\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{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}

\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),
\begin{equation}
r^{n - 1}v'(r) = \int_0^r{s^{n - 1}f(u(s))\, ds} \, > \, 0 \quad
\text{for }  r > 0 \,.
\label{eq4}
\end{equation}
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
\begin{equation}
r_{\alpha,\beta} \geq \bigl(\frac{2n\alpha}{- g(- \beta)}\bigr)^{1/2}\,.
\label{eq7}
\end{equation}
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
\begin{equation}
\beta < \int_0^{r_{\alpha,\beta}}{G_n(r_{\alpha,\beta},s)
f(u(\alpha,\beta, s))\, ds}
\label{eq8}
\end{equation}
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
\begin{equation}
u(\alpha,\gamma,r_{\alpha,\beta} + \varepsilon) < 0
\quad \text{and} \quad u'(\alpha,\gamma,r_{\alpha,\beta}  +
\varepsilon) < 0 \,,
\label{eq9}
\end{equation}
   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
\begin{equation}
u(\alpha,\gamma,r) > 0 \,, \quad r \, \in \, [0,
r_{\alpha,\beta}  +\varepsilon],\quad
u'(\alpha,\gamma,r_{\alpha,\beta}  + \varepsilon) > 0 \,,
\label{eq10}
\end{equation}
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
\begin{equation}
H(\alpha,\beta(\alpha),r(\alpha)) = 0 \quad \text{for }
\alpha > 0\,.
\label{eq11}
\end{equation}
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
\begin{equation}
\varphi(\alpha,\beta(\alpha),r(\alpha)) +
\rho(\alpha,\beta(\alpha),r(\alpha))\beta'(\alpha) = 0 \quad
\text{for }\alpha > 0\,.
\label{eq12}
\end{equation}
  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
\begin{equation}
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}
\end{equation}
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{equation}
\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}
\end{equation}
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
\begin{equation}
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}
\end{equation}
Let $\alpha \geq 1$. Using Lemma \ref{lem8}, \eqref{eq6}, (H3) and
(H6) we have
\begin{equation}
0 < v_{\alpha}(r(\alpha))
\leq a'\alpha^p\frac{r(\alpha)^2}{2n}\,,
\label{eq16}
\end{equation}
and
\begin{equation}
\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}
\end{equation}
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,
\begin{equation}
\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}
\end{equation}
Now using \eqref{eq15},
\eqref{eq16}, \eqref{eq18} an (H6) we obtain
\begin{equation}
\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}
\end{equation}
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
\begin{equation}
F(u_{\alpha}(r)) + G(v_{\alpha}(r)) =
F(\alpha) + G(- \beta(\alpha))
 +  u'_{\alpha}(r)v'_{\alpha}(r)
\label{eq20}
\end{equation}
for $r \in[0,r(\alpha)]$.  Then \eqref{eq20}  and (H7)
imply that there exist  two constants $C_1$, $C_2  > 0$ such that
\begin{equation}
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}
\end{equation}
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
\begin{equation}
u_{\alpha}(s(\alpha)) \geq
c\max(\alpha,\beta(\alpha)^{\frac{q + 1}{p + 1}})\quad \forall \,
\alpha \geq M \,.
\label{eq22}
\end{equation}
Moreover,
\begin{equation}
\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{equation}
\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
\begin{equation}
u_{\alpha_k}(s(\alpha_k)) \leq
\frac{1}{k}\alpha_k\quad \forall\, k \geq 2 \,.
\label{eq24}
\end{equation}
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
\begin{equation}
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}
\end{equation}
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
\begin{equation}
r(\alpha_k) \geq \frac{3r_0}{2}\quad \forall \, k \in \mathbb{N}\,.
\label{eq26}
\end{equation}
By Lemma \ref{lem12}, we can assume that
\begin{equation}
s(\alpha_k) \leq \frac{r_0}{2}\quad \forall \, k \in \mathbb{N}\,.
\label{eq27}
\end{equation}



\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{equation}
\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}
\end{equation}
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}

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\bibitem{sz1} J. Serrin, H. Zou;
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\bibitem{sz2} J. Serrin, H. Zou;
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\end{thebibliography}


\end{document}