\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 72, pp. 1--24.\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 2004 Texas State University - San Marcos.}
\vspace{9mm}}


\begin{document}
\title[\hfilneg EJDE-2004/72\hfil Exact multiplicity results]
{Exact multiplicity results for a $p$-Laplacian positone
problem with concave-convex-concave nonlinearities}

\author[Idris Addou \& Shin-Hwa Wang\hfil EJDE-2004/72\hfilneg]
{Idris Addou \& Shin-Hwa Wang}

\address{Idris Addou \hfill\break
D\'{e}partement de Math\'{e}matiques et Statistiques, Universit\'{e}
de Montr\'{e}al\\
C.P. 6128, Succ. Centre-ville, Montreal, Quebec, Canada, H3C2J7}
\email{addou@dms.umontreal.ca}

\address{Shin-Hwa Wang \hfill\break
Department of Mathematics, National Tsing Hua University\\
Hsinchu, Taiwan 300, Republic of China}
\email{shwang@math.nthu.edu.tw}

\date{}
\thanks{Submitted March 8, 2004. Published May 20, 2004.}
\subjclass[2000]{34B18, 34B15} 
\keywords{Exact multiplicity result, $p$-Laplacian, positone problem, bifurcation,
\hfill\break\indent
concave-convex-concave nonlinearity, positive solution, dead core
solution,  time map}

\begin{abstract}
  We study the exact number of positive solutions of a two-point Dirichlet
  boundary-value problem involving the $p$-Laplacian operator.
  We consider the case $p=2$ and the case $p>1$, when the
  nonlinearity $f$ satisfies $f(0)>0$ (positone) and has three distinct
  simple positive zeros and such that $f''$ changes sign exactly
  \emph{twice} on $(0,\infty )$. Note that we may allow  $f''$ to
  change sign more than twice on $(0,\infty )$. We also present
  some interesting examples.
\end{abstract}

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

\section{Introduction}

In this paper we present \emph{exact} multiplicity results of \emph{positive}
\emph{solutions} for the nonlinear two-point Dirichlet boundary-value
problem
\begin{equation}
\begin{gathered}
-( \varphi _{p}(u'(x))) '=\lambda f(u(x)),\; -1<x<1, \\
u(-1)=u(1)=0,
\end{gathered} \label{1.1}
\end{equation}
where $p>1$, $\varphi _{p}(y)=| y| ^{p-2}y$ and
$( \varphi_{p}(u')) '$ is the one-dimensional $p$-Laplacian$,
\lambda >0$ and $f$ is a concave-convex-concave nonlinearity. Precise
conditions are listed below.

This paper is intended as a second part of a previous paper by the present
authors \cite{a3}. In fact, whereas the previous paper was a study of
\eqref{1.1} with $f\in C^{2}[0,\infty )$ satisfying $f(0)=0$ and has \emph{two}
distinct simple positive zeros $b<c$ and such that $f''$
changes sign exactly \emph{twice} on $(0,\infty )$, here we wish to complete
the picture by studying the same sort of \eqref{1.1} but with the
nonlinearity $f\in C^{2}[0,\infty )$ satisfying instead, $f(0)>0$ (positone)
and has \emph{three} distinct simple positive zeros $a<b<c$ and such that $%
f''$ changes sign exactly \emph{twice} on $(0,\infty )$.

Note that besides being complementary to our previous paper \cite{a3}, the
present article contains an important originality which deserves to be
mentioned in this introduction. A familiar feature related to positive
solutions, say $u$, of a one-dimensional Dirichlet boundary-value problem
with the $p$-Laplacian differential operator, is that we think that the
interior zero set of the derivative $u'$ is a \emph{connected set}.
(That is, if $u\in C^{1}[-1,1]$ is a positive solution and $Z(u)=\{x\in
[ -1,1] :u'(x)=0\}$ then $Z(u)\cap (-1,1)$ is either a
single point or a closed interval.) For the particular case $p=2$, it is
easy to prove that $Z(u)\cap (-1,1)$ is indeed a connected set, but what
about the more general case $p>1$? None known results in the literature
prove or disprove this feature. This paper provide an example which
disproves this fact. Indeed, for some $p>1$, $p\neq 2$, \emph{we have
obtained some positive solutions of \eqref{1.1} such that the interior zero
set of their derivative is not a connected set.} Note that this situation is
not known even for \eqref{1.1} when $f(0)=0$ in our previous paper Addou and
Wang \cite{a3}.

For $p=2$, $( \varphi _{p}(u')) '=u''$, and \eqref{1.1} reduces to
\begin{equation}
\begin{gathered}
-u''(x)=\lambda f(u(x)),\text{ }-1<x<1, \\
u(-1)=u(1)=0,
\end{gathered} \label{1.2}
\end{equation}
and several \emph{exact} multiplicity results are known when $f$ vanishes
three times on $(0,\infty )$, see [6, 8, 9, 10, 11]. However, in all of
them, $f''$ changes sign exactly once on $(0,\infty )$. In
fact, first studies go back to Smoller and Wasserman \cite{s1} in which they
studied exact multiplicity results of (classical) positive solutions of (\ref
{1.2}) for cubic-polynomial nonlinearities $f(u)=-(u-a)(u-b)(u-c)$
satisfying $0<a<b<c$, $c>2b-a$ $( \Leftrightarrow
\int_{a}^{c}f(u)du>0) $, and a certain condition; see also Wang 
\cite{w1}. One can note that, here, $f''(u)$ changes sign exactly
once on $(0,\infty )$. Subsequently, Wang and Kazarinoff \cite{w3} and Wang
\cite{w2} studied \eqref{1.2} when $f$ is a cubic-like nonlinearity. In
particular, Wang and Kazarinoff proved the next theorem. Define 
$F(u)=\int_{0}^{u}f(t)dt$.

\begin{theorem}[{\cite[Theorem 1 and Remark 2]{w3}}]  \label{thm1.1}
Suppose $f\in C^{2}[0,\infty )$ and
there exist $0<a<b<c$ such that the following conditions are satisfied:
\begin{equation}
f(a)=f(b)=f(c)=0;  \label{1.3}
\end{equation}
\begin{equation}
\begin{gathered}
f(u)>0\quad \text{for }u\in (0,a), \\
f(u)>0\quad \text{for }u\in (b,c), \\
f(u)<0\quad \text{for }u\in (a,b)\cup (c,\infty );
\end{gathered}   \label{1.4}
\end{equation}
\begin{equation}
\int_{a}^{c}f(u)du>0;  \label{1.5}
\end{equation}
there exists a unique $\beta \in (b,c)$ defined by $\int_{a}^{\beta }f(u)du=0$
and such that $2F(a)-\beta f(\beta )<0$; \\
there exists $r\in (0,c)$ such that
$f''(u)>0$  for $0<u<r$  and $f''(u)<0$ for $r<u<c$.
\\
Then, there exists $\lambda _{0}>0$ such that
\begin{itemize}
\item[(i)]  for $0<\lambda <\lambda _{0}$, problem \eqref{1.2} has exactly
one positive solution $u_{0}$ satisfying $0<\| u_{0}\| <a$,

\item[(ii)]  for $\lambda =\lambda _{0}$, problem \eqref{1.2} has exactly
two positive solutions $u_{0}<u_{1}$ satisfying $0<\| u_{0}\|
<a<\beta <\| u_{1}\| <c$,

\item[(iii)]  for $\lambda >\lambda _{0}$, problem \eqref{1.2} has exactly
three positive solutions $u_{0}<u_{1}<u_{2}$ satisfying $0<\|
u_{0}\| <a<\beta <\| u_{1}\| <\| u_{2}\| <c$.
\end{itemize}
\end{theorem}

\begin{remark} \label{Remark1} \rm
If $f\in C[0,\infty )$ satisfies (\ref{1.3})-(\ref{1.5}),
then it can be shown that

\begin{itemize}
\item[(i)]  By the maximum principle, every classical positive solution $u$
of \eqref{1.2} satisfies either $0<\| u\| _{\infty }<a$, or $%
\beta <\| u\| _{\infty }<c$.

\item[(ii)]  Any two distinct positive solutions of \eqref{1.2} are strictly
ordered. That is, let $u$ and $\hat{u}$ be any two distinct positive
solutions of \eqref{1.2} with $0<\| u\| _{\infty }<\| \hat{u}%
\| _{\infty }$, then $u<\hat{u},\;$see e.g. \cite[Lemma 1]{w3}.
\end{itemize}
\end{remark}

Note that a similar result to Theorem 1.1 was obtained by Korman
\textit{et al.} \cite[Theorem 2.7]{k1}. For $f$ a cubic-like nonlinearity
and for \eqref{1.2} ($p=2$), similar results when $f(0)=0$ (resp. $f(0)>0$) and
$f''$ changes sign exactly \emph{once} on $(0,\infty )$ were
proved by Korman and Shi \cite{k3} (resp. Korman \textit{et al.} \cite{k2}.)

But for \eqref{1.1} with $p\neq 2$, little is known. In fact for the case
where $0=a<b<c$ we refer to Addou \cite{a2}. Problem \eqref{1.1} with $p>1$,
has been recently studied in Addou and Wang \cite{a3} for the case where
$f(0)=0$ and $f''$ changes sign exactly \emph{twice} on $(0,\infty )$.
We note that the case where $f(0)>0$ and $f''$
changes sign exactly \emph{twice} has not been studied yet.

The paper is organized as follows. Section 2 is devoted to the definitions
of the sets which contain the solutions of \eqref{1.1} and stating the main
tool used subsequently, namely, the quadrature method. Next, in Section 3,
we state our main results. In Section 4, a weakened condition and two
examples are given. Finally, in Section 5, we prove the main results.

\section{Quadrature method}

To state the main results, we first define the subsets of $C^{1}[-1,1]$
which contain the solutions of \eqref{1.1}. By a positive solution
to \eqref{1.1} we mean a positive function $u\in C^{1}[-1,1]$ with $\varphi
_{p}(u')\in C^{1}[-1,1]$ satisfying \eqref{1.1}. Recall that $%
Z(u)=\{x\in [-1,1] :u'(x)=0\}$. We note that it is easy
to show that, if $f\in C$ and $u$ is a positive solution of \eqref{1.1},
then $u\in C^{2}[-1,1] $ if $1<p\leq 2$ and $u\in C^{2}([-1,1] -Z)$ if $p>2$.
For the proof we refer to Addou \cite[Lemma 6]{a1}.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth]{fig1.pdf}
\end{center}
\caption{ Typical graph: (a) of $u\in A_{0}^{+}$;
(b) of $u\in A_{1}^{+}$;
(c) of  $u\in A_{00}^{+} $;
(d) of  $u\in A_{01}^{+}$;
(e) of  $u\in A_{10}^{+}$;
(f) of $u\in A_{11}^{+}$.}
\end{figure}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth]{fig2.pdf}
\end{center}
\caption{Typical graph: (a) of $u\in B_{0}^{+}$;
(b)  of $u\in B_{1}^{+}$;
(c)  of $u\in B_{00}^{+} $;
(d)  of $u\in B_{01}^{+}$;
(e)  of $u\in B_{10}^{+}$;
(f)  of $u\in B_{11}^{+}$.}
\end{figure}

Let $A^{+}$ (resp. $B^{+}$) be the subset of $C^{1}[-1,1]$ consisting of
the functions $u$ satisfying

\begin{itemize}
\item[(i)]  $u(x)>0$\ for all $x\in ( -1,1) $,\ $u(-1)=u(1)=0$
and $u'(-1)>0$ (resp. $u'(-1)=0$),

\item[(ii)]  $u$ is symmetrical with respect to $0$ (i.e.\textit{, }$u$ is
even).
\end{itemize}

Note that the derivative of any function $u\in A^{+}$ (resp. $B^{+}$)
satisfies $u'(0)=0$. Therefore $Z^{+}(u)$ contains at least $0$.
Also, $Z^{+}(u)$ may be connected or is an union of many connected
components. Furthermore, each connected component is either a single point
or an interval $[ \tilde{a},\tilde{b}] ,\;\tilde{a}<\tilde{b}$.
(Note that $u'$ is continuous). So, for each integer $k=1,2,\dots$.,
one can consider the subsets of $A^{+}$ (resp. $B^{+}$) which are composed
by functions $u$ such that $Z^{+}(u)$ is an union of $k$ connected
components exactly. These sets can be designed by $A_{a_{1}a_{2}\ldots
a_{k}}^{+}$ (resp. $B_{b_{1}b_{2}\ldots b_{k}}^{+}$) where for all $j\in
\{ 1,2,\ldots ,k\} $, $a_{j}=0$ (resp. $b_{j}=0$) if the $j^{th}$
connected component is a single point and $a_{j}=1$ (resp. $b_{j}=1$) if it
is an interval (not reduced to a single point). For example, $A_{0}^{+}$
(resp. $B_{0}^{+}$) is the subset of $A^{+}$ (resp. $B^{+}$) consisting of
the functions $u$ such that their derivative $u'$ vanishes once and
only once (at $0$ necessarily). An example of a function in $A_{0}^{+}$
(resp. $B_{0}^{+}$) is given by Fig. 1(a) (resp. Fig. 2(a)). Also, $%
A_{1}^{+} $ (resp. $B_{1}^{+}$) is the subset of $A^{+}$ (resp. $B^{+}$)
such that $u\in A_{1}^{+}$ (resp. $u\in B_{1}^{+}$) if and only if $u\in
A^{+}$ (resp. $u\in B^{+}$) and there exists $x_{0}\in ( 0,1) $
such that for all $x\in [0,1]$, $u'(x)=0$ if and only if $0\leq
x\leq x_{0}$ (resp. $0\leq x\leq x_{0}$ or $x=1$). An example of a function
in $A_{1}^{+}$ (resp. $B_{1}^{+}$) is given by Fig. 1(b) (resp. Fig. 2(b)).

\noindent An example of a function in $A_{00}^{+}$ (resp. $B_{00}^{+}$) is given by
Fig. 1(c) (resp. Fig. 2(c)). That is, there exists $x_{0}\in (0,1) $ such that,
for all $0\leq x\leq 1$,
\begin{equation*}
u'(x)=0\text{ if and only if }x\in \{ 0,x_{0}\}
\text{(resp. }x\in \{ 0,x_{0},1\} ).
\end{equation*}
An example of a function in $A_{01}^{+}$ (resp. $B_{01}^{+}$) is given by
Fig. 1(d) (resp. Fig. 2(d)). That is, there exist $0<x_{1}<x_{2}<1$ such
that, $\,$for all $0\leq x\leq 1$,
\begin{equation*}
u'(x)=0\text{ if and only if }x\in \{ 0\} \cup [
x_{1},x_{2}] \text{ (resp. }x\in \{ 0\} \cup [
x_{1},x_{2}] \cup \{ 1\} \text{).}
\end{equation*}
An example of a function in $A_{10}^{+}$ (resp. $B_{10}^{+}$) is given by
Fig. 1(e) (resp. Fig. 2(e)). That is, there exist $0<x_{0}<x_{1}<1$ such
that for all $0\leq x\leq 1$,
\begin{equation*}
u'(x)=0\text{ if and only if }x\in [ 0,x_{0}] \cup
\{ x_{1}\} \text{ (resp. }x\in [ 0,x_{0}] \cup \{
x_{1},1\} \text{)}.
\end{equation*}
An example of a function in $A_{11}^{+}$ (resp. $B_{11}^{+}$) is given by
Fig. 1(f) (resp. Fig. 2(f)). That is, there exist $0<x_{0}<x_{1}<x_{2}<1$
such that, for all $0\leq x\leq 1,$%
\begin{equation*}
u'(x)=0\text{ if and only if }x\in [ 0,x_{0}] \cup
[ x_{1},x_{2}] \text{ (resp. $x\in [ 0,x_{0}] \cup
[ x_{1},x_{2}] \cup \{ 1\}$ )}.
\end{equation*}

Note that if a solution $u\in A_{1}^{+}\cup B_{1}^{+}$, then it is usually
called a dead core solution of \eqref{1.1}. In this paper we extend this
terminology to the case where a solution $u\in A_{a_{1}a_{2}}^{+}\cup
B_{a_{1}a_{2}}^{+}$ for some $k=2$ and $a_{j}=1$ for some $j\in \{
1,2\} $, and call it a \emph{dead core} solution too.

First it is easy to derive an \emph{energy relation} of solutions $u$ of (%
\ref{1.1}); see e.g. \cite[p. 421]{g1} and \cite[Lemma 7]{a1}. Denote by $%
p'=p/( p-1) $ the conjugate exponent of $p$.

\begin{lemma}[Energy relation] \label{Lemma2.1}
Let $p>1$ and assume that $u$ is a
positive solution of \eqref{1.1}, then $( | u'(x)|^{p}+p'\lambda F(u(x))) '=0$
for all $x\in [-1,1]$.
\end{lemma}

\begin{lemma} \label{Lemma2.2}
Suppose $f$ satisfies conditions \eqref{1.3}--\eqref{1.5} and $u$ is a
positive solution of problem \eqref{1.1}.  Then $u\in
A_{0}^{+}\cup A_{1}^{+}\cup A_{00}^{+}\cup A_{01}^{+}$.
\end{lemma}

\begin{proof} Suppose $f$ satisfies conditions
\eqref{1.3}--\eqref{1.5}, $f(0)>0$ and $f$ changes sign exactly twice on $(0,\infty ) $.
Suppose $u$ is a positive solution of \eqref{1.1}, then $u$ is symmetrical
with respect to $0$. It can be easily proved that either
$0<\|u\| _{\infty }\leq a$ or $\beta \leq \|u\| _{\infty }\leq c$ by applying
Lemma \ref{Lemma2.1}; cf. Remark \ref{Remark1}(i). Thus
\begin{equation*}
u\in A_{0}^{+}\cup A_{1}^{+}\cup A_{00}^{+}\cup A_{01}^{+}\cup
A_{10}^{+}\cup A_{11}^{+}\cup B_{0}^{+}\cup B_{1}^{+}\cup B_{00}^{+}\cup
B_{01}^{+}\cup B_{10}^{+}\cup B_{11}^{+}.
\end{equation*}
The proof is easy but tedious, so we omit it. More precisely,

(i) Since $f(0)>0$, if $u$ is a positive solution $u$ of \eqref{1.1}
satisfying $\| u\| _{\infty }=u(0)=\eta \in ( 0,a]
\cup [ \beta ,c] $, then $u'(-1)=( p^{\prime
}\lambda F( \eta ) ) ^{1/p}>0$ by applying Lemma \ref
{Lemma2.1}. Hence $u\notin B_{0}^{+}\cup B_{1}^{+}\cup B_{00}^{+}\cup
B_{01}^{+}\cup B_{10}^{+}\cup B_{11}^{+}$.

(ii) We then show that $u\notin A_{10}^{+}\cup A_{11}^{+}$. Suppose that $%
u\in A_{10}^{+}\cup A_{11}^{+}$. Then either $\| u\| _{\infty }=c$
or $\| u\| _{\infty }=a$, and there exists $x_{1}\in (
0,1) $ such that $u'(x_{1})=0$. If $\| u\|
_{\infty }=c$, then by applying Lemma \ref{Lemma2.1}, $p'\lambda
F(c)=p'\lambda F(x_{1})$, which contradicts the fact that $%
F(c)>F(x_{1})$. So $\| u\| _{\infty }\neq c$. Similarly, $\|
u\| _{\infty }\neq a$. We conclude that $u\notin A_{10}^{+}\cup
A_{11}^{+}$.

By above (i) and (ii), we obtain that $u\in A_{0}^{+}\cup A_{1}^{+}\cup
A_{00}^{+}\cup A_{01}^{+}$.
\end{proof}

To study \eqref{1.1}, we make use of the quadrature method. Suppose $f\in
C[0,\infty )$ satisfies conditions \eqref{1.3}--\eqref{1.5}. For any $E\geq 0$ and $s>0$,
let $G(E,s):=E^{p}-p'\lambda F(s)$. It can be shown that, the
function $G(E,\cdot )$ has at most four zeros in $( 0,\infty ) $.
For any $E\geq 0$, define
\begin{equation*}
X_{1}(E)=\{ s>0:s\in \mathop{\rm dom}G(E,\cdot )\text{ and }G(E,u)>0\text{ for all }%
u\in (0,s)\}
\end{equation*}
and
\begin{equation*}
r_{1}(E)=\begin{cases}
0 & \text{if }X_{1}(E)=\emptyset , \\
\sup ( X_{1}(E))  & \text{otherwise.}
\end{cases}
\end{equation*}
Next for any $E\geq 0$, define
\begin{equation*}
X_{2}(E)=\{ s>r_{1}(E):s\in \mathop{\rm dom}G(E,\cdot )\text{ and }G(E,u)>0\text{
for all }u\in (r_{1}(E),s)\}
\end{equation*}
and
\begin{equation*}
r_{2}(E)=\begin{cases}
\infty  & \text{if }X_{2}(E)=\emptyset , \\
\sup ( X_{2}(E))  & \text{otherwise.}
\end{cases}
\end{equation*}
Note that $X_{2}(E)$ and $r_{2}(E)$ are well defined even if
$r_{1}(E)=\infty $. In fact, in this case, $X_{2}(E)=\emptyset $ and
$r_{2}(E)=\infty $.
Let
\begin{align*}
\tilde{D}_{1}=\big\{& 
E\geq 0: r_{1}(E)\in \mathop{\rm dom}G(E,\cdot ),\, G(E,r_{1}(E))=0, \\
&\ \text{and }\int_{0}^{r_{1}(E)}( E^{p}-p'\lambda F(t) ) ^{-1/p}dt<\infty
\big\} ,
\end{align*}
\begin{align*}
\tilde{D}_{2}=\big\{& E\geq 0:
r_{2}(E)\in \mathop{\rm dom}G(E,\cdot ),\, G(E,r_{2}(E))=0, \\
& \text{and }\int_{0}^{r_{2}(E)}( E^{p}-p'\lambda F(t) ) ^{-1/p}dt<\infty
 \big\} .
\end{align*}
Define the time maps
\begin{gather*}
T_{1}(E)=\int_{0}^{r_{1}(E)}( E^{p}-p'\lambda F( t)
) ^{-1/p}dt,\;E\in \tilde{D}_{1}, \\
T_{2}(E)=\int_{0}^{r_{2}(E)}( E^{p}-p'\lambda F( t)
) ^{-1/p}dt,\;E\in \tilde{D}_{2},
\end{gather*}
whenever $\tilde{D}_{1}\neq \emptyset $ (resp. $\tilde{D}_{2}\neq \emptyset $).

By Lemma \ref{Lemma2.1} and arguments in \cite{g1}, we have the following
theorem. Note that in this paper, by Lemma \ref{Lemma2.2}, we restrict
ourself on positive solutions
$u\in A_{0}^{+}\cup A_{1}^{+}\cup A_{00}^{+}\cup A_{01}^{+}$.

\begin{theorem}[Quadrature method]\label{Thm2.2}
 Consider \eqref{1.1}. Suppose $f\in C[0,\infty )$ satisfies conditions
\eqref{1.3}--\eqref{1.5}. Let $E\geq 0$. Then $T_{1}$ (resp. $T_{2}$) is a continuous
function of $E\in \tilde{D}_{1}$. (resp. $E\in \tilde{D}_{2}$). Moreover,

\begin{itemize}
\item[(i)]  Problem \eqref{1.1} has a solution $u\in A_{0}^{+}$ satisfying $%
u'(-1)=E>0$ if and only if $E\in \tilde{D}_{1}-\{ 0\} $,
$f(r_{1}(E))\geq 0$ and $T_{1}(E)=1,$\ and in this case the solution is
unique.

\item[(ii)]  Problem \eqref{1.1} has a solution $u\in A_{1}^{+}$ satisfying $%
u'(-1)=E>0$ if and only if $E\in \tilde{D}_{1}-\{ 0\} $,
$f(r_{1}(E))=0$ and $T_{1}(E)<1$, and in this case the solution is unique.

\item[(iii)]  Problem \eqref{1.1} has a solution $u\in A_{00}^{+}$
satisfying $u'(-1)=E>0$ if and only if $E\in \tilde{D}_{2}-\{
0\} $, $f(r_{1}(E))\geq 0,\;f(r_{2}(E))\geq 0$, and $T_{2}(E)=1$, and
in this case the solution is unique.

\item[(iv)]  Problem \eqref{1.1} has a solution $u\in A_{01}^{+}$ satisfying
$u'(-1)=E>0$ if and only if $E\in \tilde{D}_{2}-\{ 0\} $,
$f(r_{1}(E))=0$, $f(r_{2}(E))\geq 0$, and $T_{2}(E)<1$, and in this case the
solution is unique.
\end{itemize}
\end{theorem}

\begin{remark} \label{Remark4} \rm
In practice, we first study the variations of the real-valued
function $G(E,\cdot )$, then compute $X_{1}(E)$ and deduce $r_{1}(E)$ 
(resp. compute $X_{2}(E)$ and deduce $r_{2}(E)$). 
Next, we compute $\tilde{D}_{1}$ (resp. $\tilde{D}_{2}$). 
For this, we first compute the set
\[
D_{1}=\{ E>0:r_{1}(E)\in \mathop{\rm dom} G(E,\cdot ), \; G(E,r_{1}(E))=0,\;
f(r_{1}(E))>0 \},
\]
(resp. 
\[
D_{2}=\{ E>0:r_{2}(E)\in \mathop{\rm dom}G(E,\cdot ),\; 
G(E,r_{2}(E))=0,\; f(r_{2}(E))>0\}), 
\]
and then we deduce $\tilde{D}_{1}$ (resp. $\tilde{D}_{2}$) by observing that
$D_{1}\subset \tilde{D}_{1}-\{ 0\} \subset \overline{D}_{1}$
(resp. $D_{2}\subset \tilde{D}_{2}-\{ 0\} \subset \overline{D}_{2}
$) ; we omit the proof. (Note that $\overline{D}_{1}$ is the closure of 
$D_{1}$ (resp. $\overline{D}_{2}$ is the closure of $D_{2}$).) After that, we
define the time map $T_{1}$ on $\tilde{D}_{1}$ and then compute its limits
at the boundary points of $\tilde{D}_{1}$. We next study the variations of 
$T_{1}$ on $\tilde{D}_{1}$. For $T_{2}$, we shall show that its definition
domain $\tilde{D}_{2}$ is restricted to a single point; there is no
variation to study for $T_{2}$. We achieve our study by discussing the
number of solutions to

\begin{itemize}
\item[(i)]  Equation $T_{1}(E)=1$ and $f(r_{1}(E))\geq 0$ for 
$E\in \tilde{D}_{1}-\{ 0\} $ in case of looking for solutions $u$ in $A_{0}^{+}$.

\item[(ii)]  Inequality $T_{1}(E)<1$ and $f(r_{1}(E))=0$ for 
$E\in \tilde{D}_{1}-\{ 0\} $ in case of looking for solutions $u$ in $A_{1}^{+}$.

\item[(iii)]  Equation $T_{2}(E)=1$ and $f(r_{1}(E))\geq 0$, 
$f(r_{2}(E))\geq 0$, for $E\in \tilde{D}_{2}-\{ 0\} $ in case of
looking for solutions $u$ in $A_{00}^{+}$.

\item[(iv)]  Inequality $T_{2}(E)<1$ and $f(r_{1}(E))=0$, $f(r_{2}(E))\geq 0,
$ for $E\in \tilde{D}_{2}-\{ 0\} $ in case of looking for
solutions $u$ in $A_{01}^{+}$.
\end{itemize}
\end{remark}

\section{Main results}

We determine the exact multiplicity of positive solutions of \eqref{1.1} for
 $\lambda >0$ under hypotheses (H1)-(H5) stated below. In particular, we assume
that $f$ satisfies the $``$\emph{convexity}$"$ condition (H4) which implies
for the particular case $p=2$, that $f''$ changes sign
exactly \emph{twice} on $(0,\infty )$, i.e., $f$ is concave-convex-concave
on $(0,\infty )$. Note that if $f$ satisfies (H1)-(H3) then it satisfies
\eqref{1.3}--\eqref{1.5}. Also note that we may allow that $f''$ changes
sign more than twice, i.e., we may allow that $f$ is
concave-convex-concave-convex; see Section 4.

For $f$, recalling that $F(u)=\int_{0}^{u}f(t)dt$, we let
\begin{gather}
\theta _{p}(u):=pF(u)-uf(u), \nonumber\\
\Psi _{p}(u):=u\theta _{p}'(u)-\theta _{p}(u)=puf(u)-u^{2}f^{\prime
}(u)-pF(u), \nonumber\\
\nu _{p}:=\big\{ \int_{0}^{c}( F(c)-F(u)) ^{-1/p}du\big\}
^{p}/p'\in (0,\infty ], \nonumber\\
\alpha _{p}:=\big\{ \int_{0}^{a}( F(a)-F(u)) ^{-1/p}du\big\}
^{p}/p'\in (0,\infty ],  \label{3.3} \\
\lambda _{p}:=\big\{ \int_{0}^{\beta }( F(\beta )-F(u))
^{-1/p}\big\} ^{p}/p'\in (0,\infty ],  \label{3.4} \\
\mu _{p}:=\inf_{\beta \leq \xi \leq c}\big\{ \int\nolimits_{0}^{\xi }(
F(\xi )-F(u)) ^{-1/p}du\big\} ^{p}/p', \nonumber
\end{gather}
where $0<a<\beta <c$ are defined below. We shall show that $0<\mu
_{p}<\infty $ for $p>1$. For all $\lambda >0$, we denote $S_{\lambda }$ the
positive solution set of \eqref{1.1}.

For fixed $p>1$, suppose $f\in C^{2}[0,\infty )$ and there exist $0<a<b<c$
such that the following conditions are satisfied:
\begin{itemize}
\item[(H1)] $f(0)>0$

\item[(H2)]
$f(u)>0$ for $0<u<a$,
$f(u)<0$ for $a<u<b$,
$f(u)>0$ for $b<u<c$,
$f(u)<0$ for $u>c$

\item[(H3)] $\int_{a}^{c}f(u)du>0$, and there exists
$\beta ^{*}\in ( 0,\beta] $ such that
$\theta _{p}(\beta ^{*})=pF(\beta ^{*})-\beta^{*}f(\beta ^{*})<0$,
where $\beta \in (b,c)$ is defined by $\int_{a}^{\beta}f(u)du=0$,

\item[(H4)] There exist $0<r_{p}<s_{p}<c$ such that
\begin{gather*}
(p-2)f'(u)-uf''(u)>0\quad \text{for }0<u<r_{p}, \\
(p-2)f'(u)-uf''(u)<0\quad \text{for }r_{p}<u<s_{p}, \\
(p-2)f'(u)-uf''(u)>0\quad \text{for }s_{p}<u<\infty ,
\end{gather*}

\item[(H5)] There exists a unique $\sigma _{p}\in ( s_{p},c) $
satisfying $( p-1) f(\sigma _{p})-\sigma _{p}f'(\sigma_{p})=0$ and such
that $\Psi _{p}(\sigma _{p})\geq \Psi _{p}(r_{p})$.
\end{itemize}

\begin{remark}[Cf. Remark \ref{Remark1}]\label{Remark7} \rm
 If $f$ $\in C[0,\infty )$ satisfies (H1)-(H3), then by applying
 Lemma \ref{Lemma2.1}, it can be shown that

\begin{itemize}
\item[(i)]  Every positive solution $u$ of \eqref{1.1} satisfies $0<\|
u\| _{\infty }\leq a$ or $\beta \leq \| u\| _{\infty }\leq c
$.
\item[(ii)]  Any two distinct positive solutions of \eqref{1.1} are strictly
ordered. That is, let $u$ and $\hat{u}$ be any two distinct positive
solutions of \eqref{1.1} with $0<\| u\| _{\infty }<\| \hat{u}\| _{\infty }$,
then $u<\hat{u}$.
\end{itemize}
\end{remark}

\subsection*{Case $1<p\leq 2$}

The next theorem gives a complete description of the set $S_{\lambda }$
for $1<p\leq 2$.

\begin{theorem}[$S_{\lambda }$ for $1<p\leq 2$, see Fig. 3] \label{Thm3.5}
Assume that $1<p\leq 2$ and $f\in C^{2}[0,\infty )$\ satisfies (H1)-(H5).
Then $0<\mu_{p}<\infty $. Moreover:
\begin{itemize}
\item[(i)]  For $0<\lambda <\mu _{p}$, there exists $u_{\lambda }\in
A_{0}^{+}$ such that $S_{\lambda }=\{ u_{\lambda }\} $. Also
$0<\| u_{\lambda }\| _{\infty }<a$.

\item[(ii)]  For $\lambda =\mu _{p}$, there exist $u_{\lambda }$,
$v_{\lambda }\in A_{0}^{+}$ such that $u_{\lambda }<v_{\lambda }$ and
$S_{\lambda }=\{ u_{\lambda },v_{\lambda }\} $. Moreover,
$0<\| u_{\lambda }\| _{\infty }<a<\beta <\| v_{\lambda}\| _{\infty }<c$.

\item[(iii)]  For $\lambda >\mu _{p}$, there exist $u_{\lambda }$,
$v_{\lambda }$ and $w_{\lambda }\in A_{0}^{+}$ such that
$u_{\lambda}<v_{\lambda }<w_{\lambda }$ and
$S_{\lambda }=\{ u_{\lambda},v_{\lambda },w_{\lambda }\} $. Moreover,
$0<\| u_{\lambda}\| _{\infty }<a<\beta <\| v_{\lambda }\| _{\infty}
<\| w_{\lambda }\| _{\infty }<c$.
\end{itemize}
\end{theorem}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig3.pdf}
\end{center}
\caption{Bifurcation diagram of problem \eqref{1.1} with $f(0)>0$ and $1<p\leq 2$.}
\end{figure}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth]{fig4.pdf}
\end{center}
\caption{(a) Graph of $f_{1}(u)=-(u+1/4)(u-1)(u-2)(u-3)-0.13$. $a\approx 0.9497$,
$b\approx 2.0565$, $c\approx 2.9792$, $\beta \approx 2.9378$, $r_{2}\approx
0.7425$, $s_{2}\approx 2.1325$, $\sigma _{2}\approx 2.5787$.
\protect \linebreak (b) Graph of $\theta _{2}(u)$. $1.4751\approx \theta _{2}(
\beta ) >\theta _{2}( \sigma _{2}) \approx -0.8789$.
\protect\linebreak
(c) Graph of $\Psi _{2}(u)$. $0.8792\approx \Psi _{2}(\sigma _{2})
>\Psi _{2}(r_{2})\approx 0.5127$.}
\end{figure}

Next, we give two interesting examples of quartic polynomials of Theorem
\ref{Thm3.5} with $p=2$, of which one satisfies $\theta _{2}(\beta )>0$ and the
other satisfies $\theta _{2}(\beta )<0$.

\subsection*{Two examples of Theorem \ref{Thm3.5}}

(i) (See Fig. 4, $\theta _{2}(\beta )>0$) Let $p=2$. The function $%
f=f_{1}(u)=-(u+1/4)(u-1)(u-2)(u-3)-0.13$ satisfies all conditions (H1)-(H5)
in Theorem \ref{Thm3.5} with $a\approx 0.9497$, $b\approx 2.0565$, $c\approx
2.9792$, $\int_{a}^{c}f_{1}(s)ds\approx 0.004695>0$, $r_{2}\approx 0.7425$, $%
s_{2}\approx 2.1325$, $2.5787\approx \sigma _{2}<\beta \approx 2.9387$, $%
1.4751\approx \theta _{2}(\beta )>\theta _{2}(\sigma )\approx -0.8789$. Note
that, in (H4)-(H5), $0.8792\approx \Psi _{2}(\sigma _{2})>\Psi
_{2}(r_{2})\approx 0.5127$.

\noindent (ii) ($\theta _{2}(\beta )<0$) Let $p=2$. Let
$f=f_{2}(u)=-(u-d)(u-a)(u-b)(u-c)$ with
\begin{equation*}
d=-\frac{1}{6}<0<a=1<b=2<c.
\end{equation*}
Thus $f_{2}$ satisfies (H1), (H2) and (H4) with
\begin{equation*}
f_{2}''(u)=-12u^{2}+(17+6c)u-3-\frac{17}{3}c
\end{equation*}
and $f_{2}''(0)=-3-\frac{17}{3}c<0$. Let $r_{2}<s_{2}$ be two
positive zeros of $f_{2}''(u)$ on $( 0,c) $. So
\begin{gather*}
r_{2}=\frac{1}{24}( 6c+17-\sqrt{36c^{2}-68c+145}) , \\
s_{2}=\frac{1}{24}( 6c+17+\sqrt{36c^{2}-68c+145}) .
\end{gather*}
There exists $c_{1}\approx 2.8380$, the biggest positive zero of
$18c^{3}-49c^{2}-26c+25$, such that
\begin{equation*}
c>c_{1}\Leftrightarrow \int_{a}^{c}f_{2}(u)du=\frac{1}{360}%
(c-1)^{2}(18c^{3}-49c^{2}-26c+25)>0\text{ on }( 2,\infty ) .
\end{equation*}
So for $c>c_{1}\approx 2.8380$, there exists a unique $\beta =\beta (c)\in
(b,c)=( 2,c) $ satisfying
\begin{equation*}
\int_{a}^{\beta }f_{2}(u)du=0.
\end{equation*}
or,
\begin{equation*}
\frac{1}{360}(\beta -1)^{2}[ -72\beta^{3}+(90c+111)\beta ^{2}+(-160c+114)\beta -140c+57] =0.
\end{equation*}

Note that $\beta (c)$ can be expressed explicitly by Cartan's formulas; see
e.g. \cite{j1}. We compute that
\begin{equation*}
\theta _{2}(u)=\frac{3}{5}u^{5}-\frac{6c+17}{12}u^{4}+\frac{17c+9}{18}u^{3}+%
\frac{c}{3}u
\end{equation*}
and
\begin{equation*}
\theta _{2}(\beta (c))<0\text{ for }c>c_{2}\approx 2.9056,
\end{equation*}
where $c_{2}$ is the unique zero of $\theta _{2}(\beta (c))$ on $(
c_{1},\infty ) $. Thus $f_{2}$ satisfies (H3) for $c>c_{2}\approx
2.9056$.

Finally, we check (H5) for $c>c_{2}$. Let $\sigma _{2}=\sigma _{2}(c)$ be
the unique zero of
\begin{equation*}
\theta _{2}'(u)=3u^{4}-\frac{6c+17}{3}u^{3}+\frac{17c+9}{6}u^{2}+%
\frac{c}{3}
\end{equation*}
in $(s_{2},c)$. Note that $\sigma _{2}(c)$ can be expressed explicitly; see
e.g. \cite{j1}. We compute that
\begin{gather*}
\Psi _{2}(u)=\frac{1}{180}u^{3}[ 432u^{2}-(270c+765)u+340c+180]\,,\\
\Psi _{2}(\sigma _{2})>\Psi _{2}(s_{2})\text{ for }c>c_{2}\approx 2.9056.
\end{gather*}
We summarize above results and conclude that $f_{2}$ satisfies (H1)-(H5) for
$c>c_{2}\approx 2.9056$.

\subsection{The case $p>2$}

By Lemma \ref{Lemma2.2}, $S_{\lambda }\subset A_{0}^{+}\cup A_{1}^{+}\cup
A_{00}^{+}\cup A_{01}^{+}$. Hence
\begin{equation*}
S_{\lambda }=( S_{\lambda }\cap A_{0}^{+}) \cup ( S_{\lambda
}\cap A_{1}^{+}) \cup ( S_{\lambda }\cap A_{00}^{+}) \cup
( S_{\lambda }\cap A_{01}^{+}) ;
\end{equation*}
Theorem \ref{Thm3.6} (resp. \ref{Thm3.7}, \ref{Thm3.8}, \ref{Thm3.9}) gives
complete description of the set $S_{\lambda }\cap A_{0}^{+}$ (resp. $%
S_{\lambda }\cap A_{1}^{+},\;S_{\lambda }\cap A_{00}^{+},\;S_{\lambda }\cap
A_{01}^{+}$); see Fig. 5.


\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth] {fig5.pdf}
\end{center}
\caption{Bifurcation diagrams for Eq. \eqref{1.1} with $p>2$ and
$0<\alpha _{p}<\mu _{p}$. (a) $\mu _{p}<\nu _{p}<\lambda_{p}$;
(b) $\mu _{p}<\nu _{p}=\lambda _{p};$ (c) $\mu _{p}<\lambda _{p}<\nu_{p}$.}
\end{figure}

\begin{remark}\label{Remark8}\rm
Fig. 5 describes the solution set of \eqref{1.1} when $p>2$
and $0<\alpha _{p}<\mu _{p}$. There are two connected branches. The lower
branch bifurcates at the origin and represents solutions in $A_{0}^{+}$
until $\lambda =\alpha _{p};$ for all $\lambda >\alpha _{p}$, the branch is
horizontal and represents solutions in $A_{1}^{+}$. The upper branch is
formed by two parts: The upper horizontal curve represents solutions in $%
A_{1}^{+}$ with norm equal to $c$, and the $\subset $-shaped curve
represents solutions in $A_{0}^{+}\,$until $\lambda =\lambda _{p}$ where
there is a solution in $A_{00}^{+}$ with norm equal to $\beta $ and on its
right the lower horizontal curve which represents solutions in $A_{01}^{+}$
with norm equal to $\beta $.
\end{remark}

We shall show that, for $p>2$, $0<\mu _{p}<\lambda _{p}<\infty $ and
$0<\mu_{p}<\nu _{p}<\infty $; see Lemmas \ref{Lemma4.9}-\ref{Lemma4.10}.
We also show that, for $p>2$, $0<\alpha _{p}<\lambda _{p}<\infty $;
 see Lemma \ref{Lemma4.11}.
For $p>1$, let
\begin{gather*}
B(0,a):= \{ u\in C^{1}[-1,1]:\| u\| _{\infty }<a\} , \\
B(0,\beta ):= \{ u\in C^{1}[-1,1]:\| u\|_{\infty }<\beta \} , \\
B(0,c):= \{ u\in C^{1}[-1,1]:\| u\| _{\infty}<c\} .
\end{gather*}

\begin{theorem}[$S_{\lambda }\cap A_{0}^{+}$ for $p>2$, see Fig. 5] \label{Thm3.6}
Assume that $p>2$ and $f\in C^{2}[0,\infty )$ satisfies (H1)-(H5). Then $0<\mu
_{p}<\lambda _{p}<\infty $, $0<\mu _{p}<\nu _{p}<\infty $, and $0<\alpha
_{p}<\lambda _{p}<\infty $. $S_{\lambda }\cap A_{0}^{+}\cap ( B(0,\beta
)-\overline{B(0,a)}) =\emptyset $. Also,
\begin{itemize}
\item[(i)]  For $0<\lambda \leq \alpha _{p}$, there exists $u_{\lambda }\in
A_{0}^{+}\cap \overline{B(0,a)}$ such that $S_{\lambda }\cap A_{0}^{+}\cap
\overline{B(0,a)}=\{ u_{\lambda }\} $. Moreover, $0<\|
u_{\lambda }\| _{\infty }\leq a$, and $\| u_{\lambda }\|
_{\infty }=a$ if and only if $\lambda =\alpha _{p}$.

\item[(ii)]  For $\lambda >\alpha _{p}$, $S_{\lambda }\cap A_{0}^{+}\cap
\overline{B(0,a)}=\emptyset $.
\end{itemize}
Moreover,
\textbf{(a) If }$\mu _{p}<\nu _{p}<\lambda _{p}$, \textbf{then:}
\begin{itemize}
\item[(iii)]  For $0<\lambda <\mu _{p}$, $S_{\lambda }\cap A_{0}^{+}\cap
( \overline{B(0,c)}-B(0,\beta )) =\emptyset $.

\item[(iv)]  For $\lambda =\mu _{p}$, there exists $u_{\lambda }\in
A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) $ such that $%
\newline
S_{\lambda }\cap A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta ))
=\{ u_{\lambda }\} $ and $\beta <\| u_{\lambda }\|
_{\infty }<c$.

\item[(v)]  For $\mu _{p}<\lambda \leq \nu _{p}$, there exist $u_{\lambda
},\;v_{\lambda }\in A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta
)) $ such that $u_{\lambda }<v_{\lambda }$ and $S_{\lambda }\cap
A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) =\{
u_{\lambda },v_{\lambda }\} $. Moreover, $\beta <\| u_{\lambda
}\| _{\infty }<\| v_{\lambda }\| _{\infty }\leq c$, and $%
\| v_{\lambda }\| _{\infty }=c$ if and only if $\lambda =\nu _{p}$.

\item[(vi)]  For $\nu _{p}<\lambda <\lambda _{p}$, there exists
$u_{\lambda}\in A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) $ such that
$S_{\lambda }\cap A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta ))
=\{ u_{\lambda }\} $ and $\beta <\| u_{\lambda }\|_{\infty }<c$.

\item[(vii)]  For $\lambda \geq \lambda _{p}$, $S_{\lambda }\cap
A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) =\emptyset $.
\end{itemize}

\noindent\textbf{(b) If $\mu _{p}<\nu _{p}=\lambda _{p}$, then:}

\begin{itemize}
\item[(viii)]  For $0<\lambda <\mu _{p}$, $S_{\lambda }\cap A_{0}^{+}\cap
( \overline{B(0,c)}-B(0,\beta )) =\emptyset $.

\item[(ix)]  For $\lambda =\mu _{p}$, there exists $u_{\lambda }\in
A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) $ such that
\newline
$S_{\lambda }\cap A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta ))
=\{ u_{\lambda }\} $ and $\beta <\| u_{\lambda }\|_{\infty }<c$.

\item[(x)]  For $\mu _{p}<\lambda <\nu _{p}=\lambda _{p}$, there exist $%
u_{\lambda },\;v_{\lambda }\in A_{0}^{+}\cap ( \overline{B(0,c)}%
-B(0,\beta )) $ such that $u_{\lambda }<v_{\lambda }$ and $S_{\lambda
}\cap A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) =\{
u_{\lambda },v_{\lambda }\} $. Moreover, $\beta <\| u_{\lambda
}\| _{\infty }<\| v_{\lambda }\| _{\infty }<c$.

\item[(xi)]  For $\lambda =\nu _{p}=\lambda _{p}$, there exists $u_{\lambda
}\in A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) $ such that $%
S_{\lambda }\cap A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta ))
=\{ u_{\lambda }\} $. Moreover, $\| u_{\lambda }\|_{\infty }=c$.

\item[(xii)]  For $\lambda >\lambda _{p}=\nu _{p}$, $S_{\lambda }\cap
A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) =\emptyset $.
\end{itemize}

\noindent\textbf{(c) If }$\mu _{p}<\lambda _{p}<\nu _{p}$, \textbf{then:}

\begin{itemize}
\item[(xiii)]  For $0<\lambda <\mu _{p}$, $S_{\lambda }\cap A_{0}^{+}\cap
( \overline{B(0,c)}-B(0,\beta )) =\emptyset $.

\item[(xiv)]  For $\lambda =\mu _{p}$, there exists $u_{\lambda }\in
A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) $ such that
\newline
$S_{\lambda }\cap A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta ))
=\{ u_{\lambda }\} $. Moreover, $\beta <\| u_{\lambda}\| _{\infty }<c$.

\item[(xv)]  For $\mu _{p}<\lambda <\lambda _{p}$, there exist $u_{\lambda
},\;v_{\lambda }\in A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta
)) $ such that $u_{\lambda }<v_{\lambda }$ and $S_{\lambda }\cap
A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) =\{
u_{\lambda },v_{\lambda }\} $. Moreover, $\beta <\| u_{\lambda
}\| _{\infty }<\| v_{\lambda }\| _{\infty }<c$.

\item[(xvi)]  For $\lambda _{p}\leq \lambda \leq \nu _{p}$, there exists $%
u_{\lambda }\in A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta )) $
such that $S_{\lambda }\cap A_{0}^{+}\cap ( \overline{B(0,c)}-B(0,\beta
)) =\{ u_{\lambda }\} $ and $\beta <\| u_{\lambda}\| _{\infty }\leq c$,
and $\| u_{\lambda }\| _{\infty }=c$
if and only if $\lambda =\nu _{p}$.

\item[(xvii)]  For $\lambda >\nu _{p}$, $S_{\lambda }\cap A_{0}^{+}\cap
( \overline{B(0,c)}-B(0,\beta )) =\emptyset $.
\end{itemize}
\end{theorem}

\begin{theorem}[$S_{\lambda }\cap A_{1}^{+}$ for $p>2$, see Fig. 5]\label{Thm3.7}
Assume that $p>2$ and $f\in C^{2}[0,\infty )$\ satisfies (H1)-(H5). Then each
solution $u_{\lambda }$ of \eqref{1.1} in $A_{1}^{+}$ satisfies $\|
u_{\lambda }\| _{\infty }=a$ or $\| u_{\lambda }\| _{\infty
}=c$. Moreover:

\begin{itemize}
\item[(i)]  For $0<\lambda \leq \alpha _{p}$, $S_{\lambda }\cap
A_{1}^{+}\cap \partial \overline{B(0,a)}=\emptyset $.

\item[(ii)]  For $\lambda >\alpha _{p}$, there exists $u_{\lambda }\in
A_{1}^{+}\cap \partial \overline{B(0,a)}$ such that $S_{\lambda }\cap
A_{1}^{+}\cap \partial \overline{B(0,a)}=\{ u_{\lambda }\} $ and $%
\| u_{\lambda }\| _{\infty }=a$.

\item[(iii)]  For $0<\lambda \leq \nu _{p}$, $S_{\lambda }\cap A_{1}^{+}\cap
\partial \overline{B(0,c)}=\emptyset $.

\item[(iv)]  For $\lambda >\nu _{p}$, there exists $u_{\lambda }\in
A_{1}^{+}\cap \partial \overline{B(0,c)}$ such that $S_{\lambda }\cap
A_{1}^{+}\cap \partial \overline{B(0,c)}=\{ u_{\lambda }\} $ and $%
\| u_{\lambda }\| _{\infty }=c$.
\end{itemize}
\end{theorem}

\begin{theorem}[$S_{\lambda }\cap A_{00}^{+}$ for $p>2$, see Fig. 5]\label{Thm3.8}
 Assume that $p>2$ and $f\in C^{2}[0,\infty )$\ satisfies (H1)-(H5). Then each
solution $u_{\lambda }$ of \eqref{1.1} in $A_{00}^{+}$ satisfies $\|
u_{\lambda }\| _{\infty }=\beta $. Moreover,

\begin{itemize}
\item[(i)]  For $\lambda \neq \lambda _{p}\;$and\ $\lambda >0,\;S_{\lambda
}\cap A_{00}^{+}=\emptyset $.

\item[(ii)]  For $\lambda =\lambda _{p}$, there exists $u_{\lambda }\in
A_{00}^{+}$ such that $S_{\lambda }\cap A_{00}^{+}=\{ u_{\lambda
}\} $ and $\| u_{\lambda }\| _{\infty }=\beta $.
\end{itemize}
\end{theorem}

\begin{theorem}[$S_{\lambda }\cap A_{01}^{+}$ for $p>2$, see Fig. 5]\label{Thm3.9}
 Assume that $p>2$ and $f\in C^{2}[0,\infty )$\ satisfies (H1)-(H5). Then each
solution $u_{\lambda }$ of \eqref{1.1} in $A_{01}^{+}$ satisfies
$\|u_{\lambda }\| _{\infty }=\beta $. Moreover,

\begin{itemize}
\item[(i)]  For $0<\lambda \leq \lambda _{p},\;S_{\lambda }\cap
A_{01}^{+}=\emptyset $.

\item[(ii)]  For $\lambda >\lambda _{p}$, there exists $u_{\lambda }\in
A_{01}^{+}$ such that $S_{\lambda }\cap A_{01}^{+}=\{ u_{\lambda
}\} $ and $\| u_{\lambda }\| _{\infty }=\beta $.
\end{itemize}
\end{theorem}

\section{A weakened condition and two examples}

We point out that the \emph{convexity} condition of $\theta
_{p}(u)=pF(u)-uf(u)$ on $(0,c)$ in (H4) in Theorems \ref{Thm3.5}-\ref{Thm3.9}
can actually be weakened; cf. Remark 9 in Addou and Wang \cite{a3}, which
holds true in the case $f(0)=0$ as well as in the positone case $f(0)>0$.
More precisely, condition (H4) can be weakened as\smallskip
\begin{itemize}
\item[(H4$'$)] There exist $0\leq r_{p}<s_{p}<c$ such that
\begin{gather*}
(p-2)f'(u)-uf''(u)>0\text{ for\ }0<u<r_{p},
\text{(It is not necessary if }r_{p}=0\text{.)} \\
(p-2)f'(u)-uf''(u)<0\text{ for\ }r_{p}<u<s_{p}, \\
(p-2)f'(u)-uf''(u)>0\text{ for\ }s_{p}<u<c,
\;\text{(It can be weakened below.)\smallskip }
\end{gather*}
\end{itemize}
We note that in (H4$'$) if $r_{p}=0$ then condition (H5) is
automatically satisfied since it can be easily shown that, for $p>1$, $\Psi
_{p}(\sigma _{p})>0=\Psi _{p}(r_{p})$.

We also note that in (H4$'$) the condition
\begin{equation*}
(p-2)f'(u)-uf''(u)>0\quad \text{for }s_{p}<u<c
\end{equation*}
can actually be weakened as
\begin{equation*}
\theta _{p}'(u)=( p-1) f(u)-uf'(u)
\begin{cases}
\leq 0 & \text{for }s_{p}<u<\sigma _{p}, \\
\geq 0 & \text{for }d\leq u\leq c,
\end{cases}
\end{equation*}
\begin{equation*}
\theta _{p}''(u)=( p-2) f'(u)-uf^{\prime
\prime }(u)\geq 0\text{\ for }\sigma _{p}\leq u<d,
\end{equation*}
where $d\in (\sigma _{p},c]$ is defined by
\begin{equation}
d:=\begin{cases}
c & \text{if }\theta _{p}(c)\leq \theta _{p}(t_{p}), \\
\inf \{ u\in (\sigma _{p},c]:\theta _{p}(\xi )>\theta _{p}(t_{p})
\text{ for all }\xi \in (u,c]\}  & \text{otherwise,}
\end{cases}  \label{3.9}
\end{equation}
where $t_{p}$ is the unique zero of $\theta _{p}'(u)$ on $(r_{p},s_{p}) $.

Thus we summarize that (H4$'$) can be weakened as follows:
\begin{itemize}
\item[(H4$''$)] There exist $0\leq r_{p}<s_{p}<\sigma _{p}<d<c$ such
that
\begin{gather*}
(p-2)f'(u)-uf''(u)>0\text{ for\ }0<u<r_{p},\text{
(It is not necessary if }r_{p}=0\text{.)}\smallskip  \\
(p-2)f'(u)-uf''(u)<0\text{ for }
r_{p}<u<s_{p}, \\
\theta _{p}'(u)=( p-1) f(u)-uf'(u)
\begin{cases}
\leq 0 & \text{for }s_{p}<u<\sigma _{p}, \\
\geq 0 & \text{for }d\leq u\leq c,
\end{cases} \\
\theta _{p}''(u)=( p-2) f'(u)-uf'' (u)\geq 0\text{\ for }\sigma _{p}\leq u<d,
\end{gather*}
where $d$ is defined in (\ref{3.9}). 
\end{itemize}
For example, Theorem \ref{Thm3.5} can
be generalized as

\begin{theorem}[$S_{\lambda }$ for $1<p\leq 2$, see Fig. 3] \label{Theorem4.1}
Assume that $1<p\leq 2$ and $f\in C^{2}[0,\infty )$\ satisfies (H1)-(H3),
((H4$'$) or (H4$''$)) and (H5). Then the results in Theorem \ref{Thm3.5} hold.
\end{theorem}

Therefore, in the case that $f(0)>0$, $p=2$ and $r_{2}=0$,
Theorem \ref{Theorem4.1} generalizes \cite[Theorem 1]{w3}.
We give two examples of classes of nonlinearities of Theorem \ref{Theorem4.1}.

\begin{proposition} \label{Theorem4.2}
Let $p=2$. $f=f_{3}(u)=-(u-a)(u-b)(u-c)$ with $a=1$, $b=3$,
and
\begin{equation*}
c>2b-a=5\text{ }( \Leftrightarrow \int_{a}^{c}f_{3}(u)du>0) .
\end{equation*}
Then $f_{3}$ satisfies all conditions (H1)-(H3), (H4$'$) and (H5)
in Theorem \ref{Theorem4.1}.
\end{proposition}

\begin{proof} It is easy to see that $f_{3}(u)=-(u-1)(u-3)(u-c)$
satisfies (H1), (H2), (H5) and (H4$'$) with $r_{2}=0$ for $c>5$.
Finally, we check (H3) for $c>5$. We compute that
\begin{gather*}
\theta _{2}(u)=\frac{1}{2}u^{4}-\frac{1}{3}(c+4)u^{3}+3cu,\\
\beta =\beta (c)=\frac{1}{3}( 2c+5-2\sqrt{c^{2}-7c+10}) \in ( 3,c) ,
\end{gather*}
and $\theta _{2}(\beta )<0$ for $c>\tilde{c}\approx 5.1193$, where $\tilde{c}
$ is the unique zero of $\theta _{2}(\beta )$ on $( 5,\infty ) $.
Although for $5<c\leq \tilde{c}\approx 5.1193$, $\theta _{2}(\beta )\geq 0$,
and thus Theorem \ref{thm1.1} does not apply. We compute that
\begin{equation*}
\theta _{2}'(u)=2u^{4}-(c+4)u^{2}+3c
\end{equation*}
and find that
\begin{align*}
 \sigma _{2}=\sigma _{2}(c)
 &=\frac{1}{6}\big\{ c+4 +\frac{(c+4)^{2}}{( c^{3}+12c^{2}+114c+64
 +18\sqrt{-c^{4}-12c^{3}+33c-64c} ) ^{1/3}}\\
&\quad +\big(c^{3}+12c^{2}-114c+64+18\sqrt{-c^{4}-12c^{3}+33c-64c} \big) ^{1/3} \big\}.
\end{align*}
satisfies $0<\sigma _{2}<\beta $ and $\theta _{2}(\sigma _{2})<0$ for $%
5<c\leq \tilde{c};$ we omit the detailed numerical simulations here. So $%
f_{3}$ satisfies (H3) for $5<c\leq \tilde{c}$.

We conclude that $f_{3}(u)=-(u-1)(u-3)(u-c)$\ satisfies all conditions
(H1)-(H3), (H4$'$) and (H5) in Theorem \ref{Theorem4.1}
for $c>5$.
\end{proof}


\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth]{fig6.pdf}
\end{center}
\caption{(a) Graph of $f_{4}(u)=-\sin u+\varepsilon $ on $( 0,c) $
for $\varepsilon =0.2$, $a\approx 0.2014$, $b\approx 2.9402$,
$c\approx 6.4845$, $\beta \approx 4.7770$, $r_{2}=0$, 
$s_{2}=\pi \approx 3.1416$,  $\sigma _{2}\approx 4.4473$. 
(b) Graph of $\theta _{2}(u)$ on $( 0,c) $. $t_{2}\approx 0.8650$,
$d\approx 6.1084$.}
\end{figure}

\begin{proposition}[See Fig. 6 for $\varepsilon =0.2$]\label{Theorem4.3}
Let $p=2$. For $0<\varepsilon <1$, let
\begin{equation*}
0<a=\sin ^{-1}\varepsilon <b=\pi -\sin ^{-1}\varepsilon <c=2\pi +\sin
^{-1}\varepsilon
\end{equation*}
and $f=f_{4}(u)$ satisfy
\begin{equation*}
f_{4}(u)=\begin{cases}
-\sin u+\varepsilon  & \text{for }0<u<c, \\
<0 & \text{for }u>c.
\end{cases}
\end{equation*}
Then $f_{4}$\ satisfies all conditions (H1)-(H3), (H4$''$)
and (H5) in Theorem \ref{Theorem4.1} for $\varepsilon >0$ small
enough.
\end{proposition}

\begin{proof} For $f=f_{4}(u)$, we find that
\begin{gather*}
\theta _{2}(u)=u\sin u+2\cos u+\varepsilon u-2,\\
\theta _{2}'(u)=u\cos u-\sin u+\varepsilon .
\end{gather*}
Let $\sigma _{2}=\sigma _{2}(\varepsilon )$ be the unique zero of $\theta
_{2}'(u)$ on $( \pi ,2\pi ) \subset ( b,c) $
and $\beta =\beta (\varepsilon )$ be the unique zero of
\begin{equation*}
\int_{a}^{u}f_{4}(s)ds=\varepsilon u+\cos u-\varepsilon \sin
^{-1}\varepsilon -\sqrt{1-\varepsilon ^{2}}
\end{equation*}
on $( b,c) $.
It can be checked easily that
\begin{itemize}
\item[(i)]  $f_{4}$ satisfies (H1) and (H2) for $0<\varepsilon <1$.

\item[(ii)]  $f_{4}$ satisfies the condition $\int_{a}^{c}f_{4}(u)du>0$ in
(H3) for $0<\varepsilon <1$. Also, for $\varepsilon >0$ small enough, by
continuity, $0<\sigma _{2}(\varepsilon )<\beta (\varepsilon )$ since $\sigma
_{2}(0)\approx 4.4934<\beta (0)=2\pi $, and
\begin{equation}
\theta _{2}(\sigma _{2}(\varepsilon ))<0  \label{4.4}
\end{equation}
since $\theta _{2}(\sigma _{2}(0))\approx -6.8206<0$. So $f_{4}$ satisfies
(H3) for $\varepsilon >0$ small enough.

\item[(iii)]  We check that $f_{4}$ satisfies (H4$''$). First
\begin{equation}
\theta _{2}'(0)=f_{4}(0)=\varepsilon >0,  \label{4.5}
\end{equation}
\begin{equation}
\theta _{2}''(u)=-uf_{4}''(u)=-u\sin u
\begin{cases}
<0 & \text{for }0=r_{2}<u<s_{2}=\pi , \\
>0 & \text{for }s_{2}=\pi <u<2\pi , \\
<0 & \text{for }2\pi <u<c=2\pi +\sin ^{-1}\varepsilon .
\end{cases}   \label{4.6}
\end{equation}
Also, for $\varepsilon >0$ small enough, let $t_{2}=t_{2}(\varepsilon )$ be
the unique zero of $\theta _{2}'(u)=u\cos u-\sin u+\varepsilon $ on
$( 0,\pi ) $. It can be proved that $\lim_{\varepsilon
\to 0^{+}}t_{2}(\varepsilon )=0$. More precisely, we compute that
\begin{equation*}
t_{2}(\varepsilon )\sim (2\varepsilon )^{1/3}\text{ as }\varepsilon
\to 0^{+}
\end{equation*}
and hence
\begin{equation*}
\theta _{2}(t_{2}(\varepsilon ))\sim 2^{1/3}\varepsilon ^{4/3}\text{ as }%
\varepsilon \to 0^{+}.
\end{equation*}
Thus
\begin{equation}
\theta _{2}(2\pi )=2\varepsilon \pi >\theta _{2}(t_{2}(\varepsilon ))\text{
for }\varepsilon >0\text{ small enough.}  \label{4.7}
\end{equation}
We also find that
\begin{gather}
\theta _{2}'(2\pi )=2\pi +\varepsilon >0,  \label{4.8}\\
\theta _{2}'(c)=c\cos c=( 2\pi +\sin ^{-1}\varepsilon )
\sqrt{1-\varepsilon ^{2}}>0.  \label{4.9}
\end{gather}
So by (\ref{4.4})-(\ref{4.9}), it can be proved that there exists $d\in
( \sigma _{2},2\pi ] $ such that $f_{4}$ satisfies
\begin{gather*}
\theta _{2}''(u)=-uf_{4}''(u)=-u\sin u<0
\quad \text{for } 0=r_{2}<u<s_{2}=\pi , \\
\theta _{2}'(u)=f_{4}(u)-uf_{4}'(u)=u\cos u-\sin u+\varepsilon
\begin{cases}
<0 & \text{for }s_{2}<u<\sigma _{2}, \\
>0 & \text{for }d\leq u\leq c=2\pi +\sin ^{-1}\varepsilon ,
\end{cases} \\
\theta _{2}''(u)=-uf_{4}''(u)=-u\sin u>0\quad
\text{for }\sigma _{2}\leq u<d;
\end{gather*}
see Fig. 6(b). We omit the detailed proofs here. So $f_{4}$ satisfies (H4$''$).

\item[(iv)]  $f_{4}$ satisfies (H5) automatically for $0<\varepsilon <1$
since $r_{2}=0$.
\end{itemize}

We conclude that $f_{4}$ satisfies all conditions (H1)-(H3),
(H4$''$) and (H5) in Theorem \ref{Theorem4.1} for $\varepsilon>0 $ small enough.
\end{proof}

\section{Proofs of main results}

First, we have the next lemma which holds for nonlinearities $f\in
C[0,\infty )$ satisfying (H1), (H2) and the condition $\int_{a}^{c}f(s)ds>0$
in (H3). We omit the proof.

\begin{lemma} \label{Lemma4.7}
Assume that $f\in C[0,\infty )$ satisfies (H1), (H2) and the
condition $\int_{a}^{c}f(s)ds>0$ in (H3). Consider the function defined by
\begin{equation*}
s\longmapsto G(\lambda ,E,s):=E^{p}-p'\lambda F(s),
\end{equation*}
where $p>1$, $E\geq 0$ and $\lambda >0$ are real parameters. Then

\begin{itemize}
\item[(i)]  If $E>E_{c}:=( p'\lambda F(c)) ^{1/p}>0$,
then the function $G(\lambda ,E,\cdot )$ is strictly positive on 
$(0,\infty ) $.

\item[(ii)]  If $E=E_{c}$, then the function $G(\lambda ,E,\cdot )$ is
strictly positive on $( 0,c) $ and vanishes at $c$.

\item[(iii)]  If $E_{a}:=( p'\lambda F(a)) ^{1/p}<E<E_{c}
$, then the function $G(\lambda ,E,\cdot )$ has a unique zero $s_{1}(\lambda
,E)$ on $(\beta ,c)$ and is strictly positive on $( 0,s_{1}(\lambda
,E)) $. Moreover,

\begin{itemize}
\item[(a)]  The function $E\mapsto s_{1}(\lambda ,E)$ is $C^{1}$ on 
$(E_{a},E_{c}) $ and
\begin{equation*}
\frac{\partial s_{1}}{\partial E}(\lambda ,E)=\frac{( p-1) E^{p-1}%
}{\lambda f(s_{1}(\lambda ,E))}>0\;\text{for all }E\in (
E_{a},E_{c}) .
\end{equation*}

\item[(b)]  $\lim_{E\to E_{a}{}^{+}}s_{1}(\lambda ,E)=\beta $ and 
$\lim_{E\to E_{c}^{-}}s_{1}(\lambda ,E)=c$.
\end{itemize}

\item[(iv)]  If $E=E_{a}$, then
\[
G(\lambda ,E)\begin{cases}
>0 & \text{for }0<s<a, \\
=0 & \text{for }s=a, \\
>0 & \text{for }a<s<\beta , \\
=0 & \text{for }s=\beta , \\
<0 & \text{for }\beta <s<c.
\end{cases}
\]

\item[(v)]  If $0<E<E_{a}$, then the function $G(\lambda ,E,\cdot )$ has a
unique zero $s_{2}(\lambda ,E)$ on $( 0,a) $ and is strictly
positive on $( 0,s_{2}(\lambda ,E)) $. Moreover,

\begin{itemize}
\item[(a)]  The function $E\mapsto s_{2}(\lambda ,E)$ is $C^{1}$ on $(
0,E_{a}) $ and
\begin{equation*}
\frac{\partial s_{2}}{\partial E}(\lambda ,E)=\frac{( p-1) E^{p-1}%
}{\lambda f(s_{2}(\lambda ,E))}>0\text{ for all }E\in ( 0,E_{a}) .
\end{equation*}

\item[(b)]  $\lim_{E\to 0^{+}}s_{2}(\lambda ,E)=0$ and $%
\lim_{E\to E_{a}^{-}}s_{2}(\lambda ,E)=a$.
\end{itemize}

\item[(vi)]  If $E=0$, then  $G(\lambda ,0,s)
\begin{cases}
<0 & \text{for }0<s\leq a, \\
<0 & \text{for }\beta \leq s\leq c.
\end{cases}$
\end{itemize}
\end{lemma}
Now, for $p>1$, $\lambda >0$ and $E\geq 0$, we let
\begin{equation*}
X_{1}(\lambda ,E):=\{ s\in \mathop{\rm dom}G(\lambda ,E,\cdot )=( 0,\infty
) :G(\lambda ,E,u)>0\quad\text{for all }u\in (0,s)\} .
\end{equation*}
In view of Lemma \ref{Lemma4.7}, it follows that
\begin{equation*}
X_{1}(\lambda ,E)=
\begin{cases}
( 0,\infty ) & \text{if }E>E_{c}, \\
( 0,c] & \text{if }E=E_{c}, \\
( 0,s_{1}(\lambda ,E)] & \text{if }E_{a}<E<E_{c}, \\
( 0,a] & \text{if }E=E_{a}, \\
( 0,s_{2}(\lambda ,E)] & \text{if }0<E<E_{a}, \\
\emptyset & \text{if }E=0.
\end{cases}
\end{equation*}
Therefore,
$r_{1}(\lambda ,0):=0$, and
\begin{equation*}
r_{1}(\lambda ,E):=\sup X_{1}(\lambda ,E)=
\begin{cases}
\infty & \text{if }E>E_{c}, \\
c & \text{if }E=E_{c}, \\
s_{1}(\lambda ,E) & \text{if }E_{a}<E<E_{c}, \\
a & \text{if }E=E_{a}, \\
s_{2}(\lambda ,E) & \text{if }0<E<E_{a},
\end{cases}
\end{equation*}
Also, we let
\begin{align*}
X_{2}(\lambda ,E)&:=\big\{ s>r_{1}(\lambda ,E):
s\in \mathop{\rm dom} G(\lambda ,E,\cdot )=( 0,\infty ) , \\
&\quad G(\lambda ,E,u)>0\text{ for all }u\in (r_{1}(\lambda ,E),s) \big\}
\end{align*}
In view of Lemma \ref{Lemma4.7},
\begin{equation*}
X_{2}(\lambda ,E)=\begin{cases}
\emptyset & \text{if }E>E_{c}, \\
( c,\infty ) & \text{if }E=E_{c}, \\
\emptyset & \text{if }E_{a}<E<E_{c}, \\
(a,\beta ) & \text{if }E=E_{a}, \\
\emptyset & \text{if }0<E<E_{a}, \\
\emptyset & \text{if }E=0.
\end{cases}
\end{equation*}
Therefore,
$r_{2}(\lambda ,E):=
\begin{cases}
\beta & \text{if }E=E_{a}, \\
\infty & \text{otherwise.}
\end{cases}
$
Let
\begin{align*}
 D_{1}(p,\lambda )&:=\big\{ E>0:
r_{1}(\lambda ,E)\in \mathop{\rm dom}G(\lambda ,E,\cdot )=( 0,\infty ),\\
&\quad G(\lambda ,E,r_{1}(\lambda ,E))=0,
\text{ and }f(r_{1}(\lambda ,E))>0 \big\}\\
&=( 0,E_{a}) \cup (E_{a},E_{c})
\end{align*}
and
\begin{align*}
 D_{2}(p,\lambda )&:=\big\{ E>0:
r_{2}(\lambda ,E)\in \mathop{\rm dom}G(\lambda ,E,\cdot )=( 0,\infty ),\\
&\quad G(\lambda ,E,r_{2}(\lambda ,E))=0,\text{ and }f(r_{2}(\lambda ,E))>0 \\
&=\{ E_{a}\} .
\end{align*}

Note that the definition domains of the time maps $T_{1}$ and $T_{2}$ are
\begin{align*}
\tilde{D}_{1}(p,\lambda ):=\big\{&E\geq 0:
r_{1}(\lambda ,E)\in \mathop{\rm dom}G(\lambda ,E,\cdot )=( 0,\infty )
, G(\lambda ,E,r_{1}(\lambda ,E))=0,\\
&\text{ and }\int\nolimits_{0}^{r_{1}(\lambda ,E)}( G( \lambda ,E,u) ) ^{-1/p}du
<\infty \big\},
\end{align*}
\begin{align*}
\tilde{D}_{2}(p,\lambda ):=\big\{ &E\geq 0:
r_{2}(\lambda ,E)\in \mathop{\rm dom}G(\lambda ,E,\cdot )=( 0,\infty ) ,
G(\lambda ,E,r_{2}(\lambda ,E))=0, \\
&\text{and }\int\nolimits_{0}^{r_{2}(\lambda ,E)}( G( \lambda ,E,u) ) ^{-1/p}du
<\infty \big\}
\end{align*}

In the present case, $( 0,E_{a}) \cup ( E_{a},E_{c})
\subset \tilde{D}_{1}(p,\lambda )\subset [ 0,E_{a}] \cup [
E_{a},E_{c}] $, and $\tilde{D}_{2}(p,\lambda )\subset \{
E_{a}\} $. We define, for $E\in \tilde{D}_{1}(p,\lambda )$, the time
map
\begin{align*}
T_{1}(\lambda ,E)&:= \int\nolimits_{0}^{r_{1}(\lambda
,E)}( G(\lambda ,E,u)) ^{-1/p}du \\
&=\int\nolimits_{0}^{r_{1}(\lambda
,E)}( E^{p}-p'\lambda F(u)) ^{-1/p}du \\
&=(p'\lambda )^{-1/p}\int_{0}^{r_{1}(\lambda ,E)}( F(r_{1}(\lambda
,E))-F(u)) ^{-1/p}du,
\end{align*}
since $G(\lambda ,E,r_{1}(\lambda ,E))=E^{p}-p'\lambda
F(r_{1}(\lambda ,E))=0$. For all $\lambda >0$, $r_{1}(\lambda ,\cdot )$ is
an increasing $C^{1}$-diffeomorphism from $( 0,E_{a}] $ onto $( 0,a] $ and
from $( E_{a},E_{c}] $ onto $( \beta ,c] $. Thus $T_{1}$ may be written as
\begin{equation*}
T_{1}(p,\lambda ,E)=(p'\lambda )^{-1/p}S(p,r_{1}(\lambda ,E))
\text{ for } E\in \overline{D}_{1}(p,\lambda ),
\end{equation*}
where for all $p>1$, $S(p,\cdot )$ is defined by
\begin{equation}
S(p,\alpha ):=\int\nolimits_{0}^{\alpha }( F(\alpha )-F(u))
^{-1/p}du\text{\ for\ all\ }\alpha \in ( 0,a] \cup ( \beta ,c] .  \label{4.12}
\end{equation}
Note that $S(p,\cdot )$ takes its values in $[ 0,\infty ] $.
We define, for $E\in \tilde{D}_{2}(p,\lambda )$, the time map
\begin{equation*}
T_{2}(p,\lambda ,E):=\int\nolimits_{0}^{r_{2}(\lambda ,E)}( G(
\lambda ,E,u) ) ^{-1/p}du=(p'\lambda
)^{-1/p}S(p,r_{2}(\lambda ,E)).
\end{equation*}
Note that, if $\tilde{D}_{2}(p,\lambda )\neq \emptyset $ then $\tilde{D}%
_{2}(p,\lambda )=\{ E_{a}\} $ and $r_{2}(\lambda ,E)=\beta $.
That is why we extend the definition domain of $S(p,\cdot )$ by including
the eventual range of $r_{2}(\lambda ,\cdot );$ that is, we define $%
S(p,\cdot )$ on $( 0,a] \cup [ \beta ,c] $. On the
other hand, continuity arguments imply that if $\tilde{D}_{2}(p,\lambda
)=\{ E_{a}\} $ then
\begin{align*}
T_{2}(p,\lambda ,E_{a}) &=(p'\lambda )^{-1/p}S(p,r_{2}(\lambda
,E))=(p'\lambda )^{-1/p}S(p,\beta )\\
&=\lim_{\alpha \to \beta^{+}}(p'\lambda )^{-1/p}S(p,\alpha )
=\lim_{E\to E_{a}^{+}}(p'\lambda)^{-1/p}S(p,r_{1}(\lambda ,E))\\
&=\lim_{E\to E_{a}^{+}}T_{1}(p,\lambda ,E).
\end{align*}
So we simply study the function $\alpha \mapsto S(p,\alpha )$ for $\alpha
\in ( 0,a] \cup [ \beta ,c] $, and if $S(p,\alpha)<\infty $, we intend that
\begin{equation*}
S(p,\alpha )=
\begin{cases}
(p'\lambda )^{1/p}T_{1}(p,\lambda ,E_{\alpha }:=r_{1}^{-1}(\lambda
,\alpha )) & \text{if }\alpha \in ( 0,a] \cup ( \beta ,c] , \\
(p'\lambda )^{1/p}T_{2}(p,\lambda ,E_{a}) & \text{if }\alpha =\beta .
\end{cases}
\end{equation*}

\begin{lemma} \label{Lemma4.8}
$f'(a)<0$ and $f'(c)<0$.
\end{lemma}

The proof of Lemma \ref{Lemma4.8} is easy but tedious; we omit it.

\begin{lemma} \label{Lemma4.9}
\begin{itemize}
\item[(i)]  $S(p,0)=0$ if $p>1$.

\item[(ii)]  $S(p,a)=\infty $ if and only if $1<p\leq 2$.

\item[(iii)]  $S(p,\beta )=\infty $ if and only if $1<p\leq 2$.

\item[(iv)]  $S(p,c)=\infty $ if and only if $1<p\leq 2$.
\end{itemize}
\end{lemma}

\begin{proof}
(i) For $p>1$ and $0<\alpha <a$, we write $S(p,\alpha )$ in (\ref{4.12}) as
\begin{align*}
S(p,\alpha ) &=\int_{0}^{\alpha }( F(\alpha )-F(u)) ^{-1/p}du \\
&=\alpha ( F(\alpha )) ^{-1/p}\int_{0}^{1}\big( 1-\frac{F(\alpha t)}{F(\alpha )}
\big) ^{-1/p}dt\quad (\text{let }u=\alpha t).
\end{align*}
Applying l'Hopital's rule, it is easy to see that 
$\lim_{\alpha \to 0^{+}}\alpha (F(\alpha ))^{-1/p}=0$ and 
$\lim_{\alpha \to 0^{+}}\frac{F(\alpha t)}{F(\alpha )}=t$. 
Therefore, $S(p,0)=\lim_{\alpha\to 0^{+}}S(p,\alpha )=0\cdot 
\int_{0}^{1}(1-t)^{-1/p}dt=0\cdot p'=0$. Hence the result follows.

(ii) Recall that for $p>1$, $S(p,a)=\int_{0}^{a}( F(a)-F(u))^{-1/p}du$.
Note that $F(a)-F(u)=-\frac{1}{2}f'(a)(a-u)^{2}+o((u-a)^{2})$ near $a^{-}$
and by Lemma \ref{Lemma4.8}, $f'(a)<0$. Therefore,
\begin{equation*}
( F(a)-F(u)) ^{-1/p}\approx ( -f'(a)/2)^{-1/p}( a-u) ^{-2/p}\quad
 \text{near}\ a^{-}.
\end{equation*}
Then easy computation shows that $S(p,a)=\infty $\ if and only if $1<p\leq 2$.

(iii) We write
\begin{equation*}
S(p,\beta )=\int_{0}^{\beta }( F(\beta )-F(u)) ^{-1/p}du=\big(
\int_{0}^{a}+\int_{a}^{\beta }\big) \big( F(\beta )-F(u)\big) ^{-1/p}du.
\end{equation*}

\noindent(\textbf{Eventual singularity at } $\beta ^{-}$) Note that
$F(\beta)-F(u)=f(\beta )(\beta -u)+o(\beta -u)$\ near\ $\beta ^{-}$.
Since $f(\beta)>0$, $( F(\beta )-F(u)) ^{-1/p}\approx ( f(\beta ))
^{-1/p}(\beta -u)^{-1/p}$ near $\beta ^{-}$. Then easy computation shows
that $\int_{\beta -\varepsilon }^{\beta }( F(\beta )-F(u))^{-1/p}du<\infty $
 for $p>1$ and $\varepsilon >0$ sufficiently small.

\noindent(\textbf{Eventual singularity at }$a^{-}$) Since $F(\beta )=F(a)$ by (H3),
the same arguments as those used in the proof of part (ii) above imply that
$\int_{0}^{a}( F(\beta )-F(u)) ^{-1/p}du=\infty $ if and only if $1<p\leq 2$.

\noindent(\textbf{Eventual singularity at }$a^{+}$) Since $F(\beta )=F(a)$, the same
arguments as those used in the proof of part (ii) above imply that
$\int_{a}^{a+\varepsilon }( F(\beta )-F(u)) ^{-1/p}du=\infty $ if
and only if $1<p\leq 2$ for $\varepsilon >0$ sufficiently small.

In above analysis, $S(p,\beta )=\infty $ if and only if $1<p\leq 2$.

(iv) Recall that for $p>1$, $S(p,c)=\int_{0}^{c}( F(c)-F(u))^{-1/p}du$.
Note that $F(c)-F(u)=-\frac{1}{2}f'(c)(c-u)^{2}+o((u-c)^{2})$ near $c^{-}$
and by Lemma \ref{Lemma4.8}, $f'(c)<0$. Therefore,
\begin{equation*}
( F(c)-F(u)) ^{-1/p}\approx ( -f'(c)/2)
^{-1/p}( c-u) ^{-2/p}\quad \text{near }c^{-}.
\end{equation*}
Then an easy computation shows that $S(p,c)=\infty$ if and only if $1<p\leq 2$.
\end{proof}


\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig7.pdf}
\end{center}
\caption{Graph of $\theta _{p}(u)$}
\end{figure}

Next, we study the variations of $S(p,\alpha )$ for
$\alpha \in (0,a)\cup (\beta ,c)$. For $p>1$, $S'(p,\alpha )$ is given by
\begin{equation}
S'(p,\alpha )=\frac{1}{p\alpha }\int_{0}^{\alpha }\frac{\theta
_{p}(\alpha )-\theta _{p}(u)}{( F(\alpha )-F(u)) ^{1/p}}du\quad
\text{for }\alpha \in ( 0,a) \cup (\beta ,c),  \label{4.13}
\end{equation}
where
$\theta _{p}(u)=pF(u)-uf(u)$.
This implies
\begin{gather*}
\theta _{p}'(u)=( p-1) f(u)-uf'(u), \\
\theta _{p}''(u)=( p-2) f'(u)-uf'' (u).
\end{gather*}
Thus by (H1) and (H4),
\begin{equation} \label{4.14}
\begin{gathered}
\theta _{p}(0)=0, \\
\theta _{p}'(0)=( p-1) f(0)>0, \\
\theta _{p}''(u)
\begin{cases}
>0 & \text{for }0<u<r_{p}, \\
<0 & \text{for }r_{p}<u<s_{p}, \\
>0 & \text{for }s_{p}<u<c.
\end{cases}
\end{gathered}
\end{equation}
In addition, by (H3), (H2) and Lemma \ref{Lemma4.8},
\begin{gather*}
\theta _{p}(\beta ^{\ast })<0, \\
\theta _{p}(c)=pF(c)-cf(c)=pF(c)>0, \\
\theta _{p}'(c)=(p-1)f(c)-cf'(c)=-cf'(c)>0.
\end{gather*}
Hence there exist $t_{p}\in ( r_{p},s_{p}) $ and $\sigma _{p}\in
( s_{p},c) $ such that
\begin{gather}
\theta _{p}\quad \text{is strictly increasing on }( 0,t_{p}) , \label{4.15}\\
\theta _{p}\quad \text{is strictly decreasing on }( t_{p},\sigma _{p}), \label{4.16}\\
\theta _{p}\quad \text{is strictly increasing on }( \sigma _{p},c) .\label{4.17}
\end{gather}
In addition, there exist $\delta _{p}\in ( t_{p},\sigma _{p}) $
and $\gamma _{p}\in ( \sigma _{p},c) $ such that
\begin{equation}
\theta _{p}(\delta _{p})=\theta _{p}(\gamma _{p})=0.  \label{4.18}
\end{equation}
The typical graph of $\theta _{p}(u)$ on $[0,c]$ is depicted in Fig.
7.

\begin{lemma} \label{Lemma4.10}
For $p>1$, the following statements hold
\begin{itemize}
\item[(i)]  $S(p,\alpha )$ is strictly increasing on $( 0,a) $.
\item[(ii)]  $S(p,\alpha )$ has exactly one critical point, a minimum, on 
$(\beta ,c)$. More precisely, there exists a unique $m_{p}\in (\beta ,c)$
such that $S(p,\alpha )$ is strictly decreasing on $( \beta,m_{p}) $ and is
strictly increasing on $( m_{p},c) $.
\end{itemize}
\end{lemma}

\begin{proof} Part (i). By (\ref{4.13})-(\ref{4.15}), it suffices to show that
\begin{equation}
0<a<t_{p}\quad \text{for }p>1.  \label{Cond}
\end{equation}
Note that
\begin{equation*}
\theta _{p}(a)=pF(a)-af(a)=pF(a)>0,
\end{equation*}
and by Lemma \ref{Lemma4.8},
\begin{equation*}
\theta _{p}'(a)=(p-1)f(a)-af'(a)=-af'(a)>0,
\end{equation*}
then $a\in ( 0,t_{p}) \cup ( \gamma _{p},c) $ by \eqref{4.14}--\eqref{4.18}. 
If $a\in ( \gamma _{p},c) $ ($\subset
( \sigma _{p},c) $), then (\ref{4.17}) implies that $\theta _{p}$%
\ is strictly increasing on $( a,c) $. Hence
\begin{equation}
\theta _{p}'(u)>0\quad \text{for }u\in ( a,c) .
\label{additive}
\end{equation}
However, (H2) implies that there exists $\eta _{p}\in ( a,b) $ ($%
\subset ( a,c) $) such that $f(\eta _{p})<0$\ and\ $f^{\prime
}(\eta _{p})=0$. Therefore,
\begin{equation*}
\theta _{p}'(\eta _{p})=(p-1)f(\eta _{p})-\eta _{p}f'
(\eta _{p})=(p-1)f(\eta _{p})<0\quad \text{ for }p>1,
\end{equation*}
which leads to a contradiction with (\ref{additive}). Therefore, (\ref{Cond})
holds and hence part (i) follows.
Part (ii) follows by exactly the same arguments used to prove
\cite[Lemma 4.7]{a3}. To this end, it suffices to prove the following two
lemmas.
\end{proof}

\begin{lemma} \label{Lemma4.4a}
For $p>1$, $S''(p,\alpha )+(p/\alpha)S'(p,\alpha )>0$ for all
$\alpha \in ( \max \{ \sigma_{p},\beta \} ,c) $.
\end{lemma}

The proof of Lemma \ref{Lemma4.4a} is the same as that of \cite[Lemma 4.6]{a3};
 we omit it.

\begin{lemma} \label{Lemma4.5a}
Assume that $p>1$.
\begin{itemize}
\item[(i)]  If $p>2$ then $S'( p,c) =\infty $.

\item[(ii)]  If $\beta <\sigma _{p}$\ then $S'(p,\alpha )<0$ for
all $\alpha \in (\beta ,\sigma _{p}]$.

\item[(iii)]  If $\beta =\sigma _{p}$\ then $S(p,\beta )<\infty $ then 
$-\infty \leq S'(p,\beta )<0$.

\item[(iv)]  If $\beta >\sigma _{p}\;$and $S(p,\beta )<\infty $ then 
$S'(p,\beta )=-\infty $.
\end{itemize}
\end{lemma}

\begin{proof} The proofs of parts (i) and (iii)
follow exactly as those of parts (i) and (iii) of \cite[Lemma 4.5]{a3}; we
omit them. For part (ii) we point out that by (H3)
($\theta _{p}(\beta^{*})<0$ where $\beta ^{*}\leq \beta $)
it follows that $\delta_{p}<\beta $. Then the argument used to prove
Lemma 4.5(ii) of \cite{a3} can apply to prove $S'(p,\alpha )<0$ for all
$\alpha \in (\beta ,\sigma_{p}]$. So part (ii) holds.
Proof of part (iv). Since $F(\beta )=F(a)>0$, it follows that the
integral representing $S'(p,\beta )$, has two singularities; one at
$a$ and the other at $\beta $. So we write
\begin{equation*}
S'(p,\beta )=(p\beta) ^{-1}(I_{a^{-}}+I_{a^{+}}+I_{\beta }) ,
\end{equation*}
where
\begin{gather*}
I_{a^{-}}:= \int_{0}^{a}\frac{\theta _{p}(\beta
)-\theta _{p}(u)}{( F(\beta )-F(u)) ^{(p+1)/p}}du, \\
I_{a^{+}}:= \int_{a}^{(a+\beta )/2}\frac{\theta
_{p}(\beta )-\theta _{p}(u)}{( F(\beta )-F(u)) ^{(p+1)/p}}du, \\
I_{\beta }:= \int_{(a+\beta )/2}^{\beta }\frac{\theta _{p}(\beta )
-\theta _{p}(u)}{( F(\beta )-F(u)) ^{(p+1)/p}}\,du.
\end{gather*}
We next show
$I_{\beta }<\infty$ and $I_{a^{\pm }}=-\infty$.
First we show $I_{\beta }<\infty $. Since $(c>)$ $\beta >\sigma _{p}$, it
follows that $\theta _{p}'(\beta )>0$. Therefore,
\begin{equation*}
\frac{\theta _{p}(\beta )-\theta _{p}(u)}{( F(\beta )-F(u))
^{(p+1)/p}}\approx \frac{\theta _{p}'(\beta )}{( f(\beta
)) ^{(p+1)/p}}\frac{1}{(\beta -u)^{1/p}}\quad \text{near }\beta ^{-}.
\end{equation*}
Since $1/p<1$ and $\theta _{p}'(\beta )( f(\beta ))
^{-(p+1)/p}>0$, easy computations show that $I_{\beta }<\infty $.
We then show $I_{a^{\pm }}=-\infty $. Note that
\begin{equation*}
\frac{\theta _{p}(\beta )-\theta _{p}(u)}{( F(\beta )-F(u))
^{(p+1)/p}}\approx \frac{\theta _{p}(\beta )-\theta _{p}(a)}{(
-f'(a)/2) ^{(p+1)/p}}\frac{1}{( a-u) ^{2(p+1)/p}}\quad
\text{near }a.
\end{equation*}
Since $2(p+1)/p>1$ and
\begin{equation*}
\frac{\theta _{p}(\beta )-\theta _{p}(a)}{( -f'(a)/2)^{(p+1)/p}}=
\frac{-\beta f(\beta)}{( -f'(a)/2)^{(p+1)/p}}<0,
\end{equation*}
easy computations show that $I_{a^{\pm }}=-\infty $.
This completes the proof of Lemma \ref{Lemma4.5a}.
Therefore, the proof Lemma \ref{Lemma4.10} is also complete.
\end{proof}


\begin{lemma} \label{Lemma4.11} For $p>2$, $0<\alpha _{p}<\lambda _{p}<\infty $.
\end{lemma}

\begin{proof} By (\ref{3.4}), (\ref{4.12}) and Lemma \ref{Lemma4.9}(iii),
for $p>2$,
\begin{equation*}
\lambda _{p}=\big\{ \int_{0}^{\beta }( F(\beta )-F(u))
^{-1/p}du\big\} ^{p}/p'=\big\{ \lim_{\alpha \to \beta
^{+}}S(p,\alpha )\big\} ^{p}/p'<\infty .
\end{equation*}
We then find that
\begin{align*}
\lambda _{p} &=\{ \int_{0}^{\beta }( F(\beta )-F(u))^{-1/p}du\} ^{p}/p' \\
&=\{ \int_{0}^{a}( F(\beta )-F(u))^{-1/p}du+\int_{a}^{\beta }( F(\beta )-F(u)) ^{-1/p}du\}
^{p}/p' \\
&>\{ \int_{0}^{a}( F(\beta )-F(u)) ^{-1/p}du\}^{p}/p' \\
&=\{ \int_{0}^{a}( F(a)-F(u)) ^{-1/p}du\}^{p}/p'
\quad\text{(since }F(\beta )=F(a)\text{)} \\
&=\alpha _{p}>0\quad \text{(by (\ref{3.3})).}
\end{align*}
This completes the proof.
\end{proof}

Let $u$ be a positive solution of \eqref{1.1}, then $0<\left\Vert
u\right\Vert _{\infty }\leq a$ or $\beta \leq \left\Vert u\right\Vert
_{\infty }\leq c$. In addition, $u\in A_{0}^{+}\cup A_{1}^{+}\cup
A_{00}^{+}\cup A_{01}^{+}$ by Lemma \ref{Lemma2.2}.

By Lemma \ref{Lemma4.9}(ii)-(iv), for $p>2$,
$S(p,a)<\infty$, $S(p,\beta )<\infty$, $S(p,c)<\infty$.
In this case we have the following three statements:

\noindent\textbf{(i)}  Suppose for $\lambda =\alpha _{p}=( S(p,a))
^{p}/p'$, $u_{\alpha _{p}}$ is the corresponding solution of (\ref
{1.1}) satisfying $\| u_{\alpha _{p}}\| _{\infty }=u_{\alpha
_{p}}(0)=a$. Then
\begin{equation*}
u_{\alpha _{p}}'(x)=\{ p'\alpha _{p}[F(a)-F(u(x))] \} ^{1/p}>0\quad 
\text{ for }-1\leq x<0
\end{equation*}
by Lemma \ref{Lemma2.1}. So $u_{\alpha _{p}}\in A_{0}^{+}$. Then for each
$\lambda >\alpha _{p}$,
\begin{equation*}
u_{\lambda }(x):=\begin{cases}
u_{\alpha _{p}}\big( ( \frac{\lambda }{\alpha _{p}})
^{1/p}( | x| -1+( \frac{\alpha _{p}}{\lambda })
^{1/p}) \big)  & \text{if }1-( \frac{\alpha _{p}}{\lambda }) ^{1/p}<| x| \leq 1, \\
a & \text{if }| x| \leq 1-( \frac{\alpha _{p}}{\lambda }) ^{1/p}
\end{cases}
\end{equation*}
is a $C^{1}$ dead core solution of \eqref{1.1} satisfying 
$\|u_{\lambda }\| _{\infty }=a$,
\begin{equation*}
u_{\lambda }'(-1)=( p'\lambda F(a))^{1/p}>( p'\alpha _{p}F(a)) ^{1/p}=u_{\alpha
_{p}}'(-1)>0,
\end{equation*}
and $u_{\lambda }\in A_{1}^{+}$.

\noindent\textbf{(ii)}
  Suppose for $\lambda =\nu _{p}=( S_{p}(c))
^{p}/p'$, $u_{\nu _{p}}$ is the corresponding solution of 
\eqref{1.1} satisfying $\| u_{\nu _{p}}\| _{\infty }=u_{\nu _{p}}(0)=c$.
Then
\begin{equation*}
u_{\nu _{p}}'(x)=\{ p'\nu _{p}[
F(c)-F(u(x))] \} ^{1/p}>0\text{ for }-1\leq x<0
\end{equation*}
by Lemma \ref{Lemma2.1}. So $u_{\nu _{p}}\in A_{0}^{+}$.
Then for each $\lambda >\nu _{p}$,
\begin{equation*}
u_{\lambda }(x):=\begin{cases}
u_{\nu _{p}}\big( ( \frac{\lambda }{\nu _{p}}) ^{1/p}(
| x| -1+( \frac{\nu _{p}}{\lambda }) ^{1/p})
\big)  & \text{if }1-( \frac{\nu _{p}}{\lambda }) ^{1/p}<|
x| \leq 1, \\
c & \text{if }| x| \leq 1-( \frac{\nu _{p}}{\lambda })
^{1/p}
\end{cases}
\end{equation*}
is a $C^{1}$ dead core solution of \eqref{1.1} satisfying
$\|u_{\lambda }\| _{\infty }=c$,
\begin{equation*}
u_{\lambda }'(-1)=( p'\lambda F(c))
^{1/p}>( p'\nu _{p}F(c)) ^{1/p}=u_{\nu _{p}}^{\prime
}(-1)>0,
\end{equation*}
and $u_{\lambda }\in A_{1}^{+}$.

\noindent\textbf{(iii)}
  Suppose for $\lambda =\lambda _{p}=(S(p,\beta ))^{p}/p'$,
$u_{\lambda _{p}}\,$is the corresponding solution of \eqref{1.1}
satisfying $\| u_{\lambda _{p}}\| _{\infty }=u_{\lambda_{p}}(0)=\beta $.
Then, by Lemma \ref{Lemma2.1}, there exists a unique
negative number $-x_{0}\in (-1,0)$ such that
\begin{equation*}
u_{\lambda _{p}}(-x_{0})=a\text{ and }u_{\lambda _{p}}'(-x_{0})=0
\end{equation*}
and
\begin{equation*}
u_{\lambda _{p}}'(x)=\{ p'\lambda _{p}[ F(\beta)-F(u(x))] \} ^{1/p}>0\quad
\text{ for }x\in [ -1,0)
-\{ -x_{0}\} .
\end{equation*}
So $u_{\lambda _{p}}\in A_{00}^{+}$. Then for each $\lambda >\lambda _{p}$,
\begin{equation*}
u_{\lambda }(x):=\begin{cases}
u_{\lambda _{p}}\big( ( \frac{\lambda }{\lambda _{p}})^{1/p}| x| \big)  
& \text{if }| x| \leq ( \frac{K_{0a}}{\lambda }) ^{1/p}, \\
a & \text{if }( \frac{K_{0a}}{\lambda }) ^{1/p}\leq |x| 
\leq 1-( \frac{K_{a\beta }}{\lambda }) ^{1/p}, \\
u_{\lambda _{p}}\Big( \frac{| x| -1
+( \frac{K_{a\beta }}{\lambda }) ^{1/p}
+( \frac{K_{a\beta }}{\lambda _{p}}) ^{1/p}}{( \frac{K_{a\beta }}{\lambda }) ^{1/p}
+( \frac{K_{a\beta }}{\lambda _{p}}) ^{1/p}}\Big)  
& \text{if }1-( \frac{K_{a\beta } }{\lambda }) ^{1/p}\leq | x| \leq 1
\end{cases}
\end{equation*}
is a positive solution of \eqref{1.1} satisfying 
$\| u_{\lambda}\| _{\infty }=\beta $,
\begin{equation*}
u_{\lambda }'(-1)=( p'\lambda F(\beta ))
^{1/p}>( p'\lambda _{p}F(\beta )) ^{1/p}=u_{\lambda
_{p}}'(-1)>0,
\end{equation*}
and $u_{\lambda }\in A_{01}^{+}$, where
\begin{eqnarray*}
&&K_{0a}:= ( \int_{0}^{a}( F(\beta )-F(u))
^{-1/p}du) ^{p}/p', \\
&&K_{a\beta }:= ( \int_{a}^{\beta }( F(\beta
)-F(u)) ^{-1/p}du) ^{p}/p'.
\end{eqnarray*}


Hence Theorems \ref{Thm3.5}-\ref{Thm3.9} follow immediately by 
Theorem \ref{Thm2.2} and Lemmas \ref{Lemma4.9}-\ref{Lemma4.11}.


\subsection*{Acknowledgments}
The authors thank the anonymous referee for his/her valuable remarks. 
Much of the computation in this paper has been checked using the 
symbolic manipulator Mathcad 7 Professional for  Addou and 
Mathematica 4.0 for Wang.

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\end{document}