\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 57, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/57\hfil Existence of solutions] {Existence of solutions for a system of elliptic partial differential equations} \author[R. Dalmasso\hfil EJDE-2011/57\hfilneg] {Robert Dalmasso} \address{Robert Dalmasso \newline Laboratoire Jean Kuntzmann, Equipe EDP \\ 51 rue des Math\'ematiques, Domaine universitaire, BP 53\\ 38041 Grenoble Cedex 9, France} \email{robert.dalmasso@imag.fr} \thanks{Submitted April 25, 2011. Published May 4, 2011.} \subjclass[2010]{35J57, 34B18} \keywords{Nonlinear systems; shooting method; positive radial solutions} \begin{abstract} In this article, we establish the existence of radial solutions for a system of nonlinear elliptic partial differential equations with Dirichlet boundary conditions. Also we discuss the question of uniqueness, and illustrate our results with examples. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} In this article, we study the existence for non-trivial solutions $(u,v) \in (C^2(\overline{B_R}))^2$ of the boundary-value problem $$\begin{gathered} \Delta u = g(v) \quad \text{in } B_R \,,\\ \Delta v = f(u) \quad \text{in } B_R\,,\\ u = \frac{\partial u}{\partial \nu} = 0 \quad \text{on }\partial B_R \,, \end{gathered} \label{eq1}$$ where $B_R$ denotes the open ball of radius $R$ centered at the origin in $\mathbb{R}^n$ ($n \geq 1$), $\partial/\partial\nu$ is the outward normal derivative and $f$, $g$ satisfy the following hypotheses: \begin{itemize} \item[(H1)] $f$, $g: \mathbb{R} \to \mathbb{R}$ are $C^1$ functions; \item[(H2)] $g$ is increasing on $(0,\infty)$, $g' > 0$ on $(-\infty,0)$, $g(0) = 0$ and $\lim_{v\to -\infty}g(v)= -\infty$; \item[(H3)] $f,f' > 0$ on $(0,\infty)$. \end{itemize} Now we state our main results. \begin{theorem} \label{thm1} Let $f$, $g$ satisfy {\rm (H1)--(H3)}. Assume moreover that \begin{itemize} \item[(H4)] There exist $m$, $M > 0$ such that $m \leq f(u) \leq M$ for all $u \geq 0$. \end{itemize} Then \eqref{eq1} has at least one radial solution $(u,v) \in (C^2(\overline{\!B}_R))^2$. \end{theorem} \begin{theorem} \label{thm2} Let $f$, $g$ satisfy {\rm (H1)--(H3)}. Assume moreover that \begin{itemize} \item[(H5)] $f(0) > 0$; \item[(H6)] There exist $a$, $b$, $a'$, $b'$, $p > 0$ and $q \geq 1$ such that $pq < 1$, \begin{gather*} f(u) \geq au^p \quad \forall \, u \geq 0 \,,\quad f(u) \leq a'u^p \quad \forall \, u \geq 1 \,,\\ b|v|^q \leq |g(v)| \leq b'|v|^q \quad \forall \, v \in \mathbb{R}\,. \end{gather*} \end{itemize} Then \eqref{eq1} has at least one radial solution $(u,v) \in (C^2(\overline{\!B}_R))^2$. \end{theorem} When $n = 1$ we have the following result. \begin{theorem} \label{thm3} Assume that $n = 1$. Let $f$, $g$ satisfy {\rm (H1)--(H3)}. Assume moreover that \begin{itemize} \item[(H7)] There exist $a$, $a'$, $b$, $b' > 0$ and $p$, $q \geq 1$ such that $pq > 1$, \begin{gather*} au^{p} \leq f(u) \leq a'u^p \quad \forall \, u \geq 0\,,\\ b|v|^q \leq |g(v)| \leq b'|v|^q \quad \forall \, v \in \mathbb{R}\,, \end{gather*} \end{itemize} Then \eqref{eq1} has at least one non-trivial symmetric solution $(u,v) \in (C^2([- R,R])^2$. \end{theorem} \begin{remark} \label{rmk1}\rm Assumptions (H2) and (H4) (resp. (H5)) imply that if a solution $(u,v) \in (C^2(\overline{\!B}_R))^2$ of \eqref{eq1} exists, then $u \neq 0$ and $v \neq 0$. \end{remark} The particular case of homogeneous nonlinearities $f(u) = |u|^p$, $g(v) = |v|^{q - 1}v$ with $p$, $q > 0$ has been studied in \cite{da1}. When $n = 1$ and $g(v) = v$ the uniqueness of a non-trivial solution was proved in \cite{da2} for $f : \mathbb{R} \to [0,\infty)$ $C^1$ and satisfying the following condition: $$0 < f(u) < uf'(u) \quad \text{for } u > 0 \,.$$ An existence result was also given. Since we are interested in radial solutions the problem reduces to the one-dimensional (singular if $n \geq 2$) boundary-value problem $$\begin{gathered} \Delta u = g(v) \quad \text{in } [0,R) \,,\\ \Delta v = f(u) \quad \text{in } [0,R)\,,\\ u(R) = u'(R) = u'(0) = v'(0) = 0 \,, \end{gathered}\label{eq2}$$ where $\Delta$ denotes the polar form of the Laplacian $$\Delta = r^{1-n}\frac{d}{dr}(r^{n-1}\frac{d}{dr})\,.$$ In Section 2 we give some preliminary results. Theorems \ref{thm1} and \ref{thm2} are proved in Section 3. We also give two other existence theorems. Theorem \ref{thm3} is proved in Section 4. We examine the uniqueness question in Section 5. Finally in Section 6 we give some examples. \section{Preliminaries} Throughout this section we assume that $f$ and $g$ satisfy (H1)--(H3). When $n \geq 3$ we also assume that $f$ verifies (H5) or that $f$ and $g$ satisfy the following conditions: \begin{itemize} \item[(H8)] There exist $a$, $b$, $p$, $q > 0$ such that \begin{gather*} f(u) \geq au^p \quad \forall \, u \geq 0 \,,\quad |g(- v)| \geq bv^q \quad \forall \, v \geq 0\,, \\ \frac{1}{p + 1} + \frac{1}{q + 1} > \frac{n - 2}{n} \,. \end{gather*} \end{itemize} Notice that when $pq \leq 1$ we have $\frac{1}{p + 1} + \frac{1}{q + 1} \geq 1 \,.$ We will use a two-dimensional shooting argument as in \cite{da1}. Let $\alpha$, $\beta > 0$. We introduce the inital value problem $$\begin{gathered} \Delta u(r) = g(v(r))\,, \quad r \geq 0 \,,\\ \Delta v(r) = f(u(r)) \,,\quad r \geq 0\,,\\ u(0) = \alpha \,, \quad v(0) = - \beta \,, \quad u'(0) = v'(0) = 0\,. \end{gathered}\label{eq3}$$ \begin{lemma} \label{lem1} Let $\alpha$, $\beta > 0$ be fixed. If $(u,v) \in (C^2([0,\infty))^2$ is a solution of \eqref{eq3} such that $uu' < 0$ on $(0,\infty)$, then $v < 0$ on $(0,\infty)$. \end{lemma} \begin{proof} We have $0 < u \leq \alpha$ on $[0,\infty)$. Therefore, by (H3), $$r^{n - 1}v'(r) = \int_0^r{s^{n - 1}f(u(s))\, ds} \, > \, 0 \quad \text{for } r > 0 \,. \label{eq4}$$ Assume that the conclusion of the lemma is false. Then \eqref{eq4} implies that there exist $a$, $b > 0$ such that $$v(r) \geq a \quad \text{for } r \geq b \,.$$ With the help of (H2) we deduce that $$(r^{n - 1}u'(r))' \geq g(a) r^{n - 1} \quad \text{for } r \geq b\,,$$ hence $$r^{n - 1}u'(r) \geq g(a)\frac{r^n - b^n}{n} + b^{n - 1}u'(b)\quad \text{for } r\geq b\,,$$ which implies, using (H2) again, that $u'(r) > 0$ for $r$ large and we reach a contradiction. \end{proof} Now we define the functions $F$, $G$ and $G_n$ by \begin{gather*} F(u) = \int_0^u{f(s)\, ds}\,, \quad G(v) = \int_0^v{g(s)\, ds}\, \quad u, v \in \mathbb{R}\,,\\ G_n(r,s) = \begin{cases} r - s & \text{if } n = 1\,,\\ s\ln(\frac{r}{s}) & \text{if } n = 2\,,\\ \frac{s}{n - 2}(1 - (\frac{s}{r})^{n - 2}) & \text{if } n \geq 3\,. \end{cases} \end{gather*} \begin{lemma} \label{lem2} Let $\alpha$, $\beta > 0$ be fixed. Assume that for some $a > 0$, $(u,v) \in (C^2([0,a])^2$ is a solution of \eqref{eq3} on $[0,a]$ such that $uu' < 0$ on $(0, a)$. Then $$|v(r)| \leq \max(\beta, G^{- 1}(F(\alpha) + G(- \beta))) \,, \quad 0 \leq r \leq a \,,$$ where $G^{- 1}$ denotes the inverse of $G : [0, \infty) \to [0, \infty)$. \end{lemma} \begin{proof} We have $0 < u \leq \alpha$ on $[0,a)$. As in Lemma \ref{lem1} we deduce that $v' > 0$ on $(0,a]$. We have $$\int_0^r{(v'\Delta u + u'\Delta v)\, ds} = \int_0^r{(g(v)v' + f(u)u')\, ds}\,,$$ for $r \in [0,a]$. Since \begin{align*} \int_0^r{(v'\Delta u + u'\Delta v)\, ds} &= \int_0^r{(u'v')'\, ds} + 2(n - 1)\int_0^r{\frac{u'(s)v'(s)}{s}\, ds}\\ &= u'(r)v'(r) + 2(n -1)\int_0^r{\frac{u'(s)v'(s)}{s}\, ds} \,, \end{align*} and $$\int_0^r{(g(v)v' + f(u)u')\, ds} = G(v(r)) + F(u(r)) - G(-\beta) - F(\alpha)\,,$$ we obtain $$F(u(r)) + G(v(r)) = F(\alpha) + G(-\beta(\alpha)) + u'(r)v'(r) + 2(n - 1)\int_0^r{\frac{u'(s)v'(s)}{s}\, ds}\,,$$ for $r \in[0,a]$, which implies that $$G(v(r)) \leq F(\alpha) + G(- \beta) \, \quad 0 \leq r \leq a \,,$$ and the lemma follows. \end{proof} \begin{lemma} \label{lem3} For each $\alpha > 0$, $\beta > 0$ there exists $T > 0$ such that problem \eqref{eq3} on $[0,T]$ has a unique solution $(u,v) \in (C^2[0,T])^2$ such that $u > 0$ (resp. $v < 0$) in $[0,T]$ and $u' < 0$ (resp. $v' > 0$) in $(0,T]$. \end{lemma} \begin{proof} Let $\alpha$, $\beta > 0$ be given. Choose $T =T(\alpha,\beta) > 0$ such that $$T = \min\Big( \bigl(\frac{n\alpha}{-g(-\beta)}\bigr)^{1/2}, \bigl(\frac{n\beta}{f(\alpha)}\bigr)^{1/2}\Big) \,,$$ and consider the set of functions \begin{align*} Z = \{&(u,v) \in (C[0,T])^2; \, \alpha/2 \leq u(r) \leq \alpha \text{ and}\\ &- \beta \leq v(r) \leq - \beta/2 \text{ for } 0 \leq r \leq T\}\,. \end{align*} Clearly $Z$ is a bounded closed convex subset of the Banach space $(C[0,T])^2$ endowed with the norm $\|(u,v)\| = \max (\|u\|_{\infty},\|v\|_{\infty})$. Define $$L(u,v)(r) = (\alpha + \int_0^r{G_n(r,s)g(v(s))\, ds}, - \beta + \int_0^r{G_n(r,s)f(u(s))\, ds}) \,,$$ for $r \in [0,T]$ and $(u,v) \in (C[0,T])^2$. It is easily verified that $L$ is a compact operator mapping $Z$ into itself, and so there exists $(u,v)\in Z$ such that $(u,v) = L(u,v)$ by the Schauder fixed point theorem. Clearly $(u,v) \in (C^2[0,T])^2$ and $(u,v)$ is a solution of \eqref{eq3} on $[0,T]$. Since $f$ and $g$ are $C^1$ the uniqueness follows. Since $u > 0$ and $v < 0$ in $[0,T]$, direct integration of the system \eqref{eq3} implies that $u' < 0$ and $v' > 0$ in $(0,T]$. By Lemma \ref{lem3} for any $\alpha$, $\beta > 0$ problem \eqref{eq3} has a unique local solution: Let $[0,R_{\alpha,\beta})$ denote the maximum interval of existence of that solution ($R_{\alpha,\beta} = \infty$ possibly). Define $P_{\alpha,\beta} = \{s \in (0,R_{\alpha,\beta})\, ;\, u(\alpha,\beta, r)u'(\alpha,\beta, r) < 0 \, \forall \, r \in (0,s]\} \,$ where $(u(\alpha,\beta, .),v(\alpha,\beta, .))$ is the solution of \eqref{eq3} in $[0,R_{\alpha,\beta})$. $P_{\alpha,\beta} \neq \emptyset$ by Lemma \ref{lem3}. Set $r_{\alpha,\beta} = \sup P_{\alpha,\beta} \,.$ \end{proof} \begin{lemma} \label{lem4} We have $u'(\alpha,\beta, r) < 0$ for $r \in (0,r_{\alpha,\beta})$ and $v'(\alpha,\beta, r) > 0$ for $r \in (0,r_{\alpha,\beta}]$. \end{lemma} \begin{proof} The first assertion follows from the definition of $r_{\alpha,\beta}$. Since $u(\alpha,\beta, r) > 0$ for $r \in [0,r_{\alpha,\beta})$, integrating the second equation in \eqref{eq3} from $0$ to $r \in (0,r_{\alpha,\beta}]$ we obtain $v'(\alpha,\beta, r) > 0$ for $r \in (0,r_{\alpha,\beta}]$. \end{proof} \begin{lemma} \label{lem5} For any $\alpha$, $\beta > 0$ we have $r_{\alpha,\beta} < R_{\alpha,\beta}$. \end{lemma} \begin{proof} If not, there exist $\alpha$, $\beta > 0$ such that $r_{\alpha,\beta} = R_{\alpha,\beta}$. Suppose first that $R_{\alpha,\beta} < \infty$. Noting $u = u(\alpha,\beta,.)$ and $v = v(\alpha,\beta,.)$ we have $0 < u \leq \alpha$ in $[0, R_{\alpha,\beta})$. Then we easily deduce that $u$, $u'$, $v$ and $v'$ are bounded on $[0, R_{\alpha,\beta})$ and we obtain a contradiction with the definition of $R_{\alpha,\beta}$. Now assume that $R_{\alpha,\beta} = \infty$. We have $0 < u \leq \alpha$ in $[0,\infty)$. By Lemma \ref{lem1} $v < 0$ in $[0,\infty)$. When $n = 1$ (H2) implies that $u'' < 0$ in $[0,\infty)$ and we deduce that $u'(r) \leq u'(1) < 0$ for all $r \geq 1$, from which we obtain $u(r) \leq u(1) + u'(1)(r - 1)$ for all $r \geq 1$. Thus we can find $r > 1$ such that $u(r) < 0$ and we have a contradiction. If $n = 2$, (H2) implies that $(ru'(r))' < 0$ on $(0, \infty)$. We deduce that $ru'(r) \leq u'(1) < 0$ for all $r \geq 1$, from which we obtain $u(r) \leq u(1) + u'(1)\ln r$ for all $r \geq 1$. Thus we can find $r \geq 1$ such that $u(r) < 0$ and we obtain a contradiction. Now let $n \geq 3$. Suppose first that $f$ satisfies (H5). From the second equation in \eqref{eq3}, using (H3), we obtain $v(r) \geq - \beta + \frac{f(0)}{2n}r^2 \quad \forall \, r \geq 0 \,,$ which implies, with the help of (H5), that $v(r) \to \infty$ as $r \to \infty$ and we have a contradiction. Now suppose that $f$ verifies (H8). Let $z = - v$. We have $u$, $z > 0$ on $[0,\infty)$ and, by (H8), $- \Delta u \geq bz^q \,, \quad - \Delta z \geq au^p \quad \text{on } [0, \infty)\,.$ Since $pq < 1$ we have $\frac{1}{p + 1} + \frac{1}{q + 1} > \frac{n - 2}{n} \,.$ Then we obtain a contradiction with the help of the nonexistence results established in \cite{mi}-\cite{sz2}. \begin{proposition} \label{prop1} For each $\alpha > 0$, there exists a unique $\beta > 0$ such that $u(\alpha,\beta,r_{\alpha,\beta}) = u'(\alpha,\beta,r_{\alpha,\beta}) = 0\,.$ \end{proposition} \begin{proof} We first prove the uniqueness. Let $\alpha > 0$ be fixed. Suppose that there exist $\beta > \gamma > 0$ such that $u(\alpha,\beta, r_{\alpha,\beta}) = u'(\alpha,\beta, r_{\alpha,\beta}) = u(\alpha,\gamma, r_{\alpha,\gamma}) = u'(\alpha,\gamma, r_{\alpha,\gamma}) = 0$. In order to simplify our notations we denote $u(\alpha,\beta, r)$, $u(\alpha,\gamma, r)$, $v(\alpha,\beta, r)$ and $v(\alpha,\gamma, r)$ by $u(r)$, $w(r)$, $v(r)$ and $z(r)$. Define $b = \min(r_{\alpha,\beta},r_{\alpha,\gamma})$. Suppose that there exists $a \in(0,b]$ such that $v - z < 0$ in $[0,a)$ and $(v - z)(a) = 0$. Using (H2) we obtain $\Delta(u - w) = g(v) - g(z) < 0$ in $[0,a)$. Since $(u - w)(0) = (u - w)'(0) = 0$, we deduce that $u - w < 0$ in $(0,a]$. Using (H3) we obtain $\Delta(v - z) = f(u) - f(w) < 0$ in $(0,a]$. We have $(v - z)(0) < 0$, $(v - z)'(0) = 0$ and $(v - z)(a) = 0$. Therefore we reach a contradiction. Thus $v - z < 0$ in $[0,b]$. As before we show that $u - w < 0$ in $(0,b]$. Since $(u - w)'(0) = 0$ we deduce that $(u - w)'(b) < 0$. By Lemma \ref{lem4} we have $(u - w)'(b) = \begin{cases} u'(r_{\alpha,\gamma}) < 0 &\text{if } r_{\alpha,\beta} > r_{\alpha,\gamma}\,,\\ 0 &\text{if } r_{\alpha,\beta} = r_{\alpha,\gamma}\,,\\ - w'(r_{\alpha,\beta}) > 0 &\text{if } r_{\alpha,\beta} < r_{\alpha,\gamma}\,.\\ \end{cases}$ Therefore, $b = r_{\alpha,\gamma} < r_{\alpha,\beta}$. Now $(u - w)(b) = u(r_{\alpha,\gamma}) > 0$ and we obtain a contradiction. The case $0 < \beta < \gamma$ can be handled in the same way. Now we prove the existence. Suppose that there exists $\alpha > 0$ such that for any $\beta > 0$ $u(\alpha,\beta, r_{\alpha,\beta}) > 0$ or $u'(\alpha,\beta, r_{\alpha,\beta}) < 0$. Define the sets \begin{gather*} B = \{\beta > 0 ; \, u(\alpha,\beta, r_{\alpha,\beta}) = 0 \text{ and } u'(\alpha,\beta, r_{\alpha,\beta}) < 0\} \,, \\ C = \{\beta > 0 ; \, u(\alpha,\beta, r_{\alpha,\beta}) > 0 \text{ and } u'(\alpha,\beta, r_{\alpha,\beta}) = 0\} \,. \end{gather*} The proof of the proposition is completed by using the next two lemmas which contradict the fact that $(0,\infty) = B\cup C$. \end{proof} \begin{lemma} \label{lem6} \begin{itemize} \item[(i)] If $B \neq \emptyset$, then $\inf B > 0$. \item[(ii)] If $C \neq \emptyset$, then $\sup C < \infty$. \end{itemize} \end{lemma} \begin{lemma} \label{lem7} Sets $B$ and $C$ are open. \end{lemma} \begin{proof}[Proof of Lemma \ref{lem6}] We have \begin{gather} u(\alpha,\beta, r) = \alpha + \int_0^r{G_n(r,s)g(v(\alpha,\beta, s))\, ds} \,,\quad 0 \leq r < R_{\alpha,\beta} \,, \label{eq5} \\ v(\alpha,\beta, r) = - \beta + \int_0^r{G_n(r,s)f(u(\alpha,\beta, s))\, ds} \,,\quad 0 \leq r < R_{\alpha,\beta} \,. \label{eq6} \end{gather} (i) Let $\beta \in B$. Assume first that $v(\alpha,\beta,.) < 0$ on $[0,r_{\alpha,\beta})$. Then Lemma \ref{lem4}, Equation \eqref{eq5} and assumption (H2) imply $$r_{\alpha,\beta} \geq \bigl(\frac{2n\alpha}{- g(- \beta)}\bigr)^{1/2}\,. \label{eq7}$$ Now, if there exists $s_{\alpha,\beta} \in [0,r_{\alpha,\beta})$ such that $v(\alpha,\beta,s_{\alpha,\beta}) = 0$, Lemma \ref{lem4} implies that $- \beta \leq v(\alpha,\beta,.) < 0$ on $[0,s_{\alpha,\beta})$ and $v(\alpha,\beta,.) > 0$ on $(s_{\alpha,\beta},r_{\alpha,\beta}]$. Then from \eqref{eq5} and (H2) we obtain \begin{align*} \alpha & = -\int_0^{r_{\alpha,\beta}}{G_n(r_{\alpha,\beta},s)g(v(\alpha,\beta, s))\, ds} \\ & \leq - \int_0^{s_{\alpha,\beta}}{G_n(r_{\alpha,\beta},s) g(v(\alpha,\beta, s))\, ds} \\ & \leq - g(- \beta)\int_0^{s_{\alpha,\beta}}{G_n(r_{\alpha,\beta},s)\, ds}\\ &\leq - g(- \beta)\frac{r_{\alpha,\beta}^2}{2n} \end{align*} and \eqref{eq7} still holds. Suppose that $\inf B = 0$ and let $(\beta_j)$ be a sequence in $B$ decreasing to zero. Then $r_{\alpha,\beta_j} \to +\infty$ by \eqref{eq7} and (H2). Let $r > 0$ be fixed. We can assume that $r_{\alpha,\beta_j} > r$ for all $j$. If $v(\alpha,\beta_j,s) < 0$ for $s \in [0,r]$, using (H2) we have $u(\alpha, \beta_j, r) = \alpha + \int_0^{r}{G_n(r,s)g(v(\alpha,\beta_j, s))\, ds} \geq \alpha + \frac{r^2g(- \beta_j)}{2n} \,.$ If $v(\alpha,\beta_j, s_{\alpha,\beta_j}) = 0$ with $s_{\alpha,\beta_j} < r$ we write \begin{align*} u(\alpha, \beta_j, r) & = \alpha + \int_0^{s_{\alpha,\beta_j}}{G_n(r,s) g(v(\alpha,\beta_j, s))\, ds} + \int_{s_{\alpha,\beta_j}}^{r}{G_n(r,s)g(v(\alpha,\beta_j, s))\, ds} \\ & \geq \alpha + \int_0^{s_{\alpha,\beta_j}}{G_n(r,s)g(v(\alpha,\beta_j, s))\, ds} \\ & \geq \alpha + g(- \beta_j)\int_0^{s_{\alpha,\beta_j}}{G_n(r,s)\, ds}\\ &\geq \alpha + \frac{g(- \beta_j)r^2}{2n} \end{align*} Therefore, using Lemma \ref{lem4} we obtain $u(\alpha,\beta_j, s) \geq \alpha + \frac{g(-\beta_j)r^2}{2n} \quad \text{for } s \in [0,r] \,,$ from which we deduce with the help of (H2) that $u(\alpha,\beta_j, s) \geq \alpha/2$ for $s \in [0,r]$ and $j$ large. From \eqref{eq6}, using (H3) we obtain $v(\alpha,\beta_j, r) \geq - \beta_j + \frac{r^2}{2n}f(\frac{\alpha}{2})$ for $j$ large. Thus if we choose $r$ such that $- \beta_j + \frac{r^2}{2n}f(\frac{\alpha}{2}) \geq 1 \,,$ using Lemma \ref{lem4} we obtain $v(\alpha,\beta_j, s) \geq 1$ for $r \leq s \leq r_{\alpha,\beta_j}$ and $j$ large. There exists $k > 0$ such that $\int_r^{r_{\alpha,\beta_j}}{G_n(r_{\alpha,\beta_j},s)\, ds} \geq kr_{\alpha,\beta_j}^2$ for $j$ large. Now we write \begin{align*} \alpha & = - \int_0^{r_{\alpha,\beta_j}}{G_n(r_{\alpha,\beta_j}, s)g(v(\alpha,\beta_j, s))\, ds} \\ & = - \int_0^r{G_n(r_{\alpha,\beta_j},s)g(v(\alpha,\beta_j, s))\, ds} - \int_r^{r_{\alpha,\beta_j}}{G_n(r_{\alpha,\beta_j}, s)g(v(\alpha,\beta_j, s))\, ds} \\ & \leq - g(- \beta_j)\int_0^r{G_n(r_{\alpha,\beta_j},s)\, ds} - g(1)\int_r^{r_{\alpha,\beta_j}}{G_n(r_{\alpha,\beta_j},s)\, ds}\\ & \leq - g(- \beta_j)rr_{\alpha,\beta_j} - g(1)kr_{\alpha,\beta_j}^2 \end{align*} for $j$ large, where we have used the fact that $G_n(r_{\alpha,\beta_j},s) \leq r_{\alpha,\beta_j}$ for $0 \leq s \leq r_{\alpha,\beta_j}$. Since the last term above tends to $- \infty$ as $j \to \infty$ we obtain a contradiction. (ii) Let $\beta \in C$. We claim that $v(\alpha,\beta, r_{\alpha,\beta}) > 0$. If not, by Lemma \ref{lem4} and (H2) we have $\Delta u(\alpha,\beta, .) < 0$ on $[0,r_{\alpha,\beta})$ for some $\beta \in C$. Since $u'(\alpha,\beta, 0) = 0$, we obtain $u'(\alpha,\beta, r_{\alpha,\beta}) < 0$, a contradiction. Therefore \eqref{eq6} implies $$\beta < \int_0^{r_{\alpha,\beta}}{G_n(r_{\alpha,\beta},s) f(u(\alpha,\beta, s))\, ds} \label{eq8}$$ for $\beta \in C$. Suppose that $\sup C = \infty$ and let $(\beta_j)$ be a sequence in $C$ increasing to $\infty$. Since $0 < u(\alpha,\beta_j, r) \leq \alpha$ for $r \in [0,r_{\alpha,\beta_j}]$ , \eqref{eq8} implies that $r_{\alpha,\beta_j} \to \infty$ as $j \to \infty$. Then we can assume that $r_{\alpha,\beta_j} \geq 1$ and that $f(\alpha) \leq \beta_j$ for all $j$. From \eqref{eq6} we obtain $- \beta_j \leq v(\alpha,\beta_j, r) \leq - \frac{2n-1}{2n}\beta_j \leq - \frac{\beta_j}{2} \,, \quad \text{for } r \in [0,1] \,,$ and using \eqref{eq5} we deduce that $u(\alpha,\beta_j, 1) \leq \alpha + g(- \beta_j/2)/2n$. However by (H2), $u(\alpha,\beta_j, 1) < 0$ for $j$ large; thus we reach a contradiction. \end{proof} \begin{remark} \label{rmk2}\rm The proof above shows that, when $\beta \in C$, there exists $s_{\alpha,\beta} \in (0,r_{\alpha,\beta})$ such that $v(\alpha,\beta,.) < 0$ on $[0,s_{\alpha,\beta})$ and $v(\alpha,\beta,.) > 0$ on $(s_{\alpha,\beta},r_{\alpha,\beta}]$. When $\beta \in B$, $s_{\alpha,\beta}$ may not exist. \end{remark} \begin{proof}[Proof of Lemma \ref{lem7}] Let $\beta \in B$. We have $u(\alpha,\beta,r_{\alpha,\beta}) = 0$ and $u'(\alpha,\beta,r_{\alpha,\beta}) < 0$. Therefore, we can find $\varepsilon > 0$ such that $u(\alpha,\beta,r_{\alpha,\beta} + \varepsilon) < 0 \quad \text{and}\quad u'(\alpha,\beta,r_{\alpha,\beta} + \varepsilon) < 0 \,.$ But then by continuous dependence on initial data there exists $\eta > 0$ such that $$u(\alpha,\gamma,r_{\alpha,\beta} + \varepsilon) < 0 \quad \text{and} \quad u'(\alpha,\gamma,r_{\alpha,\beta} + \varepsilon) < 0 \,, \label{eq9}$$ for $|\gamma - \beta| < \eta$. The first inequality in \eqref{eq9} implies that there exists $x \in (0,r_{\alpha,\beta} + \varepsilon)$ such that $u(\alpha,\gamma,x) = 0$ and $u(\alpha,\gamma,r) > 0$ for $r \in [0,x)$. We claim that $x = r_{\alpha,\gamma}$. Then $(\beta - \eta,\beta + \eta) \subset B$. Thus $B$ is open. (H3) implies that $\Delta v(\alpha,\beta, r) > 0$ for $r \in [0,x)$. Then $v'(\alpha,\gamma,r) > 0$ for $r \in (0,x]$. Suppose first that $v'(\alpha,\gamma,r) > 0$ for $r \in (0,r_{\alpha,\beta} + \varepsilon)$. Then $v(\alpha,\gamma,.)$ is increasing on $[0,r_{\alpha,\beta} + \varepsilon]$. We deduce that $\Delta u(\alpha,\gamma,.)$ is increasing on $[0,r_{\alpha,\beta} + \varepsilon]$. If $\Delta u(\alpha,\gamma,r_{\alpha,\beta} + \varepsilon) \leq 0$, then $u'(\alpha,\gamma,r) < 0$ for $r \in (0,r_{\alpha,\beta} + \varepsilon]$. If $\Delta u(\alpha,\gamma,r_{\alpha,\beta} + \varepsilon) > 0$, then there exists $y \in (0,r_{\alpha,\beta} + \varepsilon)$ such that $\Delta u(\alpha,\gamma,.) < 0$ in $[0,y)$ and $\Delta u(\alpha,\gamma,.) > 0$ in $(y, r_{\alpha,\beta} +\varepsilon]$. We deduce that $r \to r^{n - 1}u'(\alpha,\gamma,r)$ is decreasing (resp. increasing) in $[0,y]$ (resp. $[y, r_{\alpha,\beta} +\varepsilon]$). Then the second inequality in \eqref{eq9} implies that $u'(\alpha,\gamma,r) < 0$ for $r \in (0,r_{\alpha,\beta} +\varepsilon]$. Therefore $x = r_{\alpha,\gamma}$ for $|\gamma - \beta| < \eta$. Suppose now that there exists $r \in(x, r_{\alpha,\beta} +\varepsilon)$ such that $v'(\alpha, \gamma, r) \leq 0$. Let $t \in(x, r_{\alpha,\beta} +\varepsilon)$ be the first zero of $v'(\alpha, \gamma,.)$. Then $v(\alpha,\gamma,.)$ is increasing on $[0,t]$. We deduce that $\Delta u(\alpha,\gamma,.)$ is increasing on $[0,t]$. If $\Delta u(\alpha,\gamma,t) \leq 0$, then $u'(\alpha,\gamma,r) < 0$ for $r \in (0,t]$ and we conclude that $x = r_{\alpha,\gamma}$. If $\Delta u(\alpha,\gamma,t) > 0$, then there exists $y \in (0,t)$ such that $\Delta u(\alpha,\gamma,.) < 0$ in $[0,y)$ and $\Delta u(\alpha,\gamma,.) > 0$ in $(y, t]$. We deduce that $r \to r^{n - 1}u'(\alpha,\gamma,r)$ is decreasing (resp. increasing) in $[0,y]$ (resp. $[y, t]$). If $u'(\alpha,\gamma,t) \leq 0$, then $u'(\alpha,\gamma, x) < 0$ and $x = r_{\alpha,\gamma}$. If $u'(\alpha,\gamma,t) > 0$, let $s \in (y,t)$ be such that $u'(\alpha,\gamma, s) = 0$. If $s > x$ then $x = r_{\alpha,\gamma}$. If $s < x$, then $u(\alpha,\gamma,r) > 0$ for $r \in [0,x]$ and we reach a contradiction. Finally if $t = x$, then $u(\alpha,\gamma,r) > 0$ for $r \in [0,x)\cup(x,t]$. We deduce that $\Delta v(\alpha,\gamma, r) > 0$ for $r \in [0,x)\cup(x,t]$ which implies that $v'(\alpha,\gamma, r) > 0$ for $r \in (0,t]$ and we reach again a contradiction. Now let $\beta \in C$. We have $u(\alpha,\beta,r_{\alpha,\beta}) > 0$ and $u'(\alpha,\beta,r_{\alpha,\beta}) = 0$. By Remark \ref{rmk2} we have $v(\alpha,\beta,r_{\alpha,\beta}) > 0$, hence $\Delta u(\alpha,\beta,r_{\alpha,\beta}) = u''(\alpha,\beta,r_{\alpha,\beta}) > 0$. Therefore we can find $\varepsilon > 0$ such that $u(\alpha,\beta,r) > 0 \,, \quad r \, \in \, [0,r_{\alpha,\beta} +\varepsilon], \quad u'(\alpha,\beta,r_{\alpha,\beta} + \varepsilon) > 0 \,.$ Then by continuous dependence on initial data there exists $\eta > 0$ such that $$u(\alpha,\gamma,r) > 0 \,, \quad r \, \in \, [0, r_{\alpha,\beta} +\varepsilon],\quad u'(\alpha,\gamma,r_{\alpha,\beta} + \varepsilon) > 0 \,, \label{eq10}$$ for $|\gamma - \beta| < \eta$. The second inequality in \eqref{eq10} implies that there exists $x \in (0,r_{\alpha,\beta} + \varepsilon)$ such that $u'(\alpha,\gamma,x) = 0$ and $u'(\alpha,\gamma,r) < 0$ for $r \in (0,x)$. Therefore, $x = r_{\alpha,\gamma}$ for $|\gamma - \beta| < \eta$ and $(\beta - \eta,\beta + \eta) \subset C$. Thus $C$ is open. \end{proof} \section{Proof of Theorems \ref{thm1} and \ref{thm2}} We use the notation introduced in Section 2. The following result clearly implies Theorems \ref{thm1} and \ref{thm2}. \begin{proposition} \label{prop2} Let $f$ and $g$ satisfy {\rm (H1)--(H3)}, and {\rm (H4)} or {\rm (H5), (H6)}. Then for any $\alpha > 0$ there exists a unique $(\beta(\alpha),r(\alpha)) \in (0,\infty)\times(0,\infty)$ such that $u(\alpha,\beta(\alpha),r(\alpha)) = u'(\alpha,\beta(\alpha),r(\alpha)) = 0$, $u(\alpha,\beta(\alpha),r) > 0$ for $r \in [0,r(\alpha))$ and $u'(\alpha,\beta(\alpha),r) < 0$ for $r \in (0,r(\alpha))$. Moreover, $\beta$, $r \in C^1(0,\infty)$, $\beta'(\alpha) > 0$ for $\alpha > 0$, $\lim_{\alpha\to 0}r(\alpha) = 0$ and $\lim_{\alpha\to\infty}r(\alpha) = \infty$. \end{proposition} \begin{proof} Let $\alpha > 0$ be fixed. The existence and uniqueness of $(\beta(\alpha),r(\alpha))$ satisfying the first part of the proposition are given by Proposition \ref{prop1}. In order to simplify our notations we denote $u(\alpha,\beta(\alpha),r)$ and $v(\alpha,\beta(\alpha),r)$ by $u_{\alpha}(r)$ and $v_{\alpha}(r)$. We begin with the following lemma. \end{proof} \begin{lemma} \label{lem8} For any $\alpha > 0$ there exists $s(\alpha) \in (0,r(\alpha))$ such that $v_{\alpha}(r) < 0$ for $r \in [0,s(\alpha))$ and $v_{\alpha}(r) > 0$ for $r \in (s(\alpha),r(\alpha)]$ \end{lemma} The proof of the above lemma follows the same arguments as in the proof of Lemma \ref{lem6} (ii). Now for $\alpha$, $\beta > 0$ define \begin{gather*} \varphi(\alpha, \beta, r) = \frac{\partial u}{\partial\alpha}(\alpha, \beta, r)\,,\quad \psi(\alpha, \beta, r) = \frac{\partial v}{\partial\alpha}(\alpha, \beta, r), \\ \rho(\alpha, \beta, r) = \frac{\partial u}{\partial\beta}(\alpha, \beta, r)\,,\quad \chi(\alpha, \beta, r) = \frac{\partial v}{\partial\beta}(\alpha, \beta, r) \end{gather*} for $r \in [0,R_{\alpha,\beta})$. Then $\varphi$, $\psi$, $\rho$ and $\chi$ satisfy the linearized equations \begin{gather*} \Delta\varphi(\alpha,\beta,r) = g'(v(\alpha,\beta,r))\psi(\alpha,\beta,r) \,, \quad 0 \leq r < R_{\alpha,\beta}\,,\\ \Delta\psi(\alpha,\beta,r) = f'(u(\alpha,\beta,r))\varphi(\alpha,\beta,r) \,, \quad 0 \leq r < R_{\alpha,\beta}\,,\\ \varphi(\alpha,\beta,0) = 1 \,, \, \psi(\alpha,\beta,0) = \varphi'(\alpha,\beta,0) = \psi'(\alpha,\beta,0) = 0 \,, \end{gather*} and \begin{gather*} \Delta\rho(\alpha,\beta,r) = g'(v(\alpha,\beta,r))\chi(\alpha,\beta,r) \,, \quad 0 \leq r < R_{\alpha,\beta}\,,\\ \Delta\chi(\alpha,\beta,r) = f'(u(\alpha,\beta,r))\rho(\alpha,\beta,r) \,, \quad 0 \leq r < R_{\alpha,\beta}\,,\\ \chi(\alpha,\beta,0) = - 1 \,, \, \rho(\alpha,\beta,0) = \rho'(\alpha,\beta,0) = \chi'(\alpha,\beta,0) = 0 \,. \end{gather*} \begin{lemma} \label{lem9} We have $\varphi$, $\psi$, $\varphi'$, $\psi' > 0$ on $(0,r_{\alpha,\beta}]$ and $\chi$, $\rho$, $\chi'$, $\rho' < 0$ on $(0,r_{\alpha,\beta}]$. \end{lemma} \begin{proof} By (H2) and (H3) we have $\Delta\psi(\alpha,\beta,0) = f'(\alpha) > 0$ and $\Delta\rho(\alpha,\beta,0)= - g'(- \beta) < 0$. Then $\psi' > 0$ and $\rho' < 0$ on $(0,\eta]$ for some $\eta > 0$ and we can define $r_0 = \sup\{r \in (0,r_{\alpha,\beta}]; \, \psi'\rho' < 0 \, \, \text{on} \, \, (0,r]\}\,.$ We have $\psi > 0$ and $\rho < 0$ on $(0,r_0]$. Since \begin{gather*} r^{n-1}\varphi'(\alpha,\beta,r) = \int_0^r{s^{n-1}g'(v(\alpha,\beta,s))\psi(\alpha,\beta,s)\, ds}, \\ r^{n-1}\chi'(\alpha,\beta,r) = \int_0^r{s^{n-1}f'(u(\alpha,\beta,s))\rho(\alpha,\beta,s)\, ds} \,, \end{gather*} using (H2) and (H3) we deduce that $\varphi' > 0$ and $\chi' < 0$ on $(0,r_0]$. Therefore, $\varphi > 0$ and $\chi < 0$ on $(0,r_0]$. Since \begin{gather*} r^{n-1}\psi'(\alpha,\beta,r) = \int_0^r{s^{n-1}f'(u(\alpha,\beta,s))\varphi(\alpha,\beta,s)\, ds}, \\ r^{n-1}\rho'(\alpha,\beta,r) = \int_0^r{s^{n-1}g'(v(\alpha,\beta,s))\chi(\alpha,\beta,s)\, ds} \end{gather*} using (H2) and (H3) we deduce that $\psi' > 0$ and $\rho' < 0$ on $(0,r_0]$. Therefore, $r_0 = r_{\alpha,\beta}$ and the lemma follows. \end{proof} Now let $D = \{(\alpha,\beta,r); \, \alpha\,, \, \beta > 0 \text{ and } r \in [0,R_{\alpha,\beta})\}$. $D$ is open in $(0,\infty)\times(0,\infty)\times[0,\infty)$. Consider the map $H: D \to \mathbb{R}^2$ defined by $H(\alpha,\beta,r) = (u(\alpha,\beta,r), u'(\alpha,\beta,r))\,.$ Then $H \in C^1(D,\mathbb{R}^2)$ and $$H(\alpha,\beta(\alpha),r(\alpha)) = 0 \quad \text{for } \alpha > 0\,. \label{eq11}$$ Using Lemmas \ref{lem8}, \ref{lem9} and (H2) we obtain $|D_{(\beta,t)} H(\alpha,\beta(\alpha),r(\alpha))| = \rho(\alpha,\beta(\alpha),r(\alpha))u''_{\alpha}(r(\alpha))\, < 0\,.$ Therefore, by the implicit function theorem $\alpha \to (\beta(\alpha), r(\alpha))$ is a $C^1$ map for $\alpha > 0$. Differentiating \eqref{eq11} with respect to $\alpha$ we obtain $$\varphi(\alpha,\beta(\alpha),r(\alpha)) + \rho(\alpha,\beta(\alpha),r(\alpha))\beta'(\alpha) = 0 \quad \text{for }\alpha > 0\,. \label{eq12}$$ From \eqref{eq12} and Lemma \ref{lem9} we deduce that $\beta'(\alpha) > 0$ for $\alpha > 0$. Now we have two cases to consider. \textbf{Case 1:} $f$ satisfies (H1), (H3) and (H4). Assume that $\beta(\alpha) \to \infty$ as $\alpha \to \infty$. Using (H4) we have $v_{\alpha}(r(\alpha)) = - \beta(\alpha) + \int_0^{r(\alpha)}{G_n(r(\alpha), s)f(u_{\alpha}(s))\, ds} \\ \leq -\beta(\alpha) + M\frac{r(\alpha)^2}{2n}\,.$ Then Lemma \ref{lem8} implies that $r(\alpha) \to \infty$ as $\alpha \to \infty$. Now suppose that $\beta(\alpha) \to c < \infty$ as $\alpha \to \infty$. Using Lemma \ref{lem8} and (H2) we can write \begin{align*} 0 & = u_{\alpha}(r(\alpha)) = \alpha + \int_0^{r(\alpha)}{G_n(r(\alpha), s)g(v_{\alpha}(s))\, ds} \\ & = \alpha + \int_{0}^{s(\alpha)}{G_n(r(\alpha), s)g(v_{\alpha}(s))\, ds} + \int_{s(\alpha)}^{r(\alpha)}{G_n(r(\alpha), s)g(v_{\alpha}(s))\, ds} \\ & \geq \alpha + g(- \beta(\alpha))\frac{r(\alpha)^2}{2n}\,, \end{align*} and again we deduce that $r(\alpha) \to \infty$ as $\alpha \to \infty$. Assume that $\beta(\alpha) \to 0$ as $\alpha \to 0$. Using (H4) we have $$v_{\alpha}(r(\alpha)) = - \beta(\alpha) + \int_0^{r(\alpha)}{G_n(r(\alpha), s)f(u_{\alpha}(s))\, ds} \\ \geq -\beta(\alpha) + m\frac{r(\alpha)^2}{2n}\,. \label{eq13}$$ Then Lemma \ref{lem2} implies that $r(\alpha) \to 0$ as $\alpha \to 0$. Now suppose that $\beta(\alpha) \to c > 0$ as $\alpha \to 0$. We claim that $r(\alpha) \to 0$ as $\alpha \to 0$. If not there exist $r_0 > 0$ and a sequence $(\alpha_k)_{k\in \mathbb{N}}$ such that $\alpha_k \to 0$ as $k \to \infty$ and $r(\alpha_k) \geq r_0$ for all $k \in \mathbb{N}$. Let $r_1 \in (0,r_0)$ be such that $- c + f(0)\frac{r_1^2}{2n} \leq -\frac{c}{2} \,.$ For $s \in [0,r_1]$ we have \begin{aligned} \lim_{k\to\infty}v_{\alpha_k}(s) &= \lim_{k\to\infty} (- \beta(\alpha_k) + \int_0^s{G_n(s,x)f(u_{\alpha_k}(x))\, dx}) \\ &= - c + f(0)\frac{s^2}{2n} \leq -\frac{c}{2} \,. \end{aligned} \label{eq14} Clearly $(v_k)_{k \in \mathbb{N}}$ converges uniformly on $[0,r_1]$. Then, for $s \in (0,r_1]$, using \eqref{eq14} and (H2) we have \begin{align*} \lim_{k\to\infty}u_{\alpha_k}(s) &= \lim_{k\to\infty}(\alpha_k + \int_0^s{G_n(s, x)g(v_{\alpha_k}(x))\, dx}) \\ &= \int_0^s{G_n(s,x)g(- c + f(0)\frac{x^2}{2n})\, dx} < 0 \,, \end{align*} and we reach a contradiction. Therefore our claim is proved and the proof of Proposition \ref{prop2} is complete in Case 1. \textbf{Case 2:} $f$ satisfies (H1), (H3), (H5) and (H6). From the proof of Lemma \ref{lem2} we obtain $$G(v_{\alpha}(r(\alpha))) = F(\alpha) + G(-\beta(\alpha)) + 2(n - 1) \int_0^{r(\alpha)}{\frac{u'_{\alpha}(s)v'_{\alpha}(s)}{s}\,ds}\,. \label{eq15}$$ Let $\alpha \geq 1$. Using Lemma \ref{lem8}, \eqref{eq6}, (H3) and (H6) we have $$0 < v_{\alpha}(r(\alpha)) \leq a'\alpha^p\frac{r(\alpha)^2}{2n}\,, \label{eq16}$$ and $$\beta(\alpha) = \int_0^{s(\alpha)}{G_n(s(\alpha),s)f(u_{\alpha}(s))\, ds} \leq a'\alpha^p\frac{s(\alpha)^2}{2n} \leq a'\alpha^p\frac{r(\alpha)^2}{2n}\,. \label{eq17}$$ Then, with the help of (H2), (H3), (H6), \eqref{eq16} and \eqref{eq17}, we can write \begin{align*} |r^{n - 1}u'_{\alpha}(r)| &= |\int_0^r{s^{n - 1}g(v_{\alpha}(s))\, ds}| \\ & \leq \frac{r^n}{n}\max(|g(-\beta(\alpha))|,g(v_{\alpha}(r(\alpha))))\\ & \leq \frac{r^n}{2^{q}n^{q+1}}b'a'^{q}\alpha^{pq}r(\alpha)^{2q} \,, \end{align*} and $|r^{n - 1}v'_{\alpha}(r)| =|\int_0^r{s^{n - 1}f(u_{\alpha}(s))\, ds}| \leq \frac{r^n}{n}f(\alpha) \leq \frac{r^n}{n}a'\alpha^p \,,$ for $r \in [0,r(\alpha)]$. Therefore, $$\big|\int_0^{r(\alpha)}{\frac{u'_{\alpha}(s)v'_{\alpha}(s)}{s}\, ds} \big|\leq \frac{b'a'^{q+1}}{2^{q+1}n^{q+2}}\alpha^{p(q+1)}r(\alpha)^{2(q+1)} \,. \label{eq18}$$ Now using \eqref{eq15}, \eqref{eq16}, \eqref{eq18} an (H6) we obtain \begin{aligned} &\frac{a'^{q+1}b'}{(q+1)2^{q+1}n^{q+1}}\alpha^{p(q+1)} r(\alpha)^{2(q+1)} \\ & \geq G(v_{\alpha}(r(\alpha)))\\ & \geq \frac{a}{p+1}\alpha^{p+1} + \frac{b}{q+1}\beta(\alpha)^{q+1} - \frac{b'a'^{q+1}}{2^{q}n^{q+1}}\alpha^{p(q+1)}r(\alpha)^{2(q+1)} \,. \end{aligned} \label{eq19} Since $pq < 1$, we deduce that $r(\alpha) \to \infty$ as $\alpha \to \infty$. Now assume that $\alpha \leq 1$. (H3) and (H5) imply that there exist $m$, $M > 0$ such that $m \leq f(u_{\alpha}(r)) \leq M \quad \forall \, \, r \in [0,r(\alpha)]\,.$ Then we show that $\lim_{\alpha\to 0}r(\alpha) = 0$ as in Case 1. \end{proof} We conclude this section with the following theorems. \begin{theorem} \label{thm4} Let $f$, $g$ satisfy {\rm (H1)--(H3)}. Assume moreover that \begin{itemize} \item[(H9)] There exist $a$, $b$, $p > 0$ and $q \geq 1$ such that $$f(u) \geq au^p \quad \forall \, u \geq 0 \,,\quad |g(- v)| \geq bv^q \quad \forall \, v \geq 0\,,$$ and $\frac{1}{p + 1} + \frac{1}{q + 1} > \frac{n - 2}{n} \quad \text{if} \quad n \geq 3 \,.$ \end{itemize} Then there exists $R > 0$ such that \eqref{eq1} has at least one non-trivial radial solution $(u,v) \in (C^2(\overline{\!B}_R))^2$. \end{theorem} Since Proposition \ref{prop1} holds, with the help of the first part of Proposition \ref{prop2}, we conclude the statement of the above theorem. \begin{remark} \label{rmk3} \rm Notice that , when $f(0) > 0$, $p$ may be less than 1. If $f(0) = 0$, necessarily $p \geq 1$ since $f$ is $C^1$. \end{remark} \begin{theorem} \label{thm5} Let $f$, $g$ satisfy {\rm (H1)--(H3)}. Moreover assume that \begin{itemize} \item[(H10)] There exist $a$, $a'$, $b$, $b' > 0$ and $p$, $q \geq 1$ such that $pq > 1$, \begin{gather*} f(u) \geq au^{p} \quad \forall \, u \geq 0 \,,\quad f(u) \leq a'u^p \quad \forall \, u \in [0,1] \,,\\ b|v|^q \leq |g(v)| \leq b'|v|^q \quad \forall \, v \in \mathbb{R}\,,\\ \frac{1}{p + 1} + \frac{1}{q + 1} > \frac{n - 2}{n} \quad \text{if } n \geq 3 \,. \end{gather*} \end{itemize} Then there exists $R_0 \geq 0$ such that for all $R > R_0$ problem \eqref{eq1} has at least one non-trivial radial solution $(u,v) \in (C^2(\overline{\!B}_R))^2$. \end{theorem} \begin{proof} Since Proposition \ref{prop1} holds we have the first part of Proposition \ref{prop2}. We also have $\beta$, $r \in C^1(0,\infty)$ and $\beta'(\alpha) > 0$ for $\alpha >0$. Now let $\alpha \in (0,1]$. \eqref{eq15}-\eqref{eq19} hold and, since $pq > 1$, we conclude that $\lim_{\alpha \to 0}r(\alpha) = \infty$. Then we take $$R_0 = \inf_{\alpha > 0} r(\alpha) \geq 0\,.$$ \end{proof} \section{Proof of Theorem \ref{thm3}} We use again the notation introduced in Section 2. The following result implies Theorem \ref{thm3}. \begin{proposition} \label{prop3} Assume that $n = 1$. Let $f$ and $g$ satisfy {\rm (H1)--(H3), (H7)}. Then for any $\alpha > 0$ there exists a unique $(\beta(\alpha),r(\alpha)) \in (0,\infty)\times(0,\infty)$ such that $u(\alpha,\beta(\alpha),r(\alpha)) = u'(\alpha,\beta(\alpha),r(\alpha)) = 0$, $u(\alpha,\beta(\alpha),r)> 0$ for $r \in [0,r(\alpha))$ and $u'(\alpha,\beta(\alpha),r)< 0$ for $r \in (0,r(\alpha))$. Moreover, $\beta$, $r \in C^1(0,\infty)$, $\beta'(\alpha) > 0$ for $\alpha > 0$, $\lim_{\alpha\to 0}r(\alpha) = \infty$ and $\lim_{\alpha\to\infty}r(\alpha) =0$. \end{proposition} \begin{proof} Let $\alpha > 0$ be fixed. The existence and uniqueness of $(\beta(\alpha),r(\alpha))$ satisfying the first part of the proposition are given by Proposition \ref{prop1}. Clearly Lemmas \ref{lem8} and \ref{lem9} also hold. Then we have $\beta$, $r \in C^1(0,\infty)$, $\beta'(\alpha) > 0$ for $\alpha > 0$. We show that $\lim_{\alpha\to 0}r(\alpha) = \infty$ as in the proof of Theorem \ref{thm5}. As in the preceding section we denote $u(\alpha,\beta(\alpha),r)$ and $v(\alpha,\beta(\alpha),r)$ by $u_{\alpha}(r)$ and $v_{\alpha}(r)$. \end{proof} Now we give some lemmas where $s(\alpha)$ is defined in Lemma \ref{lem8}. \begin{lemma} \label{lem10} There exists a constant $C > 0$ such that \begin{gather*} u'_{\alpha}(s(\alpha))^2 \leq Cu_{\alpha}(s(\alpha))(\alpha^{p + 1} + \beta(\alpha)^{q + 1})^{\frac{q}{q + 1}} \,, \\ v'_{\alpha}(s(\alpha))^2 \leq C\alpha^p(\alpha^{p + 1} + \beta(\alpha)^{q + 1})^{\frac{1}{q + 1}} \,, \end{gather*} for all $\alpha > 0$. \end{lemma} \begin{proof} Since $n = 1$ we have $$F(u_{\alpha}(r)) + G(v_{\alpha}(r)) = F(\alpha) + G(- \beta(\alpha)) + u'_{\alpha}(r)v'_{\alpha}(r) \label{eq20}$$ for $r \in[0,r(\alpha)]$. Then \eqref{eq20} and (H7) imply that there exist two constants $C_1$, $C_2 > 0$ such that $$C_1(\alpha^{p + 1} + \beta(\alpha)^{q + 1})^{\frac{1}{q + 1}} \leq v_{\alpha}(r(\alpha)) \leq C_2(\alpha^{p + 1} + \beta(\alpha)^{q + 1})^{\frac{1}{q + 1}}\,. \label{eq21}$$ Using (H1), (H2), Lemma \ref{lem4} and Lemma \ref{lem8} we have \begin{align*} \frac{u'_{\alpha}(s(\alpha))^2}{2} &= -\int_{s(\alpha)}^{r(\alpha)}{g(v_{\alpha}(s))u'_{\alpha}(s)\, ds} = \int_{s(\alpha)}^{r(\alpha)}{g'(v_{\alpha}(s))v'_{\alpha} (s)u_{\alpha}(s)\,ds}\\ &\leq u_{\alpha}(s(\alpha)) \int_{s(\alpha)}^{r(\alpha)}{g'(v_{\alpha}(s))v'_{\alpha}(s)\,ds} = u_{\alpha}(s(\alpha))g(v_{\alpha}(r(\alpha))) \end{align*} Then with the help of \eqref{eq21} and (H7) we obtain $u'_{\alpha}(s(\alpha))^2 \leq C u_{\alpha}(s(\alpha))(\alpha^{p + 1} + \beta(\alpha)^{q + 1})^{\frac{q}{q + 1}} \,,$ for another positive constant $C$. Now, using (H1), (H3) and Lemma \ref{lem8}, we write \begin{align*} \frac{v'_{\alpha}(s(\alpha))^2}{2} &= \int_0^{s(\alpha)}{f(u_{\alpha}(s))v'_{\alpha}(s)\, ds} \\ &= f(\alpha)\beta(\alpha) - \int_0^{s(\alpha)}{f'(u_{\alpha}(s))u'_{\alpha}(s)v_{\alpha}(s)\, ds}\\ &\leq f(\alpha)\beta(\alpha) \,, \end{align*} from which we obtain, using (H7), $v'_{\alpha}(s(\alpha))^2 \leq C\alpha^p(\alpha^{p + 1} + \beta(\alpha)^{q + 1})^{\frac{1}{q + 1}} \,,$ for some positive constant $C$. \end{proof} \begin{lemma} \label{lem11} There exist two constants $c \in (0,1)$ and $M > 0$ such that $$u_{\alpha}(s(\alpha)) \geq c\max(\alpha,\beta(\alpha)^{\frac{q + 1}{p + 1}})\quad \forall \, \alpha \geq M \,. \label{eq22}$$ Moreover, $$\frac{2}{a'}\frac{\beta(\alpha)}{\alpha^p} \leq s(\alpha)^2 \leq \frac{2}{ac^p}\frac{\beta(\alpha)}{\alpha^p}\quad \forall \, \alpha \geq M \,. \label{eq23}$$ \end{lemma} \begin{proof} We argue by contradiction. Suppose first that there exists a sequence $(\alpha_k)_{k \in \mathbb{N}}$ increasing to $\infty$ such that $$u_{\alpha_k}(s(\alpha_k)) \leq \frac{1}{k}\alpha_k\quad \forall\, k \geq 2 \,. \label{eq24}$$ Using Lemma \ref{lem10} and \eqref{eq24} we have \begin{align*} |u'_{\alpha_k}(s(\alpha_k))v'_{\alpha_k}s(\alpha_k))| &\leq C\frac{1}{\sqrt k}\alpha_k^{\frac{p + 1}{2}}(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{1/2} \\ &\leq C\frac{1}{\sqrt k}(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1}) \,. \end{align*} From the above inequality, (H7), Lemma \ref{lem8} and \eqref{eq20}, we obtain $u_{\alpha_k}(s(\alpha_k))^{p + 1} \geq d(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})$ for some positive constant $d$ when $k$ is large and we obtain a contradiction with \eqref{eq24}. Now suppose that there exists a sequence $(\alpha_k)_{k \in \mathbb{N}}$ increasing to $\infty$ such that $$u_{\alpha_k}(s(\alpha_k)) \leq \frac{1}{k}\beta(\alpha_k)^{\frac{q+1}{p+1}}\quad \forall\, k \geq 2 \,. \label{eq25}$$ Using Lemma \ref{lem10} and \eqref{eq25} we have \begin{align*} |u'_{\alpha_k}(s(\alpha_k))v'_{\alpha_k}s(\alpha_k))| &\leq C\frac{1}{\sqrt k}\beta(\alpha_k)^{\frac{q+1}{2(p+1)}}\alpha_k^{p/2}(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{1/2} \\ &\leq \frac{C}{2\sqrt k}(\alpha_k^p \beta(\alpha_k)^{\frac{q+1}{p+1}} + \alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})\\ &\leq C'\frac{1}{\sqrt k}(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1}) \,, \end{align*} for another constant $C' > 0$, where we have used Young's inequality. Then we obtain a contradiction as before. Now we prove \eqref{eq23}. Using (H7), we have $\beta(\alpha) = \int_0^{s(\alpha)}{(s(\alpha) - r)f(u_{\alpha}(r))\, dr} \leq \frac{a'}{2}\alpha^p s(\alpha)^2 \,,$ and, with the help of \eqref{eq22}, $\beta(\alpha) = \int_0^{s(\alpha)}{(s(\alpha) - r)f(u_{\alpha}(r))\, dr} \geq \frac{ac^p}{2} \alpha^p s(\alpha)^2 \,.$ The proof of the lemma is complete. \end{proof} \begin{lemma} \label{lem12} We have $s(\alpha) \to 0$ as $\alpha \to \infty$. \end{lemma} \begin{proof} Since $\beta'(\alpha) > 0$ for $\alpha > 0$ we have $\lim_{\alpha \to \infty}\beta(\alpha) = d \leq \infty$. If $d <\infty$ \eqref{eq23} implies that $s(\alpha) \to 0$ as $\alpha \to \infty$. If $d = \infty$, using (H7) and \eqref{eq22}, we can write \begin{align*} \beta(\alpha) &= \int_0^{s(\alpha)}{(s(\alpha) - r)f(u_{\alpha}(r))\, dr} \\ &\geq \frac{a}{2}u_{\alpha}( s(\alpha))^p s(\alpha)^2 \geq \frac{ac^p}{2} \beta(\alpha)^{p\frac{q + 1}{p + 1}}s(\alpha)^2 \,, \end{align*} from which we obtain $s(\alpha)^2 \leq \frac{2}{ac^p}\beta(\alpha)^{\frac{1 - pq}{p + 1}}\,,$ and the result follows since $pq > 1$. \end{proof} Now we show that $r(\alpha) \to 0$ as $\alpha \to \infty$. If not, there exist $r_0 > 0$ and a sequence $(\alpha_k)_{k \in \mathbb{N}}$ increasing to $\infty$ such that $$r(\alpha_k) \geq \frac{3r_0}{2}\quad \forall \, k \in \mathbb{N}\,. \label{eq26}$$ By Lemma \ref{lem12}, we can assume that $$s(\alpha_k) \leq \frac{r_0}{2}\quad \forall \, k \in \mathbb{N}\,. \label{eq27}$$ \begin{lemma} \label{lem13} There exists a constant $C > 0$ such that \begin{gather*} u'_{\alpha_k}(r_0)^2 \leq Cu_{\alpha_k}(r_0)(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{\frac{q}{q + 1}},\\ v'_{\alpha_k}(r_0)^2 \leq C\alpha_k^p(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{\frac{1}{q + 1}} \end{gather*} for all $k \in \mathbb{N}$. \end{lemma} \begin{proof} Using (H1), (H2), Lemma \ref{lem4}, \eqref{eq26} and \eqref{eq27} we have \begin{align*} \frac{u'_{\alpha_k}(r_0)^2}{2} & = - \int_{r_0}^{r(\alpha_k)}{g(v_{\alpha_k}(s))u'_{\alpha_k}(s)\, ds} \\ &= g(v_{\alpha_k}(r_0))u_{\alpha_k}(r_0) + \int_{r_0}^{r(\alpha_k)}{g'(v_{\alpha_k}(s))v'_{\alpha_k}(s) u_{\alpha_k}(s)\, ds} \\ &\leq g(v_{\alpha_k}(r_0))u_{\alpha_k}(r_0) + u_{\alpha_k}(r_0) \int_{r_0}^{r(\alpha_k)}{g'(v_{\alpha_k}(s))v'_{\alpha_k}(s)\, ds}\\ &= u_{\alpha_k}(r_0)g(v_{\alpha_k}(r(\alpha_k))) \,. \end{align*} Then with the help of \eqref{eq21} with $\alpha = \alpha_k$ and (H7) we obtain $u'_{\alpha_k}(r_0)^2 \leq C u_{\alpha_k}(r_0)(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{\frac{q}{q + 1}} \,,$ for some positive constant $C$. Now we write \begin{align*} \frac{v'_{\alpha_k}(r_0)^2}{2} & = \int_0^{r_0}{f(u_{\alpha_k}(s))v'_{\alpha_k}(s)\, ds} \\ & = f(u_{\alpha_k}(r_0))v_{\alpha_k}(r_0) + f(\alpha_k)\beta(\alpha_k) - \int_0^{r_0}{f'(u_{\alpha_k}(s))u'_{\alpha_k}(s)v_{\alpha_k}(s)\, ds} \end{align*} from which, using Lemma \ref{lem4}, Lemma \ref{lem8}, (H3), and \eqref{eq27}, we obtain \begin{align*} &\frac{v'_{\alpha_k}(r_0)^2}{2} \\ &\leq f(u_{\alpha_k}(r_0))v_{\alpha_k}(r_0) + f(\alpha_k)\beta(\alpha_k) - \int_{s(\alpha_k)}^{r_0}{f'(u_{\alpha_k}(s))u'_{\alpha_k}(s)v_{\alpha_k}(s)\, ds}\\ &\leq f(u_{\alpha_k}(r_0))v_{\alpha_k}(r_0) + f(\alpha_k)\beta(\alpha_k) - v_{\alpha_k}(r_0)\int_{s(\alpha_k)}^{r_0}{f'(u_{\alpha_k}(s)) u'_{\alpha_k}(s)\, ds}\\ &= f(\alpha_k)\beta(\alpha_k) + v_{\alpha_k}(r_0)f(u_{\alpha_k}(s(\alpha_k)))\\ &\leq f(\alpha_k)\beta(\alpha_k) + v_{\alpha_k}(r(\alpha_k))f(\alpha_k) \,. \end{align*} Then, with the help of \eqref{eq21} with $\alpha = \alpha_k$ and (H7), we obtain $v'_{\alpha_k}(r_0)^2 \leq C\alpha_k^p(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{\frac{1}{q + 1}} \,,$ for some positive constant $C$. \end{proof} \begin{lemma} \label{lem14} There exists a constant $C > 0$ such that $0 < v_{\alpha_k}(r_0) \leq C(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{\frac{1}{q(p + 1)}}\quad \forall \, k \in \mathbb{N} \,.$ \end{lemma} \begin{proof} The left hand side inequality follows from \eqref{eq27}. Now, using \eqref{eq26}, \eqref{eq27}, Lemma \ref{lem4}, Lemma \ref{lem8}, (H2) and (H7), we have $$\begin{split} u_{\alpha_k}(r_0) &= \int_{r_0}^{r(\alpha_k)}{(s - r_0)g(v_{\alpha_k}(s))\, ds} \\ &\geq \frac{(r(\alpha_k) - r_0)^2}{2}g(v_{\alpha_k}(r_0))\\ &\geq Cv_{\alpha_k}(r_0)^q \,, \end{split} \label{eq28}$$ for some positive constant $C$. With the help of (H7), \eqref{eq20} with $r = r_0$, $\alpha = \alpha_k$, lemma \ref{lem4} and \eqref{eq28} we deduce that $v_{\alpha_k}(r_0) \leq C(\alpha_k^{p + 1} + \beta(\alpha_k)^{q + 1})^{\frac{1}{q(p + 1)}} \quad \forall \, k \in \mathbb{N}\,,$ for another positive constant $C$. \end{proof} \begin{lemma} \label{lem15} There exist a constant $c \in (0,1)$ and an integer $k_0$ such that $u_{\alpha_k}(r_0) \geq c\max(\alpha_k,\beta(\alpha_k)^{\frac{q + 1}{p + 1}})\quad \forall \, k \geq k_0 \,.$ \end{lemma} \begin{proof} Lemma \ref{lem14} and (H7) imply that $\lim_{k \to \infty}G(v_{\alpha_k}(r_0))(\alpha_k^{p+ 1} + \beta(\alpha_k)^{q + 1})^{- 1} = 0\,,$ since $pq > 1$. Now the arguments are the same as in the proof of Lemma \ref{lem11} with $r_0$ in place of $s(\alpha)$, using Lemma \ref{lem13} instead of Lemma \ref{lem10}. \end{proof} \begin{proof}[Proof of Proposition \ref{prop3} completed] Using Lemma \ref{lem15} and (H7) we have \begin{align*} v_{\alpha_k}(r_0) + \beta(\alpha_k) &= \int_0^{r_0}{(r_0 - s)f(u_{\alpha_k}(s))\, ds} \\ &\geq \frac{ac^p}{2}r_0^2 \max(\alpha_k^p,\beta(\alpha_k)^{p\frac{q + 1}{p + 1}}) \end{align*} for all $k \in \mathbb{N}$. Then Lemma \ref{lem14} implies that $r_0 = 0$ since $pq > 1$ and we reach a contradiction. \end{proof} \section{The uniqueness question} Define $A = \{\alpha \in (0,\infty); \, r'(\alpha) \neq 0\}\,.$ Assume that $A \neq \emptyset$. Since $A$ is open there exists $J \subset \mathbb{N}$ such that $A = \cup_{n \in J}I_n$, where $I_n = (a_n,b_n)$. In the setting of Theorem \ref{thm1} or Theorem \ref{thm2}, Proposition \ref{prop2} implies that $A \neq \emptyset$ and that $\inf\{a_n; \, n \in J\} = 0$ and $\sup\{b_n; \, n \in J\} = \infty$. \textbf{Case 1:} $|J| = 1$. Then $A =(0,\infty)$ and for all $R > 0$, problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. \textbf{Case 2:} $|J| \geq 2$. Suppose first that there exist $j$, $k \in J$ such that $a_j = 0$ and $b_k = \infty$. Let $\gamma = \min\{r(\alpha);\, \alpha \in [b_j,a_k]\}$ and $\delta = \max\{r(\alpha);\, \alpha \in [b_j,a_k]\}$. Then for all $R \in (0,\gamma)\cup(\delta,\infty)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. Suppose that there exists $j \in J$ such that $a_j = 0$ and that $b_k \neq \infty$ for all $k \in J$. Let $\gamma = \inf \{r(\alpha);\, \alpha \geq b_j\}$. Proposition \ref{prop2} implies that $\gamma > 0$. Then for all $R \in (0,\gamma)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. Now, if there exists $k \in J$ such that $b_k = \infty$ and that $a_j \neq 0$ for all $j \in J$, we define $\delta = \sup \{r(\alpha);\, \alpha \leq a_k\}$. Proposition \ref{prop2} implies that $\delta < \infty$. Then for all $R \in (\delta,\infty)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. Otherwise we cannot give any uniqueness result. In the setting of Theorem \ref{thm3}, Proposition \ref{prop3} implies that $A \neq \emptyset$ and that $\inf\{a_n; \, n \in J\} = 0$ and $\sup\{b_n; \, n \in J\} = \infty$. \textbf{Case 1:} $|J| = 1$. Then $A = (0,\infty)$ and for all $R > 0$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. \textbf{Case 2:} $|J| \geq 2$. Suppose first that there exist $j$, $k \in J$ such that $a_j = 0$ and $b_k = \infty$. Let $\gamma = \min\{r(\alpha);\, \alpha \in [b_j,a_k]\}$ and $\delta = \max\{r(\alpha);\, \alpha \in [b_j,a_k]\}$. Then for all $R \in (0,\gamma)\cup(\delta,\infty)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. Suppose that there exists $j \in J$ such that $a_j = 0$ and that $b_k \neq \infty$ for all $k \in J$. Let $\delta = \sup \{r(\alpha);\, \alpha \geq b_j\}$. Proposition \ref{prop3} implies that $\delta < \infty$. Then for all $R \in (\delta, \infty)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. Now, if there exists $k \in J$ such that $b_k = \infty$ and that $a_j \neq 0$ for all $j \in J$, we define $\gamma = \inf \{r(\alpha);\, \alpha \leq a_k\}$. Proposition \ref{prop3} implies that $\gamma > 0$. Then for all $R \in (0, \gamma)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. Otherwise we cannot give any uniqueness result. In the setting of Theorem \ref{thm5}, the proof shows that $A \neq \emptyset$ and that $\inf\{a_n; \, n \in J\} = 0$ \textbf{Case 1:} $|J| = 1$. Then $A = (0,c)$ where $c \leq \infty$. If $c = \infty$, then for all $R > 0$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. If $c < \infty$, then for all $R > r(c)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. \textbf{Case 2:} $|J| \geq 2$. Suppose that there exists $j \in J$ such that $a_j = 0$. Let $\delta = \sup \{r(\alpha);\, \alpha \geq b_j\}$. If $\delta < \infty$, then for all $R \in (\delta,\infty)$ problem \eqref{eq2} has a unique solution $(u,v) \in (C^2([0,R])^2$. If $\delta = \infty$ we cannot give any uniqueness result. If $a_j \neq 0$ for all $j \in J$ we cannot give any uniqueness result. \section{Examples} In this section we give some examples that illustrate our results. \begin{example} \label{exa1} \rm Theorem \ref{thm1} applies when $f$ and $g$ are defined in the following six cases: \begin{itemize} \item[(1)] Let $c > \pi/2$, $m \in \mathbb{N}\backslash \{0\}$ and $f(u) =c + \arctan u^m$, $u \in \mathbb{R}$. \item[(2)] Let $f(u) = 2 - \frac{1}{1+u^2}$, $u \in \mathbb{R}$. \item[(3)] Let $q \geq 1$ and $g(v) = |v|^{q-1} v$, $v\in \mathbb{R}$. \item[(4)] Let $r$, $q > 1$ and $g(v) = \begin{cases} v^r & v \geq 0\,,\\ |v|^{q-1}v & v \leq 0 \,. \end{cases}$ \item[(5)] Let $p$, $q > 1$ and $g(v) = \begin{cases} \ln(1 + v^p) & v \geq 0\,,\\ 1 - \exp|v|^q & v \leq 0 \,. \end{cases}$ \item[(6)] Let $g(v) = \begin{cases} \arctan v^2 & v \geq 0\,,\\ v^{2}\arctan v & v \leq 0 \,. \end{cases}$ \end{itemize} \end{example} \begin{example} \label{exa2}\rm Let $p > 0$ and $q \geq 1$. For $m \in \mathbb{N}\backslash \{0\}$ define $f(u) = (1 + u^{2m})^{p/(2m)}\,, \quad u \in \mathbb{R} \,.$ Let $h \in C^1(\mathbb{R})$ be such that $h' < 0$ on $(-\infty, 0)$, $h' > 0$ on $(0,\infty)$ and $b \leq h \leq b'$ for some constants $b$, $b' > 0$. Define $g(v) = h(v)|v|^{q - 1}v\,, \quad v \in \mathbb{R} \,.$ Then Theorem \ref{thm2} applies. If $p$, $q$ satisfy the condition in (H9), then we can use Theorem \ref{thm4}. \end{example} \begin{example} \label{exa3}\rm Let $k \in C^1(\mathbb{R})$ be such that $k' > 0$ on $(0,\infty)$ and $a \leq k \leq a'$ for some constants $a$, $a' > 0$. Define $f(u) = k(u)|u|^{p - 1}u , \quad \forall u \in \mathbb{R}\,,$ where $p \geq 1$. If $g$ is as in Example \ref{exa2} and if $p$, $q$ satisfy the condition in (H9) (resp. (H10), then Theorem \ref{thm4} (resp. Theorem \ref{thm5}) applies. We can also use Theorem~\ref{thm3}. \end{example} \begin{thebibliography}{0} \bibitem{da1} R. Dalmasso; \emph{Existence and uniqueness of solutions for a semilinear elliptic system}, Int. J. Math. Math. Sci. 10 (2005), 1507-1523. \bibitem{da2} R. Dalmasso; \emph{Uniqueness of positive solutions for some nonlinear fourth-order equations}, J. Math. Anal. Appl. 201 (1996), 152-168. \bibitem{mi} E. Mitidieri; \emph{A Rellich type identity and applications}, Commun. in Partial Differential Equations, (18) (1\&2),125-151 (1993). \bibitem{sz1} J. Serrin, H. Zou; \emph{Non-existence of positive solutions of semilinear elliptic systems}, Discourses in Mathematics and its Applications, no. 3, Department of Mathematics, Texas A\&M University, College Station, Texas, (1994), 55-68. \bibitem{sz2} J. Serrin, H. Zou; \emph{Non-existence of positive solutions of the Lane-Emden system}, Differential Integral Eqs 9 (1996), 635-653. \end{thebibliography} \end{document}