon\int_{0}^{1}|x(s)|ds)\\ &\leq C_1(M+\varepsilon\| x\|_1), \end{align*} which implies $$\lim_{\| x\|_1\to\infty}\frac{\| Ax-L_{\infty}x\|_1}{\| x\|_1}=0,$$ where $C_1$ depends on bounds for $G$ and $G_{t}$. This means that the operator $A$ given in \eqref{e3.5} is Fr\'echet differentiable at $x=\infty$, and $A'(\infty)=L_{\infty}$. The proof is complete. \end{proof} Under the condition (H1), \eqref{e3.6} can be rewritten as $$x=\lambda L_{0}x+H(\lambda,x),\label{e3.7}$$ here $H(\lambda,x)=\lambda Ax-\lambda L_{0}x$, $L_{0}$ is defined as \eqref{e3.3}. Clearly by Lemma \ref{lem2.1}, $x(t)$ is the solution of \eqref{e2.1} if and only if $x(t)$ is the solution of the equation $$x=\lambda L_{0}x.\label{e3.8}$$ Therefore, the results of Lemma \ref{lem2.2} and Lemma \ref{lem2.3} satisfy \eqref{e3.8}. From Lemma \ref{lem3.3}, it can be seen that $H(\lambda,x)$ is $o(\| x\|_1)$ for $x$ near $0$ uniformly on bounded $\lambda$. \begin{lemma} \label{lem3.5} For each integer $k>0$ and each $\upsilon=+$, or $-$, there exists a continua $C_{k}^{\upsilon}$ of solutions of \eqref{e3.6} in $\Phi_{k}^{\upsilon}\cup\{(\frac{k^2\pi^2}{f_0},0)\}$, which meets $\{(\frac{k^2\pi^2}{f_0},0)\}$ and $\infty$ in $\Sigma$. \end{lemma} \begin{proof} By Lemma \ref{lem2.2} and Lemma \ref{lem2.3}, we know that for each integer $k>0$, $\frac{k^2\pi^2}{f_0}$ is a simple characteristic value of operator $L_{0}$. So with Lemma \ref{lem3.3}, \eqref{e3.7} can be considered as a bifurcation problem from the trivial solution. From Lemma \ref{lem1.1} and condition (H1) it follows that $\Sigma$ contains a component $C_{k}$ which can be decomposed into two subcontinua $C_{k}^{+}$, $C_{k}^{-}$ such that for some neighborhood $B$ of $(\frac{k^2\pi^2}{f_0},0)$, $$(\lambda,x)\in C_{k}^{+}( C_{k}^{-})\cap B,\quad\text{and}\quad (\lambda,x)\neq(\frac{k^2\pi^2}{f_0},0),$$ implies $(\lambda,x)=(\lambda,\alpha u_{k}+w)$, where $\alpha>0(\alpha<0)$ and $|\lambda-\frac{k^2\pi^2}{f_0}|=o(1), \| w\|_1=o(|\alpha|)$ at $\alpha=0$. Since $S_{k}$ is open and $u_{k}\in S_{k}$, we know $$\frac{x}{\alpha}=u_k+\frac{w}{\alpha}\in S_{k},$$ for $\alpha\neq0$ sufficiently small. Then there exists $\delta_{0}>0$ such that for $\delta\in(0,\delta_{0})$, we have $$( C_{k}\setminus\{(\frac{k^2\pi^2}{f_0},0)\})\cap B_{\delta}\subset \Phi_{k},$$ where $B_{\delta}$ is an open ball in $\Gamma$ of radius $\delta$ centered at $(\frac{k^2\pi^2}{f_0},0)$. From the assumption in section 2 and \eqref{e3.8} we know $C_{k}\setminus\{(\frac{k^2\pi^2}{f_0},0)\}\cap\partial\Phi_{k}=\emptyset$. Consequently, $C_{k}$ lies in $\Phi_{k}\cup\{(\frac{k^2\pi^2}{f_0},0)\}$. From the the same reasoning it can be seen that $C_{k}^{\vartheta}$ lies in $\Phi_{k}^{\vartheta}\cup\{(\frac{k^2\pi^2}{f_0},0)\}(\vartheta=+$, or $-$). Next we show that the alternative $(2)$ of Lemma \ref{lem1.1} is impossible. Suppose, on the contrary and without loss of generality that $C_{k}^{+}$ meets $(\frac{j^2\pi^2}{f_0},0)$ with $k\neq j$ and $\frac{j^2\pi^2}{f_0}\in\sigma(L_0)$. Then there exists a sequence $(\gamma_{m},z_{m})\in C_{k}^{+}$ with $\gamma_{m}\to\frac{j^2\pi^2}{f_0}$ and $z_{m}\to0$ as $m\to +\infty$. Notice that $$z_{m}=\gamma_{m}L_{0}z_{m}+H(\gamma_{m},z_{m}).\label{e3.9}$$ Dividing this equation by $\| z_{m}\|_1$ and noticing that $L_{0}$ is compact on $E$ and that $H(\gamma_{m},z_{m})=o(\| z_{m}\|_1)$ as $m\to\infty$, we may assume without loss of generality that $\frac{z_{m}}{\| z_{m}\|_1}\to z$ as $m\to\infty$. Thus by \eqref{e3.9} it follows that $$z=\frac{j^2\pi^2}{f_0}L_0z.$$ Since $z\neq0$, by Lemma \ref{lem2.2}, $z$ belongs to $S_{j}^{+}$ or $S_{j}^{-}$. Notice that $\|\frac{z_{m}}{\| z_{m}\|_1}-z\|_1\to0$, $S_{j}^{+}$ and $S_{j}^{-}$ are open, so $z_{m}\in S_{j}^{+}$ or $S_{j}^{-}$ for $m$ sufficiently large. This is a contradiction with $z_{m}\in S_{k}^{+}(m\geq1)$ and $j\neq k$. Hence alternative $(2)$ is impossible. Finally since $S_{k}^{+}$ and $S_{k}^{-}$ are disjoint, thus $C_{k}^{\upsilon}$ does not contain a pair of points $(\lambda,x), (\lambda,-x), x\neq 0$, which means alternative $(3)$ of Lemma \ref{lem1.1} is impossible. Therefore alternative $(1)$ of Lemma \ref{lem1.1} holds. This implies that for each integer $k>0$, and each $\upsilon=+$ or $-$, there exists a continua $C_{k}^{\upsilon}$ of solutions of \eqref{e3.6} in $\Phi_{k}^{\upsilon}\cup\{(\frac{k^2\pi^2}{f_0},0)\}$, which meets $\{(\frac{k^2\pi^2}{f_0},0)\}$ and $\infty$ in $\Sigma$. The proof is complete. \end{proof} \begin{lemma} \label{lem3.6} For each integer $k>0$ and each $\upsilon=+$, or $-$, there exists a continua $\hat{C}_{k}^{\upsilon}$ of $\Sigma$ in $\Phi_{k}^{\upsilon}\cup\{(\frac{k^2\pi^2}{f_\infty},\infty)\}$ coming from $\{(\frac{k^2\pi^2}{f_\infty},\infty)\}$, which meets $\{(\frac{k^2\pi^2}{f_0},0)\}$ or has an unbounded projection on $\mathbb{R}$. \end{lemma} \begin{proof} Let $T(\lambda,x)=\lambda Ax-\lambda L_{\infty}x$, where $L_{\infty}$ is defined as in \eqref{e3.4}. Then \eqref{e3.6} can be rewritten as $$x=\lambda L_{\infty}x+T(\lambda,x).\label{e3.10}$$ By Lemma \ref{lem3.4}, we know that $T(\lambda,x)$ is $o(\| x\|_1)$ for $x\in E$ near $\infty$ uniformly on bounded $\lambda$ intervals. Notice that $L_{\infty}$ is a compact linear map on $E$. By Lemma \ref{lem2.2} and Lemma \ref{lem2.3}, we know that for each $k>0$, $\frac{k^2\pi^2}{f_\infty}$ is a simple characteristic value of operator $L_{\infty}$. So with Lemma \ref{lem3.4}, \eqref{e3.10} can be considered as a bifurcation problem from infinity. A similar reasoning as in the proof of \cite[Theorem 2.4]{r3} shows that $\| x\|_1^{2}T(\lambda,\frac{x}{\| x\|_1^{2}})$ is compact. By Lemma \ref{lem1.2} and Lemma \ref{lem1.3}, $\Sigma$ contains a component $D_{k}$ which can be decomposed into two subcontinua $D_{k}^{+}, D_{k}^{-}$ which meets $(\frac{k^2\pi^2}{f_\infty},\infty)$. Next we show that for a smaller neighborhood $\mathcal{O}\subset\mathcal{M}$ of $(\frac{k^2\pi^2}{f_\infty},\infty)$, $(\lambda,x)\in D_{k}\cap\mathcal{O}$ and $(\lambda,x)\neq(\frac{k^2\pi^2}{f_\infty},\infty)$ implies that $x\in S_{k}$. In fact, by Lemma \ref{lem1.3} we already know that there exists a neighborhood $\mathcal{O}\subset\mathcal{M}$ of $(\frac{k^2\pi^2}{f_\infty},\infty)$ satisfying $(\lambda,x)\in D_{k}\cap\mathcal{O}$ and $(\lambda,x)\neq(\frac{k^2\pi^2}{f_\infty},\infty)$ implies $(\lambda,x)=(\lambda,\alpha u_{k}+w)$ where $\alpha>0(\alpha<0)$ and $|\lambda-\frac{k^2\pi^2}{f_\infty}|=o(1), \| w\|_1=o(|\alpha|)$ at $|\alpha|=\infty$. Since $S_{k}$ is open and $\frac{w}{\alpha}$ is small compared to $u_{k}\in S_{k}$ near $\alpha=+\infty$, and therefore $x=\alpha u_{k}+w\in S_{k}$ for $\alpha$ near $+\infty$. Therefore, $D_{k}\cap\mathcal{O}\subset(\mathbb{R}\times S_{k}) \cup(\frac{k^2\pi^2}{f_\infty},\infty)$. From the same reasoning it can be seen that $D_{k}^{\upsilon}\cap\mathcal{O}\subset(\mathbb{R} \times S_{k}^{\upsilon})\cup(\frac{k^2\pi^2}{f_\infty},\infty)$, where $\upsilon=+$, or $-$. Let $\hat{C}_{k}^{+}$ denote the maximal subcontinuum of $D_{k}^{+}$ lying in $\mathbb{R}\times S_{k}^{+}$. First suppose $\hat{C}_{k}^{+}\setminus\mathcal{O}$ is bounded. Then there exists $(\lambda,x)\in\partial \hat{C}_{k}^{+}$ with $x\in\partial S_{k}^{+}$. Hence $x$ has a double zero. By the assumption in section 2 we know $x\equiv0$. Thus there exists a sequence $(\gamma_{m},z_{m})\in\hat{C}_{k}^{+}$ satisfying \eqref{e3.10} with $z_{m}\to x\equiv0$ as $m\to+\infty$. By Lemma \ref{lem1.2}, like in the proof of Lemma \ref{lem3.5}, one can see that $\hat{C}_{k}^{+}$ meets $(\frac{k^2\pi^2}{f_0},0)$. Finally suppose $\hat{C}_{k}^{+}\setminus\mathcal{O}$ is unbounded. In this case we show $\hat{C}_{k}^{+}\setminus\mathcal{O}$ has an unbounded projection on $\mathbb{R}$. Suppose, on the contrary, that there exists a sequence $(\mu_{m},x_{m})\in\hat{C}_{k}^{+}\setminus\mathcal{O}$ with $\mu_{m}\to\mu$ and $\| x_{m}\|_1\to+\infty$ as $m\to+\infty$. Let $y_{m}:=\frac{x_{m}}{\| x_{m}\|_1},m\geq1$. From the fact that $$x_{m}=\mu_{m}L_{\infty}x_{m}+T(\mu_{m},x_{m}),$$ it follows that $$y_{m}=\mu_{m}L_{\infty}y_{m}+\frac{T(\mu_{m},x_{m})}{\| x_{m}\|_1}.\label{e3.11}$$ Notice that $L_{\infty}$ is compact. We may assume that there exists $y\in E$ with $\| y\|_1=1$ such that $\|y_{m}-y\|_1\to 0$ as $m\to+\infty$. Letting $m\to+\infty$ in \eqref{e3.11} and noticing $\frac{T(\mu_{m},x_{m})}{\| x_{m}\|_1}\to0$ as $m\to+\infty$ one obtains $$y=\mu L_{\infty}y.\label{e3.12}$$ Since $\mu\neq0$ is an eigenvalue of operator $L_{\infty}$ and $y\neq0$; that is, $\mu=\frac{i^2\pi^2}{f_\infty}$ for some $i\neq k$. Then by Lemma \ref{lem2.2} $y$ belongs to $S_{i}^{+}$ or $S_{i}^{-}$. Notice the fact that $\| y_{m}-y\|_1\to 0$ as $m\to+\infty$. Thus $x_{m}\in S_{i}^{+}$ or $S_{i}^{-}$ for $m$ sufficiently large since $S_{i}^{+}$ or $S_{i}^{-}$ are open. This is a contradiction with $x_{m}\in S_{k}^{+}(m\geq1)$. Thus $\hat{C}_{k}^{+}\setminus\mathcal{O}$ has an unbounded projection on $\mathbb{R}$. Similarly, the same argument works if $+$ is replaced by $-$ in the above cases. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm3.1}] Case 1. $\frac{k^2\pi^2}{f_0}<\lambda<\frac{k^2\pi^2}{f_\infty}$. Consider \eqref{e3.7} as a bifurcation problem from the trivial solution. To obtain the desired results we need only to show that $$C_k^\vartheta\cap(\{\lambda\}\times E)\neq\emptyset,\quad\vartheta=+, -.$$ By Lemma \ref{lem3.5} we know that $C_k^\vartheta$ joins $(\frac{k^2\pi^2}{f_0}, 0)$ to infinity $\Phi_k^\vartheta$. Therefore, there exists a sequence $(\mu_n, x_n)\in C_k^\vartheta$ such that $$\lim_{n\to \infty}(\mu_n+\|x_n\|_1)=\infty.$$ We note that $\mu_n>0$ for all $n\in\mathbb{N}$ since $(\lambda, x)=(0,0)$ is the unique solution of \eqref{e3.6} with $\lambda=0$ in $E$ and $C_k^\vartheta\cap(\{0\}\times E)=\emptyset$. If $$\lim_{n\to \infty}\mu_n=\infty,$$ then $$C_k^\vartheta\cap(\{\lambda\}\times E)\neq\emptyset.$$ Assume that there exists $M>0$, such that for all $n\in\mathbb{N}$, $$\mu_n\in(0,M].$$ In this case it follows that $\|x_n\|_1\to\infty$. We divide the equation $$x_n=\mu_n L_\infty x_n+T(\mu_n, x_n)$$ by $\|x_n\|_1$ and set $y_n=\frac{x_n}{\|x_n\|_1}$. Since $y_n$ is bounded in $E$, choosing a subsequence and relabelling, if necessary, we see that $y_n\to y$ for some $y\in E$ with $\|y\|_1=1$, and $$y=\mu L_\infty y$$ where $\mu=\lim_{n\to\infty}\mu_n$. From the proof of Lemma \ref{lem3.6} one can see $$\mu=\frac{k^2\pi^2}{f_\infty}.$$ Thus $C_k^\vartheta$ joins $(\frac{k^2\pi^2}{f_0},0)$ to $(\frac{k^2\pi^2}{f_\infty},\infty)$ which implies $$C_k^\vartheta\cap(\{\lambda\}\times E)\neq\emptyset.$$ Case 2. $\frac{k^2\pi^2}{f_\infty}<\lambda<\frac{k^2\pi^2}{f_0}$. Consider \eqref{e3.10} as a bifurcation problem from infinity. As above we need only to prove that $$\hat{C}_k^\vartheta\cap(\{\lambda\}\times E)\neq\emptyset,\quad \vartheta=+, -.$$ From Lemma \ref{lem3.6}, we know that $\hat{C}_k^\vartheta$ comes from $(\frac{k^2\pi^2}{f_\infty},\infty)$ meets $(\frac{k^2\pi^2}{f_0},0)$ or has an unbounded projection on $\mathbb{R}$. If it meets $(\frac{k^2\pi^2}{f_0},0)$, then the connectedness of $\hat{C}_k^\vartheta$ and $\frac{k^2\pi^2}{f_0}>\lambda$ guarantee that $$\hat{C}_k^\vartheta\cap(\{\lambda\}\times E)\neq\emptyset,\quad \vartheta=+, -.$$ If $\hat{C}_k^\vartheta$ has an unbounded projection on $\mathbb{R}$, notice that $(\lambda,x)=(0,0)$ is the unique solution of \eqref{e3.6}, so $$\hat{C}_k^\vartheta\cap(\{\lambda\}\times E)\neq\emptyset,\quad \vartheta=+, -.$$ The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm3.2}] Repeating the arguments used in the proof in Theorem \ref{thm3.1}, we see that for each $\vartheta\in\{+, -\}$ and each $i\in\{1,2,\dots,j\}$ $$( C_{k+i}^\vartheta\cup\hat{C}_{k+i}^\vartheta) \cap(\{\lambda\}\times E)\neq\emptyset.$$ \end{proof} \begin{remark} \label{rmk1} \rm From \cite{c3}, we know if $\beta=0$, we have also the same results as Theorem \ref{thm3.1} and Theorem \ref{thm3.2}. \end{remark} \section{Applications} In this section, we give an example to illustrate the applications of Theorem \ref{thm3.2}. \begin{example} \label{examp4.1}\rm Consider the second-order impulsive differential equation $$\begin{gathered} x''(t)+f(x(t))=0,\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{x=\frac{1}{2}}=x(\frac{1}{2}),\\ \Delta x'|_{x=\frac{1}{2}}=-x'(\frac{1}{2}-0),\\ x(0)=x(1)=0, \end{gathered}\label{e4.1}$$ where $f(x)=100\sin x+5x$. It is not difficult to see that $f$ here satisfies the assumption in section 2 and (H1) with $f_0=105, f_\infty=5$ and the eigenvalues of the boundary-value problem \begin{gather*} x''(t)+\lambda x(t)=0,\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{x=\frac{1}{2}}=\lambda x(\frac{1}{2}),\\ \Delta x'|_{x=\frac{1}{2}}=-\lambda x'(\frac{1}{2}-0),\\ x(0)=x(1)=0, \end{gather*} can be written as $\lambda_k=k^2\pi^2$, $k=1,2,\dots$. By calculation, we know that there exist $k=1$ and $j=2$ such that $$\frac{(k+j)^2\pi^2}{f_0}<1<\frac{k^2\pi^2}{f_\infty}.$$ Therefore, Theorem \ref{thm3.2} guarantees that \eqref{e4.1} has at least six nontrivial solutions. \end{example} \subsection*{Acknowledgements} We thank the anonymous referees for their suggestions. \begin{thebibliography}{00} \bibitem{a1} R. P. Agarwal, D. O'Regan; \emph{Multiple nonnegative solutions for second-order impulsive differential equations}, Appl. Math. Comput.114 (2000), 51-59. \bibitem{b1} M. S. 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