\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 55, pp. 1--23.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2016/55\hfil Global solutions to boundary-value problems] {Global structure of solutions to boundary-value problems of impulsive differential equations} \author[Y. Niu, B. Yan \hfil EJDE-2016/55\hfilneg] {Yanmin Niu, Baoqiang Yan} \address{Yanmin Niu \newline School of Mathematical Sciences, Shandong Normal University, Jinan 250014, China} \email{1398958626@qq.com} \address{Baoqiang Yan (corresponding author) \newline School of Mathematical Sciences, Shandong Normal University, Jinan 250014, China} \email{yanbqcn@aliyun.com} \thanks{Submitted January 5, 2016. Published February 25, 2016.} \subjclass[2010]{34B09, 34B15, 34B37} \keywords{Comparison arguments; eigenvalues; global bifurcation theorem; \hfill\break\indent multiple solutions; asymptotical behavior of solutions} \begin{abstract} In this article, we study the structure of global solutions to the boundary-value problem \begin{gather*} -x''(t)+f(t,x)=\lambda ax(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{t=1/2}=\beta_1 x(\frac{1}{2}),\quad \Delta x'|_{t=1/2}=-\beta_{2} x(\frac{1}{2}),\\ x(0)=x(1)=0, \end{gather*} where $\lambda\neq0$, $\beta_1\geq\beta_{2}\geq0$, $\Delta x|_{t=1/2}=x(\frac{1}{2}+0)-x(\frac{1}{2})$, $\Delta x'|_{t=1/2}=x'(\frac{1}{2}+0)-x'(\frac{1}{2}-0)$, and $f:[0,1]\times\mathbb{R}\to\mathbb{R}$, $a:[0,1]\to(0,+\infty)$ are continuous. By a comparison principle and spectral properties of the corresponding linear equations, we prove the existence of solutions by using Rabinowitz-type global bifurcation theorems, and obtain results on the behavior of positive solutions for large $\lambda$ when $f(x)=x^{p+1}$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Introduction} In this article, we study the structure of global solutions to the second-order impulsive differential equation \begin{equation} \begin{gathered} -x''(t)+f(t,x)=\lambda ax(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{t=1/2}=\beta_1 x(\frac{1}{2}),\quad \Delta x'|_{t=1/2}=-\beta_{2} x(\frac{1}{2}),\\ x(0)=x(1)=0, \end{gathered}\label{e1.1} \end{equation} where $\lambda\neq0$, $\beta_1\geq\beta_{2}\geq0$, $\Delta x|_{t=1/2}=x(\frac{1}{2}+0)-x(\frac{1}{2})$, $\Delta x'|_{t=1/2}=x'(\frac{1}{2}+0)-x'(\frac{1}{2}-0)$, and $f:[0,1]\times\mathbb{R}\to\mathbb{R}$, $a:[0,1]\to(0,+\infty)$ are continuous. Impulsive differential equations arise in the contexts of population dynamics, infectious diseases models, chemical technology and so on; see \cite{c2,f1,l3,o1,s3,t1,y2}. Because impulsive equations appear in applied mathematics, they attract a lot of attention. Many authors studied the existence, uniqueness as well as multiplicity of solutions, by using the topological degree theory and variational method; see \cite{a1,c3,l1,l2,l5,l6,n1,n2,s1,s2,t2,x1,x2,y1,y3,z1}. An important tool to study the existence of solutions for differential equations is Rabinowitz global bifurcation theory; see \cite{b1,c3,d1,m1,m2,m3,r1,r2,r3}. But up to now, just a few results have shown on the structure of global solutions for impulsive differential equations by Rabinowitz global bifurcation theorems. Liu and O'Regan \cite{l6} considered the second-order impulsive differential equation \begin{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_ix(t_i-0), \quad i=1,2,\ldots,k,\\ x(0)=x(1)=0, \end{gathered} \label{e1.2} \end{equation} in which they converted \eqref{e1.2} to the form \begin{equation} \begin{gathered} y''(t)+\frac{r}{\Pi_{00$, $m>0$ are real numbers, and give results on the behavior of positive solutions for the large $\lambda$. Some ideas come from \cite{f1}. \section{Preliminaries} Comparison argument in ODE plays an important role in analyzing some properties of solutions. Naturally, it necessary to study the relative Comparison argument to get the properties of solutions of \eqref{e1.1}. \begin{lemma} \label{lem2.1} Suppose that $y(t)$ is the solution of \begin{equation} \begin{gathered} y''+Q(t)y=0,\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta y|_{t=1/2}=\beta_1 y(\frac{1}{2}),\quad \Delta y'|_{t=1/2}=-\beta_{2} y(\frac{1}{2}), \end{gathered}\label{e2.1} \end{equation} and $w(t)$ is the solution of \begin{equation} \begin{gathered} w''+P(t)w=0,\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta w|_{t=1/2}=\beta_1 w(\frac{1}{2}),\quad \Delta w'|_{t=1/2}=-\beta_{2} w(\frac{1}{2}),\\ \end{gathered}\label{e2.2} \end{equation} where $P(t)$ and $Q(t)$ are continuous on the same interval $[0,1]$ with $Q(t)\leq P(t)$. If $\alpha$, $\beta$ are two next null points of $y(t)$, then there must be at least one null point of $w(t)$ on $[\alpha,\beta]$. \end{lemma} \begin{proof} Suppose $\alpha$, $\beta$ are two next null points of $y(t)$. We discuss two cases about them. (1) $\alpha$, $\beta\in[0,1/2)$ or $\alpha$, $\beta\in(\frac{1}{2},1]$. In this case, it is easy to obtain the results by Sturm comparison theorem in ODE. (2) $\alpha<\frac{1}{2}<\beta$. Multiply \eqref{e2.1} by $w(t)$, \eqref{e2.2} by $y(t)$ and subtract each other to obtain $$ wy''-w''y+[Q(t)-P(t)]wy=0. $$ Integrating the equation above from $\alpha$ to $\beta$ yields $$ \int_{\alpha}^{\beta}(wy''-w''y)\,dt =\int_{\alpha}^{\beta}[P(t)-Q(t)]wy\,dt; $$ therefore, \begin{equation} \begin{aligned} &\int_{\alpha}^{\frac{1}{2}-}d(wy'-w'y) +\int_{\frac{1}{2}+}^{\beta}d(wy'-w'y)\\ &=\int_{\alpha}^{\frac{1}{2}-}[P(t)-Q(t)]wy\,dt +\int_{\frac{1}{2}+}^{\beta}[P(t)-Q(t)]wy\,dt.\label{e2.3} \end{aligned} \end{equation} From the impulsive conditions in \eqref{e2.1} and \eqref{e2.2}, we have \begin{gather*} w(\frac{1}{2}+)=(1+\beta_1)w(\frac{1}{2}-),\\ w'(\frac{1}{2}+)=w'(\frac{1}{2}-)-\beta_{2} w(\frac{1}{2}-),\\ y(\frac{1}{2}+)=(1+\beta_1)y(\frac{1}{2}-),\\ y'(\frac{1}{2}+)=y'(\frac{1}{2}-)-\beta_{2} y(\frac{1}{2}-),\\ \end{gather*} which together with $y(\alpha)=y(\beta)=0$ guarantees that the left side of \eqref{e2.3} satisfies \begin{equation} -\beta_1\Big[w(\frac{1}{2}-)y'(\frac{1}{2}-) -w'(\frac{1}{2}-)y(\frac{1}{2}-)\Big] -w(\alpha)y'(\alpha)+w(\beta)y'(\beta). \label{e2.4} \end{equation} Denote $$ F(t)=w(t)y'(t)-w'(t)y(t). $$ Combining \eqref{e2.1} with \eqref{e2.2}, one has $$ F'(t)=w(t)y''(t)-w''(t)y(t)=[P(t)-Q(t)]w(t)y(t). $$ Suppose that $w(t)$ has no zero in $(\alpha,\beta)$, and without loss of generality $y(t)$ is positive in $(\alpha,\beta)$. Then two cases will be discussed according to $w(t)$. (a) $w(t)>0$, $t\in(\alpha,\beta)$. In this case, the fact that $F'(t)=[P(t)-Q(t)]w(t)y(t)>0$ for $t\in[\alpha,\frac{1}{2})$ implies that $F(t)$ increases on $[\alpha,\frac{1}{2})$. Since $$ F(\alpha)=w(\alpha)y'(\alpha)-w'(\alpha)y(\alpha)=w(\alpha)y'(\alpha)>0, $$ one has $F(\frac{1}{2}-)>F(\alpha)>0$. Then \eqref{e2.4} is $$ -\beta_1F(\frac{1}{2})-w(\alpha)y'(\alpha)+w(\beta)y'(\beta)<0, $$ while the right-hand side of \eqref{e2.3} satisfies $$ \int_{\alpha}^{1/2-}[P(t)-Q(t)]wy dt +\int_{1/2+}^{\beta}[P(t)-Q(t)]wydt>0, $$ which is a contradiction. (b) $w(t)<0$, $t\in(\alpha,\beta)$. Since $F'(t)=[P(t)-Q(t)]w(t)y(t)<0$ for $t\in[\alpha,\frac{1}{2})$, $F(t)$ decreases on $[\alpha,\frac{1}{2})$. With $F(\alpha)=w(\alpha)y'(\alpha)<0$, one has $F(\frac{1}{2}-)<0$. Then \eqref{e2.4} is $$-\beta_1F(\frac{1}{2})-w(\alpha)y'(\alpha)+w(\beta)y'(\beta)>0,$$ while the right-hand side of \eqref{e2.3} satisfies $$ \int_{\alpha}^{\frac{1}{2}-}[P(t)-Q(t)]wydt +\int_{\frac{1}{2}+}^{\beta}[P(t)-Q(t)]wydt<0, $$ which is a contradiction. The proof is complete. \end{proof} For convenience, we introduce main lemma and some symbols in global bifurcation theory. Let $\mathcal{F}:\mathcal{E}\to\mathcal{E}_1$ where $\mathcal{E}$ and $\mathcal{E}_1$ are real Banach spaces and $\mathcal{F}$ is continuous. Suppose that the equation $\mathcal{F}(U)=0$ possesses a simple curve of solutions $\Psi$ given by $\{U(t)|t\in[a,b]\}$. If for some $\tau\in(a,b)$, $\mathcal{F}$ possesses zeros not lying on $\Psi$ in every neighborhood of $U(\tau)$, then $U(\tau)$ is said to be a bifurcation point for $\mathcal{F}$ with respect to the curve $\Psi$ (see \cite{r2}). A special family of such equations has the form \begin{equation} u=G(\lambda, u),\label{e2.5} \end{equation} where $\lambda\in\mathbb{R}$, $u\in E$ is a real Banach space with the norm $\|\cdot\|$, and $G: \mathcal{E}\equiv\mathbb{R}\times E\to E$ is compact and continuous. In addition, $G(\lambda,u)=\lambda Lu+H(\lambda,u)$, where $H(\lambda,u)$ is $o(\|u\|)$ for $u$ near 0 uniformly on bounded $\lambda$ intervals and $L$ is a compact linear map on $E$. A solution of \eqref{e2.5} is a pair $(\lambda,u)\in\mathcal{E}$. The known curve of solutions $\Theta=\{(\lambda,0)|\lambda\in\mathbb{R}\}$ will henceforth be referred to as the trivial solutions. The closure of the set of nontrivial solutions of \eqref{e2.5} will be denoted by $\Sigma$. A component of $\Sigma$ is a maximal closed connected subset. If there exist $\mu\in\mathbb{R}$ and $0\neq v\in E$ such that $v=\mu Lv$, $\mu $ is said to be a real characteristic value of $L$. The set of real characteristic values of $L$ will be denoted by $\sigma(L)$. The multiplicity of $\mu\in\sigma(L)$ is the dimension of $\cup_{j=1}^\infty N((\mu L-I)^j)$ where $I$ is the identity map on $E$ and $N(P)$ denotes the null space of $P$. Since $L$ is compact, $\mu$ is of finite multiplicity. It is well known that if $\mu\in\mathbb{R}$, a necessary condition for $(\mu,0)$ to be a bifurcation point of \eqref{e2.5} with respect to $\Theta$ is that $\mu\in\sigma(L)$. \begin{lemma}[\cite{r2}] \label{lem2.2} If $\mu\in\sigma(L)$ is simple, then $\Sigma$ contains a component $\mathbf {C}_{\mu}$ that can be decomposed into two subcontinua $\mathbf {C}_{\mu}^{+}$, $\mathbf {C}_{\mu}^{-}$ such that for some neighborhood $\mathbf {B}$ of $(\mu,0)$, $$ (\lambda,u)\in\mathbf {C}_{\mu}^{+}(\mathbf {C}_{\mu}^{-})\cap\mathbf {B},\quad \text{and}\quad (\lambda,u)\neq(\mu,0) $$ implies $(\lambda,u)=(\lambda,\alpha v+w)$ where $\alpha>0 (\alpha<0)$ and $\mid{\lambda-\mu}\mid=o(1)$, $\| w\|=o(\mid\alpha\mid)$, at $\alpha=0$. Moreover, each of $\mathbf{C}_{\mu}^{+}$, $\mathbf{C}_{\mu}^{-}$ 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 $(\lambda,u)$, $(\lambda,-\mu)$, $u\neq0$. \end{itemize} \end{lemma} \section{Spectral properties for linear impulsive equations} To apply Rabinowitz global bifurcation theorems on \eqref{e1.1}, we need to study the linear equation corresponding to \eqref{e1.1}. In this section the spectral properties including its eigenvalues, along with their corresponding algebraic multiplicity, and also eigenfunctions structures will be discussed for the linear impulsive problems \begin{equation} \begin{gathered} -x''(t)=\lambda a(t)x(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{t=1/2}=\beta_1 x(\frac{1}{2}),\quad \Delta x'|_{t=1/2}=-\beta_{2} x(\frac{1}{2}),\\ x(0)=x(1)=0, \end{gathered}\label{e3.1} \end{equation} where $\lambda$, $\beta_1$, $\beta_{2}$, $a(t)$, $\Delta x$, $\Delta x'$ are introduced as in \eqref{e1.1}. Denote $PC[0,1]=\{x:[0,1]\to\mathbb{R}:x(t)$ 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 $$ \| x\|=\sup_{t\in[0,1]}| x(t)|, $$ and $PC'[0,1]=\{x\in PC[0,1]:x'(t)$ 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 $\|\cdot\|_1$. \begin{lemma}[\cite{g2}] \label{lem3.1} The solution $x(t)\in PC[I,\mathbb{R}]\cap C^{2}[I',\mathbb{R}]$ of \eqref{e3.1} is the same as the solution $x(t)\in PC'[I,\mathbb{R}]$ of the integral equation \begin{equation} x(t)=\begin{cases} \lambda\int_{0}^{1}G(t,s)a(s)x(s)ds -\big(\beta_1-\frac{1}{2}\beta_{2}\big) tx(\frac{1}{2}), &t\in[0,\frac{1}{2}],\\ \lambda\int_{0}^{1}G(t,s)a(s)x(s)ds +\big(\beta_1+\frac{1}{2}\beta_{2}\big) (1-t)x(\frac{1}{2}), &t\in(\frac{1}{2},1], \end{cases} \label{e3.2} \end{equation} where $I=[0,1]$, $I'=I\setminus{\{\frac{1}{2}\}}$, and $$ G(t,s)=\begin{cases} s(1-t), & 0\leq s\leq t\leq 1,\\ t(1-s), & 0\leq t\leq s\leq 1. \end{cases} $$ \end{lemma} \begin{lemma} \label{lem3.2} All eigenvalues of \eqref{e3.1} are in $\mathbb{R}$. \end{lemma} \begin{proof} Let $\lambda=\alpha+i\gamma$ be an eigenvalue of \eqref{e3.1} and $u(t)=u_1(t)+iu_{2}(t)$ be an eigenfunction corresponding to $\lambda$. It is sufficient to prove that $\gamma=0$. Since $\lambda=\alpha+i\gamma$ and $u(t)=u_1(t)+iu_{2}(t)$ satisfy \eqref{e3.1}, it follows that \begin{equation} -u''_1=a(t)(\alpha u_1-\gamma u_{2}),\quad -u''_{2}=a(t)(\alpha u_{2}+\gamma u_1),\label{e3.3} \end{equation} and \begin{equation} \begin{gathered} \Delta u_1|_{t=1/2}=\beta_1 u_1(\frac{1}{2}),\quad \Delta u'_1|_{t=1/2}=-\beta_{2} u_1(\frac{1}{2}),\\ \Delta u_{2}|_{t=1/2}=\beta_1 u_{2}(\frac{1}{2}),\quad \Delta u'_{2}|_{t=1/2}=-\beta_{2} u_{2}(\frac{1}{2}),\\ u_1(0)=u_1(1)=0,\quad u_{2}(0)=u_{2}(1)=0. \end{gathered}\label{e3.4} \end{equation} Multiply \eqref{e3.3} by $u_{2}(t)$ and $u_1(t)$ respectively and subtract each other to yield that $$ -u''_1u_{2}+u''_{2}u_1=-a(t)\gamma(u^{2}_1+u^{2}_{2}). $$ Integrating the equation above from $0$ to $1$ and with condition \eqref{e3.4}, we have \begin{equation} \beta_1 u'_1(\frac{1}{2}-)u_{2}(\frac{1}{2}) -\beta_1 u'_{2}(\frac{1}{2}-)u_1(\frac{1}{2}) =-\int_{0}^{1}a(t)\gamma(u^{2}_1+u^{2}_{2})dt.\label{e3.5} \end{equation} Denote $F(t)=u'_1(t)u_{2}(t)-u'_{2}(t)u_1(t)$ for $t\in[0,\frac{1}{2})$. Since $u_1$, $u_{2}$ satisfy \eqref{e3.3}, it follows that $$ F'(t)=a(t)\gamma(u^{2}_1(t)+u^{2}_{2}(t)). $$ If $\gamma>0$, by $a(t)>0$ and $u^{2}_1(t)+u^{2}_{2}(t)>0$, one has $F'(t)>0$. Then $F(0)=u'_1(0)u_{2}(0)-u'_{2}(0)u_1(0)=0$ implies that $F(\frac{1}{2}-)>0$. The left side of \eqref{e3.5} is $\beta_1 F(\frac{1}{2}-)>0$, while the right side of it is negative, which is a contradiction. On the other hand, with the similar discussion, we deduce the contradiction if $\gamma<0$. Therefore, $\gamma=0$ and Lemma \ref{lem3.2} holds. \end{proof} \begin{theorem} \label{thm3.1} Equation \eqref{e3.1} possesses an increasing sequence of eigenvalues $$ 0<\lambda_1<\lambda_{2}<\dots<\lambda_{k}<\dots, \lim_{k\to+\infty}\lambda_k=+\infty. $$ And the eigenfunction $u_k$ corresponding to $\lambda_k$ has exactly $k-1$ nodal zeros in $(0,1)$. \end{theorem} The proof of this theorem will be divided into two parts, in which the methods chosen are different due to the complexity of the impulse. Part 1 mainly shows the existence of principal eigenvalue and eigenfunction by Schauder's fixed point theorem, while Part 2 guarantees the existence of second and subsequent eigenvalues by some tools on uncontinuous Sturm-Liouville systems. \smallskip \noindent\textbf{Part 1. Existence of principal eigenvalue and eigenfunction} In \eqref{e3.2}, let $$ y(t)=\begin{cases} x(t)+(\beta_1-\frac{1}{2}\beta_{2})tx(\frac{1}{2}), &t\in[0, \frac{1}{2}],\\[4pt] x(t)-(\beta_1+\frac{1}{2}\beta_{2})(1-t)x(\frac{1}{2}),&t\in(\frac{1}{2}, 1], \end{cases} $$ and then $y(t)\in C^{1}[0,1]$. Equation \eqref{e3.2} is transformed into \begin{equation} y(t)=\lambda\int_{0}^{1}G(t,s)a(s)H(y(s))ds,\label{e3.6} \end{equation} where \begin{equation} H(y(t))=\begin{cases} y(t)-\frac{\beta_1-\frac{1}{2}\beta_{2}}{1+\frac{1}{2}\beta_1 -\frac{1}{4}\beta_{2}}y(\frac{1}{2})t, & t\in[0,\frac{1}{2}],\\[4pt] y(t)+\frac{\beta_1+\frac{1}{2}\beta_{2}}{1+\frac{1}{2}\beta_1 -\frac{1}{4}\beta_{2}}y(\frac{1}{2})(1-t), & t\in(\frac{1}{2},1]. \end{cases} \label{e3.7} \end{equation} By Lemma \ref{lem3.1}, the study of \eqref{e3.1} is the same as that of \eqref{e3.6}. Next we give the main result. \begin{lemma} \label{lem3.3} There exist $\lambda^{\ast}>0$ and nonnegative function $x^{\ast}$ satisfying \eqref{e3.1}. \end{lemma} \begin{proof} Let $D:=\{y\in C[0,1]|\int_{0}^{1}y(s)ds=1,y(t)\geq0,y(t)$ is concave and $y(0)=y(1)=0\}$ and define an operator $A$: $$ (Ay)(t)=\frac{\int_{0}^{1}G(t,s)a(s)H(y(s))ds} {\int_{0}^{1}\int_{0}^{1}G(t,s)a(s)H(y(s))\,ds\,dt},\quad \forall y\in D. $$ We show $(Ay)(t)>0$. In fact, by \eqref{e3.7}, the sign of $(Ay)(t)$ depends on the sign of $H(y(t))$. Since $y\in D$ and $\beta_1\geq\beta_{2}\geq0$, it is easy to obtain that for $t\in(\frac{1}{2},1]$ $$ y(t)+\frac{\beta_1+\frac{1}{2}\beta_{2}}{1+\frac{1}{2} \beta_1-\frac{1}{4}\beta_{2}}y(\frac{1}{2})(1-t)>0. $$ Define $$ g(t)=y(t)-\frac{\beta_1 -\frac{1}{2}\beta_{2}}{1+\frac{1}{2}\beta_1 -\frac{1}{4}\beta_{2}}y(\frac{1}{2})t, \quad t\in[0,\frac{1}{2}]. $$ Since $y(t)$ is concave, together with $g(0)=y(0)\geq0$ and $$ g(\frac{1}{2})=\frac{1}{1+\frac{1}{2}\beta_1-\frac{1}{4}\beta_{2}}y(\frac{1}{2}) \geq 0, $$ one has $g(t)\geq0$ for $t\in[0,\frac{1}{2}]$. That is to say $(Ay)(t)\geq0$ for $t\in[0,1]$. Thus, we conclude that $D$ is a closed convex set and an standard argument shows that $A:D\to D$ is continuous. Furthermore, the following two facts show that $A(D)$ is compact. (1) $A(D)$ is uniformly bounded. For $y\in D$, since $\int_0^1y(t)dt\leq \|y\|$, one has $\|y\|\geq 1$. Choose $y(t_0)=\max_{t\in[0,1]}\|y\|$. For $t\in (0,t_0)$, we have $y(t)=y(\frac{t}{t_0}t_0+(1-\frac{t}{t_0})0) \geq \frac{t}{t_0}y(t_0)+(1-\frac{t}{t_0})y(0)=\frac{t}{t_0}\|y\|$. The same argument shows that for $t\in (t_0,1)$, we have $y(t)\geq \frac{1-t}{1-t_0}\|y\|$. Hence, $$ y(t)\geq \frac{t(1-t)}{t_0(1-t_0)}\|y\|\geq t(1-t)\|y\|\geq t(1-t),\quad \forall t\in[0,1] $$ and $$ \int_0^1y(s)ds\geq\int_0^1\|y\|s(1-s)ds=\|y\|\frac{1}{6}, \text{ i.e.} \|y\|\leq 6. $$ Next we estimate $$ \int_{0}^{1}\int_{0}^{1}G(t,s)a(s)H(y(s))\,ds\,dt,\quad \forall y\in D. $$ As $a(t)>0$ is continuous, suppose $$ 00$, there exists $\delta>0$ such that $$ |G(t,s)-G(t',s)|0$ such that $\rho(\lambda)=0$ for all $\lambda<-A$; \item[(2)] $\lim_{\lambda\to+\infty}\rho(\lambda)=+\infty$ \end{itemize} \end{lemma} \begin{proof} Denote $M_{-}=\min_{t\in[0,\frac{1}{2}]}a(t)$ and $M_{+}=\min_{t\in[\frac{1}{2},1]}a(t)$. (i) When $\lambda<0$, we first consider the ODE initial value problem \begin{equation} \begin{gathered} x''(t)+\lambda M_{-}x(t)=0,\quad t\in[0,\frac{1}{2}],\\ x(0,\lambda)=0,x'(0,\lambda)=-1, \end{gathered}\label{e3.15} \end{equation} and its solution can be written as $$ x_{-}(t,\lambda)=\frac{-1}{2\sqrt{-\lambda M_{-}}}e^{\sqrt{-\lambda M_{-}}t} +\frac{1}{2\sqrt{-\lambda M_{-}}}e^{-\sqrt{-\lambda M_{-}}t}. $$ The above expression implies that $x_{-}(t,\lambda)<0$ as $\lambda\to-\infty$, which means $\rho_{-}(\lambda)=0$ by comparison theorem of ODE. Next, for the another ODE equation, \begin{equation} \begin{gathered} x''(t)+\lambda M_{+}x(t)=0,\quad t\in[\frac{1}{2},1],\\ x\big(\frac{1}{2}+,\lambda\big)=(1+\beta_1)x \big(\frac{1}{2}-,\lambda\big),x'\big(\frac{1}{2}+,\lambda\big) =x'\big(\frac{1}{2}-,\lambda\big)-\beta_{2} x\big(\frac{1}{2}-,\lambda\big), \end{gathered}\label{e3.16} \end{equation} the solution is \begin{align*} &x_{+}(t,\lambda)\\ &=\Big(\frac{1+\beta_1}{2}x_{-}\big(\frac{1}{2}-,\lambda\big) +\frac{-\beta_{2} x_{-}\big(\frac{1}{2}-,\lambda\big) +x'_{-}(\frac{1}{2}-,\lambda)}{2\sqrt{-\lambda M_{+}}}\Big) e^{\sqrt{-\lambda M_{+}}(t-\frac{1}{2})} \\ &\quad +\Big(\frac{1+\beta_1}{2}x_{-}\big(\frac{1}{2}-,\lambda\big) -\frac{-\beta_{2} x_{-}\big(\frac{1}{2}-,\lambda\big) +x'_{-}\big(\frac{1}{2}-,\lambda\big)}{2\sqrt{-\lambda M_{+}}}\Big) e^{-\sqrt{-\lambda M_{+}}(t-\frac{1}{2})}, \end{align*} from which we see $x_{+}(t,\lambda)<0$ as $\lambda\to-\infty$ and we conclude that $\rho_{+}(\lambda)=0$. \smallskip (ii) When $\lambda>0$, denote $M=\min\{M_{-},M_{+}\}$ and consider \eqref{e3.15}. A solution of it has the form $$ x_{-}(t,\lambda)=\frac{-1}{\sqrt{\lambda M_{-}}}\sin\sqrt{\lambda M_{-}}t, $$ of which the number of zeros on $(0,\frac{1}{2})$ is $[\frac{\frac{1}{2}-0}{\pi}\sqrt{\lambda M_{-}}]$ or $[\frac{\frac{1}{2}-0}{\pi}\sqrt{\lambda M_{-}}]+1$. Then we turn to \eqref{e3.16}, while the solution is \begin{align*} x_{+}(t,\lambda) &=(1+\beta_1)x_{-}\big(\frac{1}{2}-,\lambda\big) \cos\sqrt{\lambda M_{+}}\big(t-\frac{1}{2}\big)\\ &\quad -\frac{1}{\sqrt{\lambda M_{+}}}\Big(-\beta_{2}x_{-} \big(\frac{1}{2}-,\lambda\big)+x'_{-}\big(\frac{1}{2}-,\lambda\big)\Big) \sin\sqrt{\lambda M_{+}}\big(t-\frac{1}{2}\big)\\ &= C\sin\left(\sqrt{\lambda M_{+}}t+\gamma\right). \end{align*} We know the number of zeros of $x_{+}(t,\lambda)$ on $(\frac{1}{2},1)$ is $[\frac{1-\frac{1}{2}}{\pi}\sqrt{\lambda M_{+}}]$ or $[\frac{1-\frac{1}{2}}{\pi}\sqrt{\lambda M_{+}}]+1$. Notice that $$ \big[\frac{\frac{1}{2}-0}{\pi}\sqrt{\lambda M_{-}}\big] +\big[\frac{1-\frac{1}{2}}{\pi}\sqrt{\lambda M_{+}}\big] >\big[\frac{\frac{1}{2}-0}{\pi}\sqrt{\lambda M}\big] +\big[\frac{1-\frac{1}{2}}{\pi}\sqrt{\lambda M}\big]. $$ By a comparison argument, when $\lambda>M$, the number of zeros of $\zeta(t,\lambda)$ is no less than $[\frac{\frac{1}{2}-0}{\pi}\sqrt{\lambda M}] +[\frac{1-\frac{1}{2}}{\pi}\sqrt{\lambda M}]$, which prove (2). We complete the proof. \end{proof} Denote by $\mathfrak{R}_{k}$ ($k=1,2,\dots$) the point set on $\mathbb{R}$: $$ \mathfrak{R}_1=\{\lambda\in\mathbb{R}:\rho(\lambda)=0\},\quad \mathfrak{R}_{k}=\{\lambda\in\mathbb{R}:\rho(\lambda)\geq k-1>0\}. $$ According to Lemma \ref{lem3.4} and the continuity of $\zeta_{\pm}(t,\lambda)$, $\mathfrak{R}_{k}$ $(k=1,2,\dots)$ are nonempty and $\mathfrak{R}_{k}(k>1)$ are closed sets with lower bounds. \begin{lemma} \label{lem3.5} Denote $\mu_{k}=\min{\mathfrak{R}_{k}(k=2,\dots )}$. If $\zeta(1,\mu_{k})=0$, then $\zeta(t,\mu_{k})$ has exactly $k-1$ zeros in $(0,\frac{1}{2})\cup(\frac{1}{2},1)$. \end{lemma} \begin{proof} Since $\mu_{k}=\min{\mathfrak{R}_{k}(k=2,\dots )}$, it follows that $\rho(\mu_{k})\geq k-1$, which means that $\zeta(t,\mu_{k})$ has at least $k-1$ zeros, denoting by $00$ and notice \eqref{e3.17} when $k=1$. With $\lambda_1>0$, $u_1(t)>0$ for $t\in(0,1)$, two cases of $y$ will be discussed. If $y(t)$ has no zero in $(0,1)$, assume $y(t)>0$ and rewrite \eqref{e3.17} in the form \begin{equation} \begin{gathered} y''+\lambda_1a(t)\big[1+\frac{\gamma u_1(t)}{y(t)}\big]y(t)=0, \quad t\in(0,1), t\neq\frac{1}{2},\\ \Delta y|_{t=1/2}=\beta_1 y(\frac{1}{2}),\quad \Delta y'|_{t=1/2}=-\beta_{2} y(\frac{1}{2}),\\ y(0)=y(1)=0. \end{gathered}\label{e3.18} \end{equation} Inequality $1+\frac{\gamma u_1(t)}{y(t)}>1$, and Lemma \ref{lem2.1}, guarantee that $y(t)$ has a zero in $(0,1)$, which contradicts with the assumption. If $y(t)$ has a zero in $(0,1)\backslash{\frac{1}{2}}$, denote it by $t^{\ast}$, so that $y(t)>0$ for $t\in(0,t^{\ast})$ and $y(t)<0$ for $t\in(t^{\ast},1)$ (such $t^{\ast}$ always exists because it is a nodal zero). So \eqref{e3.17} on $(t^{\ast},1)$ can be transformed into \eqref{e3.18}. From $1+\frac{\gamma u_1(t)}{y(t)}<1$ and Lemma \ref{lem2.1}, $u_1(t)$ has a zero in $(t^{\ast},1)$, which yields a contradiction. So $\gamma=0$ and $y\in ker(I-\lambda_{k}K)$. We complete the proof. \end{proof} \section{Multiple solutions for nonlinear impulsive differential Equation} In this section, we consider the problem\ \begin{equation} \begin{gathered} -x''(t)+f(t,x)=\lambda ax(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{t=1/2}=\beta_1 x(\frac{1}{2}),\quad \Delta x'|_{t=1/2}=-\beta_{2} x(\frac{1}{2}),\\ x(0)=x(1)=0, \end{gathered}\label{e4.1} \end{equation} where $\lambda\neq0$, $\beta_1\geq\beta_{2}\geq0$, $\Delta x|_{t=1/2}=x(\frac{1}{2}+)-x(\frac{1}{2}-)$, $\Delta x'|_{t=1/2}=x'(\frac{1}{2}+)-x'(\frac{1}{2}-)$, and $f:[0,1]\times\mathbb{R}\to\mathbb{R}$, $a:[0,1]\to(0,+\infty)$ are continuous. In addition, suppose the following assumptions of $f(t,x)$ hold. \begin{equation} f(t,0)=0,f(t,s)=o(s) \label{e4.2} \end{equation} at the neighbourhood of $0$ and uniformly for all $t\in[0,1]$. \begin{equation} s\mapsto\frac{f(t,s)}{s}, \quad s\mapsto-\frac{f(t,-s)}{s} \label{e4.3} \end{equation} strictly increase on $\mathbb{R^{+}}$, for all $t\in[0,1]$. \begin{equation} \lim_{s\to\pm\infty}\frac{f(t,s)}{s}=+\infty, \label{e4.4} \end{equation} uniformly for all $t\in[0,1]$. Denote $\lambda_i$ be the $i^{th}$ eigenvalue of problem \begin{equation} \begin{gathered} -\phi''(t)=\lambda a(t)\phi(t),\quad t\in(0,1),\quad t\neq\frac{1}{2},\\ \Delta \phi|_{t=1/2}=\beta_1 \phi(\frac{1}{2}),\quad \Delta \phi'|_{t=1/2}=-\beta_{2} \phi(\frac{1}{2}),\\ \phi(0)=\phi(1)=0. \end{gathered}\label{e4.5} \end{equation} For the rest of this article, we suppose that the initial value problem \begin{gather*} -x''(t)+f(t,x)=\lambda ax(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x|_{t=1/2}=\beta_1 x(\frac{1}{2}),\quad \Delta x'|_{t=1/2}=-\beta_{2} x(\frac{1}{2}),\\ x(t_{0})=x'(t_{0})=0, \end{gather*} has the unique trivial solution $x\equiv0$ on $[0,1]$, for any $t_{0}\in[0,1]$. Before giving main results, it is necessary to give the lower and upper solution theorem of \eqref{e4.1}, as a useful tool in the proof of the existence of positive solutions. Now we consider \begin{equation} \begin{gathered} -x''(t)+f(t,x)=\lambda ax(t),\quad t\in(0,1),\; t\neq t_1,\\ \Delta x|_{t_1}=\beta_1 x(t_1),\quad \Delta x'|_{t_1}=-\beta_{2} x(t_1),\\ x(0)=x(1)=0, \end{gathered}\label{e4.6} \end{equation} where $f:D\to R$ is continuous, $D\subseteq[0,1]\times R$ and $t_1\in(0,1)$. The solution of \eqref{e4.6} is defined as $x(t)\in PC([0,1],R)\cap PC''((0,1),R)$ satisfying \eqref{e4.6}, where $PC([0,1],R)=\{x:[0,1]\to R:x(t)$ is continuous at $t\neq t_1$, $x(t_1-0)=\lim_{t\to t_1^{-}}x(t)$ and $x(t_1+0)=\lim_{t\to t_1^{+}}x(t)$ exist $\}$, and $PC''((0,1),R)=\{x:(0,1)\to R:x''(t)$ is continuous at $t\neq t_1$, $\lim_{t\to t_1^{-}}x''(t)$ and $ \lim_{t\to t_1^{+}}x''(t)$ exist $\}$. If $\alpha(t)\in PC([0,1],R)\cap PC''((0,1),R)$ satisfies \begin{gather*} -\alpha''(t)+f(t,\alpha(t))\leq\lambda a\alpha(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta \alpha|_{t_1}=\beta_1 \alpha(t_1),\quad \Delta \alpha'|_{t_1}=-\beta_{2} \alpha(t_1),\\ \alpha(0)\leq0,\alpha(1)\leq0, \end{gather*} it is a lower solution of \eqref{e4.6}, while the upper solution is defined as $\beta(t)$ satisfying \begin{gather*} -\beta''(t)+f(t,\beta(t))\geq\lambda a\beta(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta \beta|_{t_1}=\beta_1 \beta(t_1),\quad \Delta \beta'|_{t_1}=-\beta_{2} \beta(t_1),\\ \beta(0)\geq0,\beta(1)\geq0. \end{gather*} Denote $$ D_{\alpha}^{\beta}=\{(t,x)\in(0,1)\times R, \alpha(t)\leq x\leq\beta(t), t\in(0,1)\}, $$ where $\alpha(t)$, $\beta(t)\in PC([0,1],R)$ with $\alpha(t)\leq\beta(t)$ for all $t\in[0,1]$. Then we give the lower and upper solution theorem of \eqref{e4.6}. \begin{lemma} \label{lem4.1} Assume $\alpha(t)$, $\beta(t)$ are the lower and upper solutions respectively with $\alpha(t)\leq\beta(t)$ for $t\in(0,1)$, and $D_{\alpha}^{\beta}\subseteq D$. Moreover, for each $\lambda\in R$ there exists $ h(t,\lambda)\in C([0,1]\times R,R^{+})$ satisfying \begin{gather*} |H(t,x,\lambda)|:= |f(t,x)-\lambda a(t)x(t)| \leq h(t,\lambda),\quad \forall(t,x)\in D_{\alpha}^{\beta},\\ \int_{0}^{1}s(1-s)h(s,\lambda)\beta(t), \end{cases} $$ where $H(t,x,\lambda)=f(t,x)-\lambda a(t)x(t)$, and $$ I^{\ast}(x)=\begin{cases} I(\alpha(t_1)), &x< \alpha(t_1),\\ I(x), &\alpha(t_1)\leq x\leq\beta(t_1),\\ I(\beta(t_1)), &x>\beta(t_1), \end{cases} $$ where $I(x)=x(t_1)$. It is obvious that $$ |H^{\ast}(t,x,\lambda)|t^{\ast}: x(s)<\alpha(s), \forall s\in[t^{\ast},t], t\in(0,1)\}$. We discuss two cases about $r'$. (a) There exists a $t'\in (t^{\ast},t_1]$ satisfying $x(t')=\alpha(t')$. In this case, it's obvious that $r'\leq t_1$. Then for every $ t\in(r,r')$, $$ x(r)=\alpha(r), x(r')=\alpha(r'), x(t)<\alpha(t), H^{\ast}(t,x,\lambda)=H(t,\alpha(t),\lambda). $$ Thus, $x(t)$ satisfies $$ -x''(t)+H(t,\alpha(t),\lambda)=0,\quad t\in(r,r'). $$ On the other hand, since $\alpha(t)$ is a lower solution with $$ -\alpha''(t)+H(t,\alpha(t),\lambda)\leq0,\quad t\in(r,r'), $$ by denoting $z(t)=\alpha(t)-x(t)$, for every $ t\in(r,r')$, one has $$ z(r)=z(r')=0, z''(t)\geq0. $$ By maximum principle, $z(t)\leq0$ holds for all $t\in(r,r')$. Namely for $ t\in(r,r')$, $\alpha(t)\leq x(t)$, which contradicts $\alpha(t^{\ast})> x(t^{\ast})$. (b) For every $t\in [t^{\ast},t_1]$, $x(t)<\alpha(t)$ holds. In this case, noticing $\alpha(t_1+0)=\alpha(t_1)+\beta_1\alpha(t_1)> x(t_1) +\beta_1\alpha(t_1)=x(t_1)+\beta_1 I^{\ast}(x(t_1))=x(t_1+0)$, we have $r'>t_1$ and for every $ t\in(r,r')$, $$ x(r)=\alpha(r), x(r')=\alpha(r'), x(t)<\alpha(t), H^{\ast} (t,x,\lambda)=H(t,\alpha(t),\lambda). $$ Since $x(t)$ and $\alpha(t)$ satisfy \begin{gather*} -x''(t)+H(t,\alpha(t),\lambda)=0,\quad t\in(r,r')\setminus{t_1},\\ -\alpha''(t)+H(t,\alpha(t),\lambda)\leq0,\quad t\in(r,r')\setminus{t_1}, \end{gather*} by letting $z(t)=\alpha(t)-x(t)$, for $ t\in(r,r')$, it holds that \begin{gather*} -z''(t)\geq0,\quad t\in(r,r'),\; t\neq t_1,\\ \Delta z|_{t_1}=0,\quad \Delta z'|_{t_1}=0,\\ z(r)=z(r')=0. \end{gather*} Since $-z''(t)\geq0$ for $t\in(r,r')\setminus t_1$ and $\Delta z'|_{t_1}=0$, $z'(t)$ increases and is continuous on $(r,r')$. Lagrange mean value theorem, together with $z(r)=0$ and $z(t_1)>0$ guarantees that there exists a $t'\in(r,t_1)$ with $z'(t')>0$. Similarly, with $z(t_1+0)>0$ and $z(r')=0$, there exists a $t''\in(t_1,r')$ so that $z'(t'')<0$. This contradicts to the monotonicity of $z'(t)$. According to the discussion above, we find case (1) impossible. (2) Assume $t^{\ast}=t_1$. As before, define $r=\inf\{tt^{\ast}: x(s)<\alpha(s), \forall s\in[t^{\ast},t], t\in(0,1)\}$. The conclusion on $(r,r')$ is same with the case (b) in (1), which can be excluded. (3) Assume $t^{\ast}>t_1$. The definitions of $r,r'$ are same as before. We divide the discussion of $r$ into two cases. (a') There exists a $t'\in (t_1,t^{\ast}]$ satisfying $x(t')=\alpha(t')$ or $\alpha(t_1+0)=x(t_1+0)$. This case is similar to the $(a)$ in case (1), which deduces a contradiction. (b') For every $t\in (t_1,t^{\ast}]$, $x(t)<\alpha(t)$ holds, and so $x(t_1+0)=x(t_1)+\beta_1I^{\ast}(x(t_1))<\alpha(t_1+0) =\alpha(t_1)+\beta_1I(\alpha(t_1))$. Then it can be concluded that $x(t_1)<\alpha(t_1)$. This case is similar to $(b)$ in case (1), which is impossible. Combining cases (1), (2) and (3), we conclude that for $ t\in[0,1]$, $x(t)\geq\alpha(t)$. With the same method, $x(t)\leq\beta(t)$ is true for $t\in[0,1]$. The fact is that if $x(t)$ is the solution of \eqref{e4.7}, it must be the solution of \eqref{e4.6}, which has been proved. Next we will give the existence of solutions for \eqref{e4.7}. The solution of \eqref{e4.7} has the form (see \cite{g2}) \begin{align*} x(t) &=-\int_{0}^{1}G(t,s)H^{\ast}(s,x(s),\lambda)ds +[\beta_1-(t-t_1\beta_{2})]\sum_{0\lambda_1$, there exists an unique positive solution of \eqref{e4.1}. The mapping $\lambda\mapsto x_{\lambda}$ is continuous from $(\lambda_1,+\infty)$ to $PC''([0,1],R)$ and the branch $\{(\lambda,x_{\lambda}),\lambda\in(\lambda_1,+\infty)\}$ bifurcates from the right of trivial solutions at $(\lambda_1,0)$. Moreover, $x_{\lambda}$ increases strictly with $\lambda$: if $\lambda<\mu$, then $x_{\lambda}0$, $\rho(t)=0$ if $x(t)=0$, and $\nu=\lambda$. By condition \eqref{e4.3}, $\rho\geq0$ and $\rho\not\equiv0$ hold, so that $\lambda=\nu_1(\rho)>\nu_1(0)=\lambda_1$, which is a contradiction. We show the existence of solutions of \eqref{e4.1} for $\lambda>\lambda_1$. Let $\phi_1(t)$ be the prime eigenfunction of \eqref{e4.5} corresponding to $\lambda_1$, so $\phi_1(t)>0$ for $t\in(0,1)$. First, it will be verified that, for sufficiently small $\varepsilon>0$, $\varepsilon\phi_1$ is a lower solution of \eqref{e4.1}. Indeed, with $\lambda>\lambda_1$ and condition \eqref{e4.2}, for sufficiently small $\varepsilon$, we have \begin{align*} -\varepsilon\phi_1''(t)+f(t,\varepsilon\phi_1(t)) -\lambda a(t)\varepsilon\phi_1(t) &=\lambda_1 a(t)\varepsilon\phi_1(t)-\lambda a(t)\varepsilon\phi_1(t) +f(t,\varepsilon\phi_1(t)) \\ &=\Big[(\lambda_1-\lambda)a(t) +\frac{f(t,\varepsilon\phi_1(t))}{\varepsilon\phi_1(t)}\Big] \varepsilon\phi_1(t) <0, \end{align*} and $\Delta \varepsilon\phi_1|_{t=1/2}=\beta_1\varepsilon\phi_1(\frac{1}{2})$, $\Delta(\varepsilon\phi_1)'|_{t=1/2}=-\beta_{2} \varepsilon\phi_1(\frac{1}{2})$, and $\varepsilon\phi_1(0)=\varepsilon\phi_1(1)=0$, which infer that $\varepsilon\phi_1$ is a lower solution. Next we will find a upper solution of \eqref{e4.1}. Condition \eqref{e4.4} implies that there exists a sufficiently grand constant $M_{\lambda}$ for each $\lambda\in R$, so that $f(t,s)\geq\lambda a(t)s $ for $s>M_{\lambda}$ and $t\in[0,1]$. Define $$ x^{\ast}=\begin{cases} C_1t+C_{2}, & t\in[0,\frac{1}{2}],\\ C_{3}t+C_{4}, & t\in(\frac{1}{2},1], \end{cases} $$ where $C_1$, $C_{2}$, $C_{3}$ and $C_{4}$ can be selected by the following conditions: \begin{gather*} x^{\ast}(\frac{1}{2}+)=(1+\beta_1)x^{\ast}(\frac{1}{2}-),\\ (x^{\ast})'(\frac{1}{2}+)=(x^{\ast})'(\frac{1}{2}-) -\beta_{2} x^{\ast}(\frac{1}{2}-),\\ x^{\ast}(t)>M_{\lambda},\quad \forall t\in [0,1]. \end{gather*} Such $x^{\ast}$ defined above satisfies \begin{gather*} -(x^{\ast})''(t)+f(t,x^{\ast})\geq\lambda ax^{\ast}(t),\quad t\in(0,1),\; t\neq\frac{1}{2},\\ \Delta x^{\ast}|_{t=1/2}=\beta_1 x^{\ast}(\frac{1}{2}),\quad \Delta (x^{\ast})'|_{t=1/2}=-\beta_{2} x^{\ast}(\frac{1}{2}),\\ x^{\ast}(0)\geq0, x^{\ast}(1)\geq0, \end{gather*} which means $x^{\ast}$ a upper solution of \eqref{e4.1}. It is easy to check that Lemma \ref{lem4.1} holds, so there exists a positive solution of \eqref{e4.1} between $\varepsilon\phi_1$ and $ x^{\ast}$. Let $x_{\lambda}$ be the maximum positive solution of \eqref{e4.1} satisfying $\varepsilon\phi_1\leq x(t)\leq x^{\ast}$. Then $x_{\lambda}$ is unique. Indeed, assume $x$ be another positive solution of \eqref{e4.1} with $x0$, define $$ \Lambda(\gamma)=\lambda_1\|a\|_{C^{0}[0,1]} +\max_{t\in[0,1]}\frac{f(t,\gamma)}{\gamma}. $$ Then for all $\gamma>0$, $\Lambda(\gamma)<+\infty$ and $\Lambda(\gamma)$ increases with $\gamma$. Here $\phi_1$ is also the prime eigenfunction of \eqref{e4.5} and satisfies $\|\phi_1\|_1=1$. For all $\lambda \geq\Lambda(\gamma)$, $\gamma\phi_1$ is a lower solution of \eqref{e4.1}. In fact, condition \eqref{e4.3} implies that $f(t,\gamma\phi_1)\leq(f(t,\gamma)\setminus\gamma)\gamma\phi_1$, by which we compute $$ -(\gamma\phi_1)''+f(t,\gamma\phi_1)\leq{\lambda_1a(t) +\frac{f(t,\gamma)}{\gamma}}\gamma\phi_1\leq\Lambda(\gamma) \gamma\phi_1\leq\lambda \gamma\phi_1. $$ So $\gamma\phi_1$ is a lower solution of \eqref{e4.1} with $\gamma\phi_1\leq x_{\lambda}$, and for all $\gamma>0$ there exists $\Lambda(\gamma)>0$ telling that $$ \lambda \geq\Lambda(\gamma)\Rightarrow\gamma\phi_1\leq x_{\lambda},\quad t\in(0,1). $$ Since $\phi_1>0$ for $t\in(0,1)$, it can be deduced that $x_{\lambda}\to+\infty$, as $\lambda\to+\infty$ uniformly for $t$ in the closed interval $\subseteq (0,1)$. The proof is complete. \end{proof} For convenience of the next proof, we give some symbols here. 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 $\|\cdot\|_1$. Denote $S_{k}$ be the set of functions in $E$ which have exactly $k-1$ simple nodal zeros in $(0,1)$ (by a nodal zero we mean the function changes sign at the zeros and at a simple nodal zero, the derivative of the function is nonzero) and denote $$ S_{k}^{+}=\{x\in S_{k}; x'(0)>0\}, \quad S_{k}^{-}=-S_{k}^{+}. $$ Finally, let $\Phi_{k}^{\pm}=\mathbb{R}\times S_{k}^{\pm}$ and $\Phi_{k}=\mathbb{R}\times S_{k}$. It is easy to show that for any positive integer $k$, $ S_{k}$, $S_{k}^{+}$ and $S_{k}^{-}$ are open in $E$. With the fact that \eqref{e4.5} is the linear equation of \eqref{e4.1} at the neighbourhood of $x=0$, global bifurcation theorem of Rabinowitz can be applied to \eqref{e4.1}. \begin{theorem} \label{thm4.2} Suppose that $f$ satisfies conditions \eqref{e4.2}, \eqref{e4.3} and \eqref{e4.4}. Then for $\lambda\leq\lambda_1$, Equation \eqref{e4.1} admits trivial solution $x_{0}=0$. For $\lambda_{k}<\lambda\leq\lambda_{k+1}$, Equation $\eqref{e4.1}$ possesses at least $2k+1$ solutions: $x_{0}=0$, $x_1^{\pm}$, $\ldots x_{k}^{\pm}$, with $x_{j}^{+}\in S_{j}^{+}$ and $x_{j}^{-}\in S_{j}^{-}$. \end{theorem} \begin{proof} The existence of a solution $x\in S_{k}^{\pm}$ for $\lambda>\lambda_{k}$ is obtained easily. In fact, $x(t)$ is a solution of \eqref{e4.1} if and only if $x(t)$ is a solution of equation $x=\lambda Lx+H(t,x)$, where $L$ is defined as $$ (Lx)(t)=\begin{cases} \int_{0}^{1}G(t,s)a(s)x(s)ds -\frac{1}{\lambda}\big(\beta_1-\frac{1}{2}\beta_{2}\big)tx(\frac{1}{2}), & t\in[0,\frac{1}{2}],\\[4pt] \int_{0}^{1}G(t,s)a(s)x(s)ds+\frac{1}{\lambda} \big(\beta_1+\frac{1}{2}\beta_{2}\big)(1-t)x(\frac{1}{2}), & t\in(\frac{1}{2},1], \end{cases} $$ and $H(t,x)=\int_{0}^{1}G(t,s)f(s,x(s))ds$ satisfying $\lim_{x\to0} H(t,x)/ x=0$. The global bifurcation theorem of Rabinowitz can be applied for this problem and we just need to prove $\mathbf{C}_{k}^{\pm}\cap({\lambda}\times E)\neq\emptyset$. By Lemma \ref{lem2.2} and Lemma \ref{lem3.8}, there exists one unbounded continua $\mathbf{C}_{k}$ of solutions of \eqref{e4.1}, which bifurcates from the right of the trivial solutions at $(\lambda_{k},0)$ and satisfies $$ (\lambda_{k},0)\in\mathbf{C}_{k}\subset\Phi_{k}\cup\{(\lambda_{k},0)\}. $$ Moreover, $\mathbf{C}_{k}=\mathbf{C}_{k}^{+}\cup\mathbf{C}_{k}^{-}$ and $\mathbf{C}_{k}^{+}\cap\mathbf{C}_{k}^{-}=(\lambda_{k},0)$, while $\mathbf{C}_{k}^{\pm}$ are two unbounded continua in $R \times E$ satisfying $$ \mathbf{C}_{k}^{\pm}\subset\Phi_{k}^{\pm}\cup\{(\lambda_{k},0)\}. $$ By Theorem \ref{thm4.1}, we know that $\|x\|_1\leq M(\lambda)$ so that $\mathbf{C}_{k}^{\pm}\subset\{(\lambda,x);\lambda\geq\lambda_{k},\|x\|_1 \leq M(\lambda)\}$ and the projection of $\mathbf{C}_{k}^{\pm}$ on $R$ is unbounded belonging to $(\lambda_{k},+\infty)$. So $\mathbf{C}_{k}^{\pm}\cap({\lambda}\times E)\neq\emptyset$, and there exists at least $2k+1$ solutions of \eqref{e4.1} in $S_{k}^{\pm}$ for $\lambda\geq\lambda_{k}$. \end{proof} \section{Behavior of positive solutions of autonomous impulsive differential equation for large $\lambda$} First, we study the existence of the positive solutions to \begin{equation} \begin{gathered} -x''(t)+x^{p+1}=\lambda mx(t),\quad t\in(0,1),\; t\neq \frac{1}{2},\\ \Delta x|_{\frac{1}{2}}=\beta_1 x(\frac{1}{2}),\quad \Delta x'|_{\frac{1}{2}}=-\beta_{2} x(\frac{1}{2}),\\ x(0)=x(1)=0, \end{gathered}\label{e5.1} \end{equation} where $p>0$, $m>0$ are real numbers. This is a special case of \eqref{e1.1} with $f(t,x)=x^{p+1}$ and $a(t)=m$, so we can use the results in section 4 on it. The behavior of the positive solutions of \eqref{e5.1}, as $\lambda\to\infty$, will be discussed, on the basis of the comparison argument and the property of principal eigenvalue of linear equation of \eqref{e5.1}. By Theorem \ref{thm3.1}, we denote $\sigma_i$ the $i^{th}$ eigenvalue of \begin{equation} \begin{gathered} -x''(t)=\lambda x(t),\quad t\in(0,1),\quad t\neq \frac{1}{2},\\ \Delta x|_{\frac{1}{2}}=\beta_1 x(\frac{1}{2}),\quad \Delta x'|_{\frac{1}{2}}=-\beta_{2} x(\frac{1}{2}),\\ x(0)=x(1)=0. \end{gathered}\label{e5.2} \end{equation} From Theorem \ref{thm4.1}, \eqref{e5.1} admits a positive solution, if and only if $\lambda>\sigma_1/m$. Let $u_{\lambda, m}$ be such positive solution. In order to study the behavior of $u_{\lambda, m}$ as $\lambda\to\infty$, we use the change of variable \begin{equation} u=\lambda^{1/p}v, \label{e5.3} \end{equation} which transforms \eqref{e5.1} into \begin{equation} \begin{gathered} -\frac{1}{\lambda}v''=mv-v^{p+1},\quad t\in(0,1),\quad t\neq \frac{1}{2},\\ \Delta v|_{\frac{1}{2}}=\beta_1 v(\frac{1}{2}),\quad \Delta v'|_{\frac{1}{2}}=-\beta_{2} v(\frac{1}{2}),\\ v(0)=v(1)=0. \end{gathered} \label{e5.4} \end{equation} So the problem of analyzing the behavior of $u_{\lambda, m}$ as $\lambda\to\infty$ is equivalent to analyzing the behavior of the unique positive solution of \eqref{e5.4}, which is denoted by $\theta_{\lambda, m}$, as $\lambda\to\infty$. The behavior of $\theta_{\lambda, m}$ in the interior of $[0,1]$ is given by the following conclusion. \begin{theorem} \label{thm5.1} Let $J\subset (0,1)$ be a compact interval. Then \begin{equation} \lim_{\lambda\to+\infty}\theta_{\lambda, m}=m^{1/p} \label{e5.5} \end{equation} uniformly in $J$. \end{theorem} \begin{proof} Given $\lambda>\frac{\sigma_1}{m}$, $-\theta''_{\lambda, m}(t_{0})\geq0$ at the point $t_{0}\in I$ where $\theta_{\lambda, m}$ achieves its maximum and hence \begin{equation} \theta_{\lambda, m}\leq m^{1/p}, \quad t\in I. \label{e5.6} \end{equation} Then it is sufficient to show that given $\varepsilon>0$, there exists $\lambda(\varepsilon)$ such that \begin{equation} m^{1/p}-\varepsilon\leq \theta_{\lambda, m}(t), \quad t\in J, \label{e5.7} \end{equation} for all $\lambda\geq\lambda(\varepsilon)$. To show this, we discuss as follows. Consider $t_{0}\in J'$ and $J'=[\alpha,\beta]\subset I$. Without loss of generality, we assume $\frac{1}{2}\in J'$. Then $\sigma_1^{J'}$ (resp. $\varphi^{J'}>0$) will stand for the principal eigenvalue (resp. eigenfunction) of \eqref{e5.2} with boundary conditions $\varphi^{J'}(\alpha)=\varphi^{J'}(\beta)=0$. We consider $\varphi^{J'}(t)$ normalized so that $\|\varphi^{J'}\|_1=1$. Let $\lambda(t_{0},\varepsilon)>0$ be such that \begin{equation} \big[m-\frac{\sigma_1^{J'}}{\lambda(t_{0},\varepsilon)}\big]^{1/p} \geq m^{1/p}-\frac{\varepsilon}{2}. \label{e5.8} \end{equation} Lemma \ref{lem2.1} shows that the mapping $\lambda\to \theta_{\lambda, m}$ is strictly increasing as far as $\lambda>\frac{\sigma_1^{J'}}{m}$. Thus, \begin{equation} \theta_{\lambda, m}\geq\theta_{\lambda(t_{0},\varepsilon), m} \geq\theta_{\lambda(t_{0},\varepsilon), m}^{J'}, \quad \lambda\geq\lambda(t_{0},\varepsilon), \label{e5.9} \end{equation} where $\theta_{\lambda(t_{0},\varepsilon), m}^{J'}$ is the unique positive solution to \begin{equation} \begin{gathered} -\frac{1}{\lambda(t_{0},\varepsilon)}v'' =mv-v^{p+1},\quad t\in J', \; t\neq \frac{1}{2},\\ \Delta v|_{\frac{1}{2}}=\beta_1 v(\frac{1}{2}),\quad \Delta v'|_{\frac{1}{2}}=-\beta_{2} v(\frac{1}{2}),\\ v(\alpha)=v(\beta)=0. \end{gathered}\label{e5.10} \end{equation} It follows easily that $[m-\frac{\sigma_1^{J'}}{\lambda(t_{0},\varepsilon)}]^{1/p}\varphi^{J'}$ is a lower solution of \eqref{e5.10}. Thus, due the convexity of $v\to v^{1/p}$ and using the maximum principle, we find that $$ [m-\frac{\sigma_1^{J'}}{\lambda(t_{0},\varepsilon)}]^{1/p}\varphi^{J'} \leq\theta_{\lambda(t_{0},\varepsilon), m}^{J'}. $$ From the inequality and \eqref{e5.9} it follows that $$ \theta_{\lambda, m}\geq\big[m-\frac{\sigma_1^{J'}}{\lambda(t_{0},\varepsilon)} \big]^{1/p}\varphi^{J'}, \quad \lambda\geq\lambda(t_{0},\varepsilon). $$ Hence, from \eqref{e5.8} the following holds $$ \theta_{\lambda, m}\geq\big[m^{1/p}-\frac{\varepsilon}{2}\big]\varphi^{J'}, $$ for all $\lambda\geq\lambda(t_{0},\varepsilon)$ and $t\in J'$. Since $\varphi^{J'}(t_{0})=1$ and the local continuity of $\varphi^{J'}$, there exists $J\subset J'\subset I$ such that $$ \varphi^{J'}(t)\geq\frac{m^{1/p}-\varepsilon}{m^{1/p} -\frac{\varepsilon}{2}}, \quad t\in J. $$ Thus, $$ \theta_{\lambda, m}\geq m^{1/p}-\varepsilon, \quad t\in J, $$ for all $\lambda\geq\lambda(t_{0},\varepsilon)$. 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