\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 56, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/56\hfil Positive solutions] {Positive solutions for a class of quasilinear singular equations} \author[J. V. Goncalves \& C. A. Santos\hfil EJDE-2004/56\hfilneg] {Jos\'e Valdo Goncalves \& Carlos Alberto P. Santos} % in alphabetical order \address{Jos\'e Valdo Goncalves \hfill\break Universidade de Bras\'{\i}lia \\ Departamento de Matem\'atica \\ 70910-900 Bras\'{i}lia, DF, Brasil} \email{jv0ag@unb.br} \address{Carlos Alberto P. Santos\hfill\break Universidade Federal de Goi\'as \\ Departamento de Matem\'atica \\ Catal\~ao, GO, Brasil} \email{csantos@unb.br} \date{} \thanks{Submitted October 6, 2003. Published April 13, 2004.} \thanks{Partially supported by CNPq/Brazil} \subjclass[2000]{35B40, 35J25, 35J60} \keywords{Singular equations, radial positive solutions, fixed points, \hfill\break\indent shooting method, lower-upper solutions} \begin{abstract} This article concerns the existence and uniqueness of solutions to the quasilinear equation $$- \Delta_{p} u=\rho(x) f(u) \quad \mbox{in } \mathbb{R}^N $$ with $u > 0$ and $u(x)\to 0$ as $|x| \to \infty$. Here $1 < p < \infty$, $N \geq 3$, $\Delta_{p}$ is the $p$-Laplacian operator, $\rho$ and $f$ are positive functions, and $f$ is singular at $0$. Our approach uses fixed point arguments, the shooting method, and a lower-upper solutions argument. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} We study the existence and uniqueness of solution of the problem \begin{equation} \label{e1.1} \begin{gathered} - \Delta_{p} u=\rho(x) f(u)\quad\mbox{in } \mathbb{R}^N,\\ u > 0 \quad \mbox{in } \mathbb{R}^N, \quad \lim_{|x| \to \infty} u(x)= 0, \end{gathered} \end{equation} where $1 < p < \infty$, $N \geq 3$ and $\Delta_{p}$ is the $p$-Laplacian operator while $\rho : \mathbb{R}^N \to [0,\infty)$ is continuous and $f : (0, \infty) \to (0,\infty)$ is a $C^{1}$-function, singular at zero, for instance, in the sense that $\lim_{s\to 0 }f(s)=\infty$. The case $p=2$ has been studied by several authors. Under additional assumptions on $\rho$, Edelson \cite{Edelson} studied \eqref{e1.1} with $f(s)=s^{- \lambda}$, $\lambda \in (0,1)$. A solution was shown to exist provided $$ \int_1^{\infty} r^{(N-1)+ \lambda(N-2)} \widetilde{\rho}(r)\,dr < \infty, $$ where $\widetilde{\rho} (r):= \max_{|x|=r}\rho(x)$. That result was extended for all $\lambda > 0$, by Shaker \cite{Shaker}. Later, Lair \& Shaker \cite{LairShaker} showed existence of a solution under the condition $$ \int_0^{\infty} r \widetilde\rho(r) dr < \infty. $$ Zhang \cite{zhang} showed that \eqref{e1.1} has a solution provided that $f' < 0$ and $\lim_{s \to 0}f(s) = \infty$. Yet in the case $p=2$, Cirstea \& Radulescu \cite{CRadulesco} showed that \eqref{e1.1} is solvable under the conditions: $f$ is bounded from above near $+ \infty$, $\lim_{s \to 0} f(s)/s= \infty$, and $\frac{f(s)}{s + b}$ is decreasing for some positive constant $b$. In the present paper we shall assume that $\rho$ is radially symmetric and \begin{gather} \frac{f(s)}{s^{p - 1}} \quad\mbox{is nonincreasing in } (0,\infty),\label{e1.2}\\ \liminf_{s \to 0} f(s) > 0, \quad \lim_{s \to \infty} \frac{f(s)}{s^{p-1}}=0. \label{e1.3} \end{gather} Our main result is as follows. \begin{theorem} \label{thm1.1} Assume \eqref{e1.2}, \eqref{e1.3} and \begin{equation} \begin{gathered} 0 < \int_1^\infty r^{\frac{1}{p-1}} \rho(r)^{\frac{1}{p-1}} dr < \infty, \quad\mbox{if } 1 < p \leq 2,\\ 0 < \int_1^\infty r^{\frac{(p-2)N + 1}{p-1} } \rho(r) dr < \infty, \quad\mbox{if } p \geq 2. \end{gathered}\label{e1.4} \end{equation} Then \eqref{e1.1} has: \begin{itemize} \item[(ii)] A radially symmetric solution $u$ in $C^1(\mathbb{R}^N) \cap C^2(\mathbb{R}^N \backslash \{0\})$ if $p < N$, \item[(ii)] No radially symmetric solution in $C^1(\mathbb{R}^N) \cap C^2(\mathbb{R}^N \backslash \{0\})$ if $p \geq N$. \end{itemize} \end{theorem} \begin{remark} \label{rmk1} \rm Regarding case (i), it will be shown that $u \in C^2(\mathbb{R}^N)$ if and only if $p \leq 2$. Additionally, the solution is uniquely determined if $f(s)/(s + b)^{p-1}$ is nonincreasing for some $b > 0$. See Section 7 . \end{remark} Theorem \ref{thm1.1} improves the main existence result in Cirstea \& Radulesco \cite{CRadulesco} in the sense that we allow both a broader class of nonlinear operators as well as nonlinear singular terms $f$. Our theorem applies to the class of functions $$ f(s)=s^{-\lambda} + s^{\gamma},\quad\mbox{where}~~~ \lambda \geq 0,~ 0 \leq \gamma 0 \quad\mbox{in } B_R ,\quad u=0 \quad \mbox{in } \partial B_R , \end{gathered} \end{equation} where $B_R $ is the ball of radius $R$. \begin{theorem} \label{thm1.2} Assume \eqref{e1.2}, \eqref{e1.3} and $p < N$. Then for each sufficiently large $R$, \eqref{e1.5} has a radially symmetric solution in $C({\overline{B}_R }) \cap C^1(B_R ) \cap C^2(B_R \backslash \{0\})$. \end{theorem} \begin{theorem} \label{thm1.3} Assume \eqref{e1.2}--\eqref{e1.4} and $p < N$. Then there is a radially symmetric function $v \in C^1(\mathbb{R}^N) \cap C^2(\mathbb{R}^N \backslash \{0\})$ such that \begin{equation} \label{e1.6} \begin{gathered} - \Delta_{p} v \geq \rho(x) f(v)\quad\mbox{in } \mathbb{R}^N \backslash \{0\},\\ v > 0 \quad \mbox{in } \mathbb{R}^N, \quad \lim_{|x| \to \infty} v(x)= 0. \end{gathered} \end{equation} \end{theorem} The proof of Theorem \ref{thm1.1} will be accomplished, by at first, using Theorem \ref{thm1.3} to pick a solution $v$ of \eqref{e1.6}, (which will be referred to as an upper-solution of \eqref{e1.1}), secondly, by choosing a sufficiently large integer $j$ and applying Theorem \ref{thm1.2} to find for each integer $k > 1$, a solution say, $ u_{k}$ of $\eqref{e1.5}_{j+k}$, which after extended as zero outside $B_{j+k}$, will be shown to satisfy, $$ 0 \leq u_1 \leq u_2 \leq \dots \leq u_k \leq \dots \leq v. $$ Then we pass to the limit as $k \to \infty$, getting to a solution of \eqref{e1.1} as asserted in our main result. This kind of argument is motivated by reading Zhang \cite{zhang} and Cirstea \& Radulescu \cite{CRadulesco}. \section{Some Technical Lemmas} At first we state and prove some preliminary results, crucial in the proof of Theorem \ref{thm1.2}. As a first step in this direction, consider the initial-value problem, \begin{equation} \begin{gathered} -\Big(r^{N-1} |u'|^{p-2} u'\Big)'=r^{N-1} \rho(r) f(u(r)) \quad \mbox{in } (0,\infty),\\ u(0)=a , \quad u'(0)=0, \end{gathered}\label{e2.1} \end{equation} where $a >0$ is a parameter and note that this equation is equivalent to the integral equation, \begin{equation} u(r)=a - \int^r_0\Big[t^{1-N}\int_0^t s^{N-1} \rho(s) f(u(s)) ds \Big]^{\frac{1}{p-1}}dt. \label{e2.2} \end{equation} Moreover, a solution of \eqref{e2.2} is a fixed point of the operator, \begin{equation} \mathcal{F}u(r)=a - \int_{0}^{r} \Big[t^{1-N}\int_{0}^{t} s^{N-1} \rho(s) f(u(s))ds \Big]^{\frac{1}{p-1}}dt. \label{e2.3} \end{equation} \begin{lemma} \label{lm2.1} Assume \eqref{e1.2}. Then for each $a >0$ there is $T(a) \in (0,\infty]$ and a unique solution of \eqref{e2.1}, $u := u(\cdot,a) \in C^1([0,T(a))) \cap C^{2}((0,T(a)))$ such that $u(r) \to 0$ as $r \to T(a)$ provided $T(a) < \infty$. \end{lemma} Given $T, h > 0$ set $$ X := \left\{w \in C^1([0, T]) | w \geq h \right \}. $$ If $w_1, w_2 \in X$ let $H:[0,T] \to \mathbb{R} $ be the continuous function $$ H(s) := s^{N-1} \Big[| (w_{2}^{1/p})'|^{p-2} (w_2^{1/p})' w_2^{\frac{1-p}{p}} - |(w_1^{1/p})'|^{p-2} (w_1^{1/p})' w_1^{\frac{1-p}{p}} \Big] (w_1-w_2)(s). $$ \begin{lemma} \label{lm2.2} If $w_1, w_2 \in X$ and $0 \leq S \leq U \leq T$, then \begin{align*} &H(U) - H(S) \\ &\leq \int_S^U \Big[ \frac{(r^{N-1}| (w_2^{1/p})'|^{p-2} (w_2^{1/p})')'}{w_2^{\frac{p-1}{p}}} -\frac{(r^{N-1} |(w_1^{1/p})'|^{p-2} (w_1^{1/p})')'}{w_1^{\frac{p-1}{p}}} \Big](w_1-w_2)dr. \end{align*} \end{lemma} \begin{lemma} \label{lm2.3} Assume $a < b$ and let $ u(\cdot,a), u(\cdot,b)$ be the corresponding solutions given by Lemma \ref{lm2.1}. Then $ u(\cdot,a) < u(\cdot,b)$ in $[0,T(a))$ and moreover $T(a) \leq T(b)$. \end{lemma} \begin{lemma} \label{lm2.4} Assume \eqref{e1.2}. Let $\{a_n\} $ be a sequence in $ (0, \infty)$ such that $a_n \nearrow a $ or $ a_n \searrow a$ for some $a>0$ and let $ u(\cdot, a_n), u(\cdot, a) $ be the solutions given by Lemma \ref{lm2.1}. If $ K \in (0, \min \{T(a), \sup_{n} T(a_n) \} )$ then \[ \lim_{n \to \infty} \|u(\cdot, a_n) - u(\cdot, a)\|_{_{C([0,K])}} = 0\quad\mbox{and}\quad \lim_{n \to \infty} |u'(r,a_n)- u'(r,a)| = 0, \quad r \in [0,K]. \] \end{lemma} Next we prove results established above. The proof of Lemma \ref{lm2.1} is fairly standard and is based on Banach's Fixed Point Theorem. However we present it in detail because several related notation will be used in the rest of the paper. \section{Proofs of the Lemmas} \begin{proof}[Proof of Lemma \ref{lm2.1}] Let $a>0$. Since $f \in C^{1}$ choose $\kappa_a > 1$ such that $f$ is Lipschitz continuous on $ [a/\kappa_a,a]$. Pick $\epsilon >0$ small enough, set $$ X_{a,\epsilon} := \big\{u \in C([0, \epsilon]) : u(0)=a, a/\kappa_a \leq u(r) \leq a, r \in [0,\epsilon] \big\}\,. $$ Note that $(X_{a,\epsilon}, \|\cdot\|_{\infty})$ is a complete metric space. We claim that \begin{equation} \mathcal{F}(X_{a,\epsilon}) \subset X_{a,\epsilon} ,\quad \|\mathcal{F} (u_1) - \mathcal{F}(u_2) \|_{C([0,\epsilon])} \leq k \| u_1 - u_2 \|_{C([0,\epsilon])} \label{e3.1} \end{equation} for all $u_1, u_2 \in X_{a,\epsilon}$ and for some $k \in (0,1)$. The proof of \eqref{e3.1} is left to an Appendix. Assuming \eqref{e3.1}, $\mathcal{F}$ has an only fixed point $u \in X_{a, \epsilon}$ and so \eqref{e2.1} has a unique local solution. Setting $$ T(a) := \sup \big \{r > 0 : \eqref{e2.1}\quad\mbox{ has an only solution in } [0,r] \big\} $$ and letting $u(\cdot,a):[0,T(a)) \to \mathbb{R}$ be the solution of \eqref{e2.1}, notice that by \eqref{e2.2}, $u(\cdot,a) \in C^{1}([0,T(a)))$ and \begin{equation} u'(r,a)=- \Big[ r^{1-N}\int_0^r s^{N-1}\rho(s) f(u(s,a))ds \Big] ^{\frac{1}{p-1}}, \quad 0 < r < T(a). \label{e3.2} \end{equation} Differentiating once more, one finds that $u \in C^{2}((0,T(a)))$. Assuming $T(a) < \infty$, we claim that $u(T(a),a)=0$. Indeed, if $u(T(a),a) := \tilde{a} > 0$, then $u(r,a) \geq \tilde{a}$ for $r \in [0,T(a))$. Estimating the integral in \eqref{e3.2} and using \eqref{e1.2}, \begin{equation} \begin{aligned} \int_0^r s^{N-1} \rho(s) f(u(s,a)) ds &\leq \frac{f(\tilde{a} )}{\tilde{a} ^{p-1}} a^{p-1} T(a)^{N-1} \int_0^r s^{N-1}\rho(s)ds \\ &\leq \frac{f(\tilde{a} )}{\tilde{a}^{p-1}} a^{p-1} \int_0^{T(a)} \rho(s)ds. \end{aligned} \label{e3.3} \end{equation} Using \eqref{e3.2} and \eqref{e3.3}, $\nu :=\lim_{r \nearrow T(a)} u'(r,a)$ is defined and $\nu \in (-\infty, 0]$. Consider the problem, \begin{equation} \begin{gathered} -\big(r^{N-1} |u'|^{p-2} u'\big)'=r^{N-1} \rho(r) f(u)\quad\mbox{in } (T(a),\infty), \\ u(T(a))=\tilde{a}, \quad u'(T(a))=\nu, \end{gathered} \label{e3.4} \end{equation} whose solutions are the fixed points of, $$ \widetilde{\mathcal{F}} u(r)=\tilde{a} - \int_{T(a)}^{r} \Big\{ t^{1-N} \Big[ T(a)^{N-1} |\nu|^{p-1} + \int_{T(a)}^{t} s^{N-1} \rho(s) f(u(s))ds \Big]\Big\}^{\frac{1}{p-1}}dt. $$ Setting, $$ X_{\tilde{a},\epsilon} := \big\{u \in C([T(a), T(a)+\epsilon]) | u(T(a))=\tilde{a}, {\tilde{a}} /{\kappa_{\tilde{a}}} \leq u(r) \leq \tilde{a}, r \in [T(a), T(a)+\epsilon] \big\}, $$ we infer that, (see Appendix), \begin{equation} \widetilde{\mathcal{F}}(X_{\tilde{a},\epsilon}) \subset X_{\tilde{a},\epsilon} , \quad \|\widetilde{\mathcal{F}} (u_1) - \widetilde{\mathcal{F}}(u_2) \|_{\infty} \leq k \| u_1 - u_2 \|_{\infty} \label{e3.5} \end{equation} where $u_1, u_2 \in X_{\tilde{a},\epsilon}$ and $k \in (0,1)$. By standard fixed point arguments again, one infers the existence of a unique solution of \eqref{e2.1} on some interval $[0, T(a) + \epsilon)$ contradicting the definition of $T(a)$. Hence, $u(\cdot,a) \in C([0,T(a)])$ and $u(a, T(a))=0$. \end{proof} \begin{proof}[Proof of Lemma \ref{lm2.2}] Motivated by D\'{\i}az \& Saa \cite{dsaa} let $J: L^{1}([0,T]) \to \mathbb{R} \cup \{\infty\}$, $$ J(w) := \begin{cases} \frac{1}{p}\int^U_S s^{N-1} \big|(w^{1/p})'\big|^{p}ds, & w \in X\\ \infty, & w \not\in X, \end{cases} $$ where $0 \leq S \leq U \leq T$. It is straightforward to check that $X$ and $J$ are both convex. Letting $w_1, w_2 \in X$, $\eta=w_1 - w_2$, remarking that $w_2 + t \eta$, $ w_1 - t \eta$ are in $X$, ($0 \leq t \leq 1$), and denoting by $\left $, the directional derivative of $J$ at $w$ in the direction $\zeta $, we claim that, \begin{equation} \begin{aligned} \langle J'(w_1), -\eta \rangle &=- \frac{1}{p}U^{N-1}|(w_1^{1/p}(U))'|^{p-2}(w_1^{1/p}(U))' w_1^{\frac{1-p}{p}}(U) \eta(U)\\ &\quad + \frac{1}{p}S^{N-1}|(w_1^{1/p}(S))'|^{p-2}(w_1^{1/p}(S))' w_1^{\frac{1-p}{p}}(S) \eta(S) \\ &\quad + \frac{1}{p} \int_S^U \frac{\big(s^{N-1}|(w_1^{1/p})'|^{p-2} (w_1^{1/p})'\big)'}{w_1^{\frac{p-1}{p}}} \eta(s) ds \end{aligned} \label{e3.6} \end{equation} and \begin{equation} \begin{aligned} \langle J'(w_2), \eta \rangle &= \frac{1}{p}U^{N-1}|(w_2^{1/p}(U))'|^{p-2} (w_2^{1/p}(U))' w_2^{\frac{1-p}{p}}(U)\eta(U) \\ &\quad - \frac{1}{p}S^{N-1}|(w_2^{1/p}(S))'|^{p-2} (w_2^{1/p}(S))' w_2^{\frac{1-p}{p}}(S)\eta(S) \\ &\quad - \frac{1}{p} \int_S^U \frac{(s^{N-1}|(w_2^{1/p})'|^{p-2} (w_2^{1/p})'\big)'}{w_2^{\frac{p-1}{p}}} \eta(s) ds. \end{aligned}\label{e3.7} \end{equation} We show \eqref{e3.6} next. Note that, $$ \langle J'(w_1), -\eta \rangle=\frac{1}{p} \lim_{s \to 0} \int^U_S s^{N-1} \Big[\frac{\big| \big( (w_1-s\eta)^{1/p} \big)' \big|^p-\big| (w_1^{1/p})' \big|^p}{s} \Big]ds. $$ By computing we find \begin{equation} \langle J'(w_1), -\eta \rangle= \lim_{s\to 0} \int_S^U s^{N-1}|\theta_s|^{p-2}\theta_s \Big[ \frac{((w_1 - s\eta)^{1/p})'-(w_1^{1/p})'}{s}\Big]ds, \label{e3.8} \end{equation} where $$ \min\left\{((w_1 - s\eta)^{1/p})', (w_1^{1/p})'\right\} \leq \theta_s \leq \max\left\{((w_1 - s\eta)^{1/p})', (w_1^{1/p})'\right\}. $$ Applying Lebesgue's Theorem to \eqref{e3.8} we infer that, $$ \langle J'(w_1), -\eta \rangle=-\frac{1}{p} \int_S^U s^{N-1}|(w_1^{1/p})'|^{p-2} (w_1^{1/p})' (w_1^{\frac{1-p}{p}}\eta)'ds. $$ Computing this integral we get to \eqref{e3.6}. The verification of \eqref{e3.7} follows by similar arguments. From \eqref{e3.6} and \eqref{e3.7}, \begin{align*} &\langle J'(w_2), \eta \rangle - \langle J'(w_1), \eta \rangle\\ &=\frac{1}{p}[ H(U) - H(S)]\\ &\; - \frac{1}{p} \int_S^U \Big[ \frac{(s^{N-1}| (w_2^{1/p})'|^{p-2} (w_2^{1/p})')'}{w_2^{\frac{p-1}{p}}} -\frac{(s^{N-1} |(w_1^{1/p})'|^{p-2} (w_1^{1/p})')'}{w_1^{\frac{p-1}{p}}}\Big](w_1-w_2)ds. \end{align*} Since $J$ is convex, $\langle J'(w_1) - J'(w_2), w_1 - w_2 \rangle \geq 0$ and Lemma \ref{lm2.2} follows. \end{proof} \begin{proof}[Proof of Lemma \ref{lm2.3}] Assume, on the contrary, $u(r,a) K$. By Lemma \ref{lm2.3} again, $$ T(a_{n_K}) \leq T(a_{n}) \leq T(a) \quad\mbox{and}\quad u(\cdot,a_{n_K}) \leq u(\cdot,a_n) \leq u(\cdot,a) \leq a, $$ for $n \geq n_K$, showing that $\{u(\cdot,a_n)\}_{n=1}^{\infty}$ is equibounded. We claim that it is also equicontinuous in $C([0,K])$. Indeed, estimating as in \eqref{e3.3} we find $$ |u'(r,a_n)|^{p-1} \leq \frac{f(u(K,a_{n_K}))}{u(K,a_{n_K})^{p-1}} a^{p- 1} \int_0^{K} \rho(s) ds := \widehat{K}. $$ Let $\theta_n \in (0,K)$ such that, $$ |u(r,a_n) - u(s,a_n)|=|u'(\theta_n,a_n)| |r-s| \leq \widehat{K}^{\frac{1}{p-1}} |r - s|. $$ Then $\{u(\cdot,a_n)\}_{n=1}^{\infty}$ is equicontinuous. By the Arz\'ela-\`Ascoli theorem there is some $\widetilde{u} \in C([0,K])$ such that, up to a subsequence, $u(\cdot,a_n) \to \widetilde{u}$ uniformly in $[0,K]$. We remark that $$ s^{N-1} \rho(s) f(u(s,a_n)) \to s^{N-1} \rho(s) f(\widetilde{u}(s)) $$ and $$ s^{N-1} \rho(s) f(u(s,a_n)) \leq \frac{f(u(K,a_{n_K}))}{u(K,a_{n_K})^{p-1}} a^{p-1} s^{N-1} \rho(s) $$ for $t \in [0,K]$. By Lebesgue's theorem, $$ \int_0^r s^{N-1} \rho(s) f(u(s,a_n)) ds \to \int_0^r s^{N-1} \rho(s) f(\widetilde{u}(s)) ds $$ for each $r \in [0,K]$. This and \eqref{e3.2} amount $$ u'(r,a_n) \to - \Big( r^{1-N}\int_0^r s^{N-1} \rho(s) f(\widetilde{u}(s)) ds \Big)^{\frac{1}{p-1}} := {\overline{u}}(r) $$ so that $\int_0^r u'(t,a_n) dt \to \int_0^r {\overline{u}}(t) dt$ and hence $$ \widetilde{u}(r) - a=\int_0^r {\overline{u}}(t) dt. $$ As a consequence, $$ |{\widetilde{u}}'(r)|^{p-2} {\widetilde{u}}'(r)=- r^{1-N} \int_0^r s^{N-1} \rho(s) f(\widetilde{u}(s)) ds. $$ Hence $\widetilde{u}$ is a solution of \eqref{e2.1} and by uniqueness, provided by Lemma \ref{lm2.1}, $\widetilde{u} := u(\cdot,a)$. It has finally been shown that $u(\cdot,a_n) \to u(\cdot,a)\quad\mbox{in } C([0,K])$ and $u'(\cdot,a_n) \to u'(\cdot,a)$ pointwise in $[0,K]$. The case $a_n \searrow a$ follows by similar arguments. Lemma \ref{lm2.4} is proved. \end{proof} \section{Proof of Theorem \ref{thm1.2}} By \eqref{e1.4} pick $S > 0$ such that $\int_0^S s^{N-1} \rho(s) ds > 0$. Take $R \geq 2S$ and consider the set, $$ \mathcal{A} :=\{ a > 0 : T(a) \geq R \}. $$ We claim that $\mathcal{A} \ne \phi$. Indeed, if $T(a) < R$ for all $a > 0$, by Lemma \ref{lm2.1}, $\lim_{r \to T(a)} u(r,a) = 0$ so that $u(r_a, a)=\frac{a}{2}$ for some $r_a \in (0,T(a))$. Estimating in \eqref{e2.2} and using \eqref{e1.2}, \begin{equation} \begin{aligned} \frac{1}{2} &\leq \int_0^{r_a}\Big[t^{1-N} \int_0^t s^{N-1} \rho(s) \frac{f(u(s,a))}{u(s,a)^{p-1}} ds \Big]^{\frac{1}{p-1}} dt\\ &\leq \big(\frac{f(\frac{a}{2})}{(\frac{a}{2})^{p-1}} \big)^\frac{1}{p-1} \int_0^R \Big[t^{1-N} \int_0^t s^{N-1} \rho(s) ds \Big]^{\frac{1}{p-1}}dt. \end{aligned}\label{e4.1} \end{equation} Making $a \to \infty$ leads to a contradiction by \eqref{e1.3}(ii), showing that $\mathcal{A} \ne \phi$. We claim that $A := \inf \mathcal{A}$ is positive. Indeed, if $A=0$, it follows by Lemma \ref{lm2.3} that $u(R,a) > 0$ for all $a > 0$. Since, $$ 2 ( u(R,a) - u(\frac{R}{2}, a) )=u'(\theta_a, a),\quad\mbox{for some } \theta_a \in (\frac{R}{2}, R), $$ and $u(R,a) \leq u(\frac{R}{2}, a) \leq a$ it follows using, $$ (\theta_{a})^{N-1} |u'(\theta_a, a)|^{p-2} u'(\theta_a, a)=- \int_0^{\theta_{a}} s^{N-1} \rho(s)f(u(s, a)) ds $$ that $$ \lim_{a\to 0} \int_0^{\theta_{a}} s^{N-1} \rho(s) f(u(s, a)) ds = 0. $$ Using Fatou's lemma and \eqref{e1.3} $$ 0=\liminf_{a \to 0} \int_0^{\theta_{a}} s^{N-1} \rho(s)f(u(s, a)) ds \geq \int_0^{R/2} s^{N-1} \rho(s) \liminf_{a \to 0} f(u(s, a)) ds > 0, $$ which is impossible, showing that $A > 0$. To finish the proof of Theorem \ref{thm1.2} it suffices to show that $T(A)=R$. If $T(A) < R$, pick both $\epsilon > 0$ such that $T(A) + \epsilon < R$ and a sequence $a_n \in \mathcal{A}$ with $a_n \searrow A$. Consider further, the sequence $u(T(A) + \frac{\epsilon}{2},a_n)$ which by Lemma \ref{lm2.3} is decreasing and set $T_{\epsilon,A} :={\inf_{n}} \{ u(T(A) + \frac{\epsilon}{2},a_n) \}$. We claim that $T_{\epsilon,A} > 0$. Otherwise, it follows remarking that $u(T(A) + \epsilon, a_n) \leq u(T(A) + \frac{\epsilon}{2}, a_n)$ and, $$ 2 \Big[u(T(A) + \epsilon, a_n) - u(T(A) + \frac{\epsilon}{2}, a_n) \Big] =u'(\theta_n, a_n) \epsilon $$ for some $\theta_n \in ( T(A) + \frac{\epsilon}{2}, T(A) + \epsilon )$ that $\lim_n u'(\theta_n, a_n) = 0$. Now, by arguments as above, $$ \lim_n \int_0^{T(A)} s^{N-1} \rho(s) f(u(s, a_n) ) ds = 0. $$ On the other hand, by Lemmas \ref{lm2.3} and \ref{lm2.4} we have, for each $K \in (0,T(a))$, $$ \int_0^{K} s^{N-1} \rho(s) f(u(s, a_n) ) ds \longrightarrow \int_0^{K} s^{N-1} \rho(s) f(u(s, A)) ds, $$ showing that $\rho=0$ a.e. in $(0,T(A))$. So, by \eqref{e3.2}, $u(r,A)=A$ for $r \in [0,T(A)]$, impossible, because we are assuming $T(A) < R$ and by Lemma \ref{lm2.1} $u(T(A),A)=0$. Therefore $T_{\epsilon,A} > 0$. Choose $\delta_0 > 0$ such that $u(r,A) < \frac{T_{\epsilon,A}}{4}$ for $r \in [T(A) -\delta_0, T(A) - \frac{\delta_0}{2}]$. By Lemma \ref{lm2.4}, $$ \lim_n \|u(\cdot, a_n) - u(\cdot, A) \|_{C([0,T(A) - \frac{\delta_o}{2}])} = 0 $$ and so there is $n_0 > 1$ such that $$ | u(r, a_{n_0}) - u(r, A) | < \frac{T_{\epsilon,A}}{4}, \quad r \in [0, T(A) - \frac{\delta_0}{2}]. $$ Thus, $$ u(r, a_{n_0}) \leq | u(r, a_{n_0}) - u(r, A) | + u(r, A) < \frac{T_{\epsilon,A}}{2}, \quad r \in [T(A) -\delta_0, T(A) - \frac{\delta_0}{2}]. $$ Since $u(r, a_n) \geq T_{\epsilon,A}$ for all $n > 1$ and $r \in [0, T(A) ]$, it follows that $$ u(T(A) - \delta_0, a_{n_0}) < \frac{T_{\epsilon,A}}{2} < T_{\epsilon,A} \leq u(T(A), a_{n_0}), $$ which is impossible. Therefore $A \in \mathcal{A}$. Now we claim that \begin{equation} T(A)= R. \label{e4.2} \end{equation} Indeed, pick a sequence $a_n \nearrow A$, $a_n \in \mathcal{A}^{c}$. By Lemma \ref{lm2.3}, $T(a_n) \leq T(a_{n+1}) \leq R$ and in fact $T(a_n) \nearrow T$ for some $T > 0$. Using Lemma \ref{lm2.3} again, $T \leq T(A)$. It will be shown that $T=T(A)$. Indeed, assume by the contrary, $T < T(A)$. Setting $T_{A} := u(T,A)$ it follows that $T_{A} > 0$. So, for each $n$ large take $s_n \in (0,T)$ satisfying $u(s_n,a_n)=\frac{T_A}{4}$. Since $u(\cdot,a_n)$ is nonincreasing, consider $\tilde{s}_n \in (0, s_n)$ such that $u(\tilde{s}_n,a_n)=\frac{T_A}{2}$. We will show next that $\tilde{s}_n \to T$. Indeed, by Lemma \ref{lm2.3}, $\tilde{s}_n$ is monotone so that $\tilde{s}_n \to \tilde{T} \leq T$. If $\tilde{T} < T$ there is $n_0 > 1$ such that $T(a_{n_0}) > \tilde{T}$. Hence $u(r,a_n) \leq \frac{T_{A}}{2}$ for $n \geq n_0$ and $r \in [\tilde{T},T(a_{n_0})]$, because otherwise, there would be some $r_{n_1} \in [\tilde{T}, T(a_{n_0})]$ with $\frac{T_A}{2} < u(r_{n_1}, a_{n_1}) \leq u(\tilde{s}_{n_1}, a_{n_1}) =\frac{T_A}{2}$, which is impossible. We infer that $|u(r,a_n) - u(r, A)| \geq \frac{T_A}{2}$ for all $n \geq n_0$, $r \in [\tilde{T}, \tilde{T} + \delta)$ and for some $\delta > 0$ such that $\tilde{T} + \delta < T(a_{n_0})$. But this is impossible again, because by Lemma \ref{lm2.4}, $$ \lim_n \|u(\cdot,a_n) - u(\cdot,A) \|_{C([0,\tilde{T} + \delta])} =0. $$ Therefore, $\tilde{T}=T$. Now, noticing that, $$ u(s_n,a_n) - u(\tilde{s}_n,a_n )=u'(\theta_n,a_n) (s_n - \tilde{s}_n), \quad \tilde{s}_n < \theta_n < s_n, $$ we find, $$ \lim_n | u'(\theta_n,a_n) |=\frac{T_{A}}{ 4 | s_n - \tilde{s}_n |} = \infty, $$ which is impossible, because estimating in \eqref{e3.2} as in \eqref{e3.3}, we get, $$ |u'(\theta_n, a_n)|^{p-1} \leq \frac{f(\frac{T_A}{4})}{(\frac{T_A}{4})^{p-1}} A^{p-1} \int_0^T \rho(s) ds. $$ So, $T= T(A)=R$ showing \eqref{e4.2}. By Lemma \ref{lm2.1}, $u(R,A)=0$. As a consequence, $u(\cdot,A) \in C([0,R])$. Further on, by \eqref{e3.2}, $u(\cdot,A) \in C^1([0,R)) \cap C^2((0,R))$. The arguments above give a radially symmetric solution $u$ of \eqref{e1.5}. This proves Theorem \ref{thm1.2}. \section{Proof of Theorem \ref{thm1.3}} Let $C_1, C_2, \dots$ denote several positive constants. Next, given $a > 0$, \begin{equation} w(r)=a - \int_0^r \Big[t^{1-N} \int_0^t s^{N-1} \rho(s) ds \Big]^{\frac{1}{p-1}} dt, \label{e5.1} \end{equation} is the unique solution of the problem \begin{equation} \begin{gathered} - \big(r^{N-1} |w'|^{p-2}w'\big)'=r^{N-1} \rho(r) \quad\mbox{in } (0,\infty),\\ w(0)= a, \quad w'(0)=0, \quad w > 0\quad\mbox{in } [0,\infty). \end{gathered} \label{e5.2} \end{equation} It will be shown that \begin{equation} I(r) := \int_0^r \Big[t^{1-N} \int_0^t s^{N-1} \rho(s) ds \Big]^{\frac{1}{p-1}} dt, \label{e5.3} \end{equation} has a finite limit as $r \to \infty$. Indeed, if $1 < p \leq 2 $, by estimating the integral in \eqref{e5.3}, $$ I(r) \leq C_1 + \int_1^r t^\frac{1-N}{p- 1} \Big[\int_0^t s^{N-1} \rho(s) ds \Big]^{\frac{1}{p-1}} dt. $$ Using the assumption $N \geq 3$ in the computation of the first integral above and Jensen's inequality to estimate the last one, $$ I(r) \leq C_2 + C_3 \int_1^r t^\frac{3-N-p}{p-1} \int_1^t s^{\frac{N-1}{p-1}} \rho(s)^{\frac{1}{p-1}} ds dt. $$ Computing the above integral above, we obtain $$ I(r) \leq C_2 + C_4 \int_1^r t^\frac{1}{p-1} \rho(t)^{\frac{1}{p-1}}\,dt. $$ Applying \eqref{e1.4} in the integral above we infer that $I(r)$ has a finite limit as $r \to \infty$. On the other hand, if $p\geq 2$, set $$ H(t) :=\int_0^t s^{N-1} \rho(s) ds $$ and note that either, $H(t)\leq 1$ for $t>0$ or $H(t_0)=1$ for some $t_0>0$. In the first case, $H(t)^{\frac{1}{p-1}}\leq 1$, and hence, $$ I(r)=\int_0^r t^\frac{1-N}{p-1} H(t)^{\frac{1}{p-1}} dt \leq C_5 + \int_1^r t^\frac{1-N}{p-1}dt $$ so that $I(r)$ has a finite limit because $p < N$. In the second case, $H(s)^{\frac{1}{p-1}}\leq H(s)$ for $s\geq s_0$ and hence, $$ I(r) \leq C_6 + \int_{1}^r t^{\frac{1-N}{p-1}} \int_0^t s^{N-1} \rho(s)ds dt. $$ Estimating and integrating by parts, we obtain \begin{align*} I(r) &\leq C_6 + C_7 \int _1^r t^{\frac{1-N}{p-1}}dt + {\frac{p-1}{N-p}} \Big[\int_1^r t^{\frac{(p-2)N +1}{p-1}} \rho(t)dt - r^\frac{p-N}{p-1} \int_0^r t^{N-1} \rho(t)dt \Big]\\ &\leq C_8 + C_{9} \int_1^r t^{\frac{(p-2)N + 1}{p-1}} \rho(t)dt. \end{align*} By \eqref{e1.4} (part 2), $I(r)$ converges to some real number. Taking in \eqref{e5.2}, $$ a :=\int_0^{\infty} \Big[ t^{1-N} \int_0^t s^{1-N} \rho(s)ds \Big]^{\frac{1}{p-1}}dt=\lim_{r \to \infty} I(r), $$ gives, $\lim_{r \to infty} w(r)= 0$. In what follows, an upper-solution to \eqref{e1.1} will be constructed. First, consider the function \begin{equation} \tilde{f_{p}}(t) :=(f(t)+1)^\frac{1}{p-1}, \quad t>0, \label{e5.4} \end{equation} and note that the items below hold true, \begin{equation} \begin{gathered} \tilde{f_{p}}(t)\geq f(t)^{\frac{1}{p-1}}>0,\\ \frac{\tilde{f_{p}}(t)}{t^{p-1}}\quad \mbox{is decreasing},\\ \lim_{t \to \infty}{\frac{\tilde{f_{p}}(t)}{t}}= 0. \end{gathered}\label{e5.5} \end{equation} We claim that \begin{equation} C_{p} a \leq \int_{0}^{C_{p}^{\frac{1}{p-1}}} \frac{t^{{p -1}}}{\tilde{f_{p}}(t)}dt, \label{e5.6} \end{equation} for some $C_{p} > 0$. Indeed, by \eqref{e5.5}(iii), $$ \lim_{r\to \infty}\int_0^r{\frac{t^{p-1}}{\tilde{f_{p}}(t)}} dt=\infty, $$ and thus, $$ \lim_{r \to \infty}{\frac{ \int_0^r{\frac{t^{p-1}}{\tilde{f_{p}}(t)}}dt} {r^{p-1}}}=\frac{1}{p-1}\lim_{r \to \infty}{\frac{r}{\tilde{f_{p}}(r)}}= \infty, $$ showing \eqref{e5.6}. Now set, for $s > 0$, $$ F_{p}(s) := \frac{1}{C_{p}} \int_0^s{\frac{t^{p-1}}{\tilde{f_p}(t)}} dt, $$ and notice that, $F_{p}(0)=0$ and $F_{p}$ is increasing. Using \eqref{e5.5}(iii) it follows that, $F(s) \buildrel {s \to \infty} \over \to \infty$. Applying the Implicit Function Theorem, \begin{equation} w(r) := \frac{1}{C_{p}}\int_0^{v(r)} \frac{t^{{p-1}}}{\tilde{f_p}(t)}dt \label{e5.7} \end{equation} for some $C^{2}((0,\infty))\cap C^{1}([0,\infty))$-function $v$. It will be shown next that $v$ is an upper-solution to \eqref{e1.1}. Indeed, since $v$ is nonincreasing, it follows by \eqref{e5.8}, \eqref{e5.7} and $w(0)=a$ that, $$ \int_0^{v(r)}{\frac{t^{p-1}}{\tilde{f_p}(t)}}dt \leq\int_0^{v(0)}{\frac{t^{p-1}}{\tilde{f_p}(t)}}dt=C_{p} w(0)=C_{p} a \leq\int_0^{C_{p}^{\frac{1}{p-1}}}{\frac{t^{p-1}}{\tilde{f_p}(t)}}dt, $$ so that, \begin{equation} v(r) \leq C_{p}^{\frac{1}{p-1}}, \quad t \geq 0. \label{e5.8} \end{equation} Differentiating in \eqref{e5.7} and computing, we get to \begin{align*} \big(r^{N-1} |w'(r)|^{p-2} w'(r)\big)' &= \big(\frac{1}{C_{p}} \big)^{p-1} \big(\frac{v^{p-1}}{\tilde{f_p}(v)}\big)^{p-1} \big(r^{N-1} |v'(r)|^{p-2} v'(r)\big)'\\ &\quad + (p-1) \big(\frac{1}{C_{p}}\big)^{p-1} \big(\frac{v^{p-1}}{\tilde{f_p}(v)} \big)^{p-2} \big( \frac{d}{dv} \big(\frac{v^{p-1}}{\tilde{f_p}(v)}\big) \big) r^{N-1} |v'|^{p}. \end{align*} Now, using \eqref{e5.5}(iii), \eqref{e5.8} and \eqref{e5.5}(i), it follows that $$ \big(r^{N-1} |v'(r)|^{p-2} v'(r)\big)'\leq - \big(\frac{C_{p}}{v^{p-1}}\big)^{p-1} \tilde{f_{p}}(v)^{p-1} r^{N-1} \rho(r) \leq - r^{N-1} \rho(r) f(v(r)). $$ Remarking that by \eqref{e5.7} $v'(0)=0$ and $\lim_{r\to\infty}v(r) = 0$ it follows that $v$ is a radially symmetric solution of \eqref{e1.6}. This ends the proof of Theorem \ref{thm1.3}. \section{Proof of Theorem \ref{thm1.1}} To show (i), pick an integer $j$ sufficiently large such that \eqref{e1.5} with $R= j+k$ has, by Theorem \ref{thm1.2}, a radially symmetric solution, say $ u_{k} \in C^1([0,j+k)) \cap C([0,j+k])$ for each integer $k \geq 1$. Consider the extension to $[0,\infty)$ of $u_k$, given by $u_k(r)=0$, if $r \geq j+k$. We claim that, \begin{equation} 0 \leq u_1 \leq u_2 \leq \dots \leq u_k \leq \dots \leq v. \label{e6.1} \end{equation} We will show first that $u_k \leq u_{k+1}$. Indeed, we claim that $u_k(0) \leq u_{k+1}(0)$. Otherwise, both $u_k(r)>u_{k+1}(r)$ for $r \in [0,T)$ and $u_k(T)=u_{k+1}(T)$ for some $T \in (0, j+k)$. Arguing as in the proof of Lemma \ref{lm2.3} with the use of Lemma \ref{lm2.2} we get to, $$ \frac{|u_{k}'|^{p-2}u_{k}'}{u_{k}^{p-1}} - \frac{|u_{k+1}'|^{p-2}u_{k+1}'} {u_{k+1}^{p-1}} \geq 0, $$ which gives, $\frac{u_k }{u_{k+1}}$ is nondecreasing in $(0,T)$, and as a consequence, $$ 1<\frac{u_k (0)}{u_{k+1}(0)} \leq \frac{u_k (T)}{u_{k+1}(T)}=1, $$ which is impossible. Hence, $u_k(0) \leq u_{k+1}(0)$. Now, if $u_k(r) > u_{k+1}(r)$ for $r \in (S,U)$, for some $S,U \in (0, j+k)$ with $S 0$, pick $k_0$ such that $j + k_0 \geq r+1$ and notice that by \eqref{e6.1}, $u_k \geq u_{k_0}$ for $k\geq k_0$. Recalling that $u_{k}'$ and $v'$ are nonpositive and using \eqref{e1.2} and \eqref{e6.1}, $$ t^{N-1} \rho(t) f(u_k(t)) \leq v(0)^{p-1} \frac{f(u_{k_0}(s))}{u_{k_0}(s)^{ p-1}} t^{N-1} \rho(t), \quad t \in [0,s]. $$ Since the last function above belongs to $L^1((0,s))$, by Lebegue's theorem, $$ \int_0^s{t^{N-1} \rho(t)f(u_k(t))dt}\to \int_0^s{t^{N-1} \rho(t)f(u(t))dt}, \quad s \in [0,r], $$ and employing, once more, arguments as above, $$ \int_0^r \Big[{s^{1-N}\int_0^s{t^{N-1} \rho(t)f(u_k(t))dt} } \Big]^{\frac{1}{p-1}}ds \to \int_0^r \Big[{s^{1-N}\int_0^s{t^{N-1} \rho(t)f(u(t))dt} } \Big]^{\frac{1}{p-1}}ds. $$ Passing to the limit in \eqref{e6.2} we infer that, $$ u(r)=u(0) - \int_0^r \Big[{s^{1-N}\int_0^s{t^{N-1} \rho(t)f(u(t))dt} } \Big]^{\frac{1}{p-1}}ds. $$ Remark that \begin{equation} u''(r)=- \frac{ h(r)}{p-1} \Big[ r^{1-N} \int_0^r t^{N-1} \rho(t)f(u(t))dt \Big]^{\frac{2-p}{p-1}}, \label{e6.3} \end{equation} where \begin{equation} h(r) :=\rho(r)f(u(r)) + (1-N)r^{-N} \int_0^r t^{N-1} \rho(t)f(u(t))dt. \label{e6.4} \end{equation} Hence, $u \in C^{1}([0,\infty)) \cap C^{2}((0,\infty))$. This together with the fact that $u \leq v$, shows (i), that is, $u$ is radially symetric solution of \eqref{e1.1}. To show (ii), assume, on the contrary, that \eqref{e1.1} has a solution $u$, so that $$ r^{N-1} |u'(r)|^{p-2} u'(r)\leq -C~~ \mbox{for}~~ r\geq M, $$ where $C,M>0$ are suitable constants. As a consequence, \begin{equation} u'(r)\leq -C r^{\frac{1-N}{p-1}}, \quad r\geq M. \label{e6.5} \end{equation} Integrating from $M$ to $r$ in \eqref{e6.5} and taking into account the cases $N < p$ and $N=p-1$ and at last making $r \to \infty$ we arrive at a contradiction. This finishes the proof of Theorem \ref{thm1.1}. \section{Comments on Remark \ref{rmk1}} At this point we justify the claim in Remark \ref{rmk1}. By \eqref{e6.4}, we get $$ \lim_{r \to 0} h(r)=\frac{1}{N} \rho(0)f(u(0)). $$ On the other hand, $$ \lim_{r \to 0}\Big[ r^{1-N} \int_0^r t^{N-1} \rho(t)f(u(t))dt \Big]=0. $$ Hence, by \eqref{e6.3} $ \lim_{r \to 0} u''(r)$ exists if and only if $p \leq 2$, that is $u \in C^{2}([0,\infty))$ if and only if $p \leq 2$. Now let $u,v$ be solutions of \eqref{e1.1}. By Lemma \ref{lm2.3} we can assume $u \geq v$. Let $w_1:= (v+b)^{p}$ and $w_2:= (u+b)^{p}$ and notice that $w_1, w_2 \in X$. Taking $r > 0$ and using Lemma \ref{lm2.2} and \eqref{e1.2}, as in the proof of Lemma \ref{lm2.3}, we find, $$ \frac{|u'|^{p-2} u'}{(u+b)^{p-1}}-\frac{|v'|^{p-2} v'}{(v+b)^{p-1}}\geq 0, $$ and since $u', v' \leq 0$, we infer that, $ \frac{u+b}{v+b}$ is nondecreasing in $(0,\infty)$, so that, \begin{align*} &\int_0^r \Big[t^{1-N} \int_0^t s^{N-1} \rho(s)f(u(s))ds \Big ]^{\frac{1}{p-1}}dt\\ & \leq \frac{u(r)+b}{v(r)+b} \int_0^r \Big[t^{1-N} \int_0^t s^{N-1} \rho(s)f(v(s))ds \Big ]^{\frac{1}{p-1}}dt \end{align*} By \eqref{e2.2}, the above inequality and the fact that $\lim_{r\to\infty}u(r)=0$, we find $$ 1 \leq \frac{u(0)}{v(0)}=\lim_{r\to \infty} \frac{ \int_0^r \Big[t^{1-N} \int_0^t s^{N-1} \rho(s)f(u(s))ds \Big ]^{\frac{1}{p-1}}dt } { \int_0^r \Big[t^{1-N} \int_0^t s^{N-1} \rho(s)f(v(s))ds \Big ]^{\frac{1}{p-1} }dt} \leq 1, $$ so that, by Lemma \ref{lm2.1}, $u=v$. \section{Appendix} Recall that $\epsilon$ represents a sufficiently small positive number. Let's proof \eqref{e3.1}(i) first. Pick $u \in C([0,\epsilon])$. Using \eqref{e1.2} to estimate the integral expression in $\mathcal{F}(u)$, we obtain for $r \in [0,\epsilon]$, \begin{align*} \widehat{I}(r) & := \int_0^r \Big[ t^{1-N} \int_0^t s^{N-1} \rho(s) f(u(s)) ds \Big]^{\frac{1}{p-1}} dt\\ &\leq a \Big(\frac {f(\frac{a}{\kappa_a})}{(\frac{a}{\kappa_a}) ^{p-1}}\Big)^{\frac{1}{p-1}} r \Big(\int_0^{r} \rho(t) dt \Big)^{\frac{1}{p-1}}. \end{align*} Therefore, $\mathcal{F}(u) \in C([0,\epsilon])$ and $\widehat{I}(\epsilon) < \frac{\kappa_a - 1}{\kappa_a} a$ and as a consequence, $\frac{a}{\kappa_a} \leq \mathcal{F}(u)(r) \leq a$, showing \eqref{e3.1}(i). Next we show \eqref{e3.1}(ii). Taking $u_j \in C([0,\epsilon])$, $j= 1,2$, we find, $$ |\mathcal{F}u_1(r) - \mathcal{F} u_2(r)| \leq \int_0^r \Big|\big[ X_{u_{1}}(t) \big]^{\frac{1}{p-1}} - \big[ X_{u_{2}}(t) \big]^{\frac{1}{p-1}}\Big| dt, $$ where $$ X_{u_{j}}(t) :=t^{1-N} \int_0^t s^{N-1} \rho(s) f(u_j(s)) ds. $$ Using the inequality, \begin{equation} \big| |x|^{\sigma} x - |y|^{\sigma} y \big| \leq C_{\sigma} ( |x|^{\sigma} + |y|^{\sigma} ) |x - y| \quad x,y \in \mathbb{R} \label{e8.1} \end{equation} where ${\sigma} > -1$ and $C_{\sigma} > 0$ are constants, we find, \begin{equation} |\mathcal{F}u_1(r) - \mathcal{F} u_2(r)| \leq C_{\sigma} \int_0^r ( |X_{u_{1}}(t)|^{\sigma} + |X_{u_{2}}(t)|^{\sigma} ) |X_{u_{1}}(t) - X_{u_{2}}(t)| dt, \label{e8.2} \end{equation} where $\sigma=(2-p)/(p-1)$. We point out that \begin{equation} \begin{aligned} |X_{u_{1}}(t) - X_{u_{2}}(t)| &\leq t^{1-N} \int_0^t s^{N-1} \rho(s) |f(u_1(s)) -f(u_2(s))| ds \\ &\leq K \|u_1 - u_2 \|_{C([0,\epsilon])} t^{1-N} \int_0^t s^{N-1} \rho(s) ds. \end{aligned} \label{e8.3} \end{equation} where $K$ is the Lipschitz constant of $f$ on $[\frac{a}{\kappa_a},a]$. If $1 < p \leq 2$, using \eqref{e1.2}, \begin{equation} |X_{u_{j}}(t)|^{\sigma} \leq a^{2-p} \Big[ \frac{f(\frac{a}{\kappa_a})}{(\frac{a}{\kappa_a})^{p-1}} \Big]^{\sigma} \Big(t^{1-N} \int_0^t s^{N-1} \rho(s) ds \Big)^{ \sigma}. \label{e8.4} \end{equation} From \eqref{e8.2}, \eqref{e8.3}, and \eqref{e8.4} we find, for constant a $\widehat{K}> 0$, $$ |\mathcal{F}u_1(r) - \mathcal{F} u_2(r)| \leq \widehat{K} \epsilon \Big( \int_0^{\epsilon} \rho(s) ds \Big)^{\frac{1}{p - 1}} \|u_1 - u_2 \|_{C([0,\epsilon])}, $$ and so \eqref{e3.1}(ii) follows. The case $p > 2$ is treated as the earlier one, replacing \eqref{e8.4} by, $$ |X_{u_{j}}(t)|^{\sigma} \leq \big(\frac{a}{\kappa_a} \big)^{2-p} \Big[ \frac{f(a)}{a^{p-1}} \Big]^{\sigma} \Big(t^{1-N} \int_0^t s^{N-1} \rho(s) ds \Big)^{\sigma}. $$ This shows \eqref{e3.1}. To prove \eqref{e3.5}, we show \eqref{e3.5}(i) first. To that end let $u \in X_{\tilde{a}, \epsilon}$. Using \eqref{e1.2} we estimate the integral below, \begin{align*} &\int_{T(a)}^r \Big\{ t^{1-N} \Big[ T(a)^{N-1} |\nu |^{p-1} + \int_{T(a)}^t s^{N-1} \rho(s)f(u(s))ds \Big] \Big \}^{\frac{1}{p - 1}}dt\\ & \leq \int_{T(a)}^{\epsilon} \Big\{ t^{1-N}\Big[ T(a)^{N-1} |\nu |^{p-1} + \kappa_{\tilde{a}}^{p-1} f(\frac{\tilde{a}}{\kappa_{\tilde{a}}}) \int_{T(a)}^t s^{N-1} \rho(s)ds \Big] \Big \}^{\frac{1}{p-1}}dt\\ & \leq \frac{\kappa_{\tilde{a}} - 1}{\tilde{a}}. \end{align*} Hence, $\frac{\tilde{a}}{\kappa_{\tilde{a}}} \leq \widetilde{\mathcal{F}}u(r) \leq \tilde{a}$, for $r \in [T(a), T(a)+\epsilon]$. This shows \eqref{e3.5}(i). In order to prove \eqref{e3.5}(ii), letting $u_j \in X_{\tilde{a}, \epsilon}$ ($ j=1,2$) and using \eqref{e8.1}, \begin{equation} |\widetilde{\mathcal{F}}u_1(r) - \widetilde{\mathcal{F}} u_2(r)| \leq C_{\sigma} \int_{T(a)}^r ( | \widetilde{X}_{u_1}(t)|^{\sigma} + |\widetilde{X}_{u_2}(t)|^{\sigma} ) | \widetilde{X}_{u_1}(t) - \widetilde{X}_{u_2}(t)| dt, \label{e8.5} \end{equation} where $$\widetilde{X}_{u_j}(t) := t^{1-N}\Big[ T(a)^{N-1}|\nu |^{p-1} + \int_{T(a)}^t s^{N-1} \rho(s)f(u_j(s)) ds \Big]. $$ We remark that since $f \in C^{1}$, \begin{equation} | \widetilde{X}_{u_1}(t) - \widetilde{X}_{u_2}(t)| \leq K_{0} \|u_1 - u_2 \|_{C([T(a),T(a)+\epsilon])} t^{1-N} \int_{T(a)}^t s^{N-1}\rho(s)ds, \label{e8.6} \end{equation} where $K_{0}$ is the Lipschitz constant of $f$ on the interval $[\frac{\tilde{a}}{\kappa_{\tilde{a}}},\tilde{a}]$. Now, two further cases are considered. The first one is $\nu=0$. If $1 < p \leq 2$, using \eqref{e1.2}, \begin{equation} |\widetilde{X}_{u_j}(t) |^{\sigma} \leq \kappa_{\tilde{a}}^{2-p } f(\frac{\tilde{a}}{\kappa_{\tilde{a}}})^{\sigma} \Big[t^{1-N} \int_{T(a)}^t s^{N-1} \rho(s)ds\Big]^{\sigma}. \label{e8.7} \end{equation} By \eqref{e8.5},\eqref{e8.6} and \eqref{e8.7}, it follows that, for some constant $C>0$, \begin{align*} &|\widetilde{\mathcal{F}} u_1(r) - \widetilde{\mathcal{F}} u_2(r) | \\ &\leq C \|u_1 - u_2 \|_{C([T(a),T(a)+\epsilon])} \int_{T(a)}^{T(a) + \epsilon} \Big[ t^{1-N} \int_{T(a)}^t s^{N-1} \rho(s)ds \Big]^{\frac{1}{p-1}} \\ &\leq K_{1} \|u_1 - u_2 \|_{C([T(a),T(a)+\epsilon])}, \end{align*} where $K_{1} \in (0,1)$, showing \eqref{e3.5}(ii). If $p \geq 2$, using \eqref{e1.2} again, one obtains, \begin{equation} |\widetilde{X}_{u_j}(t) |^{\sigma} \leq (\frac{1}{\kappa_{\tilde{a}}})^{2-p} f(\tilde{a})^{\sigma} \Big[t^{1-N} \int_{T(a)}^t s^{N-1} \rho(s)ds\Big]^{\sigma}. \label{e8.8} \end{equation} Argueing as before, with \eqref{e8.8} instead \eqref{e8.7} we show \eqref{e3.5}(ii). The second case is $\nu <0$. If $1 < p \leq 2$, we get by using \eqref{e1.2}, \begin{equation} |\widetilde{X}_{u_j}(t) |^{\sigma} \leq \Big[T(a)^{- 1}|\nu |^{p-1} + T(a)^{-1}f(\frac{\tilde{a}}{\kappa_{\tilde{a}}}) \int_{T(a)}^{T(a)+ \epsilon}\rho(s)ds \Big]^{\sigma}t^{\frac{2-p}{p-1}}. \label{e8.9} \end{equation} On the other hand, if $p \geq 2$, we get \begin{equation} |\widetilde{X}_{u_j}(t) |^{\sigma} \leq \Big[ \frac{T(a)^{N-1}|\nu|^{p-1}}{(T(a) + \epsilon)^N}\Big]^{\sigma} t^{\frac{2-p}{p - 1}}. \label{e8.10} \end{equation} Proceeding as above, by replacing respectively \eqref{e8.7} and \eqref{e8.8} by \eqref{e8.9} and \eqref{e8.10}, we show \eqref{e3.5}(ii). This completes the verification of \eqref{e3.5}. \begin{thebibliography}{99} \bibitem{CRadulesco} F. Cirstea \& V. Radulescu, {\it Existence and uniqueness of positive solutions to a semilinear elliptic problem in $R^N$}, J. Math. Anal. Appl. 229 (1999) 417-425. \bibitem{dsaa} J.I. D\'{\i}az \& J. E. 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