\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 86, pp. 1--9.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/86\hfil Positive solutions and eigenvalues]
{Positive solutions and eigenvalues of nonlocal
boundary-value problems}
\author[J. Chu, Z. Zhou\hfil EJDE-2005/86\hfilneg]
{Jifeng Chu, Zhongcheng Zhou}  % in alphabetical order

\address{Jifeng Chu \hfill\break
 Department of Applied Mathematics,
College of Sciences, Hohai University\\
Nanjing 210098,  China} \email{jifengchu@hhu.edu.cn}

\address{Zhongcheng Zhou \hfill\break
School of Mathematics and Finance, Southwest Normal University\\
Chongqing 400715, China} \email{zhouzc@swnu.edu.cn}


\date{}
\thanks{Submitted April 18, 2005. Published July 27, 2005}
\subjclass[2000]{34B15}
\keywords{Nonlocal boundary-value problems;
 positive solutions, eigenvalues; \hfill\break\indent
 fixed point theorem in cones}

\begin{abstract}
 We study the ordinary differential equation
 $x''+\lambda a(t)f(x)=0$ with the boundary
 conditions $x(0)=0$ and $x'(1)=\int_{\eta}^{1}x'(s)dg(s)$.
 We characterize values of $\lambda$ for which boundary-value
 problem has a positive solution. Also we find appropriate intervals for
 $\lambda$ so that there are two positive solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}


\section{Introduction}

This paper concerns the ordinary differential equation
\begin{equation}\label{e1.1}
 x''+\lambda a(t)f(x)=0,\quad \text{a.e. }t\in[0,1]
\end{equation}
with the boundary conditions
\begin{gather}
x(0)=0 \label{e1.2}\\
x'(1)=\int_{\eta}^{1}x'(s)dg(s),\label{e1.3}
\end{gather}
where $\lambda>0$, $\eta\in(0,1)$ and the integral in \eqref{e1.3} is meant in
the sense of Riemann-Stieljes. In this paper it is assumed
that
\begin{itemize}
\item[(H1)] The function
$f:[0,\infty )\to[0,\infty )$ is continuous.
 \item[(H2)] The function $a:[0,1]\to [0,\infty )$ is continuous and
does not vanish identically on any subinterval.
 \item[(H3)] The function $g:[0,1]\to\mathbb{R}$ is increasing and such
that $g(\eta)=0<g(\eta^{+})$ and $g(1)<1$.
\end{itemize}

In recent years, nonlocal boundary-value problems of this form
have been studied extensively in the literature \cite{i1,k1,k2,k3,k4}.
 This class of problems includes, as special cases, multi-point
boundary-value problems considered by many authors (see
\cite{g1,p1} and the references therein). In fact, condition
\eqref{e1.2}-\eqref{e1.3} is the continuous version of the
multi-point condition
 \begin{equation}
x(0)=0,\quad x'(1)=\sum_{i=1}^{m}\alpha_{i}x'(\xi_{i})
\end{equation}
which happens when $g$ is a piece-wise constant function that is
increasing and has finitely many jumps, where
$\alpha_{1}$, $\alpha_{2},\dots\alpha_{m}\in \mathbb{R}$ have the same
sign, $m\geq 1$ is an integer,
$0<\xi_{1}<\xi_{2}<\dots<\xi_{m}<1$.

In the sequel, in this paper we shall denote by $\mathbb{R}$ the real line
and by $I$ the interval [0,1], $C(I)$ will denote the space of all
continuous functions $x:I\to\mathbb{R}$. Let
\[
C^{1}_{0}(I)=\{x\in C(I):x'\mbox{ is absolutely continuous on
$I$ and }x(0)=0\}.
\]
Then $C^{1}_{0}(I)$ is a Banach space when it is furnished with
the super-norm $\|x\|=\sup_{t\in I}|x(t)|$.

By a solution $x$ of \eqref{e1.1}-\eqref{e1.3} we mean $x\in C^{1}_{0}(I)$
satisfying equation \eqref{e1.1} for almost all $t\in I $ and condition
\eqref{e1.3}. By a positive solution $x$ of \eqref{e1.1}-\eqref{e1.3} if $x$ is
nonnegative and is not identically zero on $I$. If, for a
particular $\lambda$, the boundary-value problem \eqref{e1.1}-\eqref{e1.3} has a
positive solution $x$, then $\lambda$ is called an eigenvalue and
$x$ a corresponding eigenfunction. Recently, several eigenvalue
characterizations for kinds of boundary-value problems have been
carried out, for this we refer to \cite{a1,a2,c1,h1,w2,w3}.

In this paper, we will use the notation
\[
f_{0}=\lim_{x\to0^{+}}\frac {f(x)}{x},\quad
f_{\infty }=\lim_{x\to \infty }\frac{f(x)}{x}.
\]

This paper is organized as follows. In section 2, we will present
 some preliminary results, including a fixed point theorem due to
 Krasnosel'skii \cite{k5}, which is the basic tool used in this paper. We
 shall establish the eigenvalue intervals in
 terms of $f_{0}$ and $f_{\infty }$ in section 3. The investigation of the
  existence of double positive solutions is carried out in section 4.

\section{Preliminaries}

First, we present a fixed point theorem in cones due to
Krasnosel'skii, which can  be found in \cite{k5}.

\begin{theorem} \label{thm2.1}
 Let $X$ be a Banach space and $K\ (\subset X)$ be a
cone. Assume that $\Omega_1,\ \Omega_2$ are open subsets of $X$
with $0\in \Omega_1,\ \bar\Omega_1 \subset \Omega_2$, and let
$$
T: K \cap (\bar \Omega_2 \backslash \Omega_1)\to K
$$
 be a  continuous and compact operator such that either
\begin{itemize}
\item[(i)] $\| Tu \| \geq \| u \|$, $u\in K\cap \partial
     \Omega_1$ and $ \| Tu \| \leq \| u \|$, $ u\in K\cap \partial
     \Omega_2$; or

\item[(ii)] $\| Tu \| \leq \| u \|$, $ u\in K\cap \partial
     \Omega_1$ and $\| Tu \| \geq \| u \|$, $u\in K\cap \partial
     \Omega_2$.
\end{itemize}
Then $T$ has a fixed point in $K \cap ( \bar \Omega_2 \backslash
     \Omega_1)$.
    \end{theorem}

We will apply Theorem \ref{thm2.1} to find positive solutions to boundary-value problem \eqref{e1.1}-\eqref{e1.3}. To do so, we need to re-formulate the
problem as an operator equation of the form $x=T_{\lambda}x$, for
an appropriate operator $T_{\lambda}$. In fact, following from
\cite{k1}, we have:

\begin{lemma} \label{lem2.2}
A function $x\in C^{1}_{0}(I)$ is a solution of the
boundary-value problem \eqref{e1.1}-\eqref{e1.3} if and only if $x$ is a
solution of the operator equation $x=T_{\lambda}x$, where
$T_{\lambda}$ is defined by
 \begin{equation}
(T_{\lambda}x)(t)=\frac{\lambda
t}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))drdg(s)+\lambda
\int_{0}^{t}\int_{s}^{1}a(r)f(x(r))dr\,ds\,.
\end{equation}
\end{lemma}

In order to apply Theorem \ref{thm2.1}, we define
\[
K=\{x\in C^{1}_{0}(I): x(t)\geq 0,~x'(t)\geq 0 ~\mbox{and }x
\mbox{~is concave}\}.
\]
One may readily verify that $K$ is a cone in $C^{1}_{0}(I)$.
Moreover, we have the following elementary fact.

\begin{lemma} \label{lem2.3}
If $x\in K$, then, for any $\tau\in[0,1]$ it holds
$x(t)\geq \tau\|x\|$, $t\in[\tau,1]$.
\end{lemma}

\begin{theorem} \label{thm2.4}
Assume that {\rm (H1)-(H3)} hold, then
$T_{\lambda}(K)\subseteq K$ and $T_{\lambda}$ is continuous and
completely continuous.
\end{theorem}

\section{Eigenvalue intervals}

For the sake of simplicity, let
\begin{gather}
A=\frac{1}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)drdg(s)
+\int_{0}^{1}\int_{s}^{1}a(r)\,dr\,ds \label{e3.1}\\
B=\frac{1}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)drdg(s)
+\int_{\eta}^{1}\int_{s}^{1}a(r)\,dr\,ds . \label{e3.2}
\end{gather}

\begin{theorem} \label{thm3.1}
Suppose that {\rm (H1)-(H3)} hold, then the boundary-value problem
\eqref{e1.1}-\eqref{e1.3} has at least one positive solution for
each
\begin{equation}\label{e3.3}
\lambda \in (1/\eta f_{\infty }B,1/f_{0}A).
\end{equation}
\end{theorem}

\begin{proof}
We construct the sets $\Omega_{1}$ and $\Omega_{2}$ in order
to apply Theorem \ref{thm2.1}. Let $\lambda$ be given as in \eqref{e3.3} and
choose $\varepsilon>0$ such that
\[
\frac{1}{\eta(f_{\infty }-\varepsilon)B}\leq\lambda
\leq\frac{1}{(f_{0}+\varepsilon)A}.
\]
First, there exists $r>0$ such that
\[
f(x)\leq (f_{0}+\varepsilon)x,\quad 0<x\leq r.
\]
So, for any $x\in K$ with $\|x\|=r$, we have
\begin{align*}
&(T_{\lambda}x)(t)\\
&\leq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))dr\,dg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds  \\
&\leq  \frac{\lambda }{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)(f_{0}
+\varepsilon)x(r) dr\,dg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)(f_{0}+\varepsilon)x(r)\,dr\,ds  \\
&\leq  \lambda (f_{0}+\varepsilon)r \{\frac{1
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)dr\,dg(s)
+\int_{0}^{1}\int_{s}^{1}a(r)\,dr\,ds \} \\
&\leq \lambda(f_{0}+\varepsilon)Ar\leq r=\|x\|.
\end{align*}
Consequently, $\|T_{\lambda}x\|\leq\|x\|$. So, if we
set $\Omega_{1}=\{x\in K: \|x\|<r\}$, then
\begin{equation} \label{e3.4}
\|T_{\lambda}x\|\leq\|x\|,\quad \forall x\in K\cap
\partial\Omega_{1}.
\end{equation}
Next, we  choose $R_{1}$ such that
\[
f(x)\geq (f_{\infty }-\varepsilon)x, \quad x\geq R_{1}.
\]
Let
$R=\max\{2r, \eta^{-1}R_{1}\}$ and set
\[
\Omega_{2}=\{x\in K: \|x\|<R\}.
\]
If $x\in K$ with $\|x\|=R$, then
\[
\min_{t\in[\eta,1]}x(t)\geq \eta\|x\|\geq R_{1}.
\]
Thus, we have
\begin{align*}
&(T_{\lambda}x)(1)\\
&= \frac{\lambda}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))dr\,dg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds   \\
&\geq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))dr\,dg(s)+
\lambda\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds  \\
&\geq \frac{\lambda}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)(f_{\infty }
-\varepsilon)x(r)dr\,dg(s)+
\lambda\int_{\eta}^{1}\int_{s}^{1}a(r)(f_{\infty }-\varepsilon)x(r)\,dr\,ds  \\
&\geq  \lambda (f_{\infty }-\varepsilon)\eta\|x\| \{\frac{1
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)dr\,dg(s)
+\int_{\eta}^{1}\int_{s}^{1}a(r)\,dr\,ds \}  \\
&=  \lambda(f_{\infty }-\varepsilon)B\eta R\geq R=\|x\|.
\end{align*}
Hence,
\[ %3.5
\|T_{\lambda}x \|\geq\|x\|,\quad \forall x\in
K\cap\partial\Omega_{2}.
\]
 From this inequality, \eqref{e3.4}, and Theorem \ref{thm2.1}
it follows that $T_{\lambda}$ has a fixed point
$x\in K \cap ( \bar \Omega_2 \backslash
     \Omega_1)$ with $r\leq \|x\|\leq R$.
Clearly, this $x$ is a positive solution of \eqref{e1.1}-\eqref{e1.3}.
\end{proof}


\begin{theorem} \label{thm3.2}
Suppose that {\rm (H1)-(H3)} hold, then the boundary-value problem
\eqref{e1.1}-\eqref{e1.3} has at least one positive solution for
each
\begin{equation}
\lambda\in(1/\eta f_{0}B,1/f_{\infty }A). \label{e3.6}
\end{equation}
\end{theorem}

\begin{proof}
We construct the sets $\Omega_{1}$ and $\Omega_{2}$ in order
to apply Theorem \ref{thm2.1}. Let $\lambda$ be given as in \eqref{e3.6} and
choose $\varepsilon>0$ such that
\[
\frac{1}{\eta(f_{0}-\varepsilon)B}\leq\lambda\leq\frac{1}{(f_{\infty }
+\varepsilon)A}.
\]
First, there exists $r>0$ such that
\[
f(x)\geq (f_{0}-\varepsilon)x, \quad    0<x\leq r.
\]
So, for any $x\in K$ with $\|x\|=r$, we have
\begin{align*}
&(T_{\lambda}x)(1)\\
&\geq  \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))drdg(s)+
\lambda\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds  \\
&\geq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)(f_{0}-\varepsilon)x(r)drdg(s)+
\lambda\int_{\eta}^{1}\int_{s}^{1}a(r)(f_{0}-\varepsilon)x(r)\,dr\,ds  \\
&\geq  \lambda (f_{0}-\varepsilon)\eta r \{\frac{1
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)drdg(s)
+\int_{\eta}^{1}\int_{s}^{1}a(r)\,dr\,ds \}  \\
&\geq  \lambda(f_{0}-\varepsilon)B\eta r\geq r=\|x\|.
\end{align*}
Consequently, $\|T_{\lambda}x\|\geq\|x\|$. So, if we
set $\Omega_{1}=\{x\in K:\|x\|<r\}$, then
\begin{equation} \label{e3.7}
\|T_{\lambda}x\|\geq\|x\|, \quad  \forall x\in K\cap \partial\Omega_{1}.
\end{equation}
Next, we can choose $R_{1}$ such that
\[
f(x)\leq (f_{\infty }+\varepsilon)x, ~~x\geq R_{1}.
\]
Here are two cases to be considered, namely, where $f$ is bounded and
where $f$ is unbounded.

\noindent\textbf{Case 1: $f$ is bounded.} Then, there exists some
constant $M>0$ such that $f(x)\leq M, ~~x\in(0,\infty )$. Let
$R=\max\{2r, \lambda MA\}$ and set
\[
\Omega_{2}=\{x\in K:\|x\|<R\}.
\]
Then, for any $x\in K$ with $\|x\|=R$, we have
\begin{align*}
(T_{\lambda}x)(t)
&\leq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))drdg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds   \\
&\leq \lambda M\{\frac{1
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)drdg(s)+
\int_{0}^{1}\int_{s}^{1}a(r)\,dr\,ds \} \\
&\leq \lambda MA\leq R=\|x\|.
\end{align*}
Hence,
\begin{equation} \label{e3.8}
\|T_{\lambda}x \|\leq\|x\|,~~~\forall x\in
K\cap\partial\Omega_{2}.
\end{equation}

\noindent\textbf{Case 2: $f$ is unbounded.}
Then, there exists $R>\max\{2r, R_{1}\}$ such that
\[
f(x)\leq f(R),~~~0<x\leq R.
\]
For $x\in K$ with $\|x\|=R$, we have
\begin{align*}
&(T_{\lambda}x)(t)\\
&\leq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))drdg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds   \\
&\leq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(R)drdg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)f(R)\,dr\,ds  \\
&\leq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)(f_{\infty }+\varepsilon)R
dr\,dg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)(f_{\infty }+\varepsilon)R \,dr\,ds  \\
&=  \lambda(f_{\infty }+\varepsilon)RA\leq R=\|x\|.
\end{align*}
Then \eqref{e3.8} is also true in this case.

Now \eqref{e3.7}, \eqref{e3.8}, and Theorem \ref{thm2.1}
guarantee that $T_{\lambda}$ has a
fixed point $x\in K \cap ( \bar \Omega_2 \backslash \Omega_1)$ with
$r\leq \|x\|\leq R$. Clearly, this $x$ is a positive solution of
\eqref{e1.1}-\eqref{e1.3}.
\end{proof}

\subsection*{Example} %3.3
Let the function $f(x)$ in \eqref{e1.1} be
\begin{equation} \label{ef}
f(x)= x^{\alpha}+x^{\beta},
\end{equation}
then problem \eqref{e1.1}-\eqref{e1.3} has at least one
positive solution for all $\lambda\in (0,\infty )$
if $ 0<\alpha<1, 0<\beta <1$ or $\alpha>1, \beta>1$.

\begin{proof}It is easy to see that $f_{0}=\infty$, $f_{\infty }=0$ if $
0<\alpha<1$, $0<\beta <1$ and $f_{0}=0, f_{\infty }=\infty $ if
$\alpha>1, \beta>1$. Then the results can be easily obtained by
using Theorem \ref{thm3.1} or Theorem \ref{thm3.2} directly.
\end{proof}

\section{Twin positive solutions}

In this section, we establish the existence of two positive
solutions to problem \eqref{e1.1}-\eqref{e1.3}.

\begin{theorem} \label{thm4.1}
Suppose that {\rm (H1)-(H3)} hold. In addition, assume there exist
two constants $R>r>0$ such that
\begin{equation}\label{e4.1}
\max_{0\leq x\leq r}f(x)\leq r/\lambda A, \quad
\min_{\eta R\leq x\leq R}f(x)\geq R/\lambda B.
\end{equation}
Then the boundary-value problem
\eqref{e1.1}-\eqref{e1.3} has at least one positive solution
$x\in K$ with $r\leq\|x\|\leq R$.
\end{theorem}

\begin{proof}
For $x\in\partial K_{r}=\{x\in K:\|x\|=r\}$, we have
$f(x(t))\leq r/\lambda A$ for $t\in[0,1]$.
Then we have
\begin{align*}
(T_{\lambda}x)(t)
&\leq \frac{\lambda}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))dr\,dg(s)+
\lambda\int_{0}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds   \\
&\leq \frac{\lambda }{1-g(1)}\frac{r}{\lambda A
}\int_{\eta}^{1}\int_{s}^{1}a(r)dr\,dg(s)+
\lambda\frac{r}{\lambda A}\int_{0}^{1}\int_{s}^{1}a(r)\,dr\,ds
= r\,.
\end{align*}
As a result, $\|T_{\lambda}x\|\leq\|x\|,  \forall
x\in\partial K_{r}$.
For $x\in\partial K_{R}$, we have $f(x(t))\geq R/\lambda B$ for
$t\in[\eta,1]$. Then we have
\begin{align*}
 (T_{\lambda}x)(1)
&\geq \frac{\lambda
}{1-g(1)}\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))drdg(s)+
\lambda\int_{\eta}^{1}\int_{s}^{1}a(r)f(x(r))\,dr\,ds   \\
&\geq   \frac{\lambda }{1-g(1)}\frac{R}{\lambda B
}\int_{\eta}^{1}\int_{s}^{1}a(r)drdg(s)+
\lambda\frac{R}{\lambda B}\int_{\eta}^{1}\int_{s}^{1}a(r)\,dr\,ds
= R.
\end{align*}
As a result, $\|T_{\lambda}x\|\geq\|x\|$,  for all $x\in\partial K_{R}$.
Then we can obtain the result by using Theorem \ref{thm2.1}.
\end{proof}

\begin{remark} \label{rmk4.2}\rm
In Theorem \ref{thm4.1}, if condition (4.1) is replaced by
\[
\max_{0\leq x\leq R}f(x)\leq R/\lambda A, \quad
\min_{\eta r\leq x\leq r}f(x)\geq r/\lambda B.
\]
Then \eqref{e1.1} has also a solution
$x\in K$ with $r\leq\|x\|\leq R$.
\end{remark}

For the remainder of this section, we need the following
condition:
 \begin{itemize}
\item[(H4)]
 $\sup_{r>0}\min_{\eta r\leq x\leq r}f(x) >0$.
\end{itemize}

Let
\[
\lambda^{*}=\sup_{r>0}\frac{r}{A\max_{0\leq x\leq r }f(x)}, \quad
 \lambda^{**}=\inf_{r>0}\frac{r}{B\min_{\eta r \leq x\leq r }f(x)}.
\]
We can easily obtain that $0<\lambda^{*}\leq\infty $ and
$0\leq \lambda^{**}<\infty$ by using (H1) and (H4).

\begin{theorem} \label{thm4.3}
Suppose that {\rm (H1)-(H4)} hold. In addition, assume that
$f_{0}=\infty$ and $f_{\infty }=\infty$. Then the boundary-value
problem \eqref{e1.1}-\eqref{e1.3} has at least two positive
solutions for any  $\lambda\in(0,\lambda^{*})$.
\end{theorem}

\begin{proof}
Define
$$
h(r)=\frac{r}{A\max_{0\leq x\leq r}f(x)}.
$$
Using the condition (H1), $f_{0}=\infty $ and
$f_{\infty }=\infty $, we can easily obtain that
$h:(0,\infty)\to (0,\infty)$ is continuous and
$$
\lim_{r\to0}h(r)=\lim_{r\to\infty }h(r)=0.
$$
So there exists $r_{0}\in(0,\infty)$ such that
$h(r_{0})=\sup_{r>0}h(r)=\lambda^{*}$.
For $\lambda\in(0,\lambda^{*})$, there exist two constants $r_{1},
r_{2} (0<r_{1}<r_{0}<r_{2}<\infty)$ with
$h(r_{1})=h(r_{2})=\lambda $. Thus
\begin{gather}
f(x)\leq r_{1}/\lambda A,  \quad 0\leq x\leq r_{1}, \label{e4.3}\\
f(x)\leq r_{2}/\lambda A, \quad 0\leq x\leq r_{2}. \label{e4.4}
\end{gather}
 On the other hand, by using the condition
$f_{0}=\infty $ and $f_{\infty }=\infty $, there exist two constants
$r_{3} ,r_{4} (0<r_{3}<r_{1}<r_{2}<\eta r_{4}<\infty )$ with
$$
\frac{f(x)}{x}\geq\frac{1}{\lambda\eta B}, \quad
 x\in(0,r_{3})\cup(\eta r_{4},\infty).
$$
Therefore,
\begin{gather}
\min_{\eta r_{3} \leq x \leq r_{3}}f(x)\geq r_{3}/\lambda B\label{e4.5}\\
\min_{\eta r_{4} \leq x \leq r_{4}}f(x)\geq r_{4}/\lambda B.\label{e4.6}
\end{gather}
It follows from Remark \ref{rmk4.2} and \eqref{e4.3}, \eqref{e4.5}
that problem \eqref{e1.1}-\eqref{e1.3} has a solution $x_{1}\in K$
with $r_{3}\leq\|x_{1}\|\leq r_{1} $. Also, it follows from
Theorem \ref{thm4.1} and \eqref{e4.4}, \eqref{e4.6} that problem
\eqref{e1.1}-\eqref{e1.3} has a solution $x_{2}\in K$ with
$r_{2}\leq\|x_{2}\|\leq r_{4}$. As a results, problem
\eqref{e1.1}-\eqref{e1.3} has at least two positive solutions
\[
r_{3}\leq\|x_{1}\|\leq r_{1} < r_{2}\leq\|x_{2}\|\leq r_{4}.
\]
\end{proof}

\begin{theorem} \label{thm4.4}
Suppose that \rm{(H1)-(H4)} hold. In addition, assume that
$f_{0}=0$ and $f_{\infty }=0$.
  Then, the boundary-value problem \eqref{e1.1}-\eqref{e1.3}
  has at least two positive solutions for
  all $\lambda\in(\lambda^{**},\infty)$.
\end{theorem}

\begin{proof}
Define
\[
g(r)=\frac{r}{B\min_{\eta r\leq x\leq r}f(x)}.
\]
Using the conditions (H1), $f_{0}=0$ and $f_{\infty }=0$, we can
easily obtain that $g:(0,\infty)\to (0,\infty)$ is
continuous and
$$
\lim_{r\to0}g(r)=\lim_{r\to\infty }g(r)=+\infty.
$$
So there exists $r_{0}\in(0,\infty)$ such that
$g(r_{0})=\inf_{r>0}g(r)=\lambda^{**}$.
For $\lambda\in(\lambda^{**},\infty)$, there exist two constants
$r_{1},r_{2} (0<r_{1}<r_{0}<r_{2}<\infty)$ with
$g(r_{1})=g(r_{2})=\lambda $. Thus
\begin{gather}
f(x)\geq r_{1}/\lambda B, \quad  \eta r_{1}\leq x \leq r_{1}, \label{e4.7}\\
f(x)\geq r_{2}/\lambda B, \quad  \eta r_{2}\leq x \leq r_{2}. \label{e4.8}
\end{gather}
 On the other hand,  since $f_{0}=0$,  there exists a constant
$r_{3} (0<r_{3}<r_{1})$ with
\[
\frac{f(x)}{x}\leq \frac{1}{\lambda A}, \quad  x\in(0,r_{3}).
\]
Therefore,
\begin{equation} \label{e4.9}
\max_{0 \leq x\leq r_{3}}f(x)\leq r_{3}/\lambda A.
\end{equation}
Further, using the condition
$f_{\infty }=0$, there exists a constant $r  (r_{2}<r<+\infty)$ with
\[
\frac{f(x)}{x}\leq \frac{1}{\lambda A},   x\in(r,\infty).
\]
 Let
$M=\sup_{0\leq x\leq r}f(x)$ and $r_{4}\geq A\lambda M$. It is
easily seen that
\begin{equation} \label{e4.10}
\max_{0\leq x\leq r_{4} }f(x)\leq
r_{4}/\lambda A.
\end{equation}
It follows from Theorem \ref{thm4.1}, \eqref{e4.7} and
\eqref{e4.9} that \eqref{e1.1}-\eqref{e1.3} has a solution
$x_{1}\in K$ with $r_{3}\leq\|x_{1}\|\leq r_{1}$. Also, it follows
from Remark \ref{rmk4.2} and \eqref{e4.8}, \eqref{e4.10} that
problem \eqref{e1.1}-\eqref{e1.3} has a solution $x_{2}\in K$ with
$r_{2}\leq\|x_{2}\|\leq r_{4}$.
 Therefore,
problem \eqref{e1.1}-\eqref{e1.3} has two positive solutions
\[
r_{3}\leq\|x_{1}\|\leq r_{1} < r_{2}\leq\|x_{2}\|\leq r_{4}.
\]
\end{proof}

\subsection*{Example} %4.5
Assume in \eqref{ef} that $0<\alpha<1<\beta$, then problem
\eqref{e1.1}-\eqref{e1.3} has at least two positive solution for each
$\lambda\in (0,\lambda^{*})$, where $\lambda^{*}$ is some positive constant.

\begin{proof}It is easy to see that $f_{0}=\infty$, $f_{\infty }=\infty $
since $ 0<\alpha<1<\beta$. Then the result can be easily obtained
using Theorem \ref{thm4.3}.\end{proof}

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\end{document}