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\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 07, pp. 1--11.\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-2005/07\hfil Boundary-value problems]
{Semilinear elliptic boundary-value problems
on bounded multiconnected domains} 

\author[W. Gao, J. Wang, Z. Zhang\hfil EJDE-2005/07\hfilneg]
{Wenjie Gao, Junyu Wang, Zhongxin Zhang}  % in alphabetical order

\address{Wenjie Gao \hfill\break
Institute of Mathematics, Jilin University,
 Changchun 130012, China}
\email{wjgao@mail.jlu.edu.cn}

\address{Junyu Wang \hfill\break
Institute of Mathematics, Jilin University,
 Changchun 130012, China}
 
\address{Zhongxin Zhang \hfill\break
Institute of Mathematics, Jilin University,
 Changchun 130012, China}
\email{wx555555@163.net}

\date{}
\thanks{Submitted May 6, 2004. Published January 5, 2005.}
\thanks{Supported by NSFC and by Excellent Young Teacher's Foundation of MOEC.}
\subjclass[2000]{34B15}
\keywords{Semilnear elliptic boundary-value problem;
 nonnegative solution;  \hfill\break\indent
 radial solution; nonexistence; existence; multiplicity}

\begin{abstract}
 A semilinear elliptic boundary-value problem on bounded multiconnected
 domains is studied. The authors prove that under suitable conditions,
 the problem may have no solutions in certain cases and many have one
 or two nonnegative solutions in some other cases.
 The radial solutions were also studied in annular domains.
\end{abstract}

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

\section{Introduction}

This paper consists of two parts. The first part deals with a
semilinear elliptic boundary-value problem of the form
\begin{equation} \label{1.1r}
\begin{gathered}
-\Delta u=f(u, x)\quad\in \Omega,\\
u=0\quad \mbox{on } B_0:=\cup_{j=0}^{k-1}\Gamma_j,\\
u=\rho \quad\mbox{on } B:=\cup_{j=k}^{m}\Gamma_j\,,
\end{gathered}
\end{equation}
where $f(u, x)\in C(\mathbb{R}_+\times\overline\Omega; \mathbb{R})$,
$\mathbb{R}_+:=[0,+\infty)$, $\Omega$ is a bounded
multiconnected domain in $\mathbb{R}^n$, $n\ge2$, $\Gamma_0$ is its outer
boundary, $\cup_{j=1}^{m}\Gamma_j$ its inner boundary, $\Gamma_0$,
$\Gamma_1, \dots, \Gamma_m$ are all sufficiently smooth closed
surfaces so that the Green's function $G(x,y)$, for $-\Delta$ with
zero Dirichlet boundary conditions is existent (See, e.g.,
\cite[p.112]{a2}), $k\in \{1, 2, \dots, m\}$, and the constant $\rho \ge0$
is given. Some particular cases of Problem \eqref{1.1r} were
considered by Bandle and Peletier \cite{b1}, by Lee and Lin \cite{l1} and by
Hai \cite{h1}, in which $f\in(\mathbb{R}_+; \mathbb{R}_+$) and $\Omega $
is a domain with a ``hole''.

The second part is devoted to the semilinear elliptic boundary-value
problem, namely,
\begin{equation} \label{1.2r}
\begin{gathered}
-\Delta u=f(u, |x|)\quad 0<\alpha<|x|<\beta<+\infty,\\
u=0\quad \mbox{ on } |x|=\alpha,\quad
u=\rho \quad\mbox{on }|x|=\beta
\end{gathered}
\end{equation}
where $f\in C(\mathbb{R}_+\times J; \mathbb{R}_+)$,
$J=[\alpha, \beta]$, $|x|=\sqrt{x_1^2+\dots+x_n^2}$.
Problem \eqref{1.2r} was studied by Lee and Lin \cite{l1} and by Hai \cite{h1}
in the case $f\in C(\mathbb{R}_+; \mathbb{R}_+)$.

We study Problems \eqref{1.1r} and \eqref{1.2r} motivated by
the following results recently established by Hai \cite{h1}.

\begin{theorem} \label{thm1.1}
Let $\Omega\subset \mathbb{R}^n$, $n\ge2$, be a
bounded domain with a hole and $f$ in $C(\mathbb{R}_+; \mathbb{R}_+)$ satisfy
$$
\lim_{u\to0+}\frac{f(u)}{u}=0\quad\mbox{and }\quad
\lim_{u\to+\infty}\frac{f(u)}{u}=+\infty.
$$
Then there exists a positive number $\rho^*$ such that Problem
\eqref{1.1r} has a positive solution for $\rho\in (0,\rho^*)$ and
no solution for $\rho>\rho^*$.
\end{theorem}

\begin{theorem} \label{thm1.2}
 Let $f\in C(\mathbb{R}_+; \mathbb{R}_+)$ be convex and satisfy
$$
\liminf_{u\to0+}\frac{\int_0^uf(s)ds}{u^2}=0\quad\mbox{and }\quad
\lim_{u\to+\infty}\frac{f(u)}{u}=+\infty.
$$
Then there exists a positive number $\rho^*$ such that Problem
\eqref{1.2r} has at least two positive radial solutions for
$\rho\in (0,\rho^*)$, at least one for $\rho=\rho^*$ and none for
any $\rho>\rho^*$.
\end{theorem}

The above results extend and complement the corresponding results in 
\cite{b1,l1}.

As in \cite{h1}, our purpose is to extend, improve and
complement the corresponding results in \cite{b1,h1,l1}. In the first
part, we will prove three theorems. The second theorem extends and
complements Theorem \ref{thm1.1}, in which the function $f$ is allowed to
contain variable $x$ and is not necessarily nonnegative, also,
$f(u,x)$ is not necessarily superlinear at $u=0$ and at
$u=+\infty$. In the second part, we obtain a similar result as
Theorem \ref{thm1.2} where the function $f$ may depends on the variable
$|x|$ and the convexity constrain to $f$ is  removed, also $f$ is
not required to be superlinear at $u=0$ and $u=+\infty$.


\section{Results in multiconnected domains}

For the first part, we make the following assumptions:
\begin{itemize}
\item[(A1)] $f\in C(\mathbb{R}_+\times\overline\Omega;\mathbb{R})$.

\item[(A2)] $M:=\sup\{\lambda_1u-f(u,x): u\in \mathbb{R}_+, \
x\in\overline\Omega\}<+\infty$.
\end{itemize}
Here and throughout this section, $\lambda_1$ denotes the first
eigenvalue of the problem
\begin{equation} \label{e2.1}
\begin{gathered}
-\Delta \phi=\lambda \phi\quad \mbox{in } \Omega\\
 \phi=0\quad \mbox{on } \partial\Omega=B_0\cup B.
\end{gathered}
\end{equation}
The  positive eigenfunction corresponding to $\lambda_1$
is denoted by $\phi_1(x)$ with $\|\phi_1\|=1$, where  $\|\cdot\|$ stands
for the supremum norm.
\begin{itemize}
\item[(A3)]
 $$
\limsup_{u\to0+}\frac{\max\{|f(u,x)|:x\in\overline\Omega\}}{u}<\frac{1}{\|g\|},
$$
where
\begin{equation} \label{e2.2}
g(x):=\int_\Omega G(x,y)\,dy,\quad x\in\overline\Omega.
\end{equation}

\item[(A4)] $f(0,x)\ge 0$ for all $x\in\overline\Omega$ and
$$
\limsup_{u\to+\infty}\frac{\max\{|f(u,x)|:
x\in\overline\Omega\}}{u}<\frac{1}{\|g\|}.
$$
\end{itemize}
Concerning problem \eqref{1.1r}, we can establish the
following results.

\begin{theorem} \label{thm2.1}
Let {\rm (A1)} and {\rm (A2)} hold.
Then problem \eqref{1.1r} has no solution for
$$
\rho>\tilde\rho:=\frac{M\int_\Omega\phi_1(x)dx}
{\lambda_1\int_\Omega\phi_1(x)h(x)\,dx}.
\eqno{(2.3)}
$$
Here $h(x)$ is the harmonic function defined on $\overline\Omega$
satisfying
\begin{equation} \label{e2.4}
h(x)=0\quad \mbox{on }B_0 \quad\mbox{and}\quad h(x)=1\quad\mbox{on } B.
\end{equation}
\end{theorem}

\begin{theorem} \label{thm2.2}
Let {\rm (A1)}, {\rm (A2)} and {\rm (A3)} hold.
Then there exists a positive number $\rho^*$ such that problem
\eqref{1.1r} has a nonnegative solution for
$\rho\in[0,\rho^*)$ and no solution for $\rho>\rho^*$.
\end{theorem}

\begin{theorem} \label{thm2.3}
Let {\rm (A1)} and {\rm (A4)} hold.
Then problem \eqref{1.1r} has a nonnegative solution for all
$\rho\ge0$. In addition, assume that $f(0,x)\equiv0$ for all
$x\in\overline\Omega$ and there exists a $\delta>0$ such that
$f(u,x)\ge\lambda_1u$ for all $u\in[0,\delta]$ and all $x\in
\overline\Omega$. Then Problem \eqref{1.1r} with $\rho=0$
has at least one positive solution.
\end{theorem}

Theorems \ref{thm2.1} and \ref{thm2.3} are new and Theorem \ref{thm2.2}
extends and complements the corresponding theorems in \cite{b1,h1,l1}.
Before proving these theorems, we make several remarks.

\begin{remark} \label{rmk2.1} \rm
A function $u\in C(\overline\Omega; \mathbb{R})$ is a
solution to Problem \eqref{1.1r}, if and only if it is a solution
to the integral equation
\begin{equation} \label{2.5r}
u(x)=\int_\Omega G(x,y)f(u(y),y)\,dy+\rho h(x),\quad
x\in\overline\Omega.
\end{equation}
Clearly, each solution to Problem \eqref{1.1r} is positive when
$f\in C(\mathbb{R}_+\times\overline\Omega; \mathbb{R}_+)$ and $\rho>0$.
\end{remark}


\begin{remark} \label{rmk2.2} \rm
>From the maximum principle, we know that
$0<h(x)<1$ in $\Omega$.
\end{remark}

\begin{remark} \label{rmk2.3} \rm
According to \eqref{e2.1} and \eqref{2.5r}, we have
$$
0<\phi_1(x)=\lambda_1\int_\Omega G(x,y)\phi_1(x)dx\le 1\quad
\mbox{in } \Omega
$$
and hence $\lambda_1>1/\|g\|$.
\end{remark}

\begin{remark} \label{rmk2.4} \rm
It is obvious that Theorems \ref{thm2.1}, \ref{thm2.2} and \ref{thm2.3}
are still valid if the boundary conditions in \eqref{1.1r} are
replaced by $u=\rho$ on $B_0$,  $u=0$ on $B$.
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
Suppose to the contrary that Problem \eqref{1.1r} has a solution,
$$
u(x)=\int_\Omega G(x,y)f(u(y),y)\,dy +\rho h(x)\\
 =:w(x)+\rho h(x),\quad x\in \overline\Omega,
$$
for some $\rho>\tilde\rho$. In this case, we have $-\Delta
w(x)=f(u(x),x)$ in $\overline\Omega$. Consequently,
\begin{align*}
\rho\lambda_1\int_\Omega\phi_1(x)h(x)dx
&=\int_\Omega\lambda_1\phi_1(x)u(x)dx-\int_\Omega\lambda_1\phi_1(x)w(x)dx\\
&=\int_\Omega\lambda_1\phi_1(x)u(x)dx+\int_\Omega\phi_1(x)\Delta w(x)dx\\
&=\int_\Omega\phi_1(x)(\lambda_1u(x)-f(u(x),x)\,dx\\
&\le M\int_\Omega\phi_1(x)dx,
\end{align*}
i.e., $\rho\le\tilde\rho$, a contradiction. Theorem \ref{thm2.1} is thus
proved.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.2}]
To prove Theorem \ref{thm2.2}, we first define a mapping $K:E\mapsto E$ by setting
\begin{equation} \label{2.6r}
(Kw)(x):=\int_\Omega G(x,y)f^*(w(y),y)dy+\rho h(x), \quad \forall
w\in E,
\end{equation}
where $E:=C(\overline \Omega; \mathbb{R})$ and
$$
f^*(u,x):= \begin{cases}
f(0,x)\quad \mbox{if } u<0,\\
f(u,x)\quad \mbox{if } u\ge0.
\end{cases}
$$
It is easy to check that $K$ is completely continuous on $E$.
>From (A3), we know that there exists an $\epsilon>0$ such that
$$
 \limsup_{u\to0+}\frac{\max\{|f(u,x)|:
x\in\overline\Omega\}}{u}<\frac{1}{\|g\|+\epsilon}<\frac{1}{\|g\|}
$$
and hence there exists a $\sigma>0$ such that
$$
|f(u,x)|\le\frac{u}{\|g\|+\epsilon} \quad
\mbox{for all $u\in[0,\sigma]$ and all $x\in\overline\Omega$},
$$
from which it follows that
\begin{equation} \label{e2.7}
f(0,x)\equiv0\quad\mbox{for  all  }x\in\overline\Omega.
\end{equation}
Clearly, $u(x)\equiv 0$ is a trivial solution to Problem \eqref{1.1r} with
$\rho=0$.
Put
$$
D_\sigma:=\{w\in E: \|w\|\le \sigma\}\quad \mbox{and}\quad
\sigma^*:=\sigma\big(1-\frac{\|g\|}{\|g\|+\epsilon}\big).
$$
Then for each fixed $w\in D_\sigma$ and each fixed
$\rho\in[0,\sigma^*] $, we have
$$
\|Kw\|\le\frac{\|g\|}{\|g\|+\epsilon}\sigma+\sigma^*=\sigma
$$
which means that $K$ is a compactly continuous mapping from
$D_\sigma$ into itself. The Schauder fixed point theorem tells us
that $K$ has a fixed point $u\in D_\sigma$, i.e.,
\begin{gather*}
-\Delta u=f^*(u(x), x)\quad \mbox{in } \Omega,\\
u=0\quad \mbox{on } B_0,\quad  u=\rho  \mbox{on  } B.
\end{gather*}
>From \eqref{e2.7} and the maximum principle, we know that $u(x)\ge0$ on
$\overline\Omega$, which shows that the fixed point $u(x)\in
D_\sigma$ is also a nonnegative solution to Problem \eqref{1.1r}
with $\rho\in[0,\sigma^*]$.

Let
$$
\rho^*=\sup\{\rho\ge0: \mbox{ Problem \eqref{1.1r}  has a
nonnegative solution}\}.
$$
>From the previous results, $\rho^*\in[\sigma^*,\tilde\rho]$. We
are now going to prove that Problem \eqref{1.1r} has a nonnegative
solution for all $\rho\in[0,\rho^*)$.

For each given $\rho\in[0,\rho^*)$, from the definition of
$\rho^*$, we can choose a $\bar\rho\in (\rho,\rho^*)$ such that
Problem $(1.1)_{\bar\rho}$ has a nonnegative solution $\bar u(x)$.
Let $\xi(x)\equiv0$ and $\eta(x)=\bar u(x)$. Then
\begin{gather*}
-\Delta \xi(x)\equiv0\equiv f(\xi(x), x)\quad\in \Omega,\\
\xi(x)=0\quad \mbox{on }\ B_0,\quad  \xi(x)=0\le\rho \quad \mbox{on } B,
\end{gather*}
and
\begin{gather*}
-\Delta \eta(x)\equiv f(\eta(x), x)\quad\mbox{in } \Omega,\\
\eta(x)=0\quad \mbox{on } B_0,\quad \eta(x)=\bar\rho>\rho \quad
\mbox{on }B.
\end{gather*}
i.e., $\xi(x)$ is a lower solution to Problem \eqref{1.1r} and
$\eta(x)$ is an upper solution. Employing the method of upper and
lower solutions, we can find a solution $u(x)$ to Problem
\eqref{1.1r} with
$$
0\equiv \xi(x)\le u(x) \le \eta(x)=\bar u(x)\quad{on }\overline\Omega.
$$
The proof of Theorem \ref{thm2.2} is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.3}]
 From (A4), we know that there exists an $\epsilon>0$ such that
$$
\limsup_{u\to+ \infty}\frac{\max\{|f(u,x)|:
x\in\overline\Omega\}}{u}\le\frac{1}{\|g\|+\epsilon}<\frac{1}{\|g\|}
$$
and hence there exists an $N>0$ such that
$$
|f(u,x)|\le\frac{u}{\|g\|+\epsilon} \quad
\mbox{for all $\ u\ge N$  and all $x\in\overline\Omega$.}
$$
For each given $\rho\ge0$, we put
\begin{gather*}
D_\beta:=\left\{w\in E: \|w\|\le\beta \right\},\\
\beta:=\big(1-\frac{\|g\|}{\|g\|+\epsilon}\big)^{-1}
\big(N+\rho+\|g\|\max\{|f(u,x)|:
x\in\overline\Omega,\ 0\leq u\leq N\}\big).
\end{gather*}
We define a mapping $K:E\mapsto E$ by \eqref{2.6r}.  For each fixed
$w\in D_\beta$, we have
$$
\|Kw\|<\|g\|\Big(\max\{|f(u,x)|: 0\le u\le N,\
x\in\overline\Omega\}+\frac{\beta}{\|g\|+\epsilon}\Big)+\rho+N=\beta,
$$
which implies that $K$ is a compactly continuous mapping from
$D_\beta$ into itself. The Schauder fixed point theorem tells us
that $K$ has a fixed point $u\in D_\beta$, i.e.,
\begin{gather*}
-\Delta u(x)=f^*(u(x),x)\quad \mbox{in } \Omega,\\
u(x)=0\quad \mbox{on } B_0,\quad u(x)=\rho  \mbox{on  } B.
\end{gather*}
 From the maximum principle and the assumption that $f(0,x)\ge0$
for all $x\in\overline\Omega$, we know that
$$
u(x)\ge0\quad\mbox{on } \overline\Omega.
$$
This shows that the fixed point $u\in D_\beta$ is a nonnegative
solution of \eqref{1.1r}.

We now assume that $f(0,x)\equiv0$ for all $x \in\overline\Omega$
and there exists a $\delta>0$ such that
$$
f(u,x)\ge\lambda_1 u \mbox{ for all $u\in [0,\delta]$ and
 all $x\in\overline\Omega$.}
$$
 In this case, $u(x)\equiv0$ is a
trivial solution to Problem $(1.1)_0$.
Then we consider the modified boundary-value problem
\begin{equation} \label{e2.8}
\begin{gathered}
-\Delta u(x)=\bar f(u(x), x)\quad\mbox{in } \Omega,\\
u(x)=0 \quad \mbox{on } \partial\Omega,
\end{gathered}
\end{equation}
where the function
$$
\bar f(u,x):=\begin{cases}
f(\delta\phi_1(x),x)&\mbox{if  } u<\delta\phi_1(x),\\
f(u,x)&\mbox{if  } u>\delta\phi_1(x)
\end{cases}
$$
satisfies (A1) and (A4) again. From the above discussion, we know
that Problem \eqref{e2.8} has a solution $u\in D_\beta$. The maximum
principle tells us that
$$
u(x)\ge\delta \phi_1(x)\quad\mbox{on }\overline\Omega.
$$
This shows that the solution $u(x)$ is also a positive solution to
Problem \eqref{1.1r} with $\rho=0$. Theorem \ref{thm2.3} is thus proved.
\end{proof}

\section{Results in annular domains}

In this section, we restrict our attention to the multiplicity of
radial solutions to  \eqref{1.2r}. When we seek for a radial
solution to \eqref{1.2r}, the problem can be rewritten as
\begin{equation} \label{3.1r}
\begin{gathered}
-(k(t)u'(t))'=k(t)f(u(t), t),\quad \alpha<t<\beta,\\
u(\alpha)=0, \quad  u(\beta)=\rho,
\end{gathered}
\end{equation}
where $k(t)=t^{n-1}$, $n\ge2$. Concerning Problem \eqref{3.1r}, we
make the following hypotheses:
\begin{itemize}
\item[(H1)] $f\in C(\mathbb{R}_+\times J; \mathbb{R}_+)$, $J:=[\alpha, \beta]$.

\item[(H2)]
\begin{equation} \label{e3.2}
\limsup_{u\to0}\frac{\max\{f(u,t): t\in
J\}}{u}<\frac1{\|g\|},
\end{equation}
 where
\begin{equation} \label{e3.3}
\begin{gathered}
 g(t)=\int_\alpha^\beta G(t,s)ds,\quad t\in J;\\
G(t,s)=\begin{cases}
\frac{k(s)}{P(\beta)}(P(\beta)-P(t))P(s),& \alpha\le s\le t\le
\beta,\\
\frac{k(s)}{P(\beta)}(P(\beta)-P(t))P(t),& \alpha\le t\le s\le
\beta;\end{cases}\\
 P(t)=\int_\alpha^t\frac{dr}{k(r)}, \quad t\in J.
\end{gathered}
\end{equation}

\item[(H3)]  There exists an interval $[a,b]$ with  $\alpha<a<b<\beta$, such
that
$$
\liminf_{u\to+\infty}\frac{\min\{f(u,t): a\le t\le b\}}{u}>\frac1m,
$$
where
\begin{equation} \label{e3.4}
\begin{gathered}
m=\delta\max\{\int_a^b G(t,s)\,ds: a\le t\le b\},\\
\delta=\min\{q(t): a\le t\le b\},\\
q(t)=\min\big\{\frac{P(t)}{P(\beta)},\
\frac{P(\beta)-P(t)}{P(\beta)}\big\}, \quad t\in J\,.
\end{gathered}
\end{equation}

\item[(H4)]  $f(u,t)$ is nondecreasing in $u\ge0$ for each fixed
$t\in J$.

\item[(H4)*] $f(u,t)$ is locally Lipschitz continuous in $u\ge0$ for each
fixed $t\in J$.
\end{itemize}

\begin{theorem} \label{thm3.1}
 Let {\rm (H1)}, {\rm (H2)}, {\rm (H3)} and
{\rm (H4)}  (or {\rm (H4)*}) hold. Then there exists a positive
number $\rho^*$ such that Problem \eqref{1.2r} has at least two
nonnegative radial solutions for $\rho\in[0, \rho^*)$, at least
one for $\rho=\rho^*$ and none for any $\rho>\rho^*$.
\end{theorem}

\begin{remark} \label{rmk3.1} \rm
Theorem \ref{thm3.1} is still valid when the boundary
conditions in \eqref{3.1r} are replaced by
$$
u(\alpha)=\rho,\quad u(\beta)=0.
$$
\end{remark}

Clearly, Theorem \ref{thm3.1} is an extension and improvement of the
results in \cite[Theorem 1.2]{h1}.
We need the following lemmas.

\begin{lemma} \label{lem3.2}
Let $f\in C(\mathbb{R}_+\times J;\ \mathbb{R}_+)$ and
$u(t)$ a nonnegative solution to Problem \eqref{3.1r}. Then
$$
u(t)\ge\|u\|q(t),\quad t\in J\,.
$$
\end{lemma}

\begin{proof}
Note that Problem \eqref{3.1r} is
equivalent to the integral equation
\begin{equation} \label{3.5r}
u(t)=\int_\alpha^\beta G(t,s)f(u(s),s)\,ds+\rho h(t),\quad t\in J,
\end{equation}
where
\begin{equation}
 h(t):=P(t)/P(\beta), \quad t\in J. \label{e3.6}
\end{equation}
Let $\|u\|=u(t_0) $. Then $t_0\in (\alpha, \beta]$. If
$t_0=\beta$, then $\|u\|=\rho$ and hence
$u(t)=\rho h(t)\ge\|u\|q(t)$, $t\in J$.
If $t_0\in (\alpha,\beta)$, then
\begin{align*}
u(t)&=\int_\alpha^\beta
\frac{G(t,s)}{G(t_0,s)}G(t_0,s)f(u(s),s)\,ds +\rho
h(t_0)\frac{h(t)}{h(t_0)}\\
&\ge q(t)\int_\alpha^\beta G(t_0,s)f(u(s),s)\,ds +\rho h(t_0)q(t)
\\
&=q(t)\|u\|\quad \mbox{for  all } t\in J.
\end{align*}
Here we have used the fact that
$$
\frac{G(t,s)}{G(t_0,s)}\ge q(t)\quad \mbox{for  all $t, s$ in }(\alpha, \beta),
$$
the proof of which can be found in \cite{a1}.
\end{proof}


\begin{lemma} \label{lem3.3}
Let {\rm (H1)} and {\rm (H3)} hold. Then there exists a positive number $M$ such
that all solutions $u(t)$ to  \eqref{3.1r} satisfy $\|u\|<M$.
\end{lemma}

\begin{proof}
 By Lemma \ref{lem3.2}, we know that if $u(t)$ is a
nonnegative solution to Problem \eqref{3.1r}, then
$$
u(t)\ge\delta\|u\|\quad\mbox{for  all  }t\in [a,b],
$$
where the constant $ \delta$ is defined by \eqref{e3.4}.
 From (H3), we know that there exists an $\epsilon>0$ such that
$$
\liminf_{u\to+\infty}\frac{\min\{f(u,t); \
t\in[a,b]\}}{u}>\frac1{m-\epsilon}>\frac1m,
$$
and hence there exists an $M>0$ such that
\begin{equation}
f(u,t)\ge\frac{u}{m-\epsilon}\quad \mbox{for  all $u\ge\delta M$ and
all $t\in[a,b]$}. \label{e3.7}
\end{equation}

We now claim that $\|u\|<M$ for all nonnegative solution $u(t)$ to
\eqref{3.1r}, where the constant $M$ satisfies \eqref{e3.7}.
If the claim is false, then Problem \eqref{3.1r} has a nonnegative
solution $u(t)$ with $\|u\|\ge M$. In this case, we have
$$
u(t)\ge\int_a^b G(t,s)f(u(s),s)\,ds
\ge\frac{\delta\|u\|}{m-\epsilon}\int_a^bG(t,s)\,ds
$$
for all $t\in[a,b]$; i.e.,
$$
\|u\|\ge\frac{m}{m-\epsilon}\|u\|.
$$
This is a contradiction which proves the claim.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3.1}]
We first define a mapping
 $ K:E \mapsto E $ by setting
$$
(Kw)(t)=\int_\alpha^\beta G(t,s)f^*(w(s),s)\,ds+\rho h(t),
$$
where $E=C(J; \mathbb{R})$ and
$$
f^*(u,t)=\begin{cases}
f(0,t) &\mbox{if  } u<0,\\
f(u,t) &\mbox{if  } u\ge0.
\end{cases}
$$
It is easy to check that $K$ is completely continuous on $E$.
>From (H2), we know that there exists an $\epsilon>0$ such that
$$
\limsup_{u\to0+}\frac{\max\{f(u,t): t\in
J\}}{u}<\frac1{\|g\|+\epsilon}<\frac1{\|g\|}
$$
and hence there exists a $\sigma>0$ such that
$$
0\le f(u,t)\le \frac{u}{\|g\|+\epsilon}
$$
for all $u\in[0,\sigma]$ and all $t\in J$, which implies that
$f(0,t)\equiv0$ for all $t\in J$.
We now put
$$
D_\sigma=\{w\in E: \|w\|\le\sigma\}\quad \mbox{and }\quad
\sigma^*=\sigma\big(1-\frac{\|g\|}{\|g\|+\epsilon} \big).
$$
Then for each fixed $w\in D_\sigma$ and each fixed
$\rho\in[0,\sigma^*]$, we have
$$
\|Kw\|\le\frac{\|g\|}{\|g\|+\epsilon}+\sigma^*=\sigma,
$$
which implies that $K$ is a completely continuous mapping from
$D_\sigma$ into itself. The Schauder fixed point theorem tells us
that $K$ has a fixed point $u\in D_\sigma$, i.e.,
\begin{gather*}
-(k(t)u'(t))'=k(t)f^*(u(t),t),\quad \alpha<t<\beta,\\
 u(\alpha)=0,
\quad u(\beta)=\rho.
\end{gather*}

We now claim that $u(t)\ge0$ for all $t\in J$. If the claim is
false, then there exists an interval $[a,b]$, $\alpha\le a<b\le
\beta$, such that
$$
u(t)<0\quad \mbox{in }(a,b)\quad\mbox{and}\quad u(a)=u(b)=0.
$$
Consequently,
\begin{gather*}
-(k(t)u'(t))'=0,\quad a<t<b,\\
 u(a)=u(b)=0.
\end{gather*}
which implies that $u(t)\equiv 0$ on $[a,b]$. This is a
contradiction and hence the claim is true.
As a result, the fixed point $u\in D_\sigma$ is a nonnegative
solution to \eqref{3.1r}.

We now put
$$
\rho^*=\sup\{\rho\ge0: \mbox{ \eqref{3.1r} has  a nonnegative solution}\}.
$$
Then $\rho^*\in[\sigma^*, M)$. Here we have used Lemma \ref{lem3.3}.

 From the definition of $\rho^*$, we can choose a sequence
$\{\rho_j\}_{j=1}^{\infty}$ such that $\rho_j<\rho_{j+1}$,
$\rho_j\to \rho^* $ as $j\mapsto+\infty$, and Problem
\eqref{3.1r} with $\rho_J$ has a nonnegative solution $u_j(t)\in E$.
 From Lemma \ref{lem3.3} and the complete continuity of $K$ on $E$, we know that
$\{u_j(t)\}_{j=1}^{\infty}$ is uniformly bounded and
equicontinuous on $J$. Without loss of generality, we may assume
that $u_j(t)\to u^*(t)$ uniformly on $J$ as $j\to+\infty$. Note
that
$$
u_j(t)=\int_\alpha^\beta G(t,s)f(u_j(s), s)\,ds + \rho_j
h(t),\quad t\in J\,.
$$
Letting $j\to +\infty$ in the above yields
$$
u^*(t)=\int_\alpha^{\beta} G(t,s)f(u^*(s), s)\,ds +\rho^*
h(t),\quad t\in J.
$$
This shows that $u^*(t)\in E$ is a positive solution to
\eqref{3.1r} with $\rho^*$.

We are now in position to prove that  \eqref{3.1r} has
at least two nonnegative solutions for all $\rho\in [0,\rho^*)$.

For each given $\rho\in[0,\rho^*)$, we set $\xi(t)\equiv 0$ and
$\eta(t)=u^*(t)$. Then $\xi(t)$ is a lower solution to
\eqref{3.1r} and $\eta(t)$ an upper solution. Employing the method
of upper and lower solutions, we can find a solution $u(t)\in E$
with
$$
0\equiv\xi(t)\le u(t)\le \eta(t)=u^*(t)\quad\mbox{on  }J\,.
$$
We now assume that (H4)* holds. Using the local Lipschitz
continuity of $f(u,t)$ with respect to $u\in \mathbb{R}_+$ and the strong
maximum principle, we deduce that
\begin{equation}
0\le u(t)<u^*(t) \quad \mbox{for  all  }  t\in (\alpha, \beta].
\label{e3.8}
\end{equation}
If (H4) holds, i.e., $f(u,t)$ is nondecreasing in $u\in \mathbb{R}_+$ for
each fixed $t\in J$. Then
$$
u^*(t)-u(t)=\int_\alpha^\beta\big(f(u^*(s),s)-f(u(s),s)\big)\,ds
+(\rho^*-\rho)h(t)>0
$$
for all $t\in (\alpha, \beta]$. i.e., (3.8) is also valid.

To obtain the existence of a second nonnegative solution to
 \eqref{3.1r} for all $\rho\in[0,\rho^*)$, we define a
mapping $\tilde K: E\mapsto E$ by setting
$$
\big(\tilde Kw\big)(t):=\int_\alpha^\beta G(t,s)\tilde
f(w(s),s)\,ds +\rho h(t),
$$
and another mapping $K:E\times \mathbb{R}_+\mapsto E$ by setting
$$
K(w,y)(t):=\int_\alpha^\beta G(t,s) f^*(w(s),s)\,ds +y h(t),
$$
where
$$
f^*(w,t)=\begin{cases}
0 &\mbox{if  }w<0,\\
f(w,t) &\mbox{if  }w\ge0, \end{cases}
$$
and
$$
\tilde f(w,t):=\begin{cases}
f^*(w,t) &\mbox{if  }w\le u^*(t),\\
f^*(u^*(t),t) &\mbox{if  }w>u^*(t). \end{cases}
$$
Note that for each fixed $w\in E$, we have
$$
\|\tilde Kw\|<\|g\| \max\{f(u,t): 0\le u\le \|u^*\|,\; t\in J\}+\rho^*=:N^*.
$$
Picking $t_0\in (\alpha,\beta)$ such that $u^*(t_0)>0$, we define
$$
v^*(t)=\begin{cases} u^*(t_0) &\mbox{if  }\alpha\le t\le t_0,\\
u^*(t) &\mbox{if  }t_0<t\le \beta. \end{cases}
$$
We now put
\begin{gather*}
A:=\big\{ w\in E: -M-N^*<w(t)< v^*(t),\; t\in[\alpha,\beta] \big\},\\
B:=\{w\in E: \|w\|<M+N^*\}
\end{gather*}
where the constant $M$ is determined by Lemma \ref{lem3.3}. Then both $A$
and $B$ are open subsets of $E$ and $u\in A\subset B$.

Clearly, $\tilde K$ has fixed points in $A$. In fact, $u\in A$ is
a fixed point of $\tilde K$. We now consider whether $\tilde K$
has a fixed point in $B\setminus A$ or not. There are two
possibilities.

\noindent Case (i). $\tilde K$ has no fixed point in $B\setminus A$. In this
case, we have
$$
\deg (I-K(\cdot,\rho), A, 0)=\deg (I-\tilde K, A, 0)=\deg (I-\tilde K, B, 0),
$$
where $I$ denotes the identity mapping from $E$ into itself.
We now claim that $\deg (I-\tilde K, B, 0)=1$. To prove the
claim, we consider the homotopic mapping
$$
\Psi(w,\tau):=w-\tau\tilde Kw\quad \forall (w,\tau)\in E\times [0,1].
$$
For any $(w,\tau)\in\partial B\times [0,1]$, we have
$$
\|\Psi(w,\tau)\|\ge\|w\|-\|\tilde Kw\|>M+N^*-N^*=M.
$$
i.e., $\Psi(w,\tau)\neq0$ for any $(w,\tau)\in\partial B\times
[0,1]$. Consequently
$$
\deg (I-\tilde K, B, 0)=\deg (\Psi(\cdot,1), B, 0)=\deg (\Psi(\cdot,0),B,0)=\deg (I,B,0)=1.
$$
On the other hand, we know that $\deg (I-K(\cdot,y),B,0)$ is constant
for all $y\ge0$. From Lemma \ref{lem3.3},
we know that $K(\cdot, M)$ has no fixed point in $E$ and hence the
constant must be zero. Therefore,
$$
\deg (I-K(\cdot,\rho), B, 0)= \deg (I-K(\cdot,M), B, 0)=0.
$$
By the excision property of the Leray-Schauder degree, we obtain
$$
\deg (I-K(\cdot,\rho), B\setminus A, 0)=-1
$$
which implies that $K(\cdot, \rho)$ has a fixed point in
$B\setminus A $. The fixed point is a nonnegative solution to Problem
\eqref{3.1r}, of course. \smallskip

\noindent Case (ii).  $\tilde K$ has a fixed point $\bar u\in B\setminus A$.
By the maximum principle, we know that
$$
0\le \bar u(t)\le u^*(t)\quad \mbox{on }J.
$$
This means that $\bar u(t)$ is also a second solution to
\eqref{3.1r}. Since each nonnegative solution to
\eqref{3.1r} is also a nonnegative radial solution to
\eqref{1.2r}, Theorem \ref{thm3.1} is thus proved.
\end{proof}

Finally, we consider the boundary-value problem
\begin{equation}
\begin{gathered}
-\Delta u=f_j(u),\quad 0<\alpha<|x|<\beta,\\
u=0 \quad \mbox{on  }|x|=\alpha,\quad
u=\rho\quad\mbox{on } |x|=\beta,
\end{gathered} \label{3.8j}
\end{equation}
where
$$
f_1(u)=\begin{cases}
\xi u, & 0\le u\le1;\; 0\le\xi<1/\|g\|,\\
9(u-1)^{1/9} +\xi, & 1\le u\le2,\\
\eta(u-2)+9+\xi, & u\ge2;\; \eta>1/m
\end{cases}
$$
and
$$
f_2(u):=\begin{cases}
\sin^2u,& 0\le u\le 8\pi,\\
\eta(u-8\pi), & u\ge 8\pi,\; \eta>1/m,
\end{cases}
$$
the constant $m$ and the function $g(t)$ are determined by
\eqref{e3.4}
and \eqref{e3.2}, respectively.

Since $f_1(u)$ satisfies (H1), (H2), (H3) and (H4), $f_2(u)$
satisfies (H1), (H2), (H3) and (H4)*, according to Theorem \ref{thm3.1},
there exists a positive number $\rho^*$ such that Problem
\eqref{3.8j}, $j=1,\ 2$, has at least two nonnegative radial
solutions for $\rho\in [0,\rho^*)$, at least one for $\rho=\rho^*$
and none for $\rho>\rho^*$.

However, Theorem \ref{thm1.2} cannot be applied in studying \eqref{3.8j}, $j=1,\,2$.

\subsection*{Acknowledgement}
The authors want to thank  the referee for pointing out some errors in the
first version of this paper.


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\end{document}