\documentclass[twoside]{article} \usepackage{amssymb} % used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Boundary-value problems \hfil EJDE--1999/09} {EJDE--1999/09\hfil Idris Addou \& Abdelhamid Benmeza\"\i \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 1999}(1999), No.~09, pp. 1--29. \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} \\ % Boundary-value problems for the one-dimensional p-Laplacian with even superlinearity \thanks{ {\em 1991 Mathematics Subject Classifications:} 34B15, 34C10. \hfil\break\indent {\em Key words and phrases:} One-dimensional p-Laplacian, two-point boundary-value problem, \hfil\break\indent superlinear, time mapping. \hfil\break\indent \copyright 1999 Southwest Texas State University and University of North Texas. \hfil\break\indent Submitted October 28, 1998. Published March 8, 1999.} } \date{} % \author{Idris Addou \& Abdelhamid Benmeza\"\i} \maketitle \begin{abstract} This paper is concerned with a study of the quasilinear problem $$\displaylines{ -(|u'|^{p-2}u')'= |u|^p-\lambda ,\quad\mbox{in } (0,1)\,, \cr u(0) =u(1) =0\,, \cr}$$ where $p>1$ and $\lambda \in {\mathbb R}$ are parameters. For $\lambda >0$, we determine a lower bound for the number of solutions and establish their nodal properties. For $\lambda \leq 0$, we determine the exact number of solutions. In both cases we use a quadrature method. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} \label{sec1} This paper is devoted to a study of existence and multiplicity of solutions to the quasilinear two-point boundary-value problem \begin{eqnarray} &-(\varphi _p(u') )'= f(\lambda,u) ,\quad\mbox{in }(0,1)\,,& \label{AD} \\ &u(0) = u(1)=0\,, & \nonumber \end{eqnarray} where $\varphi _p(s) =| s|^{p-2}s$ and $f(\lambda,u)= |u|^p-\lambda$. Here $(\varphi _p(u') )'$ is the one-dimensional $p$-Laplacian, and $p>1$. When the differential operator is linear, i.e., $p=2$, several existence and multiplicity results, related to superlinear boundary value problems with Dirichlet boundary data, are available in the literature. Let us recall some of them for the one-dimensional case. Lupo et al \cite{Lupo} have studied the non-autonomous case \begin{eqnarray} &-u''(x) =u^2(x) -t\sin x ,\quad\mbox{in }(0,\pi )\,,& \label{lupo} \\ &u(0) =u(\pi ) =0\,.& \nonumber \end{eqnarray} Using a combination of shooting and topological arguments, they show that for any $k\in {\mathbb N}$ there exists $t_k>0$ such that for all $t\geq t_k$, problem (\ref{lupo}) admits at least $k$ solutions. Castro and Shivaji \cite{Castro and Shivaji}, using phase-plane analysis, consider the problem \begin{eqnarray} &-u''(x) =g(u(x)) -\rho (x) -t ,\quad\mbox{in }(0,1)\,, &\label{Castro}\\ &u(0) =u(1) =0\,,& \nonumber \end{eqnarray} where $\rho$ a continuous function on $[0,1]$, $g\in C^1({\mathbb R})$, $$\lim _{s\rightarrow -\infty }\frac{g(s) }s=M\in {\mathbb R},\quad\mbox{and}\quad \lim _{s\rightarrow +\infty } \frac{g(s) }{s^{1+\sigma }}=+\infty \quad\mbox{with}\quad \sigma >0\,.$$ They show that for $k\in {\mathbb N}$ there exists $t_k(M)$ such that $\lim_{k\rightarrow +\infty} t_k(M) =+\infty$, and for all $t>t_k$, problem (\ref{Castro}) has at least two solutions with $k$ nodes in $(0,1)$. The autonomous case has been studied by many authors. Let us mention some of them. Independently of Castro and Shivaji, Ruf and Solimini \cite{RufSolimini} consider the problem \begin{eqnarray} &-u''(x) =g(u(x)) -t ,\quad\mbox{in }(0,\pi )\,, &\label{Ruf} \\ &u(0)=u(\pi ) =0\,,& \nonumber \end{eqnarray} where $$g\in C^1({\mathbb R}) ,\quad \limsup_{s\rightarrow -\infty }g'(s) <+\infty , \quad\mbox{and}\quad \lim_{s\rightarrow +\infty }g'(s) =+\infty\,.$$ Using variational methods, they show that for any $k\in {\mathbb N}$ there exists $t_k\in {\mathbb R}$ such that for $t>t_k$ problem (\ref {Ruf}) has at least $k$ distinct solutions. Prior to the papers mentioned above, Scovel \cite{Scovel} obtained the same result as Ruf and Solimini \cite{RufSolimini} in the special case where $g(u) =6u^2$. He has shown that for any integer $k\geq 1$, there exist values $t_1<\cdots t_k$ problem (\ref{Ruf}) (with $g(u) =6u^2$) admits at least $k$ distinct solutions. Independently and prior to Scovel, in 1983, Ammar Khodja \cite{AmmarKhodja} obtained a complete description of the solution set of the problem \begin{eqnarray} &-u''(x) =u^2(x) -\lambda ,\quad\mbox{in }(0,1)\,, &\label{AK} \\ &u(0) =u(1) =0\,. & \nonumber \end{eqnarray} He detects all the solutions to (\ref{AK}) for any value of $\lambda \in {\mathbb R}$, and thus obtains the exact number of solutions to (\ref {AK}) for all $\lambda$. To state his result, for any integer $k\geq 1$, denote $$\displaylines { S_k^+=\left\{ u\in C_0^2[0,1] :u'(0) >0, u \mbox{ admits }k-1\mbox{ nodes in }(0,1) \right\}, \cr S_k^-=-S_k^+\quad \mbox{and}\quad S_k=S_k^+\cup S_k^-. \cr}$$ \begin{theorem} \cite{AmmarKhodja} There exists a sequence $(\lambda _k) _{k\geq 0}$ such that $$-\infty <\lambda _0<0<\lambda _1<\lambda _2<\cdots <\lambda _k<\cdots$$ and: \begin{description} \item[(i)] If $\lambda <\lambda _0$, problem (\ref{AK}) admits no solution. \\ If $\lambda _0<\lambda <0$, problem (\ref{AK}) admits exactly two solutions and they are positive. \\ If $\lambda =\lambda _0$ or $0\leq \lambda <\lambda _1$, there exists a unique positive solution. \\ If $\lambda >\lambda _1$, there is no positive solution. \item[(ii)] If $\lambda >0$, there exists a unique solution in $S_1^-$. \item[(iii)] If (and only if ) $\lambda >\lambda _k:$ \begin{itemize} \item there exist exactly 2 solutions in $S_{2k}$ \item there exists exactly one solution in $S_{2k+1}^-$ \end{itemize} \item[(iv)] There exists a sequence $(\mu _k) _{k\geq 1}$ such that $$\lambda _1<\mu _1<\lambda _2<\mu _2<\cdots <\mu _k<\lambda _{k+1}<\cdots$$ and such that: \\ if (and only if ) $\mu _k<\lambda <\lambda _{k+1}$, there exist exactly two solutions in $S_{2k+1}^+$, \\ if (and only if ) $\lambda =\mu _k$ or $\lambda >\lambda _{k+1}$, there exists a unique solution in $S_{2k+1}^+$. \end{description} \end{theorem} The objective of this paper is to extend Ammar Khodja's result to the general quasilinear case $p>1$. In particular, we will show that if $\lambda \leq 0$ the same result holds for all $p>1$, but if $\lambda >0$ and $p>2$ the situation is different from that obtained in \cite{AmmarKhodja}. So, the behavior of the solution set of problem (\ref{AD}) depends not only on the values of $\lambda$ (as was shown in \cite {AmmarKhodja}) but also on those of the parameter $p$. These changes in the behavior of the solution set when the parameter $p$ varies is not new in the literature. Guedda and Veron \cite{Guedda-Veron} consider the problem \begin{eqnarray} &-(\varphi _p(u') )' = \lambda \varphi _p(u) -f(u) ,\quad\mbox{in } (0,1)\,,& \label{GV} \\ & u(0) = u(1) = 0\,, &\nonumber \end{eqnarray} where $f$ is a $C^1$ odd function such that the function $s\mapsto f( s) /s^{p-1}$ is strictly increasing on $(0,+\infty )$ with limit $0$ at $0$ and $\lim _{s\rightarrow +\infty }f(s) /s^{p-1}=+\infty$. They denote by $E_\lambda$ the solution set of problem (\ref{GV}) and show, under some technical assumptions, that when $12$. \medskip This paper is organized as follows. In Section \ref{sec2} we introduce notation and state the main results (Theorems \ref{thm1} and \ref{thm2}). Section \ref{sec3} is devoted to explain of the method used in proving our results. In Section \ref{sec4} we prove Theorem \ref{thm1} and finally, in Section \ref{sec5}, we prove Theorem \ref{thm2}. \section{ Notation and main results} \label{sec2} In order to state the main results, for any $k\in {\mathbb N}^*$, let $$S_k^+=\left\{ \begin{array}{r} u\in C^1([\alpha ,\beta ] ) : \mbox{ u admits exactly (k-1) zeros in (\alpha,\beta )}\\ \mbox{ all simple, u(\alpha ) =u(\beta) =0 and u'(\alpha ) >0} \end{array} \right\}\,,$$ $S_k^-=-S_k^+$ and $S_k=S_k^+\cup S_k^-$. \paragraph{Definition} Let $u\in C([\alpha ,\beta ] )$ be a function with two consecutive zeros $x_11$, $u$ is $((\beta -\alpha ) /k) -$ periodic. \item Every hump of $u$ (necessarily positive) is symmetrical about the center of the interval of its definition. \item The derivative of each hump of $u$ vanishes once and only once. \end{itemize} Let $B_k^-=-B_k^+$ and $B_k=B_k^+\cup B_k^-$. The first result concerns the case $\lambda \leq 0$ and gives the {\em exact} number of solutions to (\ref{AD}). \begin{theorem}[Case $\lambda \leq 0$] \label{thm1} There exists a number $\lambda _{*}<0$ such that: \begin{description} \item[(i)] If $\lambda <\lambda _{*}$, problem (\ref{AD}) admits no solution. \item[(ii)] If $\lambda =\lambda _{*}$, problem (\ref{AD}) admits a unique solution and it belongs to $A_1^+$. \item[(iii)] If $\lambda _{*}<\lambda <0$, problem (\ref{AD}) admits exactly two solutions and they belong to $A_1^+$. \item[(iv)] If $\lambda =0$, beside the trivial solution, problem (\ref{AD}) admits a unique solution and it belongs to $A_1^+$. \end{description} \end{theorem} The second result concerns the case $\lambda >0$. \begin{theorem}[Case $\lambda >0$] \label{thm2} For any $p>1$ there exist two real numbers $J(p) >J_+(p) >0$ and for all $p>2$ there exists a positive real number $J_{-}(p) 0$) admits a solution in $A_1^+$ if and only if $0<\lambda <(2J(p) ) ^{p^2}$, and in this case, the solution is unique. \item[(iv)] Problem (\ref{AD}) admits a solution in $A_1^-$ if and only if ($10$) {\bf or} ($p>2$ and $0<\lambda <( 2J_{-}(p) ) ^{p^2}$), and in this case, the solution is unique. \item[(v)] Problem (\ref{AD}) admits a solution $u_{2n}^{\pm }$ in $A_{2n}^{\pm }$ provided $1(2nJ(p) ) ^{p^2}$ {\bf or} $p>2$ and \begin{eqnarray*} \lefteqn{ \inf \left\{ (2nJ(p) ) ^{p^2},(2n( J_{-}(p) +J_+(p) ) ) ^{p^2}\right\} }\\ <&\lambda& <\sup \left\{ (2nJ(p) ) ^{p^2},(2n(J_{-}(p) +J_+( p) ) ) ^{p^2}\right\} \end{eqnarray*} \item[(vi)] Problem (\ref{AD}) admits a solution in $A_{2n+1}^+$ provided $1(2(n+1) J(p) ) ^{p^2}$ {\bf or} $p>2$ and \begin{eqnarray*} \lefteqn{ \inf \left\{ (2(n+1) J(p) ) ^{p^2},(2((n+1) J_+(p) +nJ_{-}(p) ) ) ^{p^2}\right\} }\\ <&\lambda& <\sup \left\{ (2(n+1) J(p) ) ^{p^2},(2(( n+1) J_+(p) +nJ_{-}(p) ) )^{p^2}\right\} \end{eqnarray*} \item[(vii)] Problem (\ref{AD}) admits a solution in $A_{2n+1}^-$ provided $1(2nJ(p) ) ^{p^2}$ {\bf or} $p>2$ and \begin{eqnarray*} \lefteqn{ \inf \left\{ (2nJ(p) ) ^{p^2},(2( (n+1) J_{-}(p) +nJ_+(p) )) ^{p^2}\right\} }\\ <&\lambda &<\sup \left\{ (2nJ(p) ) ^{p^2},(2((n+1) J_{-}( p) +nJ_+(p) ) ) ^{p^2}\right\} \end{eqnarray*} \end{description} \end{theorem} \paragraph{Remark} According to Proposition \ref{artprop:1} below, if $\lambda >0$ and $p\in (1,2]$ then\\ $\tilde S\subset (\bigcup\limits_{k\geq 1}A_k) \cup (\bigcup\limits_{k\geq 1}B_k)$, where $\tilde S$ denotes the solution set of problem (\ref{AD}). \paragraph{Remark} The results obtained in \cite{AmmarKhodja}, for $p=2$, concerning solutions in $A_{2n}$, $A_{2n+1}^-$, and $A_{2n+1}^+$, are more precise than those stated in Theorem \ref{thm2}, assertions {\bf (v), (vi) }and {\bf (vii)} for $p\neq 2$. In fact, these assertions do not provide the exact number of solutions in $A_{2n}$, $A_{2n+1}^-$, and $A_{2n+1}^+$. The proof given in \cite{AmmarKhodja} uses strongly the fact that the nonlinearity $u\mapsto u^2-\lambda$ is a second degree polynomial function. We were not able to obtain the same degree of precision. \section{The method used }\label{sec3} To obtain our results, we make use of the well known time mapping approach. See, for instance, Laetsch \cite{Laetsch}, de Mottoni \& Tesei \cite{Mottoni-Tesei1}, \cite {Mottoni-Tesei2}, Smoller \& Wasserman \cite{Smoller-Wasserman}, Ammar Khodja \cite{AmmarKhodja}, Shivaji \cite{Shivaji}, Guedda \& Veron \cite {Guedda-Veron}, Ubilla \cite{Ubilla}, Man\'asevich et al \cite {Mana-Njoku-Zano}, Addou \& Ammar Khodja \cite{AddouAmmarKhodja}, Addou et al \cite{Addou-Boug-Derhab-Raffed}, Addou \& Benmeza\"I \cite{AddouBenmezai2}. To describe this method we denote by $g$ a nonlinearity and by $p$ a real parameter, and we assume one of the following conditions: \begin{eqnarray} & g\in C({\mathbb R},{\mathbb R}) \quad\mbox{and}\quad 10,\,\forall x\in {\mathbb R}^* &\label{art3} \\ &g\mbox{ is locally Lipschitzian and }10\quad \mbox{and}\quad E^p-p'G(\xi ) >0,\forall \,\xi ,0<\kappa \xi <\kappa s\right\} $$and$$ r_\kappa (E) =\left\{ \begin{array}{ll} 0 & \mbox{ if } X_\kappa (E) =\emptyset\,, \\ \kappa \sup (\kappa X_\kappa(E) ) & \mbox{ otherwise}. \end{array}\right. $$Note that r_\kappa  may be infinite. We shall also make use of the following sets:$$ D_\kappa =\left\{ E\geq 0:0<| r_\kappa (E) | <+\infty \mbox{ and }\kappa g(r_\kappa (E) ) >0\right\} $$and D=D_+\cap D_{-}. 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}(E) = n(T_+(E) +T_{-}(E) ) , \quad n\in {\mathbb N}, \quad E\in D\,, & \\ &T_{2n+1}^\kappa (E) = T_{2n}(E) +T_\kappa (E) , \quad n\in {\mathbb N}, \quad E\in D\,. \end{eqnarray*} \begin{theorem}[Quadrature method] \label{artquad} Assume that (\ref{art2}) holds. Let E\geq 0, \kappa =+,-. Then \begin{enumerate} \item Problem (\ref{art5}) admits a solution u_\kappa \in A_1^\kappa  satisfying u_\kappa'(\alpha ) =\kappa E if and only if E\in D_\kappa \cap (0,+\infty )  and T_\kappa ( E) =(\beta -\alpha ) /2, and in this case the solution is unique. \item Problem (\ref{art5}) admits a solution u_\kappa \in A_{2n}^\kappa   (n\neq 0)  satisfying u_\kappa'(\alpha ) =\kappa E if and only if E\in D\cap (0,+\infty )  and  T_{2n}(E) =(\beta -\alpha ) /2, and in this case the solution is unique. \item Problem (\ref{art5}) admits a solution u_\kappa \in A_{2n+1}^\kappa  (n\neq 0)  satisfying u_\kappa'(\alpha ) =\kappa E if and only if E\in D\cap (0,+\infty )  and T_{2n+1}^\kappa (E) =(\beta -\alpha ) /2, and in this case the solution is unique. \item Problem (\ref{art5}) admits a solution u_\kappa \in B_n^\kappa   (n\neq 0)  if and only if 0\in D_\kappa  and nT_\kappa (0) =(\beta -\alpha ) /2, and in this case the solution is unique. \end{enumerate} \end{theorem} One may observe that this result does not give information about solutions to (\ref{art5}) outside \bigcup\limits_{k\geq 1}(A_k\cup B_k). The following proposition gives some useful information. \begin{proposition} \label{artprop:1} If (\ref{art3}) holds then S\subset \left\{ 0\right\} \cup ( \bigcup\limits_{k\geq 1}A_k) . If (\ref{art4}) holds then \begin{description} \item[(i)] g(0) =0 implies S\subset \left\{ 0\right\} \cup (\bigcup\limits_{k\geq 1}A_k), \item[(ii)] g(0) \neq 0 implies S\subset ( \bigcup\limits_{k\geq 1}A_k) \cup (\bigcup\limits_{k\geq 1}B_k) . \end{description} \end{proposition} Theorem \ref{artquad} and Proposition \ref{artprop:1} are certainly well known, but we did not find a convenient reference to the precise statements used later. \section{Proof of Theorem \protect{\ref{thm1}} }\label{sec4} Since \lambda \leq 0, any solution to (\ref{AD}) is positive. In fact, if u is a solution to (\ref{AD}) then$$ u'(x) =\varphi _{p'}(\varphi _p(u'(x) ) ), \, \forall x\in (0,1) \,. $$Since x\mapsto \varphi _p(u'(x) )  is decreasing (from (\varphi _p(u') ) '(x) =-|u(x) |^p+\lambda, for all x\in (0,1) and \lambda \leq 0) and \varphi_{p'} is increasing, it follows that u' is decreasing. This shows that u is concave, and since u(0) =u(1) =0 it follows that u is positive. Moreover, the nonlinear term f(\lambda , u) =|u| ^p-\lambda  satisfies (\ref{art3}) so, from Proposition \ref{artprop:1}, it follows that any nontrivial solution is necessarily in A_1^+. Hence, we have only to define the time map T_+. In order to do this, we need the following technical lemma. \begin{lemma} \label{lemma1}Consider the equation in s\in {\mathbb R}: $$\label{Eq.1}E^p-p'F(\lambda ,s) =0 \,,$$ where p>1, \lambda \leq 0 and E\geq 0 are real parameters and F( \lambda ,s) =\int_0^sf(\lambda ,t) dt. Then for any E>0, (resp. E=0) equation (\ref{Eq.1}) admits a unique positive zero s_+=s_+(p,\lambda ,E)  (resp. a unique zero s_+=s_+(p,\lambda ,0) =0). Moreover: \begin{description} \item[(a)] The function E\longmapsto s_+(p,\lambda ,E)  is C^1 in (0,+\infty )  and$$ \frac{\partial s_+}{\partial E}(p,\lambda ,E) =\frac{ (p-1) E^{p-1}}{f(\lambda ,s_+(p,\lambda,E) ) }>0 $$for all p>1, all \lambda \leq 0, and all E>0. \item[(b)] \lim \limits_{E\rightarrow 0^+}s_+(p,\lambda,E) =0. \item[(c)] \lim \limits_{E\rightarrow +\infty }s_+(p,\lambda,E) =+\infty . \end{description} \end{lemma} \paragraph{Proof.} For a fixed p>1, \lambda \leq 0 and E\geq 0, consider the function$$ s\longmapsto M(p,\lambda ,E,s) :=E^p-p'F( \lambda ,s) =E^p-p's(\frac{| s|^p}{p+1} -\lambda ) \,, $$defined in {\mathbb R}, which is strictly decreasing and such that$$ M(p,\lambda ,E,0) =E^p\geq 0\,, \quad \mbox{and} \quad \lim \limits_{s\rightarrow +\infty }M(s) =-\infty \,. $$It is clear that (\ref{Eq.1}) admits, for any E>0, a unique positive zero, s_+=s_+(p,\lambda ,E) ; and if E=0, it admits a unique zero s_+=0. Now, for any p>1 and \lambda \leq 0, consider the real-valued function$$ (E,s) \longmapsto M_+(E,s) :=E^p-p^{\prime}s(\frac{s^p}{p+1}-\lambda ) $$defined on \Omega_+=(0,+\infty ) ^2. One has M_+\in C^1(\Omega_+)  and$$ \frac{\partial M_+}{\partial s}(E,s) =-p'f( \lambda ,s) =-p'(| s|^p-\lambda ) \quad \mbox{in }\Omega_+, $$hence$$ \frac{\partial M_+}{\partial s}(E,s) <0 \quad\mbox{in } \Omega_+ $$and one may observe that s_+(p,\lambda ,E)  belongs to the open interval (0,+\infty )  and satisfies from its definition $$\label{Eq.2}M_+(E,s_+(p,\lambda ,E) ) =0\,.$$ So, one can make use of the implicit function theorem to show that the function E\longmapsto s_+(p,\lambda ,E)  is C^1( (0,+\infty ) ,{\mathbb R})  and to obtain the expression for \frac{\partial s_+}{\partial E}(p,\lambda ,E)  given in {\bf (a)}. Hence, for any fixed p>1 and \lambda \leq 0, the function defined in (0,+\infty )  by  E\longmapsto s_+(p,\lambda ,E)  is strictly increasing and bounded from below by 0 and from above by +\infty . Thus the limit \lim \limits_{E\rightarrow 0^+}s_+(p,\lambda ,E) =l_0^+  exists as a real number and the limit \lim \limits_{E\rightarrow +\infty }s_+(p,\lambda ,E) =l_{+\infty } exists and belongs to (0,+\infty ] . Moreover$$ 0\leq l_0^+1$and$\lambda \leq 0$, the function$(E,s) \longmapsto M_+(E,s) $is continuous in$[0,+\infty ) ^2$and the function$ E\longmapsto s_+(p,\lambda ,E) $is continuous in$( 0,+\infty ) $and satisfies$($\ref{Eq.2}$)$. So, by passing to the limit in$($\ref{Eq.2}$)$as$E$tends to$0^+$one gets: $$0=\lim _{E\rightarrow 0^+}M_+(E,s_+(p,\lambda ,E) ) =M_+(0,l_0^+) .$$ Hence,$l_0^+$is a zero, belonging to$[0,+\infty ) $, of the equation in$s$:$M_+(0,s) =0$. By resolving this equation in$[0,+\infty ) $one gets:$ l_0^+=0$. The assertion {\bf (b)} is proved. Assume that$l_{+\infty }<+\infty $then by passing to the limit in (\ref {Eq.2}) as$E$tends to$+\infty $one gets: $$+\infty =p'l_{+\infty }(\frac{(l_{+\infty }) ^p}{ p+1}-\lambda ) <+\infty ,$$ which is impossible. So,$l_{+\infty }=+\infty $. Therefore, Lemma \ref{lemma1} is proved. \quad$\diamondsuit$\medskip Now we are ready to compute$X_+(p,\lambda ,E) $as defined in Section \ref {sec3}, for any$p>1,\lambda \leq 0$and$E\geq 0$. In fact,$X_+(p,\lambda ,E) =(0,s_+(p,\lambda ,E) ) $if$E>0$and$X_+(p,\lambda,0) =\emptyset $. Thus $$r_+(p,\lambda ,E) :=\sup X_+(p,\lambda ,E) =s_+(p,\lambda ,E) \mbox{ if }E>0\quad \mbox{and}\quad r_+(p,\lambda ,0) =0\,,$$ and since$f(\lambda ,s) =| s|^p-\lambda >0$,$ \forall (\lambda ,s) \in (-\infty ,0] \times {\mathbb R},(\lambda ,s) \neq (0,0)$, it follows that $$\begin{array}{ccl} D_+ & := & \left\{ E\geq 0:00\right\} \\ & = & (0,+\infty ) \,. \end{array}$$ Before going further in the investigation, we deduce from Lemma \ref{lemma1} the following: \begin{eqnarray} &\lim \limits_{E\rightarrow 0^+}r_+(p,\lambda ,E) =0\quad \mbox{and}\quad \lim \limits_{E\rightarrow +\infty }r_+(p,\lambda ,E) =+\infty \,,& \label{A} \\ &\frac{\partial r_+}{\partial E}(p,\lambda ,E) = \frac{(p-1) E^{p-1}}{f(\lambda , r_+(p, \lambda , E) ) }>0, \forall E\in D_+=(0,+\infty ) \quad \forall \lambda \leq 0\,.& \label{C} \end{eqnarray} We define, for any$p>1,\lambda \leq 0$, and$E\in D_+=(0,+\infty ) $the time map $$\label{L}T_+(p,\lambda ,E) :=\int_0^{r_+( p,\lambda ,E) }\left\{ E^p-p'F(\lambda ,\xi ) \right\} ^{-1/p}d\xi ,\;E\in D_+$$ and a simple change of variables shows that $$\label{U}T_+(p,\lambda ,E) =r_+(p,\lambda ,E) \int_0^1\left\{ E^p-p'F(\lambda ,r_+( p,\lambda ,E) \xi ) \right\} ^{-1/p}d\xi .$$ Observe that from the definition of$s_+(p,\lambda,E) $one has $$E^p-p'F(\lambda ,s_+(p,\lambda ,E) ) =0$$ and so, from the definition of$r_+(p,\lambda ,E) $, one has$E^p=p'F(\lambda ,r_+(p,\lambda ,E))$. So, (\ref{U}) may be written as \begin{eqnarray} \label{W} \lefteqn{ T_+(p,\lambda ,E) }\\ &=&r_+(p,\lambda ,E)(p') ^{-1/p}\int_0^1\left\{ F( \lambda ,r_+(p,\lambda ,E) ) -F(\lambda ,r_+(p,\lambda ,E) \xi ) \right\} ^{-1/p}\,d\xi\,. \nonumber \end{eqnarray} After some rearrangements one has \begin{eqnarray} \label{W1} \lefteqn{ T_+(p,\lambda ,E) }\\ &=&r_+^{1-\frac 1p}(p,\lambda,E) (p') ^{-1/p} \int_0^1\left\{ \frac{r_+^p(p,\lambda ,E) (1-\xi ^{p+1}) }{p+1}-\lambda (1-\xi ) \right\} ^{-1/p}\,d\xi\,. \nonumber \end{eqnarray} \begin{lemma} \label{lemma2}If$\lambda \leq 0$then one has \begin{description} \item[(i)]$\lim \limits_{E\rightarrow 0^+}T_+(p,\lambda ,E) =0$, if$\lambda <0$and$\lim \limits_{E\rightarrow 0^+}T_+(p,0,E) =+\infty $, \item[(ii)]$\lim \limits_{E\rightarrow +\infty }T_+(p,\lambda ,E) =0,\forall \lambda \leq 0$, \item[(iii)] If$\lambda <0$,$T_+(p,\lambda ,\cdot ) $admits a unique critical point,$E^*(\lambda ) $, at which it attains its global maximum value. Moreover, \begin{description} \item[(a)] The function$\lambda \mapsto T_+(p,\lambda ,E^*(\lambda ) ) $is strictly increasing in$( -\infty ,0) $. \item[(b)]$\lim \limits_{\lambda \rightarrow -\infty }T_+( p,\lambda ,E^*(\lambda ) ) =0$. \item[(c)]$\lim \limits_{\lambda \rightarrow 0^-}T_+(p,\lambda ,E^*(\lambda ) ) =+\infty $. \end{description} \item[(iv)] If$\lambda =0,(\partial T_+/\partial E) ( p,0,\cdot ) <0$in$(0,+\infty ) $. \end{description} \end{lemma} \paragraph{Proof.} {\bf (i)} If$\lambda <0$, from (\ref{W1}) one has $$0\leq T_+(p,\lambda ,E) \leq r_+^{1-\frac 1p}( p,\lambda ,E) (p') ^{-\frac 1p}\int_0^1\left\{ -\lambda (1-\xi ) \right\} ^{-1/p}\,d\xi\,.$$ So, by passing to the limit as$E$tends to$0$, one gets $$0\leq \lim _{E\rightarrow 0}T_+(p,\lambda ,E) \leq \lim _{E\rightarrow 0}r_+^{1-\frac 1p}(p,\lambda ,E) ( p') ^{-1/p}\int_0^1\left\{ -\lambda (1-\xi ) \right\} ^{-1/p}d\xi =0\,.$$ If$\lambda =0$, then from (\ref{W1}) one gets $$T_+(p,0,E) =(p') ^{-1/p}r_+^{-1/p}(p,0,E) \int_0^1(\frac{1-\xi ^{p+1}}{p+1}) ^{-1/p}\,d\xi ,$$ and from (\ref{A}) one gets$\lim \limits_{E\rightarrow 0^+}T_+( p,0,E) =+\infty $. \\ {\bf (ii)} From (\ref{W1}) one has for any$\lambda \leq 0,$$$0\leq T_+(p,\lambda ,E) \leq r_+^{-1/p}( p,\lambda ,E) (p') ^{-\frac 1p}\int_0^1\left\{ \frac{1-\xi ^{p+1}}{p+1}\right\} ^{-1/p}d\xi .$$ So, by passing to the limit as$E$tends to$+\infty $, one gets \begin{eqnarray*} 0&\leq& \lim _{E\rightarrow +\infty }T_+(p,\lambda ,E) \\ &\leq& \lim _{E\rightarrow +\infty }r_+^{-1/p}(p,\lambda ,E) (p') ^{-1/p}\int_0^1\left\{ \frac{1-\xi ^{p+1}}{ p+1}\right\} ^{-1/p}d\xi =0\,. \end{eqnarray*} {\bf (iii)} If$\lambda <0$, then from {\bf (i)} and {\bf (ii)} one deduces that$T_+(p,\lambda ,\cdot ) $admits at least one critical point. Here, we are going to prove its uniqueness. From (\ref{W}), one may observe that $$T_+(p,\lambda ,E) =(p') ^{-\frac 1p}S(p,\lambda ,\rho (p,\lambda ,E) )$$ where$\rho (p,\lambda ,E) =r_+(p,\lambda,E) $and $$S(p,\lambda ,\rho ) =\int_0^\rho \left\{ F(p,\lambda ,\rho ) -F(p,\lambda ,\xi ) \right\} ^{-\frac 1p}d\xi \,.$$ On the other hand, observe that for each fixed$\lambda <0$the function$E\mapsto \rho (p,\lambda ,E) $is an increasing$C^1$-diffeomorphism from$(0,+\infty ) $onto itself (Lemma \ref{lemma1}, assertions (a), (b) and (c)), and $$\label{bat}\frac{\partial T_+}{\partial E}(p,\lambda ,E) =(p^{\prime }) ^{-1/p}\times \frac{\partial S}{\partial \rho }( p,\lambda ,\rho (p,\lambda ,E) ) \times \frac{\partial \rho }{\partial E}(p,\lambda ,E) .$$ So, to study the variations of$E\mapsto T_+(p,\lambda ,E) $it suffices to study those of$\rho \mapsto S(p,\lambda ,\rho ) $. That is,$S(p,\lambda ,\cdot ) $attains a local maximum (resp. minimum) value at$\rho _{*}$iff$T_+(p,\lambda ,\cdot ) $does so at$\rho _{p,\lambda }^{-1}(\rho _{*})$, where$\rho _{p,\lambda }^{-1}$is the function inverse to$\rho ( p,\lambda ,\cdot ) $. From {\bf (i)} and {\bf (ii)}, it follows that$\lim \limits_{\rho \rightarrow 0}S(\rho ) =\lim \limits_{\rho \rightarrow +\infty }S(\rho ) =0$, that is,$S$admits at least a maximum value. To prove uniqueness, we first find a priori estimates on the critical points of$S(p,\lambda ,\cdot ) $. That is, for each$\lambda <0$, we look for a compact interval$J( \lambda ) $which contains all possible critical points of$S( p,\lambda ,\cdot ) $. Next, we prove that$S(p,\lambda ,\cdot ) $is concave in$J(\lambda ) $. One has $$\label{bat*}\frac{\partial S}{\partial \rho }(p,\lambda ,\rho ) =\int_0^\rho \frac{H(p,\lambda ,\rho ) -H( p,\lambda , u) }{p\rho (F(p,\lambda ,\rho ) -F(p,\lambda , u) ) ^{\frac{p+1}p}}du$$ where$H(p,\lambda , u) =pF(p,\lambda , u) -uf(p,\lambda , u) =\frac{-u^{p+1}}{p+1}-\lambda ( p-1) u$,$\forall u>0$. The variations of$u\mapsto H( p,\lambda , u) $can be described as follows.$H(p,\lambda ,\cdot ) $is strictly increasing in$(0,\rho _1( p,\lambda ) ) $and strictly decreasing in$(\rho _1(p,\lambda ) ,+\infty ) $where$\rho _1( p,\lambda ) =(-\lambda (p-1) ) ^{1/p}$. Moreover,$H(p,\lambda ,0) =H(p,\lambda ,\rho _2(p,\lambda ) ) =0$where$\rho _2(p,\lambda ) =(-\lambda (p^2-1) ) ^{1/p}>\rho _1(p,\lambda ) $. So, it follows that: $$\frac{\partial S}{\partial \rho }(p,\lambda ,\rho ) >0, \, \forall \rho \in (0,\rho _1(p,\lambda ) )$$ and $$\frac{\partial S}{\partial \rho }(p,\lambda ,\rho ) <0,\, \forall \rho \in (\rho _2(p,\lambda ),+\infty ) .$$ That is, we get the a priori estimates as follows :$\forall p>1,\forall \lambda <0,\forall \rho _{*}>0$, $$\frac{\partial S}{\partial \rho }(p,\lambda ,\rho _{*}) =0\Longrightarrow \rho _{*}\in J(\lambda ) :=[\rho _1(p,\lambda ) ,\rho _2(p,\lambda ) ] .$$ Easy computations show that for any$\rho >0$and$\lambda <0$, one has \begin{eqnarray*} \frac{\partial ^2S}{\partial \rho ^2}(p,\lambda ,\rho ) &=&\int_0^1\frac{(p+1) (H(p,\lambda ,\rho ) -H(p,\lambda ,\rho \xi ) ) ^2}{p^2\rho ( F(p,\lambda ,\rho ) -F(p,\lambda ,\rho \xi ) ) ^{\frac{2p+1}p}}d\xi \\ &&+\int_0^1\frac{p(\Psi (p,\lambda ,\rho ) -\Psi (p,\lambda ,\rho \xi ) ) (F( p,\lambda ,\rho ) -F(p,\lambda ,\rho \xi ) ) }{p^2\rho (F(p,\lambda ,\rho ) -F(p,\lambda ,\rho \xi ) ) ^{\frac{2p+1}p}}\,d\xi\,, \end{eqnarray*} where \begin{eqnarray*} \Psi (p,\lambda , u) & = & -p(p+1) F( p,\lambda , u) +2puf(p,\lambda , u) -u^2f_u^{\prime}(p,\lambda , u) \\ & = & \lambda p(p-1) u,\forall u>0\,. \end{eqnarray*} After some substitutions one gets $$\frac{\partial ^2 S}{\partial \rho ^2}(p,\lambda ,\rho ) =\int_0^1\frac{\rho (1-\xi ) ^2P(X(\xi ) ) }{p^2(F(p,\lambda ,\rho ) -F(p,\lambda ,\rho \xi ) ) ^{\frac{2p+1}p}}\,d\xi \,,$$ where $$X(\xi ) =\left\{ \begin{array}{ll} p+1 & \mbox{if }\xi =1\, \\ \frac{1-\xi ^{p+1}}{1-\xi } & \mbox{if }\xi \in [0,1) \end{array} \right.$$ and$P$is the polynomial function $$P(X) =(\frac{\rho ^{2p}}{p+1}) X^2+\frac{( p-1) (p^2+2p+2) }{(p+1) }\lambda \rho ^pX-(p-1) \lambda ^2 .$$ An easily checked fact is that$X(\xi ) \in [ 1,p+1]$, for all$\xi \in [0,1]$. In fact, the function$\xi \mapsto h(\xi ) :=\xi ^{p+1}$is convex in$(0,+\infty ) $, and $$X(\xi ) =\frac{h(1) -h(\xi ) }{1-\xi } \leq h'(1) =p+1\,, \forall \xi \in (0,1) \,.$$ So, we are interested in the sign of$P(X) $when$X\in [1,p+1] $. First, its discriminant is$\Delta =(\mu (p) /( p+1) ^2) \lambda ^2\rho ^{2p}>0$, where $$\mu (p) =(p-1) ^2(p^2+2p+2) ^2+4( p^2-1) \,,$$ and its roots are, for each$\lambda <0$and$\rho >0$, $$\displaylines{ X_1(p,\lambda ,\rho ) =\frac \lambda {2\rho ^p}( \sqrt{\mu (p) }-(p-1) (p^2+2p+2) ) <0\,, \cr X_2(p,\lambda ,\rho ) =\frac{-\lambda }{2\rho ^p}( \sqrt{\mu (p) }+(p-1) (p^2+2p+2) ) >0\,. \cr}$$ It can be verified that$\rho \mapsto X_2(p,\lambda ,\rho ) $is decreasing in$(0,+\infty ) $and one can deduce, from$ H(p,\lambda ,\rho _2(p,\lambda ) ) =0$, that $$X_2(p,\lambda ,\rho _2(p,\lambda ) ) =\frac{ \sqrt{\mu (p) }+(p-1) (p^2+2p+2) }{ 2(p^2-1) },\;\forall \lambda <0\,.$$ Hence, one can deduce that$X_2(p,\lambda ,\rho _2( p,\lambda ) ) >p+1$. (In fact, to prove this it suffices to show that $$\sqrt{\mu (p) }+(p-1) (p^2+2p+2) >2(p+1) (p^2-1)$$ which is equivalent to proving that$\mu (p) >(p( p-1) (p+2) ) ^2$, and this is (after some simple computations) equivalent to$4(p+1) p^2>0$which is true since$ p>1$). Then $$[1,p+1] \subset (X_1(p,\lambda ,\rho ) ,X_2(p,\lambda ,\rho ) ) :\forall \lambda <0,\forall \rho \in [\rho _1(p,\lambda ) ,\rho _2(p,\lambda ) ] ,$$ hence,$P(X(\xi ) ) <0$, for all$\xi \in [0,1]$, so, $$\frac{\partial ^2S}{\partial \rho ^2}(p,\lambda ,\rho ) <0\quad \forall \lambda <0,\, \forall \rho \in J(\lambda ) :=[\rho _1(p,\lambda ) ,\rho _2(p,\lambda ) ] \,,$$ which proves the uniqueness of the critical point of$S(p,\lambda ,\cdot ) $and of$T_+(p,\lambda ,\cdot ) $. {\bf (a)} Some easy computations show that \begin{eqnarray} \lefteqn{ \frac{\partial T_+}{\partial E}(p,\lambda ,E) }\label{onze}\\ &=&(p') ^{-1/p}\frac{\partial r_+}{\partial E}(p,\lambda ,E) \int_0^{r_+(p,\lambda ,E) }\frac{H(p,\lambda ,r_+) -H( p,\lambda ,\xi ) }{pr_+(F(p,\lambda ,r_+) -F(p,\lambda ,\xi ) ) ^{\frac{p+1}p}}\,d\xi \,, \nonumber \end{eqnarray} and that \begin{eqnarray} \lefteqn{ \frac{\partial T_+}{\partial \lambda }(p,\lambda ,E) }\label{douze}\\ &=& (p') ^{-1/p}\frac{\partial r_+}{\partial\lambda }(p,\lambda ,E) \int_0^{r_+(p,\lambda,E) } \frac{H(p,\lambda ,r_+) -H(p,\lambda ,\xi) }{pr_+(F(p,\lambda ,r_+) -F( p,\lambda ,\xi ) ) ^{\frac{p+1}p}}d\xi \nonumber \\ && +(p') ^{-\frac1p}\int_0^{r_+(p,\lambda ,E) }\frac{r_+( p,\lambda ,E) -\xi }{p(F(p,\lambda ,r_+) -F(p,\lambda ,\xi ) ) ^{\frac{p+1}p}}\,d\xi \,, \nonumber \end{eqnarray} and then combining (\ref{onze}) and (\ref{douze}) one gets \begin{eqnarray*} \lefteqn{ - \frac{\partial r_+}{\partial \lambda }\frac{\partial T_+}{\partial E}+ \frac{\partial r_+}{\partial E}\frac{\partial T_+}{\partial \lambda } }\\ &=&(p') ^{-1/p}\frac{\partial r_+}{\partial E} (p,\lambda ,E) \int_0^{r_+(p,\lambda,E) }\frac{r_+(p,\lambda ,E) -\xi }{p(F(p,\lambda ,r_+) -F(p,\lambda ,\xi )) ^{\frac{p+1}p}}\,d\xi \,, \end{eqnarray*} so, $$\label{treize}-\frac{\partial r_+}{\partial \lambda }\frac{\partial T_+}{ \partial E}+\frac{\partial r_+}{\partial E}\frac{\partial T_+}{\partial \lambda }>0,\quad \forall E>0\,,\ \lambda <0\,.$$ Since,$(\partial T_+/\partial E) (p,\lambda ,E^*(\lambda ) ) =0$, using (\ref{treize}) and (\ref{C}) one gets: $$\frac{\partial T_+}{\partial \lambda }(p,\lambda ,E^*( \lambda ) ) >0,\,\forall \lambda <0\,.$$ {\bf (b)} Since$H(p,\lambda ,\cdot ) $is strictly increasing on$(0,\rho _1(p,\lambda ) ) $, $$\frac{\partial T_+}{\partial E}(p,\lambda ,E) >0,\forall E\in (0,E_1(p,\lambda ) )$$ where$E_1(p,\lambda ) :=(p'F(p,\lambda ,\rho _1(p,\lambda ) ) ) ^{1/p}$. Since$ (\partial T_+/\partial E) (p,\lambda ,E^*( \lambda ) ) =0$, it follows that$E^*(\lambda ) \geq E_1(\lambda ) $, and since $$F(r_+(E^*(\lambda ) ) ) =\frac{ (E^*) ^p(\lambda ) }{p'}\geq \frac{ E_1^p(p,\lambda ) }{p'}=F(\rho _1( p,\lambda ) )$$ and$F$is continuous and strictly increasing$(\lambda <0)$, it follows that $$r_+(p,\lambda ,E^*(\lambda ) ) \geq \rho _1(p,\lambda ) =r_+(p,\lambda ,E_1( p,\lambda ) ) .$$ One has from (\ref{W1}), \begin{eqnarray*} T_+(p,\lambda ,E^*(\lambda ) ) & \leq & (p') ^{-1/p}r_+^{-1/p}(p,\lambda ,E^*(\lambda ) ) \int_0^1( \frac{1-\xi ^{p+1}}{p+1}) ^{-1/p}d\xi \\ & \leq & (p') ^{-1/p}\left\{ \rho _1( p,\lambda ) \right\} ^{-1/p}\int_0^1(\frac{1-\xi ^{p+1} }{p+1}) ^{-1/p}\,d\xi \end{eqnarray*} and by passing to the limit as$\lambda $tends to$-\infty $, one gets \begin{eqnarray*} 0&\leq &\lim \limits_{\lambda \rightarrow -\infty }T_+(p,\lambda ,E^*(\lambda ) ) \\ &\leq& (p')^{-1/p}\int_0^1( \frac{1-\xi ^{p+1}}{p+1}) ^{-1/p}d\xi \lim \limits_{\lambda \rightarrow -\infty }\left\{ -\lambda ( p-1) \right\} ^{-1/p^2}=0\,. \end{eqnarray*} {\bf (c)} For each$\lambda <0$, one has $$T_+(p,\lambda ,E^*(\lambda ) ) =\sup _{E>0}T_+(p,\lambda ,E) \geq T_+(p,\lambda,E_1(\lambda ) )$$ and from (\ref{W1}) and the fact$\rho _1(p,\lambda ) =r_+(p,\lambda ,E_1(p,\lambda ) ) $one has \begin{eqnarray*} \lefteqn{ T_+(p,\lambda ,E_1(\lambda ) ) }\\ &=& (-\lambda ) ^{-1/p^2}(p') ^{-1/p}(p-1) ^{\frac{p-1}{p^2}} \int_0^1\left\{ \frac{p-1}{p+1}(1-\xi ^{p+1}) +(1-\xi ) \right\} ^{-1/p}\,d\xi \,. \end{eqnarray*} So, $$\lim _{\lambda \rightarrow 0^-}T_+(p,\lambda ,E^*( \lambda ) ) \geq \lim _{\lambda \rightarrow 0^-}T_+( p,\lambda ,E_1(\lambda ) ) =+\infty \,.$$ {\bf (iv)} If$\lambda =0$, the function$\rho \mapsto S(p,\lambda ,\rho ) $decreases strictly on$(0,+\infty ) $, since the function$u\mapsto H(p,0, u) :=-u^{p+1}/( p+1) $does so in$(0,+\infty ) $(see (\ref{bat*})). Then, from (\ref{bat}) and (\ref{C}) it follows that$(\partial T_+/\partial E) (p,0,\cdot ) <0$in$(0,+\infty ) $. Therefore, Lemma \ref{lemma2} is proved. \quad$\diamondsuit$\paragraph{Completion of the proof of Theorem \ref{thm1}.} The proof is an easy consequence of the previous lemmas. In fact, there exists a unique$\lambda ^*<0$which satisfies$T_+( p,\lambda ^*,E^*(\lambda ^*) ) =\frac 12$, and the function$\lambda \mapsto T_+(p,\lambda ,E^*(\lambda ) ) $is strictly increasing in$(-\infty ,0) $. So, if$\lambda <\lambda ^*$, for any$E>0$and$\lambda <0,$$$T_+(p,\lambda ,E) \leq \sup _{E>0}T_+(p,\lambda,E) =T_+(p,\lambda ,E^*(\lambda ) ) < T_+(p,\lambda ^*,E^*(\lambda ^*) )=\frac 12\,.$$ Thus equation$T_+(p,\lambda ,E) =\frac 12$admits no solution. If$\lambda =\lambda ^*$,$E^*(\lambda ^*) $is the unique solution of the equation$T_+(p,\lambda ^*,E) =\frac 12$. So, problem (\ref{AD}) admits a unique positive solution and this one is in$A_1^+$. Finally, if$0>\lambda >\lambda ^*$, then$ T_+(p,\lambda ,E^*(\lambda ) ) >T_+( p,\lambda ^*,E^*(\lambda ^*) ) =\frac 12$. So, equation$T_+(p,\lambda ,E) =\frac 12$admits exactly two solutions and then problem (\ref{AD}) admits exactly two positive solutions in$A_1^+$. If$\lambda =0$,$T_+(p,0,\cdot ) $is strictly decreasing in$(0,+\infty ) $and$\lim _{E\rightarrow 0^+}T_+(p,0,E) =+\infty $and$\lim _{E\rightarrow +\infty }T_+(p,0,E) =0$. So, equation$ T_+(p,0,E) =(1/2) $admits a unique solution in$(0,+\infty ) $. Thus, Theorem \ref{thm1} is proved. \quad$\diamondsuit$\section{\bf Proof of Theorem \protect{\ref{thm2}} } \label{sec5} As for the proof of Theorem \ref{thm1}, we begin this section by some preliminary lemmas. In order to define the time-maps we need as usual the following technical lemma. \begin{lemma} \label{artlemma1}Consider the equation in the variable$s\in {\mathbb R}^*$, $$\label{art6}E^p-p'F(\lambda ,s) =0$$ where$p>1$,$\lambda >0$and$E\geq 0$are real parameters. First, if$E=0$, equation (\ref{art6}) admits a unique positive zero$s_+=s_+( p,\lambda ,0) $and a unique negative zero$s_{-}=s_{-}( p,\lambda ,0) $such that$| s_{\pm }| =(\lambda (p+1) ) ^{1/p}$. Moreover, for any$E>0$, equation ( \ref{art6}) admits a unique positive zero$s_+=s_+(p,\lambda ,E) $and this zero belongs to the open interval$((\lambda (p+1) ) ^{1/p},+\infty ) $. On the other hand, \begin{description} \item[(i)] If$E>E_{*}(p,\lambda ) :=((\frac{ pp\prime }{p+1}) \lambda ^{1+\frac 1p}) ^{1/p}$, equation (\ref{art6}) admits no negative zero. \item[(ii)] If$E=E_{*}(p,\lambda ) $, equation (\ref{art6}) admits a unique negative zero \\$s_{-}=s_{-}(p,\lambda ) =-\lambda ^{1/p}$. \item[(iii)] If$00\,, $$for all p>1, for all \lambda >0, and for all E>0. (resp. for all E\in (0,E_{*}(p,\lambda))). \item[(b)] \lim \limits_{E\rightarrow 0^+}s_+(p,\lambda ,E) =((p+1) \lambda ) ^{1/p} and \lim \limits_{E\rightarrow 0^+}s_{-}(p,\lambda,E) =0. \item[(c)] \lim \limits_{E\rightarrow +\infty }s_+(p,\lambda ,E) =+\infty  and \lim\limits_{E\rightarrow E_{*}}s_{-}(p,\lambda ,E) =-\lambda^{1/p}. \end{description} \end{lemma} \paragraph{Proof.} For a fixed p>1, \lambda>0 and E\geq 0, consider the function$$ s\longmapsto N(p,\lambda ,E,s) :=E^p-p'F( \lambda ,s) =E^p-p's(\frac{| s|^p}{p+1}-\lambda ) , $$defined in {\mathbb R}. From a study of its variations, it is clear that equation (\ref{art6}) admits, if E=0, a unique positive zero  s_+ and a unique negative zero s_{-}. Their values are obtained by simple resolution of equation (\ref{art6}). Moreover, for any E>0, equation (\ref{art6}) admits a unique positive zero, s_+=s_+( p,\lambda ,E) , and this zero belongs to the open interval  ((\lambda (p+1) ) ^{1/p},+\infty)  (since$$ N(p,\lambda ,E,(\lambda (p+1) ) ^{\frac1p}) =N(p,\lambda ,E,0) =E^p>0). $$Also, the assertions {\bf (i) (ii) }and {\bf (iii)} follow readily from the variations of N(p,\lambda ,E,\cdot ) . Now, for any p>1 and \lambda >0, consider the real-valued function$$ (E,s) \longmapsto N_{\pm }(E,s) =E^p-p^{\prime }s(\frac{(\pm s) ^p}{p+1}-\lambda ) $$defined on \Omega_+=(0,+\infty ) \times ((\lambda (p+1) ) ^{1/p},+\infty )  (resp. \Omega _{-}=(0,E_{*}(p,\lambda ) ) \times ( -\lambda ^{1/p},0) ). One has N_{\pm }\in C^1(\Omega _{\pm })  and$$ \frac{\partial N_{\pm }}{\partial s}(E,s) =-p'f(\lambda ,s) =-p'(| s|^p-\lambda ) \quad \mbox{in }\Omega _{\pm }, $$hence$$ \mp \frac{\partial N_{\pm }}{\partial s}(E,s) >0 \quad\mbox{in }\Omega _{\pm } $$and one may observe that s_{\pm }(p,\lambda ,E)  belongs to the open interval ((\lambda (p+1) ) ^{1/p},+\infty )  (resp. (-\lambda ^{1/p},0) ) and satisfies (from its definition) $$\label{art7}N_{\pm }(E,s_{\pm }(p,\lambda ,E)) =0.$$ So, one can make use of the implicit function theorem to show that the function E\longmapsto s_{\pm }(p,\lambda ,E)  is C^1((0,+\infty ) ,{\mathbb R})  (resp. C^1((0,E_{*}(p,\lambda ) ) , {\mathbb R}) ) and to obtain the expression of \frac{\partial s_{\pm }}{ \partial E}(p,\lambda ,E)  given in {\bf (a)}. Hence, for any fixed p>1 and \lambda >0, the function defined in (0,+\infty )  (resp. (0,E_{*}(p,\lambda ) ) ) by E\longmapsto s_{\pm }(p,\lambda ,E)  is strictly increasing (resp. decreasing) and bounded from below by ( \lambda (p+1) ) ^{1/p} (resp. -\lambda ^{1/p} ) and from above by +\infty  (resp. by 0). Then, the limit \lim \limits_{E\rightarrow 0^+}s_{\pm }(p,\lambda ,E) =l_0^{\pm } exists as a real number and the limit \lim \limits_{E\rightarrow +\infty }s_+(p,\lambda ,E) =l_{+\infty } (resp. \lim \limits_{E\rightarrow E_{*}}s_{-}(p,\lambda ,E) =l_{*}) exists and belongs to ((\lambda (p+1) ) ^{1/p},+\infty ]  (resp. [-\lambda ^{1/p},0] ). Moreover$$ -\infty <-\lambda ^{1/p}\leq l_{*}1$and$\lambda >0$, the function $$(E,s) \longmapsto N_{\pm }(E,s)$$ is continuous in$[0,+\infty ) \times [(\lambda (p+1) ) ^{1/p},+\infty ) $(resp.$[ 0,E_{*}(p,\lambda ) ] \times (-\infty ,0] $) and the function$E\longmapsto s_{\pm }(p,\lambda ,E) $is continuous in$(0,+\infty ) $(resp.$( 0,E_{*}(p,\lambda ) ) $) and satisfies$($\ref{art7}$ )_{\pm }$. So, by passing to the limit in$($\ref{art7}$)_{\pm }$as$E$tends to$0^+$one gets $$0=\lim _{E\rightarrow 0^+}N_{\pm }(E,s_{\pm }(p,\lambda ,E) ) =N_{\pm }(0,l_0^{\pm }) .$$ Hence,$l_0^{\pm }$is a zero, belonging to$[(\lambda ( p+1) ) ^{1/p},+\infty ) $(resp.$[-\lambda ^{1/p},0] $), to the equation in the variable$s$: $$N_{\pm }(0,s) =0\mbox{.}$$ By resolving this equation in the indicated interval one gets :$l_0^+=((p+1) \lambda ) ^{1/p}$(resp.$l_0^-=0$). The assertion {\bf (b)} is proved. Assume that$l_{+\infty }<+\infty$. Then by passing to the limit in$($\ref{art7}$)_+$as$E$tends to$+\infty $one gets $$+\infty =p'l_{+\infty }(\frac{(l_{+\infty }) ^p}{ p+1}-\lambda ) <+\infty ,$$ which is impossible. So,$l_{+\infty }=+\infty $. To prove that$l_{*}=-\lambda ^{1/p}$, it suffices to pass to the limit in$($\ref{art7}$)_{-}$as$E$tends to$E_{*}(p,\lambda ) $to get $$N_{-}(E_{*}(p,\lambda ) ,l_{*}) =0$$ and to resolve this equation in$[-\lambda ^{1/p},0] $. (To this end, one may observe that the function$s\longmapsto N_{-}( E_{*}(p,\lambda ) ,s) $is strictly increasing in$ [-\lambda ^{1/p},0] $and $$N_{-}(E_{*}(p,\lambda ) ,-\lambda ^{1/p}) =0 \mbox{).}$$ Therefore, Lemma \ref{artlemma1} is proved. \quad$\diamondsuit$\medskip Now we are ready to compute$X_{\pm }(p,\lambda ,E) $as defined in Section \ref{sec3}, for any$p>1$,$\lambda >0$and$E\geq 0$. In fact,$X_+(p,\lambda ,E) =(0,s_+(p,\lambda ,E) ) $and $$X_{-}(p,\lambda ,E) =\left\{ \begin{array}{ll} (-\infty ,0) & \mbox{ if } E>E_{*}(p,\lambda )\\[3pt] (s_{-}(p,\lambda ,E) ,0) & \mbox{if } 0\leq E\leq E_{*}(p,\lambda )\,, \end{array} \right.$$ where$s_{\pm }(p,\lambda ,E) $is defined in Lemma \ref{artlemma1}. Then $$r_+(p,\lambda ,E) :=\sup X_+(p,\lambda,E) =s_+(p,\lambda ,E)$$ and $$r_{-}(p,\lambda ,E) :=\inf X_{-}(p,\lambda ,E) =\left\{ \begin{array}{ll} -\infty & \mbox{if } E>E_{*}(p,\lambda ) \\[3pt] s_{-}(p,\lambda ,E) & \mbox{if } 0\leq E\leq E_{*}(p,\lambda ) . \end{array} \right.$$ Recall that for any$E\geq 0$,$s_+(p,\lambda ,E) $belongs to$[(\lambda (p+1) ) ^{\frac 1p},+\infty ) $. Thus $$00\,.$$ Also recall that, for any$00,\forall E\geq 0 $and $$f(\lambda ,r_{-}(p,\lambda ,E) ) =( -r_{-}(p,\lambda ,E) ) ^p-\lambda <0\leftrightarrow E\in (0,E_{*}(p,\lambda ) ) ,$$ so that $$D_+ := \left\{ E\geq 0\mid 00\right\} = [0,+\infty ) .$$ and $$D_{-} := \left\{ E\geq 0\mid -\infty 0,\,\forall E\in \mbox{int}(D_{\pm }) \,. At present, we define, for any p>1,\lambda >0, and E\in D_{\pm }, the time map $$\label{art12}T_{\pm }(p,\lambda ,E) :=\pm \int_0^{r_{\pm }(p,\lambda ,E) }\left\{ E^p-p'F(\lambda ,\xi ) \right\} ^{-1/p}d\xi ,\;E\in D_{\pm },$$ and a simple change of variables shows that $$\label{art13}T_{\pm }(p,\lambda ,E) =\pm r_{\pm }( p,\lambda ,E) \int_0^1\left\{ E^p-p'F(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) \right\} ^{-1/p}d\xi \,.$$ Observe that from the definition of s_{\pm }(p,\lambda ,E) one has E^p-p'F(\lambda ,s_{\pm }(p,\lambda ,E)) =0, and so, from the definition of r_{\pm }(p,\lambda ,E) , one has E^p=p'F(\lambda ,r_{\pm }(p,\lambda ,E)) . So, (\ref{art13}) may be written as \begin{eqnarray} \label{art14} T_{\pm }(p,\lambda ,E) &=&\pm r_{\pm }(p,\lambda,E) (p') ^{-1/p} \times \\ &&\int_0^1\left\{ F( \lambda ,r_{\pm }(p,\lambda ,E) ) -F(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) \right\} ^{-1/p}d\xi \,.\nonumber \end{eqnarray} After some substitutions one has \begin{eqnarray} \label{art15} T_+(p,\lambda ,E) &=&(r_+(p,\lambda,E) ) ^{-1/p}(p') ^{-1/p} \times \\ &&\int_0^1\left\{ \frac{1-\xi ^{p+1}}{p+1}-\lambda \frac{1-\xi }{( r_+(p,\lambda ,E) ) ^p}\right\} ^{-1/p}d\xi ,\;E\in D_+ \nonumber \end{eqnarray} and \begin{eqnarray} \label{art16} T_{-}(p,\lambda ,E) &=&(-r_{-}(p,\lambda,E) ) ^{1-\frac 1p}(p') ^{-1/p}\times \\ && \int_0^1\left\{ \lambda (1-\xi ) -\frac{(-r_{-}(p,\lambda ,E) ) ^p}{p+1} (1-\xi ^{p+1}) \right\} ^{-1/p}d\xi ,\;E\in D_{-}\,. \nonumber \end{eqnarray} Also, we define for any E\in D=D_+\cap D_{-} and n\in {\mathbb N} the time maps: \begin{eqnarray} \label{art17} & T_{2n}(p,\lambda ,E) :=n(T_+( p,\lambda ,E) +T_{-}(p,\lambda ,E) ) ,E\in D,& \\ \label{art18} & T_{2n+1}^{\pm }(p,\lambda ,E) :=T_{2n}( p,\lambda ,E) +T_{\pm }(p,\lambda ,E) ,E\in D\,. & \end{eqnarray} The limits of these time maps are the aim of the following lemmas. \begin{lemma} \label{artlemma2} For any p>1 and \lambda >0, one has T_+(p,\lambda ,E_{*}(p,\lambda ) ) =\lambda ^{-1/p^2}\times J_+(p) , where$$ J_+(p) :=(p') ^{-1/p}( p+1) ^{1/p}\theta (p) \times \int_0^1\left\{ p-( \theta (p) \xi ) ^{p+1}+(p+1) \theta ( p) \xi \right\} ^{-1/p}d\xi $$and \theta (p) >(p+1) ^{1/p} is the unique positive zero of the equation $$\label{art19}\theta ^{p+1}-(p+1) \theta -p=0\,.$$ \end{lemma} \begin{lemma} \label{artlemma3}For any p>1 and \lambda >0, let$$ J(p) :=\frac 1p(p') ^{-1/p}( p+1) ^{\frac{p-1}{p^2}}\cdot \frac{\Gamma (\frac{p-1}{p^2} ) \Gamma (\frac{p-1}p) }{\Gamma (\frac{( p-1) (p+1) }{p^2}) } $$and$$ J_{-}(p) :=(p') ^{-1/p}( p+1) ^{1/p}\int_0^1\left\{ p-(p+1) \xi +\xi ^{p+1}\right\} ^{-1/p}d\xi . $$Then one has: $$\label{art20}J_{-}(p) <+\infty \leftrightarrow p>2,$$ {\bf (i)} \lim \limits_{E\rightarrow 0^+}T_+(p,\lambda ,E) =J(p) \lambda ^{-1/p^2}, \quad {\bf (ii)} \lim \limits_{E\rightarrow 0^+}T_{-}(p,\lambda ,E) =0,\\ {\bf (iii)} \lim \limits_{E\rightarrow +\infty }T_+( p,\lambda ,E) =0, \quad {\bf (iv)} \lim\limits_{E\rightarrow E_{*}}T_{-}(p,\lambda ,E) =J_{-}( p) \lambda ^{-1/p^2}. \end{lemma} \begin{lemma} \label{artlemma4} For any p>1 and \lambda >0, one has \begin{description} \item{\bf (a)} \lim \limits_{E\rightarrow 0^+}T_{2n}(p,\lambda ,E) =nJ(p) \lambda ^{-1/p^2} \item{\bf (b)} \lim \limits_{E\rightarrow 0^+}T_{2n+1}^+(p,\lambda ,E) =(n+1) J(p) \lambda ^{-1/p^2}, \item{\bf (c)} \lim \limits_{E\rightarrow 0^+}T_{2n+1}^-( p,\lambda ,E) =nJ(p) \lambda ^{-1/p^2}, \item{\bf (d)} \lim \limits_{E\rightarrow E_{*}}T_{2n}(p,\lambda ,E) =n(J_+(p) +J_{-}(p) ) \times \lambda ^{-1/p^2}, \item{\bf (e)} \lim \limits_{E\rightarrow E_{*}}T_{2n+1}^+( p,\lambda ,E) =((n+1) J_+(p) +nJ_{-}(p) ) \times \lambda ^{-1/p^2}, \item{\bf (f)} \lim \limits_{E\rightarrow E_{*}}T_{2n+1}^-( p,\lambda ,E) =(nJ_+(p) +(n+1) J_{-}(p) ) \times \lambda ^{- 1/p^2}. \end{description} \end{lemma} \paragraph{Proof of Lemma \ref{artlemma2}.} For any p>1, let us consider the function \Theta defined in (0,+\infty ) by \Theta (\theta ) :=\theta ^{p+1}-(p+1) \theta -p. A study of its variations implies that equation (\ref{art19}) admits a unique zero in (0,+\infty ) , denoted by \theta (p) , and this zero belongs to ((p+1) ^{1/p},+\infty ) (Note that \Theta ((p+1) ^{1/p}) =-p). Furthermore, recall (Lemma \ref{artlemma1}) that, for any \lambda >0 and E>0, r_+(p,\lambda ,E) is the unique positive solution of equation (\ref{art6}). In particular, if E=E_{*}( p,\lambda ) :=((\frac{pp'}{p+1}) \lambda ^{1+\frac 1p}) ^{1/p} then r_+(p,\lambda ,E_{*}(p,\lambda ) ) is the unique positive solution of the following equation in the variable s : $$\label{art21}s^{p+1}-\lambda (p+1) s-p\lambda ^{1+\frac 1p}=0.$$ Some easy computations show that \theta (p) \lambda ^{1/p} is also a positive solution of (\ref{art21}), and since (\ref{art21}) admits a unique positive solution (which is r_+(p,\lambda ,E_{*}( p,\lambda ) ) ) it follows that$$ r_+(p,\lambda ,E_{*}(p,\lambda ) ) =\theta (p) \lambda ^{1/p},\;\forall p>1,\forall \lambda >0. $$Now, from (\ref{art15}), some simple computations show that T_+(p,\lambda ,E_{*}(p,\lambda ) ) =\lambda^{-1/p^2} J_+(p)\, where$$ J_+(p) :=(p') ^{-1/p}( p+1) ^{1/p}\theta (p) \int_0^1(p-( \theta (p) \xi ) ^{p+1}+(p+1) \theta ( p) \xi ) ^{-1/p}d\xi \,. $$Therefore, Lemma \ref{artlemma2} is proved. \quad \diamondsuit\medskip \paragraph{Proof of Lemma \ref{artlemma3}.} In order to prove the first assertion we first claim that there exists \varepsilon _0>0 (sufficiently small) such that for any \xi \in (1-\varepsilon _0,1) ,$$ \frac{p(p+1) }4(1-\xi ) ^2\leq p-(p+1) \xi +\xi ^{p+1}\leq p(p+1) (1-\xi ) ^2. $$To proof this claim, for any x>0, let$$ h_x(\xi ) :=p-(p+1) \xi +\xi ^{p+1}-x(1-\xi ) ^2,\xi \in (0,1] \,. $$Simple computations lead to$$ \frac{dh_x}{d\xi }(\xi ) =2(1-\xi ) (x-(\frac{p+1}2) \frac{1-\xi ^p}{1-\xi }) ,\xi \in ( 0,1) \,. $$Using l'H\^opital's rule one gets$$ \lim _{\xi \rightarrow 1^-}(x-(\frac{p+1}2) \frac{1-\xi ^p}{1-\xi }) =(x-\frac{p(p+1) }2) . $$So, because of continuity properties, there exists \varepsilon _1>0 (resp. \varepsilon _2>0) sufficiently small such that$$ \frac{dh_{p(p+1) }}{d\xi }(\xi ) >0,\forall \xi \in (1-\varepsilon _1,1) $$(resp. \frac{dh_{p(p+1) /4}}{d\xi }(\xi ) <0,\forall \xi \in (1-\varepsilon _2,1)). Notice that h_x(1) =0,\forall x>0, so that$$ h_{p(p+1) }(\xi ) <0,\forall \xi \in ( 1-\varepsilon _0,1) $$(resp. h_{p(p+1) /4}(\xi ) >0,\forall \xi \in ( 1-\varepsilon _0,1) ) where \varepsilon _0=\min (\varepsilon _1,\varepsilon _2) . Then the claim is proved. \medskip With this claim we are able to prove easily the first assertion of this lemma. In fact, the integral which appears in the definition of J_{-}(p) may be written as$$ \int_0^{1-\varepsilon _0}(p-(p+1) \xi +\xi ^{p+1}) ^{-1/p}d\xi +\int_{1-\varepsilon _0}^1(p-(p+1) \xi +\xi ^{p+1}) ^{-1/p}d\xi \,. $$The first integral converges because the integrand function is continuous on the compact interval [0,1-\varepsilon _0] . For the second integral, one has from the claim$$ A(p) \int_{1-\varepsilon _0}^1\frac{d\xi }{(1-\xi ) ^{\frac 2p}}\leq \int_{1-\varepsilon _0}^1(p-(p+1) \xi +\xi ^{p+1}) ^{-1/p}d\xi \leq B(p) \int_{1-\varepsilon _0}^1\frac{d\xi }{(1-\xi ) ^{\frac 2p}} $$where A(p) =(p(p+1) ) ^{-1/p} and B(p) =(p(p+1) /4) ^{-1/p}. So, from the well-known fact$$ \int_{1-\varepsilon _0}^1\frac{d\xi }{(1-\xi ) ^{\frac 2p}} <+\infty \leftrightarrow p>2 $$the first assertion follows. \paragraph{Proof of (i).} One has from (\ref{art13})$$ T_+(p,\lambda ,E) =r_+(p,\lambda ,E) \int_0^1(E^p-p'F(\lambda ,r_+(p,\lambda ,E) \xi ) ) ^{-1/p}d\xi . $$Using (\ref{art8}) one gets: \begin{eqnarray*} \lim \limits_{E\rightarrow 0^+}E^p-p'F(\lambda ,r_+(p,\lambda ,E) \xi ) & = & -p'F( \lambda ,((p+1) \lambda ) ^{1/p}\xi )\\ & = & p'((p+1) \lambda ) ^{1/p}\lambda \xi (1-\xi ^p) , \end{eqnarray*} so, some simple computations yield$$ \lim \limits_{E\rightarrow 0^+}T_+(p,\lambda ,E) =( p+1) ^{\frac{p-1}{p^2}}(p') ^{-1/p}\lambda ^{-1/p^2}\int_0^1\xi ^{-1/p}(1-\xi ^p) ^{-1/p}d\xi \,. $$To compute this integral, one can make use of the change of variables x=\xi ^p and then make use of the relationship between the Euler beta and gamma functions, see for instance \cite[Chap. VII, no 90, example 2, pp. 595-596] {Lavrentiev-Chabat}, to obtain:$$ \int_0^1\xi ^{-1/p}(1-\xi ^p) ^{-1/p}d\xi =\frac 1p \frac{\Gamma (\frac{p-1}{p^2}) \Gamma (\frac{p-1}p) }{\Gamma (\frac{(p-1) (p+1) }{p^2}) }. $$This completes the proof of {\bf (i)}. \paragraph{ Proof of (ii).} Consider the expression for T_{-}( p,\lambda ,E) given by (\ref{art16}). From (\ref{art8}) one gets \begin{eqnarray*} \lefteqn{ \lim \limits_{E\rightarrow 0^+}\int_0^1(\lambda (1-\xi ) -\frac{(-r_{-}(p,\lambda ,E) ) ^p}{p+1} (1-\xi ^{p+1}) ) ^{-1/p}d\xi }\\ & = & \int_0^1( \lambda (1-\xi ) ) ^{-1/p}d\xi = \lambda ^{-1/p} p'\,. \end{eqnarray*} So, from (\ref{art8}) and (\ref{art16}) one gets$$ \lim _{E\rightarrow 0^+}T_{-}(p,\lambda ,E) =( p') ^{-1/p}\times (0) ^{1-\frac 1p}\times \lambda ^{-1/p}\times p'=0. $$This completes the proof of {\bf (ii)}. \paragraph{Proof of (iii).} Consider the expression for T_+(p,\lambda ,E) given by (\ref{art15}). From (\ref{art8}) one gets$$ \lim _{E\rightarrow +\infty }\int_0^1(\frac{1-\xi ^{p+1}}{p+1}-\lambda \frac{1-\xi }{r_+^p(p,\lambda ,E) }) ^{-\frac 1p}d\xi =\frac 1{(p+1) ^{-1/p}}\int_0^1(1-\xi ^{p+1}) ^{-1/p}d\xi , $$and this integral may be computed by making use of the change of variables x=\xi ^{p+1} to get$$ \int_0^1(1-\xi ^{p+1}) ^{-1/p}d\xi =\frac 1{p+1}\frac{ \Gamma (\frac 1{p+1}) \Gamma (\frac{p-1}p) }{\Gamma (\frac{p-1}{p(p+1) }) }\,. $$So, from (\ref{art8}) and (\ref{art15}) one gets$$ \lim _{E\rightarrow +\infty }T_+(p,\lambda ,E) =( p') ^{-1/p}\times 0\times \frac 1{(p+1) ^{1-\frac 1p}}\times \frac{\Gamma (\frac 1{p+1}) \Gamma ( \frac{p-1}p) }{\Gamma (\frac{p-1}{p(p+1) }) } =0\,. $$This completes the proof of {\bf (iii)}. \paragraph{Proof of (iv).} Consider the expression for T_{-}( p,\lambda ,E) given by (\ref{art14}). One has \begin{eqnarray*} \lim \limits_{E\rightarrow E_{*}}(F(\lambda ,r_{-}( p,\lambda ,E) ) -F(\lambda ,r_{-}(p,\lambda ,E) \xi ) ) & = & \lim \limits_{x\rightarrow -\lambda ^{1/p}}(F(\lambda ,x) -F(\lambda ,x\xi ) ) \\ & = & \frac{\lambda ^{1+\frac 1p}}{p+1}(p-(p+1) \xi +\xi ^{p+1}) \end{eqnarray*} so that$$ \lim _{E\rightarrow E_{*}}T_{-}(p,\lambda ,E) =\lambda ^{1/p}\times (p') ^{-1/p}\times (\frac{ \lambda ^{1+\frac 1p}}{p+1}) ^{-1/p}\times \int_0^1( p-(p+1) \xi +\xi ^{p+1}) ^{-1/p}d\xi $$which is the same as$$ \lim _{E\rightarrow E_{*}}T_{-}(p,\lambda ,E) =\lambda ^{-1/p^2}\times (p') ^{-1/p}\times ( p+1) ^{1/p}\times \int_0^1(p-(p+1) \xi +\xi ^{p+1}) ^{-1/p}d\xi . $$This completes the proof of {\bf (iv) }and of Lemma \ref{artlemma3}. \quad \diamondsuit \paragraph{Proof of Lemma \ref{artlemma4}.} This proof is an immediate consequence of the two preceding lemmas and the definitions (\ref{art17}) and (\ref{art18}) of the time maps T_{2n},T_{2n+1}^{\pm }. \begin{lemma} \label{artlemma5}For any p>1,\lambda >0, one has:$$ \pm \frac{\partial T_{\pm }}{\partial E}(p,\lambda ,E) <0,\;\forall E\in D_{\pm }. $$\end{lemma} \paragraph{Proof.} From (\ref{art14}) one has \begin{eqnarray*} \lefteqn{ \pm \frac{\partial T_{\pm }}{\partial E}(p,\lambda ,E) }\\ &=&(p') ^{-1/p} \bigg\{ \frac{\partial r_{\pm }}{\partial E}(p,\lambda ,E) \int_0^1(F(\lambda ,r_{\pm }(p,\lambda ,E) ) -F(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) ) ^{-1/p}d\xi \\ &&\quad + r_{\pm }(p,\lambda ,E) \int_0^1\frac \partial {\partial E}(F(\lambda ,r_{\pm }(p,\lambda ,E) ) -F(\lambda ,r_{\pm }( p,\lambda ,E) \xi ) ) ^{-1/p}d\xi \bigg\} \\ &=&(p') ^{-1/p} \bigg\{ \frac{\partial r_{\pm }}{\partial E}(p,\lambda ,E) \int_0^1\frac{(F(\lambda ,r_{\pm }(p,\lambda ,E) ) -F(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) ) }{(F(\lambda ,r_{\pm }(p,\lambda,E) ) -F(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) ) ^{1+\frac 1p}}d\xi \\ &&-\frac 1pr_{\pm }(p,\lambda ,E) \frac{\partial r_{\pm }}{\partial E}(p,\lambda ,E) \times \\ % && \int_0^1 \frac{f(\lambda ,r_{\pm }(p,\lambda ,E) ) -f(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) \xi }{ (F(\lambda ,r_{\pm }(p,\lambda ,E) ) -F(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) ) ^{1+\frac 1p}}d\xi \bigg\} \end{eqnarray*} so that \begin{eqnarray}\label{art22} \pm \frac{\partial T_{\pm }}{\partial E}(p,\lambda ,E) &=&\frac 1p(p') ^{-1/p}(\frac{\pm \partial r_{\pm }}{ \partial E}(p,\lambda ,E) ) \times \\ && \int_0^1\frac{\pm (H(\lambda ,r_{\pm }(p,\lambda ,E) ) -H(\lambda ,r_{\pm }(p,\lambda ,E) \xi )) }{(F(\lambda ,r_{\pm }(p,\lambda ,E) ) -F(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) ) ^{1+\frac 1p}}d\xi \nonumber \end{eqnarray} where H(\lambda ,x) =pF(\lambda ,x) -xf(\lambda ,x) =\frac{-1}{p+1}| x|^px-( p-1) \lambda x. Because the function x\mapsto H(\lambda ,x) is decreasing for each fixed \lambda >0 (in fact, \frac{\partial H}{\partial x}(\lambda ,x) <0), it follows that$$ \pm (H(\lambda ,r_{\pm }(p,\lambda ,E) ) -H(\lambda ,r_{\pm }(p,\lambda ,E) \xi ) ) <0,\forall \lambda >0,\forall \xi \in (0,1) . $$Hence, the integral in (\ref{art22}) is negative. So, because of (\ref{art11}), the proof of Lemma~\ref{artlemma5} is achieved. \quad \diamondsuit \paragraph{Completion of the proof of Theorem \protect{\ref{thm2}}.} The proof is carried out by making use of the quadrature method (Theorem \ref{artquad}). We have to resolve equations of the type T(E) =\frac 12, where T designates, in each case, the appropriate time map. \paragraph{Solution in B_n^+.} Recall that r_+(p,\lambda ,0) =((p+1) \lambda ) ^{1/p}. Furthermore$$ T_+(p,\lambda ,0) =\int_0^{r_+(p,\lambda ,0) }(-p'F(p,\lambda ,\xi ) )^{-1/p}d\xi =J(p) \lambda ^{-1/p^2}\,, $$where J(p) is defined in Lemma \ref{artlemma3}. Then problem (\ref{AD}) admits a solution in B_n^+ if and only if nJ( p) \lambda ^{-1/p^2}=(1/2) , that is, if and only if \lambda =(2nJ(p) ) ^{p^2}. \paragraph{Solution in B_n^-.} Since 0\notin D_{-}=(0,E_{*}(p,\lambda ) ) , problem (\ref{AD}) admits no solution in \bigcup\limits_{n\geq 1}B_n^-. \paragraph{Solution in A_1^+.} Recall that for any p>1 and \lambda >0 the function E\mapsto T_+(p,\lambda ,E) is defined in [0,+\infty ) , is strictly decreasing (Lemma \ref{artlemma5}), and by Lemma \ref{artlemma3},$$ \lim _{E\rightarrow 0^+}T_+(p,\lambda ,E) =J( p) \lambda ^{-1/p^2},\quad \lim _{E\rightarrow +\infty }T_+(p,\lambda ,E) =0\,. $$Then, the equation T_+(p,\lambda ,E) =(1/2) in the variable E\in (0,+\infty ) admits a solution in [0,+\infty ) if and only if J( p) \lambda ^{-1/p^2}>1/2, that is, if and only if \lambda <(2J(p) ) ^{p^2}, and in this case, the solution is unique since the function T_+(p,\lambda ,\cdot ) is strictly decreasing. \paragraph{Solution in A_1^-.} {\bf Case 10, the function E\mapsto T_{-}(p,\lambda ,E) is defined in D_{-}=(0,E_{*}(p,\lambda ) ) , is strictly increasing (Lemma \ref{artlemma5}), and$$ \lim _{E\rightarrow 0^+}T_{-}(p,\lambda ,E) =0,\quad \lim _{E\rightarrow E_{*}}T_{-}(p,\lambda,E) =+\infty $$(Lemma \ref{artlemma3}, {\bf (ii)} and assertion (\ref{art20})). So, the equation T_{-}(p,\lambda ,E) =(1/2) in the variable E\in D_{-} admits a unique solution in D_{-} for any \lambda >0. {\bf Case }p>2.\ In this case, for each \lambda >0, the function E\mapsto T_{-}(p,\lambda ,E) is defined in D_{-}=( 0,E_{*}(p,\lambda ) ) , is strictly increasing (Lemma \ref{artlemma5}), and$$ \lim _{E\rightarrow 0^+}T_{-}(p,\lambda ,E) =0,\quad \lim _{E\rightarrow E_{*}}T_{-}(p,\lambda ,E) =J_{-}(p) \lambda ^{-1/p^2}<+\infty $$(Lemma \ref{artlemma3}). So, the equation T_{-}(p,\lambda ,E) =(1/2) in the variable E\in D_{-} admits a solution in D_{-} if and only if (1/2) 0, the function E\mapsto T_{2n}(p,\lambda ,E) is defined in D=(0,E_{*}(p,\lambda ) ) , and$$ \lim _{E\rightarrow 0^+}T_{2n}(p,\lambda ,E) =nJ( p) \lambda ^{-1/p^2},\quad \lim _{E\rightarrow E_{*}}T_{2n}(p,\lambda ,E) =+\infty $$(Lemma \ref{artlemma4} and Lemma \ref{artlemma3}, assertion (\ref{art20})). So, the equation T_{2n}(p,\lambda,E) =(1/2) in the variable E\in D admits a solution in D provided that ( 1/2) >nJ(p) \lambda ^{-1/p^2}, that is, provided that \lambda >(2nJ(p) ) ^{p^2}. {\bf Case }p>2. In this case, for each \lambda >0, the function E\mapsto T_{2n}(p,\lambda ,E) is defined in D=( 0,E_{*}(p,\lambda ) ) , and \begin{eqnarray*} &\lim \limits_{E\rightarrow 0^+}T_{2n}(p,\lambda ,E) =nJ(p) \lambda ^{-1/p^2}, &\\ &\lim \limits_{E\rightarrow E_{*}}T_{2n}(p,\lambda ,E) =n\lambda ^{-1/p^2}(J_{-}(p) +J_+(p)) <+\infty& \end{eqnarray*} (Lemma \ref{artlemma4} and Lemma \ref{artlemma3}, assertion (\ref{art20})). So, the equation T_{2n}(p,\lambda,E) =(1/2) in the variable E\in D admits a solution in D provided that$$ n\lambda ^{-1/p^2}\inf (J(p) ,J_{-}( p) +J_+(p) ) <\frac 120$, the function $E\mapsto T_{2n+1}^+(p,\lambda ,E)$ is defined in $D=(0,E_{*}(p,\lambda ) )$, and $$\lim _{E\rightarrow 0^+}T_{2n+1}^+(p,\lambda ,E) =( n+1) J(p) \lambda ^{-1/p^2},\quad \lim _{E\rightarrow E_{*}}T_{2n+1}^+(p,\lambda ,E) =+\infty$$ (Lemma \ref{artlemma4} and Lemma \ref{artlemma3}, assertion (\ref{art20})). So, the equation $T_{2n+1}^+(p,\lambda,E) =(1/2)$ in the variable $E\in D$ admits a solution in $D$ provided that \\ $(n+1) J(p) \lambda ^{-1/p^2}<(1/2)$, that is, provided that $\lambda >(2(n+1) J(p)) ^{p^2}$. {\bf Case }$p>2.$ In this case, for each $\lambda >0$, the function $E\mapsto T_{2n+1}^+(p,\lambda ,E)$ is defined in $D=(0,E_{*}(p,\lambda ) )$, and \begin{eqnarray*} &\lim \limits_{E\rightarrow 0^+}T_{2n+1}^+(p,\lambda ,E) =(n+1) J(p) \lambda ^{-1/p^2}, &\\ &\lim \limits_{E\rightarrow E_{*}}T_{2n+1}^+(p,\lambda ,E) =\lambda ^{-1/p^2}((n+1) J_+(p)+nJ_{-}(p) ) <+\infty & \end{eqnarray*} (Lemma \ref{artlemma4} and Lemma \ref{artlemma3}, assertion (\ref{art20})). So, the equation $T_{2n+1}^+(p,\lambda,E) =(1/2)$ in the variable $E\in D$ admits a solution in $D$ provided that \begin{eqnarray*} \lefteqn{ \lambda ^{-1/p^2}\inf ((n+1) J(p),(n+1) J_+(p) +nJ_{-}(p) ) }\\ <&\frac 12&< \lambda ^{-\frac 1{p^2}}\sup ((n+1) J(p) ,(n+1) J_+(p) +nJ_{-}(p) ) , \end{eqnarray*} that is, provided that \begin{eqnarray*} \lefteqn{ \left\{ 2\inf ((n+1) J(p) ,(n+1) J_+(p) +nJ_{-}(p) ) \right\} ^{p^2} }\\ <&\lambda& < \left\{ 2\sup((n+1) J(p) ,(n+1) J_+( p) +nJ_{-}(p) ) \right\} ^{p^2}\,. \end{eqnarray*} \paragraph{\bf Solution in $A_{2n+1}^-$.} {\bf Case }$10$, the function $E\mapsto T_{2n+1}^-(p,\lambda ,E)$ is defined in $D=(0,E_{*}(p,\lambda ) )$, and $$\lim _{E\rightarrow 0^+}T_{2n+1}^-(p,\lambda ,E) =nJ(p) \lambda ^{-1/p^2},\quad \lim_{E\rightarrow E_{*}}T_{2n+1}^-(p,\lambda ,E) =+\infty$$ (Lemma \ref{artlemma4} and Lemma \ref{artlemma3}, assertion (\ref{art20})). So, the equation $T_{2n+1}^-(p,\lambda,E) =(1/2)$ in the variable $E\in D$ admits a solution in $D$ provided that $nJ(p) \lambda ^{-1/p^2}<(1/2)$, that is, provided that $\lambda >(2nJ(p) ) ^{p^2}$. {\bf Case }$p>2.$ In this case, for each $\lambda >0$, the function $E\mapsto T_{2n+1}^-(p,\lambda ,E)$ is defined in $D=(0,E_{*}(p,\lambda ) )$, and \begin{eqnarray*} &\lim \limits_{E\rightarrow 0^+}T_{2n+1}^-(p,\lambda ,E) =nJ(p) \lambda ^{-1/p^2}, &\\ &\lim \limits_{E\rightarrow E_{*}}T_{2n+1}^-(p,\lambda ,E) =\lambda ^{-1/p^2}(nJ_+(p) +(n+1)J_{-}(p) ) <+\infty & \end{eqnarray*} (Lemma \ref{artlemma4} and Lemma \ref{artlemma3}, assertion (\ref{art20})). So, the equation $T_{2n+1}^-(p,\lambda,E) =(1/2)$ in the variable $E\in D$ admits a solution in $D$ provided that \begin{eqnarray*} \lefteqn{ \lambda ^{-1/p^2}\inf (nJ(p) ,nJ_+(p) +(n+1) J_{-}(p) ) }\\ <&\frac 12&< \lambda ^{-1/p^2}\sup (nJ(p) ,nJ_+(p) +(n+1) J_{-}(p) ) , \end{eqnarray*} that is, provided that \begin{eqnarray*} \lefteqn{ \left\{ 2\inf ((nJ(p) ,nJ_+(p)+(n+1) J_{-}(p) ) ) \right\}^{p^2} }\\ <&\lambda& <\left\{ 2\sup((nJ(p) ,nJ_+(p) +(n+1)J_{-}(p) ) ) \right\} ^{p^2}\,. \end{eqnarray*} Then the proof of Theorem \ref{thm2} is complete. \paragraph{Remark.} Theorem \ref{thm2} shows that for $12$) solutions to ( \ref{AD}) with $k\geq 1$ interior nodes exist for all $\lambda$ belonging to an interval unbounded from above (resp. a bounded interval). Hence, for $12$, and these changes in the behavior of the solution set as $p$ varies depend strongly on the nonlinearity of the problem. \begin{thebibliography}{99} \bibitem{AddouAmmarKhodja} {\sc Addou I., \& F. 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Appl. {\bf 190} (1995), pp. 611-623. \end{thebibliography} \noindent {\sc Idris Addou } \& {\sc Abdelhamid Benmeza\"\i } \\ 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}