\documentclass[twoside]{article} \usepackage{amssymb} % used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Multiplicity of solutions for boundary-value problems\hfil EJDE--1999/21} {EJDE--1999/21\hfil Idris Addou \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 1999}(1999), No.~21, pp. 1--27. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Multiplicity of solutions for quasilinear elliptic boundary-value problems \thanks{ {\em 1991 Mathematics Subject Classifications:} 34B15. \hfil\break\indent {\em Key words and phrases:} Multiple solutions, elliptic boundary-value problems. \hfil\break\indent \copyright 1999 Southwest Texas State University and University of North Texas. \hfil\break\indent Submitted January 21, 1999. Published June 16, 1999.} } \date{} % \author{Idris Addou } \maketitle \begin{abstract} This paper is concerned with the existence of multiple solutions to the boundary-value problem $$-(\varphi_p(u') ) '=\lambda \varphi_q(u) +f(u)\quad\mbox{in } (0,1)\,,\quad u(0) =u(1) =0\,,$$ where $p,q>1$, $\varphi_x(y) =|y|^{x-2}y$, $\lambda$ is a real parameter, and $f$ is a function which may be sublinear, superlinear, or asymmetric. We use the time map method for showing the existence of solutions. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} \label{sec1} We study the existence and multiplicity of solutions for the boundary-value problem \begin{eqnarray} \label{p1} &-(\varphi_p(u') ) '=\lambda \varphi_q(u) +f(u) \quad\mbox{in }(0,1)&\\ &u(0) =u(1) =0\,,& \nonumber \end{eqnarray} where $p,q>1$, $\varphi_x(y) =| y| ^{x-2}y$, $\lambda \in {\mathbb R}$, and $f$ is a continuous function such that $$a_\pm =\lim_{s\to \pm \infty }f(s) /\varphi_q(s) \quad\mbox{and}\quad a_0=\lim_{s\to 0}f(s)/\varphi_q(s)$$ exist as real numbers. Also, we assume that $m_\pm :=\inf_{\pm s\geq 0}f(s) /\varphi_q(s)$ exists in ${\mathbb R}$, and define $$m:=\inf_{u\in {\mathbb R}}f(s) /\varphi_q(s) =\min (m_{+},m_{-}) \quad\mbox{and}\quad m_k^{\pm}= \cases{ m_\pm & if k=1 \cr m & if k\geq 2.}$$ We shall consider the following three cases: Superlinear case, $a_0<\min (a_{-},a_{+})$; Sublinear case, $a_0>\max (a_{-},a_{+})$; and Asymmetric case, $a_{-}1$, we mention a recent paper by J. Wang \cite{W} where positive solutions are studied. Now we define some sets that will be used in the statement of the main results. For $k\geq 1$, let $$S_k^+=\left\{ \begin{array}{c} u\in C^1([0,1]) :u \mbox{ admits exactly }( k-1) \mbox{ zeros in }(0,1) , \\ \mbox{all are simple, }u(0) =u(1) =0\mbox{ and }u'(0) >0 \end{array} \right\}\,,$$ then $S_k^-=-S_k^+$ and $S_k=S_k^+\cup S_k^-$. If $u\in C([0,1])$ is a real-valued function vanishing at $x_1$ and $x_2$ and not between them, (with $x_11$, $\lambda_k=k^p\lambda_1$ and $$\lambda_1=(p-1) (2\int_0^1(1-t^p) ^{-1/p}dt) ^p=(p-1) (\frac{2\pi }{p\sin (\pi /p) }) ^p.$$ For fixed real constants $a_{-}$ and $a_{+}$, consider the boundary value problem \begin{eqnarray}\label{p3} &-(\varphi_p(u') ) ' = \lambda \varphi_p(u) +a_{+}\varphi_p(u^+) -a_{-}\varphi _p(u^-) ,\mbox{in }(0,1) & \\ &u(0) = u(1) =0\,,& \nonumber \end{eqnarray} where $\lambda$ is a real parameter. If $\lambda$ is such that problem (\ref{p3}) admits a nontrivial solution $u_\lambda$, then $\lambda$ is called a half-eigenvalue of (\ref{p3}). In the particular case where $p=2$, this definition goes back to Berestycki \cite{Berestycki}. For any integer $k\geq 1$, let $$b_k^\pm =\left\{ \begin{array}{ll} a_\pm & \mbox{if }k=1 \\ \min (a_{-},a_{+}) & \mbox{if }k\geq 2 \end{array} \right. \quad\mbox{and}\quad c_k^\pm =\left\{ \begin{array}{ll} a_\pm & \mbox{if }k=1 \\ \max (a_{-},a_{+}) & \mbox{if }k\geq 2\,. \end{array} \right.$$ \begin{proposition} \label{prop1} For fixed real constants $a_{-}$ and $a_{+}$, the set of half-eigenvalues of problem (\ref{p3}) consists of two increasing sequences $(\lambda_k^+)_{k\geq 1}$ and $(\lambda_k^-) _{k\geq 1}$, satisfying $h_k^\pm (\lambda_k^\pm ) =1$, for all $k\geq 1$, where $$\displaylines{ h_{2n}^\pm (\lambda ) :=\frac{\lambda_n^{1/p}}{(a_{\pm }+\lambda ) ^{1/p}}+\frac{\lambda_n^{1/p}}{(a_{\mp }+\lambda ) ^{1/p}},\mbox{ for all }\lambda >-b_{2n}^\pm ,\ n\geq 1, \cr h_{2n+1}^\pm (\lambda ) :=\frac{\lambda_{n+1}^{1/p}}{( a_\pm +\lambda ) ^{1/p}}+\frac{\lambda_n^{1/p}}{(a_{\mp }+\lambda ) ^{1/p}},\mbox{ for all }\lambda >-b_{2n+1}^\pm ,\ n\geq0\,, }$$ with the convention $\lambda_0=0$. Moreover, if $a_{\mp }c_k^\pm$ and $\max (-m_k^\pm ,\lambda _k-a_0) <\lambda_k^\pm$, problem (\ref{p1}) admits at least a solution in $A_k^\pm$ for all $\lambda$ satisfying $\max (-m_k^\pm ,\lambda_k-a_0) <\lambda <\lambda_k^\pm$. \end{enumerate} \end{theorem} The next result deals with the asymmetric case. For any integer $n\geq 1$, let $$a_{2n+1}^\pm =\frac{na_{\mp }+(n+1) a_\pm }{2n+1}\quad \mbox{and}\quad a_{2n}^\pm =\frac{na_{-}+na_{+}}{2n}=\frac{a_{-}+a_{+}}2\,.$$ \begin{theorem} \label{thm2} Assume that $q=p$ and $a_{-}-m_k^\pm$. In particular, (\ref{p1}) admits infinitely many solutions for all $\lambda >-m$. \end{theorem} The paper is organized as follows. The next section is dedicated to the proof of Proposition \ref{prop1}. In Section \ref{sec3}, we present the method used for proving the main results of this paper. In Section \ref{sec4}, we present some preliminary lemmas. In Section \ref{sec5}, we prove the main results, present some remarks. The paper ends with an appendix which contains a brief historical overview on time maps. \section{Proof of Proposition \protect{\ref{prop1}}} \label{sec2} Notice that for all $u\in {\mathbb R},\varphi _p(u) =\varphi_p(u^+) -\varphi_p(u^-)$; so, problem (\ref{p3}) may be written as \begin{eqnarray} \label{p4} &-(\varphi_p(u') ) ' = \alpha _{+}\varphi_p(u^+) -\alpha_{-}\varphi_p(u^-),\quad \mbox{in }(0,1)\,,& \\ &u(0) = u(1) =0\,,& \nonumber \end{eqnarray} with $\alpha_\pm =\lambda +a_\pm$. The set of $(\alpha _{+},\alpha_{-}) \in {\mathbb R}^2$ such that problem (\ref {p4}) has a nontrivial solution is known as the Fu\v cik spectrum for the operator $(\varphi_p(u') ) '$ under Dirichlet boundary conditions. See for instance, Dr\'abek \cite {Drabek1}, Boccardo et al. \cite{Boccardo}, Huang and Metzen \cite{Huang}. We denote this spectrum by $\sum_p$. One has $\sum_p=\bigcup_{k\geq 0}C_k^\pm$, where \begin{eqnarray*} &C_0^\pm =\left\{ (\alpha_{+},\alpha_{-}) \in {\mathbb R}^2:\alpha_\pm =\lambda_1\right\} ,& \\ &C_{2n}^\pm =\left\{ (\alpha_{+},\alpha_{-}) \in ( {\mathbb R}^+) ^2:(\lambda_n^{1/p}/\alpha_{\pm }^{1/p}) +(\lambda_n^{1/p}/\alpha_{\mp }^{1/p}) =1\right\} ,n\geq 1,& \\ &C_{2n+1}^\pm =\left\{ (\alpha_{+},\alpha_{-}) \in ( {\mathbb R}^+) ^2:(\lambda_{n+1}^{1/p}/\alpha_{\pm }^{1/p}) +(\lambda_n^{1/p}/\alpha_{\mp }^{1/p}) =1\right\} ,n\geq 0\,.& \end{eqnarray*} So, $\lambda$ is a half-eigenvalue of (\ref{p3}) if and only if $( \lambda +a_{+},\lambda +a_{-}) \in \sum_p$, that is, $h_k^{\pm }(\lambda ) =1,$ for some $k\geq 1$. Since the function $\lambda \mapsto h_k^\pm (\lambda ) ,k\geq 1$, is strictly decreasing on $(-b_k^\pm ,+\infty )$ and its limits at $-b_k^\pm$ and $+\infty$ are $+\infty$ and $0$ respectively, it follows that the equation $h_k^\pm (\lambda ) =1$ admits a unique solution $\lambda =\lambda_k^\pm$. It is easy to check that for $a_{-}-a_{-}$ one has $$h_{2n-1}^+(\lambda ) -a_{-}, k\geq 1,$$ \frac{\lambda_k^{1/p}}{(a_{+}+\lambda ) ^{1/p}}<\frac{\lambda _k^{1/p}}{(a_{-}+\lambda ) ^{1/p}},\quad\mbox{so}\quad\frac{\lambda _{2k}^{1/p}}{(a_{+}+\lambda ) ^{1/p}}0$and$\kappa =+,-$, let $$X_\kappa (E) =\left\{ s\in {\mathbb R}:\kappa s\geq 0\quad\mbox{and}\quad E^p-p'G(\xi ) >0,\,\forall \,\xi ,0\leq \kappa \xi <\kappa s\right\}$$ and $$r_\kappa (E) =\cases{ 0 & if X_\kappa (E) =\emptyset \cr \kappa \sup (\kappa X_\kappa (E) ) & otherwise. }$$ Note that$r_\kappa $may be infinite. We shall also make use of the following sets, $$D_\kappa =\left\{ E>0:0<| r_\kappa (E) | <+\infty \mbox{ and }\kappa g(r_\kappa (E) ) >0\right\}$$ and$D=D_{+}\cap D_{-}$. Also, let$D_k^\kappa :=D$if$k\geq 2$, and$D_1^\kappa :=D_\kappa $. Define the following time-maps, \begin{eqnarray*} &T_\kappa (E) = \kappa \int_0^{r_\kappa (E) }(E^p-p'G(t) ) ^{-1/p}dt, \quad E\in D_\kappa \,,&\\ &T_{2n}^\kappa (E) = n(T_{+}(E) +T_{-}(E) ) , \quad n\in {\mathbb N},\ E\in D\,,&\\ &T_{2n+1}^\kappa (E) = T_{2n}^\kappa (E) +T_\kappa (E) , \quad n\in {\mathbb N}, \ E\in D. \end{eqnarray*} \begin{theorem}[Quadrature method] \label{quad} Assume that (\ref{meth2}) holds. Let$E>0$,$k\in {\mathbb N}^{*}$,$\kappa =+,-$. If$E\in D_k^\kappa $and$T_k^\kappa ( E) =1/2$, problem (\ref{meth3}) admits at least a solution$u_k^\kappa \in A_k^\kappa $satisfying$(u_k^\kappa ) '( 0) =\kappa E$, and this solution is unique. \end{theorem} This theorem is well-known, but we did not find a convenient reference to the precise statement used later. The paper by Guedda and Veron \cite{GueddaVeron}, seems to be the first one dealing with time maps approach when the differential operator is the one dimensional p-Laplacian. An easy adaptation of the ideas contained in \cite{GueddaVeron} and in the paper by Del Pino and Manasevich \cite{DelpinoMana1}, allows one to prove Theorem \ref{quad}. Also, we mention the papers by Manasevich and Zanolin \cite{ManaZanolin} and Manasevich et al. \cite{ManasevichetAl}, where one can find time maps used when the differential operator is the one dimensional p-Laplacian. Notice that time maps were also used when the differential operator generalizes the p-Laplacian; see for example Garcia-Huidobro et al. \cite{GarciaetAl1}, \cite{GarciaetAl2}, \cite {GarciaetAl3}, \cite{GarciaetAl4}, \cite{GarciaetAl5}, Garcia-Huidobro and Ubilla \cite{GarciaUbilla}, Huang and Metzen \cite{Huang} and Ubilla \cite {Ubilla}. A brief historical overview of time maps is given in the appendix. For the sake of completeness, we dedicate the rest of this section to the proof of Theorem \ref{quad} for the case where$k=1$and$\kappa =+$. The adaptation of the other cases may be handled similarly. We shall say that$u$is a solution of (\ref{meth3}) if$u$and$\varphi _p(u') $belong to$C^1([0,1]) $and$u$satisfies $$-(\varphi_p(u'(x) ) ) ^{\prime }=g(u(x) ) \ \forall x\in (0,1)\,,\quad\mbox{with}\quad u(0) =u(1) =0\,.$$ Let us assign to each function$u\in C^1([0,1]) $the set $$Z(u') =\left\{ x\in [0,1]:u^{\prime }(x) =0\right\} .$$ \begin{lemma} \label{Lem1}Assume that (\ref{meth2}) holds and$u$is a solution of (\ref {meth3}). Then$u\in C^2([0,1]) $if$12$. \end{lemma} \paragraph{Proof.} The identity$t=\varphi_{p'}o\varphi _p(t) $for all$t\in {\mathbb R}$implies that $$u'(x) =\varphi_{p'}o(\varphi_p( u') ) (x) \mbox{\ \ for\ all\ }x\in [0,1].$$ Thus, if$12$, it follows that$11$and assume that$u$is a solution of problem (\ref {meth3}). Then $$(| u'| ^p(x) +p'G( u(x) ) ) '=0,\mbox{\ \ \ for all }x\in [0,1].$$ \end{lemma} \paragraph{Proof.} Let$x_0\in Z(u') $. One has $$(G(u))' (x_0) =g(u(x_0) ) u'(x_0) =0\,.$$ Let us prove that $$(| u'| ^p) '(x_0) =0\,. \label{Am}$$ One has for all$x\neq x_0$, $$\frac{| u'(x) | ^p-| u'( x_0) | ^p}{x-x_0}=u'(x) \times \frac{\varphi_p(u'(x) ) -\varphi_p( u'(x_0) ) }{x-x_0}.$$ Thus,$\lim_{x\to x_0}u'(x) =u'(x_0) =0$, and $$\lim_{x\to x_0}\frac{\varphi_p(u'(x) ) -\varphi_p(u'(x_0) ) }{x-x_0}=(\varphi_p(u') ) '( x_0) =-g(u(x_0) ) \in {\mathbb R}.$$ Therefore, (\ref{Am}) is proved. Regarding (\ref{AD}), Lemma \ref{Lem2} follows. \hfill$\diamondsuit$\paragraph{Remark.} In Lemmas \ref{Lem1} and \ref{Lem2}, the solutions,$u$, are arbitrary and are not necessarily in$A_1^+$. \smallskip Now, assume that$u$is a solution of problem (\ref{meth3}) belonging to$A_1^+$. Thus, $$u'>0\mbox{\ in\ }[0,(1/2) ) \mbox{\ \ and\ \ }u'(1/2) . \label{kyd}$$ It follows that $$\sup \left\{ x\in [0,1) :u'(t) >0,\forall t\in [0,x) \right\} =1/2,$$ and by the energy relation one gets $$u'(x) =\left\{ (u'(0) ) ^p-p'G(u(x) ) \right\} ^{1/p}\mbox{,\ \ for all }x\in [0,1]. \label{jet}$$ Thus, $$\sup \left\{ x\in [0,1) :(u'(0) ) ^p-p'G(u(t) ) >0,\forall t\in [0,x) \right\} =1/2,$$ or equivalently $$\sup \left\{ x\in [0,1) :(u'(0) ) ^p-p'G(z) >0,\forall z\in [0,u( x) ) \right\} =1/2,$$ which implies that $$\sup \left\{ s\geq 0:(u'(0) ) ^p-p^{\prime }G(\xi ) >0,\forall \xi \in [0,s) \right\} =u(1/2) .$$ Also, by (\ref{jet}) it follows that $$x=\int_0^{u(x) }\left\{ (u'(0) ) ^p-p'G(\xi ) \right\} ^{-1/p}d\xi \mbox{,\ \ \ for all }x\in [0,(1/2) ]. \label{qsfj}$$ Thus, the improper integral in (\ref{qsfj}) is convergent for all$x\in [0,(1/2) ]$and in particular,$u'( 0) $is such that the improper integral $$\int_0^{r_{+}(u'(0) ) }\left\{ ( u'(0) ) ^p-p'G(\xi ) \right\} ^{-1/p}d\xi$$ converges and is equal to$(1/2) $, where for all$E>0 $$r_{+}(E) =\sup X_{+}(E) \mbox{\ if }X_{+}( E) \neq \emptyset \mbox{\ \ and\ \ }r_{+}(E) =0\mbox{\ if }X_{+}(E) =\emptyset$$ and $$X_{+}(E) =\left\{ s\geq 0:E^p-p'G(\xi ) >0,\forall \xi \in [0,s) \right\} .$$ It follows that if $u$ is a solution of problem (\ref{meth3}) belonging to $A_1^+$, then there exists $E_{*}\in \tilde D_{+}$ with $$\tilde D_{+}=\big\{ E>0:00\mbox{ for all }u\in (0,r_{+}( E_{*}) ) .$$ Let $u_{+}$ be the inverse of $h_{+}$ defined by $$u_{+}(x) =h_{+}^{-1}(x) \in [0,r_{+}( E_{*}) ],\mbox{ for all }x\in [0,(1/2)]\,,$$ and let $u$ be defined on $[0,1]$ by $$u(x) =\left\{ \begin{array}{l} u_{+}(x) \mbox{\ if\ }x\in [0,(1/2) ] \\ u_{+}(1-x) \mbox{\ if }x\in [(1/2) ,1] . \end{array} \right.$$ It easy to show that this function $u$ is a solution of problem (\ref{meth3}) belonging to $A_1^+$ and satisfying $u'(0) =E_{*}$, $\max_{[0,1]}u=u(1/2) =r_{+}(E_{*})$. Let us prove its uniqueness. Assume that $v$ is also a solution of problem (\ref{meth3}) belonging to $A_1^+$ and satisfies $$v'(0) =E_{*},\max_{[0,1]}v=v(1/2) =v_{+}(E_{*}) .$$ By (\ref{qsfj}) it follows that $$x=\int_0^{u(x) }\left\{ E_{*}^p-p'G(\xi ) \right\} ^{-1/p}d\xi =\int_0^{v(x) }\left\{ E_{*}^p-p^{\prime }G(\xi ) \right\} ^{-1/p}d\xi ,$$ for all $x\in [0,1/2]$. Thus, $$\int_{u(x) }^{v(x) }\left\{ E_{*}^p-p^{\prime }G(\xi ) \right\} ^{-1/p}d\xi =0,\mbox{\ \ for all }x\in [ 0,1/2].$$ Thus, $u=v$ on $[0,1/2]$, and by symmetry it follows that $u=v$ on $[0,1]$. Therefore, because $D_{+}\subset \tilde D_{+}$, Theorem \ref{quad} is proved for the case $k=1$ and $\kappa =+$. \hfill$\diamondsuit$\smallskip \section{Preliminary Lemmas} \label{sec4}In order to define the time maps we need the following. \begin{lemma} \label{lemma1} For $s\in {\mathbb R}$, consider the equation $$E^p-p'(F(s) +\lambda \Phi_q(s) ) =0, \label{Eq.1}$$ where $p>1$, $E\geq 0$ and $\lambda \in {\mathbb R}$ are real parameters, $F(s) =\int_0^sf(t) dt$ and $\Phi _q(s) =\int_0^s\varphi_q(t) dt$. If $\lambda >-m_\pm$, then for any $E>0$, equation (\ref{Eq.1}) admits a unique positive zero $s_{+}=s_{+}(\lambda ,E)$ (resp. a unique negative zero $s_{-}=s_{-}(\lambda ,E)$) and if $E=0$ it admits no positive (resp. negative) zero beside the trivial one $s_\pm =0$. Moreover, for all $\lambda >-m_\pm ,p>1$, \begin{description} \item[(i)] The function $E\longmapsto s_\pm (\lambda ,E)$ is $C^1$ on $(0,+\infty )$, and $$\pm \frac{\partial s_\pm }{\partial E}(\lambda ,E) =\frac{\pm (p-1) E^{p-1}}{f(s_\pm (\lambda ,E) ) +\lambda \varphi_q(s_\pm (\lambda ,E) ) }>0,\ \forall E>0\,.$$ \item[(ii)] $\lim\limits_{E\to 0^+}s_\pm (\lambda ,E) =0$, $\lim\limits_{E\to +\infty }s_\pm ( \lambda ,E) =\pm \infty$. \item[(iii)] $\lim\limits_{E\to 0}\frac{E^p}{| s_\pm ( \lambda ,E) | ^q}=\frac{p'}q(a_0+\lambda )$, $\lim\limits_{E\to +\infty }\frac{E^p}{| s_{\pm }(\lambda ,E) | ^q}=\frac{p'}q(a_{\pm }+\lambda )$. \item[(iv)] $\lim\limits_{E\to 0}\frac E{| s_{\pm }(\lambda ,E) | }=\cases{ 0 & if$q-p>0$\cr ((a_0+\lambda ) /(p-1) ) ^{1/p} & if$ q-p=0$\cr +\infty & if$ q-p<0$, }$ $\lim\limits_{E\to +\infty }\frac E{| s_\pm (\lambda ,E) | }=\cases{ +\infty & if$ q-p>0$\cr ((a_\pm +\lambda ) /(p-1) ) ^{1/p} & if$q-p=0$\cr 0 & if$q-p<0$.}$ \item[(v)] $\lim\limits_{E\to 0}\frac{F(s_{\pm }(\lambda ,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_0}{a_0+\lambda },\forall t>0$, $\lim\limits_{E\to +\infty }\frac{F(s_\pm (\lambda ,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_\pm }{a_{\pm }+\lambda },\forall t>0$. \end{description} \end{lemma} \paragraph{Proof.} For $p>1$ and $E\geq0$ fixed, consider the function $$s\longmapsto G_\pm (\lambda ,E,s) :=E^p-p'( F(s) +\lambda \Phi_q(s) ) ,$$ defined in ${\mathbb R}^\pm$ and strictly decreasing on $(0,+\infty )$ (resp. strictly increasing on $(-\infty,0)$), because $$\frac{dG_\pm }{ds}(\lambda ,E,s) =-p'\varphi _q(s) (\frac{f(s) }{\varphi_q(s) }+\lambda ) \quad\mbox{and}\quad m_\pm +\lambda >0\,.$$ One has $G_\pm (\lambda ,E,0) =E^p\geq 0$, and via l'Hospital's rule, \begin{eqnarray*} \lim\limits_{s\to +\infty }G_\pm (\lambda ,E,s) & = & \lim\limits_{s\to +\infty }E^p-p'\Phi_q(s) (\frac{F(s) }{\Phi_q(s) }+\lambda ) \\ & = & E^p-p'\lim\limits_{s\to +\infty }\Phi_q( s) (\lim\limits_{s\to +\infty }\frac{f(s) }{\varphi_q(s) }+\lambda ) \\ & = & -\infty . \end{eqnarray*} So, it is clear that in the case $m_{+}+\lambda >0$ (resp. $m_{-}+\lambda >0$) for any $E>0$, (\ref{Eq.1}) admits a unique positive zero, $s_{+}=s_{+}(\lambda ,E)$, (resp. a unique negative zero, $s_{-}=s_{-}(\lambda ,E)$), and if $E=0$, it admits no positive (resp. negative) zero beside the trivial one $s=0$. \smallskip Now, for any $p>1$ and $\lambda >-m_\pm$, consider the real valued function, $$(E,s) \longmapsto G_\pm (E,s) :=E^p-p^{\prime }(F(s) +\lambda \Phi_q(s) ) ,$$ defined on $\Omega_{+}=(0,+\infty ) ^2$ (resp. $\Omega _{-}=(0,+\infty ) \times (-\infty ,0)$). One has $G_\pm \in C^1(\Omega_\pm )$ and, $$\frac{\partial G_\pm }{\partial s}(E,s) =-p'\varphi _q(s) (\frac{f(s) }{\varphi_q(s) }+\lambda ) \quad\mbox{in }\Omega_\pm \,;$$ hence, because $m_\pm +\lambda >0$, it follows that $$\pm \frac{\partial G_\pm }{\partial s}(E,s) <0,\mbox{\ \ in \ }\Omega_\pm ,$$ and one may observe that $s_\pm (\lambda ,E)$ belongs to the open interval $(0,+\infty )$ (resp. $(-\infty ,0)$) and from its definition satisfies $$G_\pm (E,s_\pm (\lambda ,E) ) =0\,. \label{Eq.2}$$ So, one can make use of the implicit function theorem to show that the function $E\mapsto s_\pm (\lambda ,E)$ is $C^1(( 0,+\infty ) ,{\mathbb R})$ and to obtain the expression of $\frac{\partial s_\pm }{\partial E}(\lambda ,E)$ given in {\bf (i)}. Hence, by $m_\pm +\lambda >0$, it follows that for any fixed $p>1$ and $\lambda >-m_\pm$, the function defined on $(0,+\infty )$ by $E\mapsto s_\pm (\lambda ,E)$ is strictly increasing (resp. strictly decreasing) and bounded from below by $0$ (resp. by $-\infty$) and from above by $+\infty$ (resp. by $0$). Then the limit $\lim_{E\to 0^+}s_\pm (\lambda ,E) =\ell_0^\pm$ exists as real number and the limit $\lim_{E\to +\infty }s_\pm (\lambda ,E) =\ell _{+\infty }^\pm$ exists and belongs to $(0,+\infty ]$ (resp. $[-\infty ,0)$). Moreover, $$-\infty \leq \ell_{+\infty }^-<\ell_0^-\leq 0\leq \ell_0^+<\ell _{+\infty }^+\leq +\infty .$$ Let us observe that, for any fixed $p>1$ and $\lambda >-m_\pm$, the function $(E,s) \mapsto G_\pm (E,s)$ is continuous on $[0,+\infty ) ^2$ (resp. $[0,+\infty ) \times (-\infty ,0]$) and the function $E\longmapsto s_\pm (\lambda ,E)$ is continuous on $(0,+\infty )$ and satisfies (\ref{Eq.2}). So, by passing to the limit in (\ref {Eq.2}) as $E$ tends to $0^+$ one obtains $$0=\lim_{E\to 0^+}G_\pm (E,s_\pm (\lambda ,E) ) =G_\pm (0,\ell_0^+) \,.$$ Hence, $\ell_0^\pm$ is a zero, belonging to $[0,+\infty )$ (resp. $(-\infty ,0]$), to the equation in $s$: $$G_\pm (0,s) =0\mbox{.}$$ By solving this equation one gets: $\ell_0^\pm =0$. Assume that $\ell_{+\infty }^\pm$ is finite, then by passing to the limit in (\ref{Eq.2}) as $E$ tends to $+\infty$ one gets, $$+\infty =p'(F(\ell_{+\infty }^\pm ) +\lambda \Phi_q(\ell_{+\infty }^\pm ) ) <+\infty ,$$ which is impossible. So, $\ell_{+\infty }^\pm =\pm \infty$. \paragraph{Proof of (iii).} Dividing equation (\ref{Eq.2}) by $| s_{\pm }(\lambda ,E) | ^q$ one gets, $$\frac{E^p}{| s_\pm (\lambda ,E) | ^q}=p^{\prime }(\frac{F(s_\pm (\lambda ,E) ) }{| s_\pm (\lambda ,E) | ^q}+\frac \lambda q) ,$$ and by passing to the limit as $E$ tends to $0^+$, (using l'Hospital's rule), $$\lim_{E\to 0^+}\frac{E^p}{| s_\pm (\lambda ,E) | ^q}=\frac{p'}q(\lim_{E\to 0^+}\frac{f(s_\pm (\lambda ,E) ) }{\varphi_q( s_\pm (\lambda ,E) ) }+\lambda ) =\frac{p'}q(a_0+\lambda ) .$$ The second limit is obtained by the same way. \paragraph{Proof of (iv).} Dividing equation (\ref{Eq.2}) by $| s_{\pm }(\lambda ,E) | ^p$ one gets, \begin{eqnarray*} \lim\limits_{E\to 0^+}\frac{E^p}{| s_\pm (\lambda ,E) | ^p} & = & \lim\limits_{E\to 0^+}p^{\prime }(\frac{F(s_\pm (\lambda ,E) ) }{| s_\pm (\lambda ,E) | ^p}+\frac \lambda q| s_{\pm }(\lambda ,E) | ^{q-p}) \\ & = & \lim\limits_{E\to 0^+}p'| s_\pm ( \lambda ,E) | ^{q-p}(\frac{F(s_\pm ( \lambda ,E) ) }{| s_\pm (\lambda ,E) | ^q}+\frac \lambda q) \\ & = & \frac{p'}q(a_0+\lambda ) \lim\limits_{E\to 0^+}| s_\pm (\lambda ,E) | ^{q-p}. \end{eqnarray*} Therefore, the first limit follows. The second one is obtained by the same way. \paragraph{Proof of (v).} Using l'Hospital's rule one gets, for any $t>0$, \begin{eqnarray*} \lim \limits_{E\to 0^+}\frac{F(s_\pm (\lambda ,E) t) }{E^p} & = & \lim \limits_{E\to 0^+} \frac{t\frac{ds_\pm }{dE}(\lambda ,E) f(s_\pm ( \lambda ,E) t) }{pE^{p-1}} \\ & = & \lim \limits_{E\to 0^+} \frac{t(p-1) E^{p-1}f(s_\pm (\lambda ,E) t) }{(f( s_\pm (\lambda ,E) ) +\lambda \varphi_q(s_{\pm }(\lambda ,E) ) ) pE^{p-1}} \\ & = & \frac t{p'}\lim \limits_{E\to 0^+} \frac{\frac{f(s_\pm (\lambda ,E) t) }{\varphi_q( s_\pm (\lambda ,E) t) }}{(\frac{f(s_{\pm }(\lambda ,E) ) }{\varphi_q(s_\pm ( \lambda ,E) ) }\frac 1{\varphi_q(t) }+\frac \lambda {\varphi_q(t) }) } \\ & = & \frac{t^q}{p'}(\frac{a_0}{a_0+\lambda }) . \end{eqnarray*} The second limit may be computed by the same way, which completes the proof of Lemma \ref{lemma1}. \hfill$\diamondsuit$\smallskip Now, for any $p>1$, $\lambda >-m_\pm$ and $E>0$ we compute $X_\pm (\lambda ,E)$ as defined in Section \ref {sec3}. In fact, for all $E>0, $$X_{+}(\lambda ,E) =(0,s_{+}(\lambda ,E) ) ,X_{-}(\lambda ,E) =(s_{-}(\lambda ,E) ,0) ,$$ where$s_\pm (\lambda ,E) $is defined in Lemma \ref{lemma1}. Then $$r_\pm (\lambda ,E) =s_\pm (\lambda ,E) \mbox{ \ for all }E>0.$$ Hence, for any$p>1,\lambda >-m_\pm $,$0<| s_\pm (\lambda ,E) | <+\infty$if and only if$E>0$. And for all$E>0$, $$\pm (f(r_\pm (\lambda ,E) ) +\lambda \varphi_q(r_\pm (\lambda ,E) ) ) = \pm \varphi_q(r_\pm (\lambda ,E) ) ( \frac{f((r_\pm (\lambda ,E) ) ) }{\varphi_q(r_\pm (\lambda ,E) ) }+\lambda ) >0.$$ So,$D_\pm (\lambda ) =(0,+\infty )$for all$\lambda >-m_\pm $and $$D(\lambda ) =D_{+}(\lambda ) \cap D_{-}( \lambda ) =(0,+\infty ) ,\forall \lambda >-m.$$ Before going further in the investigation, from Lemma \ref{lemma1}, we deduce that for any fixed$p>1$and$\lambda >-m_\pm $, $$\pm \frac{\partial r_\pm }{\partial E}(\lambda ,E) =\frac{\pm (p-1) E^{p-1}}{f(r_\pm (\lambda ,E) ) +\lambda \varphi_q(r_\pm (\lambda ,E) ) }>0,\forall E>0. \label{A}$$ $$\lim\limits_{E\to 0^+}r_\pm (\lambda ,E) =0\mbox{\ \ \ \ and\ \ \ \ }\lim\limits_{E\to +\infty }r_\pm (\lambda ,E) =\pm \infty \label{B}$$ $$\lim\limits_{E\to 0}\frac{E^p}{| r_\pm (\lambda ,E) | ^q}=\frac{p'}q(a_0+\lambda ) ,\lim\limits_{E\to +\infty }\frac{E^p}{| r_\pm ( \lambda ,E) | ^q}=\frac{p'}q(a_\pm +\lambda ) . \label{C}$$ $$\label{D} \lim \limits_{E\to 0}\frac E{| r_\pm (\lambda ,E) | }=\left\{ \begin{array}{lcc} 0 & \mbox{if} & q-p>0 \\ (\frac{a_0+\lambda }{p-1}) ^{\frac 1p} & \mbox{if} & q-p=0 \\ +\infty & \mbox{if} & q-p<0 \end{array} \right.$$ $$\label{E} \lim \limits_{E\to +\infty }\frac E{| r_\pm (\lambda ,E) | }=\left\{ \begin{array}{lcc} +\infty & \mbox{if} & q-p>0 \\ (\frac{a_\pm +\lambda }{p-1}) ^{\frac 1p} & \mbox{if} & q-p=0 \\ 0 & \mbox{if} & q-p<0 \end{array} \right.$$ $$\label{F} \lim \limits_{E\to 0}\frac{F(r_\pm (\lambda ,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_0}{a_0+\lambda },\forall t>0$$ $$\label{G} \lim \limits_{E\to +\infty }\frac{F(r_\pm (\lambda ,E) t) }{E^p}=\frac{t^p}{p'}\frac{a_\pm }{a_{\pm }+\lambda },\forall t>0.$$ At present we define, for any$p>1,$\$\lambda >-m_\pm $, and$E>0$, the time map, $$T_\pm (\lambda ,E) :=\pm \int_0^{r_\pm (\lambda ,E) }\left\{ E^p-p'(F(\xi ) +\lambda \Phi _q(\xi ) ) \right\} ^{-\frac 1p}d\xi ,E>0,$$ and a simple change of variables shows that, $$T_\pm (\lambda ,E) =| r_\pm (\lambda,E) | \int_0^1\left\{ E^p-p'(F( r_\pm (\lambda ,E) \xi ) +(\lambda /q) | r_\pm (\lambda ,E) \xi | ^q) \right\}^{-\frac 1p}d\xi , \label{U}$$ which may be written as, \begin{eqnarray}\label{H} T_\pm (\lambda ,E) &=&(| r_\pm (\lambda,E) | /E) \int_0^1\big\{ 1-p'(F(r_\pm (\lambda ,E) \xi ) /E^p \\ &&\hspace{3cm}+(\lambda \xi ^q/q) (| r_\pm (\lambda ,E) | ^q/E^p) ) \big\}^{-1/p}d\xi \,. \nonumber \end{eqnarray} Also, we define, the time maps $$\displaylines{ T_{2n}^\pm (\lambda ,E) :=n(T_{+}(\lambda ,E) +T_{-}(\lambda ,E) ) ,E>0,\lambda >-m_{2n}^\pm ,n\geq 1\,, \cr T_{2n+1}^\pm (\lambda ,E) :=T_{2n}^\pm (\lambda ,E) +T_\pm (\lambda ,E) ,E>0,\lambda >-m_{2n+1}^\pm ,n\geq 0\,. }$$ To prove Theorems \ref{thm1},\ \ref{thm2}, \ref{thm3}, it suffices to compute the limits of these time maps as$E$tends to$0^+$and$+\infty $, and then apply the intermediate value theorem. Recall that we have defined in Proposition \ref{prop1} the functions$h_k^\pm $and let us now define, $$g_k(\lambda ) =\frac{\lambda_k^{1/p}}{(a_0+\lambda ) ^{1/p}},\mbox{ for all }\lambda >-a_0,\ k\geq 1\,.$$ \begin{lemma} \label{lemma22}Assume that$p,q>1$, then for all$k\geq 1$, and all$\lambda >-m_k^\pm $, \begin{enumerate} \item$\lim\limits_{E\to 0^+}T_k^\pm (\lambda ,E) =+\infty $and$\lim\limits_{E\to +\infty }T_k^{\pm }(\lambda ,E) =0$, if$q>p$, \item$\lim\limits_{E\to 0^+}T_k^\pm (\lambda ,E) =0$and$\lim\limits_{E\to +\infty }T_k^\pm ( \lambda ,E) =+\infty $, if$qh_k^{\pm }(\lambda ) $for all$\lambda >-a_0$. \item[(ii)] If$a_0>c_k^\pm $, then$g_k(\lambda ) -b_k^\pm $. \item[(iii)] If$a_{-}-a_{-}$. \item[(iv)] If$a_{-}h_k^\pm (\lambda ) \mbox{ for all }\lambda \in (\tilde \lambda_k^\pm ,+\infty ) . \end{array} $$Moreover, the function a_0\mapsto \tilde \lambda_k^\pm (a_0)  is strictly increasing on (a_{-},a_k^\pm )  and$$ \lim_{a_0\to a_{-}}\tilde \lambda_k^\pm (a_0) =-a_{-},\mbox{and}\lim_{a_0\to a_k^\pm }\tilde \lambda_k^\pm (a_0) =+\infty . $$\end{description} \end{lemma} \paragraph{Proof.} It is immediate to prove Assertions {\bf (i)} and {\bf (ii)} of this lemma since the function a\mapsto (a+\lambda ) ^{-1/p} is strictly decreasing. Let us prove assertions {\bf (iii) }and {\bf (iv)}. We are concerned by the case k=2n+1 for some n\geq 1  and the superscript is +:(k=2n+1,+) . All the remaining cases may be handled similarly. Let y_{2n+1}^+ be the function defined on (-a_{-},+\infty )  by$$ y_{2n+1}^+(\lambda ) =(n+1) (a_{+}+\lambda ) ^{-1/p}+n(a_{-}+\lambda ) ^{-1/p}-(2n+1) (a_0+\lambda ) ^{-1/p}. $$One has, y_{2n+1}^+(\lambda ) <0 if and only if$$ ((\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1/p}+(\frac n{2n+1}) (a_{-}+\lambda ) ^{-1/p}) ^p<(a_0+\lambda ) ^{-1} $$and (y_{2n+1}^+) '(\lambda ) <0 if and only if$$ ((\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1-1/p}+(\frac n{2n+1}) (a_{-}+\lambda ) ^{-1-1/p}) ^{p/(p+1) }> (a_0+\lambda ) ^{-1}. $$Since the function t\mapsto t^p (resp. t\mapsto -t^{p/(p+1) }) is convex on (-a_{-},+\infty ) , \begin{eqnarray*} \lefteqn{ ((\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1/p}+(\frac n{2n+1}) (a_{-}+\lambda )^{-1/p}) ^p } \\ &&< (\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1}+(\frac n{2n+1}) (a_{-}+\lambda ) ^{-1} \end{eqnarray*} (resp. \begin{eqnarray*} \lefteqn{ ((\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1-1/p}+(\frac n{2n+1}) (a_{-}+\lambda ) ^{-1-1/p}) ^{p/(p+1)} }\\ && >(\frac{n+1}{2n+1}) (a_{+}+\lambda ) ^{-1}+(\frac n{2n+1}) (a_{-}+\lambda ) ^{-1}\mbox{)}. \end{eqnarray*} So, if we define on (-a_{-},+\infty )  the function x_{2n+1}^+ by$$ x_{2n+1}^+(\lambda ) =(\frac{n+1}{2n+1}) ( a_{+}+\lambda ) ^{-1}+(\frac n{2n+1}) (a_{-}+\lambda ) ^{-1}-(a_0+\lambda ) ^{-1}, $$it follows that for all \lambda >-a_{-},$$$x_{2n+1}^+(\lambda ) \leq 0\Longrightarrow y_{2n+1}^+( \lambda ) <0\mbox{\ \ \ and\ \ \ }x_{2n+1}^+(\lambda ) \geq 0\Longrightarrow (y_{2n+1}^+) '(\lambda ) <0. $$Some simple computations show that for all \lambda >-a_{-} and \kappa =+,-, one has,$$ \kappa x_{2n+1}^+(\lambda ) >0\Longleftrightarrow \kappa (\lambda (a_0-a_{2n+1}^+) -( a_{-}a_{+}-a_0a_{2n+1}^-) ) >0. $$Also, for all \lambda \in {\mathbb R} and \kappa =+,-, one has in the case where a_0>a_{2n+1}^+,$$$ \kappa (\lambda (a_0-a_{2n+1}^+) -( a_{-}a_{+}-a_0a_{2n+1}^-) ) >0\Longleftrightarrow \kappa (\lambda -\Lambda_{2n+1}^+) >0, $$and in the case where a_00\Longleftrightarrow \kappa (\lambda -\Lambda_{2n+1}^+) <0.$$ On the other hand one has, $$a_0-a_{-}\mbox{\ \ \ and\ \ \ }a_0>a_{2n+1}^+\Longrightarrow \Lambda_{2n+1}^+<-a_{-}.$$ Hence, an easy compilation of the above assertions shows that if $a_0-a_{-}) &\Longrightarrow& x_{2n+1}^+(\lambda ) \leq 0 \\ -a_{-}<\lambda <\Lambda_{2n+1}^+&\Longrightarrow& x_{2n+1}^+( \lambda ) <0 \end{eqnarray*} and if$a_0>a_{2n+1}^+$, then$\lambda >-a_{-}$implies$x_{2n+1}^+(\lambda ) >0$. It remains to study the particular case$a_0=a_{2n+1}^+$. For$\lambda >-a_{-}$, define $$\psi_\lambda (t) =(\lambda +t) ^{-1},\mbox{ for all }t>-\lambda .$$ Let us observe that$x_{2n+1}^+(\lambda ) >0$if and only if $$\psi_\lambda (a_0) <(\frac n{2n+1}) \psi_\lambda (a_{-}) +(\frac{n+1}{2n+1}) \psi_\lambda ( a_{+}) .$$ But since$\psi_\lambda $is strictly convex, one has \begin{eqnarray*} \psi_\lambda (a_0) =\psi_\lambda (a_{2n+1}^+) &=&\psi_\lambda ((\frac n{2n+1}) a_{-}+(\frac{n+1}{2n+1}) a_{+}) \\ &<& (\frac n{2n+1}) \psi_\lambda (a_{-}) +( \frac{n+1}{2n+1}) \psi_\lambda (a_{+}) , \end{eqnarray*} that is, if$a_0=a_{2n+1}^+$, then $$\lambda >-a_{-}\Longrightarrow x_{2n+1}^+(\lambda ) >0.$$ Thus, in the case where$a_0\geq a_{2n+1}^+$,$y_{2n+1}^+$is strictly decreasing on$(-a_{-},+\infty ) $and by$\lim_{\lambda \to +\infty }y_{2n+1}^+(\lambda ) =0$it follows that$y_{2n+1}^+$is strictly positive on$(-a_{-},+\infty ) $. Thus, Assertion {\bf (iii)} is proved. In the case where$a_00$, and $$\frac{\partial y_{2n+1}^+}{\partial a_0}(a_0,\tilde \lambda _{2n+1}^+(a_0) ) =\frac{2n+1}p(a_0+\tilde \lambda _{2n+1}^+(a_0) ) ^{-1-1/p}>0.$$ So, the function$a_0\mapsto \tilde \lambda_{2n+1}^+(a_0) $is strictly increasing on$(a_{-},a_{2n+1}^+) $. On the other hand, one has $$-a_{-}<\tilde \lambda_{2n+1}^+(a_0) <\Lambda _{2n+1}^+(a_0) ;\forall a_0\in (a_{-},a_{2n+1}^+) ,$$ and$\lim\limits_{a_0\mapsto a_{-}}\Lambda _{2n+1}^+(a_0) =-a_{-}$, which is easy to check. Thus,$\lim\limits_{a_0\to a_{-}}\tilde \lambda_{2n+1}^+( a_0) =-a_{-}$. Assume that the function$a_0\mapsto \tilde \lambda _{2n+1}^+(a_0) $is bounded as$a_0$tends to$a_{2n+1}^+$. Denote by$\lambda_{*}$its limit, that is,$\lim\limits_{a_0\to a_{2n+1}^+}\tilde \lambda_{2n+1}^+(a_0) =\lambda_{*}$. Since $$y_{2n+1}^+(a_0,\tilde \lambda_{2n+1}^+(a_0) ) =0,\forall a_0\in (a_{-},a_{2n+1}^+) ,$$ it follows that$y_{2n+1}^+(a_{2n+1}^+,\lambda_{*}) =0$, that is, \begin{eqnarray*} \lefteqn{ ((\frac{n+1}{2n+1}) (a_{+}+\lambda_{*}) +(\frac n{2n+1}) (a_{-}+\lambda_{*}) ) ^{-1/p} } \\ &=& (\frac{n+1}{2n+1}) (a_{+}+\lambda_{*}) ^{-1/p}+(\frac n{2n+1}) (a_{-}+\lambda_{*}) ^{-1/p}\,. \end{eqnarray*} By the strict convexity of the function$t\to t^{-1/p}$on$(0,+\infty ) $, it follows that$a_{+}+\lambda _{*}=a_{-}+\lambda_{*}$, that is$a_{-}=a_{+}$, which contradicts the hypothesis$a_{-}-m_k^\pm$$$\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac 12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to +\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm ( \lambda ) .$$ On the other hand, for all$\lambda >-a_0:g_k(\lambda ) >h_k^\pm (\lambda ) $. So, since the function$\lambda \mapsto g_k(\lambda ) $(resp.$\lambda \mapsto h_k^\pm (\lambda ) $) is strictly decreasing on$(-a_0,+\infty ) $(resp. on$(-b_k^\pm ,+\infty ) $), and $$g_k(\lambda_k-a_0) =1\mbox{ (resp. }h_k^\pm ( \lambda_k^\pm ) =1\mbox{)}\,,$$ then $$g_k(\lambda ) >1\mbox{ \ \ if and only if \ \ }-a_0<\lambda <\lambda_k-a_0$$ (resp. $$h_k^\pm (\lambda ) <1\mbox{\ \ \ if and only if\ \ \ }\lambda _k^\pm <\lambda \mbox{).}$$ Hence, one has, $$g_k(\lambda ) >1>h_k^\pm (\lambda ) \mbox{\ \ \ if and only if \ \ }\max (-a_0,\lambda_k^\pm ) <\lambda <\lambda_k-a_0.$$ Then the equation in$E>0:T_k^\pm (\lambda ,E) =1/2$admits at least a solution for all$\lambda $satisfying $$\lambda >-m_k^\pm \quad\mbox{and}\quad \max (-a_0,\lambda _k^\pm ) <\lambda <\lambda_k-a_0,$$ that is, for all$\lambda $satisfying $$\max (-m_k^\pm ,\lambda_1^\pm ) <\lambda <\lambda_k-a_0,$$ since$\max (-a_0,-m_k^\pm ) =-m_k^\pm $. If$a_0>c_k^\pm $then for all$k\geq 1$and$\lambda >-m_k^\pm $, $$\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac 12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to +\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm ( \lambda ) .$$ On the other hand, for all$\lambda \in (-b_k^\pm ,+\infty )$,$g_k(\lambda ) 1\mbox{\ \ \ if and only if\ \ \ }-b_k^\pm <\lambda <\lambda_k^\pm \mbox{).} $$Hence, one has,$$ g_k(\lambda ) <10:T_k^\pm (\lambda ,E) =1/2$admits at least a solution for all$\lambda $satisfying $$\lambda >-m_k^\pm \quad\mbox{and}\quad \max (-b_k^{\pm },\lambda_k-a_0) <\lambda <\lambda_k^\pm ,$$ that is, for all$\lambda $satisfying $$\max (-m_k^\pm ,\lambda_k-a_0) <\lambda <\lambda_k^\pm$$ since$\max (-b_k^\pm ,-m_k^\pm ) =-m_k^\pm $. The proof of Theorem \ref{thm1} is complete. \hfill$\diamondsuit$\smallskip \paragraph{Proof of Theorem \ref{thm2}.} If$a_{-}-m_\pm $, $$\lim_{E\to 0^+}T_1^\pm (\lambda ,E) =\frac 12g_1(\lambda ) \quad\mbox{and}\quad \lim_{E\to +\infty }T_1^\pm (\lambda ,E) =\frac 12h_1^\pm ( \lambda ) .$$ On the other hand, one has, $$\mbox{for all }\lambda >-a_0,g_1(\lambda ) >h_1^+( \lambda ) \mbox{\ \ \ \ (resp. for all }\lambda >-a_{-},g_1( \lambda ) -m_k^\pm =-m,$$ \lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac 12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to +\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm ( \lambda ) . $$On the other hand, one has for all \lambda >-a_{-},g_k(\lambda ) 0:T_k^{\pm }(\lambda ,E) =1/2 admits at least a solution for all \lambda satisfying$$ \max (-m,\lambda_k-a_0) <\lambda <\lambda_k^\pm . $$If a_{-}-m,$$ \lim_{E\to 0^+}T_k^\pm (\lambda ,E) =\frac 12g_k(\lambda ) \quad\mbox{and}\quad \lim_{E\to +\infty }T_k^\pm (\lambda ,E) =\frac 12h_k^\pm ( \lambda ) . $$Since the function a_0\mapsto \tilde \lambda_k^\pm (a_0) \in (-a_{-},\Lambda_k^\pm ) is strictly increasing on the interval (a_{-},a_k^\pm ) then the function a_0\mapsto h_k^\pm (\tilde \lambda_k^\pm (a_0) ) is strictly decreasing on (a_{-},a_k^\pm ) . Also, one has,$$ \lim_{a_0\to a_{-}}h_k^\pm (\tilde \lambda_k^\pm ( a_0) ) =\lim_{x\to -a_{-}}h_k^\pm (x) =+\infty , $$and$$ \lim_{a_0\to a_k^\pm }h_k^\pm (\tilde \lambda_k^{\pm }(a_0) ) =\lim_{x\to +\infty }h_k^\pm ( x) =0\,. $$Then there exists a unique \tilde a_k^\pm \in (a_{-},a_k^{\pm }) such that for all a_0\in (a_{-},a_k^\pm ) one gets, \begin{eqnarray*} h_k^\pm (\tilde \lambda_k^\pm (a_0) ) >1 &\Longleftrightarrow& a_{-}\tilde \lambda_k^\pm (a_0) ,$$ h_k^\pm (\lambda ) <10:T_k^\pm (\lambda ,E) =1/2$ admits at least a solution for all $\lambda$ satisfying $$\lambda >-m\quad\mbox{and}\quad \lambda_k^\pm <\lambda <\lambda_k-a_0\,,$$ that is, for all $\lambda$ satisfying, $\max (-m,\lambda_k^\pm ) <\lambda <\lambda_k-a_0$. Case where $\tilde a_k^\pm \tilde \lambda_k^\pm (a_0)$, $$h_k^\pm (\lambda ) 0:T_k^\pm (\lambda ,E) =1/2 admits at least a solution for all \lambda  satisfying$$ \lambda >-m\mbox{\ \ \ \ and\ \ \ }\max (-a_{-},\lambda _k-a_0) <\lambda <\lambda_k^\pm , $$that is, for all \lambda  satisfying \max (-m,\lambda _k-a_0) <\lambda <\lambda_k^\pm , this is so because \max ( -m,-a_{-}) =-m. The proof of Theorem \ref{thm2} is complete. \hfill\diamondsuit \paragraph{Proof of Theorem \ref{thm3}.} If q-p>0 (resp. q-p<0), by Lemma \ref{lemma22}, for all \lambda >-m_k^\pm ,k\geq 1,$$$\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =+\infty \mbox{ (resp. }=0\mbox{) and }\lim_{E\to 0^+}T_k^\pm (\lambda ,E) =0\mbox{ (resp. }=+\infty \mbox{).} $$So, the intermediate value theorem implies that the equation in E>0:T_k^\pm (\lambda ,E) =1/2 admits at least a solution for all \lambda >-m_k^\pm . Theorem \ref{thm3} is proved. \hfill\diamondsuit\smallskip \paragraph{Remark 1.} Some easy computations show that$$ \frac{\partial T_{+}}{\partial E}(\lambda ,E) =\frac{(p') ^{-\frac1p}}p (\frac{\partial r_{+}}{\partial E}(\lambda ,E) ) \int_0^1\frac{(H(\lambda ,r_{+}(\lambda ,E) ) -H(\lambda ,r_{+}(\lambda ,E) \xi ) ) }{(F(\lambda ,r_{+}(\lambda ,E) ) -F(\lambda ,r_{+}(\lambda ,E) \xi ) ) ^{1+\frac 1p}}d\xi , $$where H(\lambda ,x) :=p\tilde F(\lambda ,x) -x\tilde f(\lambda ,x) ,\tilde F(\lambda ,x) =\lambda \Phi_q(x) +F(x) and \tilde f( \lambda ,x) =\lambda \varphi_q(x) +f(x) . So, if q=p, one gets,$$ \frac d{dx}(\frac{f(x) }{\varphi_p(x) }) =\frac{-1}{x\varphi_p(x) }\frac{\partial H}{\partial x}(\lambda ,x) ,x>0.$$Hence, in the particular case where the function$x\mapsto f(x) /\varphi_p(x) ,(q=p) $is strictly decreasing on$(0,+\infty ) $the function$x\mapsto H(\lambda ,x) $is strictly increasing on$(0,+\infty ) $and then the time map$E\mapsto T_{+}(\lambda ,E) $is strictly increasing on$(0,+\infty ) $. So, uniqueness (if existence) of the solution of the equation in$E>0:T_{+}(\lambda ,E) =1/2$is guaranteed. The exact number of positive solution(s) in$A_1^+$should be obtained with this additional condition. The same remark works for the result of Guedda and Veron \cite{GueddaVeron}. That is, their Theorem 2.1 holds without there condition (2.7). \paragraph{Remark 2.} If$f$is an odd function,$a_{-}=a_{+}$. The statements of Theorems \ref{thm1}, \ref{thm2}, and \ref{thm3}\ may be simplified in this particular case. \paragraph{Remark 3.} Several corollaries may be deduced from Theorems \ref{thm1},\ \ref {thm2},\ and \ref{thm3} and the above remarks. In fact, one may draw some bifurcation diagrams and compute a lower bound on the number of solutions of problem (\ref{p1}) in some cases and the exact number of positive solutions in others. This would require more space and patience, and is left to the diligent, patient reader. However, a qualitative feature of the variations of the bifurcation branches as$a_0$varies is known. In fact, if$a_{-}\tilde a_k^\pm $. The case where$a_0=\tilde a_k^\pm $remains an open question. If$a_{-}>a_{+}$, one may study the asymmetric case,$a_{-}>a_0>a_{+}$, as in Theorem \ref{thm2}. \paragraph{Remark 4.} One may observe that a common feature in Theorems \ref{thm1},\ \ref {thm2},\ and \ref{thm3} is that the parameter$\lambda $is taken, in particular, such that the function$x\mapsto \lambda +f(x) /\varphi_q(x) $is strictly positive on$(-\infty ,0) $and/or$(0,+\infty ) $. Some cases where this function changes sign once are studied by Guedda and Veron \cite{GueddaVeron} and by Boucherif, Bouguima and Derhab \cite{BBD}, both for the particular case where$f$is odd. So, it is reasonable to ask the question of what happens if$f$is not necessarily odd. \section{Appendix: Historical overview on time maps} At the beginning of time maps' history, the authors used them with the one dimensional Laplacian operator, that is, with the one dimensional p-Laplacian and$p=2$. From the 1960's we can mention Opial \cite{Opial1}, \cite {Opial2}, Urabe \cite{Urabe1}, \cite{Urabe2}, \cite{Urabe3}, \cite{Urabe4}, \cite{Urabe5}, Pimbley \cite{Pimbley1}, \cite{Pimbley2}, and Gavalas \cite {Gavalas}. In the early 1970's, Laetsch \cite{Laetsch} used time maps to study positive solutions to a class of boundary-value problems with Dirichlet boundary data. Since then many authors have referred to his work. We also want to mention Chafee and Infante \cite{ChafeeInfante}, and Chafee \cite{Chafee}. Brown and Budin \cite{BrownBudin1}, \cite{BrownBudin2} used the time map approach to study positive solutions to some boundary-value problems. Independently and about the same time, De-Mottoni and Tesei \cite{DeMottoniTesei1} studied positive solutions of some other class of boundary-value problems by means of the same method. In the early 1980's Smoller and Wasserman \cite{SmollerWasserman} introduced a technique that, in some circumstances, can be used to prove uniqueness of the critical point of time maps. Their technique has been used subsequently by many authors; see for instance, Ammar Khodja \cite{AmmarKhodja}, Ramaswamy \cite{Ramaswamy}, S. H. Wang and Kazarinoff, \cite{SHWang1}, \cite{SHWang2}, S. H. Wang and F. P. Lee \cite{SHWang3}, S. H. Wang \cite{SHWang4}, \cite{SHWang5}, \cite{SHWang6}, and recently by Addou and Benmeza\"\i\ \cite{add3}. The study of sign-changing solutions by means of time maps was initiated by De-Mottoni and Tesei \cite{DeMottoniTesei2}, and independently, some years after, by Shivaji \cite{Shivaji2}. During the last two decades, time maps have been used in many publications. Besides the above mentioned papers, we want to add the following ones: Addou and Ammar Khodja \cite{Addou1}, Anuradha, Shivaji and Zhu \cite {Anuradha1}, \cite{Anuradha15}, Anuradha and Shivaji \cite{Anuradha2}, Anuradha, C. Brown and Shivaji \cite{Anuradha3}, Brown, Ibrahim and Shivaji \cite{Brown}, Brunovsky and Chow \cite{Brunovsky}, Castro and Shivaji \cite {Castro}, \cite{Castro2}, Ding and Zanolin \cite{Ding1}, \cite{Ding2}, Fernandes \cite{Fernandes}, Fonda and Zanolin \cite{Fonda}, Fonda, Gossez and Zanolin \cite{Fonda2}, Schaaf \cite{Schaaf}, Shivaji \cite{Shivaji1},\ \cite{Shivaji2},\ \cite{Shivaji3}, Smoller, Tromba and Wasserman \cite {Smoller}, Smoller and Wasserman \cite{SmollerWasserman1}, \cite {SmollerWasserman2}, \cite{SmollerWasserman3}. Notice that this list is in alphabetical order, and is not complete by any means. The differential operator in the equations studied in these papers is the$p$-Laplacian with$p=2$. 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Soc. {\bf 125} (1997), pp. 2275-2283. \end{thebibliography} \noindent{\sc Idris Addou}\\ USTHB, Institut de Math\'ematiques \\ El-Alia, B.P. no. 32, Bab-Ezzouar\\ 16111, Alger, Alg\'erie\\ E-mail address: idrisaddou@hotmail.com \end{document}