\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{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)$be the solution of \begin{gathered} (L\varphi)(x)=0,\quad 00,x\in (0,1]$; \item[(ii)] $\phi_2(x)$ is decreasing on [0,1] and $\phi_2(x)>0,x\in [0,1)$. \end{itemize} \end{lemma} Let $$k(x,y)=\begin{cases} \frac{1}{\rho}\phi_1(x)\phi_2(y), & 0\leq x\leq y\leq1,\$3pt] \frac{1}{\rho}\phi_1(y)\phi_2(x), & 0\leq y\leq x\leq1, \end{cases}\label{e2.3}$$ where \rho=\phi'_1(0). By Lemma \ref{lem2.1} we know that \phi'_1>0. Let $$K(x,y)=k(x,y)+D^{-1}\phi_1(x)\sum_{i=1}^{m-2}a_ik(\xi_i,y),\quad 0\leq x,y\leq1\label{e2.4}$$ 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 $$(A\varphi)(x)=\int_0^1K(x,y)\widetilde{p}(y)f(x,\varphi(y)) \mathrm{d}y,\ \ x\in [0,1],\label{e2.5}$$ where $$\widetilde{p}(y)=\frac{1}{p(y)}\exp\Big(\int_0^y\frac{p'(s)}{p(s)}\mathrm{d}s\Big).\label{e2.6}$$ 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{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 $$L_n\cap([\overline{\lambda},+\infty)\times (P\setminus\overline{P}_{\overline{R}}))=\emptyset,\quad n=1,2,\dots, \label{e3.3}$$ where$P_{\overline{R}}=\{\varphi\in P: \|\varphi\|<\overline{R}\}$. In fact, take a positive number$l$satisfying $$l>\Big(\rho k_1\overline{\lambda}\max_{x\in I}\int_a^bK(x,y)\widetilde{p}(y)y(1-y) \mathrm{d}y\Big)^{-1}>0, \label{e3.4}$$ where$a,b$are as the same as in (H5). Then there exists$R'>1$such that $$f(x, u)\geq lu\quad \text{for all } x\in [a,b], u>R'. \label{e3.5}$$ Choose a number$\overline{R}$with$\overline{R}>\frac{R'}{\rho k_1a(1-b)}$. It follows from the definition of cone$P$that $$\varphi(y)\geq\|\varphi\|\rho k_1y(1-y)\geq\rho k_1a(1-b)\overline{R}>R'\quad \text{for all } y\in [a,b], \varphi\in P\setminus \overline{P}_{\overline{R}}. \label{e3.6}$$ 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 $$\|\lambda A_n\varphi\|\geq\rho k_1l\overline{\lambda}\|\varphi\|\max_{x\in I}\int_a^bK(x,y)\widetilde{p}(y)y(1-y) \mathrm{d}y >\|\varphi\|, \label{e3.7}$$ 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 $$A_n\varphi(x)=\int_0^1K(x,y)\widetilde{p}(y)f_n(x,\varphi(y)) \mathrm{d}y\geq\int_0^1K(x,y)\widetilde{p}(y)\psi_{\overline{R}} (y) \mathrm{d}y,\forall\ \varphi\in \overline{P}_{\overline{R}}. \label{e3.8}$$ 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{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} which yields $$L_n\cap((\overline{\lambda},+\infty)\times P_r)=\emptyset. \label{e3.10}$$ 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 thatL_n$possesses a maximal and unbounded subcontinuum$C_n$satisfying that $$\begin{gathered} C_n\cap((0,+\infty)\times P)\setminus((\overline{\lambda},+\infty)\times \overline{P}_{\overline{R}}) \text{ is unbounded and} \\ C_n\cap((\overline{\lambda},+\infty)\times\{ P_r\cup (P\setminus \overline{P}_{\overline{R}})\})=\emptyset.\end{gathered}\label{e3.11}$$ 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 $$\lambda\leq\overline{R}\Big(\max_{x\in I}\int_a^bK(x,y)\widetilde{p}(y)\psi_{\overline{R}}(y) \mathrm{d}y\Big)^{-1}, \label{e3.12}$$ 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 $$L_n\cap([\overline{\lambda},+\infty)\times P) \ \text{is bounded }, \forall\ \overline{\lambda}>0. \label{e3.13}$$ 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 $$L_n\cap((\lambda_n^0,+\infty)\times P)=\emptyset. \label{e3.14}$$ 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{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} Thus $$C_n\cap ((0,+\infty)\times(\overline{P}_R\setminus P_r))\subset [\lambda',\lambda'']\times \overline{P}_R\setminus P_r. \label{e3.16}$$ SinceC_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 $$L_n\cap([\varepsilon,+\infty)\times P)\subset[\varepsilon,\lambda_{\varepsilon}]\times (\overline{P}_{R_{\varepsilon}}\setminus P_{r_{\varepsilon}}),\forall\ n\geq 1, \label{e3.17}$$ 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 $$C_n\cap(\{\lambda_0\}\times P)\neq\emptyset,\quad \forall\ n\geq 1. \label{e3.18}$$ On the other hand, from Corollary \ref{coro0}, one can get that there exist$\overline{R}>1>\overline{r}>0$such that $$L_n\cap (\{\lambda_0\}\times P)\subset\{\lambda_0\}\times(\overline{P}_{\overline{R}} \setminus P_{\overline{r}})\quad \text{for all } n\geq1. \label{e3.19}$$ 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 $$f_n(x,\varphi_n(x))\leq f_{\overline{r},\overline{R}}(x)\quad \text{for all } x\in (0,1), \; n\geq1. \label{e3.20}$$ 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: $$(A\varphi)(x)=\int_0^1K(x,y)\widetilde{p}(y)f(y,\varphi(y)) \mathrm{d}y \quad \text{for all } x\in I, \; \varphi\in P\setminus\{\theta\}. \label{e3.21}$$ 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 $$\Big(\bigcup_{n=1}^{+\infty}L_n\Big)\bigcap\overline{Q}_{\varepsilon} \quad \text{ are relatively compact}. \label{e3.22}$$ 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{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} Combining with \eqref{e3.22} and Lebesgue dominated convergence theorem one can get $$F_{\varepsilon}\subset L\quad \text{and}\quad (\lambda_0,\varphi^*)\in F_{\varepsilon}. \label{e3.24}$$ Now we prove thatF_{\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{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