\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 13, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/13\hfil Global structure of positive solutions] {Global structure of positive solutions for superlinear singular $m$-point boundary-value problems} \author[X. Zhang\hfil EJDE-2008/13\hfilneg] {Xingqiu Zhang} \address{Xingqiu Zhang \newline Department of Mathematics, Liaocheng University, Liaocheng 252059, China} \email{zhangxingqiu@lcu.edu.cn} \thanks{Submitted May 31, 2007. Published January 31, 2008.} \thanks{Supported by grant 10671167 from the National Natural Science Foundation of China and by \hfill\break\indent grant 31805 from Science Foundation of Liaocheng University} \subjclass[2000]{34B10,34B16} \keywords{Superlinear; singular; $m$-point boundary value problem; \hfill\break\indent global structure} \begin{abstract} Using topological methods and a well known generalization of the Birkhoff-Kellogg theorem, we study the global structure of a class of superlinear singular $m$-point boundary value problem. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \section{Introduction} We are concerned with the nonlinear second-order singular $m$-point boundary-value problem \begin{equation} \begin{gathered} -(L\varphi)(x)=\lambda f(x,\varphi(x)),\quad 00$, $\varphi\in C[I,R]\cap C^2[(0,1),R]$ satisfying \eqref{e1.1}, where $I=[0,1]$. In addition, if $\lambda>0$, $\varphi(x)>0$ holds for any $x\in (0,1)$, then $(\lambda,\varphi)$ is called a positive solution of \eqref{e1.1}. The rest of this paper is organized as follows. Section 2 gives some necessary lemmas. Section 3 is devoted to the main result and its proof. An example is worked out in Section 4 to indicate the application of our main result. \section{Preliminary Lemmas} Throughout this paper, we always suppose \begin{itemize} \item[(H1)] $p(x)\in C^1[0,1]$, $p(x)>0$, $q(x)\in C[0,1]$, $q(x)\leq 0$. \end{itemize} \begin{lemma}[\cite{z1}] \label{lem2.1} Assume that {\rm (H1)} holds. Let $\phi_1(x), \phi_2(x)$ be the solution of \begin{equation} \begin{gathered} (L\varphi)(x)=0,\quad 00,x\in (0,1]$; \item[(ii)] $\phi_2(x)$ is decreasing on [0,1] and $\phi_2(x)>0,x\in [0,1)$. \end{itemize} \end{lemma} Let \begin{equation} k(x,y)=\begin{cases} \frac{1}{\rho}\phi_1(x)\phi_2(y), & 0\leq x\leq y\leq1,\\[3pt] \frac{1}{\rho}\phi_1(y)\phi_2(x), & 0\leq y\leq x\leq1, \end{cases}\label{e2.3} \end{equation} where $\rho=\phi'_1(0)$. By Lemma \ref{lem2.1} we know that $\phi'_1>0$. Let \begin{equation} K(x,y)=k(x,y)+D^{-1}\phi_1(x)\sum_{i=1}^{m-2}a_ik(\xi_i,y),\quad 0\leq x,y\leq1\label{e2.4} \end{equation} where $D=1-\sum_{i=1}^{m-2}a_i\phi_1(\xi_i)$. \begin{lemma}[\cite{z1}] \label{lem2.2} Assume {\rm (H1)} holds. Then $k(x,y)$ defined by \eqref{e2.3} possesses the following properties: \begin{itemize} \item[(i)] $k(x,y)$ is continuous and symmetrical over $[0,1]\times[0,1]$; \item[(ii)] $k(x,y)\geq0$, and $k(x,y)\leq k(y,y)$, for all $0\leq x,y\leq 1$; \item[(iii)] There exist constants $k_1,k_2>0$ such that $$ k_1x(1-x)\leq k(x,x)\leq k_2x(1-x),x\in[0,1]. $$ \end{itemize} \end{lemma} We make the following assumptions: \begin{itemize} \item[(H2)] $\sum_{i=1}^{m-2}a_i\phi_1(\xi_i)<1$, where $\phi_1(x)$ is the unique solution of \eqref{e2.1}. \item[(H3)] $f: (0,1)\times(0,+\infty)\to R^+$ is continuous (it may be singular at $x=0,1$ and $\varphi=0$) and for any $R>r>0$, $\int_0^1K_1(y,y)f_{r,R}(y) \mathrm{d}y<+\infty$ where \[ K_1(y,y)=y(1-y)+D^{-1}\sum_{i=1}^{m-2}a_ik(\xi_i,y); \] $f_{r,R}(y):=\sup\{f(y,\varphi):\varphi\in [\rho k_1y(1-y)r,R],y\in (0,1)\}$, $k_1$ has the same meaning as in Lemma \ref{lem2.2}. \item[(H4)] For every $R>0$, there exists $\psi_R\in C[I,R^+]$ ($\psi_R\not\equiv\theta$) such that $$ f(x,\varphi)\geq\psi_R(x),\quad \text{for } x\in (0,1),\varphi\in (0,R]. $$ \item[(H5)] There exists $[a,b]\subset(0,1)$ such that $$ \lim_{\varphi\to +\infty}\frac{f(x,\varphi)}{\varphi}=+\infty \quad \text{uniformly for } x\in [a,b]. $$ \end{itemize} Set \begin{equation} (A\varphi)(x)=\int_0^1K(x,y)\widetilde{p}(y)f(x,\varphi(y)) \mathrm{d}y,\ \ x\in [0,1],\label{e2.5} \end{equation} where \begin{equation} \widetilde{p}(y)=\frac{1}{p(y)}\exp\Big(\int_0^y\frac{p'(s)}{p(s)}\mathrm{d}s\Big).\label{e2.6} \end{equation} Let \[ P=\{\varphi\in C[0,1]:\varphi(x)\geq0,\varphi(x)\geq\|\varphi\|\rho k_1x(1-x),\rho k_1<4,x\in[0,1]\}. \] where $k_1$ has the same meaning as in Lemma \ref{lem2.2}. It is easy to check that $P$ is a cone in $C[0,1]$. The following theorem is the generalization of the well known Birkhoff-Kellogg. \begin{lemma}[\cite{g1,s1}] \label{lem2.3} Let $X$ be an infinite-dimensional Banach space, $P$ a cone of $X$, and $A:P\to P$ a completely continuous operator. Suppose that there exists a bounded open set $\Omega$ in $X,\theta\in \Omega$ such that $$ \inf_{x\in P\cap\partial\Omega}\|Ax\|>0. $$ Then the closure of the set of nonzero solutions of the equation $\varphi=\lambda A\varphi$, i.e., $$ \Sigma:=\overline{\{(\lambda,\varphi):\lambda\in R_+,\varphi\in P, \varphi\neq\theta,\varphi=\lambda A\varphi\}} $$ possesses a maximal subcontinuum $C$ (i.e., a maximal closed connected subsets of $\sum$), which is unbounded and there exists $\overline{\lambda}>0$ (for example we may choose $\overline{\lambda} >\sup_{x\in P\cap\partial\Omega}\|x\|/\inf_{x\in P\cap\partial\Omega}\|Ax\|$) such that \begin{itemize} \item[(i)] $C\cap ((0,+\infty)\times P\setminus((\overline{\lambda},+\infty)\times\overline{\Omega}))$ is unbounded; \item[(ii)] $C\cap([\overline{\lambda},+\infty)\times\partial\Omega) =\emptyset$, $C\cap(\{0\}\times(P\setminus \{\theta\}))=\emptyset$; and either \item[(iii)]$C\cap([\overline{\lambda},+\infty)\times\Omega)$ is unbounded, or \item[(iii)$^*$] $C\cap([0,+\infty)\times\{\theta\})\neq\emptyset$, \end{itemize} where $\theta$ denotes zero element of $X$. \end{lemma} \section{Main Result} First, we consider the following approximating problem of BVP \eqref{e1.1} \begin{equation} \begin{gathered} -(L\varphi)(x)=\lambda f_n(x,\varphi(x)),\quad 00,\varphi\in C^2[(0,1), R^+]\cap C[I,R^+]$ and for any $ x\in (0,1),\varphi(x)>0$. It is obvious, $\varphi(x)=\lambda A_n\varphi(x)$. Take $x_0\in [0,1]$ such that $\varphi(x_0)=\|\varphi\|$. From \cite{z1}, for any $x,y\in [0,1]$ we have $k(x,y)\geq k(x_0,y)\phi_1(x)\phi_2(x)$. So, we have \begin{align*} \varphi(x)&=\lambda\int_0^1k(x,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\quad +\lambda D^{-1}\phi_1(x)\sum _{i=1}^{m-2}a_i\int_0^1k(\xi_i,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\geq\lambda\phi_1(x)\phi_2(x)\int_0^1k(x_0,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\quad +\lambda D^{-1}\phi_1(x)\sum_{i=1}^{m-2}a_i\int_0^1k(\xi_i,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\geq\lambda\phi_1(x)\phi_2(x)\Big[\int_0^1k(x_0,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\quad +D^{-1}\sum_{i=1}^{m-2}a_i\int_0^1k(\xi_i,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\Big]\\ &\geq\lambda\phi_1(x)\phi_2(x)\Big[\int_0^1k(x_0,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\quad +D^{-1}\phi_1(x_0)\sum_{i=1}^{m-2}a_i\int_0^1k(\xi_i,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\Big] \\ &=\varphi(x_0)\phi_1(x)\phi_2(x)=\|\varphi\|\rho k_1x(1-x) \end{align*} As a consequence, $\varphi\in P\setminus\{\theta\}$. \end{proof} \begin{lemma} \label{lem3.2} Assume {\rm (H1)--(H3)} hold. Then $A_n:P\to P$ is continuous for each $n\in N$. \end{lemma} The proof of the above lemma is obvious, so we omit it. Let \[ L_n:=\{(\lambda,\varphi)\in R^+\times P:\varphi=\lambda A_n\varphi\}\ \text{for all} \ n\geq1. \] \begin{lemma} \label{lem3.3} Suppose {\rm (H1)--(H4)} hold. Then for each $n, L_n$ is locally compact in $[0,+\infty)\times P$ and \[ L_n=\overline{\{(\lambda,\varphi)\in R^+\times P:\varphi=\lambda A_n\varphi,\varphi\neq\theta\}}. \] \end{lemma} \begin{proof} For every $R>0$, let $L_n^R:=\{(\lambda,\varphi)\in L_n:|\lambda|\leq R,|\varphi|\leq R\}$. If $(\lambda,\varphi)\in L_n$ and $\varphi=\theta$, then by (H4) we get $\lambda=0$. So, we need only to prove that $L_n^R$ is relatively compact and closed. In fact, for each $(\lambda,\varphi)\in L_n^R, $ from the construction of $P$ we have \[ f_n(x,\varphi(x))\leq f_{\frac{1}{n},R}(x)\ \ \text{for all}\ x\in (0,1), \] and \[ \varphi(x)=\lambda\int_0^1K(x,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y,\quad x\in [0,1]. \] Combining with (H3), it is easy to know that $\{\varphi=\varphi(x):(\lambda,\varphi)\in L_n^R\}$ are equicontinuous on $I$. Thus, from Ascoli-Arzela theorem we get that $L_n^R$ is relatively compact. On the other hand, (H3) and Lebesgue dominated convergence theorem guarantee that $L_n^R$ is closed. \end{proof} The next theorem gives the global structure of $L_n$. \begin{theorem} \label{thm3.1} Suppose that {\rm (H1)--(H5)} hold. Then for each $n\geq 1,L_n$ possesses a maximal and unbounded subcontinuum $C_n$, which comes from $(0,\theta)$ and tends to $(0,+\infty)$ eventually satisfying \begin{itemize} \item[(1)] $(0,\theta)\in C_n$; \item[(2)] There exists $\lambda_n^0\in (0,+\infty)$ such that $$ C_n \subset [0,\lambda_n^0]\times P, \quad C_n\cap(\{\lambda\}\times P)\neq\emptyset, \quad \forall\ \lambda\in [0,\lambda_n^0]; $$ \item[(3)] $C_n$ is unbounded in $[0,\lambda_n^0]\times P$; \item[(4)] $\lambda=0$ is an unique asymptotic bifurcation point of $A_n$; \item[(5)] There exists $\lambda_n^*\in (0,\lambda_n^0]$ such that for each $\lambda\in (0,\lambda_n^*)$, \eqref{e3.1} has at least two positive solution $\varphi_{n\lambda}^*$ and $\varphi_{n\lambda}^{**}$ satisfying $$ \|\varphi_{n\lambda}^*\|\leq\|\varphi_{n\lambda}^{**}\|,\quad (\lambda,\varphi_{n\lambda}^*), (\lambda,\varphi_{n\lambda}^{**})\in C_n; $$ \item[(6)] $$ \lim_{\lambda\to 0^+,\, (\lambda,\varphi_{n\lambda}^*)\in C_n} \|\varphi_{n\lambda}^*\|=0, \quad \lim_{\lambda\to 0^+,\, (\lambda,\varphi_{n\lambda}^{**})\in C_n} \|\varphi_{n\lambda}^{**}\|=+\infty. $$ \end{itemize} \end{theorem} \begin{proof} First we prove that for every $\overline{\lambda}>0$, there exists $\overline{R}>0$ such that \begin{equation} L_n\cap([\overline{\lambda},+\infty)\times (P\setminus\overline{P}_{\overline{R}}))=\emptyset,\quad n=1,2,\dots, \label{e3.3} \end{equation} where $P_{\overline{R}}=\{\varphi\in P: \|\varphi\|<\overline{R}\}$. In fact, take a positive number $l$ satisfying \begin{equation} l>\Big(\rho k_1\overline{\lambda}\max_{x\in I}\int_a^bK(x,y)\widetilde{p}(y)y(1-y) \mathrm{d}y\Big)^{-1}>0, \label{e3.4} \end{equation} where $a,b$ are as the same as in (H5). Then there exists $R'>1$ such that \begin{equation} f(x, u)\geq lu\quad \text{for all } x\in [a,b], u>R'. \label{e3.5} \end{equation} Choose a number $\overline{R}$ with $\overline{R}>\frac{R'}{\rho k_1a(1-b)}$. It follows from the definition of cone $P$ that \begin{equation} \varphi(y)\geq\|\varphi\|\rho k_1y(1-y)\geq\rho k_1a(1-b)\overline{R}>R'\quad \text{for all } y\in [a,b], \varphi\in P\setminus \overline{P}_{\overline{R}}. \label{e3.6} \end{equation} Therefore, by \eqref{e3.5} and \eqref{e3.6} for $\lambda\geq\overline{\lambda}$ and $\varphi\in P\setminus \overline{P}_{\overline{R}}$ \begin{align*} \lambda A_n\varphi(x) &=\lambda\int_0^1K(x,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\geq\overline{\lambda}\int_a^bK(x,y)\widetilde{p}(y)f(y,\varphi(y)) \mathrm{d}y\\ &\geq l\overline{\lambda}\int_a^bK(x,y)\widetilde{p}(y)\varphi(y) \mathrm{d}y \\ & \geq\rho k_1l\overline{\lambda}\|\varphi\|\int_a^bK(x,y) \widetilde{p}(y)y(1-y)\mathrm{d}y \end{align*} Combining with \eqref{e3.4}, we have \begin{equation} \|\lambda A_n\varphi\|\geq\rho k_1l\overline{\lambda}\|\varphi\|\max_{x\in I}\int_a^bK(x,y)\widetilde{p}(y)y(1-y) \mathrm{d}y >\|\varphi\|, \label{e3.7} \end{equation} for all $\lambda\geq\overline{\lambda}$, $\varphi\in P\setminus \overline{P}_{\overline{R}}$, which implies that \eqref{e3.3} holds. On the other hand, from the definition of $f_n(x,\varphi(x))$, for fixed $n\geq1$ we have $0<\varphi(y)\leq\overline{R}$, for all $\varphi\in \overline{P}_{\overline{R}}$. Consequently, by (H4) we know \begin{equation} A_n\varphi(x)=\int_0^1K(x,y)\widetilde{p}(y)f_n(x,\varphi(y)) \mathrm{d}y\geq\int_0^1K(x,y)\widetilde{p}(y)\psi_{\overline{R}} (y) \mathrm{d}y,\forall\ \varphi\in \overline{P}_{\overline{R}}. \label{e3.8}\end{equation} Let $ r<\min\Big\{R,\overline{\lambda}\max_{x\in I}\int_0^1K(x,y)\widetilde{p}(y)\psi_{\overline{R}}(y) \mathrm{d}y\Big\}$. This together with \eqref{e3.8} implies that for any $\varphi\in\overline{P}_r, \lambda>\overline{\lambda}$ \begin{equation} \begin{aligned} \|\lambda A_n\varphi\|&=\lambda\max_{x\in I}\int_0^1K(x,y) \widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &>\overline{\lambda}\max_{x\in I}\int_0^1K(x,y)\widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\geq r=\|\varphi\|, \end{aligned}\label{e3.9} \end{equation} which yields \begin{equation} L_n\cap((\overline{\lambda},+\infty)\times P_r)=\emptyset. \label{e3.10}\end{equation} Note that \eqref{e3.7} implies \begin{gather*} \inf_{\varphi\in \partial {P}_{\overline{R}}}\|A_n\varphi\|\geq \rho k_1l\overline{R}\max_{x\in I}\int_a^bK(x,y)\widetilde{p}(y)y(1-y) \mathrm{d}y>0, \\ \overline{\lambda}>\sup_{\varphi\in\partial P_{\overline{R}}}\|\varphi\|/\inf_{\varphi\in\partial P_{\overline{R}}}\|A_n\varphi\|. \end{gather*} As a consequence, by \eqref{e3.3} \eqref{e3.10} and Lemma \ref{lem2.3} we get that $L_n$ possesses a maximal and unbounded subcontinuum $C_n$ satisfying that \begin{equation} \begin{gathered} C_n\cap((0,+\infty)\times P)\setminus((\overline{\lambda},+\infty)\times \overline{P}_{\overline{R}}) \text{ is unbounded and} \\ C_n\cap((\overline{\lambda},+\infty)\times\{ P_r\cup (P\setminus \overline{P}_{\overline{R}})\})=\emptyset.\end{gathered}\label{e3.11} \end{equation} Next, for $(\lambda,\varphi)\in L_n\cap([\overline{\lambda},+\infty)\times (\overline{P}_{\overline{R}} \setminus P_r))$, noticing that $\rho k_1 rx(1-x)\leq \varphi(x)\leq\overline{R}$ for $x\in I$, by (H4) we can get $$ \varphi(x)=\lambda (A_n\varphi)(x)=\lambda\int_0^1K(x,y) \widetilde{p}(y)f_n(x,\varphi(y)) \mathrm{d}y \geq\lambda\int_0^1K(x,y)\widetilde{p}(y)\psi_{\overline{R}}(y) \mathrm{d}y $$ This means \begin{equation} \lambda\leq\overline{R}\Big(\max_{x\in I}\int_a^bK(x,y)\widetilde{p}(y)\psi_{\overline{R}}(y) \mathrm{d}y\Big)^{-1}, \label{e3.12} \end{equation} which implies $L_n\cap ([\overline{\lambda},+\infty)\times (\overline{P}_{\overline{R}}\setminus P_r))$ is bounded. This together with \eqref{e3.3} and \eqref{e3.10} guarantees that \begin{equation} L_n\cap([\overline{\lambda},+\infty)\times P) \ \text{is bounded }, \forall\ \overline{\lambda}>0. \label{e3.13} \end{equation} Thus, by \eqref{e3.11}\eqref{e3.13} we know that $C_n\cap((0,\overline{\lambda}]\times P)$ is unbounded. Furthermore, by virtue of (iii) and ${\mathrm{(iii)}}^*$ of Lemma 2.3 and \eqref{e3.11} \eqref{e3.12} one can get \[ C_n\cap ([0,+\infty)\times\{\theta\})\neq\emptyset. \] Now we show that \[ C_n\cap ([0,+\infty)\times\{\theta\})=\{(0,\theta)\}. \] Suppose $(\lambda_0,\theta)\in C_n\cap ([0,+\infty)\times\{\theta\})$, then there exist $\lambda_m\in R^+$ and $\varphi_m\in P\setminus\{\theta\},m=1,2,\dots$ such that \[ \varphi_m(x)=\lambda_m(A_n\varphi_m)(x),\ \ \lambda_m\to\lambda_0,\ \ \varphi_m\to\theta\ \ (m\to+\infty). \] Without loss of generality, assume $\varphi_m\in P_{\overline{R}}\setminus\{\theta\}$. Then \[ (A_n\varphi_m)(x)\geq\int_0^1K(x,y)\widetilde{p}(y)\psi_{\overline{R}}(y) \mathrm{d}y. \] Therefore, \[ |\lambda_m|\leq\frac{\|\varphi_m\|}{\max_{x\in I}\int_0^1K(x,y)\widetilde{p}(y)\psi_{\overline{R}}(y) \mathrm{d}y} \to 0\quad (m\to +\infty). \] So, $\lambda_0=0$, i.e., $C_n\cap ([0,+\infty)\times\{\theta\})=\{(0,\theta)\}$. As a consequence, (1) holds. By Lemma \ref{lem2.3} we know $C_n$ is a maximal and unbounded subcontinuum which comes from $(0,\theta)$. On the other hand, suppose $\lambda_0\in (0,\overline{\lambda}]$ is an asymptotic bifurcation point of the operator $A_n$. Then there exist $\lambda_m\in R^+$ and $\varphi_m\in P\setminus P_{\overline{R}}$ such that $\varphi_m=\lambda_mA_n\varphi_m$ and $\lambda_m\to\lambda_0,\|\varphi_m\| \to+\infty$ as $m\to +\infty$. From (H5), as in the proof of \eqref{e3.7}, one obtain \[ \frac{1}{\lambda_m}=\frac{\|A_n\varphi_m\|}{\|\varphi_m\|}\to +\infty\quad (\|\varphi_m\|\to+\infty). \] This means that $\lambda_0=0$ is the unique asymptotic bifurcation point. Therefore, $C_n$ tends to $(0,+\infty)$; i.e., (4) holds. Let $\mathcal{L}:=\{\lambda:\text{there exists}\ \varphi\in P\setminus\{\theta\} \ \text{such that}\ \varphi=\lambda A_n\varphi\}$. Obviously, $\mathcal{L}\neq\emptyset$. Let $\lambda_n^0:=\sup\{\lambda:\lambda\in \mathcal{L}\}$. By virtue of \eqref{e3.11} \eqref{e3.12} we know $\lambda_n^0\in (0,+\infty)$. Suppose $(\lambda_m,\varphi_m)\in L_n$ satisfying $\lambda_m\to \lambda_n^0,\ m\to\infty$. It follows from \eqref{e3.13} that $\{\varphi_m\}$ is bounded. By Lemma \ref{lem3.3} there exists $\varphi\in P\setminus\{\theta\}$ such that $(\lambda_n^0,\varphi)\in L_n$. Consequently, noticing $C_n$ is unbounded, by virtue of the connection of subcontinuum one can get (2) holds. Consequently, we have \begin{equation} L_n\cap((\lambda_n^0,+\infty)\times P)=\emptyset. \label{e3.14}\end{equation} Considering $C_n$ is unbounded and 0 is the unique asymptotic bifurcation point, it is not difficult to know from \eqref{e3.14} that (3) also holds. To get (5) and (6), noticing that for $(\lambda,\varphi)\in L_n\cap((0,+\infty)\times(\overline{P}_R\setminus P_r))$ ($R>1>r>0$), we have \begin{align*} \varphi(x)&=\lambda (A_n\varphi)(x)=\lambda\int_0^1K(x,y) \widetilde{p}(y)f_n(y,\varphi(y)) \mathrm{d}y\\ &\leq\lambda \int_0^1K(x,y)\widetilde{p}(y)f_{r,R}(y) \mathrm{d}y, \end{align*} This together with \eqref{e3.12}, we get \begin{equation} \begin{aligned} \lambda' &:=r\Big(\max_{x\in I}\int_0^1K(x,y)\widetilde{p}(y)f_{r,R}(y) \mathrm{d}y\Big)^{-1}\leq\lambda\\ &\leq R\Big(\max_{x\in I}\int_0^1K(x,y)\widetilde{p}(y)\psi_R(y) \mathrm{d}y\Big)^{-1}:=\lambda''. \end{aligned}\label{e3.15} \end{equation} Thus \begin{equation} C_n\cap ((0,+\infty)\times(\overline{P}_R\setminus P_r))\subset [\lambda',\lambda'']\times \overline{P}_R\setminus P_r. \label{e3.16} \end{equation} Since $C_n$ is a maximal and unbounded subcontinuum which comes from $(0,\theta)$ and tends to $(0,+\infty)$ eventually, for any $\lambda \in (0,\lambda ')$ from \eqref{e3.15} and \eqref{e3.16} one can get that there exist at least two points $\varphi_{n\lambda}^*$ and $\varphi_{n\lambda}^{**}\in P \setminus\{\theta\}$ such that $(\lambda,\varphi_{n\lambda}^*),(\lambda,\varphi_{n\lambda}^{**})\in C_n$ with $\|\varphi_{n\lambda}^{**}\|>R>r>\|\varphi_{n\lambda}^*\|>0$. Notice that $R$ and $r$ satisfying $R>1>r>0$ are arbitrary. Thus, it is easy to know (5) and (6) hold. \end{proof} From \eqref{e3.3} \eqref{e3.10} and \eqref{e3.12} in above Theorem \ref{thm3.1}, one can obtain the following corollary. \begin{corollary} \label{coro0} Assume {\rm (H1)--(H5)} hold. Then for every $\varepsilon>0$, there exist positive number $R_{\varepsilon}>1>r_{\varepsilon}>0,\lambda_{\varepsilon}>0$ such that \begin{equation} L_n\cap([\varepsilon,+\infty)\times P)\subset[\varepsilon,\lambda_{\varepsilon}]\times (\overline{P}_{R_{\varepsilon}}\setminus P_{r_{\varepsilon}}),\forall\ n\geq 1, \label{e3.17} \end{equation} where $R_{\varepsilon}$ and $\lambda_{\varepsilon}$ are nonincreasing and $r_{\varepsilon}$ is nondecreasing with respect to $\varepsilon\in (0,+\infty)$. \end{corollary} The next theorem gives a result for $L$ and \eqref{e1.1}. \begin{theorem} \label{thm3.2} Let {\rm (H1)--(H5)} be satisfied. Then $\overline{L}$ possesses a maximal and unbounded subcontinuum $C$, which comes from $(0,\theta)$ and tends to $(0,+\infty)$ eventually such that \begin{itemize} \item[(i)] There exists $\lambda^0>0$ satisfying $L\cap ([\lambda^0,+\infty)\times P)=\emptyset$; \item[(ii)] For each $ \overline{\lambda}>0$, $C\cap([0,\overline{\lambda}]\times P)$ is unbounded; \item[(iii)] There exist ${\lambda}^*\in (0,\lambda^0)$ such that for all $\lambda\in (0,{\lambda}^*)$, \eqref{e1.1} has at least two positive solution $\varphi_{\lambda}^1$ and $\varphi_{\lambda}^2$ satisfying $$ (\lambda,\varphi_{\lambda}^1),(\lambda,\varphi_{\lambda}^2) \in C,\quad \|\varphi_{\lambda}^2\| >\|\varphi_{\lambda}^1\|; $$ \item[(iv)] $$ \lim_{\lambda\to 0^+,\,(\lambda,\varphi_{\lambda}^1)\in C} \|\varphi_{\lambda}^1\|=0, \quad \lim_{\lambda\to 0^+,\, (\lambda,\varphi_{\lambda}^2)\in C} \|\varphi_{\lambda}^2\|=+\infty. $$ \end{itemize} \end{theorem} \begin{proof} Firstly, we prove that $L\neq\emptyset$. By Theorem \ref{thm3.1} and \eqref{e3.15}, we know that there exists $\lambda_0>0$ such that for each $n,L_n$ possesses a maximal and unbounded subcontinuum $C_n$ containing $(0,\theta)$, which satisfies \begin{equation} C_n\cap(\{\lambda_0\}\times P)\neq\emptyset,\quad \forall\ n\geq 1. \label{e3.18} \end{equation} On the other hand, from Corollary \ref{coro0}, one can get that there exist $\overline{R}>1>\overline{r}>0$ such that \begin{equation} L_n\cap (\{\lambda_0\}\times P)\subset\{\lambda_0\}\times(\overline{P}_{\overline{R}} \setminus P_{\overline{r}})\quad \text{for all } n\geq1. \label{e3.19} \end{equation} For every $n$, by \eqref{e3.18} one can take $\varphi_n\in C_n\cap(\{\lambda_0\}\times P)$. Then it follows from \eqref{e3.19} that $\varphi_n\in\overline{P}_{\overline{R}}\setminus P_{\overline{r}}$. By (H3) we know \begin{equation} f_n(x,\varphi_n(x))\leq f_{\overline{r},\overline{R}}(x)\quad \text{for all } x\in (0,1), \; n\geq1. \label{e3.20} \end{equation} Similar to the proof of Lemma \ref{lem3.3}, it is easy to know that $\{\varphi_n\}$ is uniformly bounded and equicontinuous on $I=[0,1]$. As a consequence, Ascoli-Arzela theorem generates the compactness of $\{\varphi_n\}$. So there exists a subsequence (without loss of generality, we may assume this sequence is $\{\varphi_n\}$ as well) and $ {\varphi}^*\in \overline{P}_{\overline{R}}\setminus P_{\overline{r}}$ such that $\varphi_n\to {\varphi}^*$ as $n\to+\infty$. \eqref{e3.2} and Lebesgue dominated convergence theorem guarantee $(\lambda_0,\varphi^*)\in L$, that is, $L\neq\emptyset$. Secondly, define an operator $A$ on $P\setminus\{\theta\}$ as follows: \begin{equation} (A\varphi)(x)=\int_0^1K(x,y)\widetilde{p}(y)f(y,\varphi(y)) \mathrm{d}y \quad \text{for all } x\in I, \; \varphi\in P\setminus\{\theta\}. \label{e3.21} \end{equation} By (H3), $A$ is well defined on $P\setminus\{\theta\}$. It is easy to see that to seek a positive solution of \eqref{e1.1} is equivalent to find a fixed point of $\lambda A$ on $P\setminus\{\theta\}$. Similar to Theorem \ref{thm3.1}, one can get (i) holds. To obtain (ii), noticing that for any $ \varepsilon\in (0,\lambda_0)$, it follows from Corollary \ref{coro0} that there exist $R_{\varepsilon},\lambda_{\varepsilon}$, and $r_{\varepsilon}$ such that \[ L_n\cap([\varepsilon,+\infty)\times P)\subset\overline{Q}_{\varepsilon} \quad \text{for all } n\geq 1. \] where $R_{\varepsilon},\lambda_{\varepsilon}$ are nonincreasing and $r_{\varepsilon}$ is nondecreasing functions with respect to $\varepsilon$, $Q_{\varepsilon}:=(\varepsilon,\lambda_{\varepsilon}]\times P_{R_{\varepsilon}}$. On the other hand, \[ \Big(\bigcup_{n=1}^{+\infty}L_n\Big)\bigcap\overline{Q}_{\varepsilon}\subset\Big(\bigcup_{n=1}^{+\infty}L_n\Big) \bigcap([\varepsilon,\lambda_{\varepsilon}]\times (\overline{P}_{R_{\varepsilon}}\setminus P_{r_{\varepsilon}})). \] This together with Lemma \ref{lem3.3} and its proof implies that \begin{equation} \Big(\bigcup_{n=1}^{+\infty}L_n\Big)\bigcap\overline{Q}_{\varepsilon} \quad \text{ are relatively compact}. \label{e3.22} \end{equation} Recall that a maximal subcontinuum is a maximal, closed and connected set. In what follows, we denote by $C_n^{\varepsilon}$ the subcontinuum of $C_n\cap\overline{Q}_{\varepsilon}$ containing $(\lambda_0,\varphi_n)$. Let \begin{equation} \begin{aligned} F_{\varepsilon}:=&\big\{y: \text{there exist the subsequence $\{n_k\}$ of $\{n\}$}\\ &\text{and $y_{n_k}\in C_{n_k}^{\varepsilon}$ satisfying }\lim_ {k\to+\infty}y_{n_k}=y\big\}. \end{aligned}\label{e3.23} \end{equation} Combining with \eqref{e3.22} and Lebesgue dominated convergence theorem one can get \begin{equation} F_{\varepsilon}\subset L\quad \text{and}\quad (\lambda_0,\varphi^*)\in F_{\varepsilon}. \label{e3.24} \end{equation} Now we prove that $F_{\varepsilon}$ is connected. Otherwise, there exist subsets $V_1$ and $V_2$ such that $\overline{V}_1\cap V_2=\emptyset,V_1\cap\overline{V}_2=\emptyset$ and $F_{\varepsilon}=V_1\cup V_2$. Since $F_{\varepsilon}$ is closed, $F_{\varepsilon}=V_1\cup\overline{V}_2$, and consequently, $V_2=\overline{V}_2$. Similarly, $V_1=\overline{V}_1$. Therefore, $V_1$ and $V_2$ are compact. Noticing $V_1\cap V_2=\emptyset$, there exists $\delta>0$, such that $\rho(V_1,V_2)=\delta$. Let \begin{gather*} U(V_1,\frac{\delta}{3}):=\{(\lambda,\varphi)\in R^+\times C[I,P]:d((\lambda,\varphi);V_1)< \frac{\delta}{3}\}; \\ U(V_2,\frac{\delta}{3}):=\{(\lambda,\varphi)\in R^+\times C[I,P]:d((\lambda,\varphi);V_2)< \frac{\delta}{3}\}; \end{gather*} where $d(\cdot,\cdot)$ denotes the distance between two sets in $E=R\times C[I,P]$. Without loss of generality, suppose $P_1=(\lambda_0,\varphi^*)\in V_1$, and choose $P_2\in V_2$. Obviously, $P_{1n}:=(\lambda_0,\varphi_n)\to P_1$ as $n\to +\infty$ and there exists a subsequence $\{n_k\}$ of $\{n\}$ and $P_{2,n_k}\in C_{n_k}^{\varepsilon}$ such that $\lim_ {k\to +\infty}P_{2,n_k}=P_2$. As a consequence, there exists $N>0$ such that $P_{1,n_k}\in U(V_1,\frac{\delta}{3}),P_{2,n_k}\in U(V_2,\frac{\delta}{3})$ for $n_k\geq N$. Notice that $C_{n_k}^{\varepsilon}$ is connected. Then there exists $P_{n_k}\in C_{n_k}^{\varepsilon}\cap\partial U(V_1,\frac{\delta}{3})$ for each $n_k\geq n$. Since $\{P_{n_k}\}$ are relatively compact, without loss of generality, we may assume $\lim_{k\to +\infty}P_{n_k}=P^*$ as well. Then $P^*\in\partial{U(V_1,\frac{\delta}{3})}$ and $P^*\in F_{\varepsilon}$, which contradicts $ F_{\varepsilon}\cap \partial U(V_1,\frac{\delta}{3})=\emptyset$. Consequently, $F_{\varepsilon}$ is connected . Let \[ C:=\bigcup_{0<\varepsilon<\lambda_0}F_{\varepsilon}. \] Now we are in position to show that $C$ meets our requirements. Noticing that $F_{\varepsilon}$ is connected, it follows from \eqref{e3.24} that $(\lambda_0,\varphi^*) \in F_{\varepsilon}$ for any $\varepsilon\in (0,\lambda_0)$. Thus, $C$ is connected. For every pair of positive numbers $R>r>0$, $\lambda\in (0,\lambda')(\lambda'$ is the same as in \eqref{e3.15}, $n\geq1$, by virtue of \eqref{e3.15} and the connectivity of $C_n$ there exist $\varphi_{1n},\varphi_{2n}\in P\setminus\{\theta\}$ such that \[ (\lambda,\varphi_{1n}),(\lambda,\varphi_{2n})\in C_n,\quad \|\varphi_{1n}\|\leq r \quad \text{with } \|\varphi_{2n}\|\geq R\ \text{ for each } n\geq1. \] Using Corollary \ref{coro0}, we know that $\{\varphi_{2n}\}$ is bounded. Moreover, notice that $\bigcup_{n=1}^{+\infty}L_n\bigcap(\{\lambda\}\times P)$ are relatively compact. This together with \eqref{e3.23} guarantees that there exist $\varphi_1^*$ and $\varphi_2^*$ such that \[ (\lambda,\varphi_1^*),(\lambda,\varphi_2^*)\in C,\quad \|\varphi_1^*\|\leq r,\|\varphi_2^*\|\geq R. \] Since $R$ and $r$ are arbitrary, we can easily know that $C$ is an unbounded subcontinuum. Consequently, (ii) holds. On the other hand, similar to the proof of Theorem \ref{thm3.1}, it is not difficult to see that $C$ comes from $(0,\theta)$ and tends to $(0,+\infty)$ eventually. Thus, (iii) and (iv) hold. \end{proof} \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(100,68)(0,0) \put(5,5){\line(1,0){95}} \put(99.6,4.1){$\rightarrow$} \put(10,0){\line(0,1){65}} \put(9.14,65){$\uparrow$} \qbezier(10,5)(70,5)(85,14) \qbezier(85,14)(95,20)(85,26) \qbezier(85,26)(77,31)(60,34) \qbezier(60,34)(20,38)(15,67) \multiput(90.5,5)(0,5){10}{\line(0,1){3}} \put(89,1){$\lambda_0$} \put(100,1){$R^1$} \put(4,65){$E$} \end{picture} \end{center} \caption{ Graph of continuum $C$ } \label{fig1} \end{figure} \subsection*{Example} Consider the singular $m$-point boundary-value problem \begin{equation} \begin{gathered} \varphi''(x)+\lambda f(x,\varphi)=\frac{1}{\sqrt{x(1-x)}} (1+\varphi^{\frac{3}{2}}+\frac{1}{\sqrt[3]{\varphi}}) ,\quad 0