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\markboth{\hfil Multiplicity of solutions for boundary-value problems\hfil 
EJDE--1999/21} {EJDE--1999/21\hfil Idris Addou  \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 1999}(1999), No.~21, pp. 1--27. \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} \\
%
 Multiplicity of solutions for quasilinear elliptic boundary-value problems
\thanks{ {\em 1991 Mathematics Subject Classifications:} 34B15.
\hfil\break\indent
{\em Key words and phrases:} Multiple solutions, elliptic boundary-value problems. 
\hfil\break\indent
\copyright 1999 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted January 21, 1999. Published June 16, 1999.} }
\date{}
%
\author{Idris Addou }
\maketitle

\begin{abstract} 
This paper is concerned with the existence of multiple solutions to the
boundary-value problem
  $$-(\varphi_p(u') ) '=\lambda \varphi_q(u) +f(u)\quad\mbox{in }
  (0,1)\,,\quad u(0) =u(1) =0\,,$$
where $p,q>1$,  $\varphi_x(y) =|y|^{x-2}y$, 
$\lambda $ is a real  parameter, and $f$ is a function which may be  
sublinear, superlinear, or asymmetric. 
We use the time map method for showing the existence of solutions. 
\end{abstract} 

\newtheorem{theorem}{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction} \label{sec1}
We study the existence and multiplicity of solutions for the boundary-value 
problem
\begin{eqnarray}  \label{p1}
&-(\varphi_p(u') ) '=\lambda \varphi_q(u) +f(u) \quad\mbox{in }(0,1)&\\  
&u(0) =u(1) =0\,,&  \nonumber
\end{eqnarray} 
where $p,q>1$, $\varphi_x(y) =| y| ^{x-2}y$, 
$\lambda \in {\mathbb R}$,
 and $f$ is a continuous function such that 
$$
a_\pm =\lim_{s\to \pm \infty }f(s) /\varphi_q(s) 
\quad\mbox{and}\quad a_0=\lim_{s\to 0}f(s)/\varphi_q(s) 
$$
exist as real numbers. Also, we assume that 
$m_\pm :=\inf_{\pm s\geq 0}f(s) /\varphi_q(s)$ exists in ${\mathbb R}$,
and define 
$$
m:=\inf_{u\in {\mathbb R}}f(s) /\varphi_q(s) =\min (m_{+},m_{-}) \quad\mbox{and}\quad 
m_k^{\pm}= \cases{ m_\pm  & if  $k=1$ \cr
                   m & if  $k\geq 2$.}
$$
We shall consider the following three cases:
Superlinear case, $a_0<\min (a_{-},a_{+})$;
Sublinear case, $a_0>\max (a_{-},a_{+})$;  and
Asymmetric case, $a_{-}<a_0<a_{+}$. 

In the special case when $p=2$, several authors have been interested in 
Problem (\ref{p1}), including  higher dimensions under various assumptions.
See, for instance, Amann and Zehnder \cite{az1}, Castro and Lazer \cite{cl},
Hess \cite{h}, Struwe \cite{S}, and particularly Esteban \cite{E}. For
the general case $p>1$, we mention a recent paper by J. Wang \cite{W} 
where positive solutions are studied.

Now we define some sets that will be used in the statement of the main
results.  For $k\geq 1$, let
$$
S_k^+=\left\{ 
\begin{array}{c} u\in C^1([0,1]) :u \mbox{ admits exactly }(
k-1) \mbox{ zeros in }(0,1) , \\ 
\mbox{all are simple, }u(0) =u(1) =0\mbox{ and }u'(0) >0
\end{array}
\right\}\,,
$$
then $S_k^-=-S_k^+$ and $S_k=S_k^+\cup S_k^-$. 
If $u\in C([0,1]) $ 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 hump of $u$. So, each function in $S_k^+$ has exactly 
$k$ humps such that the first one is positive, the second is negative, and
so on with alternations. 

Let $A_k^+$ $(k\geq 1) $ be the subset of $S_k^+$ consisting of the functions 
$u$ satisfying:
\begin{itemize}
\item  Every hump of $u$ is symmetrical about the center of the interval of
its definition.

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

\item  The derivative of each hump of $u$ vanishes once and only once.
\end{itemize}
Let $A_k^-=-A_k^+$ and $A_k=A_k^+\cup A_k^-$. Denote
by $(\lambda_k)_{k\geq 1}$ the eigenvalues of the one
dimensional p-Laplacian operator with Dirichlet boundary conditions, 
\begin{eqnarray}\label{p2}
&-(\varphi_p(u') ) '=\lambda\varphi_p(u) \quad\mbox{in }(0,1)&\\
&u(0) =u(1) =0\,.&\nonumber  
\end{eqnarray}
Then for each integer $k\geq 1$ and $p>1$, $\lambda_k=k^p\lambda_1$ and 
$$
\lambda_1=(p-1) (2\int_0^1(1-t^p)
^{-1/p}dt) ^p=(p-1) (\frac{2\pi }{p\sin (\pi
/p) }) ^p.
$$

For fixed real constants $a_{-}$ and $a_{+}$, consider the boundary value
problem
\begin{eqnarray}\label{p3}
&-(\varphi_p(u') ) ' =  \lambda
\varphi_p(u) +a_{+}\varphi_p(u^+) -a_{-}\varphi
_p(u^-) ,\mbox{in }(0,1) & \\ 
&u(0)  =  u(1) =0\,,& \nonumber
\end{eqnarray}
where $\lambda $ is a real parameter. If $\lambda $ is such that problem 
(\ref{p3}) admits a nontrivial solution $u_\lambda $, then $\lambda $ is
called a half-eigenvalue of (\ref{p3}). In the particular case where $p=2$,
this definition goes back to Berestycki \cite{Berestycki}. For any
integer $k\geq 1$, let 
$$
b_k^\pm =\left\{ 
\begin{array}{ll}
a_\pm  & \mbox{if }k=1 \\ 
\min (a_{-},a_{+}) & \mbox{if }k\geq 2
\end{array}
\right. \quad\mbox{and}\quad c_k^\pm =\left\{ 
\begin{array}{ll}
a_\pm  & \mbox{if }k=1 \\ 
\max (a_{-},a_{+}) & \mbox{if }k\geq 2\,.
\end{array}
\right.
$$

\begin{proposition}
\label{prop1} For fixed real constants $a_{-}$ and $a_{+}$, the set of
half-eigenvalues of problem (\ref{p3}) consists of two increasing sequences 
$(\lambda_k^+)_{k\geq 1}$ and $(\lambda_k^-)
_{k\geq 1}$, satisfying $h_k^\pm (\lambda_k^\pm ) =1$, for
all $k\geq 1$, where 
$$\displaylines{
h_{2n}^\pm (\lambda ) :=\frac{\lambda_n^{1/p}}{(a_{\pm
}+\lambda ) ^{1/p}}+\frac{\lambda_n^{1/p}}{(a_{\mp }+\lambda
) ^{1/p}},\mbox{ for all }\lambda >-b_{2n}^\pm ,\ n\geq 1, \cr
h_{2n+1}^\pm (\lambda ) :=\frac{\lambda_{n+1}^{1/p}}{(
a_\pm +\lambda ) ^{1/p}}+\frac{\lambda_n^{1/p}}{(a_{\mp
}+\lambda ) ^{1/p}},\mbox{ for all }\lambda >-b_{2n+1}^\pm ,\ n\geq0\,,
} $$ 
with the convention $\lambda_0=0$. Moreover, if $a_{\mp }<a_\pm $ then 
$$
\lambda_{2n-1}^\pm <\lambda_{2n-1}^{\mp }<\lambda_{2n}^\pm <\lambda
_{2n+1}^\pm <\lambda_{2n+1}^{\mp },\forall n\geq 1\,,
$$
and 
$$
\lambda_k-a_\pm <\lambda_k^\pm <\lambda_k-a_{\mp },\quad \forall k\geq 1.
$$
If $a_{-}=a_{+}$ then $\lambda_k^\pm =\lambda_k-a_\pm $, for all 
$k\geq 1$. 
\end{proposition}

The first result reads as follows.

\begin{theorem}
\label{thm1}Assume that $q=p$. For each integer $k\geq 1$, 
\begin{enumerate}
\item  If $a_0<b_k^\pm $ and $\max (-m_k^\pm ,\lambda_k^{\pm
}) <\lambda_k-a_0$, problem (\ref{p1}) admits at least a solution in 
$A_k^\pm $ for all $\lambda $ satisfying 
$\max (-m_k^\pm ,\lambda_k^\pm ) <\lambda <\lambda_k-a_0\,.$

\item  If $a_0>c_k^\pm $ and $\max (-m_k^\pm ,\lambda
_k-a_0) <\lambda_k^\pm $, problem (\ref{p1}) admits at least a
solution in $A_k^\pm $ for all $\lambda $ satisfying
$\max (-m_k^\pm ,\lambda_k-a_0) <\lambda <\lambda_k^\pm$.
\end{enumerate}
\end{theorem}

The next result deals with the asymmetric case. For any integer $n\geq
1$, let
$$
a_{2n+1}^\pm =\frac{na_{\mp }+(n+1) a_\pm }{2n+1}\quad
\mbox{and}\quad a_{2n}^\pm =\frac{na_{-}+na_{+}}{2n}=\frac{a_{-}+a_{+}}2\,.
$$

\begin{theorem}
\label{thm2} Assume that $q=p$ and $a_{-}<a_0<a_{+}$. Then
\begin{enumerate}
\item  If $\max (-m_{+},\lambda_1-a_{+}) <\lambda_1-a_0$, 
problem (\ref{p1}) admits at least a solution in $A_1^+$ for all $\lambda $
satisfying $\max (-m_{+},\lambda_1-a_{+}) <\lambda <\lambda
_1-a_0$. 

\item  If $\max (-m_{-},\lambda_1-a_0) <\lambda_1-a_{-}$, 
problem (\ref{p1}) admits at least a solution in $A_1^-$ for all $\lambda $
satisfying $\max (-m_{-},\lambda_1-a_0) <\lambda <\lambda
_1-a_{-}$. 

\item  For each integer $k\geq 2$, there exists $\tilde a_k^{\pm
}:a_{-}<\tilde a_k^\pm <a_k^\pm <a_{+}$ such that,

\begin{description}
\item[(i)]  if $a_{-}<a_0<\tilde a_k^\pm $ and $\max (-m,\lambda
_k^\pm ) <\lambda_k-a_0$, problem (\ref{p1}) admits at least a
solution in $A_k^\pm $ for all $\lambda $ satisfying $\max (
-m,\lambda_k^\pm ) <\lambda <\lambda_k-a_0$,

\item[(ii)]  if $\tilde a_k^\pm <a_0<a_{+}$ and $\max (-m,\lambda
_k-a_0) <\lambda_k^\pm $, problem (\ref{p1}) admits at least a
solution in $A_k^\pm $ for all $\lambda $ satisfying $\max (
-m,\lambda_k-a_0) <\lambda <\lambda_k^\pm $.
\end{description}
\end{enumerate}
\end{theorem}

\begin{theorem}
\label{thm3} Assume that $q\neq p$. Then for each integer $k\geq 1$, problem 
(\ref{p1}) admits at least a solution in $A_k^\pm $ for all $\lambda
>-m_k^\pm $. 
In particular, (\ref{p1}) admits infinitely many solutions for all 
$\lambda >-m$. 
\end{theorem}

The paper is organized as follows. The next section is dedicated to the
proof of Proposition \ref{prop1}. In Section \ref{sec3}, we present the
method used for proving the main results of this paper. In Section \ref{sec4},
we present some preliminary lemmas. In Section \ref{sec5}, we prove the
main results, present some remarks. The paper ends with an appendix which 
contains a brief historical overview on time maps.

\section{Proof of Proposition \protect{\ref{prop1}}}
\label{sec2}
Notice that for all $u\in {\mathbb R},\varphi
_p(u) =\varphi_p(u^+) -\varphi_p(u^-) $; so, problem (\ref{p3}) may be
 written as
\begin{eqnarray}  \label{p4}
&-(\varphi_p(u') ) '  =  \alpha
_{+}\varphi_p(u^+) -\alpha_{-}\varphi_p(u^-),\quad \mbox{in }(0,1)\,,& \\ 
&u(0)  =  u(1) =0\,,& \nonumber
\end{eqnarray}
with $\alpha_\pm =\lambda +a_\pm $. The set of $(\alpha
_{+},\alpha_{-}) \in {\mathbb R}^2$ such that problem (\ref
{p4}) has a nontrivial solution is known as the Fu\v cik spectrum for the
operator $(\varphi_p(u') ) '$
under Dirichlet boundary conditions. See for instance, Dr\'abek \cite
{Drabek1}, Boccardo et al. \cite{Boccardo}, Huang and Metzen \cite{Huang}. We
denote this spectrum by $\sum_p$. 

One has $\sum_p=\bigcup_{k\geq 0}C_k^\pm $, where
\begin{eqnarray*}
&C_0^\pm =\left\{ (\alpha_{+},\alpha_{-}) 
\in {\mathbb R}^2:\alpha_\pm =\lambda_1\right\} ,& 
\\ 
&C_{2n}^\pm =\left\{ (\alpha_{+},\alpha_{-}) \in (
{\mathbb R}^+) ^2:(\lambda_n^{1/p}/\alpha_{\pm
}^{1/p}) +(\lambda_n^{1/p}/\alpha_{\mp }^{1/p})
=1\right\} ,n\geq 1,& 
\\ 
&C_{2n+1}^\pm =\left\{ (\alpha_{+},\alpha_{-}) \in (
{\mathbb R}^+) ^2:(\lambda_{n+1}^{1/p}/\alpha_{\pm
}^{1/p}) +(\lambda_n^{1/p}/\alpha_{\mp }^{1/p})
=1\right\} ,n\geq 0\,.&
\end{eqnarray*}

So, $\lambda $ is a half-eigenvalue of (\ref{p3}) if and only if $(
\lambda +a_{+},\lambda +a_{-}) \in \sum_p$, that is,  $h_k^{\pm
}(\lambda ) =1,$ for some $k\geq 1$. Since the function 
$\lambda \mapsto h_k^\pm (\lambda ) ,k\geq 1$, is strictly
decreasing on $(-b_k^\pm ,+\infty ) $ and its limits at 
$-b_k^\pm $ and $+\infty $ are $+\infty $ and $0$ respectively, it follows
that the equation $h_k^\pm (\lambda ) =1$ admits a unique
solution $\lambda =\lambda_k^\pm $. 

It is easy to check that for $a_{-}<a_{+}$ and for all $\lambda >-a_{-}$
one has 
$$
h_{2n-1}^+(\lambda ) <h_{2n-1}^-(\lambda )
<h_{2n}^\pm (\lambda ) <h_{2n+1}^+(\lambda )
<h_{2n+1}^-(\lambda ) ,\quad \forall n\geq 1.
$$
Due to the fact that the function $\lambda \mapsto h_k^\pm (
\lambda ) $, $k\geq 1$, is strictly decreasing on $(
-a_{-},+\infty ) $, it follows that
$$
\lambda_{2n-1}^+<\lambda_{2n-1}^-<\lambda_{2n}^\pm <\lambda
_{2n+1}^+<\lambda_{2n+1}^-,\forall n\geq 1\,.
$$
On the other hand, if $a_{-}<a_{+}$, it follows that for all $\lambda
>-a_{-}$, $k\geq 1$, 
$$
\frac{\lambda_k^{1/p}}{(a_{+}+\lambda ) ^{1/p}}<\frac{\lambda
_k^{1/p}}{(a_{-}+\lambda ) ^{1/p}},\quad\mbox{so}\quad\frac{\lambda
_{2k}^{1/p}}{(a_{+}+\lambda ) ^{1/p}}<h_{2k}^\pm (
\lambda ) <\frac{\lambda_{2k}^{1/p}}{(a_{-}+\lambda )
^{1/p}},
$$
and then $\lambda_{2k}-a_{+}<\lambda_{2k}^\pm <\lambda_{2k}-a_{-}$.
The other cases may be handled the same way.

If $a_{-}=a_{+}$, problem (\ref{p3}) may be written as 
\begin{equation}
-(\varphi_p(u') ) '=\mu \varphi
_p(u)\quad \mbox{in }(0,1)\,,\quad u(0) =u(1) =0,  \label{p}
\end{equation}
with $\mu =\lambda +a_\pm $. So, $\lambda $ is a half-eigenvalue of (\ref
{p3}) if and only if $\lambda +a_\pm $ is an eigenvalue of (\ref{p2}),
that is, if and only if $\lambda =\lambda_k^\pm =\lambda_k-a_\pm $, 
which ends the proof of Proposition \ref{prop1}. 
\hfill$\diamondsuit$

\section{The time map method}\label{sec3}
To prove the main results, we  use the time
mapping approach as used in \cite{add1}, \cite{add2}, \cite{add3}.
To keep this paper self-contained, we describe it here.

Denote by $g$ a nonlinearity and by $p$ a real parameter, and we assume, 
\begin{equation}
g\in C({\mathbb R},{\mathbb R})\quad \mbox{and}\quad 1<p<+\infty \,.  \label{meth2}
\end{equation}
Consider the boundary value problem
\begin{equation}
-(\varphi_p(u') ) '=g(
u) \quad \mbox{in }(0,1) ,u(0) =u(1) =0,  \label{meth3}
\end{equation}
where $\varphi_p(x) =| x| ^{p-2}x$, $x\in {\mathbb R}$.
Denote by $p'=p/(p-1) $ the conjugate exponent of $p$.
Let $G(s) =\int_0^sg(t) dt$. For  $E>0$ and $\kappa =+,-$, let 
$$
X_\kappa (E) =\left\{ s\in {\mathbb R}:\kappa s\geq 0\quad\mbox{and}\quad
E^p-p'G(\xi ) >0,\,\forall \,\xi ,0\leq
\kappa \xi <\kappa s\right\}                           
$$
and
$$
r_\kappa (E) =\cases{ 0 &  if $X_\kappa (E) =\emptyset$ \cr
\kappa \sup (\kappa X_\kappa (E) ) & otherwise. }
$$
Note that $r_\kappa $ may be infinite. We shall also make use of the
following sets, 
$$
D_\kappa =\left\{ E>0:0<| r_\kappa (E) | <+\infty 
\mbox{ and }\kappa g(r_\kappa (E) ) >0\right\}$$
and $D=D_{+}\cap D_{-}$. 
Also, let $D_k^\kappa :=D$ if $k\geq 2$, and $D_1^\kappa :=D_\kappa $. 
Define the following time-maps, 
\begin{eqnarray*}
&T_\kappa (E) = \kappa \int_0^{r_\kappa (E)
}(E^p-p'G(t) ) ^{-1/p}dt, \quad E\in D_\kappa \,,&\\ 
&T_{2n}^\kappa (E)  =  n(T_{+}(E)
+T_{-}(E) ) , \quad  n\in {\mathbb N},\  E\in D\,,&\\ 
&T_{2n+1}^\kappa (E)  =  T_{2n}^\kappa (E)
+T_\kappa (E) , \quad n\in {\mathbb N}, \  E\in D.
\end{eqnarray*}

\begin{theorem}[Quadrature method]
\label{quad} Assume that (\ref{meth2}) holds. Let $E>0$, $k\in 
{\mathbb N}^{*}$, $\kappa =+,-$. If $E\in D_k^\kappa $ and $T_k^\kappa (
E) =1/2$, problem (\ref{meth3}) admits at least a solution $u_k^\kappa
\in A_k^\kappa $ satisfying $(u_k^\kappa ) '(
0) =\kappa E$, and this solution is unique.
\end{theorem}

This theorem is well-known, but we did not find a convenient
reference to the precise statement used later. The paper by Guedda and Veron 
\cite{GueddaVeron}, seems to be the first one dealing with time maps
approach when the differential operator is the one dimensional p-Laplacian.
An easy adaptation of the ideas contained in \cite{GueddaVeron} and in the
paper by Del Pino and Manasevich \cite{DelpinoMana1}, allows one to prove
Theorem \ref{quad}. Also, we mention the papers by Manasevich and Zanolin 
\cite{ManaZanolin} and Manasevich et al. \cite{ManasevichetAl}, where one
can find time maps used when the differential operator is the one
dimensional p-Laplacian.

Notice that time maps were also used when the differential operator
generalizes the p-Laplacian; see for example 
Garcia-Huidobro et al. \cite{GarciaetAl1}, \cite{GarciaetAl2}, \cite
{GarciaetAl3}, \cite{GarciaetAl4}, \cite{GarciaetAl5}, Garcia-Huidobro and
Ubilla \cite{GarciaUbilla}, Huang and Metzen \cite{Huang} and Ubilla \cite
{Ubilla}. A brief historical overview of time maps is given in the appendix.

For the sake of completeness, we dedicate the rest of this section to the
proof of Theorem \ref{quad} for the case where $k=1$ and $\kappa =+$. The
adaptation of the other cases may be handled similarly.

We shall say that $u$ is a solution of (\ref{meth3}) if $u$ and $\varphi
_p(u') $ belong to $C^1([0,1]) $ and $u$ satisfies 
$$
-(\varphi_p(u'(x) ) ) ^{\prime
}=g(u(x) ) \ \forall x\in (0,1)\,,\quad\mbox{with}\quad u(0) =u(1) =0\,.
$$
Let us assign to each function $u\in C^1([0,1]) $
the set 
$$
Z(u') =\left\{ x\in [0,1]:u^{\prime
}(x) =0\right\} .
$$

\begin{lemma}
\label{Lem1}Assume that (\ref{meth2}) holds and $u$ is a solution of (\ref
{meth3}). Then $u\in C^2([0,1]) $ if $1<p\leq 2$,
and $u\in C^2([0,1]\backslash Z(u^{\prime
}) ) $ if $p>2$. 
\end{lemma}

\paragraph{Proof.} The identity $t=\varphi_{p'}o\varphi
_p(t) $ for all $t\in {\mathbb R}$ implies that 
$$
u'(x) =\varphi_{p'}o(\varphi_p(
u') ) (x) \mbox{\ \ for\ all\ }x\in [0,1].
$$
Thus, if $1<p\leq 2$, it follows that $p'\geq 2$ and therefore 
$\varphi_{p'}\in C^1({\mathbb R}) $. By 
$\varphi_p(u') \in C^1([0,1]
) $, (since $u$ is a solution of (\ref{meth3})), it follows that 
$u'\in C^1([0,1]) $, that is $u\in
C^2([0,1]) $. If $p>2$, it follows that 
$1<p'<2$ and therefore $\varphi_{p'}\in C^1(
{\mathbb R}^{*}) $ and $\varphi_{p'}'(
0) =+\infty $. Hence, $u'$ is $C^1$ at each point $x$ where 
$\varphi_p(u'(x) ) \neq 0$. But $\varphi
_p(u'(x) ) =0$ if and only if $u^{\prime
}(x) =0$, that is, if and only if $x\in Z(u^{\prime
}) $. Therefore, Lemma \ref{Lem1} is proved. 
\hfill$\diamondsuit$\smallskip

By Lemma \ref{Lem1}, if $u$ is a solution of problem (\ref{meth3}), then 
$$
-(p-1) | u'(x) | ^{p-2}u^{\prime
\prime }(x) =g(u(x) ), \mbox{ for all }x\in [0,1]\backslash Z(u') .
$$
Multiplying both sides by $u'(x) $ one obtains for the
left hand side 
$$
-(p-1) | u'(x) | ^{p-2}u'' (x) u'(x) =-(1/p') (| u'| ^p) '(x) ,
$$
for all $x\in [0,1]\backslash Z(u')$,  and for the right hand side 
$$
g(u(x) ) u'(x) =(G(u(x) ) ) ',\mbox{\ \ for all }x\in [0,1]\backslash Z(u') .
$$
Thus, 
\begin{equation}
(| u'| ^p(x) +p'G(u(x) ) ) '=0,  \label{AD}
\end{equation}
for all $x\in [0,1]\backslash Z(u') $. 
Let us prove that (\ref{AD}) holds even for $x\in Z(u')$. 

\begin{lemma}[Energy relation]
\label{Lem2}Let $p>1$ and assume that $u$ is a solution of problem (\ref
{meth3}). Then 
$$
(| u'| ^p(x) +p'G(
u(x) ) ) '=0,\mbox{\ \ \ for all }x\in
[0,1].
$$
\end{lemma}

\paragraph{Proof.} Let $x_0\in Z(u') $. One has 
$$
(G(u))' (x_0) =g(u(x_0) ) u'(x_0) =0\,.
$$
Let us prove that 
\begin{equation}
(| u'| ^p) '(x_0) =0\,.
\label{Am}
\end{equation}
One has for all $x\neq x_0$,
$$
\frac{| u'(x) | ^p-| u'(
x_0) | ^p}{x-x_0}=u'(x) \times \frac{\varphi_p(u'(x) ) -\varphi_p(
u'(x_0) ) }{x-x_0}.
$$
Thus, $\lim_{x\to x_0}u'(x) =u'(x_0) =0$, and 
$$
\lim_{x\to x_0}\frac{\varphi_p(u'(x)
) -\varphi_p(u'(x_0) ) }{x-x_0}=(\varphi_p(u') ) '(
x_0) =-g(u(x_0) ) \in {\mathbb R}.
$$
Therefore, (\ref{Am}) is proved. Regarding (\ref{AD}), Lemma \ref{Lem2}
follows. \hfill$\diamondsuit$

\paragraph{Remark.}
In Lemmas \ref{Lem1} and \ref{Lem2}, the solutions, $u$, are
arbitrary and are not necessarily in $A_1^+$.  \smallskip


Now, assume that $u$ is a solution of problem (\ref{meth3}) belonging to 
$A_1^+$. Thus, 
\begin{equation}
u'>0\mbox{\ in\ }[0,(1/2) ) \mbox{\ \
and\ \ }u'(1/2) .  \label{kyd}
\end{equation}
It follows that 
$$
\sup \left\{ x\in [0,1) :u'(t)
>0,\forall t\in [0,x) \right\} =1/2,
$$
and by the energy relation one gets 
\begin{equation}
u'(x) =\left\{ (u'(0)
) ^p-p'G(u(x) ) \right\} ^{1/p}\mbox{,\ \ for all }x\in [0,1].  \label{jet}
\end{equation}
Thus, 
$$
\sup \left\{ x\in [0,1) :(u'(0)
) ^p-p'G(u(t) ) >0,\forall t\in
[0,x) \right\} =1/2,
$$
or equivalently 
$$
\sup \left\{ x\in [0,1) :(u'(0)
) ^p-p'G(z) >0,\forall z\in [0,u(
x) ) \right\} =1/2,
$$
which implies that 
$$
\sup \left\{ s\geq 0:(u'(0) ) ^p-p^{\prime
}G(\xi ) >0,\forall \xi \in [0,s) \right\}
=u(1/2) .
$$
Also, by (\ref{jet}) it follows that 
\begin{equation}
x=\int_0^{u(x) }\left\{ (u'(0)
) ^p-p'G(\xi ) \right\} ^{-1/p}d\xi \mbox{,\ \ \
for all }x\in [0,(1/2) ].  \label{qsfj}
\end{equation}
Thus, the improper integral in (\ref{qsfj}) is convergent for all $x\in
[0,(1/2) ]$ and in particular, $u'(
0) $ is such that the improper integral 
$$
\int_0^{r_{+}(u'(0) ) }\left\{ (
u'(0) ) ^p-p'G(\xi )
\right\} ^{-1/p}d\xi
$$
converges and is equal to $(1/2) $, where for all $E>0
$$$
r_{+}(E) =\sup X_{+}(E) \mbox{\ if }X_{+}(
E) \neq \emptyset \mbox{\ \ and\ \ }r_{+}(E) =0\mbox{\ if }X_{+}(E) =\emptyset
$$
and 
$$
X_{+}(E) =\left\{ s\geq 0:E^p-p'G(\xi )
>0,\forall \xi \in [0,s) \right\} .
$$
It follows that if $u$ is a solution of problem (\ref{meth3}) belonging to 
$A_1^+$, then there exists $E_{*}\in \tilde D_{+}$ with 
$$
\tilde D_{+}=\big\{ E>0:0<r_{+}(E) <+\infty \mbox{\ and\ }\int_0^{r_{+}(E) }\left\{ E^p-p'G(\xi )
\right\} ^{-1/p}d\xi <+\infty \big\}\,,
$$
such that 
$$
u'(0) =E_{*}\,,\quad u(1/2) =r_{+}(
E_{*}) \,,\quad\mbox{and}\quad T_{+}(E_{*}) =1/2\,,
$$
where 
$$
T_{+}(E) =\int_0^{r_{+}(E) }\left\{ E^p-p^{\prime
}G(\xi ) \right\} ^{-1/p}d\xi \mbox{,\ \ \ for all }E\in \tilde
D_{+}.
$$
Conversely, it is possible to assign to each root $E_{*}$ of the equation 
$T_{+}(E) =1/2$ in the variable $E\in \tilde D_{+}$ a unique
solution $u$ of the problem (\ref{meth3}) belonging to $A_1^+$ and
satisfying $u'(0) =E_{*}$, $\max_{[0,1]}u=u(1/2) =r_{+}(E_{*}) $. 

In fact, if $E_{*}\in \tilde D_{+}$ is such that $T_{+}(E_{*})
=1/2$, define the function $h_{+}$ on $[0,r_{+}(E_{*})
]$ by $h_{+}(u) =\int_0^u\left\{ E_{*}^p-p^{\prime
}G(\xi ) \right\} ^{-1/p}d\xi $. Notice that $h_{+}(
r_{+}(E_{*}) ) =T_{+}(E_{*}) =1/2$ and 
$$
0\leq h_{+}(u) \leq T_{+}(E_{*}) \mbox{,\ \ for all }u\in [0,r_{+}(E_{*}) ].
$$
Thus, $h_{+}$ is well defined on $[0,r_{+}(E_{*})
]$. Moreover, it is an increasing diffeomorphism from $(
0,r_{+}(E_{*}) ) $ onto $(0,T_{+}(
E_{*}) )$,
$$
h_{+}'(u) =\left\{ E_{*}^p-p'G(u)
\right\} ^{-1/p}>0\mbox{ for all }u\in (0,r_{+}(
E_{*}) ) .
$$
Let $u_{+}$ be the inverse of $h_{+}$ defined by 
$$
u_{+}(x) =h_{+}^{-1}(x) \in [0,r_{+}(
E_{*}) ],\mbox{ for all }x\in [0,(1/2)]\,,
$$
and let $u$ be defined on $[0,1]$ by 
$$
u(x) =\left\{ 
\begin{array}{l}
u_{+}(x) \mbox{\ if\ }x\in [0,(1/2) ]
\\ 
u_{+}(1-x) \mbox{\ if }x\in [(1/2) ,1]
.
\end{array}
\right. 
$$
It easy to show that this function $u$ is a solution of problem (\ref{meth3}) 
belonging to $A_1^+$ and satisfying $u'(0) =E_{*}$, 
$\max_{[0,1]}u=u(1/2) =r_{+}(E_{*}) $. 
Let us prove its uniqueness. Assume that $v$ is also a solution of 
problem (\ref{meth3}) belonging to $A_1^+$ and satisfies 
$$
v'(0) =E_{*},\max_{[0,1]}v=v(1/2) =v_{+}(E_{*}) .
$$
By (\ref{qsfj}) it follows that 
$$
x=\int_0^{u(x) }\left\{ E_{*}^p-p'G(\xi )
\right\} ^{-1/p}d\xi =\int_0^{v(x) }\left\{ E_{*}^p-p^{\prime
}G(\xi ) \right\} ^{-1/p}d\xi ,
$$
for all $x\in [0,1/2]$. Thus, 
$$
\int_{u(x) }^{v(x) }\left\{ E_{*}^p-p^{\prime
}G(\xi ) \right\} ^{-1/p}d\xi =0,\mbox{\ \ for all }x\in [
0,1/2].
$$
Thus, $u=v$ on $[0,1/2]$, and by symmetry it follows that $u=v$ on $[0,1]$. 
Therefore, because $D_{+}\subset \tilde D_{+}$, 
Theorem \ref{quad} is proved for the case $k=1$ and $\kappa =+$. 
\hfill$\diamondsuit$\smallskip

\section{Preliminary Lemmas}
\label{sec4}In order to define the time maps we need the following.

\begin{lemma}
\label{lemma1} For $s\in {\mathbb R}$, consider the equation 
\begin{equation}
E^p-p'(F(s) +\lambda \Phi_q(s)
) =0,  \label{Eq.1}
\end{equation}
where $p>1$, $E\geq 0$ and $\lambda \in {\mathbb R}$ are real
parameters, $F(s) =\int_0^sf(t) dt$ and $\Phi
_q(s) =\int_0^s\varphi_q(t) dt$. If $\lambda
>-m_\pm $, then for any $E>0$, equation (\ref{Eq.1}) admits a unique
positive zero $s_{+}=s_{+}(\lambda ,E) $ (resp. a unique
negative zero $s_{-}=s_{-}(\lambda ,E) $) and if $E=0$ it
admits no positive (resp. negative) zero beside the trivial one $s_\pm =0$. 
Moreover, for all $\lambda >-m_\pm ,p>1$, 

\begin{description}
\item[(i)]  The function $E\longmapsto s_\pm (\lambda ,E) $
is $C^1$ on $(0,+\infty ) $, and 
$$
\pm \frac{\partial s_\pm }{\partial E}(\lambda ,E) 
=\frac{\pm (p-1) E^{p-1}}{f(s_\pm (\lambda ,E)
) +\lambda \varphi_q(s_\pm (\lambda ,E) ) 
}>0,\ \forall E>0\,.
$$

\item[(ii)]  $\lim\limits_{E\to 0^+}s_\pm (\lambda
,E) =0$, $\lim\limits_{E\to +\infty }s_\pm (
\lambda ,E) =\pm \infty $. 

\item[(iii)]  $\lim\limits_{E\to 0}\frac{E^p}{| s_\pm (
\lambda ,E) | ^q}=\frac{p'}q(a_0+\lambda
)$, $\lim\limits_{E\to +\infty }\frac{E^p}{| s_{\pm
}(\lambda ,E) | ^q}=\frac{p'}q(a_{\pm
}+\lambda ) $. 

\item[(iv)]  $\lim\limits_{E\to 0}\frac E{| s_{\pm
}(\lambda ,E) | }=\cases{ 
0 & if $q-p>0$ \cr  
((a_0+\lambda ) /(p-1) ) ^{1/p} &  if $ q-p=0$ \cr 
+\infty  & if $ q-p<0$, }$

$\lim\limits_{E\to +\infty }\frac E{| s_\pm (\lambda
,E) | }=\cases{ 
+\infty  & if $ q-p>0$\cr 
((a_\pm +\lambda ) /(p-1) ) ^{1/p} & if $q-p=0$\cr  
0 & if $q-p<0$.} $ 

\item[(v)]  $\lim\limits_{E\to 0}\frac{F(s_{\pm
}(\lambda ,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_0}{a_0+\lambda },\forall t>0$, 
$\lim\limits_{E\to +\infty }\frac{F(s_\pm (\lambda
,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_\pm }{a_{\pm
}+\lambda },\forall t>0$. 
\end{description}
\end{lemma}

\paragraph{Proof.} For  $p>1$ and $E\geq0$ fixed, 
consider the function 
$$
s\longmapsto G_\pm (\lambda ,E,s) :=E^p-p'(
F(s) +\lambda \Phi_q(s) ) ,
$$
defined in ${\mathbb R}^\pm $ and strictly decreasing on 
$(0,+\infty ) $ (resp. strictly increasing on $(-\infty,0) $), because
$$
\frac{dG_\pm }{ds}(\lambda ,E,s) =-p'\varphi
_q(s) (\frac{f(s) }{\varphi_q(s) }+\lambda ) \quad\mbox{and}\quad
m_\pm +\lambda >0\,.
$$
One has $G_\pm (\lambda ,E,0) =E^p\geq 0$, and
via l'Hospital's rule, 
\begin{eqnarray*}
\lim\limits_{s\to +\infty }G_\pm (\lambda ,E,s)  & 
= & \lim\limits_{s\to +\infty }E^p-p'\Phi_q(s)
(\frac{F(s) }{\Phi_q(s) }+\lambda ) 
\\ 
& = & E^p-p'\lim\limits_{s\to +\infty }\Phi_q(
s) (\lim\limits_{s\to +\infty }\frac{f(s) }{\varphi_q(s) }+\lambda )  \\ 
& = & -\infty .
\end{eqnarray*}
So, it is clear that in the case $m_{+}+\lambda >0$ (resp. 
$m_{-}+\lambda >0$) for any $E>0$, (\ref{Eq.1}) admits a unique
positive zero, $s_{+}=s_{+}(\lambda ,E) $, (resp. a unique
negative zero, $s_{-}=s_{-}(\lambda ,E) $), and if $E=0$, it
admits no positive (resp. negative) zero beside the trivial one $s=0$. 
\smallskip 

Now, for any $p>1$ and $\lambda >-m_\pm $, consider the real
valued function, 
$$
(E,s) \longmapsto G_\pm (E,s) :=E^p-p^{\prime
}(F(s) +\lambda \Phi_q(s) ) ,
$$
defined on $\Omega_{+}=(0,+\infty ) ^2$ (resp. $\Omega
_{-}=(0,+\infty ) \times (-\infty ,0) $). One
has $G_\pm \in C^1(\Omega_\pm ) $ and, 
$$
\frac{\partial G_\pm }{\partial s}(E,s) =-p'\varphi
_q(s) (\frac{f(s) }{\varphi_q(s) 
}+\lambda ) \quad\mbox{in }\Omega_\pm \,;
$$
hence, because $m_\pm +\lambda >0$, it follows that 
$$
\pm \frac{\partial G_\pm }{\partial s}(E,s) <0,\mbox{\ \ in
\ }\Omega_\pm ,
$$
and one may observe that $s_\pm (\lambda ,E) $ belongs to
the open interval $(0,+\infty ) $ (resp. $(-\infty
,0) $) and  from its definition satisfies 
\begin{equation}
G_\pm (E,s_\pm (\lambda ,E) ) =0\,.
\label{Eq.2}
\end{equation}
So, one can make use of the implicit function theorem to show that the
function $E\mapsto s_\pm (\lambda ,E) $ is $C^1((
0,+\infty ) ,{\mathbb R}) $ and to obtain the
expression of $\frac{\partial s_\pm }{\partial E}(\lambda
,E) $ given in {\bf (i)}. Hence, by $m_\pm +\lambda >0$, it follows
that for any fixed $p>1$ and $\lambda >-m_\pm $, the function defined on 
$(0,+\infty ) $ by $E\mapsto s_\pm (\lambda ,E) 
$ is strictly increasing (resp. strictly decreasing) and bounded from below
by $0$ (resp. by $-\infty $) and from above by $+\infty $ (resp. by $0$).
Then  the limit $\lim_{E\to 0^+}s_\pm (\lambda
,E) =\ell_0^\pm $ exists as real number and the limit 
$\lim_{E\to +\infty }s_\pm (\lambda ,E) =\ell
_{+\infty }^\pm $ exists and belongs to $(0,+\infty ]$
(resp. $[-\infty ,0) $). Moreover, 
$$
-\infty \leq \ell_{+\infty }^-<\ell_0^-\leq 0\leq \ell_0^+<\ell
_{+\infty }^+\leq +\infty .
$$

Let us observe that, for any fixed $p>1$ and $\lambda >-m_\pm $, the
function $(E,s) \mapsto G_\pm (E,s) $ is
continuous on $[0,+\infty ) ^2$ (resp. $[0,+\infty
) \times (-\infty ,0]$) and the function $E\longmapsto
s_\pm (\lambda ,E) $ is continuous on $(0,+\infty
) $ and satisfies (\ref{Eq.2}). So, by passing to the limit in (\ref
{Eq.2}) as $E$ tends to $0^+$ one obtains 
$$
0=\lim_{E\to 0^+}G_\pm (E,s_\pm (\lambda
,E) ) =G_\pm (0,\ell_0^+) \,.
$$
Hence, $\ell_0^\pm $ is a zero, belonging to $[0,+\infty ) $
(resp. $(-\infty ,0]$), to the equation in $s$:
$$
G_\pm (0,s) =0\mbox{.}
$$
By solving this equation one gets: $\ell_0^\pm =0$.

Assume that $\ell_{+\infty }^\pm $ is finite, then by passing to the
limit in (\ref{Eq.2}) as $E$ tends to $+\infty $ one gets, 
$$
+\infty =p'(F(\ell_{+\infty }^\pm ) +\lambda
\Phi_q(\ell_{+\infty }^\pm ) ) <+\infty ,
$$
which is impossible. So, $\ell_{+\infty }^\pm =\pm \infty $. 

\paragraph{Proof of (iii).} Dividing equation (\ref{Eq.2}) by $| s_{\pm
}(\lambda ,E) | ^q$ one gets, 
$$
\frac{E^p}{| s_\pm (\lambda ,E) | ^q}=p^{\prime
}(\frac{F(s_\pm (\lambda ,E) ) }{|
s_\pm (\lambda ,E) | ^q}+\frac \lambda q) ,
$$
and by passing to the limit as $E$ tends to $0^+$, (using l'Hospital's
rule), 
$$
\lim_{E\to 0^+}\frac{E^p}{| s_\pm (\lambda
,E) | ^q}=\frac{p'}q(\lim_{E\to 0^+}\frac{f(s_\pm (\lambda ,E) ) }{\varphi_q(
s_\pm (\lambda ,E) ) }+\lambda ) =\frac{p'}q(a_0+\lambda ) .
$$
The second limit is obtained by the same way.

\paragraph{Proof of (iv).} Dividing equation (\ref{Eq.2}) by $| s_{\pm
}(\lambda ,E) | ^p$ one gets, 
\begin{eqnarray*}
\lim\limits_{E\to 0^+}\frac{E^p}{| s_\pm (\lambda
,E) | ^p} & = & \lim\limits_{E\to 0^+}p^{\prime
}(\frac{F(s_\pm (\lambda ,E) ) }{|
s_\pm (\lambda ,E) | ^p}+\frac \lambda q| s_{\pm
}(\lambda ,E) | ^{q-p}) \\  
& = & \lim\limits_{E\to 0^+}p'| s_\pm (
\lambda ,E) | ^{q-p}(\frac{F(s_\pm (
\lambda ,E) ) }{| s_\pm (\lambda ,E)
| ^q}+\frac \lambda q) \\  
& = & \frac{p'}q(a_0+\lambda )
\lim\limits_{E\to 0^+}| s_\pm (\lambda ,E)
| ^{q-p}.
\end{eqnarray*}
Therefore, the first limit follows. The second one is obtained by the same way.

\paragraph{Proof of (v).} Using l'Hospital's rule one gets, for any $t>0$,
\begin{eqnarray*}
\lim \limits_{E\to 0^+}\frac{F(s_\pm (\lambda
,E) t) }{E^p} & = & \lim \limits_{E\to 0^+} \frac{t\frac{ds_\pm }{dE}(\lambda ,E) f(s_\pm (
\lambda ,E) t) }{pE^{p-1}} \\ 
& = & \lim \limits_{E\to 0^+} \frac{t(p-1)
E^{p-1}f(s_\pm (\lambda ,E) t) }{(f(
s_\pm (\lambda ,E) ) +\lambda \varphi_q(s_{\pm
}(\lambda ,E) ) ) pE^{p-1}} \\ 
& = & \frac t{p'}\lim \limits_{E\to 0^+} \frac{\frac{f(s_\pm (\lambda ,E) t) }{\varphi_q(
s_\pm (\lambda ,E) t) }}{(\frac{f(s_{\pm
}(\lambda ,E) ) }{\varphi_q(s_\pm (
\lambda ,E) ) }\frac 1{\varphi_q(t) }+\frac
\lambda {\varphi_q(t) }) } \\ 
& = & \frac{t^q}{p'}(\frac{a_0}{a_0+\lambda }) .
\end{eqnarray*}
The second limit may be computed by the same way, which completes the proof
of Lemma \ref{lemma1}. \hfill$\diamondsuit$\smallskip

 Now, for any $p>1$, $\lambda >-m_\pm $ and $E>0$  we compute 
 $X_\pm (\lambda ,E) $ as defined in Section \ref
{sec3}. In fact, for all $E>0,
$$$
X_{+}(\lambda ,E) =(0,s_{+}(\lambda ,E)
) ,X_{-}(\lambda ,E) =(s_{-}(\lambda
,E) ,0) ,
$$
where $s_\pm (\lambda ,E) $ is defined in Lemma \ref{lemma1}. Then  
$$
r_\pm (\lambda ,E) =s_\pm (\lambda ,E) \mbox{
\ for all }E>0.
$$
Hence, for any $p>1,\lambda >-m_\pm $,
$0<| s_\pm (\lambda ,E) | <+\infty$ if and only if $E>0$.
And for all $E>0$, 
$$ \pm (f(r_\pm (\lambda ,E) ) +\lambda
\varphi_q(r_\pm (\lambda ,E) ) )  =
\pm \varphi_q(r_\pm (\lambda ,E) ) (
\frac{f((r_\pm (\lambda ,E) ) ) }{\varphi_q(r_\pm (\lambda ,E) ) }+\lambda
)  >0.
$$ 
So,  $D_\pm (\lambda ) =(0,+\infty )$ for all $\lambda >-m_\pm $ and 
$$
D(\lambda ) =D_{+}(\lambda ) \cap D_{-}(
\lambda ) =(0,+\infty ) ,\forall \lambda >-m.
$$

 Before going further in the investigation, from Lemma 
\ref{lemma1}, we deduce that for any fixed $p>1$ and $\lambda >-m_\pm $,  
\begin{equation}
\pm \frac{\partial r_\pm }{\partial E}(\lambda ,E) =\frac{\pm (p-1) E^{p-1}}{f(r_\pm (\lambda ,E)
) +\lambda \varphi_q(r_\pm (\lambda ,E) ) 
}>0,\forall E>0.  \label{A}
\end{equation}
\begin{equation}
\lim\limits_{E\to 0^+}r_\pm (\lambda ,E) =0\mbox{\
\ \ \ and\ \ \ \ }\lim\limits_{E\to +\infty }r_\pm (\lambda
,E) =\pm \infty   \label{B}
\end{equation}
\begin{equation}
\lim\limits_{E\to 0}\frac{E^p}{| r_\pm (\lambda
,E) | ^q}=\frac{p'}q(a_0+\lambda )
,\lim\limits_{E\to +\infty }\frac{E^p}{| r_\pm (
\lambda ,E) | ^q}=\frac{p'}q(a_\pm +\lambda
) .  \label{C}
\end{equation}

\begin{equation}  \label{D}
\lim \limits_{E\to 0}\frac E{| r_\pm (\lambda
,E) | }=\left\{ 
\begin{array}{lcc}
0 & \mbox{if} & q-p>0 \\ 
(\frac{a_0+\lambda }{p-1}) ^{\frac 1p} & \mbox{if} & q-p=0 \\ 
+\infty & \mbox{if} & q-p<0
\end{array}
\right.
\end{equation}

\begin{equation}  \label{E}
\lim \limits_{E\to +\infty }\frac E{| r_\pm (\lambda
,E) | }=\left\{ 
\begin{array}{lcc}
+\infty & \mbox{if} & q-p>0 \\ 
(\frac{a_\pm +\lambda }{p-1}) ^{\frac 1p} & \mbox{if} & q-p=0
\\ 
0 & \mbox{if} & q-p<0
\end{array}
\right.
\end{equation}
\begin{equation}  \label{F}
\lim \limits_{E\to 0}\frac{F(r_\pm (\lambda
,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_0}{a_0+\lambda },\forall t>0
\end{equation}
\begin{equation}  \label{G}
\lim \limits_{E\to +\infty }\frac{F(r_\pm (\lambda
,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_\pm }{a_{\pm
}+\lambda },\forall t>0.
\end{equation}

At present we define, for any $p>1,$\ $\lambda >-m_\pm $, and 
$E>0$, the time map, 
$$
T_\pm (\lambda ,E) :=\pm \int_0^{r_\pm (\lambda
,E) }\left\{ E^p-p'(F(\xi ) +\lambda \Phi
_q(\xi ) ) \right\} ^{-\frac 1p}d\xi ,E>0,
$$
and a simple change of variables shows that, 
\begin{equation}
T_\pm (\lambda ,E) =| r_\pm (\lambda,E) |  
\int_0^1\left\{ E^p-p'(F(
r_\pm (\lambda ,E) \xi ) +(\lambda /q)
| r_\pm (\lambda ,E) \xi | ^q) \right\}^{-\frac 1p}d\xi ,
\label{U}
\end{equation}
which may be written as, 
\begin{eqnarray}\label{H}
T_\pm (\lambda ,E) &=&(| r_\pm (\lambda,E) | /E)  
\int_0^1\big\{ 1-p'(F(r_\pm (\lambda
,E) \xi ) /E^p \\
&&\hspace{3cm}+(\lambda \xi ^q/q) (|
r_\pm (\lambda ,E) | ^q/E^p) ) \big\}^{-1/p}d\xi \,.
\nonumber
\end{eqnarray}
Also, we define, the time maps
$$ \displaylines{
T_{2n}^\pm (\lambda ,E) :=n(T_{+}(\lambda
,E) +T_{-}(\lambda ,E) ) ,E>0,\lambda
>-m_{2n}^\pm ,n\geq 1\,, \cr 
T_{2n+1}^\pm (\lambda ,E) :=T_{2n}^\pm (\lambda
,E) +T_\pm (\lambda ,E) ,E>0,\lambda
>-m_{2n+1}^\pm ,n\geq 0\,.
}$$

To prove Theorems \ref{thm1},\ \ref{thm2}, \ref{thm3}, it suffices
to compute the limits of these time maps as $E$ tends to $0^+$ and 
$+\infty $, and then apply the intermediate value theorem. Recall that we
have defined in Proposition \ref{prop1} the functions $h_k^\pm $ and let
us now define, 
$$
g_k(\lambda ) =\frac{\lambda_k^{1/p}}{(a_0+\lambda
) ^{1/p}},\mbox{ for all }\lambda >-a_0,\ k\geq 1\,.
$$

\begin{lemma}
\label{lemma22}Assume that $p,q>1$, then for all $k\geq 1$, and all 
$\lambda >-m_k^\pm $, 

\begin{enumerate}
\item  $\lim\limits_{E\to 0^+}T_k^\pm (\lambda ,E)
=+\infty $ and $\lim\limits_{E\to +\infty }T_k^{\pm
}(\lambda ,E) =0$,  if $q>p$, 

\item  $\lim\limits_{E\to 0^+}T_k^\pm (\lambda ,E)
=0$ and  $\lim\limits_{E\to +\infty }T_k^\pm (
\lambda ,E) =+\infty $,  if $q<p$, 

\item  $\lim\limits_{E\to 0^+}T_k^\pm (\lambda ,E)
=\frac 12g_k(\lambda ) $ and $\lim\limits_{E\to
+\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm (
\lambda )$, if $q=p$. 
\end{enumerate}
\end{lemma}

\paragraph{Proof.} The limits of $T_\pm (\lambda ,E) $
as $E$ tends to $0^+$ or $+\infty $ follow by making use
of (\ref{H}) and (\ref{C})--(\ref{G}), together with the fact that
$$
\lambda_k^{1/p}=2k(p-1) ^{1/p}\int_0^1(1-t^p)
^{-1/p}dt.
$$

The limits of $T_{2n}^\pm (\lambda ,E) $ and $T_{2n+1}^{\pm
}(\lambda ,E) $ as $E$ tends to $0$ or $+\infty $ follow
immediately from the definition of these maps and the limits of $T_{\pm
}(\lambda ,E) $ as $E$ tends to $0$ or $+\infty $
respectively. The proof is complete. \hfill$\diamondsuit$\smallskip

To apply the intermediate value theorem we need to know,
for each integer $k\geq 1$, which one of the limits, 
$$
\lim_{E\to 0^+}T_k^\pm (\lambda ,E) \quad\mbox{or}\quad
\lim_{E\to +\infty }T_k^\pm (\lambda ,E)
$$
is greater than the other. If $q-p\neq 0$, the answer is evident from Lemma 
\ref{lemma22}, but if $q-p=0$, a deep study is required. Let, for any
integer $n\geq 1$,
$$
\Lambda_{2n+1}^\pm =
\frac{a_{-}a_{+}-a_0a_{2n+1}^{\mp }}{a_0-a_{2n+1}^\pm }\quad\mbox{and}
\quad\Lambda_{2n}^\pm 
=\frac{a_{-}a_{+}-a_0a_{2n}^\pm }{a_0-a_{2n}^\pm }.
$$

\begin{lemma}
\label{lemma23}Let $k\geq 1$\ be an integer,

\begin{description}
\item[(i)]  If $a_0<b_k^\pm $, then $g_k(\lambda ) >h_k^{\pm
}(\lambda ) $ for all $\lambda >-a_0$. 

\item[(ii)]  If $a_0>c_k^\pm $, then $g_k(\lambda ) <h_k^{\pm
}(\lambda ) $ for all $\lambda >-b_k^\pm $. 

\item[(iii)]  If $a_{-}<a_k^\pm \leq a_0<a_{+}$, $(k\geq 2) $
then $g_k(\lambda ) <h_k^\pm (\lambda ) $ for all 
$\lambda >-a_{-}$. 

\item[(iv)]  If $a_{-}<a_0<a_k^\pm $, $(k\geq 2) $ then there
exists a unique $\tilde \lambda_k^\pm =\tilde \lambda_k^\pm (
a_0) \in (-a_{-},\Lambda_k^\pm ) $ such that, 
$$
\begin{array}{l}
g_k(\lambda ) <h_k^\pm (\lambda ) \mbox{ for all }\lambda \in (-a_{-},\tilde \lambda_k^\pm ) , \\ 
g_k(\tilde \lambda_k^\pm ) =h_k^\pm (\tilde \lambda
_k^\pm ) , \\ 
g_k(\lambda ) >h_k^\pm (\lambda ) \mbox{ for all }\lambda \in (\tilde \lambda_k^\pm ,+\infty ) .
\end{array}
$$
Moreover, the function $a_0\mapsto \tilde \lambda_k^\pm (a_0) 
$ is strictly increasing on $(a_{-},a_k^\pm ) $ and 
$$
\lim_{a_0\to a_{-}}\tilde \lambda_k^\pm (a_0)
=-a_{-},\mbox{and}\lim_{a_0\to a_k^\pm }\tilde
\lambda_k^\pm (a_0) =+\infty .
$$
\end{description}
\end{lemma}

\paragraph{Proof.} It is immediate to prove Assertions {\bf (i)} and 
{\bf (ii)} of this lemma since the function $a\mapsto (a+\lambda
) ^{-1/p}$ is strictly decreasing. Let us prove assertions
{\bf (iii) }and {\bf (iv)}.

We are concerned by the case $k=2n+1$ for some $n\geq 1
$ and the superscript is $+:(k=2n+1,+) $. All the remaining
cases may be handled similarly.

Let $y_{2n+1}^+$ be the function defined on $(-a_{-},+\infty
) $ by
$$
y_{2n+1}^+(\lambda ) =(n+1) (a_{+}+\lambda
) ^{-1/p}+n(a_{-}+\lambda ) ^{-1/p}-(2n+1)
(a_0+\lambda ) ^{-1/p}.
$$
One has, $y_{2n+1}^+(\lambda ) <0$ if and only if  
$$
((\frac{n+1}{2n+1}) (a_{+}+\lambda )
^{-1/p}+(\frac n{2n+1}) (a_{-}+\lambda )
^{-1/p}) ^p<(a_0+\lambda ) ^{-1}
$$
and $(y_{2n+1}^+) '(\lambda )
<0$ if and only if  
$$
((\frac{n+1}{2n+1}) (a_{+}+\lambda )
^{-1-1/p}+(\frac n{2n+1}) (a_{-}+\lambda )
^{-1-1/p}) ^{p/(p+1) }>  
(a_0+\lambda ) ^{-1}.
$$

Since the function $t\mapsto t^p$ (resp. $t\mapsto -t^{p/(p+1) }$)
is convex on $(-a_{-},+\infty ) $,  
\begin{eqnarray*}
\lefteqn{ ((\frac{n+1}{2n+1}) (a_{+}+\lambda )
^{-1/p}+(\frac n{2n+1}) (a_{-}+\lambda )^{-1/p}) ^p } \\ 
&&< (\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1}+(\frac
n{2n+1}) (a_{-}+\lambda ) ^{-1}
\end{eqnarray*}
(resp. 
\begin{eqnarray*}
\lefteqn{ ((\frac{n+1}{2n+1}) (a_{+}+\lambda )
^{-1-1/p}+(\frac n{2n+1}) (a_{-}+\lambda )
^{-1-1/p}) ^{p/(p+1)}  }\\  
&& >(\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1}+(\frac
n{2n+1}) (a_{-}+\lambda ) ^{-1}\mbox{)}.
\end{eqnarray*}
So, if we define on $(-a_{-},+\infty ) $ the function 
$x_{2n+1}^+$ by 
$$
x_{2n+1}^+(\lambda ) =(\frac{n+1}{2n+1}) (
a_{+}+\lambda ) ^{-1}+(\frac n{2n+1}) (a_{-}+\lambda
) ^{-1}-(a_0+\lambda ) ^{-1},
$$
it follows that for all $\lambda >-a_{-},
$$$
x_{2n+1}^+(\lambda ) \leq 0\Longrightarrow y_{2n+1}^+(
\lambda ) <0\mbox{\ \ \ and\ \ \ }x_{2n+1}^+(\lambda )
\geq 0\Longrightarrow (y_{2n+1}^+) '(\lambda
) <0.
$$
Some simple computations show that for all $\lambda >-a_{-}$ and $\kappa
=+,-$, one has, 
$$
\kappa x_{2n+1}^+(\lambda ) >0\Longleftrightarrow \kappa
(\lambda (a_0-a_{2n+1}^+) -(
a_{-}a_{+}-a_0a_{2n+1}^-) ) >0.
$$
Also, for all $\lambda \in {\mathbb R}$ and $\kappa =+,-$, one has
in the case where $a_0>a_{2n+1}^+,
$$$
\kappa (\lambda (a_0-a_{2n+1}^+) -(
a_{-}a_{+}-a_0a_{2n+1}^-) ) >0\Longleftrightarrow \kappa
(\lambda -\Lambda_{2n+1}^+) >0,
$$
and in the case where $a_0<a_{2n+1}^+$, 
$$
\kappa (\lambda (a_0-a_{2n+1}^+) -(
a_{-}a_{+}-a_0a_{2n+1}^-) ) >0\Longleftrightarrow \kappa
(\lambda -\Lambda_{2n+1}^+) <0.
$$
On the other hand one has, 
$$
a_0<a_{2n+1}^+\Longrightarrow \Lambda_{2n+1}^+>-a_{-}\mbox{\ \ \ and\ \
\ }a_0>a_{2n+1}^+\Longrightarrow \Lambda_{2n+1}^+<-a_{-}.
$$
Hence, an easy compilation of the above assertions shows that if 
$a_0<a_{2n+1}^+$, then  
\begin{eqnarray*}
\lambda \geq \Lambda_{2n+1}^+(>-a_{-}) &\Longrightarrow&
x_{2n+1}^+(\lambda ) \leq 0 \\ 
-a_{-}<\lambda <\Lambda_{2n+1}^+&\Longrightarrow& x_{2n+1}^+(
\lambda ) <0
\end{eqnarray*}
and if $a_0>a_{2n+1}^+$, then  $\lambda >-a_{-}$ implies
$x_{2n+1}^+(\lambda ) >0$.

It remains to study the particular case $a_0=a_{2n+1}^+$. 
For $\lambda >-a_{-}$, define
$$
\psi_\lambda (t) =(\lambda +t) ^{-1},\mbox{ for all }t>-\lambda .
$$
Let us observe that $x_{2n+1}^+(\lambda ) >0$ if and only if
$$
\psi_\lambda (a_0) <(\frac n{2n+1}) \psi_\lambda
(a_{-}) +(\frac{n+1}{2n+1}) \psi_\lambda (
a_{+}) .
$$
But since $\psi_\lambda $ is strictly convex, one has
\begin{eqnarray*}
\psi_\lambda (a_0) =\psi_\lambda (a_{2n+1}^+)
&=&\psi_\lambda ((\frac n{2n+1}) a_{-}+(\frac{n+1}{2n+1}) a_{+}) \\
&<& (\frac n{2n+1}) \psi_\lambda (a_{-}) +(
\frac{n+1}{2n+1}) \psi_\lambda (a_{+}) ,
\end{eqnarray*}
that is, if $a_0=a_{2n+1}^+$, then 
$$
\lambda >-a_{-}\Longrightarrow x_{2n+1}^+(\lambda ) >0.
$$
Thus, in the case where $a_0\geq a_{2n+1}^+$, $y_{2n+1}^+$ is strictly
decreasing on $(-a_{-},+\infty ) $ and by 
$\lim_{\lambda \to +\infty }y_{2n+1}^+(\lambda ) =0$ it
follows that $y_{2n+1}^+$ is strictly positive on $(-a_{-},+\infty
) $. Thus, Assertion {\bf (iii)} is proved.

In the case where $a_0<a_{2n+1}^+$, $y_{2n+1}^+$ is strictly negative on 
$[\Lambda_{2n+1}^+,+\infty ) $ and strictly decreasing on 
$(-a_{-},\Lambda_{2n+1}^+) $. By $\lim\limits_{\lambda
\to -a_{-}}y_{2n+1}^+(\lambda ) =+\infty $, it follows
that there exists $\tilde \lambda_{2n+1}^+=\tilde \lambda
_{2n+1}^+(a_0) \in (-a_{-},\Lambda_{2n+1}^+) $
such that $y_{2n+1}^+$ is strictly positive on $(-a_{-},\tilde
\lambda_{2n+1}^+) ,y_{2n+1}^+(\tilde \lambda
_{2n+1}^+) =0$, and $y_{2n+1}^+$ is strictly negative on $(
\tilde \lambda_{2n+1}^+,+\infty ) $. 

One has $y_{2n+1}^+(a_0,\tilde \lambda_{2n+1}^+(
a_0) ) =0$. So, the implicit function theorem yields 
$$
\frac{\partial \tilde \lambda_{2n+1}^+}{\partial a_0}(a_0) =-\frac{\partial y_{2n+1}^+}{\partial a_0}(a_0,\tilde \lambda
_{2n+1}^+(a_0) ) /\frac{\partial y_{2n+1}^+}{\partial
\lambda }(a_0,\tilde \lambda_{2n+1}^+(a_0) ) .
$$
One has $\frac{\partial y_{2n+1}^+}{\partial \lambda }(a_0,\tilde
\lambda_{2n+1}^+(a_0) ) <0$ since $x_{2n+1}^+(
\tilde \lambda_{2n+1}^+(a_0) ) >0$, and 
$$
\frac{\partial y_{2n+1}^+}{\partial a_0}(a_0,\tilde \lambda
_{2n+1}^+(a_0) ) =\frac{2n+1}p(a_0+\tilde \lambda
_{2n+1}^+(a_0) ) ^{-1-1/p}>0.
$$
So, the function $a_0\mapsto \tilde \lambda_{2n+1}^+(a_0) $
is strictly increasing on $(a_{-},a_{2n+1}^+) $. 
On the other hand, one has 
$$
-a_{-}<\tilde \lambda_{2n+1}^+(a_0) <\Lambda
_{2n+1}^+(a_0) ;\forall a_0\in (a_{-},a_{2n+1}^+) ,
$$
and $\lim\limits_{a_0\mapsto a_{-}}\Lambda _{2n+1}^+(a_0) =-a_{-}$,
which is easy to check. Thus,
$\lim\limits_{a_0\to a_{-}}\tilde \lambda_{2n+1}^+(
a_0) =-a_{-}$.

Assume that the function $a_0\mapsto \tilde \lambda
_{2n+1}^+(a_0) $ is bounded as $a_0$ tends to $a_{2n+1}^+$. 
Denote by $\lambda_{*}$ its limit, that is, $\lim\limits_{a_0\to
a_{2n+1}^+}\tilde \lambda_{2n+1}^+(a_0) =\lambda_{*}$. 
Since $$
y_{2n+1}^+(a_0,\tilde \lambda_{2n+1}^+(a_0) )
=0,\forall a_0\in (a_{-},a_{2n+1}^+) ,
$$
it follows that $y_{2n+1}^+(a_{2n+1}^+,\lambda_{*}) =0$,
that is, 
\begin{eqnarray*}
\lefteqn{ ((\frac{n+1}{2n+1}) (a_{+}+\lambda_{*})
+(\frac n{2n+1}) (a_{-}+\lambda_{*}) )
^{-1/p} } \\
&=& (\frac{n+1}{2n+1}) (a_{+}+\lambda_{*})
^{-1/p}+(\frac n{2n+1}) (a_{-}+\lambda_{*})
^{-1/p}\,.
\end{eqnarray*}
By the strict convexity of the function $t\to t^{-1/p}$ on 
$(0,+\infty ) $, it follows that $a_{+}+\lambda
_{*}=a_{-}+\lambda_{*}$, that is $a_{-}=a_{+}$, which contradicts the
hypothesis $a_{-}<a_{+}$. So, the function $a_0\mapsto \tilde \lambda
_{2n+1}^+(a_0) $ is unbounded, and
$$
\lim\limits_{a_0\to a_{2n+1}^+}\tilde \lambda_{2n+1}^+(
a_0) =+\infty\,. 
$$
Therefore, Assertion {\bf (iv)} is proved for the case $(k=2n+1,+) $. 

In the case $(k=2n+1,-) $ (resp. $(k=2n,\pm ) $)
one considers $y_{2n+1}^-$ and $x_{2n+1}^-$ (resp. $y_{2n}^\pm $ and 
$x_{2n}^\pm $) defined by
\begin{eqnarray*}
y_{2n+1}^-(\lambda ) &=&n(a_{+}+\lambda )
^{-1/p}+(n+1) (a_{-}+\lambda ) ^{-1/p}-(
2n+1) (a_0+\lambda ) ^{-1/p} \\ 
x_{2n+1}^-(\lambda ) &=&(\frac n{2n+1}) (
a_{+}+\lambda ) ^{-1}+(\frac{n+1}{2n+1}) (
a_{-}+\lambda ) ^{-1}-(a_0+\lambda ) ^{-1}
\end{eqnarray*}
(resp. 
\begin{eqnarray*}
y_{2n}^\pm (\lambda ) &=&(a_{+}+\lambda )
^{-1/p}+(a_{-}+\lambda ) ^{-1/p}-2(a_0+\lambda )
^{-1/p} \\ 
x_{2n}^\pm (\lambda ) &=&\frac 12(a_{+}+\lambda )
^{-1}+\frac 12(a_{-}+\lambda ) ^{-1}-(a_0+\lambda )
^{-1}\mbox{).}
\end{eqnarray*}
The same reasoning as above also works here; therefore, Lemma \ref{lemma23} 
is proved. \quad \hfill$\diamondsuit$ 


\section{Proof of the main results}
\label{sec5} 
\paragraph{Proof of Theorem \ref{thm1}.}
If $a_0<b_k^\pm $, then for all $k\geq 1$ and $\lambda
>-m_k^\pm$
$$
\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac
12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to
+\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm (
\lambda ) .
$$
On the other hand, for all $\lambda >-a_0:g_k(\lambda )
>h_k^\pm (\lambda ) $. So, since the function $\lambda \mapsto
g_k(\lambda ) $ (resp. $\lambda \mapsto h_k^\pm (\lambda
) $) is strictly decreasing on $(-a_0,+\infty ) $ 
(resp. on $(-b_k^\pm ,+\infty ) $), and 
$$
g_k(\lambda_k-a_0) =1\mbox{ (resp. }h_k^\pm (
\lambda_k^\pm ) =1\mbox{)}\,,
$$
then  
$$
g_k(\lambda ) >1\mbox{ \ \ if and only if \ \ }-a_0<\lambda
<\lambda_k-a_0
$$
(resp. 
$$
h_k^\pm (\lambda ) <1\mbox{\ \ \ if and only if\ \ \ }\lambda
_k^\pm <\lambda \mbox{).}
$$
Hence, one has, 
$$
g_k(\lambda ) >1>h_k^\pm (\lambda ) \mbox{\ \ \ if
and only if \ \ }\max (-a_0,\lambda_k^\pm ) <\lambda
<\lambda_k-a_0.
$$
Then  the equation in $E>0:T_k^\pm (\lambda ,E) =1/2$ admits
at least a solution for all $\lambda $ satisfying 
$$
\lambda >-m_k^\pm \quad\mbox{and}\quad \max (-a_0,\lambda
_k^\pm ) <\lambda <\lambda_k-a_0,
$$
that is, for all $\lambda $ satisfying 
$$
\max (-m_k^\pm ,\lambda_1^\pm ) <\lambda <\lambda_k-a_0,
$$
since $\max (-a_0,-m_k^\pm ) =-m_k^\pm $. 

If $a_0>c_k^\pm $ then for all $k\geq 1$ and $\lambda
>-m_k^\pm $, 
$$
\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac
12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to
+\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm (
\lambda ) .
$$
On the other hand, for all $\lambda \in (-b_k^\pm ,+\infty
)$,  $g_k(\lambda ) <h_k^\pm (\lambda )$. So,
since the function $\lambda \mapsto g_k(\lambda ) $ (resp. 
$\lambda \mapsto h_k^\pm (\lambda ) $) is strictly decreasing on 
$(-a_0,+\infty ) $ (resp. on $(-b_k^\pm ,+\infty
) $) and 
$$
g_k(\lambda_k-a_0) =1,\mbox{ (resp. }h_k^\pm (
\lambda_k^\pm ) =1\mbox{)}\,,
$$
it follows that 
$$
g_k(\lambda ) <1\mbox{ \ \ if and only if \ \ }\lambda
_k-a_0<\lambda
$$
(resp. 
$$
h_k^\pm (\lambda ) >1\mbox{\ \ \ if and only if\ \ \ }-b_k^\pm <\lambda <\lambda_k^\pm \mbox{).}
$$
Hence, one has, 
$$
g_k(\lambda ) <1<h_k^\pm (\lambda ) \mbox{\ \ \ if
and only if \ \ }\max (-b_k^\pm ,\lambda_k-a_0) <\lambda
<\lambda_k^\pm .
$$
Then the equation in $E>0:T_k^\pm (\lambda ,E) =1/2$ admits
at least a solution for all $\lambda $ satisfying
$$
\lambda >-m_k^\pm \quad\mbox{and}\quad \max (-b_k^{\pm
},\lambda_k-a_0) <\lambda <\lambda_k^\pm ,
$$
that is, for all $\lambda $ satisfying 
$$
\max (-m_k^\pm ,\lambda_k-a_0) <\lambda <\lambda_k^\pm 
$$
since $\max (-b_k^\pm ,-m_k^\pm ) =-m_k^\pm $. The proof
of Theorem \ref{thm1} is complete. \hfill$\diamondsuit$\smallskip

\paragraph{Proof of Theorem \ref{thm2}.}
If $a_{-}<a_0<a_{+}$ then  for all $\lambda >-m_\pm $,
$$
\lim_{E\to 0^+}T_1^\pm (\lambda ,E) =\frac
12g_1(\lambda ) \quad\mbox{and}\quad \lim_{E\to
+\infty }T_1^\pm (\lambda ,E) =\frac 12h_1^\pm (
\lambda ) .
$$
On the other hand, one has, 
$$
\mbox{for all }\lambda >-a_0,g_1(\lambda ) >h_1^+(
\lambda ) \mbox{\ \ \ \ (resp. for all }\lambda >-a_{-},g_1(
\lambda ) <h_1^-(\lambda ) \mbox{).}
$$
So, the same reasoning as in the superlinear (resp. sublinear) case leads to
the proof of the first assertion (resp. the second assertion) of Theorem \ref
{thm2}. \smallskip

If $a_{-}<a_k^\pm \leq a_0<a_{+}$ and $k\geq 2$, for all 
$\lambda >-m_k^\pm =-m$,
$$
\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac
12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to
+\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm (
\lambda ) .
$$
On the other hand, one has for all $\lambda >-a_{-},g_k(\lambda
) <h_k^\pm (\lambda ) $. So, the same reasoning as in
the sublinear case leads to the fact that the equation in $E>0:T_k^{\pm
}(\lambda ,E) =1/2$ admits at least a solution for all 
$\lambda $ satisfying
$$
\max (-m,\lambda_k-a_0) <\lambda <\lambda_k^\pm .
$$

If $a_{-}<a_0<a_k^\pm <a_{+}$ and $k\geq 2$,  for all 
$\lambda >-m$,
$$
\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac
12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to
+\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm (
\lambda ) .
$$
Since the function $a_0\mapsto \tilde \lambda_k^\pm (a_0) \in
(-a_{-},\Lambda_k^\pm ) $ is strictly increasing on the
interval $(a_{-},a_k^\pm ) $ then the function $a_0\mapsto
h_k^\pm (\tilde \lambda_k^\pm (a_0) ) $ is
strictly decreasing on $(a_{-},a_k^\pm ) $. Also, one has, 
$$
\lim_{a_0\to a_{-}}h_k^\pm (\tilde \lambda_k^\pm (
a_0) ) =\lim_{x\to -a_{-}}h_k^\pm (x)
=+\infty ,
$$
and 
$$
\lim_{a_0\to a_k^\pm }h_k^\pm (\tilde \lambda_k^{\pm
}(a_0) ) =\lim_{x\to +\infty }h_k^\pm (
x) =0\,.
$$
Then there exists a unique $\tilde a_k^\pm \in (a_{-},a_k^{\pm
}) $ such that for all $a_0\in (a_{-},a_k^\pm ) $ one
gets, 
\begin{eqnarray*}
h_k^\pm (\tilde \lambda_k^\pm (a_0) )
>1 &\Longleftrightarrow& a_{-}<a_0<\tilde a_k^\pm , \\ 
h_k^\pm (\tilde \lambda_k^\pm (a_0) )
=1 &\Longleftrightarrow& a_0=\tilde a_k^\pm , \\ 
h_k^\pm (\tilde \lambda_k^\pm (a_0) )
<1 &\Longleftrightarrow& \tilde a_k^\pm <a_0<a_k^\pm .
\end{eqnarray*}
So, one has to distinguish two cases:

Case where $a_{-}<a_0<\tilde a_k^\pm $. Then for
all $\lambda \in (-a_{-},\tilde \lambda_k^\pm (a_0)
) $,
$$
1<h_k^\pm (\tilde \lambda_k^\pm (a_0) )
=g_k(\tilde \lambda_k^\pm (a_0) ) <g_k(
\lambda ) <h_k^\pm (\lambda ) ,
$$
and for all $\lambda >\tilde \lambda_k^\pm (a_0) $,
$$
h_k^\pm (\lambda ) <1<g_k(\lambda ) \mbox{ \ \ if
and only if \ \ }\lambda_k^\pm <\lambda <\lambda_k-a_0\,,
$$
hence, the equation in $E>0:T_k^\pm (\lambda ,E) =1/2$
admits at least a solution for all $\lambda $ satisfying
$$
\lambda >-m\quad\mbox{and}\quad \lambda_k^\pm <\lambda
<\lambda_k-a_0\,,
$$
that is, for all $\lambda $ satisfying, 
$\max (-m,\lambda_k^\pm ) <\lambda <\lambda_k-a_0$.

Case where $\tilde a_k^\pm <a_0<a_k^\pm $. Then for all 
$\lambda >\tilde \lambda_k^\pm (a_0)$,
$$
h_k^\pm (\lambda ) <g_k(\lambda ) <g_k(
\tilde \lambda_k^\pm (a_0) ) =h_k^\pm (\tilde
\lambda_k^\pm (a_0) ) <1,
$$
and for all $\lambda \in (-a_{-},\tilde \lambda_k^\pm (
a_0) )$,
$$
g_k(\lambda ) <1<h_k^\pm (\lambda ) \mbox{ \ \ \
if and only if \ \ }\max (-a_{-},\lambda_k-a_0) <\lambda
<\lambda_k^\pm \,;
$$
hence, the equation in $E>0:T_k^\pm (\lambda ,E) =1/2$
admits at least a solution for all $\lambda $ satisfying
$$
\lambda >-m\mbox{\ \ \ \ and\ \ \ }\max (-a_{-},\lambda
_k-a_0) <\lambda <\lambda_k^\pm ,
$$
that is, for all $\lambda $ satisfying $\max (-m,\lambda
_k-a_0) <\lambda <\lambda_k^\pm $, this is so  because $\max (
-m,-a_{-}) =-m$. The proof of Theorem \ref{thm2} is complete. 
\hfill$\diamondsuit$

\paragraph{Proof of Theorem \ref{thm3}.}
If $q-p>0$ (resp. $q-p<0$), by Lemma \ref{lemma22}, for all $\lambda
>-m_k^\pm ,k\geq 1,
$$$
\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =+\infty \mbox{
(resp. }=0\mbox{) and }\lim_{E\to 0^+}T_k^\pm (\lambda
,E) =0\mbox{ (resp. }=+\infty \mbox{).}
$$
So, the intermediate value theorem implies that the equation in 
$E>0:T_k^\pm (\lambda ,E) =1/2$ admits at least a solution
for all $\lambda >-m_k^\pm $. Theorem \ref{thm3} is proved. 
\hfill$\diamondsuit$\smallskip

\paragraph{Remark 1.} Some easy computations show that 
$$
\frac{\partial T_{+}}{\partial E}(\lambda ,E)
=\frac{(p') ^{-\frac1p}}p (\frac{\partial r_{+}}{\partial E}(\lambda ,E) ) 
\int_0^1\frac{(H(\lambda ,r_{+}(\lambda ,E)
) -H(\lambda ,r_{+}(\lambda ,E) \xi )
) }{(F(\lambda ,r_{+}(\lambda ,E) )
-F(\lambda ,r_{+}(\lambda ,E) \xi ) )
^{1+\frac 1p}}d\xi ,
$$
where $H(\lambda ,x) :=p\tilde F(\lambda ,x)
-x\tilde f(\lambda ,x) ,\tilde F(\lambda ,x)
=\lambda \Phi_q(x) +F(x) $ and $\tilde f(
\lambda ,x) =\lambda \varphi_q(x) +f(x) $. 
So, if $q=p$, one gets, 
$$
\frac d{dx}(\frac{f(x) }{\varphi_p(x) }) =\frac{-1}{x\varphi_p(x) }\frac{\partial H}{\partial x}(\lambda ,x) ,x>0.
$$
Hence, in the particular case where the function $x\mapsto f(x)
/\varphi_p(x) ,(q=p) $ is strictly decreasing on
$(0,+\infty ) $ the function $x\mapsto H(\lambda
,x) $ is strictly increasing on $(0,+\infty ) $ and
then the time map $E\mapsto T_{+}(\lambda ,E) $ is strictly
increasing on $(0,+\infty ) $. So, uniqueness (if existence)
of the solution of the equation in $E>0:T_{+}(\lambda ,E) =1/2$
is guaranteed. The exact number of positive solution(s) in $A_1^+$ should
be obtained with this additional condition. The same remark works for the
result of Guedda and Veron \cite{GueddaVeron}. That is, their Theorem 2.1
holds without there condition (2.7).

\paragraph{Remark 2.} If $f$ is an odd function, $a_{-}=a_{+}$. The statements of
Theorems \ref{thm1}, \ref{thm2}, and \ref{thm3}\ may be simplified in this
particular case.

\paragraph{Remark 3.} Several corollaries may be deduced from Theorems \ref{thm1},\ \ref
{thm2},\ and \ref{thm3} and the above remarks. In fact, one may draw some
bifurcation diagrams and compute a lower bound on the number of solutions of
problem (\ref{p1}) in some cases and the exact number of positive solutions
in others. This would require more space and patience, and is left to the
diligent, patient reader. However, a qualitative feature of the variations
of the bifurcation branches as $a_0$ varies is known. In fact, if 
$a_{-}<a_{+}$, then the bifurcation branches are trend towards a same
direction for all $a_0<\tilde a_k^\pm $ and towards the opposite for all 
$a_0>\tilde a_k^\pm $. The case where $a_0=\tilde a_k^\pm $ remains an
open question. If $a_{-}>a_{+}$, one may study the asymmetric case, 
$a_{-}>a_0>a_{+}$, as in Theorem \ref{thm2}.

\paragraph{Remark 4.} One may observe that a common feature in Theorems \ref{thm1},\ \ref
{thm2},\ and \ref{thm3} is that the parameter $\lambda $ is taken, in
particular, such that the function $x\mapsto \lambda +f(x)
/\varphi_q(x) $ is strictly positive on $(-\infty
,0) $ and/or $(0,+\infty ) $. Some cases where this
function changes sign once are studied by Guedda and Veron \cite{GueddaVeron}
and by Boucherif, Bouguima and Derhab \cite{BBD}, both for the particular
case where $f$ is odd. So, it is reasonable to ask the question of what
happens if $f$ is not necessarily odd.

\section{Appendix: Historical overview on time maps}

At the beginning of time maps' history, the authors used them with the one
dimensional Laplacian operator, that is, with the one dimensional p-Laplacian
and $p=2$. From the 1960's we can mention Opial \cite{Opial1}, \cite
{Opial2}, Urabe \cite{Urabe1}, \cite{Urabe2}, \cite{Urabe3}, \cite{Urabe4}, 
\cite{Urabe5}, Pimbley \cite{Pimbley1}, \cite{Pimbley2}, and Gavalas \cite
{Gavalas}.

In the early 1970's, Laetsch \cite{Laetsch} used time maps to
study positive solutions to a class of boundary-value problems with
Dirichlet boundary data. Since then many authors have referred
to his work. We also want to mention Chafee and Infante \cite{ChafeeInfante}, 
and Chafee \cite{Chafee}. Brown and Budin \cite{BrownBudin1}, 
\cite{BrownBudin2} used the time map approach to study positive solutions 
to some boundary-value problems. Independently and about the same time, 
De-Mottoni and Tesei \cite{DeMottoniTesei1} studied positive solutions 
of some other class of boundary-value problems by means of the same method.

In the early 1980's Smoller and Wasserman \cite{SmollerWasserman}
introduced a technique that, in some circumstances, can be used to prove
uniqueness of the critical point of time maps. Their technique has been
used subsequently by many authors; see for instance,
Ammar Khodja \cite{AmmarKhodja}, Ramaswamy \cite{Ramaswamy}, S.
H. Wang and Kazarinoff, \cite{SHWang1}, \cite{SHWang2}, S. H. Wang and F. P.
Lee \cite{SHWang3}, S. H. Wang \cite{SHWang4}, \cite{SHWang5}, \cite{SHWang6}, 
and recently by Addou and Benmeza\"\i\ \cite{add3}.

The study of sign-changing solutions by means of time maps was initiated by
De-Mottoni and Tesei \cite{DeMottoniTesei2}, and independently,
some years after, by Shivaji \cite{Shivaji2}.

During the last two decades, time maps have been used in many publications. 
Besides the above mentioned papers, we want to add the following ones: 
Addou and Ammar Khodja \cite{Addou1}, Anuradha, Shivaji and Zhu \cite
{Anuradha1}, \cite{Anuradha15}, Anuradha and Shivaji \cite{Anuradha2},
Anuradha, C. Brown and Shivaji \cite{Anuradha3}, Brown, Ibrahim and Shivaji 
\cite{Brown}, Brunovsky and Chow \cite{Brunovsky}, Castro and Shivaji \cite
{Castro}, \cite{Castro2}, Ding and Zanolin \cite{Ding1}, \cite{Ding2},
Fernandes \cite{Fernandes}, Fonda and Zanolin \cite{Fonda}, Fonda, Gossez
and Zanolin \cite{Fonda2}, Schaaf \cite{Schaaf}, Shivaji \cite{Shivaji1},\ 
\cite{Shivaji2},\ \cite{Shivaji3}, Smoller, Tromba and Wasserman \cite
{Smoller}, Smoller and Wasserman \cite{SmollerWasserman1}, \cite
{SmollerWasserman2}, \cite{SmollerWasserman3}. Notice that 
this list is in alphabetical order, and is not complete by any means.
The differential operator in the equations studied in these papers is the
$p$-Laplacian with $p=2$. For more general differential operators, 
see references listed in Section 3 of this paper.

\paragraph{Acknowledgment.} I want to thank Professor Julio G. Dix for
his help in improving the presentation of an early version of this article.

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\noindent{\sc Idris Addou}\\
USTHB, Institut de Math\'ematiques \\
El-Alia, B.P. no. 32, Bab-Ezzouar\\
16111, Alger, Alg\'erie\\
E-mail address: idrisaddou@hotmail.com

\end{document}
