\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 03, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/03\hfil Bifurcation from intervals] {Bifurcation from intervals for Sturm-Liouville problems and its applications} \author[G. Dai, R. Ma \hfil EJDE-2014/03\hfilneg] {Guowei Dai, Ruyun Ma} % in alphabetical order \address{Guowei Dai \newline Department of Mathematics, Northwest Normal University, Lanzhou 730070, China} \email{daiguowei@nwnu.edu.cn} \address{Ruyun Ma \newline Department of Mathematics, Northwest Normal University, Lanzhou 730070, China} \email{mary@nwnu.edu.cn} \thanks{Submitted September 8, 2013. Published January 3, 2014.} \subjclass[2000]{34B24, 34C10, 34C23} \keywords{Global bifurcation; nodal solutions; eigenvalues} \begin{abstract} We study the unilateral global bifurcation for the nonlinear Sturm-Liouville problem \begin{gather*} -(pu')'+qu=\lambda au+af(x,u,u',\lambda)+g(x,u,u',\lambda)\quad x\in(0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0, \end{gather*} where $a\in C([0, 1], [0,+\infty))$ and $a(x)\not\equiv 0$ on any subinterval of $[0, 1]$, $f,g\in C([0,1]\times\mathbb{R}^3,\mathbb{R})$ and $f$ is not necessarily differentiable at the origin or infinity with respect to $u$. Some applications are given to nonlinear second-order two-point boundary-value problems. This article is a continuation of \cite{DM1}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} %\newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} %\newtheorem{definition}[theorem]{Definition} %\newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Berestycki \cite{Be} considered a class of problems involving a non-differentiable nonlinearity. More precisely, he considered the nonlinear Sturm-Liouville problem $$\label{SL0} \begin{gathered} -(pu')'+qu=\lambda au+\widetilde{F}(x,u,u',\lambda)\quad x\in(0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0, \end{gathered}$$ where $p$ is a positive, continuously differentiable function on $[0,1]$, $q$ is a continuous function on $[0,1]$, $a$ is a positive continuous function on $[0,1]$ and $b_i$, $c_i$ are real numbers such that $| b_i|+| c_i|\neq0$, $i=0,1$. Moreover, the nonlinear term has the form $\widetilde{F}=\widetilde{f}+\widetilde{g}$, where $\widetilde{f}$ and $\widetilde{g}$ are continuous functions on $[0,1]\times \mathbb{R}^3$, satisfying the following conditions: \begin{itemize} \item[(C1)] $|\frac{\widetilde{f}(x,u,s,\lambda)}{u}|\leq M$, for all $x\in[0,1]$, $0<\vert u\vert\leq1$, $\vert s\vert\leq1$ and all $\lambda\in \mathbb{R}$, where $M$ is a positive constant; \item[(C2)] $\widetilde{g}(x,u,s,\lambda)=o(\vert u\vert+\vert s\vert)$, near $(u,s)=(0,0)$, uniformly in $x\in[0,1]$ and $\lambda$ on bounded sets. \end{itemize} He obtained a global bifurcation result for \eqref{SL0}. His result has been extended by Rynne \cite{R} under the assumption that $| \widetilde{F}(x, \xi, \eta,\lambda)| \leq M_0\vert \xi\vert +M_1\vert \eta\vert,\quad (x,\xi,\eta,\lambda)\in [0, 1]\times \mathbb{R}^3,$ as either $\vert(\xi,\eta)\vert\to0$ or $\vert(\xi,\eta)\vert\to +\infty$, for some constants $M_0$ and $M_1$. Recently, Ma and Dai \cite{DM1}] improved Berestycki's result to show a unilateral global bifurcation result for \eqref{SL0}. We refer the reader to \cite{DM,D1,D2,L1,R2,SW} and their references for information on unilateral global bifurcation. Of course, the natural question is that what would happen if $a(x)$ is not strictly positive on $[0,1]$? The aim of this article is to consider this case. For this aim, we study the nonlinear Sturm-Liouville problem $$\label{SL1} \begin{gathered} -(pu')'+qu=\lambda au+af(x,u,u',\lambda)+g(x,u,u',\lambda)\quad x\in(0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0, \end{gathered}$$ where $p$, $q$, $b_i$ and $c_i$, $i=0,1$, are defined as above, $g\in C([0,1]\times \mathbb{R}^3,\mathbb{R})$ satisfies (C2), $a$ and $f$ satisfy the following assumptions: \begin{itemize} \item[(C3)] $a\in C([0, 1], [0,+\infty))$ and $a(x) \not\equiv0$ on any subinterval of $[0, 1]$; \item[(C4)] $f\in C([0,1]\times \mathbb{R}^3,\mathbb{R})$ is continuous and satisfies $\underline{f}_0, \overline{f}_0\in\mathbb{R}$, where $\underline{f}_0=\liminf_{\vert s\vert\to 0}\frac{f(x,s,y,\lambda)}{s},\quad \overline{f}_0=\limsup_{\vert s\vert\to 0}\frac{f(x,s,y,\lambda)}{s}$ uniformly in $x\in [0, 1]$, $\vert y\vert\leq 1$ and for all $\lambda\in \mathbb{R}$. \end{itemize} Under the above assumptions, we shall establish a result involving unilateral global bifurcation of \eqref{SL1}. Moreover, in line with the global bifurcation from infinity of Rabinowitz \cite{R3}], we shall also establish two results involving unilateral global bifurcation of \eqref{SL1} from infinity. Let $Lu:=-(pu')'+qu$. It is well known (see \cite{CL,I,W}) that the linear Sturm-Liouville problem \begin{gather*} Lu=\lambda au, \quad x\in(0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0 \end{gather*} possesses infinitely many eigenvalues $\lambda_1<\lambda_2<\dots <\lambda_k\to+\infty$, all of which are simple. The eigenfunction $\varphi_k$ corresponding to $\lambda_k$ has exactly $k-1$ simple zeros in $(0,1)$. In particular, if $b_0$, $c_0$, $b_1$ and $c_1$ satisfy \begin{itemize} \item[(C5)] $b_0$, $-c_0$, $b_1$, $c_1\in[0,+\infty)$ and $b_0c_1-b_1c_0+b_0b_1>0$, \end{itemize} then $\lambda_1>0$ (see \cite{MT,W}). On the basis of the unilateral global bifurcation results (Theorem \ref{thm2.1}--\ref{thm2.3}), we investigate the existence of nodal solutions for the nonlinear second-order two-point boundary-value problem $$\label{dn} \begin{gathered} Lu=r a(x)F(u), \quad x\in(0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0, \end{gathered}$$ where $a$ satisfies (C3), $r\in (0,+\infty)$, $F\in C(\mathbb{R}, \mathbb{R})$, $b_i$ and $c_i$, $i=0,1$, satisfy (C5). In this article, we assume that the nonlinear term has the form $F=f+g$, where $f$ and $g$ are continuous functions on $\mathbb{R}$, satisfying the following conditions: \begin{itemize} \item[(C6)] $\underline{f}_0, \overline{f}_0, \underline{f}_\infty, \overline{f}_\infty\in\mathbb{R}$ with $\underline{f}_0\neq \overline{f}_0$ and $\underline{f}_\infty\neq\overline{f}_\infty$, where \begin{gather*} \underline{f}_0=\liminf_{\vert s\vert\to 0}\frac{f(s)}{s},\quad \overline{f}_0=\limsup_{\vert s\vert\to 0}\frac{f(s)}{s},\\ \underline{f}_\infty=\liminf_{\vert s\vert\to +\infty}\frac{f(s)}{s},\quad \overline{f}_\infty=\limsup_{\vert s\vert\to +\infty}\frac{f(s)}{s}. \end{gather*} \item[(C7)] $g$ satisfies $g(s)s>0$ for any $s\neq 0$ and there exist $g_0, g_{\infty}\in(0,+\infty)$ such that $g_0=\lim_{|s|\to 0}\frac{g(s)}{s},\quad g_{\infty}=\lim_{|s|\to +\infty}\frac{g(s)}{s}.$ \end{itemize} In particular, we consider the special case of $g\equiv 0$ in \eqref{dn}; i.e., consider the problem $$\label{dn1} \begin{gathered} Lu=r a(x)f(u), \quad x\in(0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0. \end{gathered}$$ We shall establish the same results as those in \cite{MT}] and some new results for \eqref{dn1}. Note that the assumption (C6) is weaker than the condition (A2) in \cite{MT}] because we do not require $\underline{f}_0, \overline{f}_0, \underline{f}_\infty, \overline{f}_\infty\in[0,+\infty)$ and $f(s)s>0$ for $s\neq 0$ which are essential in \cite{MT}]. The rest of this article is arranged as follows. In Section 2, we establish the unilateral global bifurcation which bifurcates from the trivial solutions axis or from infinity of \eqref{SL1}, respectively. In Section 3, we determine the interval of $r$, in which there exist nodal solutions for \eqref{dn} or \eqref{dn1}. \section{Global bifurcation from an interval} Set $E:=\big\{u\in C^1[0,1]\big|b_0u(0)+c_0u'(0)=0,\; b_1u(1)+c_1u'(1)=0\big\}$ with the norm $\| u\|=\max_{x\in[0,1]}\vert u(x)\vert+\max_{x\in[0,1]}| u'(x)|$. Let $S_k^+$ denote the set of functions in $E$ which have exactly $k-1$ interior nodal (i.e., non-degenerate) zeros in $(0,1)$ and are positive near $x=0$, and set $S_k^-=-S_k^+$, and $S_k =S_k^+\cup S_k^-$. It is clear that $S_k^+$ and $S_k^-$ are disjoint and open in $E$. We also let $\Phi_k^{\pm}=\mathbb{R}\times S_k^{\pm}$ and $\Phi_k=\mathbb{R}\times S_k$ under the product topology. Finally, we use $\mathscr{S}$ to denote the closure in $\mathbb{R}\times E$ of the set of nontrivial solutions of \eqref{SL1}, and $\mathscr{S}_k^\pm$ to denote the subset of $\mathscr{S}$ with $u\in S_k^{\pm}$ and $\mathscr{S}_k=\mathscr{S}_k^+\cup \mathscr{S}_k^-$. \begin{theorem} \label{thm2.1} Let $I_k=[\lambda_k-\overline{f}_0,\lambda_k-\underline{f}_0]$ for every $k\in \mathbb{N}$. The component $\mathscr{D}_k^+$ of $\mathscr{S}_k^+\cup(I_k\times \{0\})$, containing $I_k\times \{0\}$ is unbounded and lies in $\Phi_k^+\cup(I_k\times \{0\})$ and the component $\mathscr{D}_k^-$ of $\mathscr{S}_k^-\cup(I_k\times \{0\})$, containing $I_k\times \{0\}$ is unbounded and lies in $\Phi_k^-\cup(I_k\times \{0\})$. \end{theorem} \begin{remark} \label{rmk2.1} \rm It is easy to verify that \cite[Lemma 2.2]{DM1} is also valid for \eqref{SL1}. So if $(\lambda, u)$ is a nontrivial solution of \eqref{SL1} under the assumptions of (C2)--(C4), then $u\in \cup_{k=1}^\infty S_k$. \end{remark} \begin{remark} \label{rmk2.2} \rm It is not difficult to see that condition (C4) is equivalent to (C1) with $M\geq\max\{| \underline{f}_0|,| \overline{f}_0|\}$. If $a(x)>0$ on $[0,1]$, applying \cite[Theorem 1]{Be}] to problem \eqref{SL1} with $\widetilde{f}=af$, we can obtain that $\widetilde{I}_k=[\lambda_k-\widetilde{M}/a_0,\lambda_k+\widetilde{M}/a_0]$, where $a_0=\min_{x\in[0,1]}a(x)$ and $\widetilde{M}=a^0M$ with $a^0=\max_{x\in[0,1]}a(x)$. It is easy to verify that $I_k\subseteq \widetilde{I}_k$. So even in the case of $a(x)>0$ on $[0,1]$, the conclusion of Theorem \ref{thm2.1} is better than the corresponding ones in \cite[Theorem 1]{Be}, and \cite[Theorem 2.1]{DM1}. \end{remark} Consider the approximate problem $$\label{SL2} \begin{gathered} -(pu')'+qu=\lambda au+af(x,u\vert u\vert^\epsilon,u',\lambda) +g(x,u,u',\lambda)\quad x\in(0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0. \end{gathered}$$ To prove Theorem \ref{thm2.1}, we need the following lemma. \begin{lemma} \label{lem2.1} Let $\epsilon_n$, $0\leq\epsilon_n\leq1$, be a sequence converging to $0$. If there exists a sequence $(\lambda_n,u_n)\in \mathbb{R}\times S_k^\nu$ such that $(\lambda_n,u_n)$ is a solution of \eqref{SL2} corresponding to $\epsilon=\epsilon_n$, and $(\lambda_n,u_n)$ converges to $(\lambda,0)$ in $\mathbb{R}\times E$, then $\lambda\in I_k$. \end{lemma} \begin{proof} By an argument similar to that of \cite[Lemma 1]{Be}, we can show that there are two intervals $(\zeta_1,\eta_1)$ and $(\zeta_2,\eta_2)$ in $(0,1)$ such that \begin{gather}\label{2.2} \int_{\zeta_2}^{\eta_2}(\lambda-\lambda_k)aw\varphi_k^\nu\,dx +\liminf_{n\to +\infty}\int_{\zeta_2}^{\eta_2}a(x)f_n(x)\varphi_k^\nu\,dx\leq0,\\ \label{2.3} \int_{\zeta_1}^{\eta_1}(\lambda-\lambda_k)aw\varphi_k^\nu\,dx +\limsup_{n\to +\infty}\int_{\zeta_1}^{\eta_1}a(x)f_n(x)\varphi_k^\nu\,dx\geq0, \end{gather} where $f_n(x)=\frac{f(x,u_n(x)| u_n(x)|^\epsilon,u_n'(x),\lambda)}{\| u_n\|}.$ Similar to that of \cite[Lemma 1]{Be}, if $w$ and $\varphi_k^\nu$ have the same sign in $(\zeta,\eta)$, we can easily show that $$\label{2.4} \underline{f}_0\int_{\zeta}^{\eta}aw\varphi_k^\nu\,dx \leq\int_{\zeta}^{\eta}af_n(x)\varphi_k^\nu\,dx \leq \overline{f}_0\int_{\zeta}^{\eta}aw\varphi_k^\nu\,dx$$ for $n$ large enough. It follows from \eqref{2.2} and \eqref{2.4} that $\int_{\zeta_2}^{\eta_2}(\lambda-\lambda_k+\underline{f}_0)aw\varphi_k^\nu\,dx \leq0,$ hence $\lambda\leq\lambda_k-\underline{f}_0$. Similarly, we from \eqref{2.3} and \eqref{2.4} we obtain $\lambda\geq\lambda_k-\overline{f}_0$. \end{proof} \begin{remark} \label{rmk2.3} \rm Note that we do not need $a(x)$ is strictly positive on $[0,1]$ in the proof of Lemma \ref{lem2.1} because our nonlinearity is different from that in \cite{Be}]. We put the same weight function $a(x)$ before $f$ while this weight function is 1 in \cite{Be,DM1}. \end{remark} \begin{proof}[Proof of Theorem \ref{thm2.1}] In view of Remark \ref{rmk2.1} and Lemma \ref{lem2.1}, by an argument similar to that in \cite[Theorem 2.1]{DM1}, we can obtain easily the desired conclusions. \end{proof} Instead of (C2) and (C4), we assume that $f$ and $g$ satisfy the following conditions: \begin{itemize} \item[(C8)] $g(x,u,s,\lambda)=o(\vert u\vert+\vert s\vert)$, near $(u,s)=(\infty,\infty)$, uniformly in $x\in[0,1]$ and on bounded $\lambda$ intervals; \item[(C9)] $f\in C([0,1]\times \mathbb{R}^3,\mathbb{R})$ is continuous and satisfies $\underline{f}_\infty, \overline{f}_\infty\in\mathbb{R}$, where $\underline{f}_\infty=\liminf_{\vert s\vert\to +\infty}\frac{f(x,s,y,\lambda)}{s},\quad \overline{f}_\infty=\limsup_{\vert s\vert\to +\infty}\frac{f(x,s,y,\lambda)}{s}$ uniformly in $x\in [0, 1]$, $\vert y\vert\geq C$ for some positive constant $C$ large enough and $\forall \lambda\in \mathbb{R}$. \end{itemize} We use $\mathscr{T}$ to denote the closure in $\mathbb{R}\times E$ of the set of nontrivial solutions of \eqref{SL1} under conditions (C3), (C8) and (C9). Applying similar methods to those in \cite[Theorem 2.2 and 2.3]{DM1}, with obvious modifications, we obtain the following two results: \begin{theorem} \label{thm2.2} Let $I_k=[\lambda_k-\overline{f}_\infty,\lambda_k-\underline{f}_\infty]$ for every $k\in \mathbb{N}$. There exists a component $\mathscr{D}_{k}$ of $\mathscr{T}\cup(I_k\times \{\infty\})$, containing $I_k\times \{\infty\}$. Moreover if $\Lambda\subset \mathbb{R}$ is an interval such that $\Lambda\cap(\cup_{k=1}^{\infty}I_k)=I_k$ and $\mathscr{M}$ is a neighborhood of $I_k\times \{\infty\}$ whose projection on $\mathbb{R}$ lies in $\Lambda$ and whose projection on $E$ is bounded away from 0, then either \begin{itemize} \item[(1)] $\mathscr{D}_k-\mathscr{M}$ is bounded in $\mathbb{R}\times E$ in which case $\mathscr{D}_k-\mathscr{M}$ meets $\mathscr{R}=\{(\lambda,0)\big|\lambda\in \mathbb{R}\}$, or \item[(2)] $\mathscr{D}_k-\mathscr{M}$ is unbounded. \end{itemize} If (2) occurs and $\mathscr{D}_k-\mathscr{M}$ has a bounded projection on $\mathbb{R}$, then $\mathscr{D}_k-\mathscr{M}$ meets $I_j\times \{\infty\}$ for some $j\neq k$. \end{theorem} \begin{theorem} \label{thm2.3} There are two subcontinua $\mathscr{D}_{k}^+$ and $\mathscr{D}_{k}^-$, consisting of the bifurcation branch $\mathscr{D}_{k}$, which satisfy the alternatives of Theorem \ref{thm2.2}. Moreover, there exists a neighborhood $\mathscr{N}\subset\mathscr{M}$ of $I_k\times \{\infty\}$ such that $(\mathscr{D}_k^\nu\cap\mathscr{N})\subset(\Phi_k^\nu\cup(I_k\times \{\infty\}))$ for $\nu=+$ and $\nu=-$. \end{theorem} \section{Applications} In this section, we shall use Theorems \ref{thm2.1}--\ref{thm2.3} to prove the existence of nodal solutions for problem \eqref{dn} under the assumptions of (C3), (C6) and (C7). The main results of this section are the following theorems. \begin{theorem} \label{thm3.1} For some $k\in \mathbb{N}$, if $g_0>-\underline{f}_0$ and $g_\infty >-\underline{f}_\infty$, either \label{arc1} \frac{\lambda_k}{g_0+\underline{f}_0}-\underline{f}_0$and$-\overline{f}_\infty-\underline{f}_0$and$g_\infty \leq-\overline{f}_\infty$, for $r>\frac{\lambda_k}{g_0+\underline{f}_0},$ then the conclusion of Theorem \ref{thm3.1} is valid. \end{theorem} \begin{theorem} \label{thm3.4} For some$k\in \mathbb{N}$, if$-\overline{f}_0-\underline{f}_\infty$, for $\frac{\lambda_k}{g_\infty+\underline{f}_\infty}-\underline{f}_\infty, for \[ r>\frac{\lambda_k}{g_\infty+\underline{f}_\infty},$ then the conclusion of Theorem \ref{thm3.1} is valid. \end{theorem} \begin{proof}[Proof of Theorem \ref{thm3.1}] Firstly, we study the bifurcation phenomena for the following eigenvalue problem $$\label{pe1} \begin{gathered} Lu=\lambda r a(x)g(u)+ra(x)f(u) \quad x\in (0,1),\\ b_0u(0)+c_0u'(0)=0,\quad b_1u(1)+c_1u'(1)=0, \end{gathered}$$ where$\lambda\in \mathbb{R}$is a parameter. Let$\zeta\in C(\mathbb{R},\mathbb{R})$be such that $$\label{pe3} g(s)=g_0s+\zeta(s)$$ with$\lim_{\vert s\vert\to0}\zeta(s)/s=0$. Let$\widetilde{\zeta}(u)=\max_{0\leq \vert s\vert\leq u}\vert \zeta(s)\vert$, then$\widetilde{\zeta}$is nondecreasing and $$\label{pe4} \lim_{ u\to 0^+}\frac{\widetilde{\zeta}(u)}{u}=0.$$ Further it follows from \eqref{pe4} that $$\label{pe5} \frac{\zeta(u)}{\| u\|} \leq\frac{\widetilde{\zeta}(\vert u\vert)}{\| u\|} \leq \frac{\widetilde{\zeta}(\| u\|_\infty)}{\| u\|} \leq \frac{ \widetilde{\zeta}(\| u\|)}{\|u\|}\to0\quad \text{as } \|u\|\to 0.$$ Hence, \eqref{pe1}, \eqref{pe3} and \eqref{pe5} imply that conditions (C2) and (C4) hold. Moreover, we have that $I_k=[\frac{\lambda_k}{rg_0}-\frac{\overline{f}_0}{g_0}, \frac{\lambda_k}{rg_0}-\frac{\underline{f}_0}{g_0}]:=I_k^0.$ Using Theorem \ref{thm2.1}, there exist two distinct unbounded components$\mathscr{D}_{k,0}^+$of$\mathscr{S}_k^+\cup(I_k^0\times \{0\})$, containing$I_k^0\times \{0\}$and lying in$\Phi_k^+\cup(I_k^0\times \{0\})$, and$\mathscr{D}_{k,0}^-$of$\mathscr{S}_k^-\cup(I_k^0\times \{0\})$, containing$I_k^0\times \{0\}$and lying in$\Phi_k^-\cup(I_k^0\times \{0\})$. Next we study the unilateral global bifurcation of \eqref{pe1} which bifurcates from infinity. Let$\xi\in C(\mathbb{R},\mathbb{R})$be such that $$\label{pei1} g(s)=g_\infty s+\xi(s)$$ with$\lim_{\vert s\vert\to+\infty}\xi(s)/s=0$. Let$\widetilde{\xi}(u)=\max_{0\leq \vert s\vert\leq u}\vert \xi(s)\vert$, then$\widetilde{\xi}$is nondecreasing. Define $\overline{\xi}(u)=\max_{u/2\leq \vert s\vert\leq u}\vert \xi(s)\vert.$ Then we can see that $$\label{tj1} \lim_{ u\to +\infty}\frac{\overline{\xi}(u)}{u}=0\quad\text{and}\quad \widetilde{\xi}(u)\leq \widetilde{\xi}(\frac{u}{2})+\overline{\xi}(u).$$ It is not difficult to verify that$\widetilde{\xi}(s)/s$is bounded in$\mathbb{R}^+$. This fact and \eqref{tj1} follows that $\limsup_{ u\to +\infty}\frac{\widetilde{\xi}(u)}{u} \leq \limsup_{ u\to +\infty}\frac{\widetilde{\xi}(u/2)}{u} =\limsup_{ t\to +\infty}\frac{\widetilde{\xi}(t)}{2t},$ where$t=u/2$. So we have $$\label{pei2} \lim_{ u\to +\infty}\frac{\widetilde{\xi}(u)}{u}=0.$$ Further from \eqref{pei2} it follows that $$\label{pei3} \frac{\xi(u)}{\| u\|} \leq\frac{ \widetilde{\xi}(\vert u\vert)}{\| u\|} \leq \frac{ \widetilde{\xi}(\| u\|_\infty)}{\| u\|} \leq \frac{ \widetilde{\xi}(\| u\|)}{\| u\|}\to0\quad \text{as } \| u\|\to +\infty.$$ Hence, \eqref{pe1}, \eqref{pei1} and \eqref{pei3} imply that conditions (C8) and (C9) hold. Moreover, we have that $I_k=[\frac{\lambda_k}{rg_\infty}-\frac{\overline{f}_\infty}{g_\infty}, \frac{\lambda_k}{rg_\infty}-\frac{\underline{f}_\infty}{g_\infty}]:=I_k^\infty.$ Using Theorem \ref{thm2.3}, we have that there are two components$\mathscr{D}_{k,\infty}^+$and$\mathscr{D}_{k,\infty}^-$of$\mathscr{S}\cup(I_k^\infty\times \{\infty\})$, containing$I_k^\infty\times \{\infty\}$, which satisfy the alternates of Theorem \ref{thm2.2}. Moreover, there exists a neighborhood$\mathscr{N}\subset\mathscr{M}$of$I_k^\infty\times \{\infty\}$such that$(\mathscr{D}_{k,\infty}^\nu\cap\mathscr{N}) \subset(\Phi_k^\nu\cup(I_k^\infty\times \{\infty\}))$for$\nu=+$and$\nu=-$. We claim that$\mathscr{D}_{k,0}^+=\mathscr{D}_{k,\infty}^+$and$\mathscr{D}_{k,0}^-=\mathscr{D}_{k,\infty}^-$. We only prove$\mathscr{D}_{k,0}^+=\mathscr{D}_{k,\infty}^+$since the proof of$\mathscr{D}_{k,0}^-=\mathscr{D}_{k,\infty}^-$is similar. As in \cite{DM1}], it suffices to show that$\mathscr{D}_{k,\infty}^+$meets some point$(\lambda_*,0)$of$\mathscr{R}$; i.e., (1) of Theorem \ref{thm2.2} occurs. Suppose on the contrary that (2) of Theorem \ref{thm2.2} occurs. Firstly, we shall show that$\mathscr{D}_{k,\infty}^+-\mathscr{M}$has a bounded projection on$\mathbb{R}$. By the same method as that of \cite{DM1}], we can show that$\mathscr{D}_{k,\infty}^+\subset\Phi_k^+$. On the contrary, we suppose that$(\mu_n, y_n) \in \mathscr{D}_{k,\infty}^+-\mathscr{M}$such that $\lim_{n\to +\infty}\mu_n=\pm\infty.$ It follows that $Ly_n=\mu_n r ag(y_n)+r af(y_n).$ Let $0<\tau(1,n)<\tau(2,n)<\dots <\tau(k-1,n)<1$ denote the zeros of$y_n$in$(0,1)$. Let$\tau(0,n=0$and$\tau(k,n)=1$. Then, after taking a subsequence if necessary, $\lim_{n\to +\infty}\tau(l,n)=\tau(l,\infty),\quad l\in\{0,1,\ldots,k\}.$ We claim that there exists$l_0\in \{0,1,\ldots,k\}$such that $\tau(l_0,\infty)<\tau(l_0+1,\infty).$ Otherwise, we have $1=\Sigma_{l=0}^{k-1}(\tau(l+1,n)-\tau(l,n))\to \Sigma_{l=0}^{k-1}(\tau(l+1,\infty)-\tau(l,\infty))=0.$ This is a contradiction. Let$(\alpha,\beta)\subset(\tau(l_0,\infty),\tau(l_0+1,\infty))$with$\alpha<\beta$. For all$n$sufficiently large, we have$(\alpha,\beta)\subset(\tau(l_0,n),\tau(l_0+1,n))$. So$y_n$does not change its sign in$(\alpha,\beta)$. In view of (C6) and (C7), we have that$\underset{n\to+\infty}\lim r(\mu_n \frac{g(y_n(x))}{y_n(x)} +\frac{f(y_n(x))}{y_n(x)}) =\pm\infty$for any$x\in(\alpha,\beta)$. If$\underset{n\to+\infty}\lim r(\mu_n \frac{g(y_n(x))}{y_n(x)}+\frac{f(y_n(x))}{y_n(x)}) =-\infty$for any$x\in(\alpha,\beta)$, applying Sturm Comparison Theorem \cite{I,W} to$y_n$and$\varphi_1$on$[\alpha,\beta]$, we can get that$\varphi_1$must change its sign in$(\alpha,\beta)$for$n$large enough. While, this is impossible. So we have that$\underset{n\to+\infty} \lim(\mu_n r \frac{g(y_n(x))}{y_n(x)}+r\frac{f(y_n(x))}{y_n(x)}) =+\infty$for any$x\in(\alpha,\beta)$. Applying Sturm Comparison Theorem \cite{I,W}] to$\varphi_1$and$y_n$, we get that$y_n$has at least one zero in$(\alpha,\beta)$for$n$large enough, and this contradicts the fact that$y_n$does not change its sign in$(\alpha,\beta)$. By an argument similar to that of \cite[Theorem 3.1]{DM1}, we can show that the case of$\mathscr{D}_{k,\infty}^+-\mathscr{M}$meeting$I_j^\infty\times \{\infty\}$for some$j\neq k$is impossible. This is a contradiction. For simplicity, we write$\mathscr{D}_{k}^+:=\mathscr{D}_{k,0}^+=\mathscr{D}_{k,\infty}^+$and$\mathscr{D}_{k}^-:=\mathscr{D}_{k,0}^-=\mathscr{D}_{k,\infty}^-$. It is clear that any solution of \eqref{pe1} of the form$(1, u)$yields a solution$u$of \eqref{dn}. While, by some simple computations, we can show that assumption \eqref{arc1} or \eqref{arc2} implies that$\mathscr{D}_{k}^+$and$\mathscr{D}_{k}^-$cross the hyperplane$\{1\}\times E$in$\mathbb{R}\times E$. \end{proof} \subsection*{Geometric meaning} The meaning of$\frac{\lambda_k}{r g_0}-\frac{\underline{f}_0}{g_0}<1 <\frac{\lambda_k}{r g_\infty}-\frac{\overline{f}_\infty}{g_\infty }$is that subsets$I_k^0\times E$and$I_k^\infty\times E$of$\mathbb{R}\times E$can be separated by the hyperplane$\{1\}\times E$, and$I_k^0\times E$on the left of$\{1\}\times E$while$I_k^\infty\times E$on the right of$\{1\}\times E$. Similarly, the meaning of$\frac{\lambda_k}{r g_\infty}-\frac{\underline{f}_\infty}{g_\infty}<1 <\frac{\lambda_k}{r g_0}-\frac{\overline{f}_0}{g_0}$is that$I_k^0\times E$on the right of$\{1\}\times E$and$I_k^\infty\times E$on the left of$\{1\}\times E$. \begin{proof}[Proof of Theorems \ref{thm3.2} and \ref{thm3.3}] The proof is similar to that of Theorem \ref{thm3.1}, we note only that the assumptions of theorems imply$\frac{\lambda_k}{r g_0}-\frac{\underline{f}_0}{g_0}<1 <\frac{\lambda_k}{r g_\infty}-\frac{\overline{f}_\infty}{g_\infty }$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm3.4} and \ref{thm3.5}] We note that the assumptions of these theorems imply$\frac{\lambda_k}{r g_\infty}-\frac{\underline{f}_\infty}{g_\infty}<1 <\frac{\lambda_k}{r g_0}-\frac{\overline{f}_0}{g_0}$. \end{proof} \begin{remark} \label{rmk3.1}\rm By some simple computations, we can show that if$g_0$and$g_\infty$satisfy one of the following two cases \begin{itemize} \item$-\overline{f}_0-\underline{f}_0$and$g_\infty >-\underline{f}_\infty$, either $\frac{\lambda_n}{g_0+\underline{f}_0}-\underline{f}_0 and -\overline{f}_\infty-\underline{f}_\infty, for \[ \frac{\lambda_n}{g_\infty+\underline{f}_\infty}0 and \underline{f}_\infty>0, either \[ \frac{\lambda_n}{\underline{f}_0}0 and \overline{f}_\infty>0\geq\underline{f}_\infty, for \[ \frac{\lambda_n}{\underline{f}_0}0 and \overline{f}_\infty\leq 0, for \[ r>\frac{\lambda_n}{\underline{f}_0},$ then the conclusion of Corollary \ref{coro3.1} is valid. \end{corollary} \begin{corollary} \label{coro3.4} For some$k,n\in \mathbb{N}$with$k\leq n$, if$\overline{f}_0>0\geq\underline{f}_0$and$\underline{f}_\infty>0$, for $\frac{\lambda_n}{\underline{f}_\infty}0, for \[ r>\frac{\lambda_n}{\underline{f}_\infty},$ then the conclusion of Corollary \ref{coro3.1} is valid. \end{corollary} \begin{remark} \label{rmk3.3} \rm If$r=1$and$n=k+j$for any$j\in\{0\}\cup \mathbb{N}$, then Corollary \ref{coro3.1} reduces to \cite[Theorem 2]{MT}. If$n=k$, then Corollary \ref{coro3.1} becomes \cite[Corollary 1]{MT}. If$n=k$,$\underline{f}_0=\overline{f}_0$,$\underline{f}_\infty=\overline{f}_\infty$,$p\equiv1$,$q\equiv0$,$b_0 = b_1 = 1$and$c_0 = c_1 = 0$, then Corollary \ref{coro3.1} reduces to \cite[Theorem 1.1]{MT1}. Note that signum condition is removed in this paper while it is essential in \cite{MT1,MT}]. \end{remark} \subsection*{Acknowledgments} This research was supported by the NNSF of China (Nos. 11261052, 11361054, 11201378). The authors are very grateful to an anonymous referees for their careful reading and valuable comments on the manuscript. \begin{thebibliography}{99} \bibitem{Be} H. Berestycki; \emph{On some nonlinear Sturm-Liouville problems}, J. Differential Equations 26 (1977), 375--390. \bibitem{CL} E. A. Coddington, N. Levinson; \emph{Theory of ordinary differential equations}, McGraw-Hill, New York, 1955. \bibitem{DM} G. Dai, R. 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