0$,
such that for $0<\lambda <\lambda _{0}$ Problem (\ref{123}) has
no nontrivial solution, it has exactly one nontrivial solution
for $\lambda =\lambda _{0}$, and exactly two nontrivial solutions
for $\lambda >\lambda _{0}.$ Moreover, all solutions lie on a
single smooth solution curve $\Sigma $, which for $\lambda
>\lambda _{0}$ has two branches denoted by $\Sigma ^{+}$ (the
upper branch) and $\Sigma ^{-}$ (the lower branch); $\Sigma ^{+}$
continues to the right up to $(\infty ,c);$ $\Sigma ^{-}$
continues to the right down to $(\infty ,g)$ for some $g\geq
\beta ;$ there exists a unique turning point on the curve, the
curve bend to the right at the turning point. (See Fig 12 in
\cite{OuyangShi}.)
{\bf{(b)}} If $f(0)<0,$ there exist
$\bar{\lambda}>\lambda _{0}>0$, such that for $0<\lambda <\lambda
_{0}$ Problem (\ref{123}) has no solution, it has exactly one
solution for $\lambda >\bar{\lambda}$ or $\lambda =\lambda _{0}$,
and exactly two solutions for $\lambda _{0}<\lambda \leq
\bar{\lambda}.$ Moreover, all solutions lie on a single smooth
solution curve $\Sigma $, which for $\lambda >\lambda _{0}$ has
two branches denoted by $\Sigma ^{+}$ (the upper branch) and
$\Sigma ^{-}$ (the lower branch); $\Sigma ^{+}$ continues to the
right up to $(\infty ,c);$ $\Sigma ^{-}$ continues to the right
down to $(\bar{\lambda},g)$ for some $g\geq \beta ;$ there exists
a unique turning point on the curve, the curve bend to the right
at the turning point. (See Fig. 10 in \cite{OuyangShi}.)
\end{theorem}
More recently Korman \cite{KormanPreprint} was able to avoid
having to prove this positivity condition, and replaced it by an
indirect argument. He shows that it is sufficient to prove that
any nontrivial solution of (\ref{atfL}) cannot vanish exactly
once. This way he considerably simplifies the proof of Ouyang and
Shi and make it more elegant.
Korman \cite{KormanPreprint} considers Problem (\ref{atf11}) for $n\geq 1.$ He assumes that
$f\in C^2( \overline{ {\mathbb R}}_+) $, $f\left( 0\right) =0$
and satisfies (\ref{124})-(\ref{128}). Next, he proves the
following
\begin{theorem}
\cite{KormanPreprint} Assume that $f( u) $ satisfies the
conditions listed above. For Problem (\ref{atf11}) there is a
critical $\lambda_{0}>0$ such that Problem (\ref{atf11}) has
exactly $0$, $1$ or $2$ nontrivial solutions, depending on
whether $\lambda <\lambda_{0}$, $\lambda =\lambda_{0}$ or
$\lambda >\lambda_{0}$. Moreover, all solutions lie on a single
smooth solution curve, which for $\lambda >\lambda_{0}$ has two
branches denoted by $0b$ for all $\lambda >\lambda_{0}$.
\end{theorem}
Other related results are available in the literature. (See for
instance, Pimbley, \cite{Pimbley}, Wang and Kazarinoff
\cite{WangKaza1}, \cite {WangKaza2}, Schaaf \cite{Schaaf}, Korman
\cite{Korman97}, Shi and Shivaji \cite{ShiShivaji}). But as we
have indicated at the beginning of this section, we shall not
attempt to make a complete historical review. Thus, we apologize
to all authors whose results are close to cubic-like
nonlinearities and which have not been either described in this
section or listed in the references.
\section{{The method used}}
\label{sec.3}We shall make use of the quadrature method. Denote
by $g$ a nonlinearity and by $p$ a real parameter, and we assume,
\begin{equation}
g\in C({\mathbb R} ) \text{\ \ and\ \ \ }10\text{ and
}E^{p}-p'G( \xi ) >0,\forall \,\xi ,0<\kappa \xi <\kappa
s\right\} \;\text{and,}
\end{equation*}
\begin{equation*}
r_{\kappa }( E) =0,\ \text{if }X_{\kappa }( E) =\emptyset \ \ \ \
\text{and}\ \ \ \ r_{\kappa }( E) =\kappa \sup ( \kappa X_{\kappa
}( E) ) \;\;\text{otherwise.}
\end{equation*}
Let
\begin{equation*}
\tilde{D}_{\kappa }=\left\{ E\geq 0:0<\left| r_{\kappa }( E)
\right| <+\infty \text{ and }\kappa \int_{0}^{r_{\kappa }( E) }(
E^{p}-p'G( t) ) ^{\frac{-1}{p}}dt<+\infty \right\}
\end{equation*}
and $\tilde{D}=\tilde{D}_+\cap \tilde{D}_-$. Also, let $\tilde{D}%
_{k}^{\kappa }:=\tilde{D}$ if $k\geq 2$, and $\tilde{D}_{1}^{\kappa }:=%
\tilde{D}_{\kappa }$. Define the following time-maps,
\begin{equation*}
T_{\kappa }( E) =\kappa \int_{0}^{r_{\kappa }( E) }( E^{p}-p'G(
t) ) ^{\frac{-1}{p}}dt,E\in \tilde{D}_{\kappa }.
\end{equation*}
\begin{equation*}
\begin{array}{rclcc}
T_{2n}^{\kappa }( E) & = & n( T_+( E)
+T_-( E) ) , & n\in {\mathbb N}, & E\in \tilde{D}, \\
T_{2n+1}^{\kappa }( E) & = & T_{2n}^{\kappa }( E) +T_{\kappa }(
E) , & n\in {\mathbb N}, & E\in \tilde{D}.
\end{array}
\end{equation*}
\begin{theorem}[Quadrature method]
\label{quad} Assume that (\ref{meth2}) holds. Let $E>0$, $\kappa
=+,-$. Then, Problem (\ref{meth3}) admits a solution
$u_{k}^{\kappa }\in A_{k}^{\kappa }$ satisfying $( u_{k}^{\kappa
}) '( 0) =\kappa E$ if and only if $E\in \tilde{D}_{k}^{\kappa
}-\left\{ 0\right\} $ and $T_{k}^{\kappa }( E) =(1/2),$\ and in
this case the solution is unique.
\end{theorem}
\begin{remark}
\label{Remark}In practice, to compute $\tilde{D}_{\kappa }$ we
first compute the set
\begin{equation*}
D_{\kappa }=\left\{ E>0:0<\left| r_{\kappa }( E) \right| <+\infty
\text{ and }\kappa g( r_{\kappa }( E) ) >0\right\}
\end{equation*}
and then we deduce $\tilde{D}_{\kappa }$ by observing that,
$D_{\kappa }\subset \tilde{D}_{\kappa }\subset
\overline{D}_{\kappa }$.
\end{remark}
\section{Some preliminary lemmas}
\label{sec.4}To apply Theorem \ref{quad} we have, first, to
determine the definition domains $\tilde{D}_+$ and $\tilde{D}_-$
of the time-maps $T_+$ and $T_-$ respectively. Lemma \ref{lemma1}
is used to. Next, we have to compute $\tilde{D}=\tilde{D}_+\cap
\tilde{D}_-$ which is the definition domain of the time-maps
$T_{k}^{\kappa }$ for all $k\geq 2$, $\kappa =+,-$. This is done
in Lemma \ref{lemma222}. Next, the aim of Lemma \ref{lemma5} is
to compare the maximum and minimum of any solution of our
problem. This comparison may be used subsequently. Next, we
define all the time-maps and make some useful transformations.
Lemmas \ref{lemma2}, \ref{lemma31}, and \ref{lemma32} are
dedicated to the computation of the limits of all the time-maps,
at each boundary point of their domains. In Lemma \ref{lemma4} we
show that under some appropriate conditions, the time-maps may be
monotonic increasing. The following step is crucial; when the
time maps are not monotonic, usually one
tries to compare there maximum and/or minimum with the real number $\frac{1}{%
2}$. This task is much simpler when $f$ is odd. Indeed, the
time-maps $T_+$ and $T_-$ are always equal if $f$ is odd. Thus;
(*) $T_{k}^{\pm }=kT_+$ for all $k\geq 2$. This way, to study the
maximum and/or minimum value of $T_{k}^{\pm }$ it suffices to
handle those of $T_+$ only. Unfortunately, in our \textit{p.h.o.}
case it seems that the identity (*) is not satisfied. To overcome
this difficulty we define two maps such that both $T_+$ and $T_-
$ are bounded from below by the first one, and from above by the
second one. Thus, $T_{k}^{\pm }$ is bounded from below by $k$
times the first map and from above by $k$ times the second one.
So, it suffices to study the two bounding maps. Moreover, these
estimates seem to be optimal in the sense that in the particular
odd case, they imply that $T_+$ and $T_- $ are equal!
In Lemma \ref{lemma6} we compare the time-maps $T_+$ and $T_-$
with the defined two maps. In Lemma \ref{lemma7} we deduces the
estimates of $T_{k}^{\pm }$ for $k\geq 2$. In Lemma \ref{lemma75}
we provide an identity which may be used in the sequel. This
identity seems to be interesting for its own right and motivates
us to ask a question in Section \ref{sec.6}. Lemma \ref{lemma8}
is dedicated to the limits of these two bounding maps.
Under appropriate conditions we provide, in Lemma \ref{lemma9},
estimates which are used to prove uniqueness of the minimum of
the time-maps $T_+$\ and $T_-$ respectively. This kind of
estimates was introduced by Smoller and Wasserman
\cite{SmollerWasserman} and was crucial in their study of
uniqueness of critical points.
\begin{lemma}
\label{lemma1}Consider the function defined in ${\mathbb R} ^{\pm
}$ by,
\begin{equation}
s\longmapsto G_{\pm }( \lambda ,E,s) :=E^{p}-p^{\prime }\lambda
F( s) , \label{Eq.1}
\end{equation}
where $E,\lambda >0$ and $p,q>1$ are real parameters, $F( s)
=\int_{0}^{s}f( t) dt$. Assume that (\ref{pq1})-(\ref{pq4}) hold.
Then,
\begin{description}
\item[(i)] If $E>E_{*}^{\pm }( p,\lambda ) :=( p^{\prime
}\lambda F( \alpha_{\pm }) ) ^{1/p}$, the function $G_{\pm }(
\lambda ,E,\cdot ) $ is strictly positive in ${\mathbb R} ^{\pm
}$.
\item[(ii)] If $E=E_{*}^{\pm }( p,\lambda ) $, the function
$G_{\pm }( \lambda ,E,\cdot ) $ is strictly positive in $(
0,\alpha_+) $ (resp. in $( \alpha_-,0) $) and vanishes at
$\alpha_{\pm }$.
\item[(iii)] If $00,
\label{der}
\end{equation}
for all $E\in ( 0,E_{*}^{\pm }( p,\lambda ) ) $.
\item[(b)] $\lim\limits_{E\rightarrow 0^+}s_{\pm }( \lambda
,E) =0$ \ and\ $\lim\limits_{E\rightarrow E_{*}^{\pm }}s_{\pm }(
\lambda ,E) =\alpha_{\pm }$,
\item[(c)] $\lim\limits_{E\rightarrow 0^+}\left| s_{\pm }( \lambda
,E) \right| /E=\left\{
\begin{array}{lcc}
+\infty & \text{if} & q-p>0 \\
( \dfrac{p-1}{\lambda a_{0}}) ^{1/p} & \text{if} & q-p=0 \\
0 & \text{if} & q-p<0,
\end{array}
\right. $
\item[(d)] $\lim\limits_{E\rightarrow 0^+}F( \kappa s_{\eta }(
\lambda ,E) \xi ) /E^{p}=\xi ^{q}/( \lambda p^{\prime })
$,\newline for all $\xi >0$ and all $( \kappa ,\eta ) \in \left\{
+,-\right\} ^{2}$.
\end{description}
\end{description}
\end{lemma}
\paragraph{Proof.} For any fixed $p>1$ and $E\geq 0$, consider the function,
\begin{equation}
s\longmapsto G_{\pm }( \lambda ,E,s) :=E^{p}-p^{\prime }\lambda
F( s) , \label{Eq1}
\end{equation}
defined in ${\mathbb R}^{\pm }$. One has, $\frac{dG_{\pm }}{ds}(
\lambda ,E,s) =-p'\lambda f( s) $. Hence, according to
(\ref{pq2}), $G_+( \lambda ,E,\cdot )$ (resp. $G_-( \lambda
,E,\cdot )$) is strictly decreasing in $( 0,\alpha_+) $ (resp. in
$( -\infty ,\alpha _-) $) and according to (\ref{pq3}), it is
strictly increasing in $( \alpha_+,+\infty ) $ (resp. in $(
\alpha_-,0) $). Moreover, according to (\ref{pq1}), $
\frac{dG_{\pm }}{ds}( \lambda ,E,\alpha_{\pm }) =0.$ Therefore,
it follows that $G_{\pm }( \lambda ,E,\cdot ) $ is strictly
positive in ${\mathbb R}^{\pm }$ for all $E>E_{*}^{\pm }:=(
p'\lambda F( \alpha _{\pm }) ) ^{1/p}$, admits a unique positive
(resp. negative) zero, $\alpha_{\pm }$, and is strictly positive
in $( 0,\alpha +) $ (resp. in $( \alpha_-,0) $) at $E=E_{*}^{\pm
}$, and finally admits a first positive (resp. negative) zero
$s_{\pm }=s_{\pm }( \lambda ,E) $ and is strictly positive in $(
0,s_+) $ (resp. in $( s_-,0) $) for all $E:01$ and $\lambda >0$, consider the real valued function,
\begin{equation*}
( E,s) \longmapsto G_{\pm }( E,s) :=E^{p}-p'\lambda F( s) ,
\end{equation*}
defined in $\Omega_+=( 0,E_{*}^+) \times ( 0,\alpha _+) $ (resp.
$\Omega_-=( 0,E_{*}^{-}) \times ( \alpha_-,0) $). One has $G_{\pm
}\in C^1( \Omega_{\pm }) $ and, $ \frac{\partial G_{\pm
}}{\partial s}( E,s) =-p'\lambda f( s) \text{\ \ in }\Omega_{\pm
}, $ hence, according to (\ref{pq2}) (resp. (\ref{pq3})), it
follows that, $ \pm \frac{\partial G_{\pm }}{\partial s}( E,s)
<0\text{\ \ \ in\ }\Omega_{\pm }, $ and one may observe that
$s_{\pm }( \lambda ,E) $ belongs to the open interval $(
0,\alpha_+) $ (resp. $( \alpha _-,0) $) and satisfies, from its
definition,
\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,E_{*}^{\pm }) ,%
{\mathbb R}
) $ and to obtain the expression of $\dfrac{\partial s_{\pm }}{%
\partial E}( \lambda ,E) $ given in \textbf{(a)}. Its sign is
given by (\ref{pq2}) (resp. (\ref{pq3}) together with the fact
that $s_{\pm }( \lambda ,E) $ belongs to $( 0,\alpha_+) $ (resp.
$( \alpha_-,0) $). Therefore, Assertion \textbf{(a)} is proved.
\paragraph{Proof of (b)}
For any fixed $p>1$ and $\lambda >0$, Assertion \textbf{(a)} of
the current lemma implies that the function defined in $(
0,E_{*}^{\pm }) $ by $E\mapsto s_{\pm }( \lambda ,E) $ is
strictly increasing (resp. strictly decreasing). It is bounded
from below by $0$ (resp. by $\alpha_-$) and from above by
$\alpha_+$ (resp. by $0$). Then, the limits
$\lim\limits_{E\rightarrow 0^+}s_{\pm }( \lambda ,E) =\ell_0^{\pm
}$ and $\lim\limits_{E\rightarrow E_{*}^{\pm }}s_{\pm }( \lambda
,E) =\ell_{*}^{\pm }$) exist as real numbers and moreover,
\begin{equation*}
\alpha_-\leq \ell_{*}^{-}<\ell_0^{-}\leq 0\leq \ell_0^+<\ell
_{*}^+\leq \alpha_+.
\end{equation*}
Let us observe that, for any fixed $p>1$ and $\lambda >0$, the
function, $( E,s) \mapsto G_{\pm }( E,s) $, is continuous in $[
0,E_{*}^+] \times [ 0,\alpha_+] $ (resp. in $[ 0,E_{*}^{-}]
\times [ \alpha_-,0] $) and the function $E\mapsto s_{\pm }(
\lambda ,E) $ is continuous in $( 0,E_{*}^{\pm }) $ and satisfies
(\ref{Eq.2}). So, by passing to the limit in (\ref{Eq.2}) as $E$
tends to $0^+$, one gets, $ 0=\lim_{E\rightarrow 0^+}G_{\pm }(
E,s_{\pm }( \lambda ,E) ) =G_{\pm }( 0,\ell_0^{\pm }) . $ Hence,
$\ell_0^{\pm }$ is a zero, belonging to $[ 0,\alpha _+] $, (resp.
$[ \alpha_-,0] $) to the equation in $s, $ $ G_{\pm }( 0,s) =0. $
By resolving this equation one gets: $\ell_0^{\pm }=0$. Also, by
passing to the limit in (\ref{Eq.2}) as $E$ tends to $E_{*}^{\pm
}$, one gets,
\begin{equation*}
0=\lim_{E\rightarrow E_{*}^{\pm }}G_{\pm }( E,s_{\pm }( \lambda
,E) ) =G_{\pm }( E_{*}^{\pm },\ell_0^{\pm }) .
\end{equation*}
Hence, $\ell_{*}^{\pm }$ is a zero, belonging to $( 0,\alpha _+]
$, (resp. $[ \alpha_-,0) $) to the equation in $s,
$\begin{equation*} G_{\pm }( E_{*}^{\pm },s) =0.
\end{equation*}
By resolving this equation one gets: $\ell_{*}^{\pm }=\alpha_{\pm
}$. Therefore, Assertion \textbf{(b)} follows.
\paragraph{Proof of (c)}
Let $\Phi_{q}( s) =\int_{0}^{s}\varphi ( t) dt=(1/q)\left|
s\right| ^{q}$. Observe that from the definition of $s_{\pm }(
\lambda ,E) $ one has
\begin{equation}
E^{p}=p'\lambda F( s_{\pm }( \lambda ,E) ) , \label{dklg}
\end{equation}
hence, dividing by $\left| s_{\pm }( \lambda ,E) \right| ^{p}$,
using l'Hopital's rule and (\ref{pq4}) one gets,
\begin{equation*}
\begin{array}{ccl}
\lim\limits_{E\rightarrow 0^+}E^{p}/\left| s_{\pm }( \lambda
,E) \right| ^{p} & = & \lim\limits_{E\rightarrow 0^+}\dfrac{%
p'\lambda }{q}\left| s_{\pm }( \lambda ,E) \right|
^{q-p}\dfrac{F( s_{\pm }( \lambda ,E) ) }{\Phi
_{q}( s_{\pm }( \lambda ,E) ) } \\
& = & \dfrac{p'\lambda }{q}\lim\limits_{E\rightarrow 0^+}\left|
s_{\pm }( \lambda ,E) \right| ^{q-p}\lim\limits_{s\rightarrow 0}%
\dfrac{f( s) }{\varphi_{q}( s) } \\
& = & \dfrac{p'\lambda }{q}a_{0}\cdot \lim\limits_{E\rightarrow
0^+}\left| s_{\pm }( \lambda ,E) \right| ^{q-p}.
\end{array}
\end{equation*}
Therefore, Assertion \textbf{(c)} follows.
\paragraph{Proof of (d)}
Remark that for all $\xi >0$ and all $( \kappa ,\eta ) \in
\left\{ +,-\right\} ^{2}$ one has,
\begin{equation*}
\frac{F( \kappa s_{\eta }( \lambda ,E) \xi ) }{E^{p}}%
=\frac{F( \kappa s_{\eta }( \lambda ,E) \xi ) }{\Phi
_{q}( \kappa s_{\eta }( \lambda ,E) \xi ) }\frac{%
\Phi_{q}( \kappa s_{\eta }( \lambda ,E) \xi ) }{%
E^{p}}.
\end{equation*}
Using l'Hopital's rule and (\ref{pq4}) one gets,
\begin{equation*}
\lim_{E\rightarrow 0^+}\frac{F( \kappa s_{\eta }( \lambda ,E) \xi
) }{\Phi_{q}( \kappa s_{\eta }( \lambda ,E) \xi )
}=\lim_{E\rightarrow 0^+}\frac{f( \kappa s_{\eta }( \lambda ,E)
\xi ) }{\varphi_{q}( \kappa s_{\eta }( \lambda ,E) \xi ) }=a_{0}.
\end{equation*}
On the other hand, since $\Phi_{q}( \cdot ) $ is an odd function
then
\begin{equation*}
\frac{\Phi_{q}( \kappa s_{\eta }( \lambda ,E) \xi )
}{E^{p}}=\frac{\Phi_{q}( s_{\eta }( \lambda ,E) \xi ) }{E^{p}}.
\end{equation*}
Thus, for all $\xi >0$, using l'Hopital's rule, (\ref{der}) and
(\ref{pq4}) one gets,
\begin{equation*}
\begin{array}{ccl}
\lim\limits_{E\rightarrow 0^+}\dfrac{\Phi_{q}( s_{\eta }( \lambda
,E) \xi ) }{E^{p}} & = & \lim\limits_{E\rightarrow 0^+}\dfrac{\xi
s_{\eta }'( \lambda ,E) \varphi
_{q}( s_{\eta }( \lambda ,E) \xi ) }{pE^{p-1}} \\
& = & \lim\limits_{E\rightarrow 0^+}\dfrac{\xi ^{q}}{\lambda p'}%
\dfrac{\varphi_{q}( s_{\eta }( \lambda ,E) ) }{%
f( s_{\eta }( \lambda ,E) ) }=\dfrac{\xi ^{q}}{%
a_{0}\lambda p'}.
\end{array}
\end{equation*}
Hence, Assertion \textbf{(d)} follows. Therefore, Lemma
\ref{lemma1} is proved. \hfill$\diamondsuit$
\paragraph{Remark.} We have used condition (\ref{pq4}) in, and only in, the
process of the proofs of assertions \textbf{(c)} and \textbf{(d)}.
\bigskip Now we are ready to compute $X_{\pm }( \lambda ,E) $
as defined in Section \ref{sec.3}, for any $E>0$ and $\lambda
>0$.\ In fact,
\begin{equation*}
X_+( \lambda ,E) =\left\{
\begin{array}{lcl}
( 0,+\infty ) & \text{if} & E>E_{*}^+ \\
( 0,\alpha_+) & \text{if} & E=E_{*}^+ \\
( 0,s_+( \lambda ,E) ) & \text{if} & 0E_{*}^{-} \\
( \alpha_-,0) & \text{if} & E=E_{*}^{-} \\
( 0,s_-( \lambda ,E) ) & \text{if} & 0E_{*}^+ \\
\alpha_+ & \text{if} & E=E_{*}^+ \\
s_+( \lambda ,E) & \text{if} & 0E_{*}^{-} \\
\alpha_- & \text{if} & E=E_{*}^{-} \\
s_-( \lambda ,E) & \text{if} & 00,\forall \lambda >0,E\in ( 0,E_{*}^{\pm }( p,\lambda
) ) , \label{eq11}
\end{equation}
\begin{equation}
\lim\limits_{E\rightarrow 0^+}r_{\pm }( \lambda ,E) =0\text{\ \ \
\ \ and \ \ \ }\lim\limits_{E\rightarrow E_{*}^{\pm }}r_{\pm }(
\lambda ,E) =\alpha_{\pm }. \label{abhu}
\end{equation}
\begin{equation}
\lim\limits_{E\rightarrow 0^+}\left| r_{\pm }( \lambda ,E)
\right| /E=\left\{
\begin{array}{lcl}
+\infty & \text{if} & q-p>0 \\
( \dfrac{p-1}{\lambda a_0}) ^{1/p} & \text{if} & q-p=0 \\
0 & \text{if} & q-p<0,
\end{array}
\right. \label{bbhu}
\end{equation}
\begin{equation}
\lim\limits_{E\rightarrow 0}F( \kappa r_\eta ( \lambda ,E) \xi )
/E^p=\xi ^q/( \lambda p') ,\;\;\text{ for all }\xi >0\text{ and
}( \kappa ,\eta ) \in \left\{ +,-\right\} ^2. \label{bbhuu}
\end{equation}
On the other hand,
\begin{equation*}
0<\left| r_{\pm }( \lambda ,E) \right| <+\infty \text{ \ \ if and
only if \ }00\text{ \ \ if and only
if \ }00:0<\left| r_{\pm }(
p,\lambda ,E) \right| <+\infty \text{\ and }\pm \lambda f(
r_{\pm }( \lambda ,E) ) >0\right\} \\
& = & ( 0,E_{*}^{\pm }) ,
\end{array}
\end{equation*}
and
\begin{equation*}
D( \lambda ) :=D_+( \lambda ) \cap D_-( \lambda ) =( 0,\inf (
E_{*}^+( \lambda ) ,E_{*}^{-}( \lambda ) ) ) .
\end{equation*}
\begin{lemma}
\label{lemma222}If $f$ satisfies (\ref{pq1})-(\ref{pq3}) and
(\ref{pho}), then for all $\lambda >0:\;E_{*}^{-}( \lambda ) \leq
E_{*}^+( \lambda ) $. Therefore, $D( \lambda ) =( 0,E_{*}^{-}(
\lambda ) ) $.
\end{lemma}
\paragraph{Remark.}
For all $\lambda >0$, $k\geq 1$ and $\kappa =+,-$, define
\begin{equation*}
E_k^\kappa ( \lambda ) =\left\{
\begin{array}{l}
E_{*}^\kappa ( \lambda ) \text{\ if }k=1 \\
E_{*}^{-}( \lambda ) \text{\ if }k\geq 2\text{\ and }\kappa \text{
arbitrary.}
\end{array}
\right.
\end{equation*}
Therefore, for all $\lambda >0$, $k\geq 1$ and $\kappa =+,-$, one
has $D_k^\kappa ( \lambda ) =(0,E_k^\kappa ( \lambda ) )$, where
$D_k^\kappa $ is defined in Section \ref{sec.3}.
\paragraph{Proof of Lemma \ref{lemma222}.}
For all $\lambda >0$, one has $E_{*}^{-}( \lambda ) \leq E_{*}^+(
\lambda ) $ if and only if, $p'\lambda F( \alpha_-) \leq
p'\lambda F( \alpha _+) $, which is equivalent to,
\begin{equation*}
-\int_{0}^{-\alpha_-}f( -t) dt\leq \int_{0}^{\alpha _+}f( t) dt,
\end{equation*}
that is,
\begin{equation*}
0\leq \int_{0}^{-\alpha_-}( f( -t) +f( t) )
dt+\int_{-\alpha_-}^{\alpha_+}f( t) dt.
\end{equation*}
The first integral is positive from (\ref{pho}) and the second
one is too from (\ref{pq2}) and (\ref{jzehjf}) (Recall that
(\ref{jzehjf}) is a consequence of (\ref{pq1}), (\ref{pq3}) and
(\ref{pho}), see the Introduction). Therefore, Lemma
\ref{lemma222} is proved.
\hfill$\diamondsuit$
\begin{lemma}
\label{lemma5}Assume that (\ref{pq1})-(\ref{pq3}) and (\ref{pho})
hold. For all $\lambda >0$ and\newline $E\in ( 0,E_{*}^{-}(
\lambda ) ) $, one has
\begin{equation*}
y( r_-( \lambda ,E) ) =r_+( \lambda ,E)
\end{equation*}
where $y(\cdot )$ is defined in (\ref{12kwdj}). Explicitly, for
all $\lambda >0 $ and $E\in ( 0,E_{*}^{-}( \lambda ) ) $ one has,
\begin{equation*}
00$ and $E\in ( 0,E_{*}^{-}(
\lambda ) ) $, one has from the definition of $r_{\pm }( \lambda
,E) :E^{p}-p'\lambda F( r_{\pm }( \lambda ,E) ) =0$. Thus,
\begin{equation}
F( r_+( \lambda ,E) ) -F( r_-( \lambda ,E) ) =0. \label{dyuk}
\end{equation}
On the other hand
\begin{equation*}
F( r_-( \lambda ,E) ) =\int_{0}^{r_-( \lambda ,E) }f( t)
dt=-\int_{0}^{-r_-( \lambda ,E) }f( -t) dt,
\end{equation*}
and
\begin{eqnarray*}
F( r_+( \lambda ,E) ) &=&\int_{0}^{r_+(
\lambda ,E) }f( t) dt \\
&=&\int_{0}^{-r_-( \lambda ,E) }f( t) dt+\int_{-r_-( \lambda ,E)
}^{r_+( \lambda ,E) }f( t) dt.
\end{eqnarray*}
Thus,
\begin{eqnarray}
F( r_+( \lambda ,E) ) -F( r_-( \lambda ,E) ) &=&\int_{0}^{-r_-(
\lambda ,E) }(
f( t) +f( -t) ) dt \label{gqw!dlk} \\
&&+\int_{-r_-( \lambda ,E) }^{r_+( \lambda ,E) }f( t) dt. \notag
\end{eqnarray}
Observe that by (\ref{eq11}) and (\ref{abhu}), $0<-r_-( \lambda
,E) <-\alpha_-$ and by (\ref{pho}) the function $t\mapsto ( f( t)
+f( -t) ) $ is positive in $( 0,-\alpha_-) $, thus the first
integral in (\ref{gqw!dlk}) is positive. Now, if we assume that
there exists at least a $\lambda_{0}>0$ and an $E_{0}\in (
0,E_{*}^{-}( \lambda_{0}) ) $ such that $r_+( \lambda_{0},E_{0})
+r_-( \lambda _{0},E_{0}) >0$, it follows that $0<-r_-( \lambda
_{0},E_{0}) 0$
which is a contradiction with (\ref{dyuk}). Therefore, for all
$\lambda >0$ and $E\in ( 0,E_{*}^{-}( \lambda ) ) $ one has,
\begin{equation*}
00$ and $E\in ( 0,E_{*}^{-}( \lambda ) ) $ one has,
\begin{equation*}
F( t) =F( r_-( \lambda ,E) ) \Longrightarrow t=r_+( \lambda ,E)
,\text{\ }\forall t\in ( 0,-r_-( \lambda ,E) ] .
\end{equation*}
But, this is immediate, since $F$ is strictly increasing in $(
0,-r_-( \lambda ,E) ] $. Therefore, Lemma \ref{lemma5} is proved.
%TCIMACRO{\TeXButton{End Proof}{\hfill$\diamondsuit$}}
%BeginExpansion
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%EndExpansion
At present, if $f$ satisfies (\ref{pq1})-(\ref{pq3}) we define,
for any $p,q>1$, $\lambda >0$ and $E\in D_{\pm }( \lambda ) $,
the time map $T_{\pm }$ \thinspace by
\begin{equation*}
T_{\pm }( \lambda ,E) :=\pm \int_{0}^{r_{\pm }( \lambda ,E) }(
E^{p}-p'\lambda F( \xi ) ) ^{-1/p}d\xi ,\;E\in D_{\pm }( \lambda
) =( 0,E_{*}^{\pm }) .
\end{equation*}
Actually, $T_{\pm }( \lambda ,E) $\ is defined for all $\lambda
>0$ and $E\in \tilde{D}_{\pm }( \lambda ) $ (see Remark \ref
{Remark}). A simple change of variables shows that,
\begin{equation}
T_{\pm }( \lambda ,E) =\left| r_{\pm }( \lambda ,E) \right|
\int_{0}^1( E^{p}-p'\lambda F( r_{\pm }( \lambda ,E) \xi ) )
^{-1/p}d\xi , \label{A}
\end{equation}
which can be written as,
\begin{equation}
T_{\pm }( \lambda ,E) =( \left| r_{\pm }( \lambda ,E) \right| /E)
\int_{0}^1( 1-p'\lambda F( r_{\pm }( \lambda ,E) \xi ) /E^{p})
^{-1/p}d\xi . \label{qze}
\end{equation}
Also, observe that one has from (\ref{dklg}), (\ref{mfrkl1}) and
(\ref {mfrkl2}), $E^{p}=\lambda p'F( r_{\pm }( \lambda ,E) ) $,
so, (\ref{A}) may be written as,
\begin{equation}
T_{\pm }( \lambda ,E) =\pm ( \lambda p') ^{-1/p}\int_{0}^{r_{\pm
}( \lambda ,E) }( F( r_{\pm }( \lambda ,E) ) -F( r_{\pm }( \lambda
,E) \xi ) ) ^{-1/p}d\xi . \label{a15}
\end{equation}
For any $p>1$ and $x\in [ 0,\alpha_+] $ (resp. $x\in [
\alpha_-,0] $ let us define $S_+( x) $ (resp. $S_-( x) $) by,
\begin{equation*}
S_{\pm }( x) :=\pm \int_{0}^{x}( F( x) -F( \xi ) ) ^{-1/p}d\xi
\in [ 0,+\infty ] .
\end{equation*}
Then, (\ref{a15}) may be written as,
\begin{equation}
T_{\pm }( \lambda ,E) =( \lambda p') ^{-1/p}S_{\pm }( r_{\pm }(
\lambda ,E) ) . \label{a16}
\end{equation}
On the other hand, we define for any $p>1$, $\lambda >0$ and
$E\in D( \lambda ) =( 0,E_{*}^{-}) $, the time maps,
\begin{equation}
T_{2n}^{\pm }( \lambda ,E) :=n( T_+( \lambda ,E) +T_-( \lambda
,E) ) ,\lambda >0,E\in ( 0,E_{*}^{-}) ,n\geq 0, \label{a18}
\end{equation}
\begin{equation}
T_{2n+1}^{\pm }( \lambda ,E) :=T_{2n}^{\pm }( \lambda ,E) +T_{\pm
}( \lambda ,E) ,\lambda >0,E\in ( 0,E_{*}^{-}) ,n\geq 0.
\label{a19}
\end{equation}
The limits of these time maps are the aim of the following Lemmas.
\begin{lemma}
\label{lemma2}Assume that \textbf{(}\ref{pq1})-(\ref{pq3}) hold.
\begin{description}
\item[(i)] If $(\ref{pq5})_{\pm }$ holds,
$\lim\limits_{E\rightarrow E_{*}^{\pm }}T_{\pm }( \lambda ,E) =(
\lambda p') ^{-1/p}S_{\pm }( \alpha_{\pm }) $ and\newline $S_{\pm
}( \alpha_{\pm }) <+\infty $ if and only if $q-p<0$.
\item[(ii)] If (\ref{pq4}) holds, $\lim\limits_{E\rightarrow 0^+}T_{\pm
}( \lambda ,E) =\left\{
\begin{array}{lcc}
+\infty & \text{if} & q-p>0 \\
\dfrac{1}{2}( \dfrac{\lambda_{1}}{\lambda a_{0}}) ^{1/p} & \text{%
if} & q-p=0 \\
0 & \text{if} & q-p<0.
\end{array}
\right. $
\item[(iii)] If (\ref{pho}) holds, for all $\lambda >0$, $r_+(
\lambda ,E_{*}^{-}( \lambda ) ) =y( \alpha _-) $, where $y( \cdot
) $ is defined in (\ref{12kwdj}) and $T_+( \lambda ,E_{*}^{-}(
\lambda ) ) =( \lambda p') ^{-1/p}S_+( y( \alpha_-) ) $.
\end{description}
\end{lemma}
\paragraph{Proof of Lemma \ref{lemma2}}
\paragraph{Proof of (i)}
The value of the limit follows by passing to the limit in
(\ref{a16}) as $E$
tends to $E_{*}^{\pm }$. In order to show the second assertion of \textbf{(i)%
} one observes that
\begin{equation*}
S_{\pm }( \alpha_{\pm }) =\pm \int_0^{\alpha_{\pm }\mp \delta
}\left\{ F( \alpha_{\pm }) -F( \xi ) \right\} ^{-1/p}d\xi \pm
\int_{\alpha_{\pm }\mp \delta }^{\alpha_{\pm }}\left\{ \cdots
\right\} ^{-1/p}d\xi ,
\end{equation*}
where $\delta >0$ is given by (\ref{pq5}). The first integral
converges because the integrand function is continuous on the
compact interval whose extremities are $0$ and $\alpha_{\pm }\mp
\delta $. For the second one, it follows from $(\ref{pq5})_{\pm
}$ that, for all $\xi $, satisfying $( \pm \xi ) \in ( \pm
\alpha_{\pm }-\delta ,\pm \alpha_{\pm }) $, one has
\begin{equation*}
\pm m_{\pm }\varphi_q( \alpha_{\pm }-\xi ) \leq \pm f( \xi ) \leq
\pm M_{\pm }\varphi_q( \alpha_{\pm }-\xi ) ,
\end{equation*}
and since for any $E$ near $E_{*}^{\pm }$ one has $\pm r_{\pm }(
\lambda ,E) \in ( \pm \alpha_{\pm }-\delta ,\pm \alpha_{\pm }) $
then for all $\xi $, satisfying $( \pm \xi ) \in ( \pm
\alpha_{\pm }-\delta ,\pm r_{\pm }( \lambda ,E) ) $, one has
\begin{gather*}
\ m_{\pm }\int_\xi ^{r_{\pm }( \lambda ,E) }\varphi_q(
\alpha_{\pm }-x) dx\leq \int_\xi ^{r_{\pm }( \lambda ,E) }f( x)
dx \leq M_{\pm }\int_\xi ^{r_{\pm }( \lambda ,E) }\varphi_q(
\alpha_{\pm }-x) dx
\end{gather*}
that is
\begin{equation*}
\frac{m_{\pm }}q\left\{ \left| \alpha_{\pm }-\xi \right|
^q-\left| \alpha _{\pm }-r_{\pm }( \lambda ,E) \right| ^q\right\}
\leq F( r_{\pm }( \lambda ,E) ) -F( \xi )
\end{equation*}
\begin{equation*}
\leq \frac{M_{\pm }}q\left\{ \left| \alpha_{\pm }-\xi \right|
^q-\left| \alpha_{\pm }-r_{\pm }( \lambda ,E) \right| ^q\right\}
\end{equation*}
then
\begin{equation*}
\pm ( \frac{M_{\pm }}q) ^{-1/p}\int_{\alpha_{\pm }\mp \delta
_{\pm }}^{r_{\pm }( \lambda ,E) }\left\{ \left| \alpha_{\pm }-\xi
\right| ^q-\left| \alpha_{\pm }-r_{\pm }( \lambda ,E) \right|
^q\right\} ^{-1/p}d\xi
\end{equation*}
\begin{equation*}
\leq \pm \int_{\alpha_{\pm }\mp \delta }^{r_{\pm }( \lambda ,E)
}\left\{ F( r_{\pm }( \lambda ,E) ) -F( \xi ) \right\} ^{-1/p}d\xi
\end{equation*}
\begin{eqnarray*}
\ \leq \pm ( \frac{m_{\pm }}q) ^{-1/p}\int_{\alpha_{\pm }\mp
\delta }^{r_{\pm }( \lambda ,E) }\left\{ \left| \alpha_{\pm }-\xi
\right| ^q-\left| \alpha_{\pm }-r_{\pm }( \lambda ,E) \right|
^q\right\} ^{-1/p}d\xi .
\end{eqnarray*}
By passing to the limit in these inequalities as $E$ tends to
$E_{*}^{\pm }$, one gets
\begin{equation*}
\pm ( \frac{M_{\pm }}q) ^{1/p}\int_{\alpha_{\pm }\mp \delta
}^{\alpha_{\pm }}\left| \alpha_{\pm }-\xi \right| ^{-q/p}d\xi
\leq \pm \int_{\alpha_{\pm }\mp \delta }^{\alpha_{\pm }}\left\{
F( \alpha _{\pm }) -F( \xi ) \right\} ^{-1/p}d\xi
\end{equation*}
\begin{equation*}
\leq \pm ( \frac{m_{\pm }}q) ^{-1/p}\int_{\alpha_{\pm }\mp \delta
}^{\alpha_{\pm }}\left| \alpha_{\pm }-\xi \right| ^{-q/p}d\xi ,
\end{equation*}
and from the well-known fact
\begin{equation*}
\pm \int_{\alpha_{\pm }\mp \delta }^{\alpha_{\pm }}\left|
\alpha_{\pm }-\xi \right| ^{-q/p}d\xi <+\infty \text{ if and only
if }p>q,
\end{equation*}
the second assertion of \textbf{(i)} follows.
\paragraph{Proof of (ii)}
By passing to the limit in (\ref{qze}) as $E$ tends to $0^+$, the
limit of $T_{\pm }( \lambda ,E) $ follows immediately from
(\ref{bbhu}), (\ref{bbhuu}) and the fact that,
\begin{equation*}
\int_0^1( 1-\xi ^q) ^{-1/p}=\frac 1qB( \frac 1q,1-\frac 1p) \in
{\mathbb R},
\end{equation*}
where $B( a,b) $ denotes the beta function. Remark that in the
particular case $q=p$ one has
\begin{equation*}
\int_0^1( 1-\xi ^p) ^{-1/p}=\frac 1pB( \frac 1p,1-\frac 1p) =\pi
/( p\sin ( \frac \pi p) ) =\frac 12( \lambda_1/( p-1) ) ^{1/p}.
\end{equation*}
Notice that condition (\ref{pq4}) was used implicitly in this
proof. In fact, to derive (\ref{bbhu}), (\ref{bbhuu}) we have
used (\ref{pq4}). See the remark located before the proof of
Lemma \ref{lemma1}.
\paragraph{Proof of (iii)}
By Lemma \ref{lemma5} it follows that for all $\lambda >0$, one
has $r_+( \lambda ,E_{*}^{-}( \lambda ) ) =y( r_-( \lambda
,E_{*}^{-}( \lambda ) ) ) $ and by (\ref{abhu}) it follows that
$r_-( \lambda ,E_{*}^{-}( \lambda ) ) =\alpha_-$. Thus, \\ $r_+(
\lambda ,E_{*}^{-}( \lambda ) ) =y( \alpha_-) $.
The formula of $T_+( \lambda ,E_{*}^{-}( \lambda ) ) $ follows
from a simple substitution in (\ref{a16}). Therefore, Lemma
\ref{lemma2} is proved.
%TCIMACRO{\TeXButton{End Proof}{\hfill$\diamondsuit$}}
%BeginExpansion
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%EndExpansion
\begin{lemma}
\label{lemma31}Assume that (\ref{pq1})-(\ref{pq4}) hold, $p,q>1$,
$\lambda
>0$ and $n\in {\mathbb N}^{*}$. Then,
\begin{equation*}
\lim\limits_{E\rightarrow 0^+}T_{n}^{\pm }( \lambda ,E) =\left\{
\begin{array}{lcc}
+\infty & \text{if} & q-p>0 \\
\dfrac{1}{2}( \dfrac{\lambda_{n}}{\lambda a_{0}}) ^{1/p} & \text{%
if} & q-p=0 \\
0 & \text{if} & q-p<0.
\end{array}
\right.
\end{equation*}
\end{lemma}
\begin{lemma}
\label{lemma32}Assume that (\ref{pq1})-(\ref{pq3}), $(\ref{pq5})_-$, and (%
\ref{pho}) hold, $p$, $q>1$, $\lambda >0$ and $n\in {\mathbb
N}^{*}$. Then,
\begin{itemize}
\item In case $q-p\geq 0$, one has $\lim\limits_{E\rightarrow
E_{*}^{-}}T_{n}^{\pm }( \lambda ,E) =+\infty $.
\item In case $q-p<0$, one has
\begin{equation*}
\begin{array}{lll}
\lim\limits_{E\rightarrow E_{*}^{-}}T_{2n}^{\pm }( \lambda ,E) &
= & ( \lambda p') ^{-1/p}( nS_-( \alpha
_-) +nS_+( \alpha_{*}) ) \\
\lim\limits_{E\rightarrow E_{*}^{-}}T_{2n+1}^+( \lambda ,E) & = &
( \lambda p') ^{-1/p}( nS_-( \alpha
_-) +( n+1) S_+( \alpha_{*}) ) \\
\lim\limits_{E\rightarrow E_{*}^{-}}T_{2n+1}^{-}( \lambda ,E) & =
& ( \lambda p') ^{-1/p}( ( n+1) S_-( \alpha_-) +nS_+( \alpha_{*})
) .
\end{array}
\end{equation*}
\end{itemize}
\end{lemma}
\paragraph{Proofs of Lemmas \ref{lemma31} and \ref{lemma32}.} These ones
follow from Lemma \ref{lemma2} and the definitions (\ref{a18})
and (\ref{a19}) of the time maps $T_k^{\pm }$. \hfill$\diamondsuit$
\begin{lemma}
\label{lemma4}For any $p$, $q>1$ and $\lambda >0$, assume that
$(\ref {njk})_{\pm }$ holds. Then,
\begin{equation*}
\frac{\partial T_{\pm }}{\partial E}( \lambda ,E)
>0,\;\forall E\in D_{\pm }( \lambda ) =( 0,E_{*}^{\pm
}( \lambda ) ) .
\end{equation*}
Therefore, if both $(\ref{njk})_+$ and $(\ref {njk})_-$ hold,
then, for all $k\geq 2$,
\begin{equation*}
\frac{\partial T_{k}^{\pm }}{\partial E}( \lambda ,E)
>0,\;\forall E\in D( \lambda ) =( 0,E_{*}^{-}(
\lambda ) ) .
\end{equation*}
\end{lemma}
\paragraph{Proof.} A simple computation shows that
\begin{eqnarray*}
\frac{\partial T_{\pm }}{\partial E}( \lambda ,E) &=&\frac{1}{p%
}( p') ^{-1/p}( \pm \frac{\partial r_{\pm }}{%
\partial E}( \lambda ,E) ) \lambda ^{-1/p} \\
&&\times \int_{0}^1\frac{H( r_{\pm }( \lambda ,E) ) -H( r_{\pm }(
\lambda ,E) \xi ) }{( F( r_{\pm }( \lambda ,E) ) -F( r_{\pm }(
\lambda ,E) \xi ) ) ^{1+( 1/p) }}d\xi .
\end{eqnarray*}
where $H( x) =pF( x) -xf( x) $. Condition $(\ref{njk})_{\pm }$
implies that
\begin{equation*}
H( r_{\pm }( \lambda ,E) ) -H( r_{\pm }( \lambda ,E) \xi )
>0,\forall \xi \in ( 0, 1) .
\end{equation*}
Thus, the integral above is positive, and by (\ref{eq11}) it
follows that $\frac{\partial T_{\pm }}{\partial E}( \lambda ,E)
>0$, for all $E\in D_{\pm }( \lambda ) $. The last assertion is
immediate from the definition of $T_{k}^{\pm }( \lambda ,\cdot )
$. The proof of Lemma \ref{lemma4} is complete.
%TCIMACRO{\TeXButton{End Proof}{\hfill$\diamondsuit$}}
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\bigskip \ When $f$ is odd, the time-maps $T_+(\lambda ,\cdot )$ and
$T_-(\lambda ,\cdot )$ are always equal. In our \emph{p.h.o.}
case, we show that both $T_+(\lambda ,\cdot )$ and $T_-(\lambda
,\cdot )$ are bounded from below and from above by a same
function respectively. These estimates imply in the particular
odd case that the two time-maps $T_+(\lambda ,\cdot )$ and
$T_-(\lambda ,\cdot )$ are equal. The following lemma is pioneer
in our analysis.
Let us define
\begin{equation*}
\Theta_+( x) =\int_{0}^{x}\{F( x) -F( -\xi ) \}^{-1/p}d\xi ,x\in
( 0,-\alpha_-) ,
\end{equation*}
and
\begin{eqnarray*}
\Theta_-( x) &=&\int_{x}^{-y( x) }\{F( x) -F( \xi ) \}^{-1/p}d\xi
+\int_{-y( x)
}^{0}\{F( x) -F( -\xi ) \}^{-1/p}d\xi , \\
\text{ for all }x &\in &( \alpha_-,0) ,\text{ where }y( x) \text{
is defined in (\ref{12kwdj}).}
\end{eqnarray*}
\begin{lemma}
\label{lemma6}Assume that (\ref{pq1})-(\ref{pq3}) and (\ref{pho})
hold. Then, for all $\lambda >0$ and $E\in ( 0,E_{*}^{-}( \lambda
) ) $, one has
\begin{description}
\item[(i)] $\Theta_+( r_+( \lambda ,E) ) \leq
S_-( r_-( \lambda ,E) ) $,
\item[(ii)] $\Theta_+( r_+( \lambda ,E) ) \leq
S_+( r_+( \lambda ,E) ) $,
\item[(iii)] $S_-( r_-( \lambda ,E) ) \leq
\Theta_-( r_-( \lambda ,E) ) $,
\item[(iv)] $S_+( r_+( \lambda ,E) ) \leq
\Theta_-( r_-( \lambda ,E) ) $.
\end{description}
\end{lemma}
\paragraph{Proof of Lemma \ref{lemma6}}
\paragraph{Proof of (i)}
For all $\lambda >0$ and $E\in ( 0,E_{*}^{-}( \lambda ) ) $, one
has
\begin{equation*}
S_-( r_-( \lambda ,E) ) =-\int_0^{r_-( \lambda ,E) }\{F( r_-(
\lambda ,E) ) -F( \xi ) \}^{-1/p}d\xi .
\end{equation*}
Using a simple change of variables one deduces,
\begin{equation}
S_-( r_-( \lambda ,E) ) =\int_0^{-r_-( \lambda ,E) }\{F( r_-(
\lambda ,E) ) -F( -\xi ) \}^{-1/p}d\xi . \label{ehjkgr}
\end{equation}
Using (\ref{dyuk}) one gets
\begin{eqnarray*}
S_-( r_-( \lambda ,E) ) &=&\Theta_+(
r_+( \lambda ,E) ) \\
&&\ +\int_{r_+( \lambda ,E) }^{-r_-( \lambda ,E) }\{F( r_-(
\lambda ,E) ) -F( -\xi ) \}^{-1/p}d\xi
\end{eqnarray*}
and by Lemma \ref{lemma5} it follows that the integral above is
positive. Therefore, Assertion \textbf{(i)} is proved.
\paragraph{Proof of (ii)}
Recall that $F$ is \textit{p.h.e. }in $[ \alpha_-,-\alpha _-] $,
(see (\ref{phe})), hence, for all $\lambda >0$, $E\in (
0,E_{*}^{-}( \lambda ) ) $, and $0<\xi 0$ and $E\in (
0,E_{*}^{-}( \lambda ) ) $ one has $y( r_-( \lambda ,E) ) =r_+(
\lambda ,E) $. Thus, for $\xi \in ( -y( r_-( \lambda ,E) ) ,0) , $
\begin{equation*}
\{F( r_-( \lambda ,E) ) -F( \xi ) \}^{-1/p}\leq \{F( r_-( \lambda
,E) ) -F( -\xi ) \}^{-1/p}.
\end{equation*}
Therefore,
\begin{eqnarray*}
&&\int_{-y( r_-( \lambda ,E) ) }^0\{F(
r_-( \lambda ,E) ) -F( \xi ) \}^{-1/p}d\xi \\
\ &\leq &\int_{-y( r_-( \lambda ,E) ) }^0\{F( r_-( \lambda ,E) )
-F( -\xi ) \}^{-1/p}d\xi .
\end{eqnarray*}
So,
\begin{eqnarray*}
S_-( r_-( \lambda ,E) ) &=&\int_{r_-( \lambda ,E) }^0\{F( r_-(
\lambda ,E) )
-F( \xi ) \}^{-1/p}d\xi \\
&\leq &\int_{r_-( \lambda ,E) }^{-y( r_-( \lambda ,E) ) }\{F(
r_-( \lambda ,E) )
-F( \xi ) \}^{-1/p}d\xi \\
&&+\int_{-y( r_-( \lambda ,E) ) }^0\{F( r_-( \lambda ,E) ) -F(
-\xi ) \}^{-1/p}d\xi
\\
&=&\Theta_-( r_-( \lambda ,E) ) .
\end{eqnarray*}
Assertion \textbf{(iii)} is proved.
\paragraph{Proof of (iv)}
For all $\lambda >0$ and $E\in ( 0,E_{*}^{-}( \lambda ) ) $ one
has
\begin{eqnarray}
\Theta_-( r_-( \lambda ,E) ) &=&\int_{r_-( \lambda ,E) }^{-y(
r_-( \lambda ,E) ) }\{F( r_-( \lambda ,E) )
-F( \xi ) \}^{-1/p}d\xi \label{Canada} \\
&&+\int_{-y( r_-( \lambda ,E) ) }^0\{F( r_-( \lambda ,E) ) -F(
-\xi ) \}^{-1/p}d\xi . \notag
\end{eqnarray}
Then by Lemma \ref{lemma5}, one has $-y( r_-( \lambda ,E) )
=-r_+( \lambda ,E) $ and $F( r_-( \lambda ,E) ) =F( r_+( \lambda
,E) ) $. Thus, using a simple change of variable, it follows that
\begin{equation*}
\int_{-y( r_-( \lambda ,E) ) }^0\{F( r_-( \lambda ,E) ) -F( -\xi
) \}^{-1/p}d\xi =S_+( r_+( \lambda ,E) ) .
\end{equation*}
On the other hand, the first integral in (\ref{Canada}) is
positive, since \newline $r_-( \lambda ,E) \leq -y( r_-( \lambda
,E) ) $. Thus, $\Theta_-( r_-( \lambda ,E) ) \geq S_+( r_+(
\lambda ,E) ) $.
Therefore, Assertion \textbf{(iv)} is proved which completes the
proof of Lemma \ref{lemma6}.
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\begin{remark}
It is well known that when $f$ is odd then
\begin{equation}
T_+(\lambda ,\cdot )=T_-(\lambda ,\cdot )\text{,\ for all }\lambda
>0. \label{R1}
\end{equation}
The estimates in Lemma \ref{lemma6} imply (\ref{R1}) in the odd
case. In fact if $f$ is odd then $\alpha_+=-\alpha_-$, $F$ is
even,
\begin{equation*}
E_{*}^+(\lambda )=E_{*}^{-}(\lambda )\text{,\ \ for all }\lambda
>0
\end{equation*}
\begin{equation*}
r_+(\lambda ,E)=-r_-(\lambda ,E),\text{\ \ for all }\lambda >0.
\end{equation*}
Also, since for all $\lambda >0$ and $E\in (0,E_{*}^{\pm
}(\lambda ))$,
\begin{equation*}
\Theta_-( r_-( \lambda ,E) ) =\int_{r_-}^{-y( r_-) }\{F( r_-) -F(
\xi ) \}^{-1/p}d\xi +\int_{-y( r_-) }^{0}\{F( r_-) -F( -\xi )
\}^{-1/p}d\xi ,
\end{equation*}
then by Lemma \ref{lemma5}, it follows that
\begin{eqnarray*}
\Theta_-( r_-( \lambda ,E) ) &=&\int_{r_-}^{-r_+}\{F( r_+) -F(
\xi ) \}^{-1/p}d\xi +\int_{-r_+}^{0}\{F( r_+) -F( -\xi )
\}^{-1/p}d\xi , \\
&=&\int_{-r_+}^{0}\{F( r_+) -F( -\xi ) \}^{-1/p}d\xi .
\end{eqnarray*}
On the other hand, since the function defined on $(-r_+,r_+)$ by
$\xi \mapsto \{F( r_+) -F( -\xi ) \}^{-1/p}$ is even then
\begin{equation*}
\int_{-r_+}^{0}\{F( r_+) -F( -\xi ) \}^{-1/p}d\xi
=\int_{0}^{r_+}\{F( r_+) -F( -\xi ) \}^{-1/p}d\xi .
\end{equation*}
Therefore,
\begin{equation}
\Theta_-( r_-( \lambda ,E) ) =\Theta_+( r_+( \lambda ,E) )
\text{,\ for all }\lambda >0\text{ and all }E\in (0,E_{*}(\lambda
)). \label{R2}
\end{equation}
Now, by the estimates of Lemma \ref{lemma6} and (\ref{R2}),
equation (\ref {R1}) follows.
\end{remark}
\begin{lemma}
\label{lemma7}Assume that (\ref{pq1})-(\ref{pq3}) and (\ref{pho})
hold. Then, for all $\lambda >0$, $E\in ( 0,E_{*}^{-}( \lambda )
) $, and for all integer $k\geq 2, $\begin{equation*} k(
p'\lambda ) ^{-1/p}\Theta_+( r_+( \lambda ,E) ) \leq
T_{k}^{\kappa }( \lambda ,E) \leq k( p'\lambda ) ^{-1/p}\Theta_-(
r_-( \lambda ,E) ) .
\end{equation*}
\end{lemma}
\paragraph{Proof.} According to (\ref{a16}) and the definition for
$T_{k}^{\kappa }( \lambda ,E) $, the proof is an immediate
consequence of Lemma \ref{lemma6}.
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\begin{lemma}
\label{lemma75}For all $\lambda >0$ and $E\in ( 0,E_{*}^{-}(
\lambda ) ) $ the following identity holds;
\begin{equation*}
\Theta_+( r_+( \lambda ,E) ) +\Theta_-( r_-( \lambda ,E) ) =S_+(
r_+( \lambda ,E) ) +S_-( r_-( \lambda ,E) ) .
\end{equation*}
\end{lemma}
\paragraph{Proof.} For all $\lambda >0$ and $E\in ( 0,E_{*}^{-}(
\lambda ) ) $, we write $\Theta_-( r_-( \lambda ,E) ) $ as follows
\begin{eqnarray*}
\Theta_-( r_-( \lambda ,E) ) &=&\int_{r_-( \lambda ,E) }^{0}\{F(
r_-( \lambda
,E) ) -F( \xi ) \}^{-1/p}d\xi \\
&&\ \ -\int_{-r_+( \lambda ,E) }^{0}\{F( r_-(
\lambda ,E) ) -F( \xi ) \}^{-1/p}d\xi \\
&&\ \ +\int_{-r_+( \lambda ,E) }^{0}\{F( r_-( \lambda ,E) ) -F(
-\xi ) \}^{-1/p}d\xi .
\end{eqnarray*}
The first integral is equal to $S_-( r_-( \lambda ,E) ) $. On the
other hand, the change of variable $\xi =-r_+( \lambda ,E) t$,
and (\ref{dyuk}) imply that the second integral is equal to
$\Theta_+( r_+( \lambda ,E) ) $. The change of variable $\xi
=-t$, and (\ref{dyuk}) imply that the third integral is equal to
$S_+( r_+( \lambda ,E) ) $. Therefore, Lemma \ref{lemma75} is
proved.
%TCIMACRO{\TeXButton{End Proof}{\hfill$\diamondsuit$}}
%BeginExpansion
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%EndExpansion
\begin{lemma}
\label{lemma8}Assume that (\ref{pq1})-(\ref{pq4}), $(\ref{pq5})_-$ and (%
\ref{pho}) hold. Then, for all $\lambda >0$, one has
\begin{description}
\item[(i)] $\lim\limits_{E\rightarrow 0^+}\Theta_{\pm }( r_{\pm
}( \lambda ,E) ) =\left\{
\begin{array}{lcc}
+\infty & \text{if} & q-p>0 \\
\dfrac{1}{2}( \dfrac{p'\lambda_{1}}{a_{0}}) ^{1/p} &
\text{if} & q-p=0 \\
0 & \text{if} & q-p<0.
\end{array}
\right. $
\item[(ii)] $\lim\limits_{E\rightarrow E_{*}^{-}( \lambda )
}\Theta_{\pm }( r_{\pm }( \lambda ,E) ) =\ell_{\pm }$ with\newline
$\left\{
\begin{array}{lcl}
\ell_{\pm }=+\infty & \text{if} & q-p\geq 0\text{ and
}-\alpha_{*}=\alpha
_- \\
\ell_+\in ( 0,+\infty ) \text{ and }\ell_-=+\infty &
\text{if} & q-p\geq 0\text{ and }-\alpha_{*}>\alpha_- \\
\ell_{\pm }\in ( 0,+\infty ) & \text{if} & q-p<0.
\end{array}
\right. $
\end{description}
\end{lemma}
\paragraph{Proof of Lemma \ref{lemma8}}
\paragraph{Proof of Assertion (i).} For all $\lambda >0$ and $E\in (
0,E_{*}^{-}( \lambda ) ) $ one has $E^p/(p^{\prime }\lambda )=F(
r_+( \lambda ,E) ) $, hence,
\begin{equation*}
\Theta_+( r_+( \lambda ,E) ) =\int_0^{r_+( \lambda ,E) }\left\{
(E^p/p'\lambda )-F(-\xi )\right\} ^{-1/p}d\xi .
\end{equation*}
The change of variable $\xi =r_+( \lambda ,E) t$, yields
\begin{equation*}
\Theta_+( r_+( \lambda ,E) ) =\frac{r_+( \lambda ,E) }E(p'\lambda
)^{1/p}\int_0^1\left\{ 1-p^{\prime }\lambda F(-r_+( \lambda ,E)
t)/E^p\right\} ^{-1/p}dt.
\end{equation*}
By (\ref{bbhu}) one has
\begin{equation}
\lim\limits_{E\rightarrow 0^+}\frac{r_+( \lambda ,E) }%
E=\left\{
\begin{array}{lcl}
+\infty & \text{if} & q-p>0 \\
( \dfrac{p-1}{\lambda a_0}) ^{1/p} & \text{if} & q-p=0 \\
0 & \text{if} & q-p<0.
\end{array}
\right. \label{Madrid}
\end{equation}
On the other hand, by (\ref{bbhuu}) one has for all $( \kappa
,\eta ) \in \left\{ +,-\right\} ^2$,
\begin{equation}
\lim\limits_{E\rightarrow 0^+}\int_0^1\left\{ 1-p'\lambda
F(\kappa r_\eta ( \lambda ,E) t)/E^p\right\}
^{-1/p}dt=\int_0^1\left\{ 1-t^q\right\} ^{-1/p} \label{Italy}
\end{equation}
with
\begin{equation}
\int_0^1\left\{ 1-t^q\right\} ^{-1/p}=\left\{
\begin{array}{lcl}
\dfrac 1qB( \dfrac 1q,1-\dfrac 1p) \in {\mathbb R}
& \text{if} & q-p\neq 0 \\
\dfrac 12( \dfrac{\lambda_1}{p-1}) ^{1/p} & \text{if} & q-p=0.
\end{array}
\right. \label{France}
\end{equation}
Therefore, the limit $\lim_{E\rightarrow 0^+}\Theta_+( r_+(
\lambda ,E) ) $ follows.
Assume that $q-p>0$. By Assertion \textbf{(ii)} of Lemma
\ref{lemma2}, one has \newline $\lim_{E\rightarrow 0^+}T_-(
\lambda ,E) =+\infty $, and by (\ref{a16}) and Assertion
\textbf{(iii)} of Lemma \ref{lemma6}, it follows that
$\lim_{E\rightarrow 0^+}\Theta_-( r_-( \lambda ,E) ) =+\infty $.
Assume that $q-p=0$. In this case, for all $\lambda >0$ and $E\in
( 0,E_{*}^{-}( \lambda ) ) $, we use the identity in Lemma
\ref{lemma75}. That is, we write $\Theta_-( r_-( \lambda ,E) ) $
as follows
\begin{equation*}
\Theta_-( r_-( \lambda ,E) ) =S_+( r_+( \lambda ,E) ) +S_-( r_-(
\lambda ,E) ) -\Theta_+( r_+( \lambda ,E) )
\end{equation*}
and we prove that each term of the right hand side tends to the
same limit; $\dfrac 12( p'\lambda_1/a_0) ^{1/p}$.
The limits of $S_-( r_-( \lambda ,E) ) $ and $S_+( r_+( \lambda
,E) ) $ follows by (\ref{a16}) and Assertion \textbf{(ii)} of
Lemma \ref{lemma2}, and the limit of $\Theta _+( r_+( \lambda ,E)
) $ was computed above. Therefore, $\lim_{E\rightarrow
0^+}\Theta_-( r_-( \lambda ,E) ) =\dfrac 12( p'\lambda_1/a_0)
^{1/p}$ which completes the proof of Assertion \textbf{(i)}.
\paragraph{Proof of Assertion (ii).} By (\ref{dyuk}) it follows that
\begin{equation*}
\Theta_+( r_+( \lambda ,E) ) =\int_0^{r_+( \lambda ,E) }\left\{
F(r_-( \lambda ,E) )-F(-\xi )\right\} ^{-1/p}d\xi .
\end{equation*}
A simple change of variable yields
\begin{equation*}
\Theta_+( r_+( \lambda ,E) ) =\int_{-r_+( \lambda ,E) }^0\left\{
F(r_-( \lambda ,E) )-F(\xi )\right\} ^{-1/p}d\xi
\end{equation*}
and thus,
\begin{equation*}
\lim_{E\rightarrow E_{*}^{-}( \lambda ) }\Theta_+( r_+( \lambda
,E) ) =\int_{-r_+( \lambda ,E_{*}^{-}( \lambda ) ) }^0\left\{
F(r_-( \lambda ,E_{*}^{-}( \lambda ) ) )-F(\xi )\right\}
^{-1/p}d\xi .
\end{equation*}
By (\ref{abhu}), one has $r_-( \lambda ,E_{*}^{-}( \lambda ) )
=\alpha_-$ and by Assertion \textbf{(iii)} of Lemma \ref {lemma2},
it follows that $r_+( \lambda ,E_{*}^{-}( \lambda ) )
=\alpha_{*}$.
Therefore,
\begin{equation*}
\lim_{E\rightarrow E_{*}^{-}( \lambda ) }\Theta_+( r_+( \lambda
,E) ) =\int_{-\alpha_{*}}^0\left\{ F(\alpha_-)-F(\xi )\right\}
^{-1/p}d\xi .
\end{equation*}
According to the definition of $\alpha_{*}$ (see (\ref{12kwdj})),
one has, $\alpha_-\leq -\alpha_{*}$. Thus, one has to distinguish
two cases:
If $-\alpha_{*}=\alpha_-$,
\begin{equation*}
\lim_{E\rightarrow E_{*}^{-}( \lambda ) }\Theta_+( r_+( \lambda
,E) ) =\int_{\alpha_-}^0\left\{ F(\alpha_-)-F(\xi )\right\}
^{-1/p}d\xi =S_-( \alpha_-) .
\end{equation*}
Thus, by Assertion \textbf{(i)} of Lemma \ref{lemma2}, it follows
that
\begin{equation*}
S_-( \alpha_-) =\left\{
\begin{array}{lcc}
+\infty & \text{if} & q-p\geq 0 \\
\ell_+\in ( 0,+\infty ) & \text{if} & q-p<0.
\end{array}
\right.
\end{equation*}
If $-\alpha_{*}>\alpha_-$, it follows that the integral
$\int_{-\alpha _{*}}^0\left\{ F(\alpha_-)-F(\xi )\right\}
^{-1/p}d\xi $ is a positive real number, since the integrand
function is continuous on the compact interval $[ -\alpha_{*},0]
$.
Therefore, the claims related to the limit $\lim_{E\rightarrow
E_{*}^{-}}\Theta_+( r_+( \lambda ,E) ) $ follows.
Assume that $q-p\geq 0$. By Assertion \textbf{(i)} of Lemma
\ref{lemma2}, one has\newline $\lim_{E\rightarrow E_{*}^{-}}T_-(
\lambda ,E) =+\infty , $ and by (\ref{a16}) and Assertion
\textbf{(iii)} of Lemma \ref{lemma6}, it follows that
$\lim_{E\rightarrow E_{*}^{-}}\Theta_-( r_-( \lambda ,E) )
=+\infty $.
Assume that $q-p<0$. In this case one has, by Assertion
\textbf{(i)} of Lemma \ref{lemma2}, $\lim_{E\rightarrow
E_{*}^{-}}S_-( r_-( \lambda ,E) ) =S_-( \alpha_-) \in (
0,+\infty ) $, and by Assertion \textbf{(iii)} of Lemma \ref{lemma2}%
,
\begin{equation*}
\lim_{E\rightarrow E_{*}^{-}}S_+( r_+( \lambda ,E) ) =S_+(
\alpha_{*}) =( \lambda p') ^{1/p}T_+( \lambda ,E_{*}^{-}( \lambda
) ) ,
\end{equation*}
and $T_+( \lambda ,E_{*}^{-}( \lambda ) ) \in ( 0,+\infty ) $,
since $E_{*}^{-}( \lambda ) \in int( D_+( \lambda ) ) =int(dom(
T_+( \lambda ,\cdot ) ) )$. Also, $\lim_{E\rightarrow
E_{*}^{-}}\Theta_+( r_+( \lambda ,E) ) =\ell _+\in ( 0,+\infty )
$ (proved above). Thus, by the identity of Lemma \ref{lemma75},
it follows that $\lim_{E\rightarrow E_{*}^{-}}\Theta _-( r_-(
\lambda ,E) ) =\ell_-\in ( 0,+\infty ) $. Therefore, Assertion
\textbf{(ii)} follows, which completes the proof of Lemma
\ref{lemma8}.
%TCIMACRO{\TeXButton{End Proof}{\hfill$\diamondsuit$}}
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%EndExpansion
\begin{lemma}
\label{lemma9}Assume that (\ref{pq1})-(\ref{pq3}) and
$(\ref{Miami})_{\pm }$ hold. Then, for any $p>1$ one has,
\begin{equation*}
r^{2}\frac{d^{2}S_{\pm }}{dr^{2}}( r) +2( p-1) r\frac{%
dS_{\pm }}{dr}( r) >0,\text{\ for all }r\in I_{\pm }( \alpha
_{\pm }) ,
\end{equation*}
where $I_+( \alpha_+) =( 0,\alpha_+) $ and $I_-( \alpha_-) =(
\alpha_-,0) $.
\end{lemma}
\paragraph{Proof.} Some easy computations show that for all $\lambda >0$ and
$r\in I_{\pm }( \alpha_{\pm }) $\begin{equation*}
r^{2}\frac{d^{2}S_{\pm }}{dr^{2}}( r) +2( p-1) r\frac{%
dS_{\pm }}{dr}( r)
\end{equation*}
\begin{eqnarray*}
\ &=&\frac{\pm 1}{p}\int_{0}^{r}\frac{\Psi ( r) -\Psi ( s) }{(
F( r) -F( s) ) ^{( p+1) /p}}ds \pm ( \frac{p+1}{p}) \int_{0}^{r}\frac{%
( H( r) -H( s) ) ^{2}}{( F( r) -F( s) ) ^{( 2p+1) /p}}ds
\end{eqnarray*}
where,
\begin{eqnarray*}
H( x) &=&pF( x) -xf( x) ,\text{for all }%
x\in I_{\pm }( \alpha_{\pm }) \\
\Psi ( x) &=&p( p-3) F( x) +2xf( x) -x^{2}f'( x) ,\text{for all
}x\in I_{\pm }( \alpha_{\pm }) .
\end{eqnarray*}
A differentiation yields
\begin{equation*}
\Psi '( x) =( p-2) ( p-1) f( x) -x^{2}f''( x) ,\text{ for all
}x\in I_{\pm }( \alpha_{\pm }) .
\end{equation*}
Thus, $(\ref{Miami})_{\pm }$ implies that $\pm \Psi $ is strictly
increasing in $I_{\pm }( \alpha_{\pm }) $. Then, $\Psi ( r) -\Psi
( s) >0$, for all $s\in I_{\pm }( r) \subset I_{\pm }(
\alpha_{\pm }) $. Therefore, Lemma \ref{lemma9} is proved.
\section{{Proof of the main results}}
\label{sec.5}To prove our main results we make use of the
quadrature method; Theorem \ref{quad}. Hence, we have to resolve
equations of the type $T( E) =(1/2)$, where $T$ designates, in
each case, the appropriate time map.
\paragraph{Proof of Theorem \ref{theorem1}}
Let us assume that $10$, the
function $E\mapsto T_{\kappa }( \lambda ,E) $ is defined in
$D_{\kappa }( \lambda ) =( 0,E_{*}^{\kappa }( \lambda ) ) $ and
\begin{equation*}
\lim_{E\rightarrow 0^+}T_{\kappa }( \lambda ,E) =0,\text{\ \ }%
\lim_{E\mapsto E_{*}^{\kappa }}T_{\kappa }( \lambda ,E) =(
\lambda p') ^{-1/p}S_{\kappa }( \alpha_{\kappa }) <+\infty .
\end{equation*}
(Lemma \ref{lemma2}). Therefore, $\tilde{D}_{\kappa }(\lambda
)=[0,E_{*}^{\kappa }(\lambda )]$ (see Remark \ref{Remark}), and
the equation $T_{\kappa }( \lambda ,E) =(1/2)$ in the variable
$E\in \tilde{D}_{\kappa }( \lambda ) -\left\{ 0\right\} $ admits a
solution in $\tilde{D}_{\kappa }( \lambda ) -\left\{ 0\right\} $
provided that
\begin{equation*}
( \lambda p') ^{-1/p}S_{\kappa }( \alpha_{\kappa }) \geq (1/2),
\end{equation*}
that is, provided that $\lambda \leq ( 2S_{\kappa }( \alpha
_{\kappa }) ) ^{p}/p'$. Furthermore, by Lemma \ref {lemma4}, if
$(\ref{njk})_{\kappa }$ holds, the function $E\mapsto T_{\kappa
}( \lambda ,E) $ is strictly increasing in $\tilde{D}_{\kappa }(
\lambda ) $. Thus, the equation $T_{\kappa }( \lambda ,E) =(1/2)$
in the variable $E\in \tilde{D}_{\kappa }( \lambda ) -\left\{
0\right\} $ admits a solution in $\tilde{D}_{\kappa }( \lambda )
-\left\{ 0\right\} $ if and only if $\lambda \leq ( 2S_{\kappa }(
\alpha_{\kappa }) ) ^{p}/p'$, and in this case the solution is
unique since $T_{\kappa }( \lambda ,\cdot ) $ is strictly
increasing.
\paragraph{Proof of Assertion (ii)}
For all $k\geq 2$, if (\ref{pq1})-(\ref{pq4}), $(\ref{pq5})_-$,
and (\ref {pho}) hold, then for each $\lambda >0$, the function
$E\mapsto T_{k}^{\kappa }( \lambda ,E) $ is defined in $D( \lambda
) =( 0,E_{*}^{-}( \lambda ) ) $ and by Lemma \ref{lemma31} its
limit at $0$ is $0$ and by Lemma \ref{lemma32} its limit at
$E_{*}^{-}( \lambda ) $ is
\begin{itemize}
\item $n( \lambda p') ^{-1/p}( S_-( \alpha
_-) +S_+( \alpha_{*}) ) $ if $k=2n$,
\item $( \lambda p') ^{-1/p}( nS_-( \alpha
_-) +( n+1) S_+( \alpha_{*}) ) $ if $k=2n+1$, and $\kappa =+$,
\item $( \lambda p') ^{-1/p}( ( n+1)
S_-( \alpha_-) +nS_+( \alpha_{*}) ) $ if $k=2n+1$, and $\kappa
=-$.
\end{itemize}
Thus, $\tilde{D}( \lambda ) =[0,E_{*}^{-}( \lambda ) ]$ and the
equation $T_{k}^{\kappa }( \lambda ,E) =\frac{1}{2}$ in the
variable $E\in \tilde{D}( \lambda ) -\left\{ 0\right\} $ admits a
solution in $\tilde{D}( \lambda ) -\left\{ 0\right\} $ provided
that,
\begin{itemize}
\item $\frac{1}{2}\leq (\lambda p')^{-1/p}(nS_-(\alpha
_-)+nS_+(\alpha_{*}))$, if $k=2n$,
\item $\frac{1}{2}\leq (\lambda p')^{-1/p}(nS_-(\alpha
_-)+(n+1)S_+(\alpha_{*}))$, if $k=2n+1$ and $\kappa =+$,
\item $\frac{1}{2}\leq (\lambda p')^{-1/p}((n+1)S_-(\alpha
_-)+nS_+(\alpha_{*}))$, if $k=2n+1$ and $\kappa =-$,
\end{itemize}
\noindent that is, provided that,
\begin{itemize}
\item $0<\lambda \leq (n\frac{2S_-(\alpha_-)}{(p')^{1/p}}+n%
\frac{2S_+(\alpha_{*})}{(p')^{1/p}})^{p}$, if $k=2n$.
\item $0<\lambda \leq (n\frac{2S_-(\alpha_-)}{(p')^{1/p}}%
+(n+1)\frac{2S_+(\alpha_{*})}{(p')^{1/p}})^{p}$, if $k=2n+1$ and
$\kappa =+$.
\item $0<\lambda \leq ((n+1)\frac{2S_-(\alpha_-)}{(p')^{1/p}}%
+n\frac{2S_+(\alpha_{*})}{(p')^{1/p}})^{p}$, if $k=2n+1$ and
$\kappa =-$.
\end{itemize}
\noindent Furthermore, by Lemma \ref{lemma4}, if $(\ref{njk})_+$
and $(\ref {njk})_-$ hold, the function $E\mapsto T_{k}^{\kappa
}( \lambda ,E) $ is strictly increasing in $\tilde{D}( \lambda ) $
so that, the equation $T_{k}^{\kappa }( \lambda ,E) =\frac{1}{2}
$ in the variable $E\in \tilde{D}( \lambda ) $ admits a solution
in $\tilde{D}( \lambda ) $ if and only if
\begin{itemize}
\item $0<\lambda \leq (n\frac{2S_-(\alpha_-)}{(p')^{1/p}}+n%
\frac{2S_+(\alpha_{*})}{(p')^{1/p}})^{p}$, if $k=2n$
\item $0<\lambda \leq (n\frac{2S_-(\alpha_-)}{(p')^{1/p}}%
+(n+1)\frac{2S_+(\alpha_{*})}{(p')^{1/p}})^{p}$, if $k=2n+1$ and
$\kappa =+$
\item $0<\lambda \leq ((n+1)\frac{2S_-(\alpha_-)}{(p')^{1/p}}%
+n\frac{2S_+(\alpha_{*})}{(p')^{1/p}})^{p}$, if $k=2n+1$ and
$\kappa =-$,
\end{itemize}
\noindent and in this case the solution is unique since the
function $E\mapsto T_{k}^{\kappa }( \lambda ,E) $ is strictly
increasing in $\tilde{D}( \lambda ) $.
Therefore, if we put $J_{\pm }:=\frac{2S_{\pm }(\alpha_{\pm
})}{(p^{\prime })^{1/p}}$ and
$J_{*}:=\frac{2S_+(\alpha_{*})}{(p')^{1/p}}$, Theorem
\ref{theorem1} is proved.
%TCIMACRO{\TeXButton{End Proof}{\hfill$\diamondsuit$}}
%BeginExpansion
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%EndExpansion
\paragraph{Proof of Theorem \ref{theorem2}}
Let us assume that $10$, the function $E\mapsto T_{k}^{\kappa }( \lambda
,E) $ is defined in $D_{k}^{\kappa }( \lambda ) =(
0,E_{k}^{\kappa }( \lambda ) ) $ and by Lemma \ref {lemma31} and
\ref{lemma32}, one has
\begin{equation*}
\lim_{E\rightarrow 0^+}T_{k}^{\kappa }( \lambda ,E) =\frac{1}{%
2}( \frac{\lambda_{k}}{\lambda a_{0}}) ^{1/p}\text{,\ \ \ \ \ }%
\lim_{E\rightarrow E_{k}^{\kappa }}T_{k}^{\kappa }( \lambda ,E)
=+\infty .
\end{equation*}
Therefore, $\tilde{D}_{k}^{\kappa }(\lambda )=[0,E_{k}^{\kappa
}(\lambda )) $ (see Remark \ref{Remark}). So, the equation
$T_{k}^{\kappa }( \lambda ,E) =\frac{1}{2}$ in the variable $E\in
\tilde{D}_{k}^{\kappa
}( \lambda ) -\left\{ 0\right\} $, admits a solution in $\tilde{D}%
_{k}^{\kappa }( \lambda ) -\left\{ 0\right\} $ provided that
\begin{equation*}
\frac{1}{2}( \frac{\lambda_{k}}{\lambda a_{0}}) ^{1/p}<\frac{1}{2%
},
\end{equation*}
that is, provided that, $\lambda >\lambda_{k}/a_{0}$.
Furthermore, by Lemma \ref{lemma4}, if $(\mathbf{K})_{k}^{\kappa
}$ holds, the function $E\mapsto T_{k}^{\kappa }( \lambda ,E) $
is strictly increasing in $\tilde{D}_{k}^{\kappa }( \lambda )
-\left\{ 0\right\} $ and then, the equation $T_{k}^{\kappa }(
\lambda ,E) =\frac{1}{2}$ in the variable $E\in
\tilde{D}_{k}^{\kappa }( \lambda ) -\left\{ 0\right\} $, admits a
solution in $\tilde{D}_{k}^{\kappa }( \lambda ) -\left\{
0\right\} $ if and only if $\lambda >\lambda_{k}/a_{0}$, and in
this case the solution is unique since $T_{k}^{\kappa }( \lambda
,\cdot ) $ is strictly increasing.
\paragraph{Proof of Theorem \ref{theorem3}}
Let us assume that $10$, the function $E\mapsto T_{\pm }( \lambda
,E) $ is defined in $D_{\pm }( \lambda ) =( 0,E_{*}^{\pm }(
\lambda ) ) $ and by Lemma \ref{lemma2},
\begin{equation}
\lim_{E\rightarrow 0^+}T_{\pm }( \lambda ,E) =\lim_{E\rightarrow
E_{*}^{\pm }}T_{\pm }( \lambda ,E) =+\infty . \label{USA}
\end{equation}
Therefore, $\tilde{D}_{\pm }( \lambda ) =D_{\pm }( \lambda ) =(
0,E_{*}^{\pm }( \lambda ) ) $, and the function $E\mapsto T_{\pm
}( \lambda ,E) $ admits at least a minimum value. Recall that for
all $\lambda >0$, the function $E\mapsto \pm r_{\pm }( \lambda
,E) $ is an increasing $C^1$-diffeomorphism from $( 0,E_{*}^{\pm
}( \lambda ) ) $ onto $( 0,\pm \alpha_{\pm }) $. So, regarding
(\ref{a16}) it follows that the local maximum and minimum values
of $T_{\pm }( \lambda ,\cdot ) $ are in a one to one
correspondence with those of $S_{\pm }(\cdot ) $ respectively.
That is, $S_{\pm }( \cdot ) $ attains a local maximum (resp.
minimum) value at $r_{\pm }^{*}\in I_{\pm }( \alpha_{\pm }) $ if
and only if $T_{\pm }( \lambda ,\cdot ) $ does so at $r_{\pm
}^{-1}( \lambda ,r_{\pm }^{*}) $, where $r_{\pm }^{-1}( \lambda
,\cdot ) $ is the function inverse to $r_{\pm }( \lambda ,\cdot )
$. Let $r_{\pm }^{*}\in I_{\pm }( \alpha_{\pm }) $ be a point
where $S_{\pm }$ attains its global minimum value in $I_{\pm }(
\alpha_{\pm }) $. (The existence of $r_{\pm }^{*}$ is guaranteed
by the limits $\lim_{r\rightarrow 0^+}S_{\pm }( r)
=\lim_{r\rightarrow \pm \alpha_{\pm }}S_{\pm }( r) =+\infty $
which follow from (\ref {a16}) and (\ref{USA})). Thus, for each
fixed $\lambda >0$, there exists a unique $\tilde{E}^{\pm
}=\tilde{E}^{\pm }( \lambda ) \in ( 0,E_{*}^{\pm }( \lambda ) ) $
such that $r_{\pm }^{*}=r_{\pm }( \lambda ,\tilde{E}^{\pm }) $
and then for all $\lambda >0$, and all $E\in ( 0,E_{*}^{\pm }(
\lambda ) ) , $\begin{equation*}
\begin{array}{ccl}
T_{\pm }( \lambda ,E) & = & ( \lambda p')
^{-1/p}S_{\pm }( r_{\pm }( \lambda ,E) ) \\
& \geq & ( \lambda p') ^{-1/p}S_{\pm }( r_{\pm
}^{*}) \\
& = & T_{\pm }( \lambda ,\tilde{E}^{\pm }) ,
\end{array}
\end{equation*}
that is, $T_{\pm }( \lambda ,\cdot ) $ attains its global minimum
value at $\tilde{E}^{\pm }( \lambda ) \in ( 0,E_{*}^{\pm }(
\lambda ) ) $. It follows that
\begin{itemize}
\item If $( \lambda p') ^{-1/p}S_{\pm }( r_{\pm
}^{*}) >(1/2)$, the equation $T_{\pm }( \lambda ,E) =(1/2) $ in
the variable $E\in ( 0,E_{*}^{\pm }( \lambda ) ) $ admits no
solution.
\item If $( \lambda p') ^{-1/p}S_{\pm }( r_{\pm
}^{*}) =(1/2)$, the equation $T_{\pm }( \lambda ,E) =(1/2) $ in
the variable $E\in ( 0,E_{*}^{\pm }( \lambda ) ) $ admits at
least a solution (Notice that $S_{\pm }$ may attains its global
minimum at two-or more-distinct points; $r_{\pm }^{*}$ and other
point(s)!).
\item If $( \lambda p') ^{-1/p}S_{\pm }( r_{\pm
}^{*}) <(1/2)$, the equation $T_{\pm }( \lambda ,E) =(1/2) $ in
the variable $E\in ( 0,E_{*}^{\pm }( \lambda ) ) $ admits at
least two solutions.
\end{itemize}
Hence, the first part of Assertion \textbf{(i)} of Theorem
\ref{theorem3} follows if we put $\mu_1^{\pm }=( 2S_{\pm }( r_{\pm
}^{*}) ) ^p/p'$. Notice that $S_{\pm }( r_{\pm }^{*}) $ is the
(unique) global minimum value of the function $S_{\pm }( \cdot )
$, and do not depends on $r_{\pm }^{*}$ which may be not unique.
\bigskip \ Now, if $(\ref{Miami})_{\pm }$ holds, let us show that the
function $S_{\pm }( \cdot ) $ admits at most a minimum value (and
no maximum one) in $I_{\pm }( \alpha_{\pm }) $. To this end,
since the set $I_{\pm }( \alpha_{\pm }) $ is an interval, that
is, a connected set, it suffices to show that $S_{\pm }( \cdot ) $
admits a minimum value at each of its critical points. This
follows if $S_{\pm }( \cdot ) $ is convex in a neighborhood of
each of its critical points, that is
\begin{equation*}
\frac{dS_{\pm }}{dr}( r) =0\Longrightarrow \frac{d^{2}S_{\pm }}{%
dr^{2}}( r) >0,\text{\ \ for all }r\in I_{\pm }( \alpha _{\pm }) .
\end{equation*}
But this holds as an immediate consequence of Lemma \ref{lemma9}.
Therefore, $S_{\pm }( \cdot ) $ admits at most a minimum value
(and no maximum one) in $I_{\pm }( \alpha_{\pm }) $, and for all
$\lambda >0$, $T_{\pm }( \lambda ,\cdot ) $ admits a unique
minimum value (and no maximum one) in $\tilde{D}_{\pm }( \lambda
) $. So, if we denote (as above) $r_{\pm }^{*}$ the point of
$I_{\pm }( \alpha_{\pm }) $ at which the function $S_{\pm }( \cdot
) $ attains its global minimum, it follows that,
\begin{itemize}
\item If $( \lambda p') ^{-1/p}S_{\pm }( r_{\pm
}^{*}) >(1/2)$, the equation $T_{\pm }( \lambda ,E) =(1/2) $ in
the variable $E\in \tilde{D}_{\pm }( \lambda ) $ admits no
solution.
\item If $( \lambda p') ^{-1/p}S_{\pm }( r_{\pm
}^{*}) =(1/2)$, the equation $T_{\pm }( \lambda ,E) =(1/2) $ in
the variable $E\in \tilde{D}_{\pm }( \lambda ) $ admits a unique
solution.
\item If $( \lambda p') ^{-1/p}S_{\pm }( r_{\pm
}^{*}) <(1/2)$, the equation $T_{\pm }( \lambda ,E) =(1/2) $ in
the variable $E\in \tilde{D}_{\pm }( \lambda ) $ admits exactly
two solutions.
\end{itemize}
Hence, the second part of \textbf{(i)} of Theorem \ref{theorem3}
follows with $\mu_1^{\pm }=( 2S_{\pm }( r_{\pm }^{*}) ) ^p/p'$.
Notice that this $\mu_1^{\pm }$ is the same as that of the first
part of Assertion \textbf{(i)} above. Therefore, Assertion
(\textbf{i}) of Theorem \ref{theorem3} is proved.
\paragraph{Proof of Assertion (ii)}
Assume that (\ref{pq1})-(\ref{pq4}), $(\ref{pq5})_-$ and
(\ref{pho}) hold. For each $\lambda >0$, the function $E\mapsto
\Theta_-( r_-( \lambda ,E) ) $ is defined in $D( \lambda ) =(
0,E_{*}^{-}( \lambda ) ) $ and by Lemma \ref{lemma8}
\begin{equation}
\lim_{E\mapsto 0^+}\Theta_-( r_-( \lambda ,E) ) =\lim_{E\mapsto (
E_{*}^{-}) ^{-}}\Theta_-( r_-( \lambda ,E) ) =+\infty .
\label{Star}
\end{equation}
Recall that for all $\lambda >0$, the function $E\mapsto r_-(
\lambda ,E) $ is a decreasing $C^1$-diffeomorphism from $(
0,E_{*}^{-}( \lambda ) ) $ onto $( \alpha _-,0) $. Thus,
(\ref{Star}) implies that
\begin{equation*}
\lim_{x\mapsto \alpha_-}\Theta_-( x) =\lim_{x\mapsto
0^{-}}\Theta_-( x) =+\infty .
\end{equation*}
Therefore, there exists at least a $x_{*}\in ( \alpha_-,0) $ at
which the function $x\mapsto \Theta_-( x) $ attains its global
minimum on $( \alpha_-,0) $, say $\Theta_-( x_{*}) $. Note that
$x_{*}$ is independent of $\lambda $. On the other hand, for each
$\lambda >0$, the function $E\mapsto T_{k}^{\kappa }( \lambda ,E)
$ is defined on $D( \lambda ) $ and by Lemmas \ref{lemma31} and
\ref{lemma32}
\begin{equation*}
\lim_{E\mapsto 0^+}T_{k}^{\kappa }( \lambda ,E) =\lim_{E\mapsto (
E_{*}^{-}) ^{-}}T_{k}^{\kappa }( \lambda ,E) =+\infty .
\end{equation*}
Therefore, $\tilde{D}(\lambda )=D(\lambda )$\ (see Remark
\ref{Remark}). Moreover, it attains its global minimum value at a
certain point in $( 0,E_{*}^{-}( \lambda ) ) $ which may depends
on $\lambda ,k$ and $\kappa $. Using Lemma \ref{lemma7} it
follows that for $\lambda
>0,
$\begin{equation*} \min_{E\in ( 0,E_{*}^{-}( \lambda ) )
}T_{k}^{\kappa }( \lambda ,E) \leq k( p'\lambda )
^{-1/p}\Theta_-( r_-( \lambda ,E) ) ,\text{%
for all }E\in ( 0,E_{*}^{-}( \lambda ) ) .
\end{equation*}
In particular,
\begin{equation*}
\min_{E\in ( 0,E_{*}^{-}( \lambda ) ) }T_{k}^{\kappa }( \lambda
,E) \leq k( p'\lambda ) ^{-1/p}\Theta_-( r_-( \lambda
,\tilde{E}_-( \lambda ) ) ) ,
\end{equation*}
where $\tilde{E}_-( \lambda ) \in ( 0,E_{*}^{-}(
\lambda ) ) $ is such that $r_-( \lambda ,\tilde{E}%
_-( \lambda ) ) =x_{*}$. The existence of $\tilde{E}%
_-( \lambda ) $ is guaranteed from the fact that $E\mapsto r_-(
\lambda ,E) $ is a $C^1$-diffeomorphism from $( 0,E_{*}^{-}(
\lambda ) ) $ onto $( \alpha _-,0) $. Thus
\begin{equation*}
\min_{E\in ( 0,E_{*}^{-}( \lambda ) ) }T_{k}^{\kappa }( \lambda
,E) \leq k( p'\lambda ) ^{-1/p}\Theta_-( x_{*}) ,\text{for all
}\lambda >0.
\end{equation*}
Therefore, if $k( p'\lambda ) ^{-1/p}\Theta_-( x_{*}) <(1/2)$, it
follows that $ \min_{E\in ( 0,E_{*}^{-}( \lambda ) )
}T_{k}^{\kappa }( \lambda ,E) <(1/2).$ Thus, the equation
$T_{k}^{\kappa }( \lambda ,E) =\frac{1}{2}$ in the variable $E\in
\tilde{D}( \lambda ) $ admits at least two solutions in
$\tilde{D}( \lambda ) $, for all $\lambda $ satisfying: $k(
p'\lambda ) ^{-1/p}\Theta_-( x_{*}) <(1/2)$, that is, for all
$\lambda >( 2k\Theta_-( x_{*}) ) ^{p}/p'$. Therefore, the
existence part of Assertion \textbf{(ii)} of Theorem
\ref{theorem3} follows by taking $\mu _{k}=( 2k\Theta_-( x_{*}) )
^{p}/p'$. (Notice that $\mu_{k}>0$ since $\Theta_-( x) >0$ for all
$x\in ( \alpha_-,0) $, in particular, $\Theta_-( x_{*}) >0$).
We have find a range of $\lambda $ where there is existence of at
least two solutions for the equation $T_k^\kappa ( \lambda ,E)
=(1/2)$. Let us, now, look for an other range where there is no
solution.
Using Lemma \ref{lemma7} it follows that for $\lambda >0,
$\begin{equation*} k( p'\lambda ) ^{-1/p}(\inf_{E\in (
0,E_{*}^{-}( \lambda ) ) }\Theta_+( r_+( \lambda ,E) ) )\leq
T_k^\kappa ( \lambda ,E) \text{,}
\end{equation*}
for all $E\in ( 0,E_{*}^{-}( \lambda ) ) $. In particular, for
all $\lambda >0$, one has
\begin{equation}
k( p'\lambda ) ^{-1/p}(\inf_{E\in ( 0,E_{*}^{-}( \lambda ) )
}\Theta_+( r_+( \lambda ,E) ) )\leq \min_{E\in ( 0,E_{*}^{-}(
\lambda ) ) }T_k^\kappa ( \lambda ,E) . \label{jhdyz}
\end{equation}
Recall that for all $\lambda >0$, the function $E\mapsto r_+(
\lambda ,E) $ is strictly increasing on the interval $D( \lambda
) =( 0,E_{*}^{-}( \lambda ) ) \subset ( 0,E_{*}^+( \lambda ) ) $.
Hence, for all $\lambda >0$, and $E\in D( \lambda ) $, $r_+(
\lambda ,E) \in ( 0,r_+( \lambda ,E_{*}^{-}( \lambda ) ) ) $. On
the other hand, by Assertion \textbf{(iii)} of Lemma \ref
{lemma2}, the quantity $r_+( \lambda ,E_{*}^{-}( \lambda ) ) $ is
independent of $\lambda >0$ and $r_+( \lambda ,E_{*}^{-}( \lambda
) ) =y( \alpha_-) $. Thus, for all $\lambda >0$ the function
$E\mapsto r_+( \lambda ,E) $ is an increasing diffeomorphism from
$( 0,E_{*}^{-}( \lambda ) ) $ onto $( 0,y( \alpha_-) ) $.
Therefore, the quantity $\inf_{E\in ( 0,E_{*}^{-}( \lambda ) )
}\Theta_+( r_+( \lambda ,E) ) $ is independent of $\lambda >0$ and
\begin{equation*}
\inf_{E\in ( 0,E_{*}^{-}( \lambda ) ) }\Theta _+( r_+( \lambda
,E) ) =\inf_{00.
\end{equation*}
Let us assume momently that $\inf_{00$. It follows that for all $\lambda $ satisfying $k(
p'\lambda ) ^{-1/p}\inf_{0(1/2)$,
one has
\begin{equation*}
\min_{E\in ( 0,E_{*}^{-}( \lambda ) ) }T_{k}^{\kappa }( \lambda
,E) >(1/2).
\end{equation*}
Hence, the equation $T_{k}^{\kappa }( \lambda ,E) =\frac{1}{2}$
in the variable $E\in \tilde{D}( \lambda ) $ admits no solution
in $\tilde{D}( \lambda ) $ for all $\lambda $ satisfying $k(
p'\lambda ) ^{-1/p}\inf_{0(1/2)$,
that is, for all $0<\lambda <( 2k\inf_{00$. By Lemma \ref{lemma8}, one has
\begin{equation}
\lim_{E\mapsto 0^+}\Theta_+( r_+( \lambda ,E) ) =+\infty \text{,\
and }\lim_{E\mapsto ( E_{*}^{-})
^{-}}\Theta_+( r_+( \lambda ,E) ) =\ell_+%
\text{ } \label{mhtsa}
\end{equation}
\begin{equation*}
\text{with }\left\{
\begin{array}{lcc}
\ell_+\in ( 0,+\infty ) & \text{if} & y( \alpha
_-) <-\alpha_- \\
\ell_+=+\infty & \text{if} & y( \alpha_-) =-\alpha_-.
\end{array}
\right.
\end{equation*}
However, since for all $\lambda >0$, the function $E\mapsto r_+(
\lambda ,E) $ is an increasing diffeomorphism from $(
0,E_{*}^{-}( \lambda ) ) $ onto $( 0,y( \alpha_-) ) $, it follows
from (\ref{mhtsa}) that
\begin{eqnarray*}
\lim_{x\mapsto 0^+}\Theta_+( x) &=&+\infty \text{,\ and} \\
\lim_{x\rightarrow y( \alpha_-) }\Theta_+( x) &=&\ell_+\text{
with }\left\{
\begin{array}{lcc}
\ell_+\in ( 0,+\infty ) & \text{if} & y( \alpha
_-) <-\alpha_- \\
\ell_+=+\infty & \text{if} & y( \alpha_-) =-\alpha_-.
\end{array}
\right.
\end{eqnarray*}
Therefore, if $y( \alpha_-) <-\alpha_-$ (resp. $y( \alpha_-)
=-\alpha_-$) there exists at least a $x^{*}\in ( 0,y( \alpha_-) ]
$ (resp. $x^{*}\in ( 0,y( \alpha_-) ) $) at which the function
$x\mapsto \Theta _+( x) $ attains its global minimum on $( 0,y(
\alpha_-) ] $ (resp. on $( 0,y( \alpha _-) ) $).
Thus $\inf_{00$ for
all $x\in ( 0,-\alpha_-) $. In particular $\Theta_+( x^{*}) >0$.
Therefore, the non existence part of Assertion \textbf{(ii)} of
Theorem \ref{theorem3} is proved, which completes the proof of
Theorem \ref{theorem3}.
%TCIMACRO{\TeXButton{End Proof}{\hfill$\diamondsuit$}}
%BeginExpansion
\hfill$\diamondsuit$%
%EndExpansion
\section{{Open questions}}
\label{sec.6}
\begin{enumerate}
\item Regarding the identity of Lemma \ref{lemma75} one may ask if there
exists a nonlinearity $\tilde{f}$ such that the corresponding
time-maps would be
\begin{equation*}
\tilde{T}_{\pm }( \lambda ,E) =( \lambda p^{\prime })
^{-1/p}\Theta_{\pm }( r_{\pm }( \lambda ,E)) .
\end{equation*}
In the affirmative, the identity of Lemma \ref{lemma75} implies
that
\begin{equation*}
\tilde{T}_{2n}( \lambda ,E) =T_{2n}( \lambda ,E) .
\end{equation*}
So, does $\tilde{T}_{2n+1}^{\pm }( \lambda ,E) =T_{2n+1}^{\pm }(
\lambda ,E) $, and if not can one compare them? On the other hand
what kind of symmetry does $\tilde{f}$ have: odd, \textit{p.h.o}.,
\textit{n.h.o.,} or something else ? A comparison of the two
solution sets corresponding to $f$ and $\tilde{f}$ would be
interesting.
Notice that this is an inverse problem. Related results are
available in the literature, see for instance Urabe \cite{Urabe},
Schaaf \cite[Chap. 4] {Schaaf}.
\item A description of the entire solution set should be interesting.
Indeed the main results of this paper describe only the solutions
which are
inside the set $\cup_{k}A_{k}$. So, how does the solution set of Problem (%
\ref{p1}) look like outside $\cup_{k}A_{k}$? (Such kinds of
descriptions can be found in Guedda and Veron \cite{Guedda} or
Addou \cite{Addou7}).
\item Open questions are numerous. In fact, in view of the known results
for $p=2$ described in Section \ref{sec.2}, one can ask to extend
each one of them to the general case where $p>1$, either in one
or higher dimensions. But a question very close to our main
results is to consider Problem (\ref
{p1}) with $f$ satisfying our conditions but with $q_{0}$ instead of $q$ in (%
\ref{pq4}) and $q_{\pm }$ instead of $q$ in $( \ref{pq5})_{\pm }$
and to give a description of the solution set with respect to all
these parameters: $p$, $q_{0}$, $q_+$, $q_->1$ when $f$ is either
\textit{p.h.o.} or \textit{ n.h.o. }We have considered only the
cases $p>1$, and $q_{0}=q_+=q_->1$, when $f$ is \textit{p.h.o.}
We believe that the same method (Theorem \ref{quad}) works, but
much more patience is required!
\end{enumerate}
\paragraph{ Acknowledgment.} It is a pleasure to thank Professor P.
Korman for sending me many of his recent papers.
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\noindent{\sc Idris Addou }\\
U.S.T.H.B., Institut de Math\'ematiques \\
El-Alia, B.P. no. 32, Bab-Ezzouar \\
16111, Alger, Alg\'erie. \\
e-mail: {\tt idrisaddou@hotmail.com}
\end{document}