\documentclass[twoside]{article}
\usepackage{amssymb, amsmath}
\pagestyle{myheadings} \markboth{\hfil Ambrosetti-Brezis-Cerami
problem \hfil EJDE--2000/66} {EJDE--2000/66\hfil I. Addou,  A.
{Benmeza\"\i}, S. M. Bouguima \& M. Derhab\hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations}, Vol.~{\bf
2000}(2000), No.~66, pp.~1--34. \newline ISSN: 1072-6691. URL:
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  Exactness results for generalized Ambrosetti-Brezis-Cerami problem and
  related one-dimensional elliptic equations
 %
\thanks{ {\em Mathematics Subject Classifications:} 34B15, 34B18.
\hfil\break\indent {\em Key words:} Exactness, p-Laplacian,
concave-convex nonlinearities, quadrature method.
\hfil\break\indent 
\copyright 2000 Southwest Texas State University. \hfil\break\indent 
Submitted January 1, 2000. Published November 2, 2000.} }
\date{}
%
\author{I. Addou,  A.  Benmeza\"\i, S. M. Bouguima \& M. Derhab}
\maketitle

\begin{abstract}
We consider the boundary-value problem
$$\displaylines{
-(\varphi _{p}(u'))' =\varphi _{\alpha }(u)+\lambda \varphi
_{\beta }(u) \quad\mbox{in }(0, 1) \cr 
u(0) =u(1)=0, }
$$ where $\varphi _{p}(x)=\left| x\right| ^{p-2}x$, $p,\alpha ,\beta >1$
and $\lambda \in \mathbb{R}^{*}$. We give the exact number of
solutions for all $\lambda$ and most values of $\alpha ,\beta,
p>1$. In the particular case where $1<\beta <p=2<\alpha $,  we
resolve completely a problem suggested by A. Ambrosetti, H.
Brezis and G. Cerami and which was partially solved by S.
Villegas.
\end{abstract}

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

\renewcommand{\theequation}{\thesection.\arabic{equation}}
\catcode`@=11 \@addtoreset{equation}{section} \catcode`@=12


\section{Introduction}
\label{sec1} The combined effects of concave and convex
nonlinearities were considered by Ambrosetti, Brezis and Cerami
in \cite{Ambrosetti1}. They consider the problem
\begin{gather}
-\Delta u = u^{\alpha -1}+\lambda u^{\beta -1},\quad x\in \Omega \nonumber\\
u  >  0, \quad x\in \Omega \label{AM} \\
u  =  0,\quad x\in \partial \Omega\nonumber
\end{gather}
with $1<\beta <2<\alpha $ and $\lambda >0$. They prove the
existence of a constant $\Lambda >0$ such that a solution
$u_{\lambda }$ of (\ref{AM}) exists if and only if $0<\lambda
\leq \Lambda $. Moreover, if the condition $\alpha \leq \alpha
^{*}$ holds, then for all $\lambda \in (0, \Lambda )$ problem
(\ref{AM}) has a second solution $v_{\lambda }>u_{\lambda }$,
where $\alpha ^{*}:=(2N)/(N-2)$ if $N\geq 3$ and $\alpha
^{*}=+\infty $ if $N=1,2$. Then several papers appeared where
concave-convex nonlinearities were involved. We refer the reader
to \cite{Ambrosetti1}-\cite{Bartsch}, \cite{OS}, \cite{Villegas}.

At the end of their paper \cite{Ambrosetti1}, Ambrosetti, Brezis
and Cerami suggested the study of the structure of the solution
set of the one-dimensional problem
\begin{gather}
-u''  = \left| u\right| ^{\alpha -2}u+\lambda \left|
u\right| ^{\beta -2}u, \;a<x<b  \label{Ville} \\
u(a)  =  u(b)=0, \nonumber
\end{gather}
with $1<\beta <2<\alpha $ and $\lambda >0$. This study was done
by S. Villegas \cite{Villegas} by means of a quadrature method.
He shows that there exist two monotone divergent sequences
$\{\varepsilon _{n}\}$ and $\{L_{n}\};$ $\varepsilon _{n}\leq
L_{n}$ satisfying:

\begin{description}
\item[i)]  If $\lambda \in (0, \varepsilon _{n})$ then (\ref{Ville}) has
exactly two pairs of opposite solutions with $n+1$ zeros.

\item[ii)]  If $\lambda \in [\varepsilon _{n}, L_{n})$ then (\ref{Ville})
has at least two pairs of opposite solutions with $n+1$ zeros.

\item[iii)]  If $\lambda =L_{n}$ then (\ref{Ville}) has at least one pair of
opposite solutions with $n+1$ zeros.

\item[iv)]  If $\lambda >L_{n}$ then (\ref{Ville}) has no solutions with
$n+1 $ zeros.
\end{description}

In the present paper, we consider the $p$-Laplacian version of
problem (\ref {Ville}), that is, we consider the boundary-value
problem
\begin{gather}
-(\varphi _{p}(u'))'  =  f(\lambda , u)\quad \text{in } (0, 1) \label{P}\\
u(0)  =  u(1)=0, \nonumber
\end{gather}
with $\varphi _{p}(x)=\left| x\right| ^{p-2}x$\ and $f(\lambda ,
u)=\varphi _{\alpha }(u)+\lambda \varphi _{\beta }(u)$ and notice
that when $p=2$, problem (\ref{P}) is reduced to problem
(\ref{Ville}). Our original interest was in answering the
question: how does the solution set of (\ref{P}) look like when
$p\neq 2$ and $1<\beta <p<\alpha $, $\lambda >0$ ? The interest in
this question comes from the fact that the structure of the
solution set of problem (\ref{P}) depends on $p>1$ for some
examples of second members $f(\lambda , u)$ (see for instance,
Guedda and Veron \cite{GueddaVeron}, Addou \cite{Addou5}) and
does not depend on $p$ for some others (see for instance, Addou
and {Benmeza\"\i} \cite{Addou2},\ for positive solutions when
$f(\lambda , u)=\lambda \exp (u)$). We shall prove that for
$f(\lambda , u)=\varphi _{\alpha }(u)+\lambda \varphi _{\beta
}(u)$, and $1<\beta <p<\alpha $, Villegas' result holds for
problem (\ref{P}) for all $p>1$ (part of Theorem \ref{Theorem1},
Assertion (C)).

Notice that Assertions (ii) and (iii), in Villegas' result,
\emph{do not provide the exact number} of solutions with $n+1$
zeros. So, when $\lambda $ belongs to the range $[\varepsilon
_{n}, L_{n}]$, the exact number of solutions with $n+1$ zeros
\emph{has yet to be studied.}

The exact number of solutions for problem (\ref{AM}) when
$\lambda $ ranges over the whole interval $(0, +\infty )$, was
given by Ouyang and Shi \cite {OS}, but under two restrictions:
$\Omega $ was taken to be the unit ball in $\mathbb{R}^{N}$ and
the space dimension $N\geq 4$. They proved the existence of some
$\Lambda >0$ such that problem (\ref{AM}) has exactly two
solutions for $\lambda \in (0, \Lambda )$, exactly one solution
for $\lambda =\Lambda $, and no solution for $\lambda >\Lambda $.
Actually their result concerns more general nonlinearities.

Next, the purpose of our investigation was to complete our study
by providing the exact number of solutions to (\ref{P}) when
$1<\beta <p<\alpha $, for all $p>1$ and all $\lambda >0$. So, in
the particular case where $p=2 $, Theorem \ref{Theorem1},
Assertion (C), completes Villegas' result and resolves completely
the Ambrosetti-Brezis-Cerami problem \cite[Section 6,
(d)]{Ambrosetti1}.

In pursuing our study, we were led quite naturally to study what
may happens if $\lambda $ is not necessarily positive or $p$ is
not necessarily between
$\alpha $ and $\beta $. A precise description of the solution set of problem (%
\ref{P}) for various values of $p, \alpha $, and $\beta $ was
given

As it is well known, exactness results are difficult to obtain.
The difficulties encountered in our study for $\lambda >0$ and
for $\lambda <0$ are of different kinds. For $\lambda >0$, we
used an idea performed by the authors Addou and {Benmeza\"\i} in
\cite{Addou3}, (see the proof of (iii) in \cite[pp.
11-13]{Addou3}).

The paper is organized as follows. The main results are stated in
Section \ref{sec2}. To prove our results we make use of a
quadrature method which is described in Section \ref{sec3}. The
main results for the case $\lambda >0$ are proved in Section
\ref{sec4}, while those for $\lambda <0$ are proved in Section
\ref{sec5}.

\section{Notation and main results}

\label{sec2}Now we define some sets that will be used in the
statements of the main results. For $k\geq 1$, let
\[
S_{k}^{+}=\left\{
\begin{array}{c}
u\in C^{1}\left( \left[ 0, 1\right] \right) :u\text{ admits
exactly }\left(
k-1\right) \text{ zeros in }\left( 0, 1\right) , \\
\text{all are simple, }u\left( 0\right) =u\left( 1\right) =0\text{ and }%
u'\left( 0\right) >0
\end{array}
\right\} ,
\]
$S_{k}^{-}=-S_{k}^{+}$ and $S_{k}=S_{k}^{+}\cup S_{k}^{-}$. If
$u\in C\left( \left[ 0, 1\right] \right) $ is a real-valued
function vanishing at $x_{1}$ and $x_{2}$ and not between them,
(with $x_{1}<x_{2}$) we call its restriction to the open interval
$(x_{1}, x_{2})$ a bump of $u$. So, each function in $S_{k}^{+}$
has exactly $k$ bumps such that the first one is positive, and
any two consecutive bumps have opposite sign.

Let $A_{k}^{+}$ $\left( k\geq 1\right) $ be the subset of
$S_{k}^{+}$ consisting of the functions $u$ satisfying:

\begin{itemize}
\item  Every bump of $u$ is symmetrical about the center of the interval of
its definition.

\item  Every positive (resp. negative) bump of $u$ can be obtained by
translating the first positive (resp. negative) bump.

\item  The derivative of each bump of $u$ vanishes once and only once.
\end{itemize}

\noindent Let $A_{k}^{-}=-A_{k}^{+}$ and $A_{k}=A_{k}^{+}\cup
A_{k}^{-}$. Denote by $\left( \lambda _{k}\right) _{k\geq 1}$ the
eigenvalues of the one dimensional p-Laplacian operator with
Dirichlet boundary conditions,
\begin{gather*}
-\left( \varphi _{p}\left( u'\right) \right) '
=  \lambda \varphi _{p}\left( u\right) \quad\text{in }\left( 0, 1\right) , \\
u\left( 0\right)  =  u\left( 1\right) =0\,.
\end{gather*}
One has for each integer $k\geq 1$ and $p>1$, $\lambda
_{k}=k^{p}\lambda _{1} $ and
\[
\lambda
_{1}=(p-1)(2\int_{0}^{1}(1-t^{p})^{-1/p}dt)^{p}=(p-1)(\frac{ 2\pi
}{p\sin (\pi /p)})^{p}.
\]
Now we define some constants we shall use them in the statement
of the main results. For all $p$, $\alpha , \beta >1$, let
\begin{equation}
J(p, \alpha , \beta ):=\frac{(p')^{-1/p}}{(\alpha -\beta )}(\frac{%
\alpha }{\beta })^{(p-\alpha )/p(\alpha -\beta )}B(\frac{(p-\beta )}{%
p(\alpha -\beta )}, \frac{p-1}{p}),  \label{JmljmLerty}
\end{equation}
where $B(\cdot ,\cdot )$ is the Beta function. Also, for all
$\beta >\alpha>1$, and $p>1$, let
\begin{equation}
K(p, \alpha , \beta )=\int_{0}^{1}\{(\frac{1-t^{\alpha }}{\alpha })-(\frac{%
1-t^{\beta }}{\beta })\}^{-1/p}dt.  \label{KsjkdytrfKh}
\end{equation}
We shall prove (see Lemma \ref{Lemma6}) that $:K(p, \alpha , \beta
)<+\infty $ if and only if $p>2$.

For all $\lambda \in \mathbb{R}$, denote $S_{\lambda }$ the
solution set of problem (\ref{P}). The main results of this paper
read as follows:

\begin{theorem}
\label{Theorem1}Let $p, \alpha , \beta >1$ and $\lambda >0$.

\begin{description}
\item[(A)]  Assume that one of the following conditions holds:

\begin{description}
\item[(a)]  $\alpha >p$ and $\beta >p$, or

\item[(b)]  $\alpha =p$ \ and\ \ $\beta >p$, or

\item[(c)]  $\alpha <p$ and $\beta <p$, or

\item[(d)]  $\alpha =p$ and $\beta <p$.
\end{description}
\noindent Then, for each integer $k=1, 2, \cdots $, there exists
$u_{k}\in A_{k}^{+}$ such that $S_{\lambda }\cap A_{k}=\left\{
u_{k}, -u_{k}\right\} $.

\item[(B)]  Assume that one of the following conditions holds:

\begin{description}
\item[(a)]  $\alpha >p$ and  $\beta =p$, or

\item[(b)]  $\alpha <p$ and $\beta =p$.
\end{description}
\noindent Then, for each integer $k=1, 2, \cdots $,
\begin{description}
\item[(i)]  If $\lambda \geq \lambda _{k}$, $S_{\lambda }\cap
A_{k}=\emptyset $.

\item[(ii)]  If $0<\lambda <\lambda _{k}$, there exists $u_{k}\in A_{k}^{+}$
such that $S_{\lambda }\cap A_{k}=\left\{ u_{k}, -u_{k}\right\} $.
\end{description}

\item[(C)]  Assume that $1<\beta <p<\alpha $. Then for each integer
$k=1, 2,\cdots $, there exist a real number $\mu _{k}>0$ such that

\begin{description}
\item[(i)]  If $\lambda >\mu _{k}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item[(ii)]  If $\lambda =\mu _{k}$, there exists $u_{k}\in A_{k}^{+}$ such
that $(S_{\lambda }\cap A_{k})=\left\{ u_{k}, -u_{k}\right\} $.

\item[(iii)]  If $\lambda \in \left( 0, \mu _{k}\right) $, there exist
$u_{k}, v_{k}\in A_{k}^{+}$ such that $u_{k}\neq v_{k}$ and
$(S_{\lambda }\cap A_{k})=\left\{ u_{k}, v_{k}, -u_{k},
-v_{k}\right\} $.
\end{description}
\end{description}
\end{theorem}

\begin{theorem}
\label{Theorem2}Let $p, \alpha , \beta >1$ and $\lambda <0$.

\begin{description}
\item[(A)]  Assume that one of the following conditions holds:

\begin{description}
\item[(a)]  $p>2, $and$\;1<\alpha <p<\beta $,\ or

\item[(b)]  $p>2, $and $1<\alpha <\beta =p$, or

\item[(c)]  $p>2$, and $1<\alpha <\beta <p$.
\end{description}
\noindent Then, there exists an increasing sequence of positive
real numbers $(\mu _{k})_{k\geq 1}$ such that for each integer
$k=1, 2, \cdots $,

\begin{description}
\item[(i)]  If $\lambda <-\mu _{k}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item[(ii)]  If $-\mu _{k}\leq \lambda <0$, there exists $u_{k}\in A_{k}^{+}$
such that $S_{\lambda }\cap A_{k}=\left\{ u_{k}, -u_{k}\right\} $.
\end{description}

\item[(B)]  Assume that $1<\beta <p<\alpha $. Then there exists an
increasing sequence of positive real numbers $(\mu _{k})_{k\geq
1}$ such that for each integer $k=1, 2, \cdots $,

\begin{description}
\item[(i)]  If $\lambda \leq -\mu _{k}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item[(ii)]  If $-\mu _{k}<\lambda <0$, there exists $u_{k}\in A_{k}^{+}$
such that $S_{\lambda }\cap A_{k}=\left\{ u_{k}, -u_{k}\right\} $.
\end{description}

\item[(C)]  Assume that $1<p\leq 2$, $\lambda <0$ and one of the following
conditions holds:

\begin{description}
\item[(a)]  $1<\alpha <p<\beta $, or

\item[(b)]  $1<\alpha <\beta =p$, or

\item[(c)]  $1<\alpha <\beta <p$, or

\item[(d)]  $1<p\leq \beta <\alpha $.
\end{description}
\noindent Then, for each integer $k=1, 2, \cdots $, there exists
$u_{k}\in A_{k}^{+}$ such that $S_{\lambda }\cap A_{k}=\left\{
u_{k}, -u_{k}\right\} $.

\item[(D)]  Assume that $1<p\leq 2$, $1<\alpha =p<\beta $, and $\lambda <0$.
Then for each integer $k=1, 2, \cdots $,

\begin{description}
\item[(i)]  If $k\geq (\frac{p^{2}}{\lambda _{1}})^{1/p}$, $S_{\lambda }\cap
A_{k}=\emptyset $.

\item[(ii)]  If $k<(\frac{p^{2}}{\lambda _{1}})^{1/p}$, there exists
$u_{k}\in A_{k}^{+}$ such that $S_{\lambda }\cap A_{k}=\left\{
u_{k}, -u_{k}\right\} $.
\end{description}

\item[(E)]  Assume that $1<\beta <\alpha =p$, then for each integer
$k=1, 2, \cdots $,

\begin{description}
\item[(i)]  If $k\leq (2J(p, \alpha =p, \beta ))^{-1}\,$or $k\geq
(p^{2}/\lambda _{1})^{1/p}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item[(ii)]  If $(2J(p, \alpha =p, \beta ))^{-1}<k<(p^{2}/\lambda
_{1})^{1/p}$, there exists $u_{k}\in A_{k}^{+}$ such that
$S_{\lambda }\cap A_{k}=\left\{ u_{k}, -u_{k}\right\} $.
\end{description}

\item[(F)]  Assume that $2<p=\alpha <\beta $, then for each integer
$k=1, 2, \cdots $,

\begin{description}
\item[(i)]  If $k<(2K(p, \alpha , \beta ))^{-1}\,$or $k\geq (p^{2}/\lambda
_{1})^{1/p}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item[(ii)]  If $(2K(p, \alpha , \beta ))^{-1}\leq k<(p^{2}/\lambda
_{1})^{1/p}$, there exists $u_{k}\in A_{k}^{+}$ such that
$S_{\lambda }\cap A_{k}=\left\{ u_{k}, -u_{k}\right\} $.
\end{description}

\item[(G)]  Assume that $1<p\leq 2$ and $p<\alpha <\beta $. Then, there
exists an increasing sequence $(\mu _{k})_{k\geq 1}$ such that
$\lim_{k\to +\infty }\mu _{k}=0$ and

\begin{description}
\item[(i)]  If $\lambda <\mu _{k}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item[(ii)]  If $\lambda =\mu _{k}$, there exists $u_{k}\in A_{k}^{+}$ such
that $(S_{\lambda }\cap A_{k})=\left\{ u_{k}, -u_{k}\right\} $.

\item[(iii)]  If $\lambda \in \left( \mu _{k}, 0\right) $, there exist
$u_{k}, v_{k}\in A_{k}^{+}$ such that $u_{k}\neq v_{k}$ and
$(S_{\lambda }\cap A_{k})=\left\{ u_{k}, v_{k}, -u_{k},
-v_{k}\right\} $.
\end{description}

\item[(H)]  Assume that one of the following conditions holds:

\begin{description}
\item[(a)]  $2<p<\alpha <\beta $ or

\item[(b)]  $1<\beta <\alpha <p$.
\end{description}

Then, there exist two increasing sequences $(\mu _{k})_{k\geq 1}$
and $(\nu _{k})_{k\geq 1}$ such that for all $k\geq 1$, $\mu
_{k}<\nu _{k}<0$, $\lim_{k\to +\infty }\mu _{k}=0$ and

\begin{description}
\item  If $\lambda <\mu _{k}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item  If $\lambda =\mu _{k}$, there exists $u_{k}\in A_{k}^{+}$ such that
$(S_{\lambda }\cap A_{k})=\left\{ u_{k}, -u_{k}\right\} $.

\item  If $\lambda \in (\mu _{k}, \nu _{k}]$, there exist $u_{k}, v_{k}\in
A_{k}^{+}$ such that $u_{k}\neq v_{k}$ and $(S_{\lambda }\cap
A_{k})=\left\{ u_{k}, v_{k}, -u_{k}, -v_{k}\right\} $.

\item  If $\lambda \in \left( \nu _{k}, 0\right) $, there exists $u_{k}\in
A_{k}^{+}$ such that $(S_{\lambda }\cap A_{k})=\left\{ u_{k},
-u_{k}\right\} $.
\end{description}
\end{description}
\end{theorem}

\section{Quadrature method}

\label{sec3} To obtain our main results, we use the well-known
quadrature method (see for instance, \cite{Addou1}-\cite{Addou5},
\cite{GueddaVeron}). In order to keep this article self-contained
we describe it here for the particular odd case. For this, some
notations are needed. Assume that $g\in C\left( \mathbb{R},
\mathbb{R}\right) , g$ is odd, and $p>1$. Consider the
boundary-value problem
\begin{equation}
-\left( \varphi _{p}\left( u'\right) \right) '=g\left( u\right)
\text{ \ in }\left( 0, 1\right) ;\;u\left( 0\right) =u\left(
1\right) =0.  \label{prop1}
\end{equation}
Denote by $p'=p/\left( p-1\right) $ the conjugate exponent of $p$.
$\left( 1/p+1/p'=1\right) $. Define $G\left( s\right)
=\int_{0}^{s}g\left( t\right) dt$. For any $E>0$, let,
\[
X\left( E\right) =\left\{ s>0:E^{p}-p'G\left( \xi \right)
 >0, \forall \xi , 0<\xi <s\right\} ,
\]
and
\[
r\left( E\right) =\left\{
\begin{array}{ll}
0 & \text{if }  X\left( E\right) =\emptyset \\
\sup X\left( E\right) &   \text{otherwise.}
\end{array}
\right.
\]
Let
\[
\widetilde{D}=\{E\in (0, +\infty ):0<r\left( E\right) <+\infty \text{ and }%
\int_{0}^{r\left( E\right) }(E^{p}-p'G(t))^{-1/p}dt<+\infty \},
\]
and define the time-map,
\[
T\left( E\right) =\int_{0}^{r\left( E\right) }(E^{p}-p^{\prime
}G(t))^{-1/p}dt, \;E\in \widetilde{D}.
\]

Due to the oddness of $g$, it follows that $u$ is a solution to (\ref{prop1}%
) in $A_{k}^{+}$ if and only if $\left( -u\right) $ is also a solution to (%
\ref{prop1}) in $A_{k}^{-}$.

In our odd case, time maps approach reads as follows. Let $E>0$,
$k\in
\mathbb{N%
}^{*}$. Then, problem (\ref{prop1}) admits a solution $u\in
A_{k}^{+}$ satisfying $u'\left( 0\right) =E$ if and only if $E\in
\widetilde{D}
$ and $kT\left( E\right) =1/2$, and in this case the solution is unique%
\footnote{%
This uniqueness means that if $v$ is also a solution to
(\ref{prop1}) in $A_k^{+}$ and satisfying $v'\left( 0\right) =E$
then $v\equiv u.$}.

\begin{remark}
\label{rempr}In practice, we first study the variations of the
real-valued function $s\mapsto E^{p}-p'G(s)\,$, then compute
$X(E)$ and deduce $r(E)$. Next, we compute $\widetilde{D}$. To
this end, we first compute the set
\[
D:=\left\{ E>0:0<r\left( E\right) <+\infty \text{ and }g\left(
r\left( E\right) \right) >0\right\}
\]
and then we deduce $\widetilde{D}$, by observing that: $D\subset \widetilde{D%
}\subset \overline{D}$. After that, we define the time map on
$\widetilde{D}$ and then compute its limits at the boundary
points of $\widetilde{D}$. Next, we study (when possible) the
variations of $T$ on $\widetilde{D}$. We achieve our study by
discussing the number of solutions to equations $kT\left(
E\right) =1/2$, for $k$ being an integer and $E\in \widetilde{D}$.
\end{remark}

\section{Proof of Theorem 2.1}

\label{sec4}This section is organized as follows. We begin by
some lemmas in the first subsection. The first lemma (Lemma
\ref{Lemma1}) is used in order to define the time map, while in
Lemma \ref{Lemma2} we compute the limits and in Lemma
\ref{Lemma3} we study the variations of the time map. Next we
dedicate a separate subsection to the proof of each assertion of
Theorem \ref {Theorem1}.

\subsection*{Preliminary lemmas}

According to the practical remark (Remark \ref{rempr}), we begin
by the following technical Lemma.

\begin{lemma}
\label{Lemma1}Consider the function defined on $\mathbb{R}^{+}$
by,
\[
s\longmapsto G(\lambda , E, s):=E^{p}-p'F(\lambda , s),
\]
where $p, \alpha , \beta >1, E>0$ and $\lambda >0$ are real
parameters, and
\[
F(\lambda , s):=\int_{0}^{s}f(\lambda , t)dt=\frac{1}{\alpha }s^{\alpha }+%
\frac{\lambda }{\beta }s^{\beta }, s\geq 0.
\]
For all $\lambda >0$ and $E>0$ there exists a unique $s(\lambda ,
E)>0$ such that the function $G(\lambda , E, \cdot )$ is strictly
positive on $(0, s(\lambda , E))$, vanishes at $s(\lambda , E)$
and is strictly negative on $(s(\lambda , E), +\infty )$.
Moreover,

\begin{description}
\item[(i)]  The function $E\mapsto s(\lambda , E)$\ is $C^{1}$ on
$(0, +\infty )$, and
\[
\frac{\partial s}{\partial E}(\lambda ,
E)=\frac{(p-1)E^{p-1}}{f(\lambda , s(\lambda , E))}>0,\text{\ for
all }E>0\text{ and all }\lambda >0.
\]

\item[(ii)]  $\lim_{E\to 0^{+}}s(\lambda , E)=0,$\
$\lim_{E\to +\infty }s(\lambda , E)=+\infty $.
\end{description}
\end{lemma}

\paragraph{Proof.}
For any fixed $p>1,$\ $E>0$ and $\lambda >0$, consider the
function
\[
s\longmapsto G(\lambda , E, s):=E^{p}-p'F(\lambda , s),
\]
defined on $[0, +\infty )$. One has
\[
\frac{\partial G}{\partial s}(\lambda , E, s)=-p'f(\lambda , s).
\]
Notice that,
\begin{equation}
f(\lambda , s)=s^{\alpha -1}+\lambda s^{\beta -1}>0,\quad\text{for all }s>0%
\text{ and all }\lambda \geq 0.  \label{Etoil}
\end{equation}
Thus, the function $G(\lambda , E, \cdot )$ is strictly
decreasing on $(0, +\infty )$. On the other hand, one has
\[
G(\lambda , E, 0)=E^{p}>0\quad \text{\ and\ \ \ \ }\lim_{s\to
+\infty }G(\lambda , E, s)=-\infty .
\]
Therefore, $G(\lambda , E, \cdot )$ admits a unique positive
zero, denoted by $s(\lambda , E)$, and it is strictly positive on
$(0, s(\lambda , E))$ and is strictly negative on $(s(\lambda ,
E), +\infty )$.

\noindent\textbf{Proof of (i)}. For any $p>1$ and $\lambda \geq
0$, consider the real-valued function,
\[
(E, s)\mapsto G(E, s):=E^{p}-p'F(\lambda , s),
\]
defined on $\Omega =(0, +\infty )^{2}$. One has $G\in
C^{1}(\Omega )$ and,
\[
\frac{\partial G}{\partial s}(E, s)=-p'f(\lambda , s)\quad\text{in
}\Omega ,
\]
hence, according to (\ref{Etoil}), it follows that
\[
\frac{\partial G}{\partial s}(E, s)<0,\quad\text{in }\Omega .
\]
Observe that for all $E>0$ and $\lambda \geq 0$, the couple $(E,
s(\lambda , E))$ belongs to $\Omega $ and one has
\begin{equation}
G(E, s(\lambda , E))=0.  \label{2Etoils}
\end{equation}
Thus, one can make use of the implicit function theorem to show
that the function $E\mapsto s(\lambda , E)$ is
$C^{1}(\mathbb{R}^{+}, \mathbb{R}^{+})$ and to obtain the
expression of $(\partial s/\partial E)(\lambda , E)$ given in
\textbf{(i)}. Its sign is given by (\ref{Etoil}) and the fact that
$s(\lambda , E)>0$ for all $\lambda \geq 0$ and $E>0$. Therefore,
Assertion \textbf{(i)} is proved.

\noindent\textbf{Proof of (ii).} For any fixed $p>1$ and $\lambda
\geq 0$,\ Assertion \textbf{(i)} of the current lemma implies
that the function defined on $(0, +\infty )$ by $E\mapsto
s(\lambda , E)$ is strictly increasing. It is bounded from below
by $0$ and from above by $+\infty $. Thus, the limits $\lim_{E\to
0^{+}}s(\lambda , E)=l_{0}$ and $\lim_{E\to +\infty }s(\lambda ,
E)=l_{+}$ exist and satisfy
\[
0\leq l_{0}<l_{+}\leq +\infty .
\]
Let us observe that for any fixed $p>1$ and $\lambda \geq 0$, the
function $(E, s)\mapsto G(E, s)$ is continuous on $[0, +\infty
)^{2}$ and the function $E\mapsto s(\lambda , E)$ is continuous
on $[0, +\infty )$ and satisfies (\ref{2Etoils}). Thus, by
passing to the limit in (\ref{2Etoils}) as $E$ tends to $0^{+}$,
one gets
\[
0=\lim_{E\to 0^{+}}G(E, s(\lambda , E))=G(0, l_{0}).
\]
Hence, $l_{0}\,$is a zero, belonging to $[0, +\infty )$, to the
equation in the variable $s:G(0, s)=0$. By solving this equation
one gets $l_{0}=0$.

Assume that $l_{+}<+\infty $, then by passing to the limit in (\ref{2Etoils}%
) as $E$ tends to $+\infty $ one gets
\[
+\infty =p'F(\lambda , l_{+})<+\infty ,
\]
which is impossible. So, $l_{+}=+\infty $. Therefore, Lemma
\ref{Lemma1} is proved. \hfill$\diamondsuit$\smallskip


Now, for any $p>1$,\ $\alpha , \beta >1$, $\lambda >0$ and $E>0$,
we compute $X(\lambda , E)$ as defined in Section \ref{sec3}. In
fact, for all $E>0$, $X(\lambda , E)=(0, s(\lambda , E))$, where
$s(\lambda , E)$ is defined in Lemma \ref{Lemma1}. Then,
$r(\lambda , E)=s(\lambda , E)$ for all $\lambda >0$ and $E>0$.
Hence, for any $p>1$, $\lambda >0, $\[ 0<r(\lambda , E)<+\infty
\text{\ if and only if }E>0.
\]
Also, for all $E>0$,
\[
f(\lambda , r(\lambda , E))=\varphi _{\alpha }(r(\lambda , E))+\frac{%
\lambda }{\beta }\varphi _{\beta }(r(\lambda , E))>0.
\]
So, $D(\lambda )=(0, +\infty )$ for all $\lambda >0$. Therefore,
$\widetilde{D}(\lambda )=(0, +\infty )$ for all $\lambda >0$.

Before going further in the investigation, from Lemma
\ref{Lemma1}, we deduce that for any fixed $p>1$ and $\lambda >0
$\begin{equation} E^{p}=p'F(\lambda , r(\lambda , E)), \text{for
all }E>0 \label{Eq1}
\end{equation}
\begin{equation}
\frac{\partial r}{\partial E}(\lambda ,
E)=\frac{(p-1)E^{p-1}}{f(\lambda , r(\lambda , E))}>0,\text{\ for
all }E>0\text{ and all }\lambda >0. \label{Eq2}
\end{equation}
\begin{equation}
\lim_{E\to 0^{+}}r(\lambda , E)=0,\ \lim_{E\to +\infty }r(\lambda
, E)=+\infty .  \label{Eq3}
\end{equation}
At present we define, for any $p>1, \lambda >0$ and $E>0$, the
time map
\[
T(\lambda , E)=\int_{0}^{r(\lambda , E)}\{E^{p}-p'F(\lambda , \xi
)\}^{-1/p}d\xi , E>0.
\]
By (\ref{Eq1}), it follows that
\[
T(\lambda , E)=(p')^{-1/p}\int_{0}^{r(\lambda , E)}\{F(\lambda ,
r(\lambda , E))-F(\lambda , \xi )\}^{-1/p}d\xi .
\]
Furthermore, a simple change of variables and a substitution yield
\[
T(\lambda , E)=(p')^{-1/p}\int_{0}^{1}\{\lambda r^{\beta
-p}(\lambda , E)(\frac{1-t^{\beta }}{\beta })+r^{\alpha -p}(\lambda , E)(%
\frac{1-t^{\alpha }}{\alpha })\}^{-1/p}dt.
\]
One may observe that
\[
T(\lambda , E)=S(\lambda , r(\lambda , E)), \text{for all }\lambda >0%
\text{, }E>0,
\]
where
\[
S(\lambda , \rho )=(p')^{-1/p}\int_{0}^{1}\{\lambda \rho ^{\beta
-p}(\frac{1-t^{\beta }}{\beta })+\rho ^{\alpha -p}(\frac{1-t^{\alpha }}{%
\alpha })\}^{-1/p}dt,
\]
for all $\lambda >0$ and $\rho >0$.

Beacuase the function $E\mapsto r(\lambda , E)$ is an increasing
$C^{1}$-diffeomorphism from $(0, +\infty )$ onto itself it
follows that if we put, for all $\lambda >0$,
\[
J_{1}(\lambda ):=\{E\in (0, +\infty ):T(\lambda , E)=1/2\},
\]
and
\[
J_{2}(\lambda ):=\{\rho \in (0, +\infty ):S(\lambda , \rho
)=1/2\},
\]
then
\[
Card(J_{1}(\lambda ))=Card(J_{2}(\lambda )),\text{\ for all
}\lambda >0.
\]
Hence, from now on, we will focus our attention in counting the
number of solution(s) of the equation $S(\lambda , \rho )=1/2$ in
the variable $\rho \in (0, +\infty )$, instead of the equation
$T(\lambda , E)=1/2$ in the variable $E\in (0, +\infty )$. In the
next lemma we shall compute the limits of $S(\lambda , \cdot
)\,$when $\lambda >0$.

\begin{lemma}
\label{Lemma2}For all $\lambda >0$, $p>1$,
\[
\lim_{\rho \to 0^{+}}S(\lambda , \rho )=0\text{,}\quad\text{if
}\alpha <p
\]
\[
\lim_{\rho \to 0^{+}}S(\lambda , \rho )=\left\{
\begin{array}{ll}
\frac{1}{2}\lambda _{1}^{1/p} & \text{if }  \alpha =p\text{ and
}\beta >p
\\
0 & \text{if }  \alpha =p\text{ and }\beta <p
\end{array}
\right.
\]
\[
\lim_{\rho \to 0^{+}}S(\lambda , \rho )=\left\{
\begin{array}{ll}
+\infty & \text{if }  \alpha >p\text{ and }\beta >p \\
\frac{1}{2}(\frac{\lambda _{1}}{\lambda })^{1/p} & \text{if } \alpha >p%
\text{ and }\beta =p \\
0 & \text{if } \alpha >p\text{ and }\beta <p
\end{array}
\right.
\]
\[
\lim_{\rho \to +\infty }S(\lambda , \rho )=\left\{
\begin{array}{ll}
0 & \text{if }  \alpha <p\text{ and }\beta >p \\
\frac{1}{2}(\frac{\lambda _{1}}{\lambda })^{1/p} & \text{if }  \alpha <p%
\text{ and }\beta =p \\
+\infty & \text{if } \alpha <p\text{ and }\beta <p
\end{array}
\right.
\]
\[
\lim_{\rho \to +\infty }S(\lambda , \rho )=\left\{
\begin{array}{ll}
0 & \text{if } \alpha =p\text{ and }\beta >p \\
\frac{1}{2}\lambda _{1}^{1/p} & \text{if } \alpha =p\text{ and
}\beta <p
\end{array}
\right.
\]
\[
\lim_{\rho \to +\infty }S(\lambda , \rho )=0\quad \text{if
}\alpha>p\,.
\]
\end{lemma}

The proof of this lemma consists of easy computations.


\begin{lemma}
\label{Lemma3}For all $p>1$, $\alpha , \beta >1$ and $\lambda >0$,

\begin{enumerate}
\item  $S(\lambda , \cdot )$ is strictly decreasing on $(0, +\infty )$
provided that
\[
(\alpha >p\text{ and }\beta \geq p)\text{ or }(\alpha =p \text{
and }\beta >p).
\]

\item  $S(\lambda , \cdot )$ is strictly increasing on $(0, +\infty )$
provided that
\[
(\alpha <p\text{ and }\beta \leq p)\text{ or }(\alpha =p%
\text{ and }\beta <p).
\]

\item  $S(\lambda , \cdot )$ is strictly increasing on $(0, \rho
_{1}(\lambda ))$ and is strictly decreasing on \newline $(\rho
_{2}(\lambda ), +\infty )$, provided that
\[
(\alpha >p\text{ and }\beta <p)\text{ or }(\alpha <p%
\text{ and }\beta >p),
\]
where
\[
\rho _{1}(\lambda ):=((-\lambda )\frac{(p-\beta )}{(p-\alpha
)})^{1/(\alpha
-\beta )}<\rho _{2}(\lambda ):=((-\lambda )\frac{(p-\beta )\alpha }{%
(p-\alpha )\beta })^{1/(\alpha -\beta )}.
\]
\end{enumerate}
\end{lemma}

\paragraph{Proof.}
For all $p>1$, $\alpha , \beta >1$ and $\lambda >0$, easy
computation yields
\[
\frac{\partial S}{\partial \rho }(\lambda , \rho )=(p^{\prime
})^{-1/p}\int_{0}^{\rho }\frac{H(\lambda , \rho )-H(\lambda ,
u)}{p\rho (F(\lambda , \rho )-F(\lambda , u))^{1+(1/p)}}du,
\]
where
\[
H(\lambda , \rho )=pF(\lambda , \rho )-\rho f(\lambda , \rho )=(\frac{%
p-\alpha }{\alpha })\rho ^{\alpha }+\lambda (\frac{p-\beta
}{\beta })\rho ^{\beta }.
\]
It follows that,
\[
\frac{\partial H}{\partial \rho }(\lambda , \rho )=(p-\alpha
)\rho ^{\alpha -1}+\lambda (p-\beta )\rho ^{\beta -1}.
\]
Thus, if $(\alpha >p$\ \ and\ \ $\beta \geq p)$ or $(\alpha =p$\
\ \ and\ \ \ $\beta >p)$, $H(\lambda , \cdot )$ is strictly
decreasing on $(0, +\infty )$ and then, for all $\lambda >0$ and
$\rho >0 $\[ H(\lambda , \rho )-H(\lambda , u)<0, \;\text{for all
}u\in (0, \rho ),
\]
and therefore,
\[
\frac{\partial S}{\partial \rho }(\lambda , \rho )<0,\quad\text{for all }%
\rho >0.
\]

If $(\alpha <p$\ and\ $\beta \leq p)$\ or\ $(\alpha =p$\ and\
$\beta <p)$, $H(\lambda , \cdot )$ is strictly increasing on $(0,
+\infty )$ and then for all $\rho >0$, $\lambda >0 $\[ H(\lambda
, \rho )-H(\lambda , u)>0, \quad\text{for all }u\in (0, \rho ),
\]
and therefore,
\[
\frac{\partial S}{\partial \rho }(\lambda , \rho )>0,\quad \text{for all }%
\rho >0.
\]

If $(\alpha >p$\ and\ $\beta <p)$\ or\ $(\alpha <p$\ and\ $\beta
>p)$, $H(\lambda , \cdot )$ is strictly increasing on $(0, \rho
_{1}(\lambda ))$ and strictly decreasing on $(\rho _{1}(\lambda
), +\infty )$. Moreover, $H(\lambda , \cdot )$ is strictly
positive on $(0, \rho _{2}(\lambda ))$ and strictly negative on
$(\rho _{2}(\lambda ), +\infty )$ and vanishes at $0$ and at
$\rho _{2}(\lambda )$. Therefore, for all $\lambda >0$ and $\rho
\in (0, \rho _{1}(\lambda )) $\[ H(\lambda , \rho )-H(\lambda ,
u)>0\quad \text{for all }u\in (0, \rho )
\]
and for all $\rho \in (\rho _{2}(\lambda ), +\infty )$,
\[
H(\lambda , \rho )-H(\lambda , u)<0\quad \text{for all }u\in (0,
\rho ).
\]
That is, $S(\lambda , \cdot )$ is strictly increasing on $(0, \rho
_{1}(\lambda ))$ and is strictly decreasing on $(\rho _{2}(\lambda
), +\infty )$.


\subsection*{Proof of Assertion A}

\textbf{Case (a).} Assume that $\alpha >p$ and $\beta >p$. By
Lemma \ref {Lemma2}, it follows that for all $\lambda >0$,
\[
\lim_{\rho \to 0^{+}}S\left( \lambda , \rho \right) =+\infty \text{%
,\ \ and\ \ }\lim_{\rho \to +\infty }S\left( \lambda , \rho
\right) =0\text{,}
\]
and by Lemma \ref{Lemma3}, $S\left( \lambda , \cdot \right) $ is
strictly decreasing on $\left( 0, +\infty \right) $. Thus, for
each integer $k=1, 2, \cdots $, the equation $kS\left( \lambda ,
\rho \right) =1/2$, in the variable $\rho >0$, admits a unique
solution for all $\lambda >0$.\ Therefore, for each integer $k=1,
2, \cdots $, problem (\ref{P}) admits a unique pair of solutions
$\left\{ u_{k}, v_{k}\right\} $ in $A_{k}$, for all $\lambda >0$.
Moreover $v_{k}=-u_{k}$.

\textbf{Case (b).} Assume that $\alpha =p$ and $\beta >p$. By
Lemma \ref {Lemma2}, it follows that for all $\lambda >0$,
\[
\lim_{\rho \to 0^{+}}S\left( \lambda , \rho \right) =\frac{1}{2}%
\lambda _{1}^{1/p}(p),\quad\text{and\ \ }\lim_{\rho \to +\infty
}S\left( \lambda , \rho \right) =0\text{,}
\]
and by Lemma \ref{Lemma3}, $S(\lambda , \cdot )$ is strictly
decreasing on $(0, +\infty )$. Thus, for each integer $k=1, 2,
\cdots $, the equation $kS(\lambda , \rho )=1/2$, in the variable
$\rho >0$, admits at least a solution in $(0, +\infty )$ if and
only if $(k/2)\lambda _{1}^{1/p}(p)>1/2 $, that is, if and only
if,
\begin{equation}
\lambda _{1}(p)>k^{-p}\text{,}  \label{Montreal}
\end{equation}
and in this case, the solution is unique. Notice that for each
integer $k=1, 2, \cdots $, $k^{-p}\leq 1$. So, (\ref{Montreal})
holds provided that
\begin{equation}
\lambda _{1}(p)>1\text{, for all }p>1\,.  \label{Canada}
\end{equation}
In the appendix, we shall prove that (\ref{Canada}) holds.
Therefore, for each integer $k=1, 2, \cdots $, problem (\ref{P})
admits a unique pair of solutions $\left\{ u_{k}, v_{k}\right\}
\,$in $A_{k}$, for all $\lambda >0 $. Moreover, $v_{k}=-u_{k}$.

The proofs of \textbf{Cases (c) }and \textbf{(d)} are similar and
then omitted. Therefore, Assertion A is proved.


\subsection*{Proof of Assertion B}

Assume that $\alpha >p$ (resp. $\alpha <p$) and $\beta =p$. By
Lemma \ref {Lemma2}, it follows that for all $\lambda >0$,
\begin{gather*}
\lim_{\rho \to 0}S(\lambda , \rho )=\frac{1}{2}(\frac{\lambda _{1}}{%
\lambda })^{1/p}\text{ (resp. =0), and }\\
\lim_{\rho \to +\infty }S(\lambda , \rho )=0
\text{ (resp. }=\frac{1}{2}(\frac{\lambda _{1}%
}{\lambda })^{1/p}\text{),}
\end{gather*}
and by Lemma \ref{Lemma3}, $S(\lambda , \cdot )$ is strictly
decreasing (resp. increasing) on $(0, +\infty )$. Thus, for each
integer $k=1, 2, \cdots $, the equation $kS(\lambda , \rho
)=1/2$, in the variable $\rho >0$, admits at least a solution in
$(0, +\infty )$ if and only if $(k/2)(\lambda _{1}/\lambda
)^{1/p}>1/2$, that is, if and only if $0<\lambda <k^{p}\lambda
_{1}(p)=\lambda _{k}(p)$, and in this case, the solution is
unique. Therefore, for each integer $k=1, 2, \cdots $, problem
(\ref{P}) admits a unique pair of solutions $\left\{ u_{k},
v_{k}\right\} $ (resp. admits no solution) in $A_{k}$, for all
$\lambda $ satisfying $0<\lambda <\lambda _{k}(p)$ (resp.
$\lambda \geq \lambda _{k}(p)$). Moreover, $u_{k}=-v_{k}$.

\subsection*{Proof of Assertion C}

Assume that $1<\beta <p<\alpha $. By Lemma \ref{Lemma2}, it
follows that for all $\lambda >0$,
\[
\lim_{\rho \to 0^{+}}S(\lambda , \rho )=\lim_{\rho \to +\infty
}S(\lambda , \rho )=0.
\]
Thus, for all $\lambda >0$, there exists a unique $M(\lambda )>0$
such that $M(\lambda )=\sup_{\rho \geq 0}S(\lambda , \rho )$.

\begin{lemma}
\label{Lemma4} Assume that $1<\beta <p<\alpha $.\ Then,

\begin{description}
\item[(a)]  $M(\cdot )$ is continuous on $(0, +\infty )$.

\item[(b)]  $M(\cdot )$ is strictly decreasing on $(0, +\infty )$.

\item[(c)]  $\lim_{\lambda \to 0^{+}}M(\lambda )=+\infty $, and
$\lim_{\lambda \to +\infty }M(\lambda )=0$.
\end{description}
\end{lemma}

\paragraph{Proof.}
Recall that for all $\lambda >0$ and $\rho >0 $\[ S(\lambda ,
\rho )=(p')^{-1/p}\int_{0}^{1}(\lambda \rho ^{\beta
-p}(\frac{1-t^{\beta }}{\beta })+\rho ^{\alpha -p}(\frac{1-t^{\alpha }}{%
\alpha }))^{-1/p}dt.
\]
For all $\lambda >0$ and $\rho >0$, let
$\bar{\rho}=\bar{\rho}(\lambda , \rho ):=\lambda ^{1/(\beta
-\alpha )}\rho $. Then, $\rho =\lambda ^{1/(\alpha -\beta
)}\bar{\rho}$ and a simple substitution yields:
\[
S(\lambda , \rho )=\lambda ^{(p-\alpha )/p(\alpha -\beta )}S(1, \bar{\rho}%
(\lambda , \rho )).
\]
Thus,
\[
M(\lambda )=\lambda ^{(p-\alpha )/p(\alpha -\beta )}\sup_{\rho \geq 0}S(1, %
\bar{\rho}(\lambda , \rho ))=\lambda ^{(p-\alpha )/p(\alpha -\beta )}\sup_{%
\bar{\rho}\geq 0}S(1, \bar{\rho}).
\]
Therefore,
\begin{equation}
M(\lambda )=\lambda ^{(p-\alpha )/p(\alpha -\beta )}M(1),\text{\ for all }%
\lambda >0.  \label{Algeria}
\end{equation}
Assertions \textbf{(a), (b)} and \textbf{(c)} are simple
consequences of formula (\ref{Algeria}). Therefore, Lemma
\ref{Lemma4} is proved. \hfill$\diamondsuit$\smallskip

By Lemma \ref{Lemma4} (or by formula (\ref{Algeria})), it follows
that the function $M(\cdot )$ admits an inverse function
$M^{-1}(\cdot )$ defined and strictly decreasing on $(0, +\infty
)$ and satisfies:
\[
\lim_{y\to 0}M^{-1}(y)=+\infty, \text{ and }\lim_{y\to +\infty
}M^{-1}(y)=0\,.
\]
Therefore, for each integer $k=1, 2, \cdots $, we define
$L_{k}:=M^{-1}(1/2k)$. Thus, $(L_{k})_{k\geq 1}$ is a strictly
increasing sequence and satisfies $\lim_{k\to +\infty
}L_{k}=+\infty $, and

\begin{itemize}
\item  $kM(\lambda )<1/2$,\ for all $\lambda >L_{k},$

\item  $kM(\lambda )=1/2$,\ for all $\lambda =L_{k}$

\item  $kM(\lambda )>1/2$,\ for all $\lambda \in (0, L_{k})$.
\end{itemize}

Therefore, the equation $kS(\lambda , \rho )=1/2$, in the
variable $\rho >0 $,

\begin{itemize}
\item  admits no solution for all $\lambda >L_{k}$,

\item  admits at least a solution for $\lambda =L_{k}$,

\item  admits at least two solutions for all $\lambda \in (0, L_{k})$.
\end{itemize}

Thus, for each integer $k=1, 2, \cdots $, problem (\ref{P}),

\begin{itemize}
\item  admits no solution in $A_{k}$, for all $\lambda >L_{k}$,

\item  admits at least one pair of solutions $\left\{ u_{k}, v_{k}\right\} $
in $A_{k}$, for $\lambda =L_{k}$. Moreover, $u_{k}=-v_{k}$.

\item  admits at least two pairs of solutions $\left\{ u_{k}, U_{k}\right\}
\cup \left\{ v_{k}, V_{k}\right\} $ in $A_{k}$, for all $\lambda
\in (0, L_{k})$. Moreover, $U_{k}=-u_{k}$ and $V_{k}=-v_{k}$.
\end{itemize} \medskip

 At present, let us prove that for each integer $k=1, 2, \cdots $,
there exists $\varepsilon _{k}\in (0, L_{k})$ such that the
equation $kS(\lambda , \rho )=1/2$, in the variable $\rho >0$,
admits exactly two solutions for all $\lambda \in (0, \varepsilon
_{k})$. To this end, it suffices to prove that for each integer
$k=1, 2, \cdots $, there exists $\varepsilon _{k}\in (0, L_{k})$
such that for all $\lambda \in (0, \varepsilon _{k}):
$\begin{equation} kS(\lambda , \rho )>1/2\text{, for all }\rho
\in [\rho _{1}(\lambda ), \rho _{2}(\lambda )]\text{,}  \label{KK}
\end{equation}
where $\rho _{i}(\lambda )$, $i=1, 2$, are defined in Lemma
\ref{Lemma3}.
In fact, assume that for all $\lambda \in (0, \varepsilon _{k})$, (\ref{KK}%
) holds, then $kS(\lambda , \rho _{1}(\lambda ))>1/2$ for all
$\lambda >0$, and by Lemma \ref{Lemma2}, $\lim_{\rho \to
0}S(\lambda , \rho )=0$, and by Lemma \ref{Lemma3}, $kS(\lambda ,
\cdot )$ is strictly increasing on $(0, \rho _{1}(\lambda )]$.
Thus, for all $\lambda >0$ there is a unique solution in $(0,
\rho _{1}(\lambda ))\,$to the equation $kS(\lambda , \rho )=1/2$,
in the variable $\rho >0$. Also, by (\ref{KK}) it follows that
$kS(\lambda , \rho _{2}(\lambda ))>1/2$ for all $\lambda >0$, and
by Lemma \ref{Lemma2}, $\lim_{\rho \to +\infty }kS(\lambda , \rho
)=0$, and by Lemma \ref{Lemma3}, $kS(\lambda , \cdot )$ is
strictly decreasing on $[\rho _{2}(\lambda ), +\infty )$. Thus,
for all $\lambda >0$ there is a unique solution in $(\rho
_{2}(\lambda ), +\infty )$ to the equation
$kS(\lambda , \rho )=1/2$, in the variable $\rho >0$. On the other hand, (%
\ref{KK}) implies that $kS(\lambda , \rho )\neq 1/2$, for all
$\lambda >0$ and all $\rho \in [\rho _{1}(\lambda ), \rho
_{2}(\lambda )]$. Thus, there is no solution in $[\rho
_{1}(\lambda ), \rho _{2}(\lambda )]\,$to the equation
$kS(\lambda , \rho )=1/2$, in the variable $\rho >0$.

Now, let us prove that for each integer $k=1, 2, \cdots $, there
exists $\varepsilon _{k}\in (0, L_{k})$ such that for all
$\lambda >0$ (\ref{KK}) holds. Simple computation shows that for
all $\lambda >0 $\begin{eqnarray*} S(\lambda , \rho _{1}(\lambda
)) &=&(p')^{-1/p}\lambda ^{(p-\alpha )/p(\alpha -\beta
)}\int_{0}^{1}\{(\frac{p-\beta }{\alpha -p})^{(\beta
-p)/(\alpha -\beta )}(\frac{1-t^{\beta }}{\beta }) \\
&&+(\frac{p-\beta }{\alpha -p})^{(\alpha -p)/(\alpha -\beta )}(\frac{%
1-t^{\alpha }}{\alpha })\}^{-1/p}dt.
\end{eqnarray*}
It follows that the function $\lambda \mapsto S(\lambda , \rho
_{1}(\lambda ))$ is continuous and strictly decreasing on $(0,
+\infty )$, and
\[
\lim_{\lambda \to 0}S(\lambda , \rho _{1}(\lambda ))=+\infty \text{%
,\ \ \ and\ \ \ }\lim_{\lambda \to +\infty }S(\lambda , \rho
_{1}(\lambda ))=0\,.
\]
Thus, for each integer $k=1, 2, \cdots $, there exists a unique
$\mu _{k}>0 $ such that
\[
kS(\mu _{k}, \rho _{1}(\mu _{k}))=\frac{1}{2}\,.
\]
Furthermore, the sequence $(\mu _{k})_{k\geq 1}$ is strictly
increasing and $\lim_{k\to +\infty }\mu _{k}=+\infty $. It is
easy to prove that for each integer $k=1, 2, \cdots
$\begin{equation} \mu _{k}\leq L_{k}.  \label{Sherbrooke}
\end{equation}
In fact, if the contrary holds, using the fact that the function
$\lambda \mapsto kS(\lambda , \rho _{1}(\lambda ))\,$is strictly
decreasing on $(0, +\infty )$, it follows that
\[
\frac{1}{2}=kS(\mu _{k}, \rho _{1}(\mu _{k}))<kS(L_{k}, \rho
_{1}(L_{k}))\,.
\]
But, for each integer $k=1, 2, \cdots $,
\[
kS(L_{k}, \rho _{1}(L_{k}))\leq k\sup_{\rho \geq 0}S(L_{k}, \rho
)=kM(L_{k})=\frac{1}{2}\,,
\]
a contradiction which proves (\ref{Sherbrooke}).

On the other hand, the function $\lambda \mapsto \rho
_{2}(\lambda )$ is continuous and strictly increasing on $(0,
+\infty )\,$and
\[
\lim_{\lambda \to 0}\rho _{2}(\lambda )=0,\text{ and }%
\lim_{\lambda \to +\infty }\rho _{2}(\lambda )=+\infty \text{,}
\]
then, for each integer $k=1, 2, \cdots $, there exists a unique
$\varepsilon _{k}>0$ such that
\[
\rho _{2}(\varepsilon _{k})=\rho _{1}(\mu _{k})\,.
\]
Using the fact that $\rho _{1}, \rho _{2}$ and $(\mu _{k})_{k\geq
1}$ are strictly increasing it follows that $(\varepsilon
_{k})_{k\geq 1}$ is also strictly increasing. Also, using the
fact that
\[
\lim_{\mu \to +\infty }\rho _{1}(\mu )=\lim_{\mu \to +\infty
}\rho _{2}(\mu )=\lim_{k\to +\infty }\mu _{k}=+\infty \text{,}
\]
it follows that: $\lim_{k\to +\infty }\varepsilon _{k}=+\infty $.

Furthermore, notice that using the fact that $\rho _{1}$ is
strictly increasing on $(0, +\infty )$ and $\rho _{1}<\rho
_{2}\,$on $(0, +\infty ) $, it follows that for each integer
$k=1, 2, \cdots $, $\varepsilon _{k}\in (0, \mu _{k})$, and by
(\ref{Sherbrooke}), it follows that
\begin{equation}
0<\varepsilon _{k}<L_{k},\text{\ for each integer }k=1, 2, \cdots
\label{Florida}
\end{equation}
Now, we believe that for each integer $k=1, 2, \cdots $, and all
$\lambda \in (0, \varepsilon _{k})$, (\ref{KK}) holds. In fact,
let $k=1, 2, \cdots $, be fixed and $\lambda _{0}\in (0,
\varepsilon _{k})$ and $\bar{\rho}\in [\rho _{1}(\lambda _{0}),
\rho _{2}(\lambda _{0})]$.

The variations of $\rho _{1}$ and $\rho _{2}\,$and the fact that
$\rho
_{1}<\rho _{2}$ on $(0, +\infty )$ imply that there exists $\bar{\lambda}%
\in [\lambda _{0}, \mu _{k})$ such that $\rho
_{1}(\bar{\lambda})=\bar{\rho}
$. Thus, $kS(\lambda _{0}, \bar{\rho})=kS(\lambda _{0}, \rho _{1}(\bar{%
\lambda}))$. But $S(\cdot , \rho )$ is decreasing on $(0, +\infty
)$.
Thus, $kS(\lambda _{0}, \rho _{1}(\bar{\lambda}))\geq kS(\bar{\lambda}%
, \rho _{1}(\bar{\lambda}))$.

On the other hand, the function $\lambda \mapsto S(\lambda , \rho
_{1}(\lambda ))$ is strictly decreasing on $(0, +\infty )$, thus,
\[
kS(\bar{\lambda}, \rho _{1}(\bar{\lambda}))>kS(\mu _{k}, \rho
_{1}(\mu _{k}))\,.
\]
Therefore,
\[
kS(\lambda _{0}, \bar{\rho})=kS(\lambda _{0}, \rho _{1}(\bar{\lambda}%
))\geq kS(\bar{\lambda}, \rho _{1}(\bar{\lambda}))>kS(\mu _{k},
\rho _{1}(\mu _{k}))=\frac{1}{2}\text{,}
\]
and hence, (\ref{KK}) holds. Therefore, a p-laplacian version of
Villegas result can be stated at this point. In fact; we can
state that if $1<\beta <p<\alpha $.\ Then, for each integer $k=1,
2,\cdots $, there exist two real numbers $\varepsilon _{k}$ and
$L_{k}$ such that $\varepsilon _{k}\leq L_{k}\,$and

\begin{description}
\item[(i)]  If $\lambda >L_{k}$, $S_{\lambda }\cap A_{k}=\emptyset $.

\item[(ii)]  If $\lambda =L_{k}$, there exists $u_{k}\in A_{k}^{+}$ such
that $(S_{\lambda }\cap A_{k})\supset \left\{ u_{k},
-u_{k}\right\} $.

\item[(iii)]  If $\lambda \in \left[ \varepsilon _{k}, L_{k}\right) $,
there exist $u_{k}, v_{k}\in A_{k}^{+}$ such that $u_{k}\neq
v_{k}$ and $(S_{\lambda }\cap A_{k})\supset \left\{ u_{k}, v_{k},
-u_{k}, -v_{k}\right\} $.

\item[(iv)]  If $\lambda \in \left( 0, \varepsilon _{k}\right) $, there
exist $u_{k}, v_{k}\in A_{k}^{+}$ such that $u_{k}\neq v_{k}$ and
$(S_{\lambda }\cap A_{k})=\left\{ u_{k}, v_{k}, -u_{k},
-v_{k}\right\} $.
\end{description}

Let us summarize. At this point, we have shown that when $1<\beta
<p<\alpha $, then for all $\lambda >0, $\[ \lim_{\rho \to
0}S(\lambda , \rho )=\lim_{\rho \to +\infty }S(\lambda , \rho )=0,
\]
hence for all $\lambda >0$, $S(\lambda , \cdot )$ admits at least
a critical point; a maximum in $(0, +\infty )$. Next, it was
proved that for all $\lambda >0$, there exist $\rho _{1}(\lambda
)$ and $\rho _{2}(\lambda )$
such that $0<\rho _{1}(\lambda )<\rho _{2}(\lambda )$ and $\frac{\partial S}{%
\partial \rho }(\lambda , \rho )>0$ on $(0, \rho _{1}(\lambda )]$ and
$\frac{\partial S}{\partial \rho }(\lambda , \rho )<0$ on $[\rho
_{2}(\lambda ), +\infty )$. So,\ the critical point belongs
necessarily to $(\rho _{1}(\lambda ), \rho _{2}(\lambda ))$.
Also, it was proved that the function $\lambda \mapsto M(\lambda
):=\sup_{0<\rho <+\infty }S(\lambda , \rho )$, is continuous,
strictly decreasing on $(0, +\infty )$ and
\[
\lim_{\lambda \to 0^{+}}M(\lambda )=+\infty ,\text{\ and \ }%
\lim_{\lambda \to +\infty }M(\lambda )=0\,.
\]

Thus, to complete the proof of Assertion (\textbf{C}), it remains
to prove that for all $\lambda >0$, $S(\lambda , \cdot )$ admits
\emph{at most one} critical point in $(\rho _{1}(\lambda ), \rho
_{2}(\lambda ))$. To this end we shall prove that $S(\lambda ,
\cdot )$ is concave on $(\rho _{1}(\lambda ), \rho _{2}(\lambda
))$ for all $\lambda >0$. Similar idea was previously used by the
authors Addou and {Benmeza\"\i} in \cite[Lemma 7, (iii)]{Addou3}.

The derivative of $S(\lambda , \cdot )$ is given by
\[
\frac{\partial S}{\partial \rho }(\lambda , \rho )=(p^{\prime
})^{-1/p}\int_{0}^{1}\frac{H(\lambda , \rho )-H(\lambda ,
u)}{p\rho (F(\lambda , \rho )-F(\lambda , u))^{(p+1)/p}}du
\]
where $F(\lambda , \rho )=\int_{0}^{\rho }f(\lambda ,
t)dt=\frac{1}{\alpha }u^{\alpha }+\frac{\lambda }{\beta }u^{\beta
}$, and

\[
H(\lambda , \rho )=pF(\lambda , \rho )-\rho f(\lambda , \rho )=(\frac{%
p-\alpha }{\alpha })\rho ^{\alpha }+\lambda (\frac{p-\beta
}{\beta })\rho ^{\beta }.
\]
Easy computations show that for all $\rho >0$ and $\lambda >0
$\begin{eqnarray*} (p')^{1/p}\frac{\partial ^{2}S}{\partial \rho
^{2}}(\lambda , \rho
) &=&\int_{0}^{1}\frac{(p+1)(H(\lambda , \rho )-H(\lambda , \rho \xi ))^{2}%
}{p^{2}\rho (F(\lambda , \rho )-F(\lambda , \rho \xi ))^{(2p+1)/p}}d\xi \\
&&+\int_{0}^{1}\frac{p(\Psi (\lambda , \rho )-\Psi (\lambda ,
\rho \xi ))(F(\lambda , \rho )-F(\lambda , \rho \xi ))}{p^{2}\rho
(F(\lambda , \rho )-F(\lambda , \rho \xi ))^{(2p+1)/p}}d\xi
\text{,}
\end{eqnarray*}
where
\begin{eqnarray*}
\Psi (\lambda , \rho ) &=&-p(p+1)F(\lambda , \rho )+2p\rho
f(\lambda
, \rho )-\rho ^{2}f_{\rho }'(\lambda , \rho ) \\
&=&\frac{(p-\alpha )(\alpha -(p+1))}{\alpha }\rho ^{a}+\lambda \frac{%
(p-\beta )(\beta -(p+1))}{\beta }\rho ^{\beta }\,.
\end{eqnarray*}
Some substitutions yield
\[
(p')^{1/p}\frac{\partial ^{2}S}{\partial \rho ^{2}}(\lambda , \rho
)=\int_{0}^{1}\frac{(1-\xi ^{\beta })^{2}P(X(\xi ))}{p^{2}\rho
(F(\lambda , \rho )-F(\lambda , \rho \xi ))^{(2p+1)/p}}d\xi
\text{,}
\]
where $X(1)=\frac{\alpha }{\beta }$ and $X(\xi )=\frac{1-\xi ^{\alpha }}{%
1-\xi ^{\beta }}$\ if \ $\xi \in [0, 1)$, and $P$ is the second
degree polynomial function
\begin{eqnarray*}
P(X)&=&\frac{(\alpha -p)\rho ^{2\alpha }}{\alpha
}X^{2}-(\frac{p(\alpha -\beta )^{2}+p(\alpha +\beta )-2\alpha
\beta }{\alpha \beta })\lambda \rho ^{\alpha
+\beta }X\\
&&+\frac{(\beta -p)}{\beta }\lambda ^{2}\rho ^{2\beta }\,.
\end{eqnarray*}
It can easily be verified that $X(\xi )\in [1, \alpha /\beta ]$
for all $\xi \in [0, 1]$. In fact, $X(0)=1$ and $\lim_{\xi \to
1}X(\xi )=\alpha /\beta $ and $X$ is strictly increasing on $(0,
1)$ since, $X'(\xi )=\xi ^{\beta -1}(1-\xi ^{\beta })^{-2}(\beta
-\alpha \xi ^{\alpha -\beta }+(\alpha -\beta )\xi ^{\alpha })$
and $\beta -\alpha \xi ^{\alpha -\beta }>\beta -\alpha >0$, for
all $\xi \in (0, 1)$. Thus $X'(\xi )>0$ for all $\xi \in (0, 1)$.

Therefore, we are interested in the sign of $P(X)$ when $X\in [1,
\alpha /\beta ]$. Its discriminant is
\[
d=(Y^{2}-4Z)\frac{\lambda ^{2}\rho ^{2(\alpha +\beta )}}{(\alpha
\beta )^{2}}\,,
\]
where
\begin{eqnarray}
Y &=&p(\alpha -\beta )^{2}+p(\alpha +\beta )-2\alpha \beta  \label{Oiuyt} \\
&=&p\alpha ^{2}+(-2(p+1)\beta +p)\alpha +p\beta (\beta +1)  \notag \\
Z &=&\alpha \beta (\alpha -p)(\beta -p)\,.  \notag
\end{eqnarray}
Notice that by our hypothesis $1<\beta <p<\alpha $, it follows
that $Z<0$, so that $d>0$. The roots of $P$ are given by
\[
X_{1}(\lambda , \rho )=\lambda \frac{Y-\sqrt{Y^{2}-4Z}}{2\beta
(\alpha -p)\rho ^{\alpha -\beta }}\leq 0,\text{\ and
}X_{2}(\lambda , \rho
)=\lambda \frac{Y+\sqrt{Y^{2}-4Z}}{2\beta (\alpha -p)\rho ^{\alpha -\beta }}%
\,.
\]
Notice that the function $\rho \to X_{2}(\lambda , \rho )$ is
strictly decreasing on $(0, +\infty )$. So, it would be perfect if
\begin{equation}
X_{2}(\lambda , \rho _{2}(\lambda ))>\frac{\alpha }{\beta }\text{
\ for all }\lambda >0\,.  \label{Mahna}
\end{equation}
In fact, it follows therefore that,
\[
\lbrack 1, \frac{\alpha }{\beta }]\subset (X_{1}(\lambda , \rho
), X_{2}(\lambda , \rho )),\text{\ }\forall \lambda >0, \forall
\rho \in (\rho _{1}(\lambda ), \rho _{2}(\lambda )).
\]
Hence, $P(X(\xi ))<0$, for all $\xi \in [0, 1]$, so,
\[
\frac{\partial ^{2}S}{\partial \rho ^{2}}(\lambda , \rho )<0,
\forall \lambda >0, \forall \rho \in (\rho _{1}(\lambda ), \rho
_{2}(\lambda )),
\]
which will prove the uniqueness of the critical point of
$S(\lambda , \cdot )$, and therefore, Theorem \ref{Theorem1}.

Let us prove the estimates (\ref{Mahna}). Notice that
\[
X_{2}(\lambda , \rho _{2}(\lambda
))=\frac{Y+\sqrt{Y^{2}-4Z}}{2\alpha
(p-\beta )}>\frac{\alpha }{\beta }\Longleftrightarrow \beta \sqrt{Y^{2}-4Z}%
>2\alpha ^{2}(p-\beta )-\beta Y\,.
\]
By taking the square of each member side we get
\[
X_{2}(\lambda , \rho _{2}(\lambda ))>\frac{\alpha }{\beta }%
\Longleftrightarrow 4\alpha ^{2}\beta (p-\beta )Y-4\beta
^{2}Z-4\alpha ^{4}(p-\beta )^{2}>0\,.
\]
Next, we substitute $Y$ and $Z$ as in (\ref{Oiuyt}) we get
\[
X_{2}(\lambda , \rho _{2}(\lambda ))>\frac{\alpha }{\beta }%
\Longleftrightarrow 4\alpha (p-\beta )Q_{\beta , p}(\alpha
)>0\text{,}
\]
where $Q_{\beta , p}(\alpha )$ is the third degree polynomial
function defined by
\[
Q_{\beta , p}(\alpha )=\alpha ^{3}(\beta (p+1)-p)+\alpha ^{2}\beta
(-2(p+1)\beta +p)+\alpha \beta ^{2}(\beta (p+1)+p)-p\beta ^{3}\,.
\]
It remains to show that $Q_{\beta , p}(\alpha )>0$ for all
$1<\beta <p<\alpha $. One can remark that $\beta $ is a root of
$Q_{\beta , p}(\cdot )$ of multiplicity at least twice. That is
to say, $Q_{\beta , p}(\beta )=Q_{\beta , p}'(\beta )=0$. Thus
there exist two constants $A$ and $B$ such that $Q_{\beta ,
p}(\alpha )=(\alpha -\beta )^{2}(A\alpha +B) $. Immediate
identification yields: $A=\beta (p+1)-p$ and $B=-p\beta $. It
remains to prove that $1<\beta <p<\alpha \Longrightarrow A\alpha
+B>0$. Notice that $1<\beta <p<\alpha $ implies that $(\beta
-1)p+\beta >\beta $ which implies that $\beta ((\beta -1)p+\beta
)^{-1}<1$ and then $-BA^{-1}<p$ and by $1<\beta <p<\alpha $ it
follows that $A\alpha +B>0$, which completes the proof of
(\ref{Mahna}). Therefore, Theorem \ref{Theorem1} is proved.

\section{Proof of Theorem 2.2}

\label{sec5}This section is organized as the previous one.

\subsection*{Preliminary lemmas}

\begin{lemma}
\label{Lemma5} Consider the function defined on $\mathbb{R}^{+}$
by
\[
s\longmapsto N(\lambda , E, s):=E^{p}-p'F(\lambda , s)\text{,}
\]
where $p, \alpha , \beta >1$,\ $\lambda <0$ and $E\in (0, +\infty
)$, are real parameters and
\[
F(\lambda , s)=\int_{0}^{s}f(\lambda , t)dt=\frac{1}{\alpha }s^{\alpha }+%
\frac{1}{\beta }s^{\beta },\text{\ }s\geq 0\,.
\]
Assume that $(\alpha -\beta )>0$, then for all $\lambda <0\,$and
$E>0$, the function $N(\lambda , E, \cdot )$ admits a unique
positive zero, $s(\lambda , E)$, and is strictly positive on $(0,
s(\lambda , E))$.

Assume that $(\alpha -\beta )<0$, then

\begin{description}
\item[(a)]  If $E>E_{*}:=(p'(-\lambda )^{\alpha /(\alpha -\beta )}(%
\frac{\beta -\alpha }{\alpha \beta }))^{1/p}$, $N(\lambda , E,
\cdot )$ is strictly positive on $(0, +\infty )$.

\item[(b)]  If $E=E_{*}$,\ $N(\lambda , E, \cdot )\,$vanishes at
$s=(-\lambda )^{1/(\alpha -\beta )}$ and is strictly positive on
$(0, (-\lambda )^{1/(\alpha -\beta )})\cup ((-\lambda
)^{1/(\alpha -\beta )}, +\infty )$.

\item[(c)]  If $0<E<E_{*}$,\ $N(\lambda , E, \cdot )$ admits a first
positive zero, $s(\lambda , E)>0$, and is strictly positive on
$(0, s(\lambda , E))$.
\end{description}

Moreover, if $(\alpha -\beta )>0$ (resp. $(\alpha -\beta )<0$),

\begin{description}
\item[(i)]  the function $E\mapsto s(\lambda , E)$ is $C^{1}$ on
$(0, +\infty )$ (resp. $(0, E_{*}(\lambda ))$) and
\[
\frac{\partial s}{\partial E}(\lambda ,
E)=\frac{(p-1)E^{p-1}}{f(\lambda , s(\lambda , E))}>0,\text{ for
all }\lambda <0 \text{ and } E>0
\]
(resp. $E\in (0, E_*(\lambda ))$).

\item[(ii)]  $\lim_{E\to 0}s(\lambda , E)=((-\lambda )\frac{\alpha
}{\beta })^{1/(\alpha -\beta )}$, (resp. $\lim_{E\to 0}s(\lambda
, E)=0$).

\item[(iii)]  $\lim_{E\to +\infty }s(\lambda , E)=+\infty $, (resp.
$\lim_{E\to E_{*}}s(\lambda , E)=(-\lambda )^{1/(\alpha -\beta
)}$).
\end{description}
\end{lemma}

The proof is the same as that of Lemma \ref{Lemma1} and then it
is omitted.
\medskip

Assume that $\alpha -\beta >0$ (resp. $\alpha -\beta <0$) then for
any $p>1$ and $E\in (0, +\infty )$ (resp. $E\in (0, E_{*}(\lambda
)$) we compute $X(\lambda , E)$ as defined in Section \ref{sec3}.
We derive from Lemma \ref{Lemma5}, for the case where $\alpha
-\beta >0,$\ $X(\lambda , E)=(0, s(\lambda , E))$ and for the
case where $\alpha -\beta <0 $\[ X(\lambda , E)=\left\{
\begin{array}{ll}
(0, +\infty ) & \text{if }  E>E_{*}(\lambda ) \\
(0, (-\lambda )^{1/(\alpha -\beta )}) & \text{if } E=E_{*}(\lambda ) \\
(0, s(\lambda , E)) & \text{if } 0<E<E_{*}(\lambda )\text{,}
\end{array}
\right.
\]
where $s(\lambda , E)$ is defined in Lemma \ref{Lemma5}. Then,
for $\alpha -\beta >0 $\[
r(\lambda , E):=\sup X(\lambda , E)=s(\lambda , E),\quad\text{for all }%
\lambda <0\text{ and }E>0
\]
and for $\alpha -\beta <0, $\[ r(\lambda , E)=\left\{
\begin{array}{ll}
+\infty & \text{if } E>E_{*}(\lambda ) \\
(-\lambda )^{1/(\alpha -\beta )} & \text{if } E=E_{*}(\lambda ) \\
s(\lambda , E) & \text{if } 0<E<E_{*}(\lambda )\,.
\end{array}
\right.
\]
Hence, for $p>1$ and $\lambda <0, $\[ 0<r(\lambda , E)<+\infty
\text{\ if and only if }\left\{
\begin{array}{ll}
E\in (0, +\infty ), & \text{if } \alpha -\beta >0, \\
0<E\leq E_{*}(\lambda ) & \text{if } \alpha -\beta <0.
\end{array}
\right.
\]
Also,
\[
f(\lambda , r(\lambda , E))>0\text{\ if and only if }\left\{
\begin{array}{ll}
E\in (0, +\infty ), & \text{if } \alpha -\beta >0, \\
0<E<E_{*}(\lambda ) & \text{if } \alpha -\beta <0.
\end{array}
\right.
\]
So, for all $\lambda <0$,
\[
D(\lambda )=\left\{
\begin{array}{ll}
(0, +\infty ) & \text{if } \alpha -\beta >0, \\
(0, E_{*}(\lambda )) & \text{if } \alpha -\beta <0.
\end{array}
\right.
\]

By $\widetilde{D}\subset (0, +\infty )$ and $\widetilde{D}\subset \overline{%
D}$, it follows that for $\alpha -\beta >0$, $\widetilde{D}=D=(0,
+\infty )$ and for $\alpha -\beta <0$, $\widetilde{D}=(0,
E_{*}(\lambda )]$ if $\int_{0}^{r(E_{*}(\lambda
))}(E^{p}-p'G(t))^{-1/p}dt<+\infty $ and $\widetilde{D}=D=(0,
E_{*}(\lambda ))$ otherwise.

Before going further in the investigation, we deduce from Lemma
\ref{Lemma5} that for any $p>1$ and $\lambda <0$,
\begin{gather}
E^{p}=p'F(\lambda , r(\lambda , E)),\text{\ for all }E\in
\widetilde{D}(\lambda )  \label{Eq11} \\
\frac{\partial r}{\partial E}(\lambda ,
E)=\frac{(p-1)E^{p-1}}{f(\lambda , r(\lambda , E))}>0,\text{\ for
all }E\in D(\lambda )
\nonumber\\
\lim_{E\to 0^{+}}r(\lambda , E)=\left\{
\begin{array}{ll}
((-\lambda )(\alpha /\beta ))^{1/(\alpha -\beta )} & \text{if }
\alpha
-\beta >0 \\
0 & \text{if }  \alpha -\beta <0\,.
\end{array}
\right.
\nonumber \\
\lim_{E\to +\infty }r(\lambda , E)=+\infty \text{ if }\alpha
-\beta
>0\,, \nonumber \\
\lim_{E\to E_{*}}r(\lambda , E)=(-\lambda )^{1/(\alpha -\beta
)}\text{ if }\alpha -\beta <0\,. \nonumber
\end{gather}
At present, we define for any $p>1$, $\lambda <0$ the time-map
\[
T(\lambda , E)=\int_{0}^{r(\lambda , E)}\{E^{p}-p'F(\lambda , \xi
)\}^{-1/p}d\xi , \;E\in \widetilde{D}(\lambda )\,.
\]
By (\ref{Eq11}), it follows that
\[
T(\lambda , E)=(p')^{-1/p}\int_{0}^{r(\lambda , E)}\{F(\lambda
, r(\lambda , E))-F(\lambda , \xi )\}^{-1/p}d\xi , \;E\in \widetilde{D}%
(\lambda )\,.
\]
Furthermore, a simple change of variable and a substitution yield
\[
T(\lambda , E)=(p')^{-1/p}\int_{0}^{1}\{\lambda r^{\beta
-p}(\lambda , E)(\frac{1-t^{\beta }}{\beta })+r^{\alpha -p}(\lambda , E)(%
\frac{1-t^{\alpha }}{\alpha })\}^{-1/p}dt.
\]
One may observe that
\[
T(\lambda , E)=S(\lambda , r(\lambda , E)),\quad\text{for all }\lambda <0%
\text{ and }E\in \widetilde{D}(\lambda ),
\]
where
\[
S(\lambda , \rho )=(p')^{-1/p}\int_{0}^{1}\{\lambda \rho ^{\beta
-p}(\lambda , E)(\frac{1-t^{\beta }}{\beta })+\rho ^{\alpha
-p}(\lambda , E)(\frac{1-t^{\alpha }}{\alpha })\}^{-1/p}dt\text{,}
\]
for all $\lambda <0$ and all $\rho \in R(\lambda )$ where, for
all $\lambda <0$, $R(\lambda )$ is the range of the function
$E\mapsto r(\lambda , E)$, defined on $\widetilde{D}(\lambda )$,
that is, for $\alpha -\beta >0$, $R(\lambda ):=(((-\lambda
)(\alpha /\beta ))^{1/(\alpha -\beta )}, +\infty )$, and for
$\alpha -\beta <0$, $R(\lambda ):=(0, (-\lambda )^{1/(\alpha
-\beta )}]$ if \\ $S(\lambda , r(\lambda , E_{*}(\lambda
))<+\infty $ and $R(\lambda ):=(0, (-\lambda )^{1/(\alpha -\beta
)})$ otherwise.

Due to the fact that the function $E\mapsto r(\lambda , E)\,$is an
increasing and continuous function from $\widetilde{D}(\lambda )$
onto $R(\lambda )$, it follows that if we put
\[
J_{1}(\lambda ):=\{E\in \widetilde{D}(\lambda ):T(\lambda , E)=\frac{1}{2}%
\}, \lambda <0,\text{ and }
\]
\[
J_{2}(\lambda ):=\{\rho \in R(\lambda ):S(\lambda , E)=\frac{1}{2}%
\}, \lambda <0,
\]
then,
\[
\text{Card}J_{1}(\lambda )=\text{Card}J_{2}(\lambda ),\text{\ for all }%
\lambda <0\,.
\]
Hence, from now on, we will focus our attention on counting the
number of solution(s) of the equation $S(\lambda , \rho )=1/2$ in
the variable $\rho \in R(\lambda )$ instead of the equation
$T(\lambda , E)=1/2$ in the variable $E\in \widetilde{D}(\lambda
)$.

\begin{lemma}
\label{Lemma6}Assume that $\lambda <0$. Then

\begin{description}
\item[(a)]  If $\alpha -\beta >0$,
\[
\lim\limits_{\rho \to ((-\lambda )(\alpha /\beta ))^{1/(\alpha
-\beta )}}S(\lambda , \rho )=\left\{
\begin{array}{ll}
(-\lambda )^{\frac{(p-\alpha )}{p(\alpha -\beta )}}J(p, \alpha ,
\beta ) &
\text{if }  1<\beta <p \\
+\infty & \text{if }  p\leq \beta .
\end{array}
\right.
\]
where $J(p, \alpha , \beta )$ is defined in (\ref{JmljmLerty}).

\item[(b)]  If $\alpha -\beta >0,
$\[ \lim\limits_{\rho \to +\infty }S(\lambda , \rho )=\left\{
\begin{array}{ll}
0 & \text{if }  p\leq \beta \\
\frac{1}{2}(\frac{\lambda _{1}}{p^{2}})^{1/p} & \text{if }
p>\beta \text{\
and\ }\alpha =p \\
0 & \text{if }  p>\beta \text{ and }\alpha >p \\
+\infty & \text{if } p>\beta \text{\ and }\alpha <p
\end{array}
\right.
\]

\item[(c)]  If $\alpha -\beta <0$,
\[
\lim\limits_{\rho \to 0}S(\lambda , \rho )=\left\{
\begin{array}{ll}
0 & \text{if } p>\alpha \\
\frac{1}{2}(\frac{\lambda _{1}}{p^{2}})^{1/p} & \text{if } p=\alpha \\
+\infty & \text{if }  p<\alpha
\end{array}
\right.
\]

\item[(d)]  If $\alpha -\beta <0$,
\[
\lim_{\rho \to (-\lambda )^{1/(\alpha -\beta )}}S(\lambda , \rho
)=(-\lambda )^{\frac{(p-\alpha )}{p(\alpha -\beta )}}\int_{0}^{1}\{(\frac{%
1-t^{\alpha }}{\alpha })-(\frac{1-t^{\beta }}{\beta })\}^{-1/p}dt
\]
and
\[
\int_{0}^{1}\{(\frac{1-t^{\alpha }}{\alpha })-(\frac{1-t^{\beta }}{\beta }%
)\}^{-1/p}dt<+\infty \Longleftrightarrow p>2.
\]
\end{description}
\end{lemma}

\paragraph{Proof.}
Assume that $\alpha -\beta >0$. Easy computation yields
\begin{eqnarray*}
\lim\limits_{\rho \to ((-\lambda )(\alpha /\beta ))^{1/(\alpha
-\beta )}}S(\lambda , \rho ) &=&(p')^{-1/p}(\frac{-\lambda }{\beta
})^{(p-\alpha )/p(\alpha -\beta )}\alpha ^{(p-\beta )/p(\alpha
-\beta
)}\times \\
&&\times \int_{0}^{1}t^{-\beta /p}(1-t^{\alpha -\beta })^{-1/p}dt.
\end{eqnarray*}
It can be shown that for all $p>1$ and $1<\beta <\alpha $\[
\int_{0}^{1}t^{-\beta /p}(1-t^{\alpha -\beta })^{-1/p}dt=\left\{
\begin{array}{ll}
+\infty & \text{if } \beta \geq p \\
\frac{1}{\alpha -\beta }B(\frac{(p-\beta )}{p(\alpha -\beta )}, \frac{p-1}{p%
}) & \text{if } 1<\beta <p.
\end{array}
\right.
\]
In fact, in the case where $\beta \geq p$, we use the estimates
\[
t^{-\beta /p}(1-t^{\alpha -\beta })^{-1/p}\geq t^{-\beta
/p},\text{\ for all }t\in (0, 1),
\]
and in the case where $1<\beta <p$, we use the change of variable
$x=t^{\alpha -\beta }$, as in Lavrentiev and Chabat \cite[pp.
595-596]{Lav}. Therefore Assertion \textbf{(a)} is proved.

Also, in the case where $\alpha -\beta >0$ and $\beta -p\geq 0
$\begin{eqnarray*}
\lim_{\rho \to +\infty }(\lambda \rho ^{\beta -p}(\frac{1-t^{\beta }%
}{\beta })+\rho ^{\alpha -p}(\frac{1-t^{\alpha }}{\alpha }))^{-1/p} &=& \\
\lim_{\rho \to +\infty }\rho ^{(p-\beta )/p}(\lambda (\frac{%
1-t^{\beta }}{\beta })+\rho ^{\alpha -\beta }(\frac{1-t^{\alpha }}{\alpha }%
))^{-1/p} &=&0\cdot 0=0\,.
\end{eqnarray*}
and an easy discussion shows that for $\alpha -\beta >0$,
\begin{eqnarray*}
\lefteqn{
\lim_{\rho \to +\infty }(\lambda \rho ^{\beta -p}(\frac{1-t^{\beta }%
}{\beta })+\rho ^{\alpha -p}(\frac{1-t^{\alpha }}{\alpha }))^{-1/p} }\\
&=&\left\{
\begin{array}{ll}
0 & \text{if } \beta \geq p \\
0 & \text{if } \beta <p\text{ and }\alpha >p \\
+\infty & \text{if } \beta <p\text{ and }\alpha <p \\
((1-t^{p})/p)^{-1/p} & \text{if } \beta <p\text{\ and }\alpha =p
\end{array}
\right.
\end{eqnarray*}
Therefore, Assertion \textbf{(b)} is proved.

In the case where $\alpha -\beta <0\,$and $\lambda <0$,
\begin{eqnarray*}
\lefteqn{ \lim_{\rho \to 0}\int_{0}^{1}(\lambda \rho ^{\beta
-p}((1-t^{\beta
})/\beta )+\rho ^{\alpha -p}((1-t^{\alpha })/\alpha ))^{-1/p}dt }\\
&=&\lim_{\rho \to 0}\rho ^{(p-\alpha )/p}\int_{0}^{1}(\lambda \rho
^{\beta -\alpha }(1-t^{\beta })/\beta +(1-t^{\alpha })/\alpha )^{-1/p}dt \\
&=&\lim_{\rho \to 0}\rho ^{(p-\alpha )/p}\int_{0}^{1}((1-t^{\alpha
})/\alpha )^{-1/p}dt\,.
\end{eqnarray*}
Notice that for all $\alpha >1 $\[
\int_{0}^{1}(1-t^{\alpha })^{-1/p}dt=\frac{1}{\alpha }B(\frac{1}{\alpha }, %
\frac{p-1}{p})<+\infty \,.
\]
This follows by making use of the change of variable $x=t^{\alpha
}$, see Lavrentiev and Chabat \cite[pp. 595-596]{Lav}. Assertion
\textbf{(c)} follows.

Also, in the case $\alpha -\beta <0$ and $\lambda <0$, easy
computation shows
\begin{eqnarray*}
\lim_{\rho \to (-\lambda )^{1/(\alpha -\beta )}}S(\lambda , \rho )
&=&(p')^{-1/p}(-\lambda )^{(p-\alpha )/p(\alpha -\beta )}\times \\
&&\int_{0}^{1}((1-t^{\alpha })/\alpha )-((1-t^{\beta })/\beta ))^{-1/p}dt%
\,.
\end{eqnarray*}
By making use of L'Hopital's rule two times, we compute
\[
\lim_{t\to 1^{-}}\frac{(1-t^{\alpha })/\alpha -(1-t^{\beta
})/\beta }{(1-t)^{2}}=\frac{\beta -\alpha }{2}>0\,.
\]
So, the integral $\int_{0}^{1}((1-t^{\alpha })/\alpha
)-((1-t^{\beta })/\beta ))^{-1/p}dt$ is convergent if and only if
the integral $\int_{0}^{1}(1-t)^{-2/p}dt$ does so. Therefore,
Assertion \textbf{(d)} follows from the well known fact that
\[
\int_{0}^{1}(1-t)^{-2/p}dt<+\infty \text{ if and only if }p>2\,.
\]
Therefore, Lemma \ref{Lemma6} is proved.


\begin{lemma}
\label{Lemma7}Assume that $p>1$, $\lambda <0$, and $\alpha \neq
\beta $, $\alpha , \beta >1$.

\begin{enumerate}
\item  If one of the following conditions holds:

\begin{description}
\item[(a)]  $\beta <p\leq \alpha $

\item[(b)]  $\beta =p<\alpha $

\item[(c)]  $p<\beta <\alpha $
\end{description}
\noindent then, $S(\lambda , \cdot )$ is strictly decreasing on
$R(\lambda )$.

\item  If one of the following conditions holds:

\begin{description}
\item[(a)]  $\alpha \leq p<\beta $

\item[(b)]  $\alpha <p=\beta $

\item[(c)]  $\alpha <\beta <p$
\end{description}
\noindent then, $S(\lambda , \cdot )$ is strictly increasing on
$R(\lambda )$.

\item  If one of the following conditions holds:

\begin{description}
\item[(a)]  $p<\alpha <\beta $

\item[(b)]  $\beta <\alpha <p<\alpha +\beta $
\end{description}
\noindent then, $S(\lambda , \cdot )$ is strictly decreasing on
$(\inf R(\lambda ), \rho _{1}]$ and is strictly increasing on
$[\rho _{2}, \sup R(\lambda )) $, where $\rho _{1}(\lambda
)\,$and $\rho _{2}(\lambda )$ are defined in Lemma \ref{Lemma3}.

\item  If $\beta <\alpha <p\,$and $\alpha +\beta \leq p$, then $S(\lambda
, \cdot )$ is strictly increasing on $[\rho _{2}, +\infty )$.
\end{enumerate}
\end{lemma}

This Lemma follows by a similar discussion as that in the proof
of Lemma \ref {Lemma3}. So, the proof of this Lemma is omitted.


Note that the third and the fourth assertions of Lemma
\ref{Lemma7} above do not provide the exact variations of the map
$S(\lambda , \cdot )$ over its entire definition domain, which
are necessary for the process of showing the exactness part in
the main result. They are the aim of the following pioneer lemma.

\begin{lemma}
\label{Lemma11*}Assume that one of the following conditions holds:

\begin{description}
\item[(c1)]  $1<p\leq 2$, and $p<\alpha <\beta ,$

\item[(c2)]  $2<p<\alpha <\beta ,$

\item[(c3)]  $1<\beta <\alpha <p$.
\end{description}
\noindent Then, for all $\lambda <0$, there exists an interior
point $\rho ^{*}(\lambda )\in int(R(\lambda ))$ such that
$S(\lambda , \cdot )$ is strictly decreasing on $(\inf R(\lambda
), \rho ^{*}(\lambda ))$ and then strictly increasing on $(\rho
^{*}(\lambda ), \sup R(\lambda ))$.
\end{lemma}

We shall prove Lemma \ref{Lemma11*} in two steps.

\paragraph{Step 1: Existence}
If
\begin{equation}
p<\alpha <\beta \text{ or }1<\beta <\alpha <p<\alpha +\beta
\label{68}
\end{equation}
the existence of $\rho ^{*}(\lambda )$ follows immediatly from
Lemma \ref {Lemma7}, Assertion 3.

Also, if
\begin{equation}
1<\beta <\alpha <\alpha +\beta \leq p,  \label{58}
\end{equation}
then, according to Lemma \ref{Lemma7}, Assertion 4, existence
follows after proving that for all $\lambda <0$, $S(\lambda ,
\cdot )$ is strictly decreasing on a right neighborhood of $\inf
R(\lambda )$. (Notice that in this case $\rho _{1}(\lambda )\leq
\inf R(\lambda )$). In fact, we shall prove:

\begin{lemma}
\label{Lemma22*}If $1<\beta <\alpha <\alpha +\beta \leq p$\ and
$\lambda <0$ then
\[
\frac{\partial S}{\partial \rho }(\lambda , (\frac{-\lambda \alpha }{\beta }%
)^{1/(\alpha -\beta )})=-\infty .
\]
\end{lemma}

\paragraph{Proof.}
The derivative of $S(\lambda , \cdot )$ is given by
\[
\frac{\partial S}{\partial \rho }(\lambda , \rho )=(p^{\prime
})^{-1/p}\int_{0}^{1}\frac{H(\lambda , \rho )-H(\lambda , \rho
u)}{p\rho (F(\lambda , \rho )-F(\lambda , \rho
u))^{1+\frac{1}{p}}}du.
\]
where
\[
F(\lambda , \rho )=\frac{1}{\alpha }\rho ^{\alpha }+\frac{\lambda }{\beta }%
\rho ^{\beta }, \rho >0\quad \text{and\ \ }H(\lambda , \rho )=(\frac{%
p-\alpha }{\alpha })\rho ^{\alpha }+\lambda (\frac{p-\beta
}{\beta })\rho ^{\beta }, \rho >0.
\]
Simple computations yield
\begin{eqnarray*}
\frac{\partial S}{\partial \rho }(\lambda , (-\lambda \frac{\alpha }{\beta }%
)^{1/(\alpha -\beta )}) &=&\frac{(p')^{-1/p}}{p}(\frac{\beta }{%
-\lambda })^{\frac{p+\alpha }{p(\alpha -\beta )}}\alpha ^{\frac{p+\beta }{%
p(\beta -\alpha )}} \\
&&\times \int_{0}^{1}\dfrac{(p-\alpha )(1-u^{\alpha })-(p-\beta
)(1-u^{\beta })}{(u^{\beta }-u^{\alpha })^{1+\frac{1}{p}}}du.
\end{eqnarray*}
The improper integral has two singularities; at $0$ and at $1$.
Then we write
\[
\frac{\partial S}{\partial \rho }(-1, (-\lambda \frac{\alpha }{\beta }%
)^{1/(\alpha -\beta )})=\frac{(p')^{-1/p}}{p}(\frac{\beta }{%
-\lambda })^{\frac{p+\alpha }{p(\alpha -\beta )}}\alpha ^{\frac{p+\beta }{%
p(\beta -\alpha )}}(I_{0}+I_{1})
\]
where
\[
I_{0}=\int_{0}^{1/2}\frac{(p-\alpha )(1-u^{\alpha })-(p-\beta )(1-u^{\beta })%
}{(u^{\beta }-u^{\alpha })^{1+\frac{1}{p}}}du,
\]
\[
I_{1}=\int_{1/2}^{1}\frac{(p-\alpha )(1-u^{\alpha })-(p-\beta )(1-u^{\beta })%
}{(u^{\beta }-u^{\alpha })^{1+\frac{1}{p}}}du.
\]
In what follows we shall prove that $I_{0}=-\infty $\ and
$I_{1}\in \mathbb{R}$.

First, we write $I_{0}$ as follows:
\[
I_{0}=\int_{0}^{1/2}\frac{-(\alpha -\beta )-[(p-\alpha )u^{\alpha
}-(p-\beta
)u^{\beta }]}{(1-u^{\alpha -\beta })^{1+\frac{1}{p}}u^{\beta (1+\frac{1}{p})}%
}du.
\]
One may observe that in a right neighborhood of $0$,
\[
\frac{-(\alpha -\beta )-[(p-\alpha )u^{\alpha }-(p-\beta )u^{\beta }]}{%
(1-u^{\alpha -\beta })^{1+\frac{1}{p}}u^{\beta (1+\frac{1}{p})}}\simeq \frac{%
-(\alpha -\beta )}{u^{\beta (1+\frac{1}{p})}},
\]
and by (\ref{58}), it follows that $\beta (1+\frac{1}{p})>1$ and
$-(\alpha -\beta )<0$, so
\[
\int_{0}^{1/2}\frac{-(\alpha -\beta )}{u^{\beta
(1+\frac{1}{p})}}du=-\infty ,
\]
and therefore $I_{0}=-\infty $.

Next, we write $I_{1}$ as follows:
\[
I_{1}=\int_{1/2}^{1}\frac{(p-\alpha )(1-u^{\alpha })-(p-\beta )(1-u^{\beta })%
}{u^{\beta (1+\frac{1}{p})}(\frac{1-u^{\alpha -\beta }}{1-u})^{1+\frac{1}{p}%
}(1-u)^{1+\frac{1}{p}}}du\,.
\]
Applying Taylor's theorem to the function $N$, defined by
\[
N(u)=(p-\alpha )(1-u^{\alpha })-(p-\beta )(1-u^{\beta }),
\]
it follows that
\begin{eqnarray*}
I_{1} &=&\int_{1/2}^{1}\frac{[(p-\alpha )\alpha -(p-\beta )\beta ](1-u)}{%
u^{\beta (1+\frac{1}{p})}(\frac{1-u^{\alpha -\beta }}{1-u})^{1+\frac{1}{p}%
}(1-u)^{1+\frac{1}{p}}}du \\
&&+\int_{1/2}^{1}\frac{-\frac{1}{2}[(p-\alpha )\alpha (\alpha
-1)-(p-\beta
)\beta (\beta -1)](1-u)^{2}+o((1-u)^{2})}{u^{\beta (1+\frac{1}{p})}(\frac{%
1-u^{\alpha -\beta
}}{1-u})^{1+\frac{1}{p}}(1-u)^{1+\frac{1}{p}}}du.
\end{eqnarray*}
Notice that $\lim_{u\to 1^{-}}\frac{1-u^{\alpha -\beta }}{1-u}%
=(\alpha -\beta )\in \mathbb{R}\mathbf{\,}$\ then, in a left
neighborhood of $1 $,
\[
u^{\beta (1+\frac{1}{p})}(\frac{1-u^{\alpha -\beta }}{1-u})^{1+\frac{1}{p}%
}(1-u)^{1+\frac{1}{p}}\simeq (\alpha -\beta
)(1-u)^{1+\frac{1}{p}}.
\]
Next, one has to distinguish two cases.

Case $[(p-\alpha )\alpha -(p-\beta )\beta ]\neq 0$. In this case
the integrand function in $I_{1}$ is equivalent in a left
neighborhood of $1$ to the function $u\mapsto \frac{(p-\alpha
)\alpha -(p-\beta )\beta }{(\alpha
-\beta )(1-u)^{1/p}}$ and since $p>1$ it follows that $\int_{0}^{1/2}\frac{%
(p-\alpha )\alpha -(p-\beta )\beta }{(\alpha -\beta
)(1-u)^{1/p}}du\in \mathbb{R}$ and therefore $I_{1}\in
\mathbb{R}$.

Case $[(p-\alpha )\alpha -(p-\beta )\beta ]=0$. In this case the
integrand function in $I_{1}$ is equivalent in a left
neighborhood of $1$ to the function $u\mapsto
(-\frac{1}{2})(p-\alpha )\alpha (1-u)^{1-\frac{1}{p}}$\ which is
a continuous function on the compact interval $[\frac{1}{2}, 1]$
then $\int_{1/2}^{1}(-\frac{1}{2})(p-\alpha )\alpha (1-u)^{1-\frac{1}{p}%
}du\in \mathbb{R}$. Then, in this case $I_{1}\in \mathbb{R}$ too.
Therefore Lemma \ref{Lemma22*} is proved.


\paragraph{Step2: Uniqueness}
First, we point out that Step 1 shows a little bit general result
than the existence. In fact, for all $\lambda <0$, it was proved
that $\rho ^{*}(\lambda )$ exists and belongs necessarily to
$(\rho _{1}(\lambda ), \rho _{2}(\lambda ))$ if (\ref{68}) holds
and belongs to $(\inf R(\lambda ), \rho _{2}(\lambda ))$ if
(\ref{58}) holds. So, to prove uniqueness, we shall restrict
ourselves to $(\rho _{1}(\lambda ), \rho _{2}(\lambda ))$ (resp.
to $(\inf R(\lambda ), \rho _{2}(\lambda ))$). That is, we shall
prove that $S(\lambda , \cdot )$ admits \emph{at most one}
critical point in $(\rho _{1}(\lambda ), \rho _{2}(\lambda ))$
(resp. in $(\inf R(\lambda ), \rho _{2}(\lambda ))$). To this end
we shall prove that for all $\lambda <0$, $S(\lambda , \cdot )$
is convex in a neighborhood of each of its critical points lying
in $(\rho _{1}(\lambda ), \rho _{2}(\lambda ))$ (resp. in $(\inf
R(\lambda ), \rho _{2}(\lambda ))$). Similar idea was previously
used in \cite{Addou5}.

This step follows by two lemmas. The first one is technical but
the second one is the heart of this step.

\begin{lemma}
\label{Lemma1*}Let $p, \alpha , \beta >1$. If $\alpha \neq \beta $, then $(%
\frac{\beta }{\alpha })^{1/(\alpha -\beta )}<1$. If one of the
following conditions holds: (a) $p<\alpha <\beta $ or (b) $\beta
<\alpha <p$, then for all $\lambda <0$, the function defined on
the interval $[0, \rho _{2}(\lambda )]$ by $\rho \mapsto \Psi
(\lambda , \rho ):=(p-\alpha )\rho ^{\alpha }+\lambda (p-\beta
)\rho ^{\beta }$, is strictly decreasing on $[0, \rho
_{1}(\lambda )(\frac{\beta }{\alpha })^{1/(\alpha -\beta )}]$ and
is strictly increasing on $[\rho _{1}(\lambda )(\frac{\beta }{\alpha }%
)^{1/(\alpha -\beta )}, \rho _{2}(\lambda )]$. Moreover, $\Psi
(\lambda , 0)=\Psi (\lambda , \rho _{1}(\lambda ))=0$, for all
$\lambda <0$.
\end{lemma}

The proof is very simple and therefore omitted. For all $\lambda
<0$,\ let $\rho _{*}(\lambda ):=\max \{\rho _{1}(\lambda ), \inf
R(\lambda )\}$.

\begin{lemma}
\label{Lemma2*}Let $p, \alpha , \beta >1$. Assume that one of the
following conditions holds: (a) $p<\alpha <\beta $, or (b) $\beta
<\alpha <p $. Then for all $\lambda <0, $\[ \frac{\partial
^{2}S}{\partial \rho ^{2}}(\lambda , \rho )+(\frac{p+1}{\rho
})\frac{\partial S}{\partial \rho }(\lambda , \rho )>0,\text{\ for all }%
\rho \in (\rho _{*}(\lambda ), \rho _{2}(\lambda ))\,.
\]
\end{lemma}

\paragraph{Proof.}
Notice that for all $\lambda <0$, $(\rho _{*}(\lambda ), \rho
_{2}(\lambda ))\subset R(\lambda )=:domS(\lambda , \cdot )$. The
second derivative of $S(\lambda , \cdot )$ is given by
\begin{eqnarray*}
\frac{\partial ^{2}S}{\partial \rho ^{2}}(\lambda , \rho )
&=&(p^{\prime })^{-1/p}\int_{0}^{1}\frac{(p+1)(H(\lambda , \rho
)-H(\lambda , \rho u))^{2}}{p^{2}\rho (F(\lambda , \rho
)-F(\lambda , \rho u))^{(2p+1)/p}}du
\\
&&+(p')^{-1/p}\int_{0}^{1}\frac{\Phi (\lambda , \rho )-\Phi
(\lambda , \rho u)}{p\rho (F(\lambda , \rho )-F(\lambda , \rho
u))^{(p+1)/p}}du\text{,}
\end{eqnarray*}
where
\begin{eqnarray*}
\Phi (\lambda , \rho ) &=&-p(p+1)F(\lambda , \rho )+2p\rho
f(\lambda
, \rho )-\rho ^{2}f_{\rho }'(\lambda , \rho ) \\
&=&\frac{(p-\alpha )(\alpha -(p+1))}{\alpha }\rho ^{a}+\lambda \frac{%
(p-\beta )(\beta -(p+1))}{\beta }\rho ^{\beta }\,.
\end{eqnarray*}
It follows that
\begin{eqnarray*}
\lefteqn{ (p')^{1/p}p\rho \{\rho \frac{\partial ^{2}S}{\partial
\rho ^{2}} (\lambda , \rho )+(p+1)\frac{\partial S}{\partial \rho
}(\lambda , \rho
)\} }\\
&=&
\int_{0}^{\rho }\frac{\Psi (\lambda , \rho )-\Psi (\lambda , \xi )}{%
(F(\lambda , \rho )-F(\lambda , \xi ))^{(p+1)/p}}d\xi  \\
&&+(\frac{p+1}{p}%
)\int_{0}^{\rho }\frac{(H(\lambda , \rho )-H(\lambda , \xi ))^{2}}{%
(F(\lambda , \rho )-F(\lambda , \xi ))^{(2p+1)/p}}du\,,
\end{eqnarray*}
where
\[
\Psi (\lambda , \rho ):=\Phi (\lambda , \rho )+(p+1)H(\lambda ,
\rho )=\rho \frac{\partial H}{\partial \rho }(\lambda , \rho
)=(p-\alpha )\rho ^{\alpha }+\lambda (p-\beta )\rho ^{\beta }\,.
\]
By Lemma \ref{Lemma1*}, it follows that for all $\lambda <0$ and
all $\rho \in (\rho _{*}(\lambda ), \rho _{2}(\lambda ))$,
\[
\Psi (\lambda , \rho )-\Psi (\lambda , \xi )>0,\text{\ for all
}\xi \in (0, \rho )\,.
\]
Therefore
\[
\int_{0}^{\rho }\frac{\Psi (\lambda , \rho )-\Psi (\lambda , \xi )}{%
(F(\lambda , \rho )-F(\lambda , \xi ))^{(p+1)/p}}d\xi >0,\text{ for all }%
\lambda <0\text{ and all }\rho \in (\rho _{*}(\lambda ), \rho
_{2}(\lambda ))
\]
and Lemma \ref{Lemma2*} is proved, which ends the proof of Lemma
\ref {Lemma11*}.  \hfill$\diamondsuit$ \medskip

 Notice that if one of the hypothesis of Assertions A through F
holds then $S(\lambda , \cdot )$ is monotonic, the proofs follow
by an elementary discussion as in Assertion A or B of Theorem
\ref{Theorem1}. Therefore the proofs of Assertions A, through F
are omitted.

Concerning the remaining assertions, the same ideas performed for
Assertion C of Theorem \ref{Theorem1} apply. For this, it
suffices to use Lemma \ref {Lemma11*} and the following

\begin{lemma}
\label{Lemma11**}By Lemma \ref{Lemma11*} let $m(\lambda
):=\inf_{\rho \in R(\lambda )}S(\lambda , \rho )$,\ $\forall
\lambda <0$. Then,

\begin{description}
\item[(a)]  $m(\cdot )\,$is continuous on $(-\infty , 0)$

\item[(b)]  $m(\cdot )$ is strictly decreasing on $(-\infty , 0)$

\item[(c)]  $\lim_{\lambda \to -\infty }m(\lambda )=+\infty $ and
$\lim_{\lambda \to 0}m(\lambda )=0$.
\end{description}
\end{lemma}

\begin{lemma}
\label{Lemma11***}The function $\lambda \mapsto \ell (\lambda
):=\lim_{\rho \mapsto \inf R(\lambda )}S(\lambda , \rho )\,$is
either infinite on the whole set $(-\infty , 0)$ or satisfies
Assertions (a), (b), and (c) of Lemma \ref{Lemma11**}.
\end{lemma}

\begin{lemma}
\label{Lemma4*}The function $\lambda \mapsto L(\lambda
):=\lim_{\rho \mapsto \sup R(\lambda )}S(\lambda , \rho )\,$is
(independently of $\ell (\lambda ) $) either infinite on the
whole set $(-\infty , 0)$ or satisfies Assertions (a), (b), and
(c) of Lemma \ref{Lemma11**}.
\end{lemma}

\paragraph{Proof of Lemma \ref{Lemma11**}.} Recall that for all $\lambda <0$
and $\rho \in R(\lambda ), $\[ S(\lambda , \rho
)=(p')^{-1/p}\int_{0}^{1}\{\lambda \rho ^{\beta
-p}(\frac{1-t^{\beta }}{\beta })+\rho ^{\alpha -p}(\frac{1-t^{\alpha }}{%
\alpha })\}^{-1/p}dt\,.
\]
For all $\lambda <0$ and $\rho \in R(\lambda )$, let $\bar{\rho}=\bar{\rho}%
(\lambda , \rho ):=(-\lambda )^{1/(\beta -\alpha )}\rho $. Then,
$\rho =(-\lambda )^{1/(\alpha -\beta )}\bar{\rho}$ and a simple
substitution yields
\[
S(\lambda , \rho )=(-\lambda )^{(p-\alpha )/p(\alpha -\beta )}S(-1, \bar{%
\rho}(\lambda , \rho ))\,.
\]
Thus,
\begin{eqnarray*}
m(\lambda ) & = & (-\lambda )^{(p-\alpha )/p(\alpha -\beta
)}\inf\limits_{\rho \in R(\lambda )}S(-1, \bar{\rho}(\lambda , \rho )) \\
& = & (-\lambda )^{(p-\alpha )/p(\alpha -\beta
)}\inf\limits_{\rho \in R(-1)}S(-1, \rho )
\end{eqnarray*}
So,
\[
m(\lambda )=(-\lambda )^{(p-\alpha )/p(\alpha -\beta )}m(-1),
\forall \lambda <0\,.
\]
Therefore, Lemma \ref{Lemma11**} is proved. \hfill$\diamondsuit$

\paragraph{Proof of Lemma \ref{Lemma11***}.} By Lemma \ref{Lemma6} it follows
in the case where $1<\beta <\alpha <\alpha +\beta \leq p$ that
$\lim_{\rho \to \inf R(\lambda )}S(\lambda , \rho )=(-\lambda
)^{(p-\alpha )/p(\alpha -\beta )}J(p, \alpha , \beta )<+\infty $,
where $J(p, \alpha , \beta )$ is defined in (\ref{JmljmLerty}).
Hence, Lemma \ref{Lemma11***} is proved in this case. The other
cases are similar or easier.

The proof of Lemma \ref{Lemma4*} is similar to that of Lemma
\ref{Lemma11***}.


\section{Remarks}

By the study of Problem (\ref{P}) with $\lambda \in \mathbb{R}$,
it can be deduced the structure of the solution set of a quite
general problem
\begin{gather}
-(\varphi _{p}(u'))'  =  \mu \varphi _{\alpha
}(u)+\lambda \varphi _{\beta }(u)\quad\text{in }(0, 1) \label{Madison} \\
u(0)  =  u(1)=0\,, \nonumber
\end{gather}
when $p, \alpha , \beta >1$, $\mu >0$ and $\lambda \in
\mathbb{R}$. In fact, if $v$ is a solution of (\ref{P}) for
$\lambda =\lambda _{0}\in \mathbb{R}$ and $\alpha \neq p$, then
for all $\mu _{0}>0$, the function $u:=\mu _{0}^{1/(p-\alpha )}v$
is a solution to (\ref{Madison}) with $\mu =\mu _{0}$ and
$\lambda =\lambda _{0}\mu _{0}^{(\beta -p)/(\alpha -p)}$. (If
$\alpha =p\,$ and $\beta \neq p$, similar change of variable
works). Conversely, if $u $ is a solution of (\ref{Madison}) with
$\mu =\mu _{0}>0$, $\lambda =\lambda _{0}\in \mathbb{R\,}$ and
$\alpha \neq p$, then $v:=\mu _{0}^{1/(\alpha -p)}u$ is a
solution to (\ref{P}) with $\lambda =\lambda _{0}\mu _{0}^{(\beta
-p)/(\alpha -p)}$.

The structure of the solution set of problem (\ref{Madison}) when
$\mu <0$ and $\lambda \in \mathbb{R}$ can be deduced from that of
the problem
\begin{gather*}
-(\varphi _{p}(u'))'  =  -\varphi _{\alpha }(u)+\lambda
\varphi _{\beta }(u)\quad\text{in\ }(0, 1) \\
u(0)  =  u(1)=0,
\end{gather*}
which is not treated here. However, upon completing our paper,
the work \cite {Diaz} by D\'{i}az and Hern\'{a}ndez appeared.
Positive solutions to problem (\ref{Madison}) with $\mu <0$,
$\lambda >0$ and $1<\alpha <\beta \leq p$ are treated there.

After completing this work, e-mail correspondence between the
first author (I. Addou) and Professor Pedro Ubilla from Chile,
revels that simultaneously and independently of the present
authors, Professors J. S\'{a}nchez and P. Ubilla from Chile, were
studying problem (\ref{P}) with $\lambda >0$. That is to say,
they resolved the p-Laplacian version of the
Ambrosetti-Brezis-Cerami problem. To do so, they provide
essentially the same proof as that of Theorem \ref{Theorem1}
above by making use of the same idea performed in \cite[Lemma 7,
(iii)]{Addou3}. Their work was presented, under the title: ''The
exact number of positive solutions for an elliptic equation with
concave and convex nonlinearities'' by Professor P. Ubilla at the
''USA-Chile Workshop on Nonlinear Analysis'' meeting which held in
Valpara\'{i}so in Chile on 17-21, January 2000. Also, it was
published in this volume of this journal, see \cite{Ubilla}.

Also, after submitting this work for publication, e-mail
correspondence between the first author (I. Addou) and Professor
Shin-Hwa Wang from Taiwan (R. O. China) revels that he has write
(independently of J. Sanchez and P. Ubilla and independently of
the present authors) a paper \cite{Wang2} in which he resolves
(for $p=2$) the Ambrosetti-Brezis-Cerami problem \cite[Sect. 6,
(d)]{Ambrosetti1} (among many other interesting things), by
making use of the quadrature technique. To deal with the
difficult step (uniqueness of the maximum of the time map), he
used an interesting argument which is comparable to that of
\cite[Lemma 7, (iii)]{Addou3} and used by him previously in
\cite[Proof of Theorem 7]{Wang1}. (See, also \cite
{SmollerWasserman}).

\paragraph{Acknowledgments} The author I. Addou is deeply grateful to
Professors P. Ubilla (from Chile) and S.-H. Wang (from Taiwan,
R.O.C.) for the interesting and fruitful e-mail correspondence he
had with them and for sending him some of their works. Also, the
authors would like to thank the anonymous referee for his/her
relevant suggestions.

\section{Appendix}

In the process of our proofs, we have used the fact that
\[
\lambda _{1}(p)=(p-1)(\frac{2\pi }{p\sin (\pi /p)})^{p}>1, \forall p>1\text{%
.}
\]
In this appendix we shall prove the following:

\begin{description}
\item[(A1)]  $\lambda _{1}(p)>1, \forall p\in (1, 2)$.

\item[(A2)]  $\lambda _{1}(2)=\pi ^{2}$.

\item[(A3)]  $\lambda _{1}(p)>4, \forall p\in (2, +\infty )$.

\item[(A4)]  $\lim_{p\to 1^{+}}\lambda _{1}(p)=2$,\ and\
$\lim_{p\to +\infty }\lambda _{1}(p)=+\infty $.
\end{description}

\noindent\textbf{Proof of (A1).} Observe that for all $p\in (1,
2)$, $\lambda _{1}(p)\,$can be written as
\begin{equation}
\lambda _{1}(p)=(\frac{2}{p})^{p}(p-1)^{(1-p)}(\frac{\pi
}{\frac{\sin (\pi /p)-\sin (\pi /1)}{p-1}})^{p}\,.  \label{D}
\end{equation}
The function $p\mapsto \theta (p):=(\frac{2}{p})^{p}$ is strictly
decreasing on $(1, 2]$ and $\theta (2)=1$. Thus,
\begin{equation}
(\frac{2}{p})^{p}>1,\text{\ }\;\forall p\in (1, 2).  \label{A}
\end{equation}
The function $p\mapsto K(p):=(p-1)^{(1-p)}$ is strictly
increasing on $(1, 1+\exp (-1)]$ and is strictly decreasing on
$[1+\exp (-1), 2)$. Thus, $K(p)>\min \{K(2)$, \\ $\lim_{p\to
1^{+}}K(p)\}=1$, for all $p\in (1, 2)$. Therefore,
\begin{equation}
(p-1)^{(1-p)}>1, \forall p\in (1, 2).  \label{B}
\end{equation}
Notice that for all $c\in (1, 2):c^{2}>1>-\cos (\pi /c)>0$. Then,
$\frac{\pi }{-(\pi ^{2}/c)\cos (\pi /c)}>1$, for all $c\in (1,
2)$. Therefore,
\begin{equation}
(\frac{\pi }{-(\pi ^{2}/c)\cos (\pi /c)})^{p}>1,\;\forall (p,
c)\in (1, 2)^{2}\,.  \label{California}
\end{equation}
On the other hand, for all $p\in (1, 2)$, there exists $c=c_{p}\in
(1, p)\subset (1, 2)$, such that
\[
\frac{\sin (\pi /p)-\sin (\pi /1)}{p-1}=-\frac{\pi }{c^{2}}\cos \frac{\pi }{c%
}\,.
\]
Therefore, for all $p\in (1, 2)$, there exists $c=c_{p}\in (1,
p)\subset (1, 2)$, such that
\[
(\frac{\pi }{\frac{\sin (\pi /p)-\sin (\pi /1)}{p-1}})^{p}=(\frac{\pi }{%
-(\pi ^{2}/c)\cos (\pi /c)})^{p}
\]
and by (\ref{California}), it follows that
\begin{equation}
(\frac{\pi }{\frac{\sin (\pi /p)-\sin (\pi /1)}{p-1}})^{p}>1,\text{\ }%
\forall p\in (1, 2)_{.}  \label{C}
\end{equation}
Now, by (\ref{A}),\ (\ref{B}), (\ref{C}), and (\ref{D}), Assertion \textbf{%
(A1)} follows.

\noindent\textbf{Proof of (A2).} Simple computation.

\noindent\textbf{Proof of (A3).} Observe that for all $p>2$,
$\lambda _{1}(p)$ can be written as
\begin{equation}
\lambda _{1}(p)=(p-1)\cdot 2^{p}\cdot (\frac{\sin (\pi /p)-\sin
(0)}{(\pi /p)-0})^{-p}\,.  \label{Z}
\end{equation}
It is clear that for all $p>2 $\begin{equation} (p-1)>1\text{ and
}2^{p}>2^{2}=4\,.  \label{X}
\end{equation}
On the other hand, for all $p>2$, there exists $c=c_{p}$ such that
\[
0<\frac{\sin (\pi /p)-\sin (0)}{(\pi /p)-0}<\cos c<1\,.
\]
Therefore,
\begin{equation}
(\frac{\sin (\pi /p)-\sin (0)}{(\pi /p)-0})^{-p}>1, \forall p>2\,.
\label{Y}
\end{equation}
Now, by (\ref{X}), (\ref{Y}), and (\ref{Z}), Assertion
\textbf{(A3)} follows.

\noindent\textbf{Proof of (A4).} It is clear that
\[
\lim_{p\to 1^{+}}\theta (p)=2,\text{\ }\lim_{p\to
1^{+}}K(p)=1,\text{\ }\lim_{p\to 1^{+}}\pi ^{p}=\pi \text{, and}
\]
\[
\lim_{p\to 1}(\frac{\sin (\pi /p)-\sin (0)}{p-1})^{p}=(-\frac{\pi }{%
1^{2}}\cos \frac{\pi }{1})^{1}=\pi \,.
\]
Therefore, using expression (\ref{D}) of $\lambda _{1}(p)$, it
follows that $\lim_{p\to 1^{+}}\lambda _{1}(p)=2$. The
computation of the second limit is straightforward, which
completes the proof of Assertion \textbf{(A4)}.

\begin{thebibliography}{99}

\bibitem{Addou1}  \textsc{Addou, I. S. M. Bouguima, M. Derhab \& Y.\ S.\
Raffed},\textit{\ On the number of solutions of a quasilinear
elliptic class of B.V.P. with jumping nonlinearities,} Dynamic
Syst. Appl. \textbf{7} (4) (1998), pp. 575-599.

\bibitem{Addou2}  \textsc{Addou, I. \& A. Benmeza\"\i}\textit{, Exact
number of positive solutions for a class of quasilinear boundary
value problems,} Dynamic Syst. Appl. \textbf{8} (1999), pp.
147-180.

\bibitem{Addou3}  \textsc{Addou, I. \& A. Benmeza\"\i}\textit{, Boundary
value problems for the one dimensional p-Laplacian with even
superlinearity,} Electron. J. Diff. Eqns. \textbf{1999} (1999),
No. 09, pp. 1-29.

\bibitem{Addou4}  \textsc{Addou, I.,}\textit{\ Multiplicity of solutions for
quasilinear elliptic boundary value problems,} Electron. J. Diff.
Eqns. \textbf{1999} (1999), No. 21, pp. 1-27.

\bibitem{Addou5}  \textsc{Addou, I.,}\textit{\ Exact multiplicity results
for quasilinear boundary value problems with cubic-like
nonlinearities,} Electron. J. Diff. Eqns. \textbf{2000} (2000),
No. 01, pp. 1-26. Addendum pp. 27-29.

\bibitem{Addou6}  \textsc{Addou, I.,}\textit{\ Multiplicity results for
classes of one-dimensional p-Laplacian boundary-value problems
with cubic-like nonlinearities,} Electron. J. Diff. Eqns.
\textbf{2000} (2000), No. 52, pp. 1-42.

\bibitem{Ambrosetti1}  \textsc{Ambrosetti, A.,\ H. Brezis, and G. Cerami,}%
\textit{\ Combined effects of concave and convex nonlinearities
in some elliptic problems,} J. Funct. Anal. \textbf{122} (1994),
pp. 519-543.

\bibitem{Ambrosetti2}  \textsc{Ambrosetti, A., J. G.\ Azorero, and I.\ Peral}%
, \textit{Quasilinear equations with a multiple bifurcation,
}Diff. Integral Equations, \textbf{10} (1997), 37-50.

\bibitem{Ambrosetti3}  \textsc{Ambrosetti, A., J. G.\ Azorero, and I.\ Peral}%
, \textit{Multiplicity results for some nonlinear elliptic
equations}, J.\ Funct. Anal.. \textbf{137} (1996), pp. 219-242.

\bibitem{Bartsch}  \textsc{Bartsch, T., and M. Willem,}\textit{\ On an
elliptic equations with concave and convex nonlinearities}, Proc.
Amer. Math. Soc. \textbf{123} (1995), 3555-3561.

\bibitem{Diaz}  \textsc{D\'{i}az, J. I., and J. Hern\'{a}ndez}\textit{,
Global bifurcation and continua of nonnegative solutions for a
quasilinear elliptic problem,} C. R. A. S. Paris, \textbf{t.
329}, S\'{e}rie I, (1999), pp. 587-592.

\bibitem{GueddaVeron}  \textsc{Guedda, M., \& L. Veron,}\textit{\
Bifurcation phenomena associated to the p-Laplace operator},
Trans. Amer. Math. Soc. \textbf{310} (1988), pp. 419-431.

\bibitem{Lav}  \textsc{Lavrentiev, M., \& B. Chabat}\textit{, }%
''M\'{e}thodes de la th\'{e}orie des fonctions d'une variable
complexe'', Edition MIR, Moscou, 2e \'{e}dition, 1977.

\bibitem{OS}  \textsc{Ouyang, T., and J. Shi,}\textit{\ Exact multiplicity
of solutions and global bifurcation of }$\Delta u+\lambda f(u)=0$,
Proceeding of US-Chinese Conference on Differential Equations and
Applications, Hangzhou, China, 1996, pp. 356-363.

\bibitem{Ubilla}  \textsc{Sanchez, J. \& P. Ubilla,}\textit{,
One-dimensional elliptic equation with concave and convex
nonlinearities}, Electron. J. Diff. Eqns. \textbf{2000} (2000),
No. 50, pp. 1-9.

\bibitem{SmollerWasserman}  \textsc{Smoller, J. A. \& A. G. Wasserman, }%
\textit{Global bifurcation of Steady-State solutions,} J. Diff.
Eqns. \textbf{39} (1981),\ pp. 269-290.

\bibitem{Villegas}  \textsc{Villegas, S.,}\textit{\ On the structure of the
solutions of a BVP with concave and convex nonlinearity, }Proc.
of the International Conference on Differential Equations, held
in Lisbon (Portugal) in 24-29 July, 1995, Word Scientific, pp.
541-545.

\bibitem{Wang1}  \textsc{Wang, S.-H., }\textit{Bifurcation of positive
solutions of generalized nonlinear undamped pendulum problems},
Bulletin Institut Math. Acad. Sinica \textbf{21}, No. 3 (1993),
pp. 211-227.

\bibitem{Wang2}  \textsc{Wang, S.-H.,} \textit{On the bifurcation curve of
positive solutions of some boundary value problems}, submitted
(September 1999).
\end{thebibliography}

\noindent{\sc Idris Addou} \\
119-8230, Rue Sherbrooke Est,\\
Montr\'{e}al, Qu\'{e}bec, H1L-1A9,  Canada \\
email: idrisaddou@yahoo.com
\smallskip

\noindent{\sc Abdelhamid Benmeza\"\i} \\
U.S.T.H.B.-Institut de Mathematiques\\
El-Alia, BP 32, Bab-Ezzouar\\
16111, Alger, Algeria\\
email: abenmzai@hotmail.com
\smallskip

\noindent{\sc Sidi Mohammed Bouguima }\\
Department of Mathematics, Faculty of sciences\\
University of Tlemcen\\
B.P.119, Tlemcen 13000, Algeria \\
email: bouguima@yahoo.fr
\smallskip

\noindent{\sc Mohammed Derhab} \\
Institut des Sciences Exactes,
Universit\'{e} Djilali Liabes\\
BP. 89, Sidi Bel Abbes\\
22000, Algeria\\
email: derhab@yahoo.fr

\end{document}